Thermal Engineering - 1

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THERMAL ENGINEERING - I (For B.E./B.Tech., Mechanical Engineering Students)

(As per New Syllabus of Leading Universities)

Dr. S.Ramachandran, M.E., Ph.D., Professor and Research Head Dr. A. Anderson, M.E., Ph.D., Asst. Professor Faculty of Mechanical Engineering Sathyabama University Jeppiaar Nagar Chennai - 600 119

AIR WALK PUBLICATIONS (Near All India Radio) 80, Karneeshwarar Koil Street Mylapore, Chennai - 600 004. Ph.: 2466 1909, 94440 81904 Email: [email protected], [email protected] www.airwalkpublications.com

First Edition: 2nd Feb 2016

ISBN :978-93-84893-66-8

SYLLABUS Thermal Engineering - I Chapter 1: I.C. ENGINES Four & Two stroke engine - SI & CI engines - Valve and Port Timing Diagrams - Fuel injection Systems for SI engines - Fuel injection systems for CI engines- Ignition - Cooling and Lubrication system - Fuel properties and Combustion Stoichiometry.

Chapter 2: Combustion in SI and CI Engine Combustion in SI and CI Engine: Normal combustion and abnormal combustion in SI Engine - Importance of flame speed and effect of engine variables - Abnormal combustion - Pre Ignition and Knocking in SI Engine - Fuel requirement and fuel rating - anti knock additives - combustion chamber - Requirement - types of SI Engine - Four stages of combustion in CI Engine - Delay period and its Importance - Effect of engine variables - Diesel knock - Need for air movement, suction, compression and combustion induced turbulence in Diesel engine - Open and divided combustion chambers and fuel Injection - Diesel fuel requirements and fuel rating.

Chapter 3: Testing and Performance of Engines Indicator diagram and properties - Pressure Transducer- Brake power measurements. Dynamometers - Performance calculations -Morse test - Air consumption - Fuel consumption Exhaust gas compositon - Heat balance sheet.

Chapter 4: Air Compressors Classification of Compressors - Fans, blowers and compressors - Positive displacement and dynamic types - Reciprocating and rotary types Rotary, Dynamic and Axial Flow (Positive Displacement type): Roots Blower, Vane sealed compressor, Lysholm compressor mechanical details and principle of working - efficiency considerations. Centrifugal compressors: Mechanical details and principle of operation - Velocity and pressure variation. Energy transfer - impeller

blade shape - losses - slip factor, power input factor, pressure coefficient and adiabatic coefficient - velocity diagrams - power. Axial flow compressors: Mechanical details and principle of operation - velocity triangles and energy transfer per stage, degree of reaction, work done factor - Isentropic efficiency - pressure rise calculation - polytropic efficiency.

Chapter 5: Refrigeration Mechanical Refrigeration and types - Units - Air refrigeration system, Details, Principle of operations application - Vapour compression refrigeration - Calculation of COP - Effect of super heating and sub cooling - Desired properties of refrigerants and Common refrigerants - Vapour absorption system - Mechanical Details, Working and Principle - Use of p-h chart for calculation.

Chapter 6: Actual Cycles and their Analysis Introduction, Comparison of air standard and actual cycles, Time loss factor, Heat loss factor, Exhaust Blowdown - Loss due to Gas exchange processes, Volumetric efficiency. Loss due to Rubbing friction, Actual and fuel - air cycles of CI engine.

Contents C.1

Contents 1. I.C. ENGINES 1.1 Introduction. .....................................................................

1.1

1.2 Basic Terms Connected With I.C. Engines .................

1.1

1.3 Four Stroke SI (Petrol) Engine .....................................

1.4

1.4 Four Stroke CI Engine ...................................................

1.6

1.5 Working of Two Stroke Cycle Engine ..........................

1.9

1.6 Working of Two Stroke Petrol Engine ......................... 1.10 1.7 Classification of IC Engines ........................................... 1.12 1.8 IC Engine Components – Functions And Materials .. 1.13 1.9 Valve Timing And Port Timing Diagrams................... 1.24 1.9.1 Valve Timing Diagram for a 4-Stroke Cycle Petrol Engine.................................................................. 1.26 1.9.2 Valve Timing Diagram of 4-Stroke Diesel Engine .... ........................................................................ 1.29 1.9.3 Port Timing Diagram of Two Stroke Cycle Petrol Engine.................................................................. 1.29 1.9.4 Port timing Diagram for Two Stroke Cycle Diesel Engine ................................................................. 1.32

1.10 Fuel Supply System ...................................................... 1.34 1.10.1 Fuel injection Systems for S.I. Engines.......... 1.34 1.10.2 Different types of Fuel Systems ....................... 1.35 1.10.3 Fuel Injection System In SI Engines.............. 1.36 1.10.4 Fuel Pump (for S.I. Engine) ............................ 1.36

1.11 Carburetor . ..................................................................... 1.37 1.12 Air Fuel Ratio ............................................................... 1.40 1.13 Various Compensation In Carburetors ....................... 1.40 1.14 Types of Carburetors..................................................... 1.43 1.15 Fuel Injection Systems For C.I Engines .................... 1.53 1.15.1 Fuel Pump (C.I. Engine) .................................. 1.53

1.16 Fuel Injection System ................................................... 1.55

C.2 Thermal Engineering - I

1.16.1 Fuel Injector ....................................................... 1.56

1.17 Ignition Systems. ........................................................... 1.57 1.17.1 Spark Plug ......................................................... 1.57 1.17.2 Types of Electronic ignition systems: .............. 1.65

1.18 Lubrication System ........................................................ 1.69 1.18.1 Purpose of Lubrication...................................... 1.69 1.18.2 Systems of Lubrication...................................... 1.70

1.19 Cooling System............................................................... 1.73 1.19.1 Purpose of Cooling............................................. 1.73 1.19.2 Methods of Cooling IC Engine......................... 1.74

1.20 Fuels And Combustion .................................................. 1.77 1.20.1 Introduction ........................................................ 1.77 1.20.2 Classification of fuels: ....................................... 1.78

1.21 Solid Fuels ..................................................................... 1.79 1.22 Gaseous Fuels: ............................................................... 1.80 1.23 Fuel Properties ............................................................... 1.81 1.23.1 Determination of Calorific value of fuel by Bomb Calorimeter.......................................................... 1.83 1.23.2 Determination of Higher Calorific value of Gaseous fuel. .................................................................. 1.85

1.24 Combustion Stoichiometry ............................................ 1.88 1.25 Excess Air And Air Fuel Ratio Calculation .............. 1.95

2. Combustion in SI and CI Engine 2.1 Normal Combustion ......................................................... 2.1.1 Factors affecting normal combustion in S.I Engines... ........................................................................

2.1

2.1.2 Flame Front Propagation....................................

2.3 2.4

2.3 Importance of Flame Speed And Effect of Engine Variables ............ .....................................................................

2.5

2.3.1 Factors affecting flame speed .............................

2.5

2.2 Abnormal Combustion .....................................................

2.8

2.2.1 Pre-ignition ...........................................................

Contents C.3

2.2.2 Knocking (or) Detonation (or) Pinking.............. 2.2.3 The phenomenon of knock in SI Engine .......... 2.2.4 Effects of knocking in SI Engine ...................... 2.2.5 Effect of engine variables on knock ..................

2.8 2.9 2.10 2.13 2.14

2.3 Fuel Requirement And Fuel Rating ............................. 2.18 2.3.1 Important properties of fuel in SI Engine ....... 2.18 2.3.2 Important characteristics of SI Engine fuel..... 2.18 2.3.5 Octane Number (ON) .......................................... 2.21

2.4 Anti-Knock Additives ....................................................... 2.23 2.4.1 Anti-knock Agents ................................................ 2.23 2.4.2 Effects of Anti knock additives .......................... 2.24 2.4.3 Factors affecting Detonation and Remedies ..... 2.25

2.5 Types of Combustion Chamber In SI Engine ............. 2.26 1. Overhead valve (or) I - Head combustion chamber.. ........................................................................ 2.26 2. T-Head combustion chamber ................................... 2.27 3. L-head combustion chamber .................................... 2.27 4. F-head combustion chamber .................................... 2.28

2.6 Combustion in CI Engines ............................................. 2.28 1. Ignition Delay period ............................................... 2.29 2. Period of Rapid Combustion (or) Uncontrolled Combustion ..................................................................... 2.32

2.7 Factors That Affect Delay Period In Diesel Engine .. 2.32 2.7.1 Effect of variables on the Delay period ............ 2.34

2.8 Knocking (or) Diesel Knock............................................ 2.35 2.8.1 The phenomenon of knock in CI engine........... 2.36 2.8.2 Comparison of knock on SI and CI Engines .. 2.38 2.8.3 Characteristics Tending to Reduce Detonation 2.40

2.9 Need For Air Movement In Diesel Engine.................. 2.40 2.10 Combustion Chamber Design For Compression Ignition Engine . ..................................................................... 2.42

C.4 Thermal Engineering - I

1. Open combustion chamber ....................................... 2.42 (a) Shallow depth chamber .......................................... 2.43 (b) Hemispherical chamber........................................... 2.43 (c) Cylindrical chamber ................................................ 2.43 (d) Toroidal chamber .................................................... 2.43 2. Divided combustion chamber type .......................... 2.45 (a) Turbulent combustion chamber ............................. 2.45 (b) Pre combustion chamber......................................... 2.46 (c) Energy-cell chamber................................................. 2.47

2.11 Diesel Fuel Requirement : For Compression Ignition Engines ..................................................................... 2.50 1. Knocking characteristics........................................... 2.50 2. Volatility of the fuel ................................................. 2.51 3. Starting characteristics of the fuel ......................... 2.51 4. Smoke produced by fuel and its odour ................. 2.51 5. Viscosity of fuel......................................................... 2.51 6. Corrosion and wear .................................................. 2.51 7. Easy to handle .......................................................... 2.51 2.11.1 Cetane Number (CN)......................................... 2.52 2.11.2 Fuel Rating for CI Engine ............................... 2.52

3. Testing and Performance of Engines 3.1 Performance Calculations (or) Performance Test On I.C. Engines ....... ..................................................................... 1. Indicator Diagram .................................................... 2. Indicated Power (IP) ................................................ 3. Mean Effective Pressure P m  .................................. 5. Brake Power (BP) ..................................................... 6. Different Arrangements used to find Brake Power...... ........................................................................ 7. Friction Power (FP) .................................................. 8. Specific Fuel Consumption (S.F.C).........................

3.1 3.1 3.2 3.2 3.6 3.6 3.9 3.9

9. Mechanical Efficiency mech  .................................. 3.10

Contents C.5

10. Thermal Efficiency .................................................. 3.10 11. Indicated Thermal Efficiency indicated .............. 3.10 12. Brake Thermal Efficiency Brake  ........................ 3.11 12. Relative Efficiency or Efficiency Ratio ................. 3.11 14. Volumetric Efficiency volumetric ........................ 3.11

3.2 Dynamometer.... ............................................................... 3.12 3.2.1 Hydraulic Dynamometer ..................................... 3.13 3.2.2 Eddy Current Dynamometer............................... 3.15

3.3 Measurement of Indicated Power of Multicylinder Engine Morse Test . ............................................................... 3.21 3.4 Measurement of Air Consumption ................................ 3.31 3.5 Fuel Consumption ........................................................... 3.34 3.5.1 Volumetric type .................................................... 3.35 3.5.1.1 Burette method .................................................. 3.35 3.5.1.2 Automatic volumetric flow meter .................... 3.37 ......................................................................................... 3.5.2 Gravimetric fuel flow measurement................... 3.38 Construction.................................................................... 3.39 3.5.3 Measurement of Heat Carried Away by Cooling Water................................................................. 3.40

3.6 Exhaust Gas Composition............................................... 3.41 3.6.1 Oxides of Nitrogen............................................... 3.41 3.6.2 Carbon monoxide (CO)........................................ 3.43 3.6.3 Unburnt hydrocarbons......................................... 3.45 3.6.4 Aldehydes .............................................................. 3.47

3.7 Visible Emissions . ........................................................... 3.47 3.7.1 Smoke .................................................................... 3.47 3.7.1.1 Comparison method .......................................... 3.47 3.7.1.2 Obscuration method.......................................... 3.48 3.7.2 Measurement of Heat Carried Away by Exhaust Gas ................................................................... 3.52

3.8 Heat Balance Sheet ......................................................... 3.52

C.6 Thermal Engineering - I

Solved Problems............................................................. 3.53-3.96

4. Air Compressors 4.1 Introduction. .....................................................................

4.1

4.2 Classification of Air Compressors..................................

4.2

4.3 Single Acting Reciprocating Air Compressor ...............

4.3

4.4 Double Acting Air Compressor ......................................

4.4

4.5 Single Stage Compressor ................................................

4.4

4.6 Multi Stage Compressor .................................................

4.5

4.7 Working Principle of Reciprocating Air Compressors

4.5

4.7.1 Workdone during isothermal compression PV  c without clearance volume...............................

4.7

4.7.2 Workdone during polytropic compression [ PV n  constant ] without clearance volume ..............

4.8

4.7.3 Workdone During Isentropic Compression PV   constant Without Clearance Volume............... 4.10

4.8 Minimum Workdone ........................................................ 4.11 4.9 Power Required To Run The Compressor ................... 4.11 Solved Problems............................................................. 4.12-4.15 4.9.1 Clearance Volume ................................................ 4.16

4.10 Workdone By Reciprocating Air Compressor With Clearance Volume ... ............................................................... 4.16 4.11 Isothermal Efficiency of A Reciprocating Air Compressor......... ..................................................................... 4.18 4.12 Volumetric Efficiency in A Reciprocating Air Compressor......... ..................................................................... 4.18 4.12.1 Factors affecting volumetric efficiencies .......... 4.20

4.13 Important Technical Terms .......................................... 4.20 1. Volumetric Efficiency  vo l ........................................ 4.20 2. Clearance Ratio ‘k’ .................................................... 4.21 3. Volume and Volume Rate........................................ 4.21 4. Indicated Power (I.P) and Brake Power (B.P) ..... 4.22

Contents C.7

5. Free Air Delivered (FAD) ........................................ 4.22 6. Mean Effective Pressure m.e.p .............................. 4.22 8. Swept Volume Vs .................................................... 4.23 9. Piston speed ............................................................. 4.23

4.14 Single Stage Solved Problems...................................... 4.23-4.66 4.15 Two Stage Compression ................................................ 4.64 4.15.1. Complete (or) Perfect Intercooling................... 4.65 4.15.2 Incomplete (or) Imperfect Intercooling............. 4.65 4.15.3 Workdone when perfect and imperfect intercooling. .................................................................... 4.66 4.15.4 Minimum Work Required for 2 Stages and multi stages .................................................................... 4.66

4.16 Multistage Reciprocating Compressors ....................... 4.67 4.17 Various Types of Compressors Rotary Positive Displacement Compressors ....................................................4.112 4.18 Roots Blower... ...............................................................4.112 4.19 Vane Type Blower of Compressor ...............................4.117 4.20. Centrifugal Compressor................................................4.125 4.21 Velocity And Pressure Variation .................................4.128 4.22 Static Temperature And Total Head (or) Stagnation Temperature ........................................................4.129 4.23 Steady-flow Energy Equation .......................................4.132 4.24 Euler’s Equation - (Energy Transfer) .........................4.133 4.24.1 Velocity components at the entry and exit of the rotor ..................................................................... 4.134

4.25 Impeller Blade Shape - Backward, Radial And Forward Blade Impellers .....................................................4.137 4.26 Velocity Triangle And Work Done By The Centrifugal Compressor..........................................................4.140 4.27 Important Formulae ......................................................4.142 1. Power required ........................................................ 4.142 2. Width of Blades of Impeller and Diffuser ............ 4.144

C.8 Thermal Engineering - I

3. Isentropic Efficiency of the Compressor ................. 4.145 4. Slip factor .................................................................. 4.147 5. Work factor ................................................................ 4.149 6. Pressure Co-efficient p .......................................... 4.149 7. Stage Work ................................................................ 4.149 8. Stage Pressure Rise .................................................. 4.151 9. Enthalpy-Entropy Diagram...................................... 4.151 10. Degree of Reaction .................................................. 4.153 11. Mass Flow Rate ...................................................... 4.153 Solved Problems............................................................. 4.154-4.165

4.28 Axial Flow Compressors ...............................................4.166 4.28.1 Working Principles of a Compressor Stage .... 4.167 4.28.2 Stage Velocity Triangles.................................... 4.169 4.28.3 Blade Loading, Flow Coefficients and Specific Work.................................................................. 4.171 4.28.4 Static Pressure Rise in a Stage....................... 4.171

4.29 Degree of Reaction.........................................................4.175 4.30 Infinitesimal Stage Efficiency (or) Polytropic Efficiency ............ .....................................................................4.176 4.31 Finite Stage Efficiency ..................................................4.178 4.38 Important Formulae ......................................................4.180 1. Flow coefficient f ................................................... 4.180 2. Head or work coefficient h  .................................. 4.180 3. Deflection co-efficient def ...................................... 4.181 4. Pressure co-efficient p ........................................... 4.181 5. Pressure ratio ............................................................ 4.181 6. Stagnation pressure ratio......................................... 4.181 7. Number of stages ...................................................... 4.181

4.32 Losses In Axial Flow Compressor Stage....................4.182 1. Profile losses on the surface of the blades:........... 4.183 2. Skin friction loss on the annulus walls: ............... 4.183 3. Secondary flow losses: .............................................. 4.183

Contents C.9

4.33 Surging ...... .....................................................................4.184 4.34 Stalling ...... .....................................................................4.186 4.35 Comparison Between Reciprocating And Centrifugal Compressors ........................................................4.187 4.36 Comparison Between Reciprocating And Rotary Air Compressors .....................................................................4.188 4.37 Comparison Between Centrifugal And Axial Flow Compressors ....... .....................................................................4.189 Solved Problems............................................................. 4.191-4.208

5. Refrigeration 5.1 Introduction. .....................................................................

5.1

5.1.1 Fundamentals of Refrigeration:.......................... 5.1.5 Unit of Refrigeration: (Ton of Refrigeration) ...

5.1 5.3 5.3

5.2 Air Refrigeration System ................................................

5.6

5.2.1 Air Refrigeration Cycles ......................................

5.7

5.3 Reversed Carnot Cycle ..................................................

5.7

5.1.4 Types of Mechanical Refrigeration system........

5.4 Bell-Coleman Cycle .......................................................... 5.17 5.5 Vapour Compression Refrigeration ................................ 5.39 5.5.1 Different Conditions of the Vapour ................... 5.42

5.6 PH Chart..... ..................................................................... 5.56 Solved Problems............................................................. 5.57-5.65

5.7 Subcooling or Undercooling ............................................ 5.66 5.7.1 Effect of Subcooling............................................. 5.67

5.8 Super Heating .. ............................................................... 5.69 5.8.1 Effect of Super heating ....................................... 5.69 Solved Problems............................................................. 5.70-5.95

5.9 Performance Calculation - Factors Affecting Performance of A Vapour Compression System ................ 5.99 5.10 Vapour Absorption System ...........................................5.102 5.10.1 Working principle of vapour absorption ......... 5.103

C.10 Thermal Engineering - I

5.10.2 Practical vapour absorption system................. 5.104

5.11 Refrigerant .....................................................................5.106 5.11.1 Characteristics of good refrigerants................. 5.106 5.11.2 Refrigerants Number ......................................... 5.107

5.12 Ammonia - Water Absorption System ........................5.108 5.12.1 Absorption Refrigeration: (Electrolux Refrigerator).................................................................... 5.108 5.12.2 Lithium Bromide Absorption System .............. 5.109

5.13 Gas Liquefaction System ..............................................5.111 5.13.1 Hampson-Linde Gas Liquefaction System ...... 5.111 5.13.2 Claude System for Liquefying Gases .............. 5.112 5.13.3 Advantages and Limitations of vapour absorption system: ......................................................... 5.113

5.14 Comparision Between Vapour Compression And Vapour Absorption System ....................................................5.114 5.15 Application of Cryogenic ...............................................5.114 (i) Application of Liquid Oxygen................................. 5.114 (ii) Application of Liquid Nitrogen ............................. 5.115 (iii) Application of Carbon-di-oxide ............................. 5.115 (iv) Application of Inert Gases .................................... 5.115

6. Actual Cycles and Their Analysis 6.1 Introduction. .....................................................................

6.1

6.2 Comparison of Air-standard and Actual Cycles ..........

6.2

(a) The working medium.............................................. (b) The nature of working substance ..........................

6.3 6.3

6.3 Time Loss.... .....................................................................

6.4

6.4 Heat Loss Factors ...........................................................

6.6

6.5 Exhaust Blowdown...........................................................

6.8

6.6 Loss due to Gas Exchange Process ..............................

6.9

6.7 Volumetric Efficiency ....................................................... 6.10 6.7.1 The amount of air-fuel ratio .............................. 6.10

Contents C.11

6.7.2 The design of intake and exhaust manifold and port design ............................................................. 6.11 6.7.3 Compression ratios............................................... 6.11 6.7.4 The intake and exhaust valve timings ............. 6.11

6.8 Loss due to Rubbing Friction ........................................ 6.14 6.8.1 Surface finish ....................................................... 6.14 6.8.2 Lubricant properties ............................................ 6.14 6.8.3 Engine power........................................................ 6.15 6.8.4 Heat Dissipation .................................................. 6.15 6.8.5 Other miscellaneous components ........................ 6.15

6.9 Blowby Losses .. ............................................................... 6.16 6.9.1 Effects of blowby loss on efficiency ................... 6.16 6.9.2 Crankcase ventilation .......................................... 6.17

6.10 Actual and Fuel-Air Cycles of CI Engines ................ 6.19

Chapter 1

I.C. ENGINES Four & Two stroke engine - SI & CI engines - Valve and Port Timing Diagrams - Fuel injection Systems for SI engines Fuel injection systems for CI engines- Ignition - Cooling and Lubrication system - Fuel properties and Combustion Stoichiometry.

1.1 INTRODUCTION Heat engines: Heat engine is a type of engine or machine which derives heat energy from the combustion of fuel or any other source and converts this into mechanical energy. Heat engines are generally classifed as 1. External combustion engines (E.C. Engines) 2. Internal combustion engines (I.C. Engines) If the combustion of fuel takes place inside the cylinder,

then

the

engine

is

known

an

Internal

combustion engine (or) IC engine. Example: Petrol engines.

engine,

Diesel

engine,

Oil

and

Gas

If the combustion of fuel takes place outside the cylinder, then

engine is known External combustion

engine. Example: Steam engine, Steam turbine.

1.2 BASIC TERMS CONNECTED WITH I.C. ENGINES 1. Bore. The inside diameter of the cylinder is known as the bore and it is measured in mm.

1.2

Thermal Engineering - I

2. Stroke. It is the distance travelled by the piston between two dead centre positions. 3. Dead Centers. They correspond to the position occupied by the piston at the end of its stroke where the centre lines of the connecting rod and crank are in the same straight line. These conditions arise for two positions of the piston. For vertical engine these are known as Top Dead Centre (TDC) position and Bottom Dead Centre (BDC) position and for horizontal engines, these positions are known as Inner Dead Centre (IDC) position and Outer Dead Centre (ODC) position (Refer Fig 1.1 (a)&(b))

C O N N EC TIN G R O D

C Y L IN D ER B O R E -d

C R AN K R AD IU S,r

IDC

O D C

S W EP T V O LU M E

VS C L E AR AN C E V O LU M E V C

S TR OK E LE N G TH L=2r

B O R E -d

Fig.1.1 (a) P iston in O D C positio n

IDC

O D C S TR OK E LE N G TH L=2r

Fig. 1.1 (b)

P iston in ID C po sition

IC Engines 1.3

4. Top Dead Centre (TDC). The top most position of the piston towards the cover end side of the cylinder of a vertical engine is called Top Dead centre (TDC) position. 5. Bottom Dead Centre (BDC). The lowest position of the piston towards the crank end side of the cylinder of a vertical engine is known as Bottom Dead Centre (BDC) position. 6. Crank Throw or Crank Radius. The distance between the centre of main shaft and centre of crank pin is known as crank throw or crank radius. This distance will be equal to half of the stroke length. 7. Piston Displacement or Swept Volume. It is the volume through which the piston sweeps for its one stroke. Mathematically, it is equal to the area of cross-section of the piston multiplied by its stroke length. 8. Clearance Volume. It is the volume included between the piston and the cylinder head when the piston is at its Top Dead Centre (in-vertical engine) or Inner Dead Centre (in horizontal engine). The piston never enters this portion of the cylinder during its travel. The clearance volume is generally expressed as percentage of the swept volume and is denoted by vc. 9. Compression Ratio. It is the ratio of the total cylinder volume to the clearance volume. V s  Swept Volume V c  Clearance volume r  Compression ratio

1.4

Thermal Engineering - I

Then, the total cylinder volume  Vs  Vc and Compression ratio, r 

Vs  Vc Vc

For petrol engines the value of compression ratio varies from 5:1 to 9:1 and for Diesel engines from 14:1 to 22:1. 10. Piston Speed. It is the distance travelled by piston in one minute. If RPM of engine shaft

 N and

Length of stroke

 L metre,

Then piston speed

 2 LN metre/min.

1.3 FOUR STROKE SI (PETROL) ENGINE Working Of Four Stroke Spark Ignition Engine The number of strokes required to complete the reciprocating engine cycle is four and hence this name four stroke cycle. The work is obtained only during one stroke out of these for a single cylinder engine or for every cylinder individually for a multi-cylinder engine. These strokes are as follows (Refer Fig. 1.2) 1. Suction or Induction Stroke. [Fig. 1.2 (a)]. During this stroke, the inlet valve stays open and the exhaust valve is closed. The piston is moved downward from Top Dead Centre (TDC) by means of crankshaft which is revolved by the momentum of the flywheel or by power generated by the electric starting motor. This movement increases the size of combustion space thereby reducing the pressure in it with the result that the higher pressure of the outside atmosphere forces the air into the combustion space.

IC Engines 1.5 S park P lug In le t Valve In le t p ort

E xha u st Valve

E xha u st port C ylind er P iston C ra nk Ca se

C ra nk Sh aft

(a )S u ction stro ke

(c) P o wer stroke

C onn ecting R od R (b ) C o mp ressio n stroke

(d ) E xha ust stroke

Fig. 1.2 Four S troke S park Ignition Engine

A carburetor is put in the passage of incoming air which supplies a metered quantity of fuel to this air. This air fuel mixture thus comes into the engine cylinder. 2. Compression Stroke. [Fig 1.2 (b)]. The air fuel mixture sucked in during the suction stroke is compressed during this upward stroke. The compression forces the fuel into closer combination with air. The heat produced during this compression stroke aids the combustion of fuel. Just a little before the end of compression stroke, the mixture is ignited by a spark produced by the spark plug. Both the inlet and exhaust valves remain closed during this stroke.

1.6

Thermal Engineering - I

3. Working or Power Stroke. [Fig. 1.2 (c)]. Both inlet and exhaust vales remain closed during this stroke. The mixture of fuel and air which burns at the end of compression stroke expands due to the heat of combustion. It exerts pressure in the cylinder and on the piston and under this impulse, the piston moves downward thus doing useful work. 4. Exhaust stroke. [Fig 1.2 (d)]. The inlet valve remains closed while the exhaust valve opens. The greater part of the burnt gases escape because of their own expansion. The upward movement of the piston pushes the remaining gases out of the open exhaust valve. This cycle or series of events takes place over and over again thus delivering power. Theoretically speaking, the inlet valve opens at the beginning of suction stroke and closes at the end of this stroke. Both the valves remain closed at compression and power strokes. The exhaust valve opens at the beginning of exhaust stroke and closes at the end of same stroke.

1.4 FOUR STROKE CI ENGINE WORKING OF A FOUR STROKE DIESEL ENGINE This engine works on Diesel cycle or constant pressure cycle. Heavy motor vehicles, stationary power plants, big industrial units and ships mostly employ this engine. The Diesel or Compression ignition engine which mostly uses Diesel oil [light and heavy] as fuel differs from a petrol engine in that in the latter the air fuel mixture after being compressed in the engine cylinder to a high pressure, is ignited by an electric spark from a spark plug

IC Engines 1.7

p 3 ATM O SP H E R IC P R E SSU R E

2

b 5

a

4 1

v

o

Fig. 1.3 Actual pv- diagram of four stroke cycle Diesel engine

2

p

3

4 E X HA U ST

1

S U CTIO N o

v Fig . 1.4 Hypoth etical pv- d iag ram of a fo ur stro ke D iesel cycle eng in e

while in the former the fuel is ignited by being injected into the engine cylinder containing air compressed to a very high pressure, the temperature of this air is sufficiently high to ignite the fuel. There is no spark plug in a Diesel engine. The temperature of the air compressed itself is

1.8

Thermal Engineering - I

sufficient to ignite the fuel which in the form of very fine spray is injected at a controlled rate so that the combustion proceeds at constant pressure. 1. Suction Stroke (5-a-3). The piston moves down from the top centre Position. The air is drawn into the cylinder through the inlet valve which closes at the end of this stroke. The exhaust valve remains closed during this stroke. 2. Compression Strokes (3-4). The piston moves up from the bottom dead centre position. The inlet valve is also now closed. The air drawn into the cylinder in the previous stroke is entrapped inside the cylinder and compressed with the upward movement of the piston. As the compression ratio used in this engine is high (14 to 22), the air is finally compressed to a pressure as high as 40 bar at which its temperature is high - as high as 1000 C enough to ignite the fuel. 3. Constant Pressure Stroke (4-1). As the piston moves after reaching top dead centre the fuel is injected into the hot compressed air where it starts burning, maintaining the pressure constant. At the point 1 the fuel supply is cut-off. Theoretically, the fuel is injected at the end of compression stroke and injection continues till the point of cut-off. In actual practice, the ignition starts before the end of compression stroke to take care of ignition lag. 4. Power or Working Stroke (1-2). Both inlet and exhaust valves remain closed during the stroke. The hot gases and air now expand adiabatically to the point 2, in the engine cylinder pushing the piston down and hence doing work. The piston finally reaches the bottom dead centre.

IC Engines 1.9

5. Exhaust Stroke (3-b-5). The piston now moves up once again. The inlet and fuel valves are closed but the exhaust valve opens. A greater part of the burnt gases escape due to their own expansion. The upward movement of the piston pushes the remaining gases out through the open exhaust valve. Only a small quantity of exhaust gases stay in the combustion chamber. The exhaust valve closes at the end of exhaust stroke. The cycle is thus completed.

1.5 WORKING OF TWO STROKE CYCLE ENGINE There is one working stroke in one cycle of four stroke cycle engine, ie. in two revolutions of the crankshaft. The desire to have one working stroke per cylinder for every revolution of the crankshaft had led to the development of two stroke cycle engines. Two stroke cycle engines are very widely employed for small power required for motor pads (auto cycles) scooters and motor cycles because of compactness and ease in manufacture and being simpler although the specific fuel consumption (S.F.C.), i.e., the fuel consumption per bp kW hour is more, In a two stroke cycle engine, the suction and exhaust strokes are eliminated. Here the burnt exhaust gases are forced out through the exhaust port by a fresh charge of the fuel which enters the cylinder nearly at the end of the working stroke through inlet port. The process is termed as scavenging. There are no inlet and exhaust valves as in a four stroke engine but we have inlet and exhaust ports only.

1.10

Thermal Engineering - I

1.6 WORKING OF TWO STROKE PETROL ENGINE Fig.1.5 shows a two stroke petrol engine commonly used in motor cycles. It has no valves but consists of the inlet or induction port (IP), exhaust port (EP) and a third port called the transfer port (TP). Referring to Fig.1.5 (a) let the piston be nearing the completion of its compression stroke as shown. The ignition starts due to the spark given by the spark plug and the piston is pushed down [Fig.1.5 (b) and S p ark P lu g

S p ark P lu g . . ....... . . .. ..... ...... ..... .......... ............. . ..... .......

C o m p re ss ed .. ............... . .... C h arg e ............ ........... ......... . . ........... . ........... ....... ... ... P iston Tra nsfe r P o rt

E xha u st P o rt In le t P ort O pe n

P iston Tra nsfe r P o rt

E xha u st P o rt In le t P ort O pe n

C ra nk C a se

(a )

(b )

... .. .......................................... .... ............. ....... ........ ............. .. .... ..... ....... .................. .... ................ . .. . . .. . .. .. . Tra nsfe r ......................................................... E xha u st P o rt O p en P o rt O p en ............... . .... ........... . Tra nsfe r In le t P ort P o rt

(c)

(d ) Fig. 1.5 Tw o stro ke petrol en gin e

.. . . ........ .. . .... .......................... ...... . .. . ... ........ ...... ... .. . ................... ......... . . . ........ ....... . .. . ................ . .. E xha u st P o rt In le t P ort

IC Engines 1.11

(c)] performing the working stroke and in doing so, the air fuel mixture already drawn from the inlet port inside crank case in the previous stroke is compressed to a high pressure. When about four-fifth of this stroke is completed, the exhaust port (EP) is uncovered slightly and some of the charge of burnt gases escape to the atmosphere. Immediately afterwards as the exhaust port is uncovered by the further downward movement of the piston, the transfer port (IP) which is only very slightly lower than (EP) is also uncovered as shown in Fig.1.5 (c) and a charge of compressed fuel air mixture enters the cylinder and further pushes out the burnt gases out of the exhaust port (EP). The top of the piston is made of a particular shape that facilitates the deflection of the fresh charge upwards and thus avoids its escape along with the exhaust gases. After reaching the bottom dead centre, when the piston moves up it first closes the transfer port (TP) and then the exhaust port EP. The charge of fuel which previously entered the cylinder, is now compressed. Simultaneously there is a fall of pressure in the crank case creating a partial vacuum. When the piston is nearing its upward movement, the inlet port opens and a fresh charge of air fuel mixture from the carburetor enters the crank case. After the ignition of the charge takes place, the piston moves down for the power stroke and the cycle is repeated as before. 2. Two Stroke Cycle Compression Ignition Engine In a two smoke cycle compression ignition engine, all the operations are exactly the same as those in the spark ignition except that in this case, only air is taken in instead of air fuel mixture and the fuel is injected at the end of

1.12

Thermal Engineering - I

compression stroke, a fuel injector being fitted instead of a spark plug.

1.7 CLASSIFICATION OF IC ENGINES IC Engines are classified as follows 1.

According to type of fuel used (a) Petrol engine (b) Diesel engine (c) Gas engine.

2.

According to the ignition method used (a) Spark Ignition engine (SI engine) (b) Compression Ignition engine (CI engine).

3.

According to number of strokes per cycle (a) Four stroke engine (b) Two stroke engine.

4.

According to Air standard cycle (a) Otto cycle engine (b) Diesel cycle engine (c) Dual cycle engine.

5.

According to number of cylinders used (a) Single cylinder engine (b) Multi cylinder engine.

6.

According to position of cylinder (a) Horizontal engine (b) Vertical engine (c) Radial engine (d) ‘V’ engine.

IC Engines 1.13

7.

According to fuel supply system (a) Carburetor engine (b) Air-injection engine (c) Airless injection (or) Solid injection engine.

8.

According to cooling system used (a) Air cooled engine (b) Water cooled engine.

9.

According to speed of engine (a) Slow speed engine (b) Medium speed engine (c) High speed engine.

10.

According to location of valves (a) Overhead valve engine (b) Side valve engine. (c) L-head type engine (d) T-head type engine (e) F-head type engine

11.

According to their uses (a) Aircraft engines (b) Marine engines (c) Automobile engines etc.

1.8 IC ENGINE COMPONENTS – FUNCTIONS AND MATERIALS A cross section of water cooled IC engine with principal parts is shown in the Fig. 1.6. The major components of IC engine are as follows. I. Components common to both Petrol and Diesel engines

1.14

Thermal Engineering - I

Va lv e R o c k e r P u sh rod Va lv e s p rin g

Va lv e

S p a rk p lu g

C om bu stio n ch a m be r

G u d g eo n p in W a te r ja c k e t P isto n

C on ne c

P isto n ring s

tin g ro d

C ylin d er

Ta p pe t

C am

Cr

k an

C am s h aft C ra nk pin C ra nk c as e Fig. 1 .6 C ros s se c tio na l of w a ter c o ole d S .I E ng in e

1. 3. 5. 7. 9. 11. 13.

Cylinder 2. Piston Cylinder head 4. Piston rings Gudgeon pin 6. Connecting rod Crank 8. Crankshaft Bearings 10. Crank case Flywheel 12. Governor Valves or port mechanisms.

II. Components for petrol engine only (a) Spark plug (b) Carburetor (c) Fuel pump

IC Engines 1.15

III. Components for Diesel engines only (i) Fuel pump (ii) Fuel injector 1. Cylinder The cylinder is the closed part which contains the gas mixture under pressure and guides the piston. The combustion of the gas mixture takes place inside the cylinder. The ideal form of cylinder consists of a plain cylinder barrel with a liner material in which the piston slides. The movement of the piston or stroke is longer than the diameter or bore of cylinder. The ratio of stroke to bore is called stroke bore ratio. The upper end consists of a combustion or clearance space in which the ignition and combustion of the charge takes place. Since a large amount of energy is generated during combustion, therefore it has to be cooled. Generally IC engines are water cooled or air cooled engines. The cylinder is made up of hard grade cast iron and is usually cast in one piece. 2. Cylinder head One end of cylinder is closed by means of a removable cylinder head which usually contains the inlet valve for admitting the gas mixture and exhaust valve for discharging the products of combustion. These two valves are operated by means of cam mechanisms geared to the engine shaft. The passage in the cylinder head leading to and from the valves are called ports. The pipes which connect the inlet ports of the various cylinders to a common intake pipe for the engine is called the inlet manifold. If

1.16

Thermal Engineering - I

the exhaust ports are similarly connected to a common exhaust system, this system is called exhaust manifold. The main purpose of the cylinder head is to seal the working ends of the cylinders and not to permit entry and exit of gases to engine. The inside cavity of cylinder head is called the combustion chamber. Its shape controls the direction and rate of combustion. The cylinder head is usually made up of cast iron or aluminium. 3. Piston A piston is fitted to each cylinder to receive gas pressure and transmit the force to the connecting rod. The piston must provide the following (a) Give tight seal to the cylinder through bore; (b) Slide freely; (c) Be light; (d) Be strong. The piston wall called skirt must be strong enough to withstand the side thrust. Pistons are made up of cast iron or aluminium alloy for lightness. Light alloy pistons expand more than cast iron and therefore need large clearance to the bore. 4. Piston rings Pistons are generally loose fit with cylinders to avoid the sticking of piston to cylinder during expansion on combustion. To provide a good sealing fit between piston and cylinder, pistons are equipped with piston rings. They are split at one point so that they can be expanded and slipped over the end of the piston and into rings grooves which have cut in the piston. During installation, the rings are compressed into the grooves and split ends come almost

IC Engines 1.17

together. The rings fit tightly against the cylinder wall and against the sides of the ring grooves in the piston. These rings can expand or contract as they heat and cool and still make a good seal. Thus they are free to slide up and down the cylinder wall. Small two stroke cycle engines have two rings on the piston. Both are compression rings. Four stroke cycle engines have an extra ring called the oil control ring. Four stroke cycle engines are so constructed that they get much oil in the cylinder wall than two stroke cycle engines. These rings are usually made up of cast iron or alloy steel. 5. Gudgeon pin (Or Wrist pin or Piston pin) These are hardened steel parallel spindles fitted through the piston bosses and the small end bushes or eyes to allow the connecting rods to swivel. Gudgeon pins are press fit and heated in hot oil or water bath to remove it. Gudgeon pins are made hollow for lightness. 6. Connecting Rod Connecting rod transmits the piston load to the crank, causing the latter to turn, thus converting the reciprocating motion of the piston into rotary motions of crankshaft. Connecting Rods are made up of Nickel, Chromium and Vanadium steels. 7. Crank The reciprocating motion of the piston is converted to rotatory motion on the wheels by using connecting rod and crank assembly. The connecting rod connects the piston to crank. The crank is made up of steel forging.

1.18

Thermal Engineering - I

8. Crankshaft

1

4

Crank is a part of the crankshaft. The crankshaft of an IC engine receives efforts via cranks supplied by the pistons through the Fig. 1.7 2 3 connecting rods. All the engines auxiliary mechanisms with mechanical transmission are geared in one way or the another to the crankshaft. The crankshaft converts the reciprocating motion of piston to rotary motion of wheels. The crankshaft are mounted in bearings which encircle the journals so it can rotate freely. The Fig. 1.7 shows a typical crankshaft layout for a 4 cylinder engine. It is usually a steel forging or cast iron like spheroidal graphite or nickel alloy castings. 9. Engine Bearings The crankshaft is supported by bearings. The connecting rod big end is attached to the crank pin on the crank of the crankshaft by a bearing. A piston pin at the small end is used to attach the rod to the piston. The piston pin rides in bearings. Bearings are used to support the moving parts. The purpose of bearings is to reduce the friction and allow the parts to move easily. Bearings are lubricated with oil to make the relative motion easier. Usually sleeve bearings are used for most engine applications. Some engines use ball and roller bearings to

IC Engines 1.19

support the crankshaft and for the connecting rod and piston-pin bearings. 10. Crank Case The main body of engine to which the cylinders are attached and which contains the crankshaft and its bearings is called crank case. This member also holds other parts in alignment and resists the explosion and inertia forces. It also protects the parts from dirt etc and serves as a part of lubricating system. 11. Fly wheel A flywheel is secured on the crankshaft. It is made up of steel or cast iron disc. It performs the following functions: It stores energy required to rotate the shaft during preparatory strokes and provides uniform crankshaft rotation. 12. Governor A governor is defined as a device for regulating automatically the output of a machine by regulating the supply of working fluid to the variation of loads. When the load on the engine increases, the speed reduces and there is an increase in supply of fluid and when the load on the engine reduces, the speed increases and the fluid supply is reduced. This is controlled by governor. Thus the function of a governor is to control the fluctuations of engine speed due to changes in load. 13. Valves and Valve Operating Mechanisms The inlet and exhaust of internal combustion engines are controlled by poppet valves. These valves are held to their seating by strong springs, and as the valves usually open inwards, the pressure in cylinder helps to keep them

1.20

Thermal Engineering - I

closed. The valve gear mechanisms are shown in the Fig. 1.6. It consists of poppet valve, the steam bushings or guide, valve spring, spring retainer, lifter or push rod, camshaft and half speed gear for a 4 stroke engine. The timing of the valves, i.e. their opening and closing with respect to the travel of the piston is very important thing for efficient working of the engines. The drive of the camshaft is arranged through gears or chain and sprocket called timing gear. 1.8.1 COMPARISON OF FOUR-STROKE AND TWO STROKE CYCLE ENGINES S. No 1.

2.

3.

4.

Title/Aspect Cycle

4 Stroke Engine The cycle gets

The cycle is

completed in 4 strokes or in two

completed in 2 strokes or in one

revolutions of crank shaft. Power produced The power produced is small since one

Flywheel size

Cooling and Lubrication

2 Stroke Engine

revolution of crankshaft. The power produced is large since one

power stroke for 2 revolutions of

power stroke for one revolution.

crankshaft. Heavier flywheel is required since non

Lighter flywheel is required since

uniform turning moment. Since one power

uniform turning moment. One power stroke for

stroke per two revolution less cooling

one revolution - high cooling and

and lubrication is required.

lubrication system is required.

IC Engines 1.21 S. No 5.

6.

7. 8.

Title/Aspect

4 Stroke Engine

2 Stroke Engine

Valve and Valve

Valve and valve mechanisms are

Port mechanisms are present in 2 stroke.

Mechanism Initial cost

present in 4 stroke engine. Initial cost is high

Initial cost is low

Volumetric Efficiency Part load

but running cost is low. High volumetric efficiency. Better than that of

but running cost is more. Low volumetric efficiency. Poor than that of 4

2 stroke engine. High thermal efficiency. Used in cars, buses,

stroke engine. Comparatively lower thermal efficiency. Used in scooters,

trucks, industrial engines, aeroplane,

motor cycles, lawn movers etc.,.

generators etc. Heavy and High

Compact engine.

efficiency Thermal efficiency 10. Applications 9.

11. Compactness

engine.

1.8.2 COMPARISON BETWEEN AND DIESEL ENGINE Petrol Engine (a)

(b) (c)

PETROL

ENGINE

Diesel Engine

During suction stroke, the mixture of air and petrol is

During suction stroke, only air is sucked in the engine

sucked in the engine cylinder. The petrol engine works on

cylinder. The diesel engine works on

Otto cycle In petrol engine, spark plug is used to ignite the charge

diesel cycle In diesel engine, fuel injector is used. The fuel burns by

with an electric spark.

the heat of compressed air.

1.22

Thermal Engineering - I Petrol Engine

(d)

(e)

Diesel Engine

Here carburetor is used which supplies the mixture

Here fuel injector is used to inject the diesel

of air and petrol in correct proportion The compression ratio is

The compression ratio in

upto 11. Average compression ratio in petrol

diesel engine varies from 12:1 to 22:1

engine varies from 5:1 to 9:1

(f) Due to lower compression

(g)

ratio, petrol engines are

Due to higher compression ratio, diesel engines are

light and less stronger than diesel engine. There is a chance of

heavier and stronger than petrol engines As only air is compressed

pre-ignition

during compression stroke, there is no chance of

(h)

Air standard efficiency is

pre-ignition Air standard efficiency is

(i)

lower due to lower compression ratio The initial cost of petrol

higher due to higher compression ratio. The initial cost of diesel

engine is less but the running cost is high because

engine is high but the running cost is low as the

the cost of petrol is more than diesel. Petrol engines are used in

cost of diesel is less.

(j)

Diesel engines are used in

cars, scooters and motorcycles heavy duty vehicles like trucks, buses and locomotive (k) (l)

It occupies less space. Maintenance cost is less.

(m) Thermal efficiency upto 25%

engines. It occupies more space. It is having more parts. So maintenance cost is high. Thermal efficiency upto 40%

IC Engines 1.23

1.8.3 Advantages of two stroke cycle engines over four stroke cycle engines 1.

The two stroke cycle engine gives one working stroke for each revolution of the crankshaft. So the power developed by two stroke cycle engine is twice that developed by four stroke cycle engine for the same engine speed and cylinder volume.

2.

The turning moment on the crankshaft is more even in 2 stroke engine and so a lighter flywheel is required in it.

3.

A two stroke engine is more compact, light and requires less space than a four stroke cycle engine for the same power. So it is more suitable for auto, motor cycles and scooters.

4.

Weight of engine is less due to absence of valves.

5.

It has high mechanical efficiency due to the absence of cams, cam shaft and rockers etc.

6.

It requires fewer spare parts due to its simple design.

1.8.4 Disadvantages 1.

The air standard efficiency of two stroke cycle engine is less than four stroke cycle engine because the compression ratio of the two stroke cycle engine is less than that of four stroke cycle engine.

2.

A portion of the fresh air-petrol mixture (in case of S.I. engine) escapes unused through the exhaust port. So, the specific fuel consumption is higher. So, overall efficiency is less.

Thermal Engineering - I

1.24

3.

In two stroke cycle engines, the piston gets over heated due to firing in each revolution and oil cooling of the piston is necessary.

4.

The consumption of lubricating oil is large in a two stroke cycle engine because of high operating temperature.

5.

The fresh charge is mixed by the burnt gases due to incomplete scavenging.

6

There is greater wear and tear of moving parts.

1.9 VALVE TIMING AND PORT TIMING DIAGRAMS Review of Actual Cycle (Actual Indicator Diagram) for 4-stroke Petrol Engine

p

The actual cycle (actual indicator diagram) for a 4-stroke cycle petrol engine is shown in Fig.1.8 (a). This diagram is drawn for one complete cycle ie for two complete revolutions of the crank. The theoretical P-V diagram of a 4-stroke petrol engine (otto cycle) is also shown in Fig. 1.8 (b) for comparison.

p

3

Exp

2 co m

ex ha ust 5

pres

b

ans

3

A tm os phe ric pre ssure

io n 2

s io n

a

4 suction

ex ha ust 5

1 v

O (a) A c tual p -v d ia gram o f fo ur stroke O tto cyc le en gin e Fig. 1.8

suction O

4 1 v

IC Engines 1.25

1. Suction

Stroke 5  a  1: The

suction

stroke

is

shown by the line 5  a  1. The piston moves down from the Top Dead Centre (TDC). The pressure inside the cylinder is below the atmospheric pressure. So the air-fuel mixture flows into the engine cylinder through the inlet valve. At end of this stroke, the inlet valve closes. The exhaust valve remains closed during this suction stroke. 2. Compression Stroke 1  2: The piston moves up from BDC to TDC. The inlet valve closes (IVC) at the end of suction stroke (or) a little before compression stroke starts. The air is compressed with the upward movement of the piston. At the end of this stroke, the air attains high pressure. 3. Expansion (or) working Stroke 3  4: A short time before the end of the compression stroke i.e., (TDC) the air-fuel mixture is ignited (IGN) by the spark plug. During ignition, the volume remains constant. Both valves are closed. Due to this ignition, the pressure and temperature of the charge is enormously increased. This explosion pushes the piston to BDC. A little before BDC, the exhaust valve opens. (EVO). 4. Exhaust Stroke 1  b  5: The burnt gases are exhausted to the atmosphere in this stroke. The piston moves from BDC to TDC to send the burnt gases out of the cylinder, through exhaust valve. During this stroke, the inlet valve is in closed position. In actual cycle, the corners are rounded off, because the inlet and outlet valves do not open and close suddenly but they take sometime to do so. Because of the resistance of the inlet valve to the entering charge, the actual pressure

1.26

Thermal Engineering - I

inside the cylinder during suction stroke is slightly less than the atmospheric pressure. Similarly, because of the resistance of the exhaust valve to the burnt gases leaving the cylinder, the actual pressure inside the cylinder during exhaust stroke is slightly higher than the atmospheric pressure. The hatched area called as negative loop 5  a  1  b  5 is shown in actual cycle. This negative loop is subtracted from the larger loop (called positive loop) to get the net workdone during a cycle. 1.9.1 Valve Timing Diagram for a 4-Stroke Cycle Petrol Engine The actual valve timing diagram is shown in Fig.1.9. At the end of exhaust stroke (or) before the beginning of the suction stroke, the inlet valve opens (IVO) 10 to 20 in advance of the TDC position to facilitate the inflow of fresh charge and outflow of burnt gases. When the piston reaches the TDC, the suction stroke starts. The piston moves down and reaches BDC position and then it moves up. A little beyond the BDC, the inlet valve closes (IVC). i.e. The inlet valve closes (IVC) at 30 to 40 after BDC. Now during further upward movement, the piston compresses the charge (with both valves closed). The spark is produced at 30 to 40 before the TDC position to give time for the fuel to ignite. (IGN) Due to ignition, the pressure rises up, the burnt gases pushes the piston downwards during expansion stroke. Now the exhaust valve open (EVO) before the piston reaches the BDC position. i.e. The exhaust valve opens

IC Engines 1.27

TD C

I

ON

T AR ST I.V.O

IG

35

SU o

N

M P R E S SI ON

IO

20

CT

E X PA N

CO

EX HA

o

IG N

US

T

NI

T

E .V.C

o

SIO

35 35

o

N

I.V.C

E .V.O

Fig. 1.9 Actual valve tim ing diagram for Petrol (Otto) four stroke engine

(EVO) at 30 to 50 before BDC to enable the burnt gases leave the cylinder. Now the piston reaches BDC and then it moves up to perform the exhaust stroke. The exhaust valve opens till the piston reaches TDC and then it moves down for suction stroke (i.e. till the piston is about 8  10 past the TDC). So the exhaust valve closes (EVC) at 8  10 past the TDC.

1.28

Thermal Engineering - I

But before starting of suction stroke, the inlet valve opens (IVO) at 15 10  20 before TDC. For a fraction of time, both inlet and exhaust valve are open. The angle between the position of the crank at the inlet valve opening and that at the exhaust valve closing is known as the angle of valve overlap or simply valve overlap. Review of Actual Cycle for Four Stroke Cycle Diesel Engine (Diesel Cycle) The Fig.1.10 (i) and (ii) shows the theoretical and actual cycle (P-V diagram) for four stroke cycle diesel engine. The actual cycle differs from the theoretical one. In actual cycle, the corners are rounded off because the inlet and exhaust valve do not open and close all of a sudden but they take sometime to do so. Due to the resistance of the inlet valve to the entering fresh air, the actual pressure inside the cylinder during suction stroke is slightly less than the atmospheric pressure.

2

3

p ressure

p ress ure

2

4 5 Ex hau st

b 5 Su ctio n a Vo lum e

1

3

Ex hau st b 5

Su ctio n a

Vo lum e (ii) Fig.1.10 (i) (i) Theoretical and (ii) Actual P-V diagram for a four stroke diesel engine

4

IC Engines 1.29

Similarly, due to the resistance of the exhaust valve to the exhaust gases leaving the cylinder, the actual pressure inside the cylinder during exhaust stroke is slightly higher than the atmospheric pressure. This gives small hatched area 5  a  1  b  5 in the form of a loop (called negative loop). This negative loop area is subtracted from the area of the larger loop (positive loop) to get net workdone during a cycle. 1.9.2 Valve Timing Diagram of 4-Stroke Diesel Engine The actual valve timing diagram of a 4-stroke diesel engine is shown in Fig.1.11. The inlet valve opens (IVO) at 10 to 25 before TDC position and inlet valve closes (IVC) after BDC position. Fuel Injection Valve (FIVO) opens at 5 to 10 before TDC position and fuel injection valve closes (FIVC) at 15 to 25 after TDC position. Exhaust valve opens (EVO) at 30 to 50 before BDC position and exhaust valve closes (EVC) at 10 to 15 after TDC position. 1.9.3 Port Timing Diagram of Two Stroke Cycle Petrol Engine The corners are rounded off since inlet port and exhaust port are covered and uncovered not in all of a sudden. Covering and uncovering ports take sometime. Port Timing Diagram Step 1: Transfer port closes when crank angle is at 50 after BDC (Suction stroke is accomplished during 120 of crank rotation. i.e. The angle between “transfer port opens” and “Transfer port closes” is 120.)

1.30

Thermal Engineering - I

T.D.C

E.V.C(Exha ust Valve C loses) e V a lv

on e c ti l In j ( F u e n s ) F.I.V.O O pe

F.I.V.C (Fu el Injection Va lve C loses)

o

Ex pa

o

C om pres sion

on

ion

c ti

o

25

E xh au st

ns

25

Su

I.V.O (In le t Va lve O pe ns)

o 5 15

45 30

o

o

I.V.C (In le t Valve C loses)

E.V.O (Exha ust Valve O p ens)

B.D.C

Fig. 1.11 Valve tim ing diagram of a four stro ke diesel en gine

Step 2: Exhaust port closes when crank angle is at 60 after BDC. i.e. A little time after transfer port closes, the exhaust port closes. Step 3: As soon as exhaust port closes, the compression starts. Then the inlet port opens at 50 before TDC. i.e. The crank angle between ‘Inlet port opens’ and ‘Inlet port closes’ is 100.

IC Engines 1.31

3

3 2

2

P ressure

Pressure

4 1

4 1

Volum e

Volum e (b) A ctual cy cle

(a) Theoritical cycle T.D.C

S te p 4 50

I.P.Closes

o

E xpa

5 0o

C om p ressio n

S te p 3 I.P.O pens

n s io n

1 00 o

S te p 5 E.P.O pens

S te p 2 E.P.Closes T.P.Closes S te p 1

T.P.O pens S te p 6

1 40 o 1 20

50

o

70o

o

60

o

80o Exh

aus

t

B.D.C (c.) P ort timing diagram o f a tw o stro ke petrol en gin e Fig. 1.12

Step 4: At the end of compression stroke, the spark plug produces spark. Due to combustion of air-fuel mixture, a large force is acted upon the piston moving downward. During downward movement of the piston, the inlet port is

1.32

Thermal Engineering - I

covered by the piston. i.e. The inlet port is closed at 50 after TDC. Step 5: Further downward movement of the piston, during expansion stroke, uncovers the exhaust port. i.e. Exhaust port is uncovered at 80 before BDC. (or) The angle between ‘Exhaust port opens’ and ‘Exhaust port closes’ is 140. The exhaust stroke starts when exhaust port is uncovered. Step 6: A little time after ‘Exhaust port opens’, the ‘transfer port opens’. As we have already seen, the angle between ‘Transfer port opens’ and ‘Transfer port closes’ is 120. i.e. The transfer port opens at 70 before BDC. 1.9.4 Port timing Diagram for Two Stroke Cycle Diesel Engine In actual cycle, the corners are rounded off, since inlet port and exhaust port are covered and uncovered not in all of a sudden. Covering and uncovering of ports takes sometime. Port Timing Diagram The actual valve timing diagram for two stroke diesel engine is similar to that of two stroke petrol engine. The difference between them are given below. 1.

In diesel engine, the air is sucked inside the cylinder instead of air-fuel mixture in case of petrol engine.

2.

In diesel engine, the fuel injector injects fuel instead of spark plug produces spark in case of petrol engine. Slight variations of crank angles for EPO, TPO, etc. which are discussed below:

IC Engines 1.33 2

3

F.I.V.O

8

1 Volum e (b) A ctua l cy cle

n (1 2 C o m p re s s io

o

2 0o ) o n (1

50

n si

o

50

pa

I .P .C

Ex

0 )o

St

O IP

E .P.C S te p 4

4

ep

S te p 5

F.I.V.C

P res s ure

20 o

S te

p

7

TDC F ue l s u pp ly S te p 6 o 15

E .P.O

T.P.C

T.P.O

S te p 1

E xh aus t S te p 3 60

o

45 o

45

o

60

o

(c .)

S te p 2

BDC

Fig . 1 .13 P ort tim in g diagra m for tw o s tro ke die sel engine

Step 1: Exhaust port opens (EPO) at 60 before BDC position. Step 2: Transfer port opens (TPO) at 45 before BDC position. Step 3: Transfer port closes (TPC) at 45 after BDC position. Step 4: Exhaust port closes (EPC) at 60 after BDC position.

1.34

Thermal Engineering - I

Step 5: Inlet port opens at 50 before TDC position. Step 6: Fuel injection valve opens (FIVO) at 15 before TDC position. Step 7: Fuel injection valve closes (FIVC) at 20 after TDC position. Step 8: Inlet port closes (IPC) at 50 after TDC. By seeing port timing diagram, the following points are noted. The compression stroke occurs for angle of 120 of crank rotation. The fuel supply occurs for angle of 35 of crank rotation. The suction stroke occurs for 100 of crank rotation. The expansion stroke occurs for 120 of crank rotation. The exhaust stroke occurs for angle of 120 of crank rotation.

1.10 FUEL SUPPLY SYSTEM 1.10.1 Fuel injection Systems for S.I. Engines To run S.I. engine, the petrol from the fuel tank must reach cylinder. The petrol vaporize easily at atmospheric condition, therefore the engine suction is sufficient to vaporize petrol. In petrol engine, the petrol from the fuel tank reaches through the fuel pump, filter and carburetor to the cylinder. Thus, the fuel feed system of a petrol engine consists of the following components.

IC Engines 1.35

1. Fuel tank, 2. Fuel pump, 3. Fuel filter 4. Carburetor, 5. Intake manifold, 6. Fuel tubes for necessary connections, 7. Gauge to indicate to the driver the fuel level in the fuel tank. The fuel system is used for the following reasons.   

To store fuel in the fuel tank To supply fuel to the required amount and proper condition To indicate to the driver the fuel level in the fuel tank.

1.10.2 Different types of Fuel Systems The fuel from the fuel tank is supplied to the engine cylinder by the following systems: (a) Gravity system, (b) Pressure system, (c) Vacuum system, (d) Pump system, (e) Fuel injection system In gravity system, the fuel tank is placed above the carburetor. The fuel flows from the tank to the carburetor due to the gravitational force. Thus the system does not have fuel pump. This system is cheap and simple one. The fuel tank is directly connected to the carburetor. Motor cycles and scooters use this system. In pressure system, the pressure is created inside the tank by means of a pump, and the fuel flows to the

1.36

Thermal Engineering - I

carburetor. In this system the tank can be placed above or below the carburetor. In a vacuum system, the engine suction is used for sucking the petrol from the main tank to the auxiliary fuel tank and then it flows by gravity to the carburetor. In pump system, a fuel feed pump is used to feed the petrol from the fuel tank to the carburetor. In this system the fuel tank can be placed at any suitable position in the vehicle. In fuel injection system, a fuel injection pump is used in place of carburetor. The fuel is atomized by means of a nozzle and then delivered into an air stream. 1.10.3 FUEL INJECTION SYSTEM IN SI ENGINES A schematic diagram of fuel supply system is shown in Fig. Here, the storage tank is located below the carburetor the fuel pump sucks the petrol from tank and pumps it to carburetor through fuel filter. Filter is used to prevent the dust and other materials going along with petrol. A Schematic diagram of fuel injection system

Storage Tank

Fu el Pum p

Fu el filter

Carburetor

Engin e

Fig 1.14 A Schem atic diagram of fuel supply system .

1.10.4 Fuel Pump (for S.I. Engine) This type of pump is used in petrol engine when the cam shaft rotates, it pushes the lever in upward direction.

IC Engines 1.37

O utlet valve S traine r

D ia p hra gm

P um p C ha mb er

S pring H in g ed po int G la ss b ow l C am Fig.1.15 Fuel pu mp for S I Engin e

This upward movement pulls the diaphragm downward. It creates a vacuum in the pump chamber and the petrol comes to pump chamber from the glass bowl. Strainer is used to prevent the impurities of the fuel coming along with fuel. On the return stroke, the spring pushes the diaphragm in the upward direction and the petrol is forced to carburetor. 1.11 Carburetor Carburetor is a device which is used for atomizing and vapourizing the fuel (petrol) and mixing it with the air in varying proportions, to suit the changing operating conditions of the engine. Atomization is the breaking up the liquid fuel (petrol) into very small particles so that it is properly mixed with the air. But vaporization is the change of state of the fuel from liquid to vapour. Carburetor performs both the process i.e., atomization of the fuel and vaporization of the fuel.

1.38

Thermal Engineering - I

The working of all modern carburetors are based upon Bernoulli’s theorem. 1.11.1 SIMPLE CARBURETOR A simple carburetor consists of following 1. float and float chamber, 2. venturi and throttle valves and 3. choke valve. 1. Float and Float chamber The petrol is supplied to the float chamber from the fuel tank through the filter and fuel pump. When the petrol in float chamber reaches a particular level, the needle valve blocks the inlet passage and thus cuts off the petrol supply. On the fall of the petrol level in the float chamber, the float descends down and inlet passage opens. The petrol is supplied to the chamber again. Thus a constant fuel (petrol)

To E n g ine T h rottle Va lv e

F u e l Inle t M ixtu re Ve n t

N e e d le va lv e

Fuel Je t Ve n tu ri

F lo at 2

x

2

F lo at ch a m b er C h o ke Va lv e Fig.1.16 S im ple C arburetor

A ir

IC Engines 1.39

level is maintained in the float chamber. The float chamber supplies the petrol to the main discharge jet placed in venturi tube. The level of fuel in the float chamber is kept slightly below the top of the jet to prevent the leakage when not operating. 2. Venturi and Throttle valve The carburetor consists of a narrower passage at its centre, called venturi. One end of the carburetor is connected with the intake manifold of the engine. During the suction stroke, vacuum is created inside the cylinder. Due to vacuum, the air is sucked to the carburetor. The velocity of the air increases as it passes through the venturi where the area of cross section is minimum. Due to increased velocity of air at the venturi, the pressure at the venturi decreases. Therefore a low pressure zone is created in the venturi. So the jet (nozzle) located at the venturi is in the zone of low pressure. The fuel comes out from jet (nozzle) in the form of fine spray. This fuel spray is mixed with air and the mixture is supplied to the intake manifold of the engine. The throttle valve is placed between the jet (nozzle) and the intake manifold of the engine. The quantity of the mixture is controlled by means of throttle valve. 3. Choke Valve While starting in cold weather the engine needs extra rich mixture. So a choke valve is introduced in the air passage before the venturi. When the choke valve is closed it creates high vacuum near the fuel jet and small quantity of air is allowed, to get rich mixture. The fuel flow increases as the vacuum near the jet increases.

1.40

Thermal Engineering - I

1.12 AIR FUEL RATIO Oxygen is very much necessary to burn the fuel. This oxygen is taken from atmospheric air. The proper proportion of air and fuel mixture should be obtained for complete combustion of fuel. For complete combustion, the Air-Fuel ratio should be approximately 15:1 by weight. This is known as chemically correct or stoichiometric air fuel ratio. The normal range of Air fuel ratio is in between 20:1 to 8:1 approximately. Air fuel ratio during starting is approximately 10:1 i.e., very rich mixture. Air fuel ratio during idling speed (low speed) is approximately 12:1 - i.e., rich mixture. Air fuel ratio during normal running condition, is approximately 15:1 neither rich nor lean mixture. Air fuel ratio for economic running (medium load), is approximately 17:1 - economic mixture. Air fuel ratio during overtaking, is approximately 12:1 - rich mixture.

1.13 VARIOUS COMPENSATION IN CARBURETORS A simple carburetor can not supply different air-fuel ratio according to the speeds and loads of the engine. To supply correct airfuel ratio to meet the existing condition is known as the compensation in carburetor. The various compensations in carburetor are given below. 1. Auxiliary (or) extra air valve compensation

IC Engines 1.41

2. Restricted air bleed compensation 3. Compensating jet compensation 4. Economiser needle in metering jet. 1. Auxiliary (or) Extra air valve compensation A ir E xtra a ir va lve A ir P a rt o f floa t ch a m b er Th ro ttle Fig .1.17 To in d uctio n m an ifo ld A uxilia ry (o r) Extra air va lve co m pensa tio n

An extra air valve is provided to the carburetor to supply extra air to mixture, when the throttle valve is opened more and more. So the air-fuel ratio (mixture strength) is maintained constant. 2. Restricted air-bleed compensation Here, a jet tube having openings at its periphery is provided in the carburetor. A restricted air bleed opening is connecting the main air passage to the outer enclosure of the jet tube. During starting and slow speed, more quantity of fuel flows into venturi to give rich mixture. During high speed, the throttle valve opens more and the vacuum in the venturi become more. So more fuel is

1.42

Thermal Engineering - I

A ir

R e stricted a ir ble ed o pe nin g O u te r en clo ser

Je t Tu be

P art o f flo at ch am be r

T hrottle To ind uction m anifold R estricted air-blee d co m pe nsatio n

F ig .1.18

drawn and sprayed by nozzle. But at this stage, the air bubbles start bleeding through the jet-tube and make the mixture lean. 3. Compensating Jet Compensation In this system, main jet and compensating jet are provided in the carburetor. The main jet is connected to float chamber directly. The compensating jet is connected to float chamber through tube C whose top end is open to atmosphere. For normal throttle valve openings, both the jets supply fuel to venturi. But when the throttle opens more and more, the fuel supply from main-jet increases and the fuel supply from compensating jet decreases due to falling level of fuel in tube C. Because of atmospheric pressure acting in this tube C, the richness of mixture decreases.

IC Engines 1.43

To in du ction m an ifo ld T hrottle Op en to atmo sp here

P art of floa t cha m b er

c

A M ain jet B Com pensating jet

A ir

Fig.1.19 Com pensating jet Com pensation

4. Economiser needle in metering jet The flow of fuel is controlled by changing the area of the metering nozzle supplying fuel from float chamber to the main jet. The area is changed by means of a needle operated by linkage with accelerator pedal.

1.14 TYPES OF CARBURETORS There are three important types of carburetor 1. Zenith carburetor 2. Solex Carburetor 3. Amal Carburetor 1.14.1 Simple Calculations involved in Carburetors We assume that air is incompressible fluid, and the flow is expressed by using Bernoullis energy equation. Here ‘Suffix a’ stands for air and ‘Suffix f’ stands for fuel.

1.44

Thermal Engineering - I

 Mass of air flowing in kg/sec  m a  a A a V a

where a  density of air constant in kg/m 3 V a  Ve locity of air in m/sec A a  Cross sectional area of throat in m 2 V 2  Ve locity of air at throat 1

V a  Ca V 2  C a

   P2    2 RT 1  1    P 1  1 

  

 ma Air fuel ratio   mf  where m f  Mass of fuel flow in kg /sec  ma Aa Ca Air fuel ratio    mf A f C f

    pa  x  

 m a in kg/sec  A a C a

2 a pa  

 Air/fuel  m f  Aa Ca

2  a  pa  

 m f  Af C f

a  p a

f

 f 2 f pa  x  9.81    

V a critical  C a

 

2x  9.81 f a

f

IC Engines 1.45

where C a  Coefficient of discharge for airflowin g venturi throat C f  Coefficient of discharge for fuel jet pa  Pressure of air in N/m2 A f  Cross sectional area fuel jet x  height of jet above the float chambe r level . It is called as lip of jet in m. pa  Drop in pressure causing air flow f  Density of fuel

PROBLEMS IN CARBURETOR Problem 1.1: A petrol engine has a fuel consumption of 20 lits per hour. The air fuel ratio supplied through the carburetor is 15. The choke has a diameter of 20 mm. Determine the diameter of fuel jet of carburetor if the top of the jet is 5 mm above the fuel level in the float chamber. The barometer reads 750 mm of Hg and the temperature is 32C. Neglect the compressibility of air. Assume Ca  0.85 and Cf  0.7; f  700 kg/m3

Solution: A/F  15; Fuel consumption = 20 lit /hr or volume flow

of fuel in m 3/sec , Vf 

20  5.5556  10  6 m 3/sec 1000  3600

f  700 kg/m 3; C a  0.85 ; C f  0.7 Choke dia d a  0.02 m ; dia of jet df  ?; x  0.005 m

1.46

Thermal Engineering - I

Atmospheric pressure 

750  1.01325  0.99992 bar 760

 0.99992  10 5 N/m 2 Atmo spheric Temperature  32  273  305 K

Density of air: a P Va  Ra T Va 

a 

Ra T P

and a 

P Ra T

0.99992  10 5  1.1423 kg/m 3 287  305

Drop in pressure causing air flow:  pa  m a  Aa Ca

pa 2 a    

and

  m a  A/F  mf   m 3 kg  Mass flow m f  Vf  f in sec m 3  5.5556  10  6  700

 m f  3.88889  10  3 kg/sec   M ass flow ma of air  A/F  mf  15  3.88889  10  3  0.058333 kg/sec  Also m a  Aa Ca  2  a  pa 

IC Engines 1.47

  0.058333    0.02 2   0.85  4

2  1.1423   pa  

p   a  144.524 pa  20,887.2

To find Af Area of a jet  m f  Af Cf  2 f pa  x f  3.88889  10  3  A f  0.7

2  700 20887.2  0.005  700   

A f  1.0282  10  6 m 2 

 2 d 4 f

d f  1.1442  10  3 m Diameter of fuel jet d f  1.1442 mm Problem 1.2: A single jet simple carburetor is to supply 6 kg/min of air and 0.45 kg/min of petrol whose density is 740 kg/m3. The air is initially at 1.027 bar and 27C. Calculate the throat diameter of the venturi throat if the speed of air is 92 m/sec, assuming velocity coefficient of 0.8. Assume adiabatic expansion and  for air is 1.4. If the pressure drop across the fuel metering orifice is 0.75 of that at the throat, calculate orifice diameter assuming Cf  0.6, Ca  0.8

Solution: T 1  27  273  310 K

1.48

Thermal Engineering - I

To find Dia of throat

Va  Ca

 

92  0.8

 

 P 2   1/   2 R T1  1     P 1  1   0.4

 P2  1.4   2  1.4  287  310  1     0.4 P  1  

0.4

 P 2  1.4  P   0.97876  1 P2 P1

 0.927629

P2  1.027  0.927629  0.952675 bar 1

 P2     P T1  1

T2

T2  310

 0.4    0.927629  1.4 

 303.42 K  ma   a A a V a

a 

P2 R T2



0.95 2675  10 5  1.094 kg/m 3 287  303.42

a  1.094 kg/m 3  m a  a Aa Va

IC Engines 1.49

6 kg/sec  1.094  A a  92 60 A a  9.93549  10  4m 2  2 d  9.93549  10  4 4 a da  0.035567 m Diameter of throat  3.5567 c m

To find Diameter of jet ‘d f’ Pressure drop at venturi P 1  P 2  1.027  0.9527  0.0743 bar Pressure drop at jet  0.75  0.0743  0.055725 bar  m f  Af C f  2 f pressure drop a t jet  0.45 kg/sec  Af  0.6 60

2  740  0.055725  10 5 

A f  4.35265  10  6 

 2 d 4 f

d f  2.3541  10  3 m  2.354 mm Problem 1.3: A carburetor consumes 6 Kg of fuel/hr. The density of fuel is 700 kg/m3. The level of fuel in the float chamber is 3 mm below the top of jet. Ambient conditions are 1.01325 bar and Temperature 17C. The jet diameter is 1.2mm

1.50

Thermal Engineering - I

and its discharge coefficient is 0.6. The coefficient of discharge for air is 0.85. A/F ratio is 15. Determine the critical air velocity and throat diameter. Express the pressure depression in mm of water. Neglect compressibility of air.

Solution: Given  mf 

6  1.6667  10  3 kg/sec 3600

a  700 k g/m 3 ; x  0.003 m P 1  1.01325 bar ; T 1  17  273  290 K d f  2mm  0.002 m; C f  0.6 ; Ca  0.85 ; A/F  15

Density of air a a 

P1 R T1



1.01325  10 5  1.21741 kg/m 3 287  290

a  1.21741 kg /m3 (Compressibility of air is neglected) To find critical velocity Critical Velocity V a  C a

 

 0.85

2x  9.81 f a

 

2  0.0003  9.81  700 1.21741

V a  4.945 m/sec

IC Engines 1.51

[critical velocity is the minimum velocity of air at the throat of venturi at which the fuel begins to flow.] 1mm of H 2O  9.81 N/m 2

So 3 mm height  0.003  9.81 To find pressure drop  f m f  Af Cf  2 f pa  x  9.81     1.6667  10  3    0.00132   0.6   4 2  700 pa  0.003  9.81  700   pa  20.601   4309.03 pa  4329.03 N/m 2 1mm of H 2O  9.81 N/m 2 pa 4329.03 N/m 2 

4329.03  441.3 mm o f water 9.81

To find throat dia ‘d a’   m a  A /F  m f  15  1.6667  10 3  25  10  3kg/sec  m a  A a C a 2 a pa 25  10  3  A a  0.85

2  1.21741  4329.03  

A a  0.28647  10  3  d a  0.0191 m

 2 d 4 a

1.52

Thermal Engineering - I

d a  19.1 mm Problem 1.4: A four cylinder - 4 stroke engine has bore 10 cm and stroke length 12 cm. It is running at 2000 rpm. Its carburetor throat dia is 8cm. Determine the suction pressure at

the

throat.

Assume

volumetric

efficiency

=

70%

3

a  1.3 kg/m ; Ca  0.8; Initial pressure  P1  1.027 bar Solution:  Stroke volume V s   D2  L  4 4 

  0.1 2  0.12  4  3.7699  10  3 m 3 4

Actual volume  0.7  V s  2.6389  10  3 m 3 Actua l volume in m 3/sec  2.6389  10  3 

N for 4 stroke 2  60

engine  2.6389  10  3 

2000 2  60

V a  0.043982 m 3/sec   M ass of air flow m a  V a  a  0.043982  1.2  m a  0.0528 kg /sec  Also m a  A a C a 0.0528 

2  a  pa  

  0.082  4

2  1.2  pa  

IC Engines 1.53

pa  71.835 N /m 2  71.835  10  5 bar Suction pressure a t throat  P 1  P 2  1.027  71.835  10  5  1.02628 bar

1.15 FUEL INJECTION SYSTEMS FOR C.I ENGINES The fuel supply system of a diesel engine (CI Engine) consists of 1. fuel tank, 2. fuel filter, 3. injection pump, 4. injector, 5. fuel lines for necessary connections, and 6. fuel gauge. In petrol engines, carburetor and spark plug are used. In Diesel engine, fuel injector is used. Remaining elements are same for both types of engines. The fuel from the fuel tank flows to pump. It then passes out to the inlet side of the main fuel filter. Then filtered fuel proceeds to the inlet side of the fuel injection pump. From the injection pump it flows under pressure in the feed pipes leading to the fuel injectors. 1.15.1 Fuel Pump (C.I. Engine) The plunger is driven by cam and tappet mechanism. The plunger is reciprocating inside a barrel. The plunger has a vertical groove at its top side and helical groove at its bottom side. Once the diesel in the barrel gets enough pressure, it lifts off the delivery valve from its seat. When diesel pressure is down, the spring pushes the delivery valve down on its seat.

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Thermal Engineering - I

The top passage is connected to fuel injector. Diesel is supplied to barrel through supply port. When the plunger is at its bottom, the diesel enters through supply port. When the plunger moves up, both supply port and spill port are closed and the diesel gets compressed. Now the pressure of diesel increases. The high diesel lifts delivery valve, fuel injector passage.

pressure off the flows to through

Pa ssa ge (P )

Sp rin g (S)

D e live ry valve (V )

Ba rre l (B ) Sp ill P ort (SP ) Su pp ly p ort

Plun ge r (L ) R o ck(R .)

Fig 1.20 Fu el p um p (fo r C .I. Engine )

When the plunger moves up further, the spill port is uncovered. i.e., the spill port is connected to the diesel in the top of plunger through vertical groove and helical groove at its side. So the pressure of the diesel suddenly comes down. The delivery valve comes to its seat and stops the diesel flow to injector. Now the diesel flows through spill port. The plunger is rotated by the action of Rack and Pinion arrangement which is activated by governor. By rotating the plunger, the angular position of helical groove to supply port is changed. It changes the length of stroke. So the quantity of diesel supplied to engine is also changed

IC Engines 1.55

accordingly. So the rack and pinion arrangement is used to control the quantity of diesel supplied to engine. Requirements of Fuel Injection System (a)

The fuel injection should occur at the correct moment, rate and quantity as required by the engine at different load conditions.

(b)

The fuel should be injected in a finely atomized condition.

(c)

The fuel should be distributed uniformly inside the combustion chamber.

(d)

The beginning and end of injection should take place sharply.

1.16 FUEL INJECTION SYSTEM There are two methods of fuel injection in compression ignition engines. 1. air injection, 2. airless or solid or mechanical injection. 1. Air Injection System This system was developed by Dr.Rudolph Diesel. In this method, air is first compressed to a very high pressure. A blast of this air is then injected carrying the fuel along with it into the cylinder. The high pressure air requires a multistage compressor. The compressor consumes more power developed by the engine, thus decreasing the net output of the engine. This method of fuel injection is costly and complicated. Therefore it is not used in nowadays. 2. Airless or Solid Injection In this system the fuel is supplied at a very high pressure about (150 bar) from the fuel pump to the fuel injector and from there it is injected to the combustion chamber. It burns due to the heat of compressed air. This

1.56

Thermal Engineering - I

method requires a fuel pump. This method is used in all types of diesel engines. 1.16.1 Fuel Injector The fuel injector is used (a)

To atomize the fuel to the required extent.

(b)

To distribute the fuel such that there is complete mixing of fuel and air.

(c)

To start and stop fuel injection instantaneously. S pring loaded fuel injector Lo ck n u t

Le ak-off co nne ctio n

A d ju s tin g S crew

S p ring

F u el inlet S p in d le

F u el d uct

C a p n ut

S tea m N o zzle bo dy

Fig.1 .21

Va lve

Fuel Injecter or A to m iser

IC Engines 1.57

It consists of a needle valve fitted on its seat in the nozzle body by a spindle. The spring controls the pressure on the spindle by which the needle valve opens. The nozzle is attached to the body by means of a capnut. The fuel enters the nozzle through holes in the injector body. When the needle valve is raised from its seat by the pressure of the fuel, the injection of the fuel into the combustion chamber takes place. When the injection pressure falls below the spring pressure, the valve closes.

1.17 IGNITION SYSTEMS In case of petrol engines during suction operation, charge of air and petrol fuel will be taken in. During compression, this charge is compressed by the upward moving piston. And just before the end of compression, the charge of air and petrol fuel will be ignited by means of the spark produced by means of spark plug. And the ignition system does the function of producing the spark in case of spark ignition engines. 1.17.1 SPARK PLUG

Term ina l

Spark plug is used in SI engines (Petrol engines) to produce electric spark to ignite the compressed air fuel mixture inside the engine cylinder.

C en tral e le ctrod e

The spark plug consists of three main parts: (a)

A central electrode.

(b)

A metal screw with a ground electrode.

P orce la in in su la to r

M e ta l S crew

S pa rk g a p or G ro u nd e le ctro de A ir g a p Fig 1.22 Spark plug

1.58

(c)

Thermal Engineering - I

An porcelain insulator separating the two electrodes.

The central electrode in the spark plug is covered by a porcelain insulator. The central electrode extends for a short length from the bottom of the insulator. The upper end of the central electrode is connected to the cable from the ignition coil. A metal screw surrounds the bottom part of the insulator. The lower portion of the screw is attached to a ground electrode and bent towards the central electrode so that there is a gap between the two electrodes. The air gap is generally kept between 0.6 mm to 1 mm. The high voltage current is given to the terminal of the central electrode. This current jumps in the air gap between the central electrode and ground electrode. Too large or too small air gap reduces the efficiency of the entire ignition system. A spark plug will fail in its function due to the following reasons: (a)

The plug may fail due to engine oil entering the combustion chamber.

(b)

Plug fouled by too rich mixture.

(c)

Plug gap is too large (or) too small.

(d)

Plug gap filled with carbon deposits.

(e)

Burned electrode.

(f)

Cracked or broken insulator sealing.

Fig. 1.23 shows a typical spark plug used for petrol engines. It mainly consists of a central electrode and metal tongue. Central electrode is covered by means of porcelain insulating material. Through the metal screw, the spark plug is fitted in the cylinder head. When the high tension voltage of the order of 30000 volts is applied across the

IC Engines 1.59

C o nta ct

P o rcela in insulator

C e ntral elec tro de

G as tig ht se al S e alin g w as he r M etal scre w M etal to ng ue S p ark g ap Fig. 1.23 Spark plu g

electrodes, current jumps from one electrode to another producing a spark. Whereas in case of diesel (Compression Ignition-CI) engines, only air is taken in during suction operation and is compressed during compression operation and just before the end of compression, when diesel fuel is injected, it gets ignited due to heat of compressed air. Once the charge is ignited, combustion starts and products of combustion expand, i.e. they force the piston to move downwards i.e. they produce power and after producing the power, the gases are exhausted during exhaust operation. Basically Conventional Ignition systems are of 2 types (a) Battery or Coil Ignition System, and (b) Magneto Ignition System. Both these conventional ignition systems work on mutual electromagnetic induction principle. Battery ignition

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Thermal Engineering - I

system was generally used in 4-wheelers, but now-a-days it is more commonly used in 2-wheelers also (i.e. Button start, 2-wheelers like Pulsar, Kinetic Honda, Honda-Activa, Scooty, Fiero, etc.). In this case 6 V or 12 V batteries will supply necessary current in the primary winding. Magneto ignition system is mainly used in 2-wheelers, kick start engines. (Example, Bajaj Scooters, Boxer, Victor, Splendor, Passion, etc.). In this case magneto will produce and supply current to the primary winding. So in magneto ignition system, magneto replaces the battery. (a) Battery or Coil Ignition System Fig.1.24 shows line diagram of battery ignition system for a 4-cylinder petrol engine. It mainly consists of a 6 or 12 volt battery, ammeter, ignition switch, auto-transformer (step up transformer), contact breaker, capacitor, distributor rotor, distributor contact points, spark Prim ary w ind in g (2 00 -30 0 tu rns o f 2 0 g au g e w ire) Ig nitio n sw itch

C o il

Se co n dary w in d in g (2 10 0 turn s of 4 0 g au g e w ire)

(2 00 00 -30 0 00 V) D istribu to r con ta cts

1 1

2

C o nta ct b rea ke r

Am m eter

2 3

4

C a pa citor

3 4 Sp ark p lug s

Ba tte ry (6 or 12 V)

C o nta ct b rea ker o pe rating cam

D istribu to r

Fig. 1.24: Schem atic diagram of coil/b attery ign itio n system

IC Engines 1.61

plugs, etc. Note that the Fig.1.24 shows the ignition system for 4-cylinder petrol engine, here there are 4-spark plugs and contact breaker cam has 4-corners. (If it is for 6-cylinder engine it will have 6-spark plugs and contact breaker cam will be a perfect hexagon). The ignition system is divided into 2-circuits : (i)

Primary Circuit: It consists of 6 or 12 V battery, ammeter, ignition switch, primary winding having 200-300 turns of 20 SWG (Sharps Wire Gauge) gauge wire, contact breaker and capacitor.

(ii)

Secondary Circuit: It consists of secondary winding. Secondary Ignition Systems winding consists of about 21000 turns of 40 (SWG) gauge wire. Bottom end of which is connected to bottom end of primary coil and top end of secondary winding is connected to centre of distributor rotor. Distributor rotors rotate and make contacts with contact points and are connected to spark plugs which are fitted in cylinder heads (engine earth).

Working When the ignition switch is closed and engine is cranked, the contact breaker closes and a low voltage current will flow through the primary winding. The contact beaker cam opens and closes the circuit 4-times (for 4 cylinders) in one revolution. When the contact breaker opens, (breaks the contact), the magnetic field begins to collapse. Because of this collapsing magnetic field, current will be induced in the secondary winding. And because of more turns (21000 turns) of secondary winding, voltage goes upto 28000-30000 volts.

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Thermal Engineering - I

This high voltage current is brought to centre of the distributor rotor. Distributor rotor rotates and supplies this high voltage current to proper spark plug depending upon the engine firing order. When the high voltage current jumps the spark plug gap, it produces the spark and the air-fuel charge is ignited-combustion starts-products of combustion expand and produce power. Note: (a)

The function of the capacitor is to reduce arcing at the contact breaker (CB) points. Also when the CB opens, the magnetic field in the primary winding begins to collapse. When the magnetic field is collapsing, capacitor gets fully charged and then it starts discharging and helps in building up of voltage in secondary winding.

(b)

Contact breaker cam and distributor rotor are mounted on the same shaft.

In 2-stroke cycle engines, these are operated at the same engine speed. And in 4-stroke cycle engines, they are operated at half the engine speed. Advantages of Battery Ignition System (a)

It gives better spark during starting speed and idling speed.

(b)

Its initial cost is low. For this reason this system is used in cars and commercial vehicles.

(c)

The maintenance cost is less.

(d)

The mechanism of distributor drive is simple.

Disadvantages of Battery Ignition System (a)

The engine cannot be started if the battery is low (or) discharged.

IC Engines 1.63

(b)

It occupies more space.

(c)

It has a complicated wiring.

(d)

The spark intensity falls as the engine speed increases.

(b) Magneto Ignition System In this case, magneto will produce and supply the required current to the primary winding. Here, a rotating magneto with fixed coil or a rotating coil with fixed magneto is set up for producing and supplying current to primary, remaining arrangement is same as that of a battery ignition system. D istribu to r contact points D istribu to r ro to r 1 2 3

C o il

4 Sp ark p lu gs

Cam

N S

Prim ary w ind in g Se con dary w in ding

C o ntact b rea ker

R o tating m a gn et tw o pole sh ow n

Ig nition sw tich C a pacitor Fig. 1.25 :Schematic diagram of Magneto Ig nition system

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Thermal Engineering - I

Both these conventional ignition systems work on mutual electromagnetic induction principle. Following are the drawbacks of conventional ignition systems: (a)

The arcing and pitting of contact breaker points will lead to regular maintenance problems.

(b)

Poor starting: After few thousands of kilometers of running, the timing becomes inaccurate, which results in poor starting (Starting trouble).

(c)

At very high engine speed, performance is poor because of inertia effects of the moving parts in the system.

Advantages of Magneto Ignition System (a) It occupies less space. (b) Wiring is simple. (c) Spark intensity improves as the engine speed increases. (d) It is used in motor cycles, scooters, and racing cars. Disadvantages of Magneto Ignition System (a) Starting is difficult. (b) Cost is more. (c) It does not give better spark during starting speed and idling speed.

IC Engines 1.65

Comparison between Battery and Magneto ignition system S. No.

1. 2.

3. 4. 5.

Battery Ignition

Magneto Ignition

Battery is must No need of battery A good spark is available During starting the at low speed also. quality of spark is poor due to slow speed. Occupies more space Very much compact Battery maintenance is No battery maintenance required problems. Mostly used in car and bus Used on motorcycles, for which it is required to Scooters etc., crank the engine

1.17.2 Types of Electronic ignition systems: (a) Capacitance Discharge Ignition system (b) Transistorized coil Ignition system (c) Piezo-electric Ignition system (d) The Texaco Ignition system (a) Capacitance Discharge Ignition System It mainly consists of 6-12 V battery, ignition switch, DC convertor, charging resistance, tank capacitor, Silicon Controlled Rectifier (SCR), SCR-triggering device, step up transformer and spark plugs. A 6-12 volt battery is connected to DC converter i.e. power circuit through the ignition switch, which is designed to increase the voltage to 250-350 volts. This high voltage is used to charge the tank capacitor (or condenser) to this voltage through the charging resistance. The charging resistance is also so designed that it controls the required current in the SCR.

Thermal Engineering - I DC con vertor (C h arg ing resista nce ) R SCR

To spa rk plug

P rim ary

Ignition sw itch

2 50 V 3 50 V B attery 6 -12 V

Tan k cap acitor C o r condenser

S econda ry

1.66

SCR trigg ering d evice

Fig.1.26 C apacitance d ischarge ignition system

S tep -up transfo rm er

Depending upon the engine firing order, whenever the SCR triggering device sends a pulse, then the current flowing through the primary winding is stopped, and the magnetic field begins to collapse. This collapsing magnetic field will induce or step up high voltage current in the secondary winding, which while jumping through the spark plug gap produces the spark, and the charge of air fuel mixture is ignited. Advantages of capacitance discharge ignition system: (a)

Moving parts are absent-so no maintenance.

(b)

Contact breaker points are absent-so no arcing.

(c)

Spark plug life increases by 50% and they can be used for about 60000 km with out any problem.

(d)

Better combustion in combustion chamber, about 90-95% of air fuel mixture is burnt compared with 70-75% with conventional ignition system.

(e)

More power output.

(f)

More fuel efficiency.

IC Engines 1.67

(b) Transistorized Assisted Contact (TAC) Ignition System This system incorporates normal mechanical breakers, which drives a transistor to control the current in the primary circuit. Since a very small breaker current is used, erosion of the contacts is eliminated so that good coil output is maintained. Also it provides accurate spark timing for a much longer period. When a low inductive coil and ballast resistor are used with this system, excessive contact arcing produced by the high primary current is also eliminated. The basic principle of a breaker-triggered, inductive, semiconductor ignition system is illustrated in Fig.1.27. Here, a transistor works as a contact breaker, by acting as power switch to make and break the primary circuit. The transistor also performs as a relay, which is operated by the current supplied by a cam-operated control switch and thereby called as breaker-trigger. A small control current passes through the base-emitter of the transistor when the contact breaker is

Ig nitio n sw tich P

S C o nta ct b reake r

Tra ns is tor R1

R2

Fig. 1.27 T.A.C Ign itio n S ystem .

1.68

Thermal Engineering - I

in closed condition. This switches-on the collector-emitter circuit of the transistor and allows full current to flow through the primary circuit to energize the coil. The flow of current, at this stage, in the control circuit and transistor base is governed by the total and relative values of the resistors R1 and R 2. These resistance values are chosen to provide a control current of about 0.3A, which is sufficient to provide a self-cleaning action of the contact surfaces without overloading the breaker. When the spark is required, the cam opens the contact to interrupt the base circuit, which causes the transistor to switch-off. With sudden opening of the primary circuit, a high voltage is induced into the secondary, which produces a spark at the spark plug. This sequence is repeated to provide the required number of sparks per each revolution of the cam. The T.A.C. arrangement provides a quicker break of the circuit compared with a non- transistorised system, and, as a result, a more rapid collapse of the magnetic flux takes place. Consequently a high HT secondary voltage is obtained. The components of this ignition system are similar to those used with a conventional system except for the extra control module containing the power transistor. Advantages (a)

Breaker point contact surfaces last the life of the engine.

(b)

Improved cold starting due to faster rise time offered by transistor switching.

(c)

Improved consistency and repeatability of secondary voltage energy and waveform without degradation over time.

IC Engines 1.69

(d)

Ability to fire partially fouled plugs aforementioned rise time improvement.

due

to

Disadvantages (a)

As in the conventional system, mechanical breaker points are necessary for timing the spark.

(b)

The cost of the ignition system is increased.

(c)

The voltage rise-time at the spark plug is about the same as before.

1.18 LUBRICATION SYSTEM Lubrication is necessary for proper maintenance of a motor vehicle. Use of lubricating oil between the moving parts is known as lubrication. (The moving parts are subjected to wear and tear due to continuous rubbing action of one part on another.) The main parts to be lubricated in case of IC engines are: (a) cylinder walls, (b) crankshaft main bearing, (c) big end bearing of connecting rod, (d) small end bearing of connecting rod, (e) cam faces where they engage with the tappets, (f) push rod guides, (g) rocker arm pin, (h) valve guides, (i) timing gears, and (j) cam shaft bearings. 1.18.1 Purpose of Lubrication (a)

To reduce friction between the mating parts.

(b)

To reduce wear and tear of the moving parts.

(c)

To keep the engine parts clean.

(d)

To absorb shock between bearing and other engine parts.

(e)

To reduce noise and to increase engine life.

(f)

To act as a cooling medium for removing heat.

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Thermal Engineering - I

(g)

To form a good seal between piston rings and cylinder walls.

(h)

To absorb and carry away harmful substances resulting from incomplete combustion.

(i)

To prevent deposition of carbon, soot and lacquer.

(j)

To prevent metallic components from corrosive attack due to acid formation during combustion process.

1.18.2 Systems of Lubrication The different systems for lubricating the engine are: (a) petroil system, (b) splash system, (c) pressure system, (d) semi-pressure system, (e) dry pump system. (a) Petroil System This system of lubrication is generally used in two stroke petrol engines like scooters and motor cycles. In this system oil pump is not required for the purpose of lubrication. The lubricating oil is mixed into the petrol itself in a specific ratio. When the fuel goes into the crank chamber during the engine operation, the oil particles go deep into the bearing surfaces and lubricates them. The piston rings, cylinder walls, piston pin etc are lubricated in the same way. If the engine remains unused for a long period, the lubricating oil separates off from petrol and cloggs the passage of the carburettor, resulting in the starting trouble of the engine. This is the main disadvantage of this system. (b) Splash System In this method, the lubricating oil is stored in an oil sump. A dipper or scoop is made in the lowest part of the connecting rod. When the engine runs, the dipper dips in

IC Engines 1.71

G udge on pin

O il gaug e

C onne cting rod O il hole

Journa l

Scoop

oil sum p

~

Pum p

Pressu re releas e valve

Fig.1.28 Sp lash Lub rication

the oil once in every revolution of the crankshaft and the oil is splashed on the cylinder walls. In this way the engine wall, piston pin, piston rings, crankshaft bearings and big end bearings are lubricated. This system works in connection with pressure system. (c) Pressure System (Forced Lubrication) In this system the lubricating oil is stored in a separate tank or sump and the pump is immersed in the lubricating oil. The oil pump pumps the oil through the strainer and delivers it through a filter to the main oil gallery at a pressure of 200400 kN/m2. The oil from the main gallery goes to the main bearings and then through a hole to the crank pin. From the crank pin it goes to the piston pin through a hole in the connecting rod web, where it lubricates the piston pin bearings and piston rings. For lubricating the cam shaft and timing gears the oil is led through a separate oil line from the oil gallery. The valve

1.72

Thermal Engineering - I

Lo w pressu re o il to cam sha ft O il tan k S caven ge pu m p s O il c ooler

P re ssure pu m p

Filter H igh p re ssure relief valve

Fig.1.29 Lo w pressu re relief valv e Fo rce d lu brication

tappets are lubricated by connecting the main oil gallery to the tappet guide surfaces through drilled holes. An oil pressure gauge is inserted in the oil circuit to indicate the pressure of oil in the system. (d) Semipressure System This system is the combination of splash system and pressure system. All four stroke engines are lubricated by this system. (e) Dry Sump System In this system the lubricating oil is kept in a separate tank from where it is fed to the engine. The oil which falls into the oil sump after lubrication, is sent back to the oil tank by a separate delivery pump. This system is used where the vehicle has to change its position continuously like in aircrafts. The advantage of this system is that there

IC Engines 1.73

is no chance of breakdown of the oil supply during up and down movement of the vehicle.

1.19 COOLING SYSTEM 1.19.1 Purpose of Cooling The maximum temperature attained inside the engine cylinder is in the range of 2000 C to 2500 C. The large amount of heat so produced is absorbed by the cylinder walls, piston, cylinder head and engine valves. So the temperature of these parts increase. This high temperature will break the lubricating oil film between the moving parts. So, this high temperature must be reduced by cooling system so that the engine can work efficiently. The overheating of these parts over 250C to 300C may cause the following defects: 1.

At high temperature the lubricating oil decomposes and carbon deposits is formed in the cylinder.

2.

Piston may breakdown.

3.

High temperature reduces the strength of the piston and piston ring.

4.

The uneven expansion of cylinder and piston may seize the piston.

5.

High temperature around the valve may cause the burning of valves and valve seats.

6.

There is a tendency for detonation to increase.

7.

There is a chance of pre-ignition.

To avoid all the above effects it is necessary to cool the engine. The cooling system is designed to remove about 30 to 35% of the total heat produced in the engine cylinder.

1.74

Thermal Engineering - I

1.19.2 Methods of Cooling IC Engine There are two methods of cooling IC engines: 1. Air cooling, and 2. Water cooling. 1. Air Cooling The heat from the cylinder walls is dissipated directly to the air. For this purpose fins and flanges are provided on the outer surfaces of the cylinder and cylinder head. An air stream is flowing continuously over the heated surface of the engine from where heat is to be removed. The amount of heat dissipated depends upon the following factors: (a)

Surface area of metal in contact with air.

(b)

Rate of air flow.

(c)

Temperature difference between the heated surface and the air.

(d)

Conductivity of the metal and convective heat transfer coefficient between air and metal.

Advantages of Air Cooled Engines (a)

Light weight due to the absence of radiator, cooling jacket and coolant.

(b)

Simple design and less costly.

(c)

There is no chance of leakage of coolant.

(d)

Anti-freeze not required.

(e)

Engine warms up faster than that of water cooled engines.

(f)

Can be operated in cold climate where water may freeze.

(g)

It does not depend on any coolant.

IC Engines 1.75

Disadvantages of Air Cooled Engines (a)

Less efficient cooling because the convective coefficient of heat transfer for air is less than that of water.

(b)

Cooling is not even all around the cylinder.

(c)

More noisy operation.

(d)

Limited use as in the case of motor cycles and scooters where the cylinders are exposed to air stream.

2. Water Cooling In this method the water is circulated through water jackets around each of the combustion chambers, cylinders, valve seats and valve stems. The circulating water takes heat of the combustion. When it passes through the radiator it is cooled by air drawn through the radiator by a fan and by air flow developed by the forward motion of the vehicle. After passing through the radiator, the cooled water is again circulated. Systems of Water Cooling There are various systems of water cooling: 1. Thermo-syphon cooling 2. Forced (or) Pump cooling 3. Cooling with thermostatic repulator 4. Pressurized cooling Now we shall discuss about the first two types in detail. Thermosyphon system or natural circulation system: In this system of cooling, the circulation of water is

1.76

Thermal Engineering - I

obtained due to the difference of densities of hot and cold regions of the cooling water. There is no pump to circulate the water. The hot water from F in the engine jacket rises up in the hose pipe as it is lighter and goes to the radiator from C yc lind er the top. Then it is cooled Fig. 1.30 Air cooling there and goes down to the bottom of the radiator. From there it goes again in the engine jacket. This system is quite simple and cheap, but cooling is rather slow. To maintain continuity of the water flow, the water must be maintained up to a certain Fig. 1.31 Water cooling (Thermosyphon system ) minimum head. If the water level falls down, the circulation will discontinue and the cooling system will fail. Forced circulation system: In this system of water cooling the circulation of water is obtained by a pump which is driven by a V-belt from a pulley on the engine crankshaft. The water is kept continuously in motion. After passing through the engine jackets in the cylinder block and heads, the water passes through radiator. In radiator, the water is cooled by atmospheric air drawn by fan through radiator. After passing through the radiator, the water is drained and delivered to the water pump. The water is again circulated through the engine jackets. This

IC Engines 1.77

system is more effective. The circulation of water becomes faster as the engine speed increases. It is not necessary to maintain the water up to a particular level. Advantages of Water Cooling System (a)

The specific fuel consumption is low as compared to air cooled engine.

(b)

Uniform cooling is possible.

(c)

It is not necessary to place the engine in the front of the vehicle as in case of air cooled engine.

(d)

Compact engine design with minimum frontal area is possible.

Disadvantages of Water Cooling System (a)

It is dependent on water supply.

(b)

The engine may be damaged in case of failure of cooling system.

(c)

The pump takes considerable power.

(d)

Initial and maintenance costs are higher.

(e)

There is a danger of leakage and freezing of coolant.

(f)

The design of the system is complicated and more costly.

1.20 FUELS AND COMBUSTION 1.20.1 Introduction Fuel is a substance which is used to liberate heat while burning. It starts burning continuously when it is raised to its ignition temperature. The combustion of fuels is a Chemical combination of oxygen with hydrocarbons. (Most of the fuels belong to hydrocarbon family - i.e., containing hydrogen and carbon).

1.78

Thermal Engineering - I

1.20.2 Classification of fuels: Fuels are classified as (i) Solid fuels (ii) Liquid fuels (iii) Gaseous fuel (i) Solid fuels: The following are the important solid fuels. Among all solid fuels, coal is mostly used for power generation. 1. Wood 2. Peat (mixture of water and decayed vegetable products) 3. Lignite 4. Bituminous coal 5. Anthracite coal. (ii) Liquid fuels: Liquid fuels are mainly derived from the crude oil (natural petroleum). These crude oil can be taken from various parts of the world mainly in Arabian countries. The important liquid fuels are 1. Gasoline (or) Petrol

2. Paraffin

3. Diesel

4. Crude oil (Heavy oil)

5. Kerosene.

6. Benzol

7. Shale oil (iii) Gaseous fuel Commonly used gaseous fuels are given below 1. Natural gas

2. Liquified petroleum gas

3. Refinery oil gas 4. Coal gas

IC Engines 1.79

5. Producer gas

6. Blast furnace gas

7. Blue water gas

8. Carbureted water gas

9. Acetylene C 2H2 11. Ethylene C 2H4

10. Methane CH 4 12. Ethane

C 2H6

1.21 SOLID FUELS Coal passes through different stages given below during its formation. Plant debris - peat - Lignite - Bituminous coal Anthracite coal. Peat: It is a spongy and humidified substance. It is the first stage in formation of coal. It has low heat value, and high moisture content. It is difficult for handling and storing. So it is not used for power generation. Lignite: It has brown fibrous structure. It has also low heat value and high moisture content. So it is also not used for power generation. It is an intermediate product between peat and bituminous coal. Bituminous coal: Next stage to lignite is Bituminous coal. It has little moisture content. So it is mostly used for power generation. Bituminous coal has two characteristics which are ‘Caking’ and ‘Non-Caking’. The Caking coal is useful for gas manufacturing process, while non-caking is mostly used for steam generation. Anthracite: It is the final stage in coal formation. It is exceptionally good for domestic purposes since it gives smokeless combustion, low ash content and cleanliness in

1.80

Thermal Engineering - I

handling. It has high calorific value and it is mostly used for steam power plant.

1.22 GASEOUS FUELS: Natural Gas: It consists of methane CH4 and ethane C 2H 6. It has high calorific value of 20,000 kJ/m 3 and is

used for both domestic and industrial heating processes. Coal gas: It is obtained by carbonisation by heating the bituminous coal. It consists of hydrogen, carbon monoxide and various hydrocarbons. It is generally used in gas fired boilers and for commercial purposes. Producer Gas: It is obtained by the incomplete combustion of coal, coke, anthracite coal (or) charcoal in a current of air. It contains more amount of N 2, so it has less heating value. Blue water gas: It is obtained by passing steam through red hot coke. Since it burns with blue flame, it is called ‘Blue water gas’. Blast furnace Gas: It is a by-product in the production of pig iron in the blast furnace. It contains large percentage of N 2 and has low heating value. Requirements of a Good fuel: 1.

Good fuel should have high calorific value.

2.

It should have low ignition temperature.

3.

It should burn freely with high efficiency.

4.

It should not produce any harmful gases.

5.

It should produce very less smoke.

6.

It should be economical.

IC Engines 1.81

7.

It should be easily stored and transported.

1.23 FUEL PROPERTIES Brief explanation of fuel properties are given below. 1. Viscosity of Fuel Viscosity is the resistance offered by the fuel itself to its flow. Viscosity decreases when the temperature of fuel increases and vice versa. Good fuel should have proper viscosity. 2. Pour Point of Fuel The pour point (freezing point) of fuel must be less than the lowest climate temperature of atmosphere. In cold climate days, the fuel should be in liquid state. So its pour point should be less sufficiently. 3. Sulphur Content in the Fuel Sulphur present in the fuel is dangerous to engine. During combustion, the sulphur in the fuel become sulfuric acid. This acid cause corrosion of engine parts. So the sulphur content in the fuel should be removed (or) sulphur content should be kept as minimum as possible. 4. Volatility The ability to evaporate is called volatility. If the fuel evaporates in low temperature, then it has high volatility. The petrol and diesel should have adequate volatility. 5. Flash Point and Fire Point Flash point is the minimum temperature of fuel when the fuel gives a momentary flame (or) flash. Fire point is the minimum temperature of fuel when the fuel starts continuously burning.

Thermal Engineering - I

1.82

The flash point and fire point of fuels should be adequate so that it is used in IC engine without any problem. 6. Calorific Value of Fuels: The amount of heat liberated by burning 1 kg (or 3

1 m  of fuel is known as Calorific value of fuel (or Heating

value of fuel). For solid fuel, the unit for calorific value is expressed in kJ/kg. For liquid and gaseous fuel, the unit is kJ/m 3 measured in S.T.P. condition (i.e., Standard Temperature and Pressure  15 C and 760 mm of mercury). Higher Calorific value: The amount of heat obtained by the complete combustion of 1 kg of fuel, when the products of combustion are cooled down to the temperature of the surroundings is known as Higher Calorific Value HCV of the fuel. Here the water vapour formed by combustion is condensed and the entire heat of steam is recovered from the products of combustion. Dulong’s formula is used to determine HCV of a fuel. O2   KJ HCV  33800 C  144000  H 2    9270 S 8  kg 

where C, H 2, S an d O 2 are the fractions of mass of carbon, hydrogen and sulphur and oxygen in 1 kg of fuel). Lower Calorific Value (LCV) The amount of heat obtained by the combustion of 1 kg of fuel, when the product of combustion is not

IC Engines 1.83

sufficiently cooled down to condense the steam formed during combustion is known as Lower Calorific Value (LCV) of the fuel. So, LCV of the fuel  H.C.V  Enthalpy of evaporation of steam formed  H.C.V.  2466  steam formed  K J/k g   H.C.V.  2466  9H 2

where 2466 KJ/kg is the specific enthalpy of evaporation of steam at 15 C . 1.23.1 Determination of Calorific value of fuel by Bomb Calorimeter Dulong formula is used for calculating H.C.V (Higher Calorific Value) of fuel (solid and liquid fuel) Dulong’s formula H.C.V 

1  O   33800 C  144000  H    9270 S  kJ  100  8   

Thus the H.C.V is got by using chemical analysis. But for determining H.C.V in the laboratory, the bomb calorimeter is used. Bomb Calorimeter: Since the shape of the calorimeter resembles the bomb, it is named as bomb calorimeter. The crucible is filled by 1 gm of sample fuel. We should ensure that the fuse wire has close contact with the fuel. The bomb is now supplied with oxygen and with a pressure of 25 atmosphere through oxygen valve. The

1.84

Thermal Engineering - I Th ermo m eter C o pp er C a lorim eter O2 O xyg en valve to allo w O 2 R e lease valve to release exhaust g as

Bo m b

w a ter (m esured quantity)

Fu se w ire

.. . . . C rucible fille d w ith 1gm of fuel

Wa te r Se al to preve nt w a ter ente rin g

M ains

R h eo stat Fig. 1.32

Leads to fuse

Bom b calorimeter

calorimeter is filled with measured quantity of water and the water is stirred for uniform temperature distribution. Now due to combustion of fuel, heat is released. This heat is used to heat the water in the calorimeter. The thermometer indicates the steady temperature 1 rise of fuel. The temperature readings are noted every 2 min until it reaches the maximum temperature. Now the bomb is taken out from calorimeter. The products of combustion are released with the help of release valve. It is allowed to dry. Unburnt fused wire if any, is

IC Engines 1.85

collected and weighed. By using above observations, the temperature - time curve is plotted. The heat liberated by the fuel on combustion is absorbed by the water and calorimeter. By equating, Heat released by the sample fuel  Heat received by water and calorimeter. By using above equation, we can find out Higher Calorific Value H.C.V of solid and liquid fuel. 1.23.2 Determination of Higher Calorific value of Gaseous fuel. By using Juncker’s Gas Calorimeter This is similar to Bomb Calorimeter Heat liberated by the gas  Heat gained by surrounding cold water. The above equation is used for calculating calorific value of gas. The Gas meter is used for allowing measured quantity of gas by recording its volume. The pressure regulator is used for measuring the pressure of gas and allowing limited pressure of gas. When the gas burns, the heat is liberated. This heat is used to heat the surrounding water. The hot products of combustion travel upwards in chamber and then downwards through flues and finally goes through outlet. During its travel, the heat is transferred to the water. The exit temperature of travelling gas should be very close to the atmospheric temperature, so that all the heat liberated by gaseous fuel should be transferred to cold

Thermal Engineering - I co ld w ater

T herm om e te r

1.86

Th erm om eter To F low m eter to m easu re qu antity of w a te r H ot p ro duc ts of com bu stion (i.e.,exha ust ga ses)

w ater

Th erm om eter E xhau st ( G ase s) to atm osp here

G as m eter P re ssure regu la to r Fig . 1.33

C ond ens ate

Ju nker’s g as calo rim e ter

water. Thermometers are fitted at various places to measure the temperatures at various points. By knowing quantity of gas and water and temperature of inlet and outlet water, we can find the calorific value of gaseous fuel. Atomic mass and Molecular mass: The atomic mass of Hydrogen is taken as unity. The atomic mass of other elements are determined with respect to atomic mass of hydrogen. For example, the atomic mass of nitrogen is 14 ie Nitrogen atom is 14 times heavier than hydrogen atom. Similarly, the atomic mass of oxygen is 16. ie Oxygen atom is 16 times heavier than hydrogen atom. Molecular mass: (or) Molecular weight The molecular mass of a substance is the number of times a molecule of that substance is heavier than the hydrogen atom.

IC Engines 1.87

For example, one molecule of oxygen consists of two atoms of oxygen. So molecular weight of oxygen  2  16  32 . i.e., The molecule of oxygen is 32 times heavier than hydrogen atom. The following table gives the symbols, atomic weights (atomic mass) molecular weights (molecular mass) of various elements and compounds (molecules).

Substance

Elements Hydrogen Oxygen Nitrogen Carbon Sulphur Compounds (Molecules) Carbon monoxide Carbon dioxide Sulphur dioxide Water (or) steam Methane Ethane Acetylene Ethylene

Symbol

H2 O2 N2 C S

CO CO 2 SO 2 H 2O CH 4 C 2H 6 C 2H 2 C 2H 4

Atomic Weight (Atomic mass)

Molecular weight (Molecular mass)

1 16 14 12 32

2

-

32 (ie 2  16 28 ie 2  14  12 32

12  16  28 12  32  44 32  32  64 2  16  18 12  4  16 24  6  30 24  2  26 24  4  28

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Thermal Engineering - I

1.24 COMBUSTION STOICHIOMETRY The following chemical equations are used to find the amount of oxygen required and the amount of gases produced by the combustion of fuel. Combustion means burning. 1. For Complete Combustion of Carbon: When carbon burns by taking sufficient amount of oxygen in the air, the carbon dioxide is produced. During this reaction, a large amount of heat is released. C arbon  O xygen  Ca rbon dioxide C  O 2  CO 2

Molecular Weights: Divide by 12:

12  32  44 1 

i.e., 1 kg of carbon requires

8 11  3 3

8 kg of oxygen for 3

complete combustion and produces

11 kg of carbon 3

dioxide. i.e., 1 kg C 

11 8 kg of O 2  kg of C O2 3 3

2. For Incomplete combustion of carbon: If the air supplied is insufficient, then sufficient oxygen will not be available. Then the combustion of carbon will be incomplete. In this case, the carbon monoxide will be produced.

IC Engines 1.89

ie Carbon  Oxygen  Carbon monoxide  O2  2CO

2C

2  12  32  2  28  56

Molecular Weights: Divide by 24

:



1

7 4  3 3

So 1 kg of carbon requires produces

4 kg of oxygen and 3

7 kg of carbon monoxide. 3

ie 1 kg of C 

7 4 kg of O 2  kg of CO 3 3

3. Combustion of Carbon monoxide: When carbon monoxide is burnt further, it produces carbon dioxide. i.e., 2CO  O2  2CO 2 2CO  O 2  2CO 2

Molecular weights: Divide by 56

:

2  28  32  2  44 1



11 4  7 7

So 1 kg of carbon monoxide requires oxygen and produces ie 1 kg of CO 

4 kg of 7

11 kg of carbon dioxide. 7

11 4 kg of CO2 kg of O2  7 7

1.90

Thermal Engineering - I

4. For combustion of sulphur: During combustion, sulphur in the fuel burns with oxygen and produces sulphur dioxide. Sulphur  Oxygen  Sulphur dioxide

Molecular weights: Divide by 32

:

S



O 2  SO 2

32



32  64

1



1



2

So 1 kg of sulphur requires 1 kg of oxygen for complete combustion and produces 2 kg of sulphur dioxide. ie 1 kg of S  1 kg of O 2  2 kg of SO 2 5. For combustion of hydrogen: During combustion, hydrogen in the fuel combines with the oxygen and produces water vapour (or) steam. Hydrogen  Oxygen  Water vapour 2H 2  O 2  2H 2O Molecular weights: 2  2  32  2  18  36 Divide by 4 :

1



8  9

So 1 kg of hydrogen requires 8 kg of oxygen and produces 9 kg of water vapour (or) steam. ie 1 kg of H 2  8 kg of O 2  9 kg of H 2O 6. For combustion of methane (or) marsh gas: During combustion, methane burns with oxygen and produces carbon-di-oxide and water vapour. Methane  Oxygen  Carbon dioxide  Water vapour

IC Engines 1.91

CH 4  Molecular weights:

16

Divide by 16:

1

2O 2

 CO 2  2H 2O

 2  32 

4

 2  18



11 4





4

9 4

So, 1 kg of methane requires 4 kg of oxygen and 9 11 kg of carbon-dioxide and kg of water produces 4 4 vapour. i.e., 11 9 kg of CO 2  kg of H 2O 1 kg of CH 4  4 kg of O 2  4 4

7. For combustion of ethylene: Ethylene  Oxygen  Carbondioxide  Water vapour C 2H 4  Molecular weights: Divide by 28:

3O2

 2H 2O

28  3  32  2  44  2  18 1 

24 7

So 1 kg of ethylene requires produce

 2CO 2



22 7



9 7

24 kg of oxygen to 7

9 22 kg of carbon-di-oxide and kg of water 7 7

vapour i.e., 1 kg of C 2H 4 

22 24 kg of O 2  kg of CO 2 7 7 

9 kg of water vapour 7

Atmospheric Air: Atmospheric air contains oxygen, nitrogen, a little amount of carbon-di-oxide and very very

1.92

Thermal Engineering - I

little amount of neon, argon and kryton etc. The oxygen is very much necessary for complete combustion of fuel (any fuel). Normally, the composition of atmospheric air is given below. Nitrogen  77 % and Oxygen  23% By weight or mass Nitrogen  79 % and Oxygen  21% By volume

Minimum mass of (Stoichiometric) air required for complete combustion of solid and liquid fuels: The theoretical (or) minimum mass of Oxygen required for complete combustion of 1 kg of fuel can be determined by the chemical analysis of the fuel. Now consider 1 kg of a fuel. It contains Carbon Hydrogen Oxygen Sulphur

-

C kg H 2 kg O 2 kg and S kg

For complete combustion 1 kg of carbon requires

8 kg of O 2 3

So, C kg of carbon requires

8 C kg of O 2 3

Similarly 1 kg of Hydrogen requires 8 kg of O 2 So, H 2 kg of hydrogen requires 8H 2 kg of O 2 Similarly, 1 kg of sulphur requires 1 kg of O 2 So, S kg of sulphur requires S kg of O 2.

IC Engines 1.93

 Total oxygen required for complete combustion of

1 kg of fuel 

8 C  8 H 2  S kg 3

If some amount of oxygen (say O 2 kg ) is already available in the fuel, then Net oxygen required for complete combustion of 1 kg  8 of fuel   C  8H 2  S   O 2 kg 3   For 23 kg of oxygen  100 kg of air is required. . . [ . Composition of air in mass: N 2  77 % and O 2  23% ]

For  8    3 C  8 H 2  S   O 2  Kg of oxygen, the air required    

100  8  C  8H 2  S  O 2  kg of air 23  3 

 Theoretical (minimum (or) stoichiometric) mass of

air required for complete combustion of 1 kg of a fuel is given by m air m in mair min 

100  8  C  8H 2  S  O 2  kg 23  3 

Theoretical (minimum) volume of air required for complete combustion: Consider 1 kg of a gaseous fuel. It contains Carbon monoxide -

CO m 3

Hydrogen

-

H2 m3

Methane

-

CH 4 m 3

Thermal Engineering - I

1.94

Ethylene

C 2H 4 m 3

-

For complete combustion of the gas 1 m3 carbon monoxide requires 0.5 m3 of O 2 So CO m 3 of carbon monoxide requires 0.5 CO m 3 of O 2 Similarly, 1 m3 of hydrogen requires 0.5 m3 of O 2 So H 2 m 3 of hydrogen requires 0.5 H 2 m 3 of O2 Similarly, 1 m3 of methane requires 2 m 3 of O 2 CH 4 m 3 of methane requires 2 CH 4 m 3 of O 2

Similarly, 1 m3 of ethylene requires 3 m 3 of O 2 C 2H 4 m 3 of ethylene requires 3 C2H 4 m 3 of O 2

Total

volume

of

oxygen

required

for

complete

3

combustion of 1 m of fuel  0.5 CO  0.5 H 2  2CH 4  3 C2H4 m 3

If some amount of oxygen (say O 2 m 3) is already available in the fuel, then Net oxygen required for complete combustion of 1m

3

of fuel  0.5 CO  0.5 H 2  2CH 4  3C 2H 4  O 2 m 3

We know that oxygen present in the air is 21% by volume. So for 21 m 3 of oxygen  100 m 3 of air is required [ 0.5 CO  0.5 H 2  2CH 4  3C 2H 4  O 2 ] m 3 oxygen, the air required is

For

of

IC Engines 1.95



100 [ 0.5 CO  0.5 H 2  2CH 4  3C 2H 4  O 2 ] m 3 21

 Minimum (Theoretical) volume of air required for

combustion of 1 m3 of fuel, V min air V min air 

100 [ 0.5 CO  0.5 H 2  2CH 4  3C 2H 4  O 2 ] m 3 21

1.25 EXCESS AIR AND AIR FUEL RATIO CALCULATION If just minimum (theoretical) amount of air is supplied, some amount of fuel may be unburnt. It is due to the fact, that all the air supplied does not have intimate contact with the fuel particles. So, the excess air is supplied to ensure complete and rapid combustion of fuel. The amount of excess air supplied depends on the quantity of fuel, rate of combustion, firing conditions etc. Normally, 25% to 50% excess air is supplied. The excess air gives cooling effect. But this can be avoided by preheating the air. Mass of excess air supplied can be calculated by the mass of unused oxygen available in the flue gases. 100 kg of air is required for 1 kg of oxygen. 23  Excess air supplied 

100  Mass of oxygen in flue gases 23

Excess air supplied (approximate) per kg of fuel 

79  O 2  C 21  33 CO 2  CO

1.96

Thermal Engineering - I

where O 2, CO 2 and CO represents % of them by volume and C represents % of it by mass. Total air supplied  minimum Theoretical air  E xcess air m total  m airmin  mexcess

Total air supplied (approximate) per kg of fuel 

N2  C 33 CO2  CO

where N 2 is the % of nitrogen in flue gases by volume. This is also known as Air fuel ratio. Mass of carbon in flue gases: The mass of carbon in flue gases, can be calculated from the mass of CO 2 and CO present in them. During combustion, 1 kg of carbon produces

11 kg of CO 2 3

3  1 kg of CO 2 contains kg of carbon 11

Similarly, 1 kg of carbon produces  1 kg of CO contains 

7 kg of CO. 3

3 kg of carbon. 7

3 3 Ma ss of carbon  kg of CO 2  kg of CO in flue gases 11 7

where CO 2 and CO are the amounts of CO 2 and CO present in 1 kg of flue gases respectively.

IC Engines 1.97

Mass of flue gases per kg of fuel burnt: Since there is no loss of carbon during combustion, the mass of flue gases can be obtained by comparing the mass of carbon present in the flue gases with the mass of carbon present in the fuel. Mass of flue gases per kg of fuel 

Mass of carbon in 1 kg of fuel kg of flue gas Mass of carbon in 1 kg of flue gas kg of fuel

Actual mass of air supplied (or Air fuel ratio) By volumetric composition of flue gases and mass of carbon in 1 kg of fuel, we can calculate, The a ctual mass of air or  Air fuel ratio N2  kg of air   3.03 C   kg of fuel  CO 2  CO 

where C  mass of carbon in 1 k g of fuel N 2  % by volume of N 2 in flue gases CO 2  % by volume of CO 2 in flue gases CO  % by volum e of CO in flue gases

By Gravimetric composition of flue gases and mass of carbon in 1 kg of fuel, we can calculate The actu al air sup plied or Air fuel ratio 

100  N 2  C

kg of air 21 CO 2  33 CO kg of fuel

where C  Mass of carb on in 1 kg of fuel N 2  % by mass of N 2 in flue gases

1.98

Thermal Engineering - I

CO 2  % by mass of CO 2 in flue gases CO  % by mass of CO in flue gases

Note: If the fuel contains appreciable amount of nitrogen, then the above two formulae should not be used. % of excess air Actual A  F ratio  Stoichiometric A  F ratio  Stoichiometric A  F ratio Problem 2.5: A sample of coal has the following composition by mass. C 70% ; H2 8%, O2 10% ; N2 3% ; sulphur S 2% and ash 7%. Determine its higher calorific value and lower calorific value per kg of fuel.

Solution: C  0.7 kg ; H 2  8%  0.08 kg ; O 2  10%  0.1 kg ; N 2  3% unnecessary data); Ash  7% unnecessary data); S  2%  0.02 kg

Higher Calorific Value H.C.V O2   H.C.V  33800 C  144000  H 2    9270 S 8   0.1    9270  0.02  33800  0.7  144000  0.08  8    33565.5 kJ/kg

Lower Calorific Value L.C.V L.C.V  H .C.V  9H 2  2466   33565.5  9  0.08  2466  31789.88 kJ/kg

IC Engines 1.99

Problem 2.6: A fuel consists of 85% Carbon, 12% of hydrogen, 3% of residual matter by mass. Find the Higher Calorific Value and Lower Calorific Values per kg of fuel.

Solution: C  85%  0.85 kg ; H 2  12%  0.12 kg

Residual matter  3% [This data is not necessary for solving problem]. Higher Calorific Value H.C.V  33800 C  144000 H 2  33800  0.85  144000  0.12  46010 kJ/kg

Lower Calorific Value L.C.V  H.C.V  9H 2  2466   46010  9  0.12  2466   43,347 kJ/kg

. . [ . O2  0 ; S0]

Chapter 2

Combustion in SI and CI Engine Combustion in SI and CI Engine: Normal combustion and abnormal combustion in SI Engine - Importance of flame speed and effect of engine variables - Abnormal combustion - Pre Ignition and Knocking in SI Engine - Fuel requirement and fuel rating anti knock additives - combustion chamber - Requirement - types of SI Engine - Four stages of combustion in CI Engine - Delay period and its Importance - Effect of engine variables - Diesel knock - Need for air movement, suction, compression and combustion induced turbulence in Diesel engine - Open and divided combustion chambers and fuel Injection - Diesel fuel requirements and fuel rating.

2.1 NORMAL COMBUSTION A high intensity spark is produced by a spark plug. This spark travels through the air fuel mixture. This spark leaves a thin thread of flame behind it. The air-fuel mixture enveloped around the thin thread of flame gets ignited and combustion commences. Since the air fuel mixture is in turbulent condition, the surface area of heat transfer is more and combustion is speeded up enormously. In P-V diagram (Fig. 2.1), we can see the stages of normal combustion. LNQM is the normal compression curve. At point N, the ignition starts [N is the point 35 before TDC]. At point Q, pressure rise can be noticed. From point M , sudden pressure rise occurs. Ignition lag: The time period between first igniting fuel and commencement of main phase of combustion is called

2.2 Thermal Engineering - I 2

P (kg f/cm ) For b est pe rfo rm ance o o a t 1 0 to 12

M ax.P r. 40

Ig nitio n a dvance

C om p re ssio n

N

E xpa nsion

Q

M S

L o

B D C 1 50 1 20

o

90

o

o

60 30

o

TD C 3 0 Fig.2.1

o

60

o

90

o

o

1 20 1 50

o

BDC



ignition lag (or) period of incubation. The ignition lag is normally 0.0015 sec. (Pre-ignition  Detonation  Engine failure) Ignition Advance: The ignition actually starts at about 35 before TDC. This angle of crank is called ignition advance. Maximum pressure: The maximum pressure inside the cylinder is attained at about 10to 12 after TDC. After Burning: Once it reaches its maximum pressure, the ignition stops. But at this point the whole heat of the fuel is not liberated. So the remaining heat in the fuel is burnt after this maximum pressure point. This is called ‘after burning’. The different steps of normal combustion is shown in following Fig. 2.2.

Combustion in SI and CI Engine 2.3 Step 1

Step 2

Sp ark pro duce d

Step 3

Step 4

C om bu stion C om bu stion spreads starts Fig. 2.2.

C om bu stion com pleted

2.1.1 Factors affecting normal combustion in S.I Engines 1. Induction pressure As the pressure falls, delay period increases, and the ignition must be earlier at low pressures.

Tem p era tu re

Id ea l C om bu stio n

M ax

C o m b ustion w ith D isso cia tio n

W ea k

R ich A ir Fuel R atio .

Fig. 2.3.

2. Engine speed When the engine speed increases, the delay period time needs more crank angle and ignition should take place earlier.

2.4 Thermal Engineering - I

3. Ignition timing If the ignition takes place too early, then the peak pressure will occur early and work transfer falls. If the ignition takes place too late, then peak pressure will be low and the work transfer falls. 4. Fuel choice The calorific value and enthalpy of vaporisation will affect the temperature achieved. The induction period of the fuel will affect the delay period. 5. Combustion chamber The combustion chamber should be designed to give shorter flame path to avoid knocking and it should give proper turbulence. 6. Compression ratio When the compression ratio increases, it increases the maximum pressure and the work transfer. 7. Mixture strength The rich mixture is necessary for producing the maximum work transfer. 2.1.2 Flame Front Propagation The concept of flame propagation speed is important in SI engines, as it may lead to detonation. Flame front is the front surface of the flame that separates the burnt charges from the unburnt one. The rate of movement of flame front across the combustion chamber is based on reaction rate and transposition rate. The reaction rate is the result of chemical reaction occurring within a region where unburnt mixture is heated and converted into products.

Combustion in SI and CI Engine 2.5

The transposition rate is due to the movement of flame frant relative to the cylinder wall. It is also the result of pressure differences existing between the burning and unburnt gases in the combustion chamber.

2.3 IMPORTANCE OF FLAME SPEED AND EFFECT OF ENGINE VARIABLES Flame speed Flame speed is the speed at which the flame travels. Flame speed affects the combustion phenomena, pressure developed and power produced. Burning rate of mixture depends on the flame speed and shape of combustion chamber. 2.3.1 Factors affecting flame speed 1. Turbulence It helps in mixing and boosts the chemical reaction. A lean mixture can be burnt easily without any difficulties. The flame speed is quite low in non-turbulent mixture and increases with increase in turbulence. Turbulence consisting of many minute swirls increases the rate of reaction and produces a higher flame speed than that of larger and fewer swirls. 2. Engine speed When engine speed increases, flame speed also increases due to the turbulence inside the engine cylinder. The crank angle required for the flame propagation during the entire phase of combustion, will remain constant at all speeds.

2.6 Thermal Engineering - I

3. Engine size The time taken for flame propagation is smaller in small engines when compared to larger engines. In larger engines, the time required for complete combustion is more because the flame has to travel a longer distance. 3. Compression ratio A higher compression ratio increases the pressure and temperature of mixture. This reduces the initial phase of combustion and hence less ignition advance is needed. High pressure and temperature of the compressed mixture also speed up the second phase of combustion. Increased compression ratio reduces the clearance volume. Thus engines having higher compression ratio have higher flame speed. A further increase in the peak pressure and temperature results in the increase in the tendency of the engine to detonate. 4. Inlet temperature and pressure When the inlet temperature and pressure increases, it results in better homogenous mixture which helps to increase the flame speed. 5. Fuel-Air ratio The highest flame speed obtained with slightly rich mixture gives complete combustion. Lean mixtures have low thermal energy and hence have low flame speed. A rich mixture burns readily and completely, resulting in higher flame speeds. A stoichiometric air fuel ratio is usually

Combustion in SI and CI Engine 2.7

chosen to prevent compromise on flame speed and air fuel ratio.

0 .00 4

0 .00 2

S tio c hio m e tric M ixture

Tim e in S e con ds

0 .00 6

A 60 L ea n M ixtu re

1 00

1 40 R ich M ixtu re

1 80

Fig. 2.4 Effect of m ixture stren gth on flam e pro paga tion tim e

6. Engine output When the engine output is increased, the cycle pressure also increases. With the increased throttle opening the cylinder gets filled to a higher density of mixture. This results in increased flame speed. When the output is decreased by throttling, the initial & final pressure decreases. Poor combustion at low loads and the necessity of mixture enrichment causes wastage of fuel and discharge of products like carbon monoxide etc. in the atmosphere which are the main disadvantages of SI engines.

2.8 Thermal Engineering - I

2.2 ABNORMAL COMBUSTION The abnormal combustion deviates from the normal behavior resulting in loss of performance and physical damage to the engine. There are two types of Abnormal combustion. 1. Pre-ignition 2. Knocking (or) Detonation (or) Pinking 2.2.1 Pre-ignition

.. . .. ... . .. ... . . . . .. . .. . . . . . . . . . . . ... . . ..

. . .. . ..... .. .. ... . .. ... . . .. .. .. .. ... . . .. .

. . . . .. . . . .... . ..

.... . . .

.

Sp ark p ro duced Bo th flam es Bo th flam es Ignition started by S pa rk Plug S o spread fast C o llide d at left side b y ho t regular ig nition carbon deposits starts from the insid e the right side. com b ustion cham ber Ignition b ecause o f ho t d ep osits also sp read from the left side . Fig. 2.5. Pre ignition

Pre-ignition High temperature carbon deposits formed inside the combustion chamber ignite the airfuel mixture before normal ignition occurs by spark plug. This ignition due to hot carbon deposits is called pre-ignition. After some time of Pre-ignition, the normal ignition starts and both the flames get collided. If Pre-ignition occurs much early in the compression stroke, the work to compress the charge will be increased.

Combustion in SI and CI Engine 2.9

So the net power output will be reduced. Also this may cause crank failure due to high load to compress charge. Pre-ignition causes very high pressure and temperature. It causes the detonation. Pre-ignition is considered as abnormal combustion. 2.2.2 Knocking (or) Detonation (or) Pinking

Sp ark pro duces

Com bustion starts

Very h igh temp. flam e com press the remaining charge

Detonation

Fig. 2.6. D etonation

There are two general theories of detonation: 1. The auto-ignition theory 2. Detonation theory A sudden and violent noise (knock) experienced inside the engine cylinder is known as Detonation. This detonation is due to high pressure waves striking the cylinder walls, cylinder head and piston with loud noise. When spark occurs, the combustion of fuel near the spark plug commences. The flame travels through combustion chamber with high speed. The high pressure and high temperature gases produced by this ignition compress the fresh charge in front of the moving flame. Hence the temperature and pressure of fresh charge is

2.10 Thermal Engineering - I

increased beyond the limit and a spontaneous ignition takes place in far away from spark plug. This zone, far away from spark plugs where spontaneous ignition takes place is called ‘detonating zone’. This auto ignition spreads throughout the air-fuel mixture making its temperature and pressure rise further and produces loud pulsating sound called ‘pinking’ or ‘knocking’ or ‘hammer-blow’. The temperature in the detonating zone is higher than the non-detonating zone. More heat is lost in the surface of the combustion chamber and as a result, the output of engine is decreased. In mild detonation, the engine surface will be heated up. In severe detonation, fracture may occur on the engine. Due to detonation, carbon may be deposited inside the combustion chamber. When this carbon deposit gets heated, its temperature will be very high to preignite the fresh charge which is known as pre-ignition. Detonation occurs after sparking and pre-ignition occurs before sparking. One of the causes for pre-ignition is detonation. The detonation can be reduced by properly designing the combustion chamber so that there is always a turbulence of mixture. 2.2.3 The phenomenon of knock in SI Engine In spark-ignition engine, the combustion using spark-plug electrodes which spread mixture across the chamber. A flame front separate the fresh mixture from the product of

is initiated combustible is used to combustion.

Combustion in SI and CI Engine 2.11

In combustion chamber, burnt part of mixture has higher pressure & temperature than the unburnt mixture. To maintain a pressure equilization, the burnt mixture will expand and compress the unburnt mixture adiabatically there by increasing its pressure and temperature. The flame front propagates completely till the end of the cylinder, thereby leaving the unburnt mixture at an increased pressure and temperature. The temperature of unburnt mixture exceeds the self-ignition temperature during preflame reaction and hence spontaneous ignition occurs at various points inside the engine. This phenomenon is called knocking. An important fact about knocking is that it is very much dependent on the properties of the fuel. Knocking does not occur when the unburnt charge does not reach the auto ignition temperature, or in other words, in ignition lag period, if the flame front takes more time to burn the unburnt charge, no knocking occurs. But if the flame front takes less time to burn the unburnt charge, knocking occurs [since the end charge will detonate]. Hence, fuels with high auto ignition temperature and a long ignition lag are often used as fuels for S.I engines, inorder to avoid detonation. In summary, during auto ignition, two different cases are encountered. 

A large amount of mixture gets autoignited leading to a very rapid increase in pressure throughout the combustion chamber and there will be a direct blow on the engine structure. This results in the thudding sound and consequent

2.12 Thermal Engineering - I



noise from the free vibration of the moving parts. These noises can be detected by human ears. A large pressure difference may exist in the combustion chamber and the resulting gas vibrations force the walls of the chamber to vibrate in the same frequency as that of the gas. In this case, an audible sound may be evident.

Normally knocking combustion in an engine is often detected by a distinct audible sound. But a scientific method of detecting the phenomenon of knocking involves the use of a ‘Pressure Transducer’.

Ig nitio n

C o m p re

P re ss ure

The output of this pressure transducer is connected to a cathode ray oscilloscope. The pressure-time traces obtained due to the presence or absence of knock are shown in Fig. 2.7.

Po

s s io n

we

r

TD C

BDC

BDC

Ig nitio n

C

ess om pr

BDC

P ressure

Tim e (a) N orm al C o m bu stion

Po

io n

we

r

TD C

BDC

Tim e (b) K no ck ing C o m bu stion Fig. 2.7. R esu lts Plo tted b y P ressure Transd ucer.

Combustion in SI and CI Engine 2.13

2.2.4 Effects of knocking in SI Engine 1. Noise and Roughness Knocking produces a loud pulsating noise and pressure waves. These waves vibrates back and forth across the cylinder. The presence of this vibratory motion causes crank shaft vibration and the engine runs roughly. 2. Mechanical Damage 1.

The high pressure wave generated during knocking can increase rate of wear of parts in combustion chamber. Severe erosion of piston crown, cylinder head and small holes created on inlet and outlet valves may result in complete damage of the engine.

2.

Due to Detonation, high noise level occurs in engine. In small engines, the noise can be easily detected and corrective measures can be taken, but in large engines, it is difficult to detect knocking noise and hence corrective measures cannot be taken which results in complete damage of the piston.

3. Carbon deposits Detonation leads to a huge deposition at the engine outlet.

amount

of

carbon

4. Increase in heat transfer Knocking is accompanied with the increase in rate of heat transfer across the combustion chamber walls. 5. Decrease in power output and efficiency Due to increase in the rate of heat transfer, the power output as well as efficiency of the engine decreases.

2.14 Thermal Engineering - I

6. Pre-Ignition The increase in heat transfer on the walls causes local overheating of the spark plug which may reach a temperature high enough to ignite the charge before the passage of spark, thus leading to pre-ignition. An engine detonating over a long period of time often results in preignition which is the real danger of detonation. 2.2.5 Effect of engine variables on knock It has already been established that the knocking of an engine typically depends upon either the quantity of the charge inside the chamber, the temperature of the chamber or the time of detonation. Hence, the different variables which affect knocking can be classified into   

Density factors Time factors Composition factors

1. Density factors Density factors deal with the basic mass properties of the charge present inside the cylinder. The properties include different thermodynamic variables like the temperature of the charge, pressure, volume of charge, density etc. It is evident that the auto ignition can be prevented if the temperature of the charge entering the cylinder is minimum. Similarly, a charge at lower pressure is less likely to cause a knock. This is due to the reduced energy of the charge, disabling it from combusting automatically. The different density factors which affect the knocking/phenomenon are discussed below. Compression Ratio: Higher compression ratio simply implies that the pressure of the air-fuel mixture is quite

Combustion in SI and CI Engine 2.15

high. Hence, the temperature of the gases at the end of compression is also high. Therefore, upon combustion, there is a considerable decrease in ignition delay. This directly increases the possibility of a knock. Hence, to prevent knocking, it is always wise to limit the compression ratio to a lower value, but not low enough to drastically decrease the efficiency of the engine. Charge temperature: An increased inlet temperature of the air-fuel mixture causes it to rise above the normal temperature at the end of the compression stroke. Due to this increased temperature, the ignition delay is decreased, resulting in knocking of the engine. However, a low inlet temperature could result in vapourization and starting problems in an engine. Mass of fuel injected: A reduced amount of charge experiences lower pressure and has lower energy when compared to normal levels. Thus, the temperature of the reduced amount of charge at the end of the compression, is not high enough to cause knocking. Hence, the possibility of a knock is directly proportional to the mass of the charge inside the cylinder. Cylinder wall temperature: The combustion chamber is continuously subjected to several frictional and thermal stresses during operation. Hence, the walls of the chamber may develop minute hotspots which could ignite a fuel before the anticipated time, thereby resulting in knocking. Hence, uniform cooling of the walls using an efficient coolant is of paramount importance. Moreover, the exhaust valves and the spark plugs are the most hottest regions inside the cylinder. Hence, the concentration of the

2.16 Thermal Engineering - I

compression against these regions, is to be avoided to reduce knocking. Horse power: High powered engines operate at high temperatures and pressures. Thus the chances of a knock to occur in a high powered engine is greater than that of a low powered engine. 2. Time factors Time factors play an important role in determining the chances of a knock in an engine. Some common time factors are flame speed, velocity of the charge, engine speed etc. The effect of different time factors on the knock of an engine is discussed below. Velocity of the charge: A turbulent charge ignites much faster than a non-turbulent charge. Thus, the flames propagate much faster, leaving little margin for the end charge to auto ignite. Thus, the chances of a knock is reduced effectively by increasing the velocity of the charge, above its turbulent level. Engine speeds: At higher engine speeds, the turbulence of the charge increases greatly. This results in reduced knocking, as discussed above. Flame travel distance: It has been well established that a faster flame reduces knocking possibilities when compared to a slower flame. Hence, if the time taken for the flame to travel across the chamber is reduced, knocking can be prevented. This can be done by either decreasing the combustion chamber size, or by repositioning the spark plug appropriately.

Combustion in SI and CI Engine 2.17

A centrally placed spark plug, or usage of two or more plugs, can effectively reduce the knocking of an engine. Combustion chamber configuration A combustion chamber should be designed in such a way that it promotes the turbulence of the particles inside. Moreover, the chamber should be made as spherical as possible with the least possible height. These two factors can effectively reduce the flame travel time, thereby preventing knocking. 3. Composition factor Composition factor deals with the flammability of the charge present inside the cylinder. Air-fuel ratio and the octane number of the fuel are the most important composition factors pertaining to the knocking phenomenon. (i) Air-fuel ratio: Flame speeds depend upon the air-fuel ratio. It varies as per the type of fuel used. The flame temperatures and the reaction time also vary based on the air-fuel ratio. If a specific ratio can cause low reaction time, then this ratio can give way to increased chances of knocking. (ii) Octane value Knocking can be reduced by either increasing the self-igniting temperature of a fuel or by reducing its pre-flame reactivity. In general, Aromatic hydrocarbons have the minimum tendency to knock an engine, where as the paraffin series are more likely to knock an engine. Any appropriate compound with a compact molecular structure is less prone to knock an engine.

2.18 Thermal Engineering - I

2.3 FUEL REQUIREMENT AND FUEL RATING 2.3.1 Important properties of fuel in SI Engine The fuel characteristics that are important for the performances of internal combustion engines are      

Volatility of the fuel. Detonation characteristics. Good thermal properties like heat of combustion and heat of evaporation. Sulphur content. Aromatic content. Cleanliness of fuel.

2.3.2 Important characteristics of SI Engine fuel Every engine is designed for a particular fuel according to its desired qualities. For good performance of SI engine, the fuel used must have the proper characteristics like,       

It should readily mix with air to make an uniform mixture at inlet. It must be knock resistant. It should not pre-ignite easily. It should not tend to decrease the volumetric efficiency of the engine. Its sulphur content should be low. It must have adequate calorific value. It must have proper viscosity.

Combustion in SI and CI Engine 2.19

2.3.4 Fuel properties Brief explanation of fuel properties are given below. 1. Viscosity of Fuel Viscosity is the resistance offered by the fuel to its own flow. Viscosity decreases when the temperature of fuel increases and vice versa. Good fuel should have proper viscosity. 2. Pour Point of Fuel The pour point (freezing point) of fuel must be less than the lowest climate temperature of atmosphere. In cold climate days, the fuel should be in liquid state. So its pour point should be less sufficiently. 3. Sulphur Content in the Fuel Sulphur present in the fuel is dangerous to engine. During combustion, the sulphur in the fuel become sulfuric acid. This acid causes corrosion of engine parts. So the sulphur content in the fuel should be removed (or) sulphur content should be kept as minimum as possible. 4. Volatility The ability to evaporate is called volatility. If the fuel evaporates in low temperature, then it has high volatility. The petrol and diesel should have adequate volatility. 5. Flash Point and Fire Point Flash point is the minimum temperature of fuel when the fuel gives a momentary flame (or) flash. Fire point is the minimum temperature of fuel when the fuel starts continuously burning. The flash point and fire point of fuels should be adequate so that it is used in IC engine without any problem.

2.20 Thermal Engineering - I

6. Calorific Value of Fuels: The amount of heat liberated by burning 1 kg (or 3

1 m  of fuel is known as Calorific value of fuel (or Heating

value of fuel). For solid fuel, the unit for calorific value is expressed in kJ/kg. For liquid and gaseous fuel, the unit is kJ/m3 measured in S.T.P. condition (i.e., Standard Temperature and Pressure  15 C and 760 mm of mercury). Higher Calorific value: The amount of heat obtained by the complete combustion of 1 kg of fuel, when the products of combustion are cooled down to the temperature of the surroundings is known as Higher Calorific Value HCV  of the fuel. Here the water vapour formed by combustion is condensed and the entire heat of steam is recovered from the products of combustion. Dulong’s formula is used to determine HCV of a fuel. O2   kJ HCV  33800 C  144000  H 2    9270 S 8 kg  

where C , H2, S and O 2 are the fractions of mass of carbon, hydrogen, sulphur and oxygen in 1 kg of fuel. Lower Calorific Value (LCV) The amount of heat obtained by the combustion of 1 kg of fuel, when the product of combustion is not sufficiently cooled down to condense the steam formed during combustion is known as Lower Calorific Value (LCV) of the fuel.

Combustion in SI and CI Engine 2.21

So, LCV of the fuel  H.C.V  Enthalpy of evaporation o f steam fo rmed  H.C.V.  2466  steam fo rmed  kJ/kg  H.C.V.  2466  9 H2

where 2466 kJ/kg is the specific enthalpy of evaporation of steam at 15 C. 2.3.5 Octane Number (ON) Octane Number (gaseous fuel) indicates the anti-knock properties of a fuel, based on the comparison of mixtures of Iso octane and normal heptane. Fuel rating for SI engine Octane value is for SI engines Octane Number: (Applicable for SI Engine) This is a number to rate the petrol fuel according to its detonating tendency. If the fuel has the tendency to detonate less, then it has high octane number and vice versa.  

Iso-octane is a high rating fuel (i.e. detonation is less). Normal heptane is a low rating fuel (i.e. detonation is more).

Iso-octane and normal heptane are mixed together and this sample mixture is used for running a test engine. The octane number of the fuel is the percentage of octane in this sample mixture which detonates in similar way as the fuel under the same condition. High octane fuel’s number is 100. This type of fuel will not have tendency to detonate. We can make given fuel

2.22 Thermal Engineering - I

into octane number 90 to 100 by adding tetra ethyl lead. But this addition will reduce the engine life. Fuels with a higher octane ratings are used in high performance gasoline engines that require higher compression ratio. Fuels with lower octane number are ideal for diesel engines, because diesel engines do not compress the fuel but rather compress only air and then inject the fuel. Two methods that are employed for measuring octane number are Research Octane Number (RON) and Motor Octane Number (MON). The octane numbers measured under two different engine conditions in a standard “Cooperative Fuels Research (CFR)” engine has a variable compression ratio. Research Octane Number (RON) The most common type of octane rating is Research Octane Number (RON). RON is determined by using the fuel in a test engine running at 600 rpm with the variable compression ratio under controlled condition, and comparing the results with the mixture of iso-octane and n-heptane. Motor Octane Number Motor Octane Number is determined at 900 rpm engine speed instead of 600 rpm used in RON. MON testing uses a similar test engine used in RON testing but with a preheated fuel mixture, higher engine speed and variable ignition timing.

Combustion in SI and CI Engine 2.23

Anti-knock Index 

RON  MON 2

Advantages of High-Octane Fuel: 1.

We can increase the compression ratio without detonation.

2.

Engine efficiency detonation.

3.

Super charging can be done without detonation.

can

be

increased

without

So totally, the unwanted detonation can be reduced.

2.4 ANTI-KNOCK ADDITIVES Anti knock additives are used to reduce engine knocking and to increase the fuel’s octane rating by raising the temperature and pressure at which auto ignition occurs. The widely used antiknock agents are:      

Tetraethyl lead [TEL] CH 3CH 24 Pb Methylcyclopentadienyl manganese (MMT) CH 3C 5H 4MnCO3

tricarbonyl

Ferrocene Fe C5H52 Iron pentacarbonyl Toluene Iso octane

2.4.1 Anti-knock Agents Anti-knock agents are classified into high-percentage additives like alcohol and low-percentage additives based on heavy elements. Internal combustion engine discharges various substances to the atmosphere. Some of these emissions are

2.24 Thermal Engineering - I

harmful to the environment such as Carbon monoxide, Nitrogen oxides, unburnt hydrocarbons and certain compounds of lead. The catalytic converter is used to oxidize the unburnt hydrocarbons and carbon monoxide to carbon dioxide and to decompose nitrogen oxides into nitrogen and oxygen. High percentage additives are those organic compounds that do not contain metals, but require high blending ratios, such as 20-30% for benzene and ethanol. Ethanol is inexpensive, and widely available but being corrosive in nature, it is not used. Tetra ethyl lead (TEL) CH3CH 24 Pb is a main additive and it is a common anti knock agent. Adding a small amount of Tetra ethyl load (TEL) improves the anti-knock quality of fuel. 2.4.2 Effects of Anti knock additives 



The main problem in using Tetra ehtyl lead is the lead content in it since lead is extremely toxic and poisonous. A manganese carrying additive like methylcyclopentadienyl manganese tricarbonyl (MMT) directly affects the humans.

The exposure of MMT results in eye irritation, giddiness, headache and it causes difficulties in breathing. 

Ferrocene

Fe C 5H 22

is

an

organometallic

compound of iron. The iron contents in ferrocene forms a conductive coating on the spark plug.

Combustion in SI and CI Engine 2.25

2.4.3 Factors affecting Detonation and Remedies 1.

2.

3.

4.

5.

Factors Remedies The type of fuel used is Fuel like alcohol and the reason for detonation benzol do not cause detonation. Addition of a small quantity of tetraethyl lead with petrol will suppress the detonation. (This process is called doping). The position of spark plug Less distance reduces the in the combustion chamber chances of detonation. A determines the distance spark plug placed the flame travels to reach centrally will reduce the detonating zone. More detonation. distance causes detonation High temperature The cooling system should combustion chamber raise be proper to maintain the temperature of cylinder cylinder wall temperature wall and also detonating optimum. zone. The compression ratio is The compression ratio the cause for detonation. should not be increased More compression ratio beyond the limit. will overheat the engine. The presence of carbon Good quality fuel should deposits promote be used. detonation.

2.26 Thermal Engineering - I

6.

Factors Excessive temperature detonation

Remedies sparking Ignition system voltage promotes should be limited to produce spark with sufficient temperature to ignite.

2.5 TYPES OF COMBUSTION CHAMBER IN SI ENGINE 1. Overhead valve (or) I - Head combustion chamber In this type of combustion chamber, both the valves are located on the cylinder head, so it is called over head valve. This type of combustion chamber has two forms. Bath-tub form This type of combustion chamber, consists of oval shaped chamber with both valves mounted overhead. The spark plug is mounted at the side. Wedge form This type of combustion chambers also consist of oval shaped chamber with both valves mounted overhead at its side with slight inclination. The spark plug is mounted centrally. A few features of this combustion chamber are listed below: S park p lu g

front in let valve (back exh aust valve is hid den )

(a) Bath - tub form o f com bustio n chamb er

Fig 2.8

(b) Wed ge form o f com bustio n chamb er

Combustion in SI and CI Engine 2.27

1.

Less heat loss because of less surface to volume ratio.

2.

Less flame travel length and greater freedom from knock.

3.

High volumetric cylinder.

4.

By keeping the hot exhaust valve in the cylinder head instead of cylinder block, it reflects in confinement of thermal failure to cylinder head.

efficiency

from

larger

valve

2. T - Head combustion chamber In this type of combustion chamber, two valves are placed on either side of the cylinder which requires two camshafts. Fig. 2.9 (a) In a manufacturing point of view, providing two camshafts is not recommended.

Exhaust Valve T - H ead Types Fig 2.9. (a)

The distance across the combustion chamber is very long so the knocking tendency is high in this type of engine. L - He ad Types

3. L-head combustion chamber In this L - head type two valves are provided on the same side of the cylinder which can be operated by a single camshaft. In this type, it is easy to lubricate the valve mechanism, with the

Fig 2.9. (b )

L - He ad Types

2.28 Thermal Engineering - I

detachable head provision. The cylinder head can be removed without disturbing valves, gears etc. In Fig. 2.9 (b) the air flow has to travel a longer distance to enter the cylinder. This causes loss of velocity head and loss in turbulence level. This design reduces knocking tendency by reducing the flame travel length. This chamber type gives additional turbulence during compression stroke. F - head combustion chamber The F - head type, exhaust valve is in cylinder head and the inlet valve is in cylinder block. In this type, the valves are actuated by two camshafts which is a disadvantage. C ylinder he ad

Sp ark p lu g EV

IV C ylinder block R e cipro cating Piston

Fig. 2.10 F-head com bu stion cham ber.

2.6 COMBUSTION IN CI ENGINES In CI engine, combustion occurs because of the high temperature of the compressed air. Since the fuel is ignited with the high temperature of compressed air, it is called auto-ignition. For the auto ignition, compression ratio should be maximum (about 12). It requires heavier construction. So CI engines are heavier and bigger.

Combustion in SI and CI Engine 2.29

The air is compressed and the fuel is injected with high pressure in the form of fine spray near the end of the compression. This leads to delay period. This is also called ignition lag. The fuel which is in atomized form is slightly colder than the hot compressed air in the cylinder. An appreciable time elapses before the actual combustion starts. This elapsed time is called delay period or ignition lag. Four stages of combustion in CI Engine. 1.

Ignition delay period.

2.

Period of rapid or uncontrolled combustion.

3.

Period of controlled combustion.

4.

After burning.

1. Ignition Delay period A period in between the start of injection and start of combustion is called delay period. [shown in Fig 2.11) 80

60

S tart of com bustio n

50

Inje ctio n starts

C om pre ssion p ressure

2

P (kg f / cm )

70

40 0 .00 1se c 30 20 10

1 2 3 inje ctio n A tm ospheric

6 0 4 0 2 0 TD C 2 0 4 0 6 0 8 0 1 00 1 20 1 20 1 00 8 0 Tim e , D eg ree s of C ra nksha ft ro tation ( ) Fig 2.11. Combustion in C.I Eng ines

2.30 Thermal Engineering - I

The fuel will not ignite immediately after the injection of fuel into the combustion chamber. There is a definite period of inactivity between the time when the fuel hits the hot compressed air in the combustion chamber and the time it starts burning. The ignition delay period is also called the preparatory phase during which some fuel already gets admitted but not yet ignited. If the delay period is more, more fuel will be injected inside the cylinder and more will be the pressure rise. This causes diesel knock. Some delay period is needed to dispense and atomise the fuel in the air for complete combustion. So we have to keep the delay period as short as possible. The delay period in the CI engine is a very great incluencing factor on both engine design and performance. It influences the following: (i) The combustion rate (ii) Knocking (iii) Starting ability (iv) The presence of smoke in the exhaust This delay period is divided into physical delay period and chemical delay period. (i) Physical delay The physical delay is the time period, in between the beginning of injection and the starting of chemical reaction. During the period, the fuel is atomized, vaporized, mixed

Combustion in SI and CI Engine 2.31

with air and its temperature is raised to its self ignition temperature. This physical delay normally depends on the type of fuel i.e for light viscous fuel, the physical delay is less while for heavy viscous fuels, the physical delay is more. The physical delay can be greatly reduced by using high injection pressure, high combustion chamber temperature and high turbulence in order to break the jet and improve evaporation. (ii) Chemical delay During the chemical delay period, reactions start slowly and then gets accelerate till ignition takes place. Generally, the chemical delay period is longer than the physical delay period. It depends on the temperature of the surrounding. At high temperature, the chemical reactions are faster leading to less chemical delay. The ignition lag in SI Engine is a similar phenomenon like the chemical delay in CI Engine. Factors Affecting Delay Period 1.

Temperature and pressure in chamber at the time of injection.

the

combustion

2.

Air-fuel ratio.

3.

Turbulence of air.

4.

Presence of residual gases.

5.

Rate of fuel injection.

6.

The extent of atomization and vaporization and fineness of fuel spray.

2.32 Thermal Engineering - I

2. Period of Combustion

Rapid

Combustion

(or)

Uncontrolled

After delay period, this period starts. This period is counted from the end of delay period to the point of maximum pressure on the indicator diagram. In this stage, the pressure rise is rapid. About one-third of heat is released at this stage. The rate of pressure raised in this stage depends on 1.

the amount of fuel sprayed in the delay period.

2.

the degree of turbulence.

3.

fineness of fuel-spray.

3. Period of Controlled Combustion: This is the third stage starting after rapid combustion period. The fuel injected in the stage is burnt immediately and any further pressure rise can be controlled by injection rate. The period of controlled combustion is coming to an end at maximum cycle temperature. 4. After Burning: The unburnt fuel particles will get inflamed even after fuel injection is over. This is called after burning. This after burning may continue in the expansion stroke upto 70 to 80 of crank angle from TDC.

2.7 FACTORS

THAT

AFFECT

DIESEL ENGINE (i) Compression ratio (ii) Intake temperature (iii) Intake pressure (iv) Engine size

DELAY

PERIOD

IN

Combustion in SI and CI Engine 2.33

(v) Fuel temperature (vi) Injection time (vii) Output 1. Compression ratio The increase in compression ratio reduces ignition lag. Due to increased intensity of compressed air and the closer contact of molecules, the time of action is reduced when fuel is injected. 2. Intake temperature Increase in intake temperature, increases the air temperature after compression, resulting in reduced delay period. 3. Intake pressure (super charging) Increase in intake pressure reduces the auto ignition temperature and hence reduces the delay period. The peak pressure will be higher as the compression pressure increases with intake pressure. 4. Engine size The engine size has little effect on the delay period in the order of milliseconds. Large engines operate at low revolutions per minute because of inherent stress limitations. The delay period in terms of crank angle is smaller and hence less fuel enters the cylinder during this period. 5. Fuel temperature Increase in fuel temperature will reduce both physical and chemical delay.

2.34 Thermal Engineering - I

2.7.1 Effect of variables on the Delay period Effect on delay period

Increase in variable

Reason

Compression ratio

Reduces

Increases air temperature and pressure and reduces auto ignition temperature

Intake temperature

Reduces

Increases temperature

Injection pressure

Reduces

Reduces physical delay due to greater surface volume ratio

Cetane fuel

Number

Injection advance

of Reduces

timing Reduces

Reduces self-ignition temperature.

air

the

Reduced pressure and temperature when the injection begins.

Fuel temperature

Reduces

Increase chemical reaction due to better vapourization.

Engine speed

Reduces in Reduces loss of heat. the order of milliseconds

Combustion in SI and CI Engine 2.35

2.8 KNOCKING (OR) DIESEL KNOCK If the delay period is prolonged, a large amount of diesel will be injected in the chamber. Combustion of the large amount of fuel may cause high pressure rise and this high pressure rise cause knocking. The methods to prevent knocking 1.

By reducing the delay period by doping. Note: Doping is the process of adding 1% ethyl nitrate to accelerate the combustion and as a result, we can reduce knocking.

2.

By raising the compression ratio, we can raise the temperature of air much higher than that required for auto ignition of the fuel. By doing so, we can reduce knocking. Note: In petrol engine, detonation occurs if we increase compression ratio. Here in CI engine, we can prevent knocking by increasing compression ratio.

3.

By increasing the turbulence of the compressed air, we can prevent knocking.

4.

By adjusting the fuel injector so as to inject only a small quantity of fuel in beginning.

5.

By super charging, we can reduce knocking. Note: Super charging is the process of increasing the inlet pressure of air. But super charging will increase the tendency of detonation in SI engine.

6.

By increasing the injector pressure, we can atomize the fuel efficiently to avoid knocking.

2.36 Thermal Engineering - I

2.8.1 The phenomenon of knock in CI engine In CI engine, the injection of fuel takes place for a definite interval of time.

Inje ction o f fuel End

Inje ction of fu el Start D e lay P eriod

P re ssure

Total Injection Time

If the, ignition delay of fuel being injected is short, the actual burning of first few droplets of fuel will commence in relatively short time. After injection only the small amount of fuel is accumulated in the chamber. When the burning commences, the rate of rise in pressure will exert a smooth force on the piston. Fig. 2.12. S ta rt of Injection of fuel S tart of C o mb ustion TDC

(a)

Time Fig. 2.12.

(b)

As the ignition is further delayed, relatively large amount of fuel gets accumulated inside the chamber. Hence upon combustion, there is a rapid rise in rate of pressure, which results in rough engine operation.

Sta rt of Injection o f fue l S tart of C o m bustio n TDC

Time

Fig. 2.14. Injection of fu el End

Total Inje ction Tim e

(c)

Inje ction of fu el Start D elay P eriod

P ressure

Sta rt of Injection o f fue l S ta rt of C om bustio n TD C

Injection of fu el End

Total Inje ction Tim e

Injection o f fuel S tarts D elay P eriod

P ressure

Combustion in SI and CI Engine 2.37

Time Fig. 2.13.

If the ignition delay is quite longer, large amount of fuel gets accumulated in chamber. Hence upon combustion, the instantaneous rise in pressure takes place, which results in vibration of engines called knocking.

2.38 Thermal Engineering - I

In CI engine, knocking occurs near the beginning of combustion where in SI Engine, knocking occurs near the end of combustion. 2.8.2 Comparison of knock on SI and CI Engines Knocking is a phenomenon which may occur in any type of engine when excess heat is generated. The amount of heat generated to cause knocking may very depending upon the engine. However, in CI and SI engine, knock predominantly occurs due to auto ignition of the air-fuel mixture. The Fig. 2.15 shows a graph which traces the path of the knocking phenomenon. Although an auto ignition leads to knocking of both CI and SI engines, a few significant differences can be observed with in the processes.

Pressu re

Start of C om bustion TD C

S tart of In jection

TD C

Pressu re

Start o f ign itio n

Time: In an SI engine, the auto ignition occurs near the end of the combustion which leads to knocking. It is evident from the graph that auto ignition starts only after the peak

Time S I Eng ine

Fig. 2.15.

Time C I E ngine

Combustion in SI and CI Engine 2.39

pressure in the SI engine. Hence, to avoid knocking, the auto ignition of the end gas needs to be avoided. On the contrary, in a CI engine an autoignition occurs before the peak pressure, at the start of the combustion. Hence to prevent the knock in a CI engine, autoignition before the combustion needs to be eliminated. Preignition: Since the air-fuel mixture is sent together for the combustion process, preignition could play a major role in the knocking of a spark-ignition engine. In a CI engine, only compressed air is taken in and the fuel is injected only before the Top dead centre. Thus there is no possibility for a pre ignition in a compression-ignition engine. Intensity of knock: A spark-ignition engine is more likely to undergo a detonation process during a knock, when compared to a compression ignition engine. This is due to the explosive auto ignition of the homogeneous air-fuel mixture in a SI engine, where as in a CI engine the intensity of knock is less severe as the air-fuel mixture is not homogeneous. Pressure and other factors: A CI engine operates at higher pressure limits when compared to an SI engine. During a normal cycle in a CI engine, an audible noise is always present. Due to operations at high pressures and excess heat, the fuel is ignited even before the Top dead centre. Hence when the audible noise prevails and causes heavy vibrations in the engine, it is said that the engine is knocking. Also the factors which prevent knock in a SI engine may promote knocking in a CI engine. A few factors which reduce knocking in both the engines have been given below.

2.40 Thermal Engineering - I

2.8.3 Characteristics Tending to Reduce Detonation S. No.

Characteristics

SI Engine

CI Engine

1.

Ignition temperature of fuel High

Low

2.

Ignition delay

Long time

Short time

3.

Compression ratio

Low

High

4.

Inlet temperature

Low

High

5.

Inlet pressure

Low

High

6.

Cylinder size

Small

Large

7.

Combustion wall temperature Low

High

2.9 NEED FOR AIR MOVEMENT IN DIESEL ENGINE The performance of a diesel engine interms of power and mileage can be greatly influenced by alternating the motion of air within the cylinder. A proper mixture of the compressed air and the injected fuel can lead to increased efficiency during the combustion process. A well mixed air fuel solution burns completely and minimizes the amount of unburnt fuel, thereby reducing emission. The formation of a proper mixture of air and fuel predominantly depends upon the motion of air, which in turn depends upon the geometry and configuration of the combustion chamber and its various elements. (i) Swirl: It is defined as the helical path traced by the discharge of air about the axis of the cylinder. It assists in the process of mixing the air with the injected fuel, thereby affecting the combustion process significantly. A simple initial angular momentum along the side of the cylinder, when supplied to the discharge, causes the

Combustion in SI and CI Engine 2.41

swirl motion. It is generated during the intake process by altering the intake port and intensifies during the compression stroke based on the configuration of the cylinder. (ii) Suction swirl: The air from the inlet valve is sent tangentially inside the cylinder. After subsequent deflections from the walls, it assumes a helical path about the axis of the cylinder. This is known as suction swirl. The tangential motion of air can be provided by   

Partial masking of the inlet valve. Providing a lip in the port on one of the sides. Positioning the inlet port in the desired tangential angle.

(iii) Compression swirl: The suction swirl is usually intensified in the compression process. As the piston approaches the Top dead centre, the swirling air is forced into the piston bowl. For reduced diameters of the piston bowl, the rotational force is magnified. Thin and deep bowls usually have higher swirl rate. (iv) Squish: The squish motion occurs within the small gap or recess between the piston and the cylinder head as the piston approaches TDC. The whole volume of air is compressed into this small recess just before the combustion stroke and this compression causes an inward radial movement of air called squish by Ricardo. This recess is usually created by either the crown on the piston or the configuration of the cylinder head. The crown on the piston, generally preferred as the cylinder head, is continuously subjected to heat loss by the flow of coolant.

2.42 Thermal Engineering - I

(v) Turbulence: A local fluctuation in the flow field of air often results in increased rate of momentum of the particles, thereby resulting in a turbulent flow. A turbulent motion of air vastly contributes to the dispersion and mixing of the air and fuel. A turbulent flow of air has higher rates of heat and mass transfer than a laminar flow or a molecular diffusion process. This instantaneous increase in momentum and heat and mass transfer is an essential ingredient in the satisfactory working of an internal combustion engine.

2.10 COMBUSTION CHAMBER COMPRESSION IGNITION ENGINE

DESIGN

FOR

The combustion chamber characteristics have to be such as 1.

To avoid maximum cylinder pressure.

2.

To avoid excessive pressure rise.

3.

It is to be designed so that the fuel should be burnt fully in the expansion stroke.

Types of combustion chamber 1. Open (or) Direct combustion chamber 2. Divided (or) Indirect combustion chamber 1. Open combustion chamber The open combustion chamber is the simplest form of chamber. It is suitable for only slow speed four-stroke cycle engine. In the open chamber, the fuel is injected from the space on top of the cylinder. The combustion space is formed by the top of the piston and the cylinder head. It

Combustion in SI and CI Engine 2.43

is shaped to provide swirling action of the air when the piston comes up on the compression. This type of chamber requires a higher injection pressure and a greater degree of fuel atomization than the other combustion chamber, to obtain proper fuel mixing. To equalize combustion in the combustion chamber, multiple orifice type injector tip is used. In this kind of chamber design, there is a possibility to attain ignition lag. Open combustion chambers are further divided as follows: (a) Shallow depth chamber In shallow depth chamber, the depth of cavity provided is quite small. This chamber is used for large engines which is running at low speed. In this type, the squish produced is negligible, since the cavity diameter is very large. (b) Hemispherical chamber The hemispherical chamber gives better performance, since the depth to diameter ratio for a cylindrical chamber can be varied to give any desired squish. (c) Cylindrical chamber In cylindrical chamber by varying the depth, squish can also be varied. The swirl was produced by masking the valve for nearly 180 circumference. (d) Toroidal chamber The toroidal chamber is designed to provide a powerful squish along with the air movement. This will give better utilisation of oxygen due to powerful squish and the mask needed on inlet valve is small.

2.44 Thermal Engineering - I

Combustion Chambers for CI Engines

(a) Shallow Depth Cham ber

(b) Hem ispherical Cham ber

Direct - In jection (DI) Combustio n Cham ber

( c ) Cylindrical Cham ber (d) Toro idal C ham ber Open Com bustio n Cham bers Fig.2.16.

Advantages of open combustion chamber 

 

Because the surface area to volume ratio is lower, the loss of heat during compression is minimum. This enhances better efficiency. No cold starting problem. Fine atomization is achieved because of multiple hole nozzle.

Combustion in SI and CI Engine 2.45

Disadvantages  

High pressure is required for fuel injection leading to complex design of fuel injection pump. For small engines, it is necessary for accurate metering of fuel by the injection system.

2. Divided combustion chamber type In this type, the combustion chamber is divided into two or more compartments. In between these divided compartments, restrictions or throats are provided, which are so small so that considerable pressure difference occurs between them during the combustion process. These chambers are further classified into   

Turbulent combustion chamber Pre combustion chamber Energy cell

(a) Turbulent combustion chamber This chamber consists of a spherical-shaped chamber separated from engine cylinder and located in cylinder head. The air is passed into the chamber during compression stroke, which produces a rotary motion to the compressed air. Hence, the turbulence of the air is increased. When the fuel is injected into the rotating air, it is partially mixed and the combustion starts. The pressure buildup in the chamber forces the burning and unburnt air-fuel mixture to flow back into main chamber thus, increasing the turbulence. This causes considerable heat loss to the walls of the chamber. It can be reduced by employing heat - insulated

2.46 Thermal Engineering - I

Sp herica l Sh aped C h am ber

Sp raying N o zzle R e ciprocating Piston

Fig.2.1 7. Tu rbu len t C o m b us tion C ham ber

chamber. The heat loss in this chamber is greater than the open combustion chamber. (b) Pre combustion chamber Pre-combustion chamber consists of anti chamber which produces rotary motion to compressed air, which is connected to main chamber through number of small holes. This chamber is located at the cylinder head and is connected to the engine cylinder by small holes. During the compression stroke, piston forces the air from the main cylinder to enter into the pre combustion chamber. At this point, the fuel is injected into the precombustion chamber and combustion begins. It results in high pressure and the flaming fuel droplets along with considerable quantity of air are forced through the small holes into the main cylinder. The bulk of combustion where

Combustion in SI and CI Engine 2.47

about 80% of energy is released into the main combustion chamber. The pre combustion chamber has multi fuel capability. The rate of pressure rise is low when compared to open type chamber.

S p raying N o zzle P re c ha m b e r / A n tich am b er

O rifice

R e cipro c ating P iston

Fig.2.18. Precom bu stion C ham ber

(c) Energy-cell chamber In this energy cell chamber, the clearance volume in the cylinder head is divided in two parts, the main cylinder and the energy cell. Energy cell is further sub-divided into

2.48 Thermal Engineering - I

two parts, major and minor chamber which are separated from each other and from the main chamber by narrow artifice. During compression, pressure difference occurs between the main chamber and energy cell due to the restricted passage between them. During combustion, the fuel is injected by pintle type nozzle, part of fuel passes across the main chamber and enters the energy cell. Combustion starts initially in main chamber, where the temperature is higher but the rate of burning of air is low due to the absence of air motion. In the major and minor chamber of energy cell, fuel is well mixed with air because of good turbulence resulting in complete combustion. The design of energy cell is to reduce ignition lag and to run hot. M ain C om bu stion C h am be r M inor C ell S p rayin g N ozzle

C ylind er H e ad M ajor C ell

R e cipro cating P iston

Fig.2.19. Energy Cell

Combustion in SI and CI Engine 2.49

Advantages of Indirect-injection combustion chamber   

No complex design is needed for fuel-injection pump since the injection pressure required is very low. Direction of spraying is not very important. Higher engine speeds can be achieved since burning is continuous in pre chamber.

Disadvantages  



Heater plugs are required, because of poor cold starting performance. Specific fuel consumption is high as there is a loss of pressure due to air motion through the duct and heat loss due to large heat transfer area. The increase in temperature and pressure on the part of the piston leads to cracking and distortion.

2.10.1 Open and Divided combustion chambers Open combustion chamber 1. It requires multiple hole injection nozzles. 2. It can consume good ignition quality fuels. 3. Open combustion chamber is more efficient.

Divided combustion chamber 1. It requires single hole injection nozzle. 2. It consumes poor ignition quality fuels. 3. Divided combustion chamber leads to pressure loss and heat losses during compression and expansion. So this type is not efficient. 4. It requires high injection 4. It requires moderate pressure. injection pressure. 5. Cylinder construction is 5. Cylinder construction is simple and less cost. hard and more expensive.

2.50 Thermal Engineering - I

2.11 DIESEL FUEL REQUIREMENT COMPRESSION IGNITION ENGINES

:

FOR

The fuel used in Compression Ignition (CI) engine is diesel, which is a type of hydrocarbon. Fuel for CI engines should have certain qualities to be qualified as an ideal fuel for these engines. Diesel is used as the fuel it CI Engine because it possesses the quality that are desired for the CI engine. Some of the desired characteristics of Diesel: 1. Knocking characteristics In case of CI engine, the burning of fuel occurs due to compression of air. It is desired that as soon as the fuel is injected into the cylinder, it starts burning, but in practical situations, this never happens immediately, there is always a time lag between the injection and burning of the fuel. As the duration of ignition lag increases, more and more amounts of fuel gets accumulated in the cylinder head. When the fuel is burnt, large amount of energy is released, which produces extremely high pressure inside the engine. This causes the knocking sound in the engine. Thus the engine should have short ignition lag so that the energy is produced uniformly inside the engine and can avoid the abnormal sound in engine. The knocking capacity of the fuel is measured in terms of cetane rating of the fuel. The fuel used in CI engine should have high cetane number to avoid knocking of engine.

Combustion in SI and CI Engine 2.51

2. Volatility of the fuel The fuel should be volatile in nature. Within the operating temperature range in the cylinder head, it gets converted into its gaseous state when mixed with compressed air. 3. Starting characteristics of the fuel For smooth starting of vehicle, the fuel should have good volatility so that it mixes with the air uniformly and it should have high cetane number so the ignition of fuel will be fast. 4. Smoke produced by fuel and its odour The exhaust gases produced from the fuel should not have too much smoke and odour. 5. Viscosity of fuel The fuel should have a low viscosity so it can easily flow through the fuel system and it should not be frozen at the lowest working temperature. 6. Corrosion and wear The fuel used in CI Engine should not cause corrosion to the components of the engine before (or) after combustion. 7. Easy to handle Large quantities of fuel for a CI engine have to be transported and so it should be easy to handle and transport. The fuel should have high flash point and high fire point to avoid the catching of fire during transport.

2.52 Thermal Engineering - I

2.11.1 Cetane Number (CN) Cetane Number (diesel fuel) refers to the ease at which diesel fuel ignites at a relatively low temperature. Lower cetane number will result in cold starting problems. Higher cetane number results in faster ignition. Diesel Index The Diesel index indicates the ignition quality of the fuel. It is found to correlate, approximately to the cetane number of commercial fuels. Diesel index and cetane number are usually about 50. Lower value will result in smoky exhaust. 2.11.2 Fuel Rating for CI Engine Cetane Number (applicable for CI engine) Octane number is for rating Petrol. Cetane number is for rating Diesel. The cetane number is a number to rate diesel fuel’s ability to auto ignite quickly when it is injected into the high pressure, high temperature air in the cylinder. Higher the cetane number, lesser is the ‘Diesel knocking’ tendency. Procedure for finding Cetane Number 

Cetane C 10 H34  has high ignitabilty.



methyl-napthalene C 11 H10  has low ignitability.



Both are mixed together and this sample mixture is used for running a test engine.

Combustion in SI and CI Engine 2.53

The cetane number of the diesel is the percentage of cetane in this sample mixture which knocks in a similar way as the diesel under the same condition.   

Lower cetane number, higher are the hydrocarbon emission in the exhaust gases. Lower cetane number, higher are the noise level. Lower cetane number, increases the ignition delay.

Note: Generally, fuel having higher octane number means it has low cetane value.

Chapter - 3

Testing and Performance of Engines Indicator diagram and properties - Pressure TransducerBrake power measurements. Dynamometers - Performance calculations -Morse test - Air consumption - Fuel consumption Exhaust gas compositon - Heat balance sheet.

3.1 PERFORMANCE CALCULATIONS (OR) PERFORMANCE TEST ON I.C. ENGINES IC Engines can be tested for their performance. During this test, we can find out the following quantities. 1. Indicated power, 2. Indicated Mean Effective Pressure (MEP), 3. Brake power, 4. Mechanical efficiency, 5. Indicated thermal efficiency, 6. Brake thermal efficiency, 7. Relative efficiency (or) Efficiency ratio, 8. Volumetric efficiency, 9. Air consumption, 10. Fuel consumption, 11. Heat balance sheet. 1. Indicator Diagram An indicator diagram is a P-V diagram traced by the indicator which is attached to the piston. The P-V diagram represents the work done by a = A rea re pre se n tin g the engine in one cycle. P d

The power developed inside the engine cylinder is known as indicated power. This is measured by indicator diagram.

w o rkd on e in on e cy cle E q uiva le nt 1 re ctan gle o f a rea = a d

2 V Fig. 3.1

3.2 Thermal Engineering - I

We can measure ad (area of indicator diagram) by using planimeter. Now we can draw an equivalent rectangle whose area is equal to the area of indicator diagram. And the height of this rectangle gives the Mean effective pressure Pm. 2. Indicated Power (IP) IP 

P m A LN or N /2  n 60

kW

where P m  Mean effective pressure in kN/m 2 or K Pa A  Area of piston in m 2 L  Length of stroke in m N  Speed of the engine in r.p.m. . . N for 2 stroke engine [ . In two stroke engine, the

cycle is completed in two strokes of the piston or in one revolution of the crankshaft.] . . N/2 for 4 stroke engine [ . In four stroke engine,

the cycle is completed in 4 strokes of the piston (or) in two revolutions of the crankshaft.] and n or k No. of cylinders in the engine. 3. Mean Effective Pressure Pm The mean effective pressure ‘P m’ can be calculated from the following formula. Pm 

ad ld

S

Testing and Performance of Engines 3.3

where ad  Area of indicator diagram (or) rectangle in m 2. ld  length of the diagram in m. S  Spring constant (or) Spring number used in

engine indicator - unit in N/m 2/m or bar/m. MEASUREMENT OF CYLINDER PRESSURE Cylinder pressures can be easily obtained using several analytical equations relating to the temperature and the volume of the working fluid. Indicator diagrams are one of the most common tools needed to compute the pressure of a cycle. However, several electronic and mechanical components are being used to verify the pressure at each stage manually. A typical pressure measuring component consists of the following:    

A pick-up transducer Amplifier Recorder Display unit

TRANSDUCER A transducer is a device capable of converting one form of energy into another. A pressure transducer typically converts the pressure exerted on an object into noticeable output like displacement, electrical signals etc. Since the cylinder of an engine is subjected to various thermal and shear stresses, the pressure transducer must be capable of withstanding all the stresses. Most transducers are likely to fail when subjected to the enormous heat and forces from

3.4 Thermal Engineering - I

an engine. One of the main challenges in measuring a cylinder pressure is to identify the transducer which is capable of working accurately under such conditions. A piezo electric transducer is a device which can withstand the energy from an engine, and also determine the pressure more accurately. It works on the principle of piezo electric effect by which certain materials can generate an electric charge in response to an applied mechanical stress. It consists of a piezo electric crystal resting upon a diaphragm. The ends of the crystal are connected to an electronic device. The external pressure which is to be measured is received by the diaphragm which transfers it to the crystal. Due to applied mechanical stress, the crystal gives out amperage which can be measured using the electronic device. The amount of current flowing through the circuit is directly proportional to the applied mechanical stress. Pressu re C rystal

O utput +

+

+







O utput

D ia phra gm

Pressu re (a) Piezoe le ctric effect

Externa l Pre ssure (b) Piezoe le ctric pressure tra nsducer

Fig. 3.2 Piezoelectric transd ucer

Since the electrical output from the crystal is small, an amplifier is used to scale up the signals appropriately. Then comes the recorder circuit which records the variation

Testing and Performance of Engines 3.5

in pressure over time. The main drawback in this system is to calibrate the output according to the varying pressure. Strain gauges are also nearly effective in determining the pressure. However, due to several inertia forces acting on it, the usage of strain gauges is not preferred.

C ylind e r Pre ssu re ( B a r )

Now that the pressure has been noted and the output is usually displayed through a cathode ray oscilloscope (CRO). Since, CRO provides an inertia free recording and the electrical signals can be directly used to manipulate the beams, a CRO can accurately display the variation of

80 60 40 20

0

C ylin d e r P res sure ( B a r )

BDC

TD C (a)

2 00

3 60



C ra nk a ng le (  )

4 3 2 TDC

1

300

(b)

500 F ig . 3 .3 A Typical P- d iagram

BDC

700 C ran k a n g le (  )

3.6 Thermal Engineering - I

pressure with respect to time and with respect to the crank angle. Typical CRO unit consists of an input from the transducer and another set of inputs from the time recording device and the crank angle measuring device. The outputs are plotted in the form of a graph with pressure along the y-axis and crank angle along the x-axis. 5. Brake Power (BP) The power available at the crankshaft of the engine is known as brake power. The brake power is measured by some brake mechanism, hence the name brake power. 6. Different Arrangements used to find Brake Power (i)

Rope brake arrangement

(ii)

Prony brake arrangement

(iii)

Band brake arrangement.

(i) Rope Brake Arrangement The rope makes one complete turn around the flywheel keyed to the crankshaft of the engine. One end of the rope carries a dead load and the other end is connected to spring balance. Here the flywheel is rotating clockwise and the pull of the dead load makes anticlockwise torque. The engine is started with no load. Then gradually load is increased by adding weights in the dead load hanger.

Sp rin g ba la nce

RO PE BR AKE A rra n g e m en t Fig 3.4

At any steady condition, The effective radius R at which the net resisting force is acting

Testing and Performance of Engines 3.7

R

Dd m 2

Resisting torque acting on the brake wheel is given by T  W  S  R in kNm

Brake Power  B.P 

2 NT in kW 60

2N  D  d   W  S   in k W 60  2 

where W  Dead load in kN

S

S  Spring balance reading in kN R

D  Dia. of Brake drum in m d  Dia. of rope in m

Fig. 3.5

N  Speed of the engine in rpm.

W

(ii) Prony Brake Arrangement It consists of brake shoes which touch on the rim of the brake wheel by means of bolts, nuts and springs. The N ut Loa d lever

Brake shoe

Spring

Brake drum W

Fig. 3.6

L

3.8 Thermal Engineering - I

pressure on the rim can be varied by adjusting the nut in the arrangement. When the brake drum is rotating in anticlockwise direction, the dead load produce clockwise torque. So the resisting torque (clockwise) ‘T’  W  L where W  Weight on load carrier in kN. L  Distance from the centre of the brake drum to

the point of load in m. Brake Power B.P  B.P 

2NT in kW 60 2NW L in kW 60

(iii) Bank Brake Arrangements A flat belt is put around the rim of the brake wheel. The two ends of the flat belt is attached with W 1 and W 2. Another belt is integrated with the flat belt and its other end is connected to the spring balance. Now the resisting torque ‘T’ is given below.

S

S p rin g b alan ce

B e lt w2 B a nd b rake Fig.3.7

T  [ W 1  S  W 2 ]  R

where W 1  Wt. on right side load carrier in kN W 2  Wt. on left side load carrier in kN S  Spring balance reading in kN R  Effective radius of the brake drum in m.

w1

Testing and Performance of Engines 3.9

Brake Power  B.P  B.P 

2NT in k W 60

2N [ [ W 1  S ]  W 2]  R in kW 60

7. Friction Power (FP) The Power available in the engine flywheel (crankshaft end) is less than the power developed inside the engine. i.e. The BP is less than the IP. Because, there is a loss of power due to friction between the moving parts. The Power lost in this way is known as friction power. So, F.P  I.P  B.P The difference between the indicated power and brake power is known as friction power. 8. Specific Fuel Consumption (S.F.C) in kg/kW hr It is defined as the amount of fuel consumed per unit of power developed per hour.  mf The ratio is known as specific fuel B.P or I.P  consumption per kW per hour. [Here m f  mass of fuel consumed in kg/hr.] BSFC 

ISFC 

 mf B.P  mf I.P

kg/kW  hr

kg/kWhr

where, BSFC - Brake Spepcific fuel consumption. ISFC - Indicated Specific fuel consumption)

3.10 Thermal Engineering - I

9. Mechanical Efficiency mech  Mechanical efficiency is defined as the ratio of the power available at the engine crankshaft and power developed inside the engine cylinder. Mechanical Efficiency, mech 

Brake Power Indicated Power

10. Thermal Efficiency The ratio of B.P (or) I.P to the energy supplied by fuel during the same interval of time is known as thermal efficiency. If it is based on I.P, then it is known as Indicated thermal efficiency. If it is based on B.P, then it is known as Brake thermal efficiency. 11. Indicated Thermal Efficiency indicated  indicated 

I.P in kW  3600  mf  C.V.

where C.V  Calorific value of fuel in kJ/kg  m f  Mass of fuel in kg/hr

If C.V is given in kJ/m 3, then  indicated thermal 

I.P in kW  3600  V f  C.V.

 where Vf  Volume of gas fuel supplied in m 3/hr

Testing and Performance of Engines 3.11

12. Brake Thermal Efficiency Brake  Brake 

B.P in kW  3600  mf  C.V.

where C.V in kJ/kg. If C.V is in kJ /m 3, then  Brake 

B .P in kW  3600  V f  C.V.

 Brake   indicated  mech

Brake thermal efficiency is also known as overall efficiency. i.e.,  Brake   overall 12. Relative Efficiency or Efficiency Ratio The ratio of the indicated thermal efficiency or the brake thermal efficiency to the air standard efficiency is known as relative efficiency or efficiency ratio. Relative efficiency,  relative 

Indicated o r Brake thermal efficiency Air standard efficiency

14. Volumetric Efficiency volumetric The ratio of the actual volume of the charge admitted into the cylinder to the swept volume of the piston is known as volumetric efficiency. volumetric 

Volume of charge admitted at NTP condition Sw ept vo lume

NTP = Normal temperature (1.01325 bar) condition.

0C

and

pressure

3.12 Thermal Engineering - I

3.2 DYNAMOMETER Dynamometer is used to measure the brake power. The brake power is measured by coupling the brake dynamometer to the engine shaft. Let W  Load measured in the dynamometer in N R  Arm length in m

Resisting torque  T  WR in N-m Brake Power 

2NT in kW 60  1000



2NWR in kW 60  1000

In the hydraulic dynamometer, the arm length R is fixed. So the factor 2R/60  1000  is constant and is known as reciprocal of dynamometer constant. So, B.P 

WN K

 60  1000  where K  Dynamometer constant    2R  

Dynamometer can be classified into two types (i) Absorption dynamometers (ii) transmission dynamometer (i) Absorption dynamometer These types of dynamometers are used to measure and absorb the power output of the engine to which they are coupled. The power absorbed is usually released as heat

Testing and Performance of Engines 3.13

(or) by other forms of energy. Eg., Hydraulic, Eddy current dynamometer etc. (ii) Transmission Dynamometer Transmission dynamometer is also called as Torque meter. The purpose of these meters is to simply sense the torque. It doesnt supply (or) receive any energy. Torque meters employ normal measuring units like a strain gauge to directly determine the torque acting on a shaft. Usually, more strain gauges are fixed on a rotating shaft and the angular strain of the shaft obtained from the gauges is directly proportional to the torque acting on the shaft. These dynamometers have high accuracy, and are often employed in automatic units. Absorption Dynamometers 3.2.1 Hydraulic Dynamometer The water brake is of hydraulic nature and it is the simplest example for hydraulic dynamometer. Generally the water brake is used for large capacity systems as compared to prony brake system, because large amount of heat is dissipated to the water in water brake system. 

This device uses fluid friction and not dry friction.

Description The main parts of this system are shown in Fig. 3.8 

 

A rotating disk is fixed to the driving shaft. Semi-elliptical grooves are provided on the disc through which a stream of water flows. A casing is stationary in which the disc rotates. The casing is mounted on anti-friction bearings and it carries a braking arm and a balance

3.14 Thermal Engineering - I

C ham be r D isc

W ater in

A ntifriction B earin gs

D rivin g sha ft

W ater W ater ou t

C asin g

Fig : 3.8 Water Brake or Hydraulic Dyn am om eter

system. Therefore, the casing can rotate freely, but its movement can be restricted by the arm. Operation 



 

When the driving shaft rotates, water flows in a helical path in the chamber. Due to vortices and eddy-currents setup in the water, the casing tends to rotate in the same direction as that of the driving shaft. By varying the amount of water and its pressure, the braking action can be initiated. Braking can also be provided by varying the distance between the rotating disk and the casing. The heat developed due to the power is cooled by the continuous supply of working fluid. Power absorption is approximately the cube of rotational speed and the fifth power of disc diameter

Testing and Performance of Engines 3.15



The housing is constrained by a force-measuring load cell placed at the end of the arm of radius r. Torque T  F  r where F  force m easured at radius r Power P 

2  NT 60

3.2.2 Eddy Current Dynamometer It is an absorption type dynamometer. It works on the basis of Faraday’s Laws of induction, by which loops of opposing currents are induced in a conductor under a changing magnetic field. B e aring S tato r E n co d er

M otor

R o tor R o tatin g m em b er

W a tt m e te r L oa d c e ll

Fig. 3.9 Ed dy C urre nt dy nam om eter

The arrangement consists of a stator, made up of electromagnets and a rotor coupled to the shaft of the engine. The rotor is made of copper or steel or other conducting materials. When the rotor rotates, Eddy currents are induced in the stator. This current acts in the opposite direction, thereby intending to stop the change in the magnetic flux. Due to this opposing force, a torque acts on the shaft, thereby loading the engine. This load is measured using a moment arm. As the eddy currents

3.16 Thermal Engineering - I

produce a significant amount of heat, the system requires a cooling arrangement for the purpose of heat dissipation. Performance Curves The following are the performance curves. B.P. vs speed

2.

I P vs speed

3.

SFC vs speed

4.

Mechanical efficiency vs speed

80 70 60 50 40 30 20 10

0 .4

M ax P ow er

B ra ke S FC

IP

0 .5 BP

B P & IP

1.

0 .3 0 .2 0 .1 5 00

5 00

1 50 0 2 50 0 3 50 0 S pee d (rpm ) (a) BP & IP Vs Speed Cu rves

1 50 0 2 50 0 3 50 0 S pee d (rpm )

(b) Brake SFC Vs Speed

70 60 50 40 5 00

1 50 0 2 50 0 S pe e d (rpm ) (c)

m ec h V s

S p ee d

al rm he

90 80

40 35 30 25 20 15 10 5

Bt

Bra ke T he rm al E fficie ncy

M echa n ic a l E ffic ien cy

Fig 3.10

5 00

1 50 0

2 50 0 3 50 0

S pee d (rpm )

(d) B ra ke T herm al E fficien cy V s S p ee d

Fig 3.11

Testing and Performance of Engines 3.17

5.

Brake Thermal efficiency vs speed

The sample curves for the above are shown here in the Fig 3.10 and 3.11 PROBLEMS ON PERFORMANCE CALCULATIONS OF IC ENGINES Problem 3.1: During test on single cylinder oil engine, working on the four stroke cycle fitted with a rope brake, the following readings are taken. Effective diameter of Brake wheel = 600 mm. Dead load on brake = 200 N; spring balance reading = 30 N; speed = 450 r.p.m; Area of indicator diagram  400 mm2; length of indicator diagram = 60 mm; spring scale = 1.1 bar per mm. Bore = 100 mm; stroke = 150 mm; Quantity of oil = 0.815 kg/hr. Calorific value of oil = 42000 kJ/kg. Calculate the brake power, indicated power, mechanical efficiency, brake thermal efficiency and brake specific fuel consumption and Indicated thermal efficiency.

Solution: Given: Effective Radius R 

600  300 mm  0.3 m 2

(Dead load)W  200 N, S  30 N, N  450 r.p.m. ad  400 mm 2, ld  60 mm , s  1.1 bar/mm,

Bore dia. D  100 mm  0.1 m, L  150 mm  0.15 m,  Mass of fuel m f  0.815 kg/hr, C.V  42,000 kJ/kg. Brake power (BP) B.P 

2NW  S  R 60

3.18 Thermal Engineering - I



2  450  200  30   0.3 60

B.P  2403.32 W  2.403 kW

Indicated Power (IP) Before that, we have to find mean effective pressure P m Pm 

ad  s ld



400  1.1  7.333 bar 60

P m  7.333  10 2KPa

Area of cylinder A 

   D2   0.1 2 4 4 A  7.854  10  3m 2

Indicated Power I.P 

P m AL N /2  n 60 . . [ . N/2 for 4 stroke engine]

I.P. 

7.333  10 2  7.854  10  3  0.15  450/2  1 . 60 . [ . n  1 for single cylinder]

 3.2396 kW

Mechanical Efficiency mech   mech 

B.P 2.403  I.P 3.2396

 0.74175  74.175 %

Testing and Performance of Engines 3.19

Brake Thermal Efficiency  Brake  

B.P.  3600  m f  C.V 2.403  3600  0.25273 0.815  42,000

 25.273 %

Indicated Thermal Efficiency  indicated  

I.P  3600  m f  C.V 3.2396  3600  0.3407 0.815  42,000

 34.07 %

Brake specific fuel consumption SFCBrake S FCBrake  

 mf B.P. 0.815 kg 2.403 kWhr

 0.3392

kg kWhr

Problem 3.2: A Four stroke four cylinder gasoline engine has a bore of 60 mm and a stroke of 100 mm. On test, it develops a torque of 66.5 Nm when running at 3000 rpm. If the clearance volume in each cylinder is 60 cc, the relative efficiency with respect to break thermal efficiency is 0.5 and calorific value of the fuel is 42 MJ/kg, determine the fuel consumption in kg/h and the break mean effective pressure. (JNTU - Jan/Feb - 2015)

3.20 Thermal Engineering - I

Given data: 4 stroke; 4 cylinder; D  0.06 m ; L  0.1 m ; T  66.5 Nm ; N  3000 rpm V c  60 cm 3  60  10  6 m 3 ; relative  0.5 ; CV  42  10 3 kJ/kg

Solution: Com pression ratio  Air std   Brake thermal   BP  mf  Pmbrake

To find r Vs 

r

   D 2  L   0.06 2  0.1  2.83  10  4 m 3 4 4 V s  V c 2.83  10  4  60  10  6   5.712 Vc 60  10  6

Air standard efficiency  1 

1

1 r  1 1 5.7120.4

 0.5

[Since it is gasoline engine, it is considered as petrol engine. So otto cycle  ] Brake therm al efficency Relative efficiency  Air standard efficiency Brake thermal efficiency  0.5  0.5  0.25 Brake power BP 

2  NT 2   3000  66.5   20891.6 watts 60 60  20.892 kW

Testing and Performance of Engines 3.21

 To find fuel consumption mf

We know, brake thermal efficiency 

B.P  3600  m f  Cv

 20.892  3600 mf  0.25  42  10 3  7.16 kg /hr

To find brake mean effective pressure Pmbrake P m brake 

B.P  60 N LA   n  2  N [ for 4 stroke; n  4 for 4 cylinder] 2 20.892  60

 0.1 

  3000   0.062   4 4  2 

 738.9 kPa  7.4 ba r

3.3 MEASUREMENT OF INDICATED POWER OF MULTICYLINDER ENGINE MORSE TEST This method is used to measure the indicated power without the use of indicator diagram in multicylinder engines. The brake power of the engine is measured by cutting off each cylinder in turn. The cylinder of a petrol engine is cut off by shorting the spark plug and in case of diesel engine, this is done by cutting off the diesel supply to the required cylinder. For example, consider a 4 cylinder engine. First of all, measure the brake power of the engine when all the

3.22 Thermal Engineering - I

cylinders are in operation. Then cylinder 1 is cut-off so that it does not develop any power. The speed of the engine decreases. In order to attain the initial speed back, the load on the engine is reduced. Now, the brake power is measured with this new condition which gives the brake power of the remaining three cylinders. Similar way, we can cut-off each cylinder one by one and measure the brake power of the remaining three cylinders by maintaining the engine speed as original speed. Let I1, I2, I3 and I4  Indicated power of cylinder 1, 2, 3 and 4 respectively. F 1, F 2, F 3 and F 4  Frictional power of cylinder 1, 2, 4 and 4 respectively.

When all the cylinders are in operation the total brake power B.P. simply B is given as follows. B  Total indicated po wer  Total Friction P ower ... (1) B  I1  I2  I3  I4  F 1  F 2  F3  F 4

when cylinder 1 is cut off, I1  0, but the frictional losses of the cylinder 1 remain the same  Brake power of the remaining three cylinders  B 1 B 1  0  I2  I3  I4  F 1  F 2  F3  F 4

...(2)

Subtracting the equation (2) from equation (1), we get B  B 1  I1 (or)

Testing and Performance of Engines 3.23

Indicated power of the first cylinder, I1 I1  B  B 1

Similarly, I.P. of 2nd cylinder I2 I2  B  B 2

IP of 3rd cylinder, I3 I3  B  B 3

I.P. of 4th cylinder, I4 I4  B  B 4

and the total indicated power IP  I1  I2  I3  I4 Problem 3.3: In a test with a four cylinder, four stroke petrol engine, the following results were found: B.P with all cylinders working = 24.25 kW B.P with cylinder No:1 cut off = 16.53 kW B.P with cylinder No:2 cut off = 17.2 kW B.P with cylinder No:3 cut off = 17.34 kW B.P with cylinder No:4 cut off = 17.8 kW Estimate the indicated power of the engine and its mechanical efficiency

Solution: Given B.P. or simply B = 24.25 kW, B 1  16.53 kW; B 2  17.2 kW ; B 3  17.34 kW; B 4  17.8 kW I1  B  B 1  24.25  16.53  7.72 kW I2  B  B 2  24.25  17.2  7.05 kW I3  B  B 3  24.25  17.34  6.91 kW

3.24 Thermal Engineering - I

I4  B  B 4  24.25  17.8 6.45 kW Total Indicated Power IP  I1  I2  I3  I4  7.72  7.05  6.91  6.45  28.13 kW

Mechanical efficiency  mech  

B.P I.P 24.25  0.8621 28.13

 86.21 % Problem 3.4: During a brake on a four cylinder, four stroke engine coupled to a hydraulic dynamometer at constant speed, the following readings were obtained. B.P. with all cylinders working = 14.7 kW B.P. with cylinder No.1. cut off = 10.14 kW B.P. with cylinder No.2. cut off = 10.3 kW B.P. with cylinder No.3. cut off = 10.36 kW B.P. with cylinder No.4. cut off = 10.21 kW Petrol consumption = 5.5 kg/hr Calorific value of petrol = 44,000 kJ/Kg Dia of cylinder  8cm Stroke of piston  10 cm Clearance volume 0.1 litre Calculate (1) Mechanical efficiency (ii) Relative efficiency on the basis of IP.

Solution: Given: B.P (or) simply B = 14.7 kW; B 1  10.14; B 2  10.3;  B 3  10.36 ; B 4  10.21 ; m f  5.5 kg/hr

Testing and Performance of Engines 3.25

C.V = 42,000 kJ/Kg; D = 0.08 m; L = 0.1 m V c  Clearance volume  0.1 litre  0.1  10  3 m 3

. . [ . 1000 lit  1 m 3; So 1 lit 

1 m 3] 1000

I1  B  B 1  14.7  10.14  4.56 kW I2  B  B 2  14.7  10.3  4.4 kW I3  B  B 3  14.7  10.36  4.34 kW I4  B  B 4  14.7  10.21  4.49 kW Total I.P.  I1  I2  I3  I4  4.56  4.4  4.34  4.49  17.79 kW

1. Mechanical efficiency: mech 

14.7 B.P.  17.79 I.P

 0.8263  82.63%

Indicated thermal efficiency: indicated  indicated  

I.P.  3600  m f  C.V 17.79  3600  0.27724 5.5.  42,000

 27.725 %

Air Standard efficiency: Air standard  air stand ard  1 

1 r  1

[Since it is a petrol engine the otto cycle efficiency is used]

3.26 Thermal Engineering - I

where r  compression ratio

Vs  V c Vc

where Vs  Swept volum e an d V c  Clearance volume Vs 

   D2  L   0.08 2  0.1  5.027  10  4 m 3 4 4

V c  0.1  10  3 m 3 given  So, r 

V s  Vc Vc



5.027  10  4  0.1  10  3 0.1  10  3

 6.027  air standard  1 

1 6.0271.4  1

 0.51252

 51.252 %

Relative efficie ncy    indicated 0.27724 or  0.541   0.51252 Efficiency ratio  air standard on the ba sis of IP   Relative  54.1% Problem

3.5: A four stroke petrol engine 8 cm bore and 10

cm stroke is tested at full throttle at constant speed. The fuel supply is fixed at 0.065 kg per minute and the spark plugs of four cylinders are successively short circuited without change of speed. load being adjusted accordingly. The Brake powers are measured and given below. With all cylinders working = 12 kW With cylinder No:1 Short circuited (not firing) = 8.46 kW With cylinder No:2 Short circuited = 8.56 kW With cylinder No:3 Short circuited = 8.6 kW

Testing and Performance of Engines 3.27

With cylinder No:4 Short circuited = 8.5 kW Determine the indicated power of the engine under these condition. Determine the indicated thermal efficiency if the calorific value of the fuel in 43500 kJ/kg. Find the relative efficiency, if the clearance volume of one cylinder being 70 cm3

Solution: Total B.P. with all cylinders firing = B = 12 kW I1  B  B 1  12  8.46  3.54 kW I2  B  B 2  12  8.56  3.44 kW I3  B  B 3  12  8.6  3.4 kW I4  B  B 4  12  8.5  3.5 kW Total IP  I1  I2  I3  I4  3.54  3.44  3.4  3.5  13. 88 kW Indicated thermal efficie ncy 

I.P  3600  m f  C.V

 m f  0.065  kg/min  0.065  60  3.9 kg /hr

indicated 

13.88  3600  0.29454 3.9  43500

 29.454%

Air standard efficiency : air standard  Swept volume V s   D 2  L for on e cylinder 4 

  0.08 2  0.1  5.0265  10  4m 3 4

3.28 Thermal Engineering - I

Clearance volume V c  69.5 c m3  69.5  10  6 m 3 compression ra tio  r 

V s  V c 5.0265  10  4  69.5  10  6  Vc 69.5  10  6  8.2324

 air standard  1 

1 r

1

[Otto cycle efficiency formula is used since it is a petrol engine] 1

1 8.2324 1.4  1

 0.56968

 56.97% Relative efficiency on the basis of IP 



indicated air standard 0.29454  0.51702 0.56968

 51.702 % Problem 3.6: In a test of four cylinders four stroke petrol engine of 75 mm bore and 100 mm stroke, the following results were obtained at full throttle at a constant speed and with a fixed fuel supply of 0.082 kg/min BP with all cylinder working  15.24 kW BP with cylinder number 1 cut off  10.45 kW BP with cylinder number 2 cut off  10.38 kW BP with cylinder number 3 cut off  10.23 kW BP with cylinder number 4 cut off  10.45 kW Estimate the Indicated power of the engine under this

Testing and Performance of Engines 3.29

condition. If the calorific value of the fuel is 44,000 kJ/kg, find the Indicated thermal efficiency of the engine. Compare this with the air standard efficiency. The clearance volume of one cylinder being 115 cc,

(JNTU - Jan/Feb - 2015 - Set 2)

Solution: Given data: B  15.24 kW ; B 1  10.45 kW ; B 2  10.38 kW B 3  10.23 kW ; B 4  10.45 kW

To find the Indicated Power (IP)  I1  I2  I3  I4 I1  B  B 1  15.24  10.45  4.79 kW I2  B  B 2  15.24  10.38  4.86 kW I3  B  B 3  15.24  10.23  5.01 kW I4  B  B 4  15.24  10.45  4.79 kW

IP (Indicated Power)  I1  I2  I3  I4  4.79  4.86  5.01  4.79  19.45 kW

Calorific value of fuel  44000 kJ/kg Dia of Bore  75 mm  0.075 m Stroke length 100 mm  0.1 m (i) Indicated thermal efficiency 

I.P  3600  m f  C.V

 m f  0.082 kg/min  0.082  60  4.92 kg/hr

3.30 Thermal Engineering - I

Indicated 

19.45  3600  0.32345 4.92  44000  32.345%

Air standard efficiency: air standard Swept volume V s  

  D 2  L for one cylinder 4

  0.075 2  0.1 4

V s  4.4178  10  4 m3 V c  115 cm 3  115  10  6 m 3

Compression ratio  r 

Vs  Vc Vc



4.4178  10  4  115  10  6

115  10  6

 4.8415

[Otto cycle efficiency formula is used since it is a petrol engine]  air standard  1  1

1 r

1

1 4.8415 1.4  1

 0.4679

 46.79%

Relative efficiency on the basis of IP  

ind icated air standatd 0.32345  0.6912 0.4679

 69.12%

Testing and Performance of Engines 3.31

3.4 MEASUREMENT OF AIR CONSUMPTION Orifice chamber method is used in laboratory for measuring the consumption of air. The arrangement of this system is shown in Fig. 3.12. It consists of an air tight chamber in which a sharp edged orifice is fitted. The orifice is situated away from the suction connection of the engine. A rubber diaphragm is provided to further reduce the pressure pulsations. There is a pressure depression due to the suction of the engine which causes the flow through the orifice for obtaining a steady flow. The volume of the chamber should be sufficiently large as compared with the swept volume of the cylinder. Generally 500 to 600 times the swept volume. The pressure difference which causes the flow through the orifice is measured with the help of manometer fitted in the airbox. The pressure difference is kept to 10 cm of water to make the compressibility effect negligible. Let A  Are a of orifice in m 2 hw  Head of water in m d  D iameter of orifice in m w  Density of water kg/m 3 a  Density of air in kg/m 3 C d  Coefficient of discharge of orifice

Head in terms of air in m is given by H  hw

w

M ea su re m e nt of A ir C o nsum p tion th rou gh the orifice cha m b er m eth od

. . ;  . H  a  h w w  a  

3.32 Thermal Engineering - I

The velocity of air passing through the orifice is given by V  2 g H m/s  The volume of air passing through the orifice is given   by V a  A  V  C d  C d  A   2gH

Mass of air passing through the orifice is given by,   m a  V a  a kg/sec The density of atmospheric air is given by Pa Va  m a R a Ta

a 

ma Va



Pa Ra Ta



Pa

287  T a

The volumetric efficiency of the engine is given by Vol. 



Actual volum e of air tak en in m 3/sec D isplacement volume in m 3/sec   Cd  A   2gH 2

D  N  L or N  2   No . of cylinders 4 60  

when the volumetric analysis of the exhaust gas is known, then the mass of air supplied per kg of fuel is given by m a/kg of fuel 

NC 33 C 1  C 2

where N  Percentage of nitrogen by volume in exhaust gas C  Pe rcen tage of carbon in fuel by weigh t C 1  Percentage of CO 2 by volume in exhaust gas C 2  Percentage of CO by volum e in exhaust gas

Testing and Performance of Engines 3.33

PROBLEM IN MEASUREMENT OF AIR SUPPLIED Problem 3.7: Following readings were obtained during the test on a single cylinder, 4 stroke IC engine. Dia of orifice of the air tank = 21 mm; pressure causing air flow through the orifice = 10 cm of water. Find the quantity of air consumed per second, if its density under atmospheric conditions is 1.15 kg/m3. Take coefficient of discharge for the orifice as 0.7.

Solution: h w  0.1m of water ; a  1.15 kg/m 3; C d  0.7

Area of orifice A 

  0.021 2  3.464  10 4m 2 4

Head causing flow of air through orifice (H) w

H  hw 

a 1000 1.15

 0.1 

 . . .   1000 kg/m 3  w  

H  86.957 m of air column    2gH Quantity of air flow in m 3/sec  V a  C d A   0.7  3.464  10 4

2  9.81  86.957  

 V a  0.010016 m 3/sec

 Mass of air flow in kg/sec  V a  a  0.010016  1.15  0.01152

3.34 Thermal Engineering - I

 0.01152

m 3 kg or kg/s ec  sec m 3

 m a 0.01152 kg/sec

3.5 FUEL CONSUMPTION Fuel consumption can be specified in terms of volume or by weight of fuel consumed. For automobiles it is expressed in terms of kilometer per litre. Measurement of fuel consumption is very important in engine testing. Fuel consumption measurement is a tedious process as it is affected by a number of factors: (i)

Vapour bubbles formed in the fuel line could cause a back flow in the movement of fuel. Some flow meters cannot distinguish between a back flow and a forward flow.

(ii)

In the case of turbine type flow meter, if there is any swirl in the fuel flow, then it is registered as a higher flow rate.

(iii)

The density of fuel may vary according to the temperature thereby causing errors in the measurements.

(iv)

Some flowmeters use light beam where the measurement may be affected by the colour of the fuel. The basic flow measuring methods are (i) Volumetric type (ii) Gravimetric type

Testing and Performance of Engines 3.35

3.5.1 Volumetric type Time taken by the engine to consume the required volume is measured by using stop watch. Volumetric flow rate is defined as the rate of consumption of fuel over time V olumetric flow rate 

Consumption o f fuel Time

There are two methods to measure fuel consumption based on volumetric type. 1. Burette method 2. Automatic volumetric flow meter 3.5.1.1 Burette method A simple arrangement for measuring fuel consumption rate is shown in Fig 3.15(a). A small glass tube burette is fitted to the main fuel tank. When the fuel consumption rate is to be measured, the valve is closed and the fuel is consumed from the burette as shown in the figure. For a known value of fuel consumption, the time is measured and the fuel consumption rate is calculated as under. F ue l consumption kg/hr 

V cc  Sp.gravity of fu el 1000  time

In practical, the burette method is modified as shown in Fig. 3.13.

3.36 Thermal Engineering - I

Fu el Stora ge Tank

3 - w a y cork Index

1 00 cc

2 00 cc

Index

3 - w a y cork

To E ngin e Fig. 3.13(b). Burette M ethod of M easuring Fuel Con su mption

This method consists of two spherical glass bulbs having 100 cc and 200 cc capacities. They are connected to three way corks, where one is used to feed the engine while the other is being filled. The volume of fuel lost in one of the bulb is equal to the volume of fuel gained in the other. Glass bulbs of different volumes are used so as to account to the small

Testing and Performance of Engines 3.37

variations in the fuel head which is highly essential in case of carburetor engines. Another reason to use the glass bulbs of different capacities is to maintain the test timing as a constant regardless of the engine load. To avoid the error, the calibrations on the burette are illuminated using photo cells. A stop watch is used to determine the time taken for fuel consumption and the volumetric fuel flow is thus determined. 3.5.1.2 Automatic volumetric flow meter This automatic volumetric type fuel flow measuring instrument is commercially available. It consists of measuring volume chamber (V) which has photocell (P) and lighting source (S) fitted in tubular housings. These housings are put opposite to each other at an angle so that a point of light is formed on the axis of measuring volume (V). Light source are also placed on the lower and upper portions of the measuring cylinder. An equalization chamber (E) is connected to the measuring tube through the airtube (F) and, Magnetic valve (M) and an equalization pipe (G) which are used to provide an air cushion. This cushion is used to maintain the supply line pressure and to store fuel during measurement. WORKING On pressing the start button, the lamps in the two photoelectric systems light up so that the Magnetic valve stops the flow on the lower end. The fuel level in the measuring volume chamber starts falling down at a rate depending on the engine consumption. At the same time,

3.38 Thermal Engineering - I Po sition 1 E

Po sition 2 E

F

P

S

N o rm al flow E

F

P

S

U G

G

M easuring Volum e

M easuring Volum e P S

S

P

S

U

U

P

M M eter

G

M easuring Volum e

P

S

L

L

F

L

M

M

M eter

M eter

Fig. 3.14. Autom atic Vo lum etric Flow M eter.

an equal amount of fuel flows through equalization tube to equalization chamber. When the fuel level reaches the upper measuring level (U) inside the measuring chamber, the focussed beam of light is reflected upon the photocell at the lower end which in turn converts it into an electrical signal. This signal opens the valve and starts the timer counter. When the fuel level falls to lower measuring level (L) the corresponding signal stops the timer counter. Thus the time period of consumption of fuel can be accurately determined. 3.5.2 Gravimetric fuel flow measurement Volumetric fuel flow meters are often used to determine the time taken for fuel consumptions in liters. Hence it is necessary to relate the output to the specific

Testing and Performance of Engines 3.39

Fuel Tan k

To E ngin e Valve 1

Valve 2

Flask

Pa n of W eighing Ba lance Fig. 3.15. G ravim etric Fuel Flow Measuring Unit Schematic Diagram

gravity of the fuel in order to determine the efficiency of the engine. Due to various external factors, this may lead to several errors. Since the efficiency of an engine is directly related to the mass of fuel and not its volume, gravimetric fuel flow measuring device provides higher accuracy relatively. This method is used for weighing the fuel supplied to the engine by an arrangement shown in Fig. 3.15. Construction It consists of a fuel tank which is connected to the engine through a one way valve V1. The measuring unit is a flask which is placed on a balance. A part of the fuel

3.40 Thermal Engineering - I

can enter the flask through a two way valve V2 as shown in the figure. Working When the measurement is not needed V1 is opened and V2 is closed, so that the fuel flows directly to the engine. To measure the fuel consumed, V2 is now opened and a part of the fuel from V1 is sent to the flask. The weight of the fuel is recorded. V1 is then closed and the fuel from the flask is syphoned off to the engine. The time taken to completely consume the fuel from the flask is recorded using a timer. This effectively determines the mass of fuel consumed per unit time. Thus the fuel consumption in gravimetric units are obtained. 3.5.3 Measurement of Heat Carried Away by Cooling Water The heat carried away by cooling water is measured by measuring the quantity of water flowing through the jacket and the rise in temperature of the cooling water. The quantity of water flowing through the jacket is measured by collecting it in a bucket for a specified time or directly with the help of a flow meter. The inlet and outlet temperatures of the water are measured by thermometers which are inserted in the pockets provided. The heat carried away by cooling water is given by  Q w  m w  C pw To  Ti w here m w  mass of wa ter/minute C pw  Specific heat of water generally taken as 4.187 kJ/kg To  Outlet temperature of water in C T i  Inlet temperature of w ater in C

Testing and Performance of Engines 3.41

3.6 EXHAUST GAS COMPOSITION Emissions can be defined as the unburnt fuel and other by-products exiting the combustion chamber. Based on the visibility, emissions can be broadly classified into  

Visible emissions Invisible emissions

Some common emissions found are: (i) Varbon dioxide (ii) Water vapour (iii) Unburnt hydrocarbons (iv) Oxides of nitrogen (v) Aldehydes (vi) Carbon monoxide (vii) Smoke (viii) Particulate matter Based on the type of emission, it’s effect on the environment can vary. 3.6.1 Oxides of Nitrogen Oxides of Nitrogen occur only in the engine exhaust and are a combination of nitric oxide (NO) and nitrogen dioxide NO 2. Nitrogen and oxygen react only at high temperatures. More amount of nitric oxide (NO) is formed when the proper amount of oxygen is available and when the combustion temperature is high. The NO x concentration in exhaust is affected by the mode of vehicle operation and the engine design. Air-fuel ratio and spark advance also play a key role in it. The concentration of NO x is low at rich and lean air fuel

3.42 Thermal Engineering - I

Sp an G a s inle t

Sa m ple G a s in let Air in le t

P

P Flow M eter

NOX / NO C o nvertor

O 3 G e n erator

3 W a y Valve

Scrubb er

R e action C ham be r

Ph oto M ultip lier Tube

Fig. 3.16. M easu ring O xides of Nitro gen b y Chemilum inescence M eth od

Testing and Performance of Engines 3.43

mixtures. The maximum NO x is formed at air-fuel ratios between 14:1 and 16:1. If the ignition advance is increased, it will result in lower peak combustion temperatures and high exhaust temperature and thus high NO x concentration is obtained in the exhaust. Chemiluminescence analyzer is a method used for measuring Oxides of Nitrogen. The principle of measurement is based a chemiluminescence reaction between ozone and NO which results in the formation of excited NO 2. This excited NO 2 emits light of intensity proportional to the concentration of NO. Photo multiplier tube is used to amplify and measure the light intensity. Thus the concentration of NO x can be measured effectively. However, Nitrogen dioxides NO 2 are not measured by the analyser. To analyse all the oxides of Nitrogen, a converter is fitted in reaction chamber to convert oxides of nitrogen into nitric oxide. 3.6.2 Carbon monoxide (CO) Carbon monoxide is emitted in engine exhaust as a result of the incomplete combustion either due to insufficient amount of air in air-fuel mixture, or insufficient time in the cycle to complete the combustion. Theoretically the gasoline engine exhaust can be made free from CO if the air-fuel mixture ratio is greater than 16:1. When the engine is in idling conditions, due to low oxygen concentrations, production of CO is high. Upon acceleration, due to enough oxygen supply, the formation of CO is reduced. The amount of CO produced is minimum

3.44 Thermal Engineering - I

during accelerations and at constant speeds. During decelerations, the supply of oxygen is minimized by closing the throttle valve. Hence a high amount of CO is produced at the exhaust. The instrument used for measuring of CO is Non Dispersive Infra-red Analyzer. This instrument is commonly used for testing and legal certification of some automotive exhaust emissions. This method of detection is based on the principle of attenuation of light to the properties of the material through which the light is travelling. NDIR Method (Non Dispersive infrared method) An NDIR setup typically consists of a sampling chamber, a reference chamber, an infra-red light source and an infra-red detector (Fig. 3.17). A chopper wheel is mounted between the chambers and the light source, and it is controlled by using a DC motor. The detector is placed at the end of the chambers, directed towards the chamber. The sampling chamber consists of the gas which is to be detected (Carbon Monoxide emissions) while the reference chamber consists of a known gas, typically Fo c usin g L en s

E m itter

S a m ple C h am be r

D e tec to r R e feren ce C h op pe r C h am be r W h e el Fig.3.17. S ch em atic D ia gram o f N D IR

Testing and Performance of Engines 3.45

Nitrogen. When the IR light is turned on, the radiations excite the gases present in the chambers. These gases absorb a portion of the radiation. The remaining unabsorbed part of radiation is sent to the detector. The wavelength of radiation produced in the sampling chamber is compared with the wavelength from the reference chamber and thus the amount of CO present in the sampling chamber can be determined. 3.6.3 Unburnt hydrocarbons In the case of incomplete combustion, the unburnt hydrocarbon emission occurs. The induction system design and combustion chamber design are the two key factors which affect the amount of hydrocarbon emitted. Other variables like air-fuel ratio, speed, load, mode of operations and maintenance also play a major role in the emission conditions. The air-fuel ratio of the engine mainly depends upon the configuration of the induction chamber. Since unburnt hydrocarbons are a result of improper mixing of air and fuel, the induction chamber directly influences these emissions. Engine maintenance also determines how long an engine will operate in the designated air-fuel ratio. Piston ring wears, coolant effectiveness, lubrication, deposits and other factors regarding wear are also more likely to affect the air-fuel ratio, thereby influencing the emissions. The design of combustion chamber is important. A portion of the air-fuel mixture in the combustion chamber comes into direct contact with chamber walls where it is quenched due to cooling action and do not burn completely.

3.46 Thermal Engineering - I

In exhaust stroke, this unburnt air-fuel mixture is forced out from the chamber. Thus a sizable amount of unburnt fuel exits the engine, thereby leading to high emission of unburnt hydrocarbons. Hydrocarbon emissions are usually measured using a flame ionization detector Fig. 3.18. It works on the principle that electrically charged particles move towards the respective electrodes after ionization. The emitted hydrocarbons are ionized to a plasma state inside a chamber. The movement of ionized carbon atoms is obtained in the form of micro ampheres. This is directly proportional to the concentration of hydrocarbons.

H 2 + Air Fla m e

A m p lifie r C ollector P late s

E xit G a ses

A ir D istrib utor

H eated M etal B lock Flam e C ham b er

A ir

Fuel G a s C apilla ry C olu mn Fig. 3.18. Flam e ion ization metho d for unburn t hydrocarbon detection

Testing and Performance of Engines 3.47

3.6.4 Aldehydes The fuels which contain aldehydes can lead to higher level of oxygenated hydrocarbon emissions. Emission of odorous hydrocarbons from the engine is generally known as carcinogenic. These aldehydes are responsible for the pungent smell from the engine exhaust. Based on wet chemical principle, the measurement for aldehydes are done by following methods. (i) Dinitrophenol hydrozine method (ii) Iodine titration method (iii) Chronotropic acid method

3.7 VISIBLE EMISSIONS Visible emissions are often considered as a nuisance. For several diesel engine like cars, trucks, ships, buses, motor cycles etc. smoke is one of the visible emissions. 3.7.1 Smoke In combustion process, due to incomplete combustion, smoke is formed and carried out from the engine exhaust. Smoke in diesel engines are of blue, white and black in colour. Smoke measurement can be done using comparison method and obscuration method. 3.7.1.1 Comparison method The regulation of smoke emissions is based on estimation of density of smoke from exhaust. Among comparison methods, ringelmann chart is one of the commonly used methods. The amount of dense black smoke from exhaust is directly correlated to the combustion

3.48 Thermal Engineering - I

2 0% B lack C h art 1

4 0% B lack 6 0% B lack C h art 2 C h art 3 Fig. 3.19. Ringelm ann Ch art

8 0% B lack C h art 4

efficiency. If the smoke is darker, then the engine has poor efficiency. Ringelmann developed a chart, to categorize the density of black smoke into four shades of gray, pure white background and all black grids. To remove the difficulties, the shades are built in black lines of varying widths. By placing the charts nearly 10 m away from the exhaust, the grids appear as shades of grey. Comparing the shades of smoke with the corresponding shade on the charts the density of the smoke can be determined. The charts are numbered from 1 to 5 with 1 being clean smoke and 5 being most dense and black. 3.7.1.2 Obscuration method This method can be divided into (i) Light extinction type (ii) Continuous filtering type (iii) Spot filtering type

Testing and Performance of Engines 3.49

(i) Light extinction type In this method, the intensity of light beam, is reduced due to presence of smoke, which is a measure of smoke intensity.

P h otoe le ctric c ell ( D e te ctor )

Illu m ina to r ( L ig ht S o urce ) E n gine E xha u st

To Im ag e S e nso r

In dica ting M eter

Fig. 3.2 0. O bscu ration Metho d F or M easuring Smo ke

An exhaust sample of smoke is continuously passed through a tube of 50 cm in length. One end of the tube has a source of light while the other end has a photocell. The intensity of light falling on the photocell produces a relative amount of photocurrent. The amount of photo current produced is inversely proportional to the intensity of the smoke. Smoke level or smoke density is defined as the ratio of electric output from the photocell when sample is passed through the tube to the electric output when clean air is passed through it. (ii) Continuous Filtering Type This type provides provision for continuous checking of the quality of the exhaust. The measurement of smoke intensity is carried out by passing the exhaust gas

3.50 Thermal Engineering - I

continuously through a moving strip of filter paper. The filter paper collects the particles emitted from it. Van Brand Smokemeter works on this principle. E ngine E xhau st

Filte r B locks

Filte r Ta pe R o ll Filte r Ta pe D rive

Fig. 3.21. Con tin uou s F ilter Type Sm oke M eter.

The exhaust gas sample is passed continuously at constant rate through a strip of filter paper which is moving at preset speed. While moving the strip, a stain due to emission is imprinted on the paper. The intensity of the stain is directly proportional to the intensity of the smoke produced. Amount of light passing through the paper is usually used to determine the intensity of the stain. (iii) Spot filtering type The smoke intensity is measured by filtering the quantity of exhaust gas through a fixed filter paper Filte r Ta pe P um p

E ngine E xh au st

Fig. 3.22. Sp ot Filterin g Type Sm oke Meter.

Testing and Performance of Engines 3.51

Bosch smokemeter is based on this concept. From the exhaust, a definite quantity of exhaust gas is passed through a fixed filter paper. The amount of smoke stains on filter paper are evaluated by using a light source. Now this type is modified by using a pneumatically operated sampling pump and an electronic unit (Photo cell) to measure the intensity of smoke stain. The intensity of the smoke stain is directly proportional to the intensity of the smoke.

La m p P hoto Ele ctric D e te ctor R o llin g Filter P aper d isc

P ow e r S ource

R e ciproca tin g P iston

0 10

P neum a tic Tripp in g D ev ice

B ellow s S etting K nob Fig. 3.23. B osch S moke M eter.

M icrom eter sc aled 0 to 1 0 sm oke num ber (M ea su ring U n it)

3.52 Thermal Engineering - I

3.7.2 Measurement of Heat Carried Away by Exhaust Gas The mass of air supplied per kg of fuel is calculated by the equation  ma 

 NC o r m a can be measured by an orifice 33 C 1  C 2

(The mass of exhaust gases can be obtained by adding together the mass of fuel supplied and the mass of air supplied) The heat carried away by exhaust gas per kg of fuel is given by Q g  m g C pg  T g  T a  m a  1  C pg Tg  Ta  where m a  1   mass of exhaust gas formed per kg of fuel supplied C pg  Specific heat of exhaust gases T g  Temperature of exhaust gases coming out of the engine in C

Ta  Ambient Temperature in C

3.8 HEAT BALANCE SHEET A heat balance sheet is an account of heat supplied, heat utilised and dissipated in different ways in a system. The performance of the engine is obtained from the heat balance sheet. A heat balance account includes the following items.

Testing and Performance of Engines 3.53

 Heat supplied by the fuel to the engine  mf  L.C.V  where m f is the mass of fuel supplied per minute and L.C.V

is the lower calorific value of the fuel. 1. Heat equivalent of brake power = Brake power  60 kJ/min . where Brake power is in kW.

2. Heat lost to jacket cooling water  m w C pw T o  T i kJ/min 3. Heat lost to exhaust gases  m g C pg  Tg  Ta  kJ/min 4. The remaining heat is lost by convection and radiation. This cannot be measured and so this is known as unaccounted loss. This is calculated by the difference of heat supplied and the sum of (1) + (2) + (3). i.e Q ua  Q s  [Q I.P or B.P  Q w  Qg kJ/hr] A heat balance sheet is shown in Table below Heat Balance Sheet Heat supplied per minute Heat supplied by

kJ —

the combustion of fuel

Total

%

Heat expenditure per minute

100 1. Heat equivalent of brake power (or) Indicated Power 2. Heat lost to jacket cooling water 3. Heat lost to exhaust gases 4. Unaccounted heat = Heat supplied –

—

[(1) + (2) + (3)] 100 Total

kJ

%

—

—

—

—

—

—

—

—

—

100

3.54 Thermal Engineering - I

Problem 3.8: A gas engine, working on 4 stroke constant volume cycle (Otto cycle) gave the following results when loaded by friction brake during a test of an hour’s duration. Cylinder diameter = 250 mm; Stroke length = 500 mm Clearance volume  4450  10 6 m3 Effective circumference of the brake wheels = 3.86 m; Net load on brake 1260 N at overall speed of 230 rpm. Average explosion/min 77; m.e.p = 7.5 bar. Gas used 13m3/hr at 15C and 771 mm of Hg. L.C.V (Lower Calorific Value) of gas 49500 kJ/m3 at NTP; Cooling jacket water 660 kg raised to 35C. Heat lost to exhaust gases 10%. Calculate 1. I.P. 2. B.P.; 3. Indicated thermal efficiency 4. Relative efficiency; 5. Also draw a heat balance sheet for the engine.

Solution: 1. To find I.P I.P. 

P m AL n.e. 

60

where n.e. = no of explosion/min I.P 

7.5  10 2  /4  0.25 2  0.5  77 60

 23.623 kW

2. To find B.P B.P. 

2  NT 2  N[W  S  R ]  60 60

where R = effective radius; (W-S) Net load on brake = 1260 N Effective circumference  d  3.86 m d  1.2287 m

Testing and Performance of Engines 3.55

R  0.6143

B.P 

2   230  1.26   0.6143 60

 18.6437 kW

3. To find indicated

the

Indicated

thermal

efficiency:

I.P  3600 indicated   [V f  C.V]at N.T.P

 First of all we have to find out V f at N.T.P condition [i.e., N.T.P means Normal temperature 0C and pressure 760 mm of Hg] C.V  49,500 kJ /m3 V f  V olume of fuel ga s at NTP in m 3/hr TNTP  0C  273  273 K P NTP  760 mm of Hg  1.01325 bar Gas used V 1  13 m 3/hr; T 1  15  273  288K; P 1  771 mm of Hg P NTP V NTP P 1 V 1  T1 TNTP 771  13 273  288 T1 P NTP 760   12.50127 m 3/hr  V f

V NTP 

P 1 V1

 indicated 



TNTP



23.623  3600  0.13743 12.50127  49500

 13.743 %

3.56 Thermal Engineering - I

4. To find Relative efficiency Relative efficiency on the basis of indicated thermal  

Indicated thermal efficiency Air standard efficiency

 air standard  Air standard efficiency  1 

1 r

1

Total volume  Vs  Vc  where r  compression ratio    Vc  Clearance volume    Swept volume V s   D2  L   0.25 2  0.5 4 4  0.02454 m 3 Clearance volume V c  4450  10  6 m 3 r

Vs  Vc

Vc



0.02454  4450  10  6

 air standard  1 

4450  10  6 1 6.515 1.4  1

Relative efficiency 

 6.515

 0.5275  52.75 % [For air   1.4]

0.13743  0.260554 0.5275

 26.06 %

To draw Heat Balance Sheet [in min basis]  V f  C.V Heat supplied by fuel Qs  60 

12.50127  49500 60

 10313.55 kJ/min

Testing and Performance of Engines 3.57

Q s  10,313.55 kJ/min

To find Q IP Heat utilised for I.P.  23.623 kW  23.623  60  1417.38 kJ/min Q I.P.  1417.38 kJ /min Q I.P in% 

Q I.P Qs

 100 

1417.38  100  13.74% 10313.55

Heat carried out by cooling water  Q w  m w Cpw t2  t1  660 Kg/min  11 kg/min m w  660 kg/hr  60 C pw  4.187 kJ/kg K t2  t1   t  35 C Q w  11  4.187  35  1612 kJ/min Q w  1612 kJ/min Q w in% 

Qw Qs

 100 

1612  100  15.63% 10313.55

Heat lost through exhaust gases Q g  10% of Q s  0.1  10,313.55  1031.36 kJ/min Q g  1031.36 kJ /m in Q g in% 

Qg Qs

 100 

1031.36  100  10% 10313.55

3.58 Thermal Engineering - I

Unaccou nted heat loss  Q s  [QI.P  Q w  Q g]  10313.55  [1417.38  1612  1031.36 ]  6252.81 kJ/min Q unaccounted  6253 kJ/min Q ua in% 

Q ua Qs

 100 

6253  100  60.63% 10313.55

Now we can draw the heat balance sheet. Heat Balance Sheet CREDIT Heat supplied

kJ/min

DEBIT %

per minute Heat supplied by the

Heat expenditure per min

10313.55 100% 1. Heat utilized for

kJ min

%

1417.38

13.74%

1612

15.63 %

1031.36

10%

6252.81

60.63%

I.P. QI.P.

combus- tion of fuel

2. Heat carried out by cooling water Qw 3. Heat lost through exhaust gases Qg 4. Unaccounted heat loss Qunacc Total

10313.55 100%

10313.55 100%

Testing and Performance of Engines 3.59

Problem 3.9: During the trial of a single cylinder 4-stroke oil engine, the following results were obtained. Cylinder dia 20 cm; stroke 40 cm, MEP 6 bar, torque 407 N-m, speed 250 rpm, Oil consumption 4 kg/h, Calorific value 43 MJ/kg, cooling water flow rate 45 kg/min, air used per kg of fuel 30 kg, rise in cooling water temperature 45C, temperature of exhaust gases 420C, room temperature 20C, mean specific heat of exhaust gas 1 kJ/kgK. Find the indicated power, brake power, and draw the heat balance sheet for the test in kJ/hr. (JNTU - August 2014 set (2))

Solution: To find Indicated Power I.P 

Pm  L  A 

N 2

60

. .N   . 2 for 4 stroke   

 250  600  0.4    0.2 2   2 4     15.71 kW 60

To find Brake Power BP 

2  NT 2   250  0.407  10.65 kW  60 60

To draw Heat Balance Sheet [in hour basis]  Heat supplied by fuel Q s  m f  C.V  4  43  10 3  172  10 3 kJ/hr Q ip

Heat utilized for I.P  15.71  3600  56.556  10 3 kJ /hr

3.60 Thermal Engineering - I

Qw

Heat carried out by cooling water  Q w  m w Cpw t2  t1  4.5  60  4.187 45  50.872  10 3 kJ/hr

Heat lost through exhaust gases ‘Q g’  Q g  m g  C pg   tg A : F  30 : 1  m g  Air  Oil consum ption/hr  30  4  4  124 kg/hr Q g  124  1  420  20  49.6  10 3 kJ/hr

To find Unaccounted loss ‘Q U ’ Q u  Q s  [Q IP  Q w  Q g]  172  10 3  [56.56  10 3  50.872  103  49.6  10 3] Q  14.972  10 3 kJ/hr

Now we can draw the heat balance sheet.

Testing and Performance of Engines 3.61

Heat Balance Sheet CREDIT

DEBIT

Heat supplied per hour Heat

Heat kJ/hr

%

172  103

expenditure per hour 1. Heat

kJ hr

%

56.56  103 32.9.%

utilized for

supplied by the combus-

I.P. QI.P.

tion of fuel 2. Heat

50.87  103 29.6 %

carried out by cooling water Qw 3. Heat lost through

49.6  103

28.8%

exhaust gases Qg 14.97  103 8.7%

4. Unaccounted heat loss Qu Total

172  10

3

100%

172  103

100%

Problem 3.10: In a test of oil engine, under full load, the following results were obtained. IP 33 kW, Brake power 27 kW, oil consumption 8 kg/h, Calorific value 43 MJ/kg, cooling water flow rate 7 kg/min, rate of flow for water through gas calorimeter 12 kg/h, rise in cooling water temperature 60C, final temperature of exhaust gases 80C, room temperature 17C, Air-fuel ratio as mass basis 20, rise in water temperature through exhaust gas calorimeter 40C, mean specific heat of exhaust gas 1 kJ/kg. Draw the heat balance sheet and find thermal and mechanical efficiencies.

(JNTU - Aug 2014 - Set 3)

3.62 Thermal Engineering - I

Solution: Indicated thermal  

33  3600 IP  3600   m f  CV 8  43  10 3

 0.3453  34.53% Brake thermal



27  3600 B.P  3600   m f  CV 8  43  10 3

 28.26%

Mechanical



B.P 27  81.82%  33 I.P

To draw Heat balance sheet  Heat supplied by fuel Q s  mf  C.V  8  43  10 3  344  10 3 kJ/kg Q IP

Heat utilized for IP Q IP  33  3600  118.8  10 3 kJ/hr Qw

Heat carried out by cooling water  Q w  m w Cpw  Tw  7  60   4.187  60  105.51  10 3 kJ/hr

Testing and Performance of Engines 3.63

Q cal

Heat lost through gas calorimeter Q cal  12  4.187  40  2  10 3 kJ/hr

Heat lost through exhaust gases ‘Q g’  Q g  m g  C pg  Tg A : F  20 : 1  m g  Air  Fuel co nsumption/hr  8  20  8  168 kg/hr Q g  168  1  80  17  10.58  10 3 kJ /hr

To find Unaccounted loss Q U  Q S   Q IP  Q w  Q cal  Q g     344  10 3  [118.8  10 3  105.51  10 3  2  10 3  10.58  10 3]  107.11  10 3 kJ/hr

Now we can draw the heat balance sheet

3.64 Thermal Engineering - I

Heat Balance Sheet CREDIT Heat supplied

DEBIT

kJ/hr

%

per hour Heat

Heat expenditure per hour

344  103 100

supplied by the combus-

1. Heat

kJ hr

%

118.8  103

34.53

utilized for I.P. QI.P.

tion of fuel 2. Heat carried out by

105.51  103 30.67

cooling water Qw 3. Heat lost through

10.58  103

3.08

2  103

0.006

exhaust gases Qg 4. Heat lost through gas calorimeter Qcal 5. Unaccounted

107.11  103 31.14

heat loss Qu Total

344  10

3

100

344  103

100%

Problem 3.11: Following data are available for a four stroke 4 cylinder petrol engine: Air-fuel ratio (by weight) 15.5:1 Calorific value of fuel 45,000 kJ/kg

mech  80%; air standard  53% ; Relative efficiency  70% Volumetric efficiency 80%; stroke/bore ratio 1.25 Suction condition 1 bar and 27 C; Speed - 2400 rpm Power at brakes = 75 kW: Calculate (1) compression ratio

Testing and Performance of Engines 3.65

(2) indicated (3) Brake specific fuel consumption (4) Bore and stroke

Solution: 1. To find compression ratio: Since it is petrol engine, we can use air standard efficiency for otto cycle. A ir stan dard efficiency  1 

0.53  1  1 0.4

r

1 r

1

1 r

. . [ .   1.4 for air]

1.4  1

 1  0.53  0.47

r0.4 

1 0.47

r  6.603

2. Indicated thermal efficiency: Relative efficiency 

Indicated thermal  Air standard 

Indicated thermal   Relative   Air standard   0.7  0.53  0.371  37.1 %

3. To find brake specific fuel consumption

 First of all, we have to find fuel consumption m f in

kg/hr  To find m f in kg/hr

We can use this formula Indicated thermal efficiency 

I.P  3600  mf  C.V

3.66 Thermal Engineering - I

To find IP Mech   I.P 

B .P I.P. B.P 75   93.75 kW  mec h 0.8  mf 



I.P.  3600  indic ated  C.V. 93.57  3600  20.216 kg/hr 0.371  45,000

 5.6155  10  3 kg/sec B.S.F.C 

 mf B.P



20.216  0.26954 kg/kW.hr 75

To find Bore and Stroke For every 1 kg of fuel, mass of air-fuel mixuture  1  15.5  16.5 kg  For 5.6155  10  3 kg/sec of fuel, the mass of air-fuel

mixture  16.5  5.6155  10  3  0.092655 kg/sec

 To find volume of air-fuel mixture in m 3/sec V    P V  m RT   m RT 0.092655  287  27  273  V P 1  10 5  0.07978 m 3/sec

Testing and Performance of Engines 3.67

Volumetric efficiency 

Actua l volume in m 3/sec

S wept volu me in m3/sec   V 0.07978 3  Swept Vo lume in m /sec V s   0.8 vol  0.09972 m 3/sec    N n Vs    D2  L    2  60 4

[

N for 4 stroke cycle engine; 2 n  No of cylinders = 4; L  1.25; L  1.25 D ] D

0.09972 

 2400  D 2  1.25 D   4  78.54 D 3 4 2  60

B ore D  0.1083 m S troke L  1.25 D  1.25  0.1083  0.1354 m Problem 3.12: Calculate the bore and storke of a four stroke single

cylinder

diesel

engine

designed

to

the

following

particulars. B.P = 25 kW when running on diesel fuel having composition by mass C = 85% and H2  15% and lower calorific value of 41000 kJ/Kg. The fuel oil is burnt with 25% excess air

and

volumetric

efficiency

calculated

on

atmospheric

condition of 1.01325 bar and 10C is 80%. Assume mechanical efficiency of 0.9, indicated thermal efficiency of 0.35. Assume R = 0.287 kJ/kg K and bore to stroke ratio of 1:12. Assume speed of the engine 300 rpm.

3.68 Thermal Engineering - I

To find quantity of air required Minimum air required per kg of fuel 

O2    100  8  C  8  H2  S 8  23  3  



100  8   0.85  8 [0.15  0]  0  23  3 

 15.0724 kg of air/kg of fuel

Since the fuel is burnt with 25% excess air quantity of air required  1.25  15.0724  18.841 kg of air/kg of fuel

To find mass of air in kg/sec I.P 

25 B.P   27.78 kW  mec h 0.9

I.P  3600 indicated   m f  C.V.

where

 m f  mass of fuel in kg/hr  mf 



I.P  3600 indic ated  C.V 27.78  3600  6.969 kg/hr 0.35  41000

To burn 1 kg of fuel  18.841 kg of air required to burn 6.969 kg/hr of fuel  18.841  6.969  131.296 kg/hr

Testing and Performance of Engines 3.69

So to burn 6.969 kg of fuel/hr, 131.296 kg of air/hr is required. The mass of air required = 131.296 kg/hr. To find volume of air in m 3/sec   P V  m RT  131.296  287  10  273  105.25 m 3/hr V 5 1.01325  10  105.25  0.029235 kg/sec V 3600  To find swept volume: V s  Actual volume in m 3/sec Swept vo lume V s  vol  V 0.029235   0.8  vol    N n Also swept volume V s    D 2  L    2  60 4 D  1.2; L  1.2 D L

   300 1 V s  0.03654    D2  1.2 D   4  60 2    2.3562 D 3 D  0.2494 m L  1.2 D  0.29925 m

3.70 Thermal Engineering - I

Problem 3.13: A nine cylinder petrol engine of bore 150 mm and stroke 200 mm has a compression ratio 6:1 and develops 360 kW at 2000 rpm when running on a mixture of 20% rich. The fuel used has a calorific value of 43 MJ/kg and contains 85.3% carbon and 14.7% hydrogen. Assuming volumetric efficiency of 70% at 17C and mechanical efficiency of 90%, find the indicated thermal efficiency of the engine. (April/May - 2013 (set-4))

Solution: To find minimum air required per kg of fuel 

O2    100  8  C  8  H2  S 8  23  3  



100  8   0.853  8 [0.147  0]  0  23  3 

 15 kg of air/kg of fuel

To find Indicated thermal  indicated 

I.P  3600  m f  CV

 So we have to find IP and m f

To find IP IP 

360 BP   400 kW mech 0.9

 To find m f (Actual mass of fuel consumed in kg/hr) Assume P  1.01325 bar  101.325 kPa  V  Volume of mixture sucked inside the cylinders in m 3/s

Testing and Performance of Engines 3.71

 N or N/2 V  V   vol   No.of cylinders 60

Where V 

 2 N D  L and 2 4

for 4 stroke and n  9

cylinder.    1  2000  V    0.15 2  0.2  0.7     60  9  2 4      PV  m

 0.3711 m 3/s   m RT  P V 102.325  0.3711   0.452 kg/s  RT 0.287  290

15 kg of air is required for 1 kg of fuel Mass of fuel used 

0.452  0.03012 kg of fuel/sec 15

Since it is 20% rich mixture, Actual amount of fuel used  0.03012  1.2  0.03614 kg of fuel/sec  m f  mass of fuel/hr  0.03614  3600  130.11 kg /hr

To find indicated indicated 

400  3600 I.P  3600   m f  CV 130.11  43  10 3  0.2574  25.74%

Problem 3.14: Air consumption for a four stroke petrol engine is measured by means of a circular orifice of diameter 3.5 cm.

3.72 Thermal Engineering - I

The Cd of orifice is 0.6 and the pressure across the orifice is 14 cm of water. The barometer reads 760 mm of Hg. Temperature

of

air

in

the

room

is

24 C. The

piston

3

displacement volume is 1800 cm . The compression ratio is 6.5. The fuel consumption is 0.13 kg/min of calorific value 44000 kJ/kg. The brake power developed at 2500 rpm is 28 kW. Determine 1. Air fuel ratio, 2. the volumetric efficiency on the basis of air alone 3. the brake mean effective pressure 4. the relative efficiency on the basis of brake thermal efficiency

Solution: To find air consumption in kg/sec Head of air column causing flow H  h w

w a

h w  14 cm  0.14 m

w  1000 kg/m 3 where a 

1.01325  10 5 P  RT 287  24  273  1.18872 kg /m3 H  0.14 

1000  117.774 m o f air 1.18872

 2gH V  A ir consu mption in m 3/sec  C d A    C d  0.6 Area of orifice A 

  0.035 2  9.62112  10  4 m 2 4

 V  0.6  9.62112  10  4

2  9.81  117.774 

Testing and Performance of Engines 3.73

 0.02775 m 3/sec

  m  Air consumption in kg/sec  V  a  0.02775  1.18872  0.03299 kg/sec

Air fuel ratio Air fuel ratio 

0.03299 15.23 Mass of air in kg/sec   Mass o f fuel in kg /sec 0.00216 1

Air consumption in one stroke (V)  N VV 2  60  V  2  60 0.02775  2  60  V 2500 N  1.332  10 3 m 3 V s  Sw ept volum e  piston displaceme nt volum e given  1800 cm 3  1800  10 6m 3

To find volumetric efficiency  volumetric 



Actual volume in m 3 Swept volume in m 3 1.332  10  3 1800  10  6

 0.74

 Volumetric  74%

To find Brake mean effective pressure I.P 

P m  LA 60

N n 2

. . [ . N/2 fo r 4 stroke ]

3.74 Thermal Engineering - I

Also B.P 

Pmbrake  LA N/2  n 60

B.P  60 So P mBrake  LA  N/2  n LA  D isplaceme nt volume  1800  10  6 m 3

B.P  28 kW P mBrake 

28  60 1800  10

6

  2500 /2  1 [n  1 for single cylinder]

 746.667 kP a  7.47 bar

To find Brake thermal efficiency  brake 

B.P  3600  m f  C.V

 m f  m ass of fuel in kg/hr  0.13  60 kg/hr C.V  44000 kJ/kg brake 

28  3600  0.29371  29.37 % 0.13  60  44000

To find Air Standard efficiency [for otto cycle since petrol engine] [compression ratio r = 6.5 given] air standard  1 

1 r

1

1

1 6.5

1.4  1

 0.527  52.7%

Testing and Performance of Engines 3.75

To find relative efficiency on the basis of brake Relative  

brake air standard



0.2937  0.5573  55.73 % 0.527

Problem 3.15: The following data were recorded during a test on a 4 stroke cycle gas engine. Area of indicator diagram  90 cm2 ; Length of indicator diagram  7 cm ; Spring scale  0.3 bar/mm [  3 bar/cm] Diameter of piston  20 cm Length of stroke  25 cm Speed  300 rpm Determine (i) Indicated mean effective pressure (ii) Indicated Power.

JNTU - January 2014 - Set 1

Solution: To find Indicated P m indicated P m indicated 

ad  s ld

mean



effective

pressure

90  3 7

 38.57 bar To find IP N P m in  AL  n  2  IP  60     300  2 2  38.57  10   0.2  0.25   1 2 4     60 . .N for 4 stroke] [ . 2

3.76 Thermal Engineering - I

[n  1 for single cylinder]  75.73 kW Problem 3.16: A four stroke cycle gas engine has a bore of 20 cm and a stroke of 40 cm. The compression ratio is 6. In a test on the engine, the IMEP is 5 bar, the air to gas ratio in 6:1 and calorific value of the gas is 12 MJ/m3 at NTP. At the beginning of the compression stroke, the temperature is 77C and pressure 0.98 bar. Neglecting residual gases, find the indicated power and the thermal efficiency of the engine at 250 rpm.

(JNTU - August 2014 - Set 4)

Solution: Given

data: 4

stroke;

D  0.2 m ; L  0.4 m; r  6 ;

P mindicated  5 bar ; A : F  6, Cv  12  10 3 kJ/m 3 at NTP. t1  77C; T 1  350 K ; P 1  0.98 bar N Pm  LA  n  2  Indicated power IP  60  250  1 5  10 2  0.4    0.2 2   2  4 I.P  60  13.09 kW . . [ . N/2 for 4 stroke and n  1 for single cylinder]  To find V f   V 1  V s   D 2  L   0.22  0.4  0.013 m3 4 4  N /2 0.013  250/2   0.0271 m 3/s Vs  Vs  60 60

Testing and Performance of Engines 3.77

 V s  Volume of gas  Volume of air     Vf  6 V f  7V f  0.0271 m 3/s  0.0271  3.87  10  3 m 3/s Vf  7

 To find V f at NTP condition

NTP means T  0  273  273 K;P  1.01325 bar  P NTP V NTP P 1 V 1  T1 T NTP 101.325  V NTP 273



98  3.87  10  3 350

 V NTP  2.92  10  3 m 3/s  V f at NTP

To find indicated thermal efficiency IP  3600 13.09  indicated   Vf  CVNTP 2.92  10  3  12  10 3  0.3736  37.36%  [V f is given in m 3/s ]

3.78 Thermal Engineering - I

Problem 3.17: The following data refer to an oil engine working on Otto four-stroke cycle:  14.7 kW

Brake power Suction pressure Mechanical efficiency Ratio of compression Index of compression curve Index of expansion curve Maximum explosion pressure Engine speed Ratio of stroke: bore

 0.9 ba r  80% 5  1.35  1.3  24 bar  1000 r.p.m  1.5

Find the diameter and stroke of the piston. [JNTU - January - 2014 set (4)]

Solution: Refer Fig. P 3

Ex

pa

2 Co

mp

re s

ns

io n

PV

1 .3

=C o s io

nP V

1 .3

ns

t.

5

=C on

4 s t.

1 V2 = V3

O TTO C ycle

V1 = V4

V

Testing and Performance of Engines 3.79

B.P.  14.7 kW, P 1  0.9 ba r,  mech  80%, r  5, P 3  24 bar N  1000 r.p.m ,

L  1.5; D  ?, L  ? D

Compression ratio r 

V1 V2



V4 V3

To find P 2: (compression process 1-2)  P 2 V 1.35 P 1 V 1.35 2 1 1.35

or

 V1  P2    V  2



P 2  P1  8.78  0.9  8.78  7.9 bar

 P 1  51.35  0.9

To find P 4: (Expansion process 3-4) 1.3 P 3 V 1.3 3  P4 V 4

 V4  [P 3/P 4]    V  3 P4 

P3 8.1



1.3

 51.3  8.1

24  2.96 bar 8.1

Work done/cycles  Area 1  2  3  4  Area under the curve 3  4  area under the curve 1  2 



P 3 V 3  P 4 V4 1.3  1



P 2 V 2  P 1V 1 1.35  1

10 2 24V 3  2.96 V 4 10 2 7.9V 3  0.9V 4  0.3 0.35 . . [ . V1  V4 and V2  V3]

3.80 Thermal Engineering - I

 [80 V 3  9.87 V 4  22.57 V 3  2.57V 4]  10 2  57.43 V 3  7.3 V 4  10 2   . . V4  . V 5 3  

 57.43 V3  7.3  5V 3  10 2  2093 V 3 kN  m

Mean effective pressure Pm  

Work done/cycle Stroke volume V s 2093 V 3 V 4  V 3

Now, mech   I.P. 



2093 V 3 5V 3  V 3

 523.25 kPa  5.23 bar

B.P. I.P.

14.7 B.P.  18.37 kW  0.8 m ech

To find D and L: N n P m LA  2   I.P  60 . .N [ . for 4 stroke n  1 for single cylinder] 2  1000  523  1.5D    D 2   1 4 2   18.37  60 D 3  3.5795  10  3 D  0.152 m L  1.5D  1.5  0.152  0.229 m

Testing and Performance of Engines 3.81

Problem 3.18: A 2 cylinder 4 stroke engine runs at 240 rpm developing a torque of 5 kN-m. The bore and stroke of cylinder are 30 cm and 60 cm respectively. Engine runs with gaseous fuel having calorific value of 16.8 MJ/m3. The gas and air mixture is supplied in proposition of 1:7 by volume. The volumetric efficiency is 0.85. Determine, (i) Brake Power (ii) The piston speed in m/s. (iii) The brake mean effective pressure (iv) The brake thermal efficiency. (JNTU - Apr/May 2013 - Set 1)

Solution: (i) Brake Power B.P 

2  NT 2   240  5   125.67 kW 60 60

 LN  (ii) Mean piston speed  2   60  vp 

2  0.6  240  4.8 m/s 60

(iii) Pmb Brake mean effective pressure

B.P 

Pmb

P mb  LA 

N n 2

60

. .N [ . for 4 stroke n  2 cylinder] 2 125.67  60    240   0.6    0.3 2    2 4   2    740.95 kPa  7.41 b ar

3.82 Thermal Engineering - I

(iv) Brake thermal efficiency  brake  brake 

B.P   3600 V f  CV

 To find V f Vs 

   D2  L   0.3 2  0.6  0.0424 m3 4 4

Actual volume V  0.0424  0.85  0.036 m 3 in 1 cycle  Actual volume of mixture in m 3/s  V  V 

N n 2  60

. .N [ . for 4 stroke n  2 for 2 cylinder] 2  240 2 V  0.036  2  60  0.144 m 3/s    V  Volume of mixture  V a  V f  0.1442  [V f  Volume of fuel in m 3/s ]     7 V f  V f  8 Vf  0.1442  0.1442  0.018 m 3/s Vf  8  V f in m 3/hr  64.8 m 3/hr

Brake thermal   

B.P   3600 V f  CV 125.67  3600 64.8  16.8  10 3

 41.56%

 0.4156

Testing and Performance of Engines 3.83

Problem 3.19: A four stroke four cylinder diesel engine running at 300 rpm

produces 250 kW of brake power. The

cylinder dimensions are 30 cm bore and 60 cm stroke. Fuel consumption rate is 1 kg/min while air fuel ratio is 10. The average

indicated

Determine

mean

indicated

effective

power,

pressure

mechanical

is

0.8

MPa.

efficiency,

brake

thermal efficiency and volumetric efficiency of the engine. The CV is 43 MJ/kg. The ambient conditions are 1.013 bar and 27C

(JNTU-Apr/May -2013 (set 2))

Solution: (i) To find Indicated Power I.P 

Pm  L  A  60

N n 2

. .N for 4 stroke, n  4 for 4 cylinder] [ . 2  300  0.3 2  4 2 4 60

0.8  10 3  0.6  

 339.3 kW

(ii) Mechanical efficiency  mech 

250 B.P   0.737  73.7% 339.3 I.P

(iii) Brake thermal   brake 



B.P  3600  m f  CV  [m f  1  60  60 kg/hr]

250  3600 60  43  10 3

 0.349  34.9%

3.84 Thermal Engineering - I

(iv) Volumetric  vol 

V actual Vs

Mass of air fuel mixture  1  10  11 kg/min  0.183 kg/s

  P V  m RT

 0.183  0.287  300  0.156 m 3/s V 101.3  N n V  V actual per cycle  2  60  V  2  60 V actual per cycle  Nn 

0.156  2  60 300  4

V actual  0.156 m 3/cycle Vs 

    D 2  L    0.3 2  0.6  4  4  0.0424 m3

vol 

V actual Vs



0.0156  0.368  36.8% 0.0424

Problem 3.20: A 4 stroke cycle automobile engine is tested while running at 3600 rpm. Inlet air temperature is 18C and the pressure is 101.36 kN/m2. The engine has eight in-line cylinders with a total piston displacement of 4000 cc. The

Testing and Performance of Engines 3.85

airfuel ratio is 15 and the bsfc is 0.39 kg/kWh. Dynamometer readings show a power output of 89 kW. Find the volumetric efficiency.

(JNTU - May 2013)

Solution: Brake specific fuel consumption BSFC  0.39 kg/kWh  mf   mf  0.39  89  34.71 kg/hr BSFC  B.P Mass of air/hr  15  34.71  520.65 kg/hr Total air fuel mixture  555.36 kg/hr   m  m af  0.154 kg /s   PV  m RT   m RT 0.154  0.287  291   0.127 m3/s V P 101.36  Actual volume of mixture V  0.127 m 3/s V s swept volume in 1 cycle  4000 cm 3  4000  10  6 m 3 N  2   n V s in m 3/s  V s   60 

4000  10  6  3600  8 2  60

 0.96 m 3/s

. . [ . n  8 cylinder]  V 0.127  0.1323  13.23%  vol    Vs 0.96

3.86 Thermal Engineering - I

Problem 3.21: A four stroke petrol engine with a compression ratio

of 3

5.2  10

6.5

to

1

and

total

piston

displacement

of

3

m develops 100 kW brake power and consumes 33

kg of petrol per hour of calorific value 44300 kJ/kg at 3000 r.p.m.

Find.(i) Brake

mean

effective

pressure

(ii) Brake

thermal efficiency (iii) Air standard  and (iv) Air fuel ratio by mass. Assume volumetric efficiency of 80%. One kg of petrol vapour occupies 0.26 m3 at 1.013 bar and 15C. Take R for air 287 J/kgK.

(JNTU - April/May-2013) (set 3) )

Solution: (i) Brake mean effective pressure B.P  60 P mb  N AL   n  2  . .N [ . for 4 stroke n  1 for single cylinder] 2 P mb 

100  60  3000  [5.2  10  3]   1  2 

[AL  Total piston displacement  769.23 kPa  7.7 bar

 5.2  10 3 m 3  V s ]

(ii) To find Brake thermal efficiency: brake brake 

B.P  3600 100  3600    0.2463  24.63% 33  44300 m f  CV

(iii) To find Air standard efficiency air standard otto  1 

1 r

1

Testing and Performance of Engines 3.87

1

1 6.50.4

 0.527  52.7%

(iv) Air Fuel Ratio by mass  N Vs  Vs  2  60  5.2  10  3 

3000 2  60

 V s  0.13 m 3/s

 Actual volume of air fuel mixture V  vol  VS  V  0.8  0.13  0.104 m 3/s     P V 101.3  0.104   0.127 kg/s P V  m RT  m  RT 0.287  288

Mass of air fuel mixture  m  0.127  3600  459 kg/hr  Mass of air/hr m a  (mass of air fuel mixture – mass of fuel)  459  33  426 kg /hr Air Fuel ratio on mass basis 

Air in kg/hr 426  Fuel in kg/hr 33  12.9

Problem 3.22: A four cylinder, four stroke cycle petrol engine 80 mm bore, 130 mm stroke develops 29 kW brake power while running at 1500 rpm and using a 20% rich mixture. If the volume of air into the cylinder when measured at 15.5C and 760 mm of mercury is 70% of the swept volume, the theoretical air fuel ratio in 15, the heating value of the petrol used is

3.88 Thermal Engineering - I

44000 kJ/kg and the mechanical efficiency of the engine is 90%, find (a) Indicated thermal efficiency and (b) Brake mean effective pressure. Take R = 0.287 kJ/kg K.

Solution: Given : No of cylinders = 4; Type 4 stroke; Bore = 80mm = 0.08 m; L = 0.130 m; B.P = 29 kW; N = 1500 rpm; Mixture strength = 20% Rich; Volume of air at 15.5C and 760 mm of Hg = 70% of V s. Theoretical Air fuel ratio = 15; C.V = 44,000 kJ/kg. mech  0.9; R  0.287 kJ/kg K To find Indicated thermal  indicated 

I.P  3600  m f  C .V

 So we have to find IP and m f To find I.P. I.P 

B.P 29  32.22 kW  mec h 0.9

 To find m f (Actual mass of fuel consumed in kg/hr) 760 mm of Hg  1.01325 bar  V  Volume of mixture sucked inside the cylinders in m 3/sec

 N or N/2  No . o f cylinder V  V  vol  60 where V 

 N  D 2  L and for 4 stroke. 2 4

  1500 4 So V   0.08 2  0.13  0.7  4 2  60  0.022871 m 3/sec

Testing and Performance of Engines 3.89

  P V  m RT  5  P V 1.01325  10  0.022871  m RT 287  15.5  273  m  0.02798 kg /sec

Theoretically 15 kg of air is required for 1kg of fuel So mass of fuel used 

0.02798 15

 1.8653  10  3 kg o f fuel/sec

Since it is 20% rich mixture Actual amount of fuel used  1.8653  10  3  1.2  2.2384  10  3 kg of fuel/sec  m f  m ass of fuel/hr  2.2384  10 3  3600  8.05824 kg of fuel /hr

To find in dicated indicated  

I.P  3600  m f  C.V 32.22  3600 8.05824  44000

 0.32714  32.714 %

To find Brake Mean effective pressure I.P 

S im ilarly B.P. 

P m AL N/2 60

 No. of cylinders

Pmb AL N/2 60

 No. of cylinders

3.90 Thermal Engineering - I

P mb 



B.P.  60 AL  N/2  4 29  10 3  60  1500 2 4   0.08   0.13  2 4 

 887595 N/m 2

Break mean effective pressure

P mb   8.87595 bar

Problem 3.23: A two stroke cycle 20 cm bore  30cm stroke, single cylinder oil engine gives the following results. Speed 350 rpm, Net brake load = 600 N; Diameter of brake drum = 1m Oil consumption = 4.25 kg/hr. Indicated mean effective pressure = 275 kN/m2; C.V = 43000 kJ/kg; A.F = 32; Temp of room air = 20 C; Temp of exhaust gases = 370C. Calculate 1. Indicated power; 2. Brake power; 3. indicated ; 4. brake and 5.

% heat lost to exhaust gases. The Cp for exhaust gases

in 1.

Solution: To find Indicated Power I.P 

Pm  L  A  N 60

 No .of cylinders

. . [ . N fo r 2 strok e

n  1 for single cylinder)]   275  0.3    0.2 2   350 4   1512 kW  60

To find Brake power B.P 

2  N W  S R 60  1000

Testing and Performance of Engines 3.91



2  350 600  0.5 60  1000

 10.996 kW

To find in dicated indicated 

I.P  3600  m f  C.V

 m f  4.25 kg/hr ; C.V  43,000 kJ /kg

indicated thermal 

15.12  3600  0.29785  29.79 % 4.25  43,000

To find Brake Brake

thermal



B .P  3600 10.996  3600   0.21661  m f  C .V 4.25  43,000  21.661 %

To find % heat lost in exhaust gases Mass of fuel consumed in 1 hr = 4.25 kg For 1 kg of fuel, mass of air required = 32 kg  For 4.25 kg of fuel,

Mass of air required  32  4.25  136 kg  Total mass of exhaust gases = Mass of fuel + Mass of

air consumed.  4.25  136  140.25 kg/hr

Heat carried out by exhaust gases in 1 hr  m e C p t2  t1  140.25  1  370  20  49087.5 kJ/hr

3.92 Thermal Engineering - I

% heat lost in exhaust gases 

Heat carried away by exha ust gas Heat supplied by fuel



49087.5  0.2686  26.86 % 4.25  43,000

Problem 3.24: A single cylinder 4 stroke diesel engine gave the following while running on full load. Area of Indicator diagram  3 cm2; Length of the diagram = 4 cm; spring constant  1000 kN/m2/cm; speed of engine = 400 r.p.m; Load on the brake = 400 N; spring balance reading = 50 N; Diameter of the brake drum = 1.2 m; Fuel consumption = 3 kg/hr; C.V = 42,000 kJ/kg; Dia. of cylinder = 17 cm; stroke = 20 cm. Calculate (1) Indicated mean effective pressure; (2) Indicated power; (3) Brake mean effective pressure; (4) Brake specific fuel consumption and (5) Brake thermal and indicated thermal efficiencies.

Solution: To Find (1) Indicated Mean Effective Pressure Pm  

Area of in dicator diagram  spring constant Length of indicator diagra m 3  1000  750 KN/m 2 4

2. To Find Indicated Power I.P 

P m L AN/2 60

 . .N  . 2 for 4 strokes   

Testing and Performance of Engines 3.93

  400 750  0.2    0.17 2   2 4   11.35  60 I.P  11.35 k W

3. To Find Brake Power B.P 

2NW  S  R 60  1000 

2  400  400  50   0.6 60  1000

B.P  8.7965 kW

To Find Brake Mean Effective Pressure Pmb  N P mb  LA    2  B.P  60 P mb 



B.P  60 N LA    2  8.7965  60  400  0.2    0.17 2   2 4  

 581.32 kN/m 2

To Find Brake Specific Fuel Consumption (B.S.F.C) B.S.F.C  

Fuel consumed in kg/hr B.P

3  0.341 kg/kWhr 8.7965

 0.341 kg/kWhr

3.94 Thermal Engineering - I

To find in dicated thermal  indicated thermal 



I.P  3600  m f  C.V 11.35  3600  0.3243 3  42,000

 32.43%

To Find brake therm al  brake thermal 



B.P  3600  m f  C.V 8.7965  3600  0.25133 3  42,000

 25.133% Problem

3.25: The

compression

curve

on

the

indicator

1.3

 constant. At two points on the curve at 1/4 stroke and 3/4 stroke, the diagram from a gas engine follows the law pv

respectively. 140 kN/m2 and 360 kN/m2 Determine the compression ratio of the engine. Calculate the

pressures

are

thermal efficiency and the gas consumption per kW hour on indicated power basis if the relative efficiency is 0.4 and the gas has the C.V of 18,840 kJ/m3.

Solution: V s  Swept volume in m 3 V c  Clearance volume in m 3

Assume the point (1) is at 1/4th of stroke and the point (2) is at 3/4th of stroke on the PV diagram.

Testing and Performance of Engines 3.95

To Find Compression Ratio 1

V1  Vc  Vs  Vs  Vc  0.75 Vs ... (i) 4 V2  Vc  Vs 

3 V  Vc  0.25 Vs ... (ii) 4 s

1.3 P 1V1.3 1  P 2V 2

 V1  V   2

1.3



B 2

P2

P2

 360   V 2  140 

1.3

=C

1

P1

P1

V1

PV

V C= C le ara nce Vo lu m e

1 1.3

 2.068

V1  2.068 V 2

VC

0 .25 V S 0 .75 V S V S = S w ep t Vo lu m e

... (iii)

from (i) and (ii), substitute the value of V1 & V2 in (iii) Vc  0.75 Vs  2.068 Vc  0.25V s  2.068 V c  0.517 V s 0.233 V s  1.068 V c Vs Vc

A

 4.583

Compression ratio  

Total volume Clearance volume

Vc  Vs Vc

1

Vs Vc

 1  4.583  5.583

To Find Thermal Efficiency  air stand ard  1 

1 r

1

 49.74%

1

1 5.583 1.4  1

3.96 Thermal Engineering - I

Re lative efficiency 

Indicate d thermal  A ir standard 

Indicated thermal    relative  air standard  0.4  0.4978  0.1989  19.89 %

To Find Gas Consumption in m 3/kWhr indicated thermal  Substitute

I.P  3600  V f  C.V

I.P  1 kW

 Then Vf is in m 3/kWhr  Vf 

So

1  3600  0.9605 m 3/kWhr 0.1989  18840

 V f  0.9605 m 3/kWhr Problem 3.26: Calculate the relative efficiency based on indicated power and A:F ratio for a four stroke gas engine working on otto cycle from the following data: Brake power = 5 kW; Speed = 180 r.p.m; Volumetric efficiency=85%; Clearance volume  1500 cm3; Swept volume  6500 cm3; mech  80%;

Fuel

consumption

C.V  17,000 kJ/m3.

Solution: To Find relative  relative 

Indicated thermal  Air standard 

4

m3/hr

Testing and Performance of Engines 3.97

I.P 

5 B.P   6.25 kW  mech 0.8

Indicated thermal  

I.P   3600 V f  C.V 6.25  3600  0.331  33.1% 4  17,000

 Air standard  1 

1 r  1

where r  Compression ratio    Air standard  1 

 relative 

Vc  Vs Vc 1500  6500  5.333 1500

1 5.333 0.4

indicated thermal air std



 0.4881  48.81%

0.331  0.6782 0.4881

 67.82% V cycle in m 3  V s  vol

[If vo l is not given, then vo l  1 ]

 N o r N/2  No . of cylinders V in m 3/sec  Vc yc le  60 [N for 2 stroke and N /2 for 4 stroke] Volume of mixture admitted into the cylinder per cycle V cycle in m 3  V s  vol  6500  0.85  5525 cm 3  5525  10  6m 3

3.98 Thermal Engineering - I

Volume of fuel consumed per cycle  

1 4  60 N/2 4 1  60 90

 7.4074  10  4m3

So volume of air per cycle  5525  10  6  7.4074  10  4  4.7842  10  3m 3

A:F ratio by volume



4.7842  10  3 7.4074  10  4

 6.4588 :1 Problem 3.27: Determine the bore and stroke of a single cylinder 4 stroke oil engine from the following data: Brake power = 18.5 kW; Engine speed = 250 r.p.m. At suction, Volumetric   80%; Pressure at the end of suction = 100 kPa; Temp.

at

the

end

of

suction

 10C

mech  88%;

indicated thermal  35%. % of C and H in the fuel used 85% and 15%. C.V. of fuel used = 42,000 kJ/kg; Excess air supplied = 24%; Stroke-bore ratio = 1.5.

Solution: Indicated Power 

Brake pow er 18.5  21.023  0.88 mech

 indicated thermal  0.35 

I.P  3600 21.023  3600    m f  42,000 m f  C.V

 21.023  3600 mf   5.15 kg/hr 0.35  42,000

Testing and Performance of Engines 3.99

Mass of fuel per stroke 

5.15  60  N /2

5.15  250  60     2 

 6.867  10  4kg C  O 2  CO 2 and 2H 2  O 2  2H 2O 12  32  44 and 4  32  36

Actual amount of air required per kg of fuel 32  100 32   0.15   1.24  18.69 kg   0.85   23 4  12   Air required per stroke  6.867  10  4  18.69  0.012834 kg

Volume of air required per stroke V

mR T 0.012834  287  10  273  10.42  10  3m 3  3 P 100  10

Volume of cylinder 



10.42  10 3  vol 10.42  10  3 3 m 0.8

 13.03  10  3 

   D 2  L   D2  1.5D 4 4

D  0.223 m ; L  1.5 D  0.3342 m

Chapter 4

Air Compressors Classification of Compressors - Fans, blowers and compressors - Positive displacement and dynamic types Reciprocating and rotary types Rotary, Dynamic and Axial Flow (Positive Displacement type): Roots Blower, Vane sealed compressor, Lysholm compressor - mechanical details and principle of working - efficiency considerations. Centrifugal compressors: Mechanical details and principle of operation - Velocity and pressure variation. Energy transfer impeller blade shape - losses - slip factor, power input factor, pressure coefficient and adiabatic coefficient - velocity diagrams power. Axial flow compressors: Mechanical details and principle of operation - velocity triangles and energy transfer per stage, degree of reaction, work done factor - Isentropic efficiency pressure rise calculation - polytropic efficiency.

4.1 INTRODUCTION Air compressors are used to compress the air and to raise its pressure. The air compressor sucks the air to the cylinder from atmosphere, compresses it and then delivers the same under a high pressure. 4.1.1 Applications of Compressed Air 1.

It is used to inflate the automobile tyres.

2.

It is used to inject fuel in air injection Diesel engines.

3.

It is used for spray painting.

4.2

Thermal Engineering - I

4.

It is used to operate pneumatic circuits. (such as pneumatic drills, hammers, hoist, air brakes, pile drivers and blast furnaces)

5.

It is used to start IC engine and for supercharging IC engines.

6.

It is used for gas turbine plants.

7.

It is used for cleaning purposes.

8.

It is used in the processing of food and farm maintenance.

4.2 CLASSIFICATION OF AIR COMPRESSORS 1.

According to working (a) Reciprocating air compressors (b) Rotary compressors.

2.

According to action (a) Single acting; (b) Double acting.

3.

According to number of stages (a) Single stage; (b) Multistage compressors.

4.

Air compressors may be further classified as (i) Air pumps and exhausters (ii) Blowers and superchargers (iii) Air Boosters.

5.

According to method of cooling (i) Air cooled (ii) Water cooled

6.

According to number of air cylinders (i) Simplex (ii) Duplex (iii) Triplex

Air Compressors

7.

4.3

According to power drive (i) Direct drive (ii) Belt drive (iii) Chain drive

8.

According to applications (i) Rock drill compressor; (ii) Pneumatic land tool compressor; (iii) Trench digging compressor; (iv) Sand blasting compressor; (v) Spray painting compressor etc.

9.

According to principle of operation (i) Positive displacement compressors (eg) reciprocating, rotary compressor (ii) Dynamic compressors (eg) centrifugal, axial flow compressor

10.

According to nature of installation (i) Portable (ii) Semifixed (iii) Fixed

4.3 SINGLE ACTING RECIPROCATING AIR COMPRESSOR

A ir

Fig. 4.1

In this compressor, only one side of the piston is used to suck the air, compress it and to deliver the air. So, for Single acting, y  1

cycle revolution

4.4

Thermal Engineering - I

4.4 DOUBLE ACTING AIR COMPRESSOR In this compressor, both sides of the piston are used to suck the air, compress it and to deliver the air when suction is occurred in one side, compression and delivery will be taken on the other side. So simultaneously, two cycles will be completed in one revolution of the crank.

$LU $LU

Fig.4.2

So for Double acting, y  2

cycles revolution

4.5 SINGLE STAGE COMPRESSOR Initial Pressure

Final Pressure

S ingle stage

Air Com pressor Fig. 4.3

The compression of air from initial pressure to final pressure is carried out in one cylinder.

Air Compressors

4.5

4.6 MULTI STAGE COMPRESSOR The compression of initial pressure to final pressure is carried out in more than one cylinder. Interm ediate pressure Inter cooling

Air Com pressor Stage 1

Air Com pressor Stage 2

Fig 4.4

4.7 WORKING PRINCIPLE OF RECIPROCATING AIR COMPRESSORS Fig 4.5 (a). The piston is moving downward. The inlet valve opens. The fresh air enters the cylinder. So suction stroke is completed. During this stroke delivery D elivery valve closed

C om pressed Air delivery Inlet valve T open T

T

Inlet valve closed T

C ylinder Air Inlet

Piston C onnecting rod

C rank (a)

(b)

Fig.4.5 W orking of Reciprocating Air Com pressor

4.6

Thermal Engineering - I

valve is closed. This stroke is completed in 180 of crank revolution (1/2 rev.). During the return stroke, the piston moves upward. Both valves are closed. So the compression of air takes place. When the air is compressed to required pressure the delivery valve opens (Fig 4.5 (b). So the compressed air is delivered through delivery valve. So the return stroke is completed by partly compression and partly delivery. This stroke is completed in remaining 180 of crank revolution (remaining 1/2 revolution). So the suction, compression and delivery completes in one cycle. Refer Fig. During return stroke, the air is compressed by its major part (i.e compression stroke 1-2). 2

3 3

com pression

2

2 1

A B W ork done on 4 delivery

B

C

W ork done on com pression

1 4

1

A

C

A

B

C

V4

V2

V1

D elivery valve S uction valve 3

2 D elivery

1 C om pression

Fig 4.6

W ork done by suction

Air Compressors

4.7

The compression continues till the pressure reaches P 2 which is sufficient to open the delivery valve. Once the delivery valve opens, no more compression takes place during remaining part of the return stroke. During this time, the compressed air is delivered till the piston reaches the cylinder end. Thus the cycle is completed. Again the fresh air is sucked into the cylinder in the next cycle. The compression of air may be isothermal or polytropic or isentropic process. We can find the workdone to compress the air in one cycle for the three cases. 4.7.1 Workdone during isothermal PV  c without clearance volume

compression

Workdone  Area 1-2-3-4  V1  W  P 2V2  P 2V 2 ln    P 1V 1 V  2  V1   V1   P 2V 2 ln    P 1V 1 ln   V V  2  2

3

2

W = Area o f

w hich gives w ork d one per cycle w ithout clearance volum e.

1

4 3

2

3

2

2 (+ )

= 4

. . [ . P 1V 1  P2V 2 ]

1

1

4

1

A

C

(-) Fig 4.7

A

B

B

C

4.8

Thermal Engineering - I

 V1   mRT ln   V  2

4.7.2 Workdone

. . [ . P1V1  mRT ]

during

polytropic

compression

n

[ PV  constant ] without clearance volume

If the compression is done by polytropic process, then W  P 2V 2 



P 2V 2  P 1V 1 n1

 P 1V 1

n  1 P 2V 2  P 2V 2  P 1V 1  n  1 P 1V1 n1



n P V  P 1V 1 n1 2 2



 P 2V 2  n 1 P 1V 1  P V n1  1 1 

... (i)

For polytropic compression, P 1V n1  P2Vn 2 1

n

V1  P 2   P1     or  V2 V1 P P  2  1 V2

Substitute the value of

W

V2 V1

1 n

in eqn. (i)

   P 2   P 1  1 n n P 1V 1   1  P P n1   1  2  n1

  P2  n n P 1V 1  W  P n1  1

 1 

... (ii)

Air Compressors n1

  P2  n n W mRT 1   P n1  1

 1 

4.9

... (iii)

Note:  P2  Also,    P1 

n1 n



T2 T1

P1V1  m RT1 P1V1 T1

T2 T1 







P 2V2

T2 P 2V 2

P1V1 1 . . V 2  P 1  n  .  V1  P2 

 P 1  1 n   P P1  2

P2

 P 2  1n   P P1  1

P2

n1

 P2  n   P T1  1 T2

So,

W



Workdone

W 

  T2 n mRT 1 1 T n1   1  T2  T1  n mR T1  T1 n1   n mRT 2  T 1 n1

... (iv)

Thermal Engineering - I

4.10

4.7.3 Workdone

During

Isentropic

Compression



PV  constant Without Clearance Volume

Here the polytropic index n can be changed by isentropic index  and we get the workdone in the same way as already discussed. Workdone in one cycle 1

  P2    W  P 1V 1   P 1  1

 1 

1

  P2     mRT 1  W  P 1  1 W

  mR T 2  T 1 1 CP

The ratio of sp. heats

i.e., C v 

CP



Cv

 1 

... (v)

... (vi)

... (vii)



and C P  C v  R CP 

i.e.,

CP



R

1  Cp 1    R    1 Cp R   

Substitute R value in (vii) W

 1  m C p  T  T 1 1    2

W  mC pT 2  T 1

... (viii)

Air Compressors

4.11

4.8 MINIMUM WORKDONE P

Isotherm al( P V = C ) Polytropic( P V n = C ) 21 22 2 n tr op ic ( PV 

) =C

The isothermal process requires compression very slow to maintain the temperature constant. Practically, it is not possible. So isothermal process can be approached but not obtained.

Is e

The workdone on air is minimum when the air is compressed in isothermal process.

1

V

Fig 4.8

The isothermal process can be approached (i) by air cooling or water cooling during compression, (ii) by inter cooling in multistage compressors.

4.9 POWER REQUIRED TO RUN THE COMPRESSOR P  Work done pe r cycle 

N 2N or 60 60

N  for single acting 2N  for double acting. Problem 4.1: A single stage reciprocating air compressor is required to compress 1 kg of air from 1 bar to 5 bar. The initial temperature is 27C. Compare the work requirement in the following cases: 1. Isothermal process, 2. Polytropic compression with PV1.2  constant and 3. Isentropic compression.

4.12

Thermal Engineering - I

Solution: Isothermal Compression  P2   P2  W  P1V 1ln   mRT1ln   P P  1  1

.. [ . P1V 1  mRT1 ]

5  1  0.287  27  273 ln  1 .. [. R  138.573 kJ

for air = 0.287 kJ/kg K]

. . [ . T1  27  293  300 K]

Polytropic Compression PV 1.2  C n1

  P2  n n  mRT1   W  P n1  1

 1  0.2

1.2  5   1  0.287  300    1.2  1   158.94 kJ 0.2 1 

Isentropic Compression 1

  P2     mRT 1  W  P 1  1

 1  0.4

1.4 5    1  0.287  300    1.4  1  0.4 1     175.93 kJ

We can see that workdone is minimum in isothermal process and maximum in isentropic process. Problem 4.2: A single stage reciprocating air compressor is required to compress 60 m3 of air from 1 bar to 8 bar at

Air Compressors

4.13

22C. Find workdone by the compressor, if the compression of air is (i) isothermal (ii) adiabatic (iii) polytropic with index as 1.25. (Ap. 2007 - AU)

Given: V 1  60 m 3; P 1  1 bar; P 2  8 bar; T 1  22  273  295K.

Solution: Case (i) Isothermal  P2  W  mRT1 ln   P  1 P 1V 1  mRT1  m 

P 1V 1 RT1



1  10 5  60  70.87 kg 287  295

8 W  70.87  0.287  295 ln   1  12,477 kJ

Case (ii) Adiabatic (i.e. Reversible adiabatic (or) Isentropic) So, n    1.4 1

  P2     m  R  T 1  W  P 1  1

 1  0.4

1.4  8    70.87  0.287  295    1.4  1  0.4 1     17,041 kJ

4.14

Thermal Engineering - I

Case (iii) Polytropic with index n  1.25 n1

  P2  n n  mRT1  W  n1   P1 

 1  0.25

1.25  8   70.87  0.287  295    1.25  1  1 0.25     15,472 kJ Problem 4.3: A single stage single acting reciprocating air compressor with 0.3 m bore and 0.4 m stroke runs at 400 rpm. The suction pressure is 1 bar at 300 K and the delivery pressure is 5 bar. Find the power required to run it, if the compression is isothermal, adiabatic and compression follow pv1.3  C. Also find the isothermal efficiency. (JNTU - Apr/May 2013)

Given Data D  0.3 m; L  0.4 m; N  400 rpm ; P 1  1 bar; T 1  300 K; P 2  5 bar; Vs  V1 

  2 D  L   0.3 2  0.4  0.0283 m 3 4 4

Isothermal compression work  P2  5 W  P 1 V 1 ln    100  0.0283  ln   P1 1    4.55 kJ/kg

Power  W 

N 400  4.55  60 60

 30.33 kW

Air Compressors

Adiabatic compression work n    1.4 1

   P    2 W  P 1 V 1    P 1  1

   1  

 5 1.4   100  0.02873     1 0.4  

0.4 1.4

  1  

 5.78 kJ /kg  5.78 

Power

400  38.55 kW 60

(iii) Polytropic compression work with n  1.3 n1

  1  

n  P  n 2  P1 V 1    W  P1  n1   0.3 1.3  5 1.3   100  0.0283     1 0.3    5.52 kJ /kg

Power

 5.52 

400  36.97 kW 60

(iv) Isothermal efficiency isothermal 

I.Pisothermal I.Ppolytropic

 82.48%



30.33  0.8248 36.77

  1  

4.15

4.16

Thermal Engineering - I

4.9.1 Clearance Volume V s  Swept volume (or) Stroke volume (or)

Displacement volume 

This swept volume is defined as the volume swept by the piston in one stroke. V c  Clearance volume



  

Clearance volume is required so that the piston does not hit the cylinder end at the end of the stroke. Clearance volume is essential to give some space for valve movements: In general, clearance volume is expressed as a percentage of stroke volume Vs. Clearance volume is the volume of air which remains in the cylinder after the piston has reached the end of its return stroke.

4.10 WORKDONE

BY

RECIPROCATING

AIR

COMPRESSOR WITH CLEARANCE VOLUME Refer Fig 4.9. From point 1, the piston starts compressing the air till it reaches the point 2. (i.e. till it reaches the pressure P 2). At point 2, the delivery valve opens and the piston is further moving inside so that the compressed air is delivered through valve. At point 3, the delivery of compressed air stops. Vc is the clearance volume and it is filled with some amount of compressed air. From the point 3, the piston starts moving in the opposite direction (i.e. in the right side direction). This makes the trapped air in the clearance volume expand. This trapped high pressure air will expand according to PVn  C. The

Air Compressors

P P2 6 3

4.17

2 n

P V =c

P1 5

V 3 =V C

C learance volum e = V 3 =V C

4 V4

1 V1

V

E ffective sw ept Volum e = V 1 -V 4 S w ept Volum e =V 1 -V 3 = V s Total Volum e =V 1 Fig.4.9

expansion occurs till the piston reaches point 4 (i.e. till the pressure comes down to intake pressure). At this point 4, the suction valve (inlet valve) opens and the fresh atmospheric air enters the cylinder. So the actual volume of air taken inside the cylinder in one stroke is V1  V4. This volume is called effective swept volume. So, Vactual  V effective  V1  V4. Workdone/cycle  Net area 1 2 3 4 1  Area 1 2 3 6 5 1  Area 3 6 5 4 3 n1

  P2  n n   P1V1   P n1  1

 1  n1

  P2  n n  P 1V 4    P n1  1

 1 

4.18

Thermal Engineering - I n1

  P2  n n  P 1V1  V4   W/cycle   P n1  1 n1

  P2  n n  mRT1  W/cycle   P n1  1

 1 

 1 

where V 1  V 4 and m are the actual volume and mass of air sucked by piston per cycle respectively.

4.11 ISOTHERMAL

EFFICIENCY

OF

A

RECIPROCATING AIR COMPRESSOR Isothermal efficiency of a reciprocating air compressor is defined as the ratio of isothermal workdone to the actual workdone Isothermal efficiency 

Isotherm al workdone Actual workdone

Isothermal compression cannot be achieved in practice but an attempt is made to approach isothermal case by cooling the compressor either by addition of cooling fins or a water jacket to the compressor cylinder. Thus the higher the isothermal efficiency, the more nearly has the actual compression approached the ideal isothermal compression.

4.12 VOLUMETRIC EFFICIENCY IN A RECIPROCATING AIR COMPRESSOR The volumetric efficiency of a compressor is the ratio of free air delivered to the displacement of the compressor. It is also the ratio of effective swept volume to the swept volume.

Air Compressors

Volumetric efficiency 

4.19

Effective swept volume V1  V 4  Swept volume V1  V3

Because of the presence of clearance volume, volumetric efficiency is always less than unity. It varies from 60% to 85%. The clearance ratio 

Vc V3 Clearance volume   k Sw ept volum e V1  V3 Vs

As a percentage k varies from 4% to 10% Volumetric efficiency  vol 



V1  V4 V1  V3 V 1  V 3  V 3  V 4 V 1  V3

1

V3 V1  V3



V4 V1  V3

V4 V 3 V 4/V 3  vo l  1  k  1k V1  V3 V 1  V3  1  k  k.

  . . V3 k  . V 1  V3  

V4

V3 1

 P3  n  vol  1  k  k   P  4 1

1

 . . V 4  P3  n    .   V3  P4   

 P2  n  vol  1  k  k   1k P  1

1 n

    P2    1   P1     . . [ . P 3  P 2 ; P 4  P 1]

4.20

or

Thermal Engineering - I

  vol  1  k  k     vol  1  k  k  

V1   V2  V4   V3 

  V1 Also  vol   1  k  k  V   2

1

 . .  P 2  n V1      .  V2   P1  

 Pi T a    P aT i   1

or

 P 2  n  P i Ta   vol   1  k  k    P T P  1  a i 

a : Ambient condition i : Inside condition

Note: This efficiency should not be used for finding out the dimensions of the cylinder. For finding out the dimensions of the cylinder, volumetric efficiency based on suction condition only should be used. 4.12.1 Factors affecting volumetric efficiencies (i) Very high speed (ii) Very large clearance volume (iii) Leakage past piston (iv) Obstruction at the inlet valves (v) Overheating of the air (vi) Inertia effect of air at suction pipe.

4.13 IMPORTANT TECHNICAL TERMS 1. Volumetric Efficiency vol It is the ratio of free air delivered to the displacement of the compressor. It is the ratio of effective swept volume to the swept volume.

Air Compressors

 vol 

i.e.,

4.21

Effective s wept volume V1  V 4  Swept vo lume V1  V 3

V actual

Vswept



Va Vs



V1  V4 V 1  V3

2. Clearance Ratio ‘k’  P  2  vol  1  k      P1  

1

n

where k  Clearance ratio 

1

  P2  n  11kk  P   1  Vc Vs

Note: If k is not given, k  0; then  vol  1. 3. Volume and Volume Rate  V  Volume rate in m 3/s ec V  Volume in m3 v  Specific volume in m 3/kg

Suffix ‘a’ stands for actual (or) effective. Suffix ‘s’ displacement.

stands

for

swept

(or)

stroke

 N rev. VV  yZ  m 3   m 3/sec 60 sec

where y  1 single acting y  2 for double acting Z  number of cylinders (or) number of stages.   m 3 kg   m 3/sec Also V  vm  kg sec

(or)

4.22

Thermal Engineering - I

4. Indicated Power (I.P) and Brake Power (B.P)  I.P  m W where W in kJ/kg  kg/sec 

mech 

kJ kJ  kW  kg sec

I.P B.P

Note: Whenever  mech is given, the power involved will be B.P. 5. Free Air Delivered (FAD) Free air delivered is defined as the actual volume rate of air reduced to atmospheric condition and expressed in m 3/min or m3/sec P atm V FAD

Tatm atm



 P 1V a T1

  mR

- means atmospheric condition (or) Ambient

condition. Suffix ‘1’ compression.

-

means

Note: If atmospheric  FAD  V a .

suction

condition

is

(or)

not

beginning

given,

of

then

6. Mean Effective Pressure m .e.p It is found mathematically - dividing the workdone per cycle by the stroke volume. MEP 

W I.P   in kPa (or) Pa V s Vs

Air Compressors

4.23

7. Isothermal Efficiency: It is the ratio of isothermal work to polytropic work. isothermal 

W isothermal W polytropic



I.P isothermal I.Ppolytropic

 P2  Wisothermal  mRT1 ln   in kJ/kg P  1

 P2   I.Pisothermal  mR T1 ln   in k W  P1  8. Swept Volume V s Vs 

 2 D  L in m 3 4

 N     3  V s  V s  60  yz in m /sec     

where D  Bore (or) Dia. of cylinder. L  Stroke (or) Stroke length. Note: 1000 litre  1 m3 9. Piston speed 

2LN in m/sec 60

4.14 SINGLE STAGE n1

  P2  n n  P 1V1  V4   W/cycle   P n1  1 n1

  P2  n n  mRT1  W/cycle   P n1  1

 1 

 1 

where V 1  V 4 and m are the actual volume and mass of air sucked by piston per cycle respectively.

4.24

Thermal Engineering - I

Workdone/cycle n1

  P2  n n  P 1V 1  V 4     P n1  1

 1 

To Find I.P Theoretical power (Indicated power) n1

  P2  n  n  m RT 1   P n1  1

 1 

1 n    P2   vol    1  k  k   P1      

  P1 T a   P atm  T1 

Problem 4.4: A single stage, single acting reciprocating air compressor has a bore of 200 mm and a stroke of 300 mm. It receives air at 1 bar and 25C and delivers it at 6 bar. If the compression follows the law PV1.3  constant and clearance volume is 5% of the stroke volume, determine 1. mean effective pressure and 2. the power required to run the compressor, if the speed is 800 r.p.m.

Given: D  0.2 m ; L  0.3 m ; P1  1 bar ; T 1  25  273  298 K P2  6 bar Solution: Vs  Stroke volume (or) Swept volume 

 2 D L 4

Air Compressors



  0.2 2  0.3 4

4.25

p P 2 =P 3 3

2

 9.425  10  3m 3

Clearance volume V c  0.05  V s

P 1 =P 4 Vc

 0.05  9.425  10  3  4.7124  10  4m 3  V3  P3    P V3  4 V4

1n

 P2     P1 

 P2  V 4  V 3  P  1

Vs

V 3 = Vc Fig

1 n

1n

 4.7124  10

6 1  

 4

1 1.3

 1.8699  10  3m3 V actual  Veffective  V 1  V 4 V 1 V c  V s  4.7124  10  4  9.425  10  3  9.89624  10  3m 3 V eff  V 1  V 4  9.89624 10  3 1.8699  10  3  8.02634  10  3m 3

1

4

v V4

V eff

V1

4.26

Thermal Engineering - I

Workdone/cycle n1

  P2  n n  P 1V 1  V 4     P n1  1

 1  0.3

1.3  6  1  10 5  8.02634  10 3    1.3  1   1 1.3  1    W  1781.04 J  1.781 kJ

To Find Mean Effective Pressure (MEP) MEP 



Work done /cycle Stroke volume 1781.04 9.425  10  3

 188969.61 N/m 2  1.89 bar

To Find Power P 

WN 60 1781.04  800 60

 23.747 kW Problem 4.5: A single acting compressor has zero clearance, stroke 200 mm and piston diameter 150 mm. When the compressor is operating at 250 rpm and compressing air from 10 N/cm2, 25 C, to 40 N/cm2, find (i) the volume of air handled (ii) the ideal power required. (JNTU Jan/Feb - 2015 Set - 2)

Air Compressors

4.27

Given Data L  0.2 m; D  0.15 m; N  250 rpm; P 1  10 N/cm 2  1 bar; T1  25  273  298 K;P 2  40 N/cm 2  4 bar

Solution: Volume of air handled in m 3/min    Vs   D2  L  N  y  Z 4 y  1 for single acting Z  1 for single stage   V s   0.15 2  0.2  250  1  1 4  0.8836 m 3/min  V s  0.8836 m 3/min

Ideal power Mnimum power (Ideal power) can be obtained by using Isothermal process.   P2  Isothermal power  P 1 V1 ln    P1   100 

0.8836 4  ln   60 1

   V  V in m 3/s  s   1

 2.04 kW Problem 4.6: A single stage, single acting reciprocating air compressor has a bore of 20 cm and a stroke of 30 cm. The compressor runs at 500 rpm. The clearance volume is 4% of the swept volume and index of expansion and compression is

4.28

Thermal Engineering - I

1.3. The suction conditions are 0.97 bar and 27C and delivery pressure is 5.6 bar. The atmospheric conditions are at 1.013 bar and 17C. Determine (i) the free air delivered in m3/min (ii) the volumetric efficiency referred to the free air conditions (iii) the indicated power.

(Nov/Dec - 2011 - AU)

Given D  20 cm, L  30 cm; N  500 rpm ; V c  4% Vs P 1  0.97 bar ; n  1.3 ; P 2  5.6 bar ; P a  1.013 bar T a  17C ; T1  27 C

Solution: Vs 

  2 D  L  0.202  0.30  9.42  10  3 m 3 4 4

vol 

Va Vs 1

n    P2  vol   1  k  k    P  1 

  P1 T a   P a T1  1

1.3    5.6    1  0.04  0.04     0.97  

  0.97  17  273    1.013  27  273  

vol  0.82  82% 0.82 

Va

9.42  10  3 V a  7.728  10 3 m3

Air Compressors

4.29

Free air delivered  Va   Va  y  Z  N  7.728  10 3  1  1  500  0.0644 m 3/s . .  . y  1 for single acting   z  1 for single stage  

Indicated Power   P2  n P1 Va   W  P n1  1

n1 n

 1  1.3  1

1.3   5.6  1.3 W  0.97  10 2  7.728  10  3    1.3  1   0.97 

 1 

W  1.62 kJ IP 

W  N 1.62  500   13.5 kW 60 60

IP  13.5 kW Problem

4.7: A single stage single acting reciprocating air

compressor delivers 15 m3/min of free air from 1 bar to 8 bar at 300 rpm. The clearance volume is 5% of the stroke volume and compression and expansion follow the law PV1.3  constant. Calculate the diameter and the stroke of the compressor. Take L/D  1.5. The temperature and pressure of the air at suction are the same as atmospheric air. (JNTU - Apr/May 2013 - Set 4) (Apr/May - 2008 - AU)

Given  15 m 3  0.25 m 3/sec FAD  Va  60 sec

P1  1 b ar; P 2  8 ba r; N  300 rp m

4.30

Thermal Engineering - I

k  5%  0.05; n  1.3;

L  1.5 D

Solution: 1 n     P2   Volumetric efficiency vol  1  k   1   P1     1 1.3  8  1  0.05     1    0.802    Va Va vol    V s  Vs vol

Also

Swept volume in

m3  0.25 ,V   0.3115 m 3/s sec s 0.802

 N We known V s  V s   yZ 60

Here

y  1 for single acting Z  1 for single stage  60 Vs  Vs  N  0.3115 

We know, VS  0.0623 

60  0.0623 m 3 300

 2 D L 4  2 D  1.5 D 4

  1  

Air Compressors

4.31

D  0.375 m L  1.5 D  0.563 m

Diameter D  0.375 m Stroke

L  0.563 m

Problem 4.8: A single stage single-acting air compressor delivers 0.6 kg of air per minute at 6 bar. The temperature and pressure at the end of suction stroke are 30C and 1 bar. The bore and stroke of the compressor are 100 mm and 150 mm respectively. The clearance is 3% of the swept volume. Assuming the index of compression and expansion to be 1.3, find: (i) Volumetric efficiency of the compressor, (ii) Power required if the mechanical efficiency is 85%, and (iii) Speed of the compressor.

(May/June- 2012 - AU)

Given Mass of air delivered

m  0.6 kg/min

Delivery pressure,

P 2  6 bar

Suction pressure,

P 1  1 bar

Suction temperature

T1  30  273  303 K

Bore

D  100 mm  0.1 m

Stroke length,

L  150 mm  0.15 m

Clearance volume,

V c  0.03 V

Mechanical efficiency, mec h  85%

4.32

Thermal Engineering - I

Solution: P (bar)

P 2= 6

T2

3

2

1.3

Exp

pV =C

ansi on

Co

m

pr

es

s io

n

T 1 =303 K P 1= 1

1

4 3

V(m ) Vc= 0.03V s

Vs

(i) Volumetric efficiency of compressor, vol   P2  vol  1  k    P  1

Where k 

Vc Vs

1/ n

 1 

 0.03

6  vol  1  0.03    1

1 1.3

  1   0.91096  91.096% 

Air Compressors

4.33

(ii) Power required to drive the compressor

Indicated power 



  P2   n m RT 1    P n1  1

1.3 0.6 6   0.287  303    1.3  1 60 1

n1    n 

1.3  1 1.3

 1 

  1   1.93 kW 

 Power required to drive the compressor 

1.93 1.93   2.27 kW mech 0.85

(iii) Speed of the compressor, N: Free air delivery, F.A.D.  m RT 1 0.6  0.287  303   0.5218 m3/min  2 P1 1  10

Displacement volume  F.A.D 0.5218   0.5728 m3/min Vs  0.91096 vol Also,

or

 2 D LN 4 . .  . y  1 for singleacting com pressor   Z  1 for single s tage    0.5728   0.1 2  0.15  N 4

0.5728 

 Speed of compressor N

0.5728  4   0.1 2  0.15

 486.2 r.p.m

4.34

Thermal Engineering - I

Problem 4.9: A single stage double acting air compressor running at 300 rpm delivers 15 m3 of free air per min at 700 kPa, 200C. Clearance volume is 8% of stroke volume and the index of compression and expansion are same. Find the volumetric efficiency, swept volume referred to (i) intake condition (ii) free air conditions of 101 kPa and 15C (JNTU - April/May 2013-Set 1) (Oct-2000, Madras University)

Given : N = 300 rpm; y = 2 cycles/rev (double acting); Z = 1 (single

stage);

FAD

=

15 m 3/min ;

P 2  700 kPa ;

T2  200  273  473 K ; k  8%  0.08

Solution: Case (1) Swept Volume refered to Suction Condition = Intake condition  15  0.25 m 3/sec FAD  V a  15m 3/min  60 Assume n    1.4 [since n is not given we can assume n  ] P 1  1 bar [since initial pressure is not given, assume P 1  1bar]   P 2 1/  1 vo l  1  k    P   1   7 1/1.4   1  0.08    1 1     0.758833  75.88%  Va Also vol  0.758833   Vs

Air Compressors

4.35

  Va 0.25  0.3295 m3/sec  Vs  vol 0.758833

  N  But V s  V s   yZ  60  y  2 for dou ble acting ; Z  1 (for single stage)  Vs 0.3295   0.032945 m 3 Vs  N/60  y  Z 300 /60  2  1

Case (ii) Swept volume referred to Atmospheric conditions Given P atm  101 kPa; T atm  15  273  288 K 1

0.4

 P1   1  ; T1  473   1.4  271.3 K  P2 T2 7    P 1 V a P atm FAD  Tatm T1 T1

 1  10 5  V a 271.3



101  10 3  0.25 288

 V a  0.238 m 3/sec   P 2 1/  vol  1  k   1  P  1    7 1/1.4   1  0.08     1   0.7588326 1     V a 0.238 vol  0.7588326     Vs Vs

4.36

Thermal Engineering - I

 V s  0.31364 m 3/sec  V s  V s N/ 60  y Z Vs 

0.31364  0.031364 m 3 300/ 60  2  1

Problem 4.10: An air compressor takes in air at 0.98 bar and 20 C and compresses it according to the law pv1.2  C. It is then delivered to a receiver at constant pressure of 9.8 bar. Determine (i) the temperature at the end of compression: (ii) the work done per kg of air: (iii) the heat transferred during the compression; and (iv) the work done during delivery. Take R  287 J/kg K and   1.4. (JNTU - August 2014 - Set (4))

Given Data P 1  0.98 ba r; T 1  20  273  293 K; n  1.2; P 2  9.8 bar; T2  ? ; m  1 kg

Solution: To find T 2 n1  n

 P2   P T1  1 T2

n1

 P2  n  T2  T1   P  1 0.2

 9.8  1.2  430 K T 2  293    0.98  T 2  430 K

Air Compressors

4.37

(ii) To find workdone per kg of air   P2  n  m RT 1   W  n1   P1 



n1 n

 1 

1.2   9.8   1  0.287  293    0.2   0.98 

0.2 1.2

 1 

 236.03 kJ /kg

(iii) Heat transfered during compression Q  m C n T 2  T 1

 n  C n  Polytropic specific heat  C v   1n  1.4  1.2   0.718     0.718 kJ /kg K  1  1.2  Q  1   0.718  430  293   98.37 kJ/kg  sign indicates heat lost to surroundings

(iv) Workdone delivery

during

Area of 2  3  A  B  2 gives W on delivery

P 2

3

W = P 2V 2

W  P 2 V 2  mRT 2  1  0.287  430  123.41 kJ /kg

4 A

1 B

V

Problem 4.11: A single acting of single stage air compressor with 5% of clearance volume compresses air from 1 bar to 5 bar. Find the change in volumetric efficiency if the exponent of

4.38

Thermal Engineering - I

expansion process changes from 1.25 to 1.4. (Apr-2006, Anna University)

Given: k  5%  0.05 ; P 1  1 bar; P 2  5 bar Solution: Case (i) n  1.25    P 2  1 n  vo l  1  k    1 P   1 1

5   1  0.05    1.25  1   0.8688  86.88 % 1   

Case (ii) n    1.4   P 2 1/   vo l  1  k   1  P  1  1    5  1.4  1  0.05     1   0.8922  89.22%  1

Change in  vol 

Fina l  vol Initial vol

1

0.8922  1  0.02688  2.7% 0.8688

Problem 4.12: A single acting 14 cm  10 cm reciprocating compressor is operating at P1  1 bar, T1  20  C, P2  6 bar and T2  180 C. The speed of compressor is 1200 rpm and shaft power is 6.25 kW. If the mass of air delivered is 1.7 kg/min, calculate the actual volumetric efficiency, the indicated power, the isothermal efficiency, the mechanical efficiency and the overall efficiency. (JNTU - January 2014 - Set 4) (Nov/Dec - 2010 - AU)

Air Compressors

4.39

Given: Suction pressure

P 1  1 bar

Suction temperature

T1  20  273  293 K

Discharge pressure

P 2  6 bar

Discharge temperature

T2  180  273  453 K

Speed of the compressor N  1200 rpm Shaft power P shaft  6.25 kW  m a  1.7 kg  min

Solution: (i) Actual volumetric efficiency  here    V s  D2 L  N  y  Z  y  1 4 Z1  

    

P 3

 2   0.14    0.10   1200 4

P V n =C 4

FAD 

vol 

P1



1.7  0.287  1000  293 5

1  10

FAD 1.4295   100  77.38%  Vs 1.8473

(ii) Indicated Power n1

IP 

1 V

 1.8473 m2  min

 m RT1

2

  P2  n  n m RT 1    n1   P1 

 1 

 1.4295 m3  min

4.40

Thermal Engineering - I n1

 P2  n   T1 P  1 T2

Taking ln on bothsides. T2 P2 n1  ln ln n P1 T1  ln  n1   n  ln  

T2   453   ln  293  T1    P2  6 ln    P1 1  n  1.32

IP 

1.7 1.32  6    0.287  293    1.32  1 60  1 

1.32  1 1.32

 1 

IP  5.346 kW

(iii) Isothermal efficiency  P2   Isothermal power  m RT1 ln    P1  

Iso 

17 6  0.287  293 ln    4.269 kW 60 1 4.269  100  79.85% 5.346

IV. The Mechanical efficiency mech 

Indicated Power 5.346  100  85.5%  S haft Power 6.25

Air Compressors

4.41

(v) Overall Isothermal efficiency  over iso    Isothermal Power 4.269   100  68.3%  overIso  Shaft Power 6.25 Problem 4.13: A single stage, double acting compressor has a free air delivery of 14 m3/min, measured at 1.013 bar and 15C. The pressure and temperature in the cylinder during induction are 0.95 bar, 15C. The delivery pressure is 7 bar and index of compression and expansion is 1.3. The clearance volume is 5% of the swept volume. Calculate (i) Indicated power required (ii) Volumetric efficiency. (JNTU - Apr/May 2013 - Set 3)

Given data: y  2;

Z  1;

FAD  14 m 3/m in;

P atm  1.013 bar;

Tatm  15 C; P 1  0.95 bar; T 1  15 C; P 2  7 bar; n  1.3; k  0.05

Volume of air required at suction condition:  P atm FAD  P1 Va  T1 T atm  1.013 14   0.249 m3/s Va  60 0.95

IP 



   P2  n P1 Va    P n1  1

n1 n

1.3  7   95  0.249    0.3   0.95 

 60 kW

 1  0.3 1.3

 1 

[ Tatm  T1]

4.42

Thermal Engineering - I 1 1 n   1.3    P2   7    1   1  0.05   1 vol  1  k      P1    0.95     

 0.8176  81.76% Problem 4.14: Determine the size of the cylinder for a double acting air compressor of 40 kW indicated power, in which air is drawn at 1 bar and 15  C and compressed according to the law PV1.2 constant to 6 bar. The compressor runs at 100 rpm with average piston speed of 152.5 m/min. Neglect clearance.

(JNTU - January - 2014 - Set (4))

Given Data I.Power  40 kW; P 1  1 bar; T 1  15  273  288 K PV 1.2  C; P 2  6 bar; Double acting; N  100 rpm;

Piston speed  152.5 m/min; Piston speed  L

2LN 152.5  m/s 60 60

152.5  0.7625 m 2  100

Indicated power   n1  V a  IP  n

   P2  n P1 V a    P n1  1

n1 n

1

  P  2 P1    P1  

n1 n

  1  

 1 

Air Compressors

 0.2 V a  40   1.2

4.43

1 0.2

 1.2 6  100     1  

  1  

 0.192 m 3/s   Since Vol  is not given, Vs  Va  0.192 m3/s  N Vs  Vs  yZ 60

0.192  V s 

 y  2 for double acting   Z  1 for sin gle stage   

100 21 60

V s  0.0575 m 3/cycle

and

Vs 

0.0575 

 2 D L 4

  D 2  0.7625 4 D  0.309 m

Size: D  0.309 m L  0.7625 m Problem 4.15: A single stage single acting reciprocating air compressor has a bore of 200 mm and stroke of 300 mm. It receives the air at 1 bar and 20C and delivers it at 5.5 bar. If the compression follows the law PV1.3  C and the clearance volume is 5% of the stroke volume, determine the power required to drive the compressor, if it runs at 500 r.p.m. (Oct-97, Madras University)

4.44

Thermal Engineering - I

Given: D  0.2 m; L  0.3 m; P 1  1 bar; T1  293 K; P 2  5.5 bar ; n  1.3; k  5%  0.05; N  500 r.p.m.

Solution: 1   P 2  1n     5.5  1.3  vo l  1  k      1 1 1 0.05    1  P1      

 0.864442

Swept volume V 

   D 2  L   0.2 2  0.3 4 4

 9.4248  10  3m3  N yZ Vs  Vs  60 500  9.4248  10  3   1  1  0.07854 m 3/sec 60    Va  vo l   So, V a  vol  Vs  0.864442  0.07854 Vs  0.067893  V a  0.067893 m 3/sec

  P 1V a 1  10 5  0.067893    P 1Va  mRT1 So, m  RT 1 287  293  0.080738 kg/sec n1

  P2  n  n m RT1  Power   P n1  1

 1  0.3

1.3    5.5  1.3  0.080738  287  293     1 1 0.3     14181.14 W  14.18114 kW

Air Compressors

4.45

Problem 4.16: A single stage reciprocating air compressor has clearance volume 5% of stroke volume of 0.05 m3/sec. The intake conditions are 95 kN/m2, 300 K. The delivery pressure is 720 kN/m2. Determine the volumetric efficiency referred to (i) intake conditions (ii) atmospheric conditions of 100 kN/m2 and 290 K (iii) FAD and (iv) power required to drive the compressor, if the ratio of actual to indicated power is 1.5. Take index of compression and expansion as 1.3. (Nov/Dec-2010 - AU)

Given Data A single stage Reciprocating Air compressor V c  5% V s  V a  0.05 m 3  sec ; P 1  95 kN  m 2 ; T1  300 K P 2  720 kN  m 2

Solution: (i) Volumetric efficiency P

(a) Intake condition  P2  vol  1  k  k   P  1

1n

2

3 W

5 V 100 s   0.05 k Vs Vs

1

4

Vc

1

720  1.3

  vol  1  0.05  0.05    95 

vol  81.3%

VC

VS V

4.46

Thermal Engineering - I

(b) Atmospheric condition of volumetric efficiency 1

P 1 Ta  P2  n    vol    1  k  k    P     1   P atm  T 1

Patm  100 kN  m 2 given 1

  720   vol   1  0.05  0.05    a tm    95 

95  290  1.3  0.7462    300 100 

 74.62%   vol  atm 

(ii) Free Air Delivered FAD   0.05 m3  sec (iii) Power required to Drive the Compressor    P2  n P1 V a   P  P n1  1

n1

n

 1  1.3  1

1.3   720  1.3  0.95  10 2  0.05     1.3  1   95 

 1 

P  12.26 kW Problem

4.17: A four cylinder, double acting compressor is

required to compress 30 m3/min of air at 1 bar and 25C to a pressure of 15 bar. Determine the capacity of motor required and cylinder dimensions if the following data are given: Speed of the compressor 310 r.p.m.; Clearance volume 3.5%; Stroke to bore ratio 1.2; mech  82%; Index of compression 1.3. Assume no pressure change in suction valves and the air gets heated by 12C during suction stroke. (Apr.98, Madras University)

Air Compressors

4.47

Given Hint: Here atmospheric condition and suction condition are different. So the given volume rate is FAD. Z  4 cylin ders; Double acting so y  2; FAD  30 m3/min 

30 3 m /sec  0.5 m 3/sec ; 60

P atm  1 bar; Tatm  25  273  298 K; n  1.3 N  310 r.p.m. ; k  3.5%  0.035 ;  mech  0.82; P 1  Patm  1 bar

[Atmospheric

pressure

and

suction

pressure are same]. T1  Tatm  12  298  12  310 K. Solution:  To Find Va P atm FAD Tatm

P P2 5 3



 P 1V a T1

2 n

P V = C on sta n t

P1 6 Vc

1 V4

4

S u ction Volum e (V a ) S w ept Volu m e (V s )

Fig.

v

4.48

Thermal Engineering - I

[atm - atmospheric conditions; suffix 1 stands for suction condition]  5 1  105  0.5 1  10  Va  298 310  V a  0.52013 m 3/sec  To Find m   P 1V a  mRT1  P 1V a 1  10 5  0.52013    0.58462 kg/sec m RT 1 287  310

To Find I.P Theoretical power (Indicated power) n1

  P2  n  n  m RT 1   P n1  1

 1  0.3

1.3   15  1.3   0.58462  0.287  310      1 1 0.3     195.67 kW

To Find B.P Brake power (Actual power required for motor) B.P 

I.P mech



195.67 0.82

i.e. Capacity of motor  238.62 kW To Find Cylinder Dimensions   P 2 1n   vol  1  k    1 P   1

Air Compressors

4.49

1    15  1.3 1  1  0.035    1   

 0.75397  Va Also vol   Vs   Va 0.52013   0.689856 m 3/sec Vs  vol 0.75397

 N yZ But V s  V s  60  Vs 0.689856  0.01669 m 3  Vs  N/60  y  Z 310/60  2  4 Vs  

  D 2  L  0.01669 m 3 4   D2  1.2D  0.01669 m 3 4  . . L  . D  1.2   

D  0.260653 m

L  1.2D  0.312783 m Problem 4.18: The free air delivery of a single stage reciprocating air compressor is 2.5 m3/min. The ambient air is at N.T.P condition and the delivery pressure is 7 bar. The clearance

volume

is

5% of

stroke 1.25

compression and expansion is pv

volume

and

law

of

 C. If L  1.2 D and the

compressor runs at 150 rpm, determine the dimensions of the cylinder.

(April / May - 2011 - AU)

4.50

Thermal Engineering - I

3s

P 2 =P 3

2

P

W

P1

4

1 VS

VC

V

Given

  FAD  Va  V 1  2.5 m 3/m in

NTP P 1  1.013 bar ; T 1  0  273  273 K   k  0.05 ; P 1 V 1  m RT1

Solution:

 P 1 V 1 1.013  10 5  2.5   m RT1 287  273  60 m  0.054

kg sec

1

   P2  n vol  1  k    1 P1    1

  7  1.25   1  0.05    1   0.8153    1.013  

Air Compressors

4.51

Displacement Volume  V a  Vs  y  Z  N  vol VS 

VS 

Va  volume  N  y  Z

. .  . y  1 single a cting   Z  1 single stage  

 2.5  D2  1.2 D 150  0.8153 4

D  0.2789 m  27.89 cm L  1.2 D  1.2  0.2789  33.47 cm Problem 4.19: A single stage double acting air compressor delivers air at 7 bar. The pressure at the end of suction stroke is 1 bar and temperature 25C. It delivers 2.2 m3 of free air per minute when the compressor is running at 310 r.p.m. The clearance volume is 5% of stroke volume. The pressure and temperature of atmospheric air are 1.03 bar and 20C. Determine volumetric efficiency, B.P. of compressor if mechanical efficiency is 85% and diameter and stroke of the cylinder if both are equal. Assume index of compression and expansion are 1.3 (Oct. 98, Madras University)

Given: y  2 for double acting; FAD  2.2 m 3/min; P2  7 ba r; P 1  1 b ar; T1  25  273  298 K; N  310 r.p.m ; Vc  0.05V s; P atm  1.03 bar; Tatm  20  273  293 K; mech  85 %; D  L; n  1.3.

4.52

Thermal Engineering - I

Solution: P atm FAD  T atm



 P 1V a T1

 5 1.03  10 5  2.2/60 1  10  V a  293 298  V a  0.038411 m 3/sec   P 1 V a  m RT 1  P 1V a 1  10 5  0.038411   0.044912 kg/sec  m RT 1 287  298

To Find vol    P 2  1 n vo l  1  k    1 P   1 1

7   1  0.05    1.3  1   0.82662 1    P P2 5 3

2 n

P V = C o n sta n t

P1 6 V4 Vc

4

S u c tio n Vo lu m e (V a ) S w e p t Vo lu m e (V s )

Fig.

v

Air Compressors

4.53

To Find Cylinder Dimensions  Va  vo l   Vs  0.038411 Vs   0.04647 m 3/sec 0.82662  N y But V s  Vs  60  Vs Vs    N  yZ   60     4.496862  10  3 

Z

0.04647  310   60   2  1  

  D2  L 4

. . [ . DL]

D  0.1789 m L  0.1789 m

To Find Theoretical Power: (I.P) n1

  P2  n  n m RT 1   Power   P n1  1

 1  0.3

1.3 7    0.044912  0.287  298    1.3  1  0.3 1   9.4349053 kW

To Find B.P Brake Power 

I.P 9.4349053   11.0999 kW 0.85  mech

Problem 4.20. A single stage double acting air compressor delivers 15 m3/min of air measured at 1.013 bar, 27C. The air is delivered at 7 bar. The conditions at the end of suction stroke

4.54

Thermal Engineering - I

are pressure 0.98 bar and temperature 35C. The clearance volume is 4% of stroke volume, the L/D ratio is 1.3 and the compressor

runs

at

300

rpm.

Calculate

the

volumetric

efficiency, cylinder dimensions and isothermal efficiency of the compressor. Take index of expansion and compression as 1.3 and R  0.287 kJ/kg.K

(Apr/May - 2011 - AU)

Given:

 P 1  0.98 bar ; T1  35  273  308 K ; Va  15 m 3/min

P atm  1.013 bar ; T atm  27  273  300 K ; P 2  7 bar ; k  0.04;

L  1.3; N  300 rpm D

Solution: 1

  P2  n v  1  k    1 P  1

1

   7  1.3  1   1  0.04    0.98    

v  85.85%

P

3

2 PV

1.3

=C

Free air 0.98 bar

4

1 V

Air Compressors

4.55

Volume of air Required at Inlet Conditions  P atm FAD  P 1 V a  Tatm T1  P atm FAD   T1 Va  Tatm  P 1  15 308 1.013 m3   0.2653 Va   0.98 sec 60 300

Displacement Volume  Volume rate 0.2653 m3  VS   0.3090 sec 0.8585 v . .    . y  2 for double acting  300 1  V S  D2  1.3 D  2  Z  1 for single stage  60 4 

0.3090 

 3 300 D  2.6  4 60

D  0.31163 m or 31.16 c m L  0.405 m or 40.5 cm n1

   P2  n n IP  P1 Va    P n1 1  

 1  0.3

1.3    7  1.3 1   0.2653  0.98  10 2     1.3  1    0.98  IP  64.68 kW

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Thermal Engineering - I

  P2  Isothermal work  P 1Va ln   P  1  7   0.98  10 2  0.2653 ln    0.98   51.12 kW

Isothermal efficiency 

51.12  79.03% 64.68

Problem 4.21: An air compressor takes in air at 5 bar and 30C and compress it according to law PV1.5  C. It is then delivered at a constant pressure of 20 bar. R  0.287 kJ/kg K. Determine (i) Temperature at end of compression (ii) Workdone and heat transfer during compression per kg of air.

Given: T 1  30  273  303 K ; T 2  ? ; P1  5 bar P 2  20 bar ; R  0.287 kJ/kg K ; PV1.5  C

Solution: (i) Temperature at end of compression. We know that 1.5  1

n1

n1

 P2  n  P2  n  20  1.5  303   481 K  or T2  T1     P T1 P  5   1  1 T2

 Temperature T2  481 K.

(ii) Workdone during compression  Workdone W   m RT1

n1

 n   P2  n 1   n  1   P1  

Air Compressors

 1.5    20   1  0.287  303      1.5  1    5 

1.5  1 1.5

4.57

 1 

Workdone W  153.24 kJ/kg of air Heat transferred during compression Q  P 1V 1  P2V2 Q  W  U   C V T2  T1 n1 

R T 1  T 2  n1

 C V T 2  T 1 

R    T 2  T 1  CV  n  1   0.287    481  303  0.718   1  1.5  Q  25.632 kJ/kg Problem

4.22: A

single

acting

reciprocating

compressor

24 cm  20 cm has following parameters under testing. Suction  pressure  2 ; Suction temperature  30C; Discharge pressure  12 bar; Discharge temperature  280C; Speed N  1200 rpm; Shaft power P  10 kW; mass of air delivered  2.7 kg/min. Calculate (i) the actual volumetric efficiency, the indicated power, the isothermal efficiency, the mechanical efficiency, the overall isothermal efficiency.

Given: P 1  2 bar ; P 2  12 bar ; T1  30  273  303 K, D  0.24 m L  0.20 m ; T 2  280  273  553 K  N  1200 rpm ; Power P  10 kW ; m  2.7 kg/min

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Thermal Engineering - I

Solution: (i) Actual volumetric efficiency vol We know that vol 

F.A.D 

Vd 

F.A.D Free air delivered  100  100   Displacement volum e Vd

 m RT 1 P1



2.7  0.287  303

2  10

2

 1.174 m 3/min

 2  D L  N    0.24 2   0.20   1200 4 4

D  0.24 m L  0.2 m  10.858 m 3/min FAD 1.174  100   100  10.81 %  vo l  Vd 10.858

(ii) Indicated power (I.P) We know that indicated power (I.P) n1

  P2  n   n  1 mRT1    P n1  1 

 P2   Also,  T1  P 1  T2

n1 n

T akin g ln on both sides,

 T2   P2  n1  ln  ln    n P T  1  1

Air Compressors

4.59

n  1 ln T 2/T1  n ln P 2/P 1 1

 ln T2/T 1  1 1 ln T 2/T1  or   1    n ln P 2/P 1 n  ln P 2/P 1   ln 553 /303  1 1  n  ln 12/2  

1  1  0.3357 n n

1  1.505 1  0.3357 1.505  1

1.505 2.7   12  1.505  0.287  303    I.P    1.505  1 60  2 

 1 

I.P  9613 Watts or 9.613 kW

(iii) Isothermal Efficiency  Isothermal power  m RT1 ln P 2/P 1 

2.7  12   0.287  303 ln    7.012 kW 60  2 

Isothermal power  7.012 kW isothermal Efficiency  

Isotherm al power Indicated power 7.012  100  73.04% 9.6

 iso  73.04 %

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Thermal Engineering - I

(iv) Mechanical efficiency me ch:  mech 

9.6 I.P  100  96%  100  10 S haft power

(v) Overall isothermal efficiency over iso  overiso 

Isothermal power 7.012  100   100  70.12 % Shaft power 10

Problem 4.23: A single stage single acting air compressor delivers 2.6 kg of air per minute at 10 bar. The temperature and pressure at the end of suction stroke are 40C and 1 bar. The bore and stroke of the compressor are 200 mm and 350 mm respectively. The clearance is 2% of swept volume. Assuming the index of compression and expansion to be 1.5. Find (i) vol (ii) power required if mech is 85% (iii) speed of compressor in rpm.

Solution:  Given: m  2.6 kg/min ; P2  10 bar ; P1  1 bar ; T1  40  273  313 K ; D  200 mm  0.2 m ; L  350 mm  0.35 m; n  1.5 ;  mech  85 % ; k  0.02

(i) Volumetric efficiency of compressor, vol 1

 P2  n We know that vol  1  k  k   P  1 1

 10  1.5  vol  1  0.02  0.02   0.927 or 92.7 %   1  (ii) Power required to drive the compressor n1

  P2  n  n m RT 1   Indicated Power   P n1  1

 1 

Air Compressors 1.5  1

2.6 1.5   10  1.5   287  313     60 1.5  1  1 

4.61

 1 

Indicated power  13481 J /s or 13.481 kW Power required to drive the compressor 

13.481 13.481   15.86 kW 0.85 mech

(iii) Speed of compressor  m RT1 2.6  287  313  F.A.D  P1 1  10 5  2.3356 m 3/min

Displacement volume  Vd 

 2 D L  N For s ingle acting 4   0.2 2  0.35  N 4

We know that vol   0.927 

or

F.A.D Displa ce me nt volume 2.3356   0.2 2  0.35  N 4

N  229.1 rpm

Speed of compressor 229.1 rpm Problem 4.24: A single stage double acting air compressor delivers air at 7.5 bar. The pressure and temperature at the end of suction stroke are 1 bar and 25C. It delivers 2.2 m3 of free air per minute when the compressor is running at 310 rpm. The Vc  0.05 Vs. The pressure and temperature of

4.62

Thermal Engineering - I

ambient air are 1.03 bar and

20C. Determine (i) vol (ii)

Dimensions of cylinder if L/D  1 (iii) I.P and B.P if mec h is 85%. Take nc  1.25 ; ne  1.3

Given: P 1  1 bar ; P 2  7.5 bar ; T1  25  273  298 K,  FAD  Vamb  2.2 m 3/min ; N  310 rpm; Vc  0.05 V s ; P amb  1.03 bar ; Tamb  20  273  293 K ,  mech  0.85, nc  1.25, n e  1.3

Solution: The P.V diagram is shown (i) Volumetric Efficiency vol  1

   P2  n vol  1  k    e1 P   1 1  1.3

  7.5  1  0.05     1 

P P 2 =P 3 3

2

 1 

 0.814  81.44% or V1  V4 We know that V c  Vs

Also we know that

P 1 =P 4 Vc V3 = Vc F ig

PV

1.25

PV

1.3

=C

=C

4 V s=V 1 -V 3

V1 V4 Veff=V 1 -V 4

1.3 (from diagram) P 2V1.3 3  P1 V 4

 P2   V4  V3   P  1

1/1.3

1

1

 7.5  1.3 . .  4.71 Vc [ . V3  Vc]  Vc   1  

 V 4  4.71 V c  4.71  0.05 V s  0.2355 Vs

V

Air Compressors

vol  

V1  V4 Vs



4.63

V c  V s  0.2355 Vs Vs

0.05 V s  V s  0.2355 Vs Vs

 vol  1.05  0.2355 

Vs Vs

 0.8145 or 81.45 %

(ii) Dimensions of cylinder L and D We know that   P amb V amb T1 1.03  2.2  298   2.305 m 3/mi n V1  P 1 Tamb 1  293 Volume of air delivered per min  V 1  V s   vol  N  2  1

. .  . y2  Z  1  

2.305  V s  0.8145  310  2 2.305  Vs   0.00456 m 3 0.8145  310  2   V s  D 2  L  0.00456 4 or

 . .  D 3  0.00456  . D  L 4 or D 3  5.805  10  3 or D  0.1797 m or D  17.97 cm D  L  17.97 cm

4.64

Thermal Engineering - I

(iii) I.P and B.P Workdone per cycle of operation   P2    n1 P 1V 1   W2  P  1   n1  1

n1  1 n1

 1  n2  1

 n2   P3  n  P 4V 4    2 P n 1  4  2

  1   

P 3  P2 and P 4  P 1 ; n 1  1.25 ; n2  1.3 V 1  Vc  V s  0.05 V s  V s  1.05 V s ; V 4  0.2355 V s 1.25  1

  7.5  1.25   1.25 5 1  10  1.05  0.00456    W/cycle  2    1.25 1   1  

 1 

1.3  1

  7.5  1.3   1.3  1  10 5  0.2355  0.00456    1  1 1.3  1    

 2 1188.08  275.47   1825.7 J/cycle  Indicated power 

W.D/cycle  N 1825.7  310  9.43 kW  60  1000 60  1000

Brake power B.P 

I.P

mech



9.43  11.09 kW 0.85

4.15 TWO STAGE COMPRESSION When the air is required to be compressed to a very high pressure, we cannot go for single stage compression. The air should be compressed in two or more stages.

Air Compressors

Inte rcoo ler

3

P2V T 2

P V2 T 3 H.P.C ylinder

Ist Stage P 1’ T 1

Cooling water

2

L.P.Cylinder

4.65

2nd S tage 1 Air in

4 P 3’ T4 Air out

Fig.4.10 Two stage reciprocating air com pressor w ith intercooler

Intercooling: Intercooler is introduced in between two cylinders to reduce the temperature of the air leaving the compressor at very high pressure. Also the temperature of air is cooled by intercooler to maintain the temperature. T3  T1 4.15.1. Complete (or) Perfect Intercooling If the temperature of air leaving the intercooler T 3 is equal to the original inlet temperature of air T 1, then the intercooling is known as perfect intercooling. If T 3 is not given, assume T 3  T 1. Then the intermediate  P 1P 3 pressure P 2  

4.15.2 Incomplete (or) Imperfect Intercooling If the temperature of air leaving the intercooler T 3 is greater than the original inlet temp. of air T1, then the intercooling is imperfect intercooling.

4.66

Thermal Engineering - I

Here T 3  T 1 4.15.3 Workdone intercooling.

when

perfect

and

imperfect

W  W for LP cylinder + W for HP cylinder. n1

n1

    P2  n   P3  n n n 1   P1 V1   1  P 2V 2     P P n1   n1  1  2

when the intercooling is perfect, T 3  T 1 i.e., P 1V1  P 2V 2 n1

  P2  n n P 1V 1   So, W   P n1  1

n1

 P3  n   P  2

n1

  P2  n n  mRT 1    P n1  1  P atmFAD  P 1VaLp  Also, T1 Tatm

 2 

n1

 P3  n   P  2

 2 

 Q  Heat removed in the intercooler  mC pT2  T 3  If m is not given, then q  CpT2  T3 4.15.4 Minimum Work Required for 2 Stages and multi stages The atmospheric pressure is fixed i.e. P 1 is fixed. The delivery pressure is fixed i.e. P 3 is fixed. So the workdone required will be minimum, if we get the intermediate pressure optimum.  i.e., if P 2   P1P3 then W is minimum.

Air Compressors

4.67

 Substitute P 2   P1P3 in the W equation, then

Minimum W for 2 stages   P3  n  P1V 1   2  P n1  1

n1 2n

  1  for 2 stages. 

Minimum W for 3 stages   P4  n  P1V 1   3  P n1  1

n1 3n

 1 

Minimum W for Z stages   Pz  1  n Z  P1V 1    P1 n1  

n1 zn

 1 

4.16 MULTISTAGE RECIPROCATING COMPRESSORS Multistage compression is a series arrangement of cylinders in which compressed air from the cylinder 1, becomes the intake air for the cylinder 2 which follows as shown in Fig. 4.11. H igh pressure air

Inter cooler

LP Cylinder 1 d 1 A ir in take

IP Cylinder 2

d 2

H P C ylinder 3

Inter cooler

Fig 4.11 Th ree stage com pressor

d 3

4.68

Thermal Engineering - I

The low pressure ratio in the low-pressure cylinder means that the clearance air expansion is reduced and the effective swept volume of this cylinder is increased since this cylinder controls the mass flow through the machine, because it is the cylinder which introduces the air into the machine, so there is greater mass flow through the multistage arrangement than the single stage machine. If an intercooler is installed between cylinders, in which the compressed air is cooled between cylinders, then the final delivery temperature is reduced. This reduction in temperature means a reduction in internal energy of the delivered air and since this energy must have come from the input energy required to drive the machine, this results in a decrease in input work requirement for a given mass of delivered air. Two stage and three stage compressors are common. The complexity of the machinery limits the number of stages. The diameters d 1  d 2  d3  gets reduced as the pressure goes on increasing the volume gets reduced. A General P.V diagram is shown Fig 4.12. Cycle 8156 -: Single stage compressor P D elivery 6 pressure HP P 3 or P d

3

Iso the rm al Inte rm ediate process 7 pressure P 2 Inta ke pressure P 1 or P S Fig 4.12

8

W ork saved 5 n PV = C P erfe ct intercooling 2

4

LP 1 v

Air Compressors

4.69

Cycle 8147,7456 -: Two stage without Intercooling Cycle 8147,7236 -: Two stage with perfect Intercooling Perfect intercooling means that after the initial compression in L.P cylinder,the air is cooled in an intercooler to its original intake temperature T1  T2. Case 1. Single - Stage Compressor Cycle 8156 is that of a single stage compressor neglecting clearance, for this cycle. n1

Workdone

  P5  n n W   P1V 1    P n1  1

 1 

n1

 P5  n Delivery Temperature T 5  T 1   P  1

Case 2. Two - Stage Compressor (i) Without Intercooling Cycle 8147 is low pressure compression Cycle 7456 is high pressure compression  The workdone n1

  P4  n n P 1V 1   W    P n1  1

 1  n1

  P5  n n  P4V 4    P n1  4

 1 

(ii) With Perfect intercooling Cycle 8147 is low pressure compression Cycle 7236 is high pressure compression

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Thermal Engineering - I

  P4  n Workdone W   P1V1    n1   P1 

n1 n

 1 

n1

  P3  n n  P V   n  1 2 2   P2 

 1 

Delivery temperature n1

n1  n

 P3 T3  T2   P  2

 P3  n  T1   P  2

. . [ . T1  T2]

Now since T 1  T 2 ; P 2V2  P 1V1 also P 2  P 4  Workdone n1

  P2  n n P 1V 1   W  P n1  1

 P3    P  2

n1 n

 2 

...(1)

4.16.3 Conditions for Minimum Workdone Differentiating (1) w.r.t to P 2 and equating to zero. n1

  P2  n d  P  1

dW  dP 2



n1

 P3  n   P  2

dP2 n1

1

P 1 

n1 n

  

0 

1  n1   P 2 n    n  n1

 P 3  n

n1  n1  10   P 2  n  n  

Air Compressors





P2

n1  1   n 

n1 P 1  n   1   P 2 n 

n1  P2   n  1 

 P2



1 n



 P 1P 3

 P 1P 3 n

2n  2 n

 P 1 P 3  n

n1  n 

 P 1P 3 n

 P 22 



  n1 1  n     P2

n1 n

 2n  1    P n 

 P 2 

or

 P3

n1    n 

4.71

n1

n1

n1

P 22  P1P 3

  Intermediate pressure P 2   P 1P 3  P2   P 1P 3 ; P 2  P4 ; P 2V2  P 1V 1 equation (1), we can get minimum workdone.

Substituting

n1

  P 3  2n  2n 1 P 1V 1   W min   P n1  1 

Case (iii) Multistage Compressor We know that for single stage compressor n1

  P2  n n P 1V 1   W  P n1  1

 1 

For double stage compressor

...(2) in

4.72

Thermal Engineering - I n1

   P 3  2n 2n 1 W P 1V 1    P n1   1   For three stage compressor n1

   P 4  3n 3n 1 P 1V 1   W  n1    P1   For Z stage compressor n1

  P Z  1   Zn Zn 1 P 1V 1   W  n1   P1   Z 4.16.4 Intermediate Pressures for compressor Running under Ideal Condition P2 P1



P3 P2



P4 P3



PZ  1 PZ

Stage

 y say

P 2  yP1; P 3  yP2  y2P1  P Z  1  yP Z  yZP 1

4.16.5 Heat Rejected per stage per kg of air For a single stage, the heat rejected is given by W

n C  C V T 2  T 1 [Without Intercooling] n1 P

Heat rejected per stage with perfect intercooling.  n  W  [C P  C V  ] T 2  T 1 per kg of air. n1

Air Compressors

4.16.6 The Change Compression

in

Entropy

in

First

4.73

Stage

 T2  n C p Cv ln  S2  S1   T n1  1

4.16.7 For Two Stage Compression S 2  S 1  C p  C v ln P 3/P 1  Cp  Cv   P3  S2  S1    ln   P 2    1 condition.

for minimum work

Problem 4.25: A two stage compressor takes in 3 m3 of air per minute at a pressure of 1 bar and temperature of 25C. It delivers the air at 9 bar. The compression is carried out in each cylinder according to the law PV1.2  C. The air is cooled to its initial temperature in intercooler. Neglecting clearance, find the minimum power required to drive the compressor.

Given:  V 1  3 m 3/min; P 1  1 b ar; P 3  9 bar; n  1.2. T1  25  273  298 K;

Solution: Since the air is cooled to its initial temperature, it is  P 1 P3 perfect intercooling. So, P 2     1 9  3 bar

To Find Minimum Work n1

   P3   n 2n  P 1V 1   1 W min  2   P n1   1

4.74

Thermal Engineering - I 0.2

1.2 3   9  1.2  2  2 1  1  102  0.2 60   1    12.0562 kW Problem 4.26: A single acting, two stage reciprocating air compressor with complete intercooling delivers 0.175 kg/sec of air at 16 bar from 1.013 bar. The compression and expansion follows the law PV1.3  constant. Calculate the power required to drive the compressor. Take R  0.287 kJ/kgK (Nov/Dec 11 - AU) P 1  1.013 bar  1.013  10 5 N/m 2 P 3  16 bar  16  10 5 N/m 2 T1  273 K   P3  n WZ  P1 V 1    P n1  1

n1 1  n Z

  1 

Z  No. of stages  2 P 1 V 1  mRT 1 W2

1.3  0.175  0.287  273  1.3  1 1.3  1

  16  1.3   1.013   



1 2

  1 

 44.62 kJ/sec P  44.62 kW Problem

4.27: 2 kg/s of air enters the LP cylinder of a two

stage, reciprocating air compressor. The overall pressure ratio is 9. The air at inlet to compressor is at 100 kPa and 35C.

Air Compressors

4.75

The index of compression in each cylinder is 1.3. Find the intercooler pressure for perfect intercooling. Also, find the minimum power required for compression, and percentage saving over single stage compression. Take R  0.287 kJ/kg K and Cp  1 kJ/kg K

(Nov/Dec - 2011 -AU)

Given: m  2 kg/sec ; P 1  100 kPa  1 bar T 1  35C  35  273  308 K ; n  1.3 P3 P1

 9 given

Solution: Intercooling Pressure P2 1 9  3 bar  P2    P1 P3   Minimum workdone 1

n1

  P3  Z  n n  P 1 V1   WZ  P n1  1

 1 

P 1 V 1  mRT 1 1

1.3 9  2  2  0.287  308    2 1.3  1 1

1.3  1 1.3

 1 

W  442.132 kJ/sec W min  442.132 kW

(ii) % Saving by two stage instead of Single Stage   P3  n W P1 V 1    P n1  1

n1 n

 1 

4.76

Thermal Engineering - I

P 1 V 1  mRT 1 1.3 9   2  0.287  308    1.3  1 1

1.3  1

1.3

 1 

W single  505.92 kW

% saving 

W single  W min W single



505.92  442.132 505.92

 0.126  12.6% Problem 4.28: A two stage air compressor compresses air from 1 bar, 20C to 42 bar. If the law of compression is PV1.35  constant and intercooling is perfect, find per kg of air (i) the work done in compression. (ii) the mass of cooling water necessary for abstracting the heat in the intercooler, if the temperature rise of the cooling water is 25C. (JNTU April/May 2013 - Set 2)

Given Data: 2 stage; P 1  1 bar; T1  20  273  293 K P 3  42 bar; n  1.35 ; Perfect intercooling; m  1 kg

Solution: Work done W2

2

n1  2n

  P3 n m RT1    P n1  1

 1 

1.35   42   1  0.287  293    0.35  1 

0.35 2  1.35

 1 

Air Compressors

4.77

 404.4 kJ/kg Q rejected in intercooler: Q  m C p T2  T3 P1 P3   P2      6.48 bar 1  42  n1  n

 P2   T1 P  1

T2

n1  n

 P2  T2  T1   P  1 0.35

 6.48  1.35  475.7 K T2  293    1  Q  1  1.005  475.7  293  183.56 kJ/kg

[ T3  T1  293 K ]

This heat Q  183.5 kJ/kg is taken away by cooling water. Q  183.56  mw  C pw Tw mw  

183.56 C pw  Tw 183.56  1.754 kg/s 4.187  25

Mass of cooling water m w  1.754 kg/s Problem 4.29: A single acting two stage reciprocating air compressor with complete inter cooling delivers 8 kg/min at 16 bar pressure. Assume an intake condition of 1 bar and 15C and the compression and expansion processes are polytropic with 1.35. Calculate (i) power required (ii) the isothermal efficiency.

(JNTU - H - May 2013)

4.78

Thermal Engineering - I

Given Data:

 2 stage; m  8 kg/min; P 3  16 bar

P 1  1 bar; T1  15  273  288 K; n  1.35

Solution:

Power 



n1  2n

  P3 2n  mRT 1    P n1  1

 1 

2  1.35 8   16   0.287  288     0.35 60  1 

0.35 2  1.35

 1 

Polytropic power  36.76 kW  P3   Isothermal power  m RT1 ln   P  1 

8  16   0.287  288  ln   60  1 

 30.56 kW

Isothermal

 

Isothermal pow er Po lytropic po wer

30.56  0.8312 36.76 isothermal  83.12%

Problem

4.30: A

two-stage

single

acting

reciprocating

3

compressor takes in air at the rate of 0.2 m /s. The intake pressure and temperature of air are 0.1 MPa and 16C. The air is compressed to a final pressure of 0.7 MPa. The intermediate pressure is ideal and intercooling is perfect. The

Air Compressors

4.79

compression index in both the stages is 1.25 ad the compressor runs at 600 r.p.m. Neglecting clearance, determine (i) The intermediate pressure, (ii) The total volume of each cylinder, (iii) the power required to drive the compressor, and (iv) the rate

of

heat

rejection

in

the

intercooler.

Take

Cp  1.005 kJ/kg K and R  287 J/kgK. (JNTU - August - 2014 - (Set 3))

Given Data:

  2 stage; V a  0.2 m3/s  V 1;

P 1  0.1 MPa  1 bar; T 1  16  273  289 K; P 3  0.7 MPa  7 bar; n  1.25 ; N  600 rpm;

Solution: (i) To find P 2 P2 

P1 P3  

 1 7  2.65 bar 

[Since intermediate pressure is ideal and perfect intercooling] (ii) Total volume of each cylinder Swept volume for LP cylinder    . . [ .  vol is not given ] V a  V1  V s  0.2 m 3/s  0.2 V   0.02 m 3/cycle Vs  N/60 600/60 V s LP  0.02 m3/cycle

Swept volume for HP cylinder V s LP V s HP



P2 P1

 2.65

4.80

Thermal Engineering - I

V s HP 

V s HP 2.65



0.02  7.55  10  3 m 3/cycle 2.65

V s HP  0.00755 m 3/cycle n1

   P 3  2n  2n (iii) Power  P1 Va   1  n1   P1   

2  1.25 7  100  0.2    0.25 1

0.25 2  1.25

 1 

 42.96 kW  To find m   P 1 V a  m R T1  P1 Va  100  0.2 m   0.2411 kg/s RT 1 0.287  289

(iv) Rate of heat rejection in intercooler  Q  m C p T2  T1 n1

n1

 P2  n   P T1  1 T2

 P2  n  T2  T1   P  1 0.25

 2.65  1.25 T 2  289   351.2 K   1  Q  0.2411  1.005 351.2  289  15.1 kW Q  15.1 kW Problem 4.31: A two-stage single acting reciprocating air compressor draws in air at a pressure of 1 bar and 17C and

Air Compressors

4.81

compresses it to a pressure of 60 bar. After compression in the L.P. cylinder, the air is cooled at constant pressure of 8 bar to a temperature of 37C. The low pressure cylinder has a diameter of 150 mm and both the cylinders have 200 mm stroke. If the law of compression is pv1.35  C, find the power of the compressor, when it runs at 200 r.p.m. Take R = 287 J/kg K.

(JNTU - August 2014 (Set - 2))

Given Data: 2 stage; P 1  1 bar; T1  17  273  290 K; P 3  60 bar; P 2  8 bar; T3  37  273  310 K; D LP  0.15 m; L HP  L LP  0.2 m; PV 1.35  C; N  200 rpm; R  0.287 kJ/kgK; IP  ?

Solution: Swept volume of LP cylinder  Vs  

 2 D L 4

  0.152  0.2 4

 3.53  10  3 m3  Swept volume rate of LP cylinder  Vs  N yZ V s  Vs  60

[y  1 for single acting; Z  1 for single LP cylinder]  200 11 V s  3.53  10  3  60  0.0118 m 3/s

4.82

Thermal Engineering - I

Since volumetric  is not given,   V a  V s  0.0118 m 3/s  To find mass flow rate of air : m   P 1 V a  m RT 1  100  0.0118  0.0142 kg/s m 0.287  290

To find indicated power   Power  W LP  W HP for imperfect intercooling . . [ . T1  T3] n1



  P2  n  m RT1    P n1  1

n

 1  n1  n

  P3  n m RT 3    P n1  2



1.35 8  0.0142  0.287  290    0.35 1 

0.35 1.35

 1 

 1 

1.35   60   0.0142  0.287  310    0.35  8 

0.35 1.35

 1 

 3.257  3.343  6.6 kW Problem 4.32: A two stage air compressor with perfect intercooling takes in air at 1 bar pressure and 27C. The law of compression in both the stages is pv1.3  constant. The compressed air is delivered at 9 bar from the H.P cylinder to an air receiver. Calculate per kg of air (i) the minimum work

Air Compressors

4.83

done, and (ii) The heat rejected to intercooler. (JNTU - August 2014 - Set 1)

Given Data: 2 stage; P 1  1 b ar; T1  27  273  300 K PV 1.3  C; P3  9 bar;

Solution: To find minimum work: W min  2 

n1   P 3  2n

n mRT 1    P n1  1

 1  0.3

2

1.3 9  1  0.287  300     0.3 1

2  1.3

 1 

[m  1 kg ]

 215.32 kW To find optimum intermediate pressure P2 P2 

P1 P3  

  1 9  3 bar

To find Q in the intercooler: Q  m Cp T2  T3

Here T 3  T 1 since it is perfect intercooling. Hence Q  m Cp T 2  T 1 To find T 2  P2    P T1  1 T2

n1 n

n1

 P2  n  T2  T1   P  1

4.84

Thermal Engineering - I 0.3

2

 3  1.3  386.57 K T 2  300   1 Q  m C p T 2  T1  1  1.005  386.57  300  87 Q  87 kJ /kg

Heat rejected in the intercooler Problem 4.33: A three stage reciprocating air compressor compresses air from 1 bar and 17C to 35 bar. The law of compression is PV1.25  constant and is same for all stages of compression. Find the minimum power required to compress 15 m3/min of free air. Also find the intermediate pressures. Assume perfect intercooling and neglect the clearance. (Apr/May-2008 - AU) (JNTU - Jan/Feb - 2015 Set- 3)

Given: 3 stage; P 1  1 bar; t1  17C P 4  35 bar; n  1.25; FAD  15 m 3/min 

15 3 m /sec 60

Solution: P 2  ?; P 3  ?; perfect inter cooling

Minimum work for 3 stages

3

  P4  n  P1 V1    P n1  1

n1 3n

 1 

To find Intermediate pressures,

Air Compressors

P2 P3 P4 for perfect     intercooling  P 1 P 2 P3 

P4 P3



P3 P2



P2 P1



P P4

PV

P3

P4  P2     P1  P1 

P2

3

3 2

1

P1

1

 P4   P P1  1

3

V 1

 P4  3 P2  P 1   P  1 1

 35  3 1   3.27 bar  1  P3 P2



P4 P3

P 23  P 4 P 2 P3 

P4 P2  

 35  3.27   P 3  10.7 bar

Minimum power required for 3 stages

    

n1

   P 4  3n  n  1 3  P 1 V1    n1    P1  0.25

3

15   35  3  1.25

1.25  1  10 2  60   1  0.25

 100.3 kW

=C

4

P4 P1

P2

1 .2 5

4.85

 1 

Thermal Engineering - I

4.86

Problem 4.34: A 3 stage compressor is used to compress air from 1 bar and 27C to 20 bar. The compression index in all the stages is n  1.5. Neglecting clearance and assuming perfect intercooling, find out the I.P required to deliver 20 m3 of air per min measured at inlet conditions and intermediate pressures also. Take R  287 J/kg K

Solution: Given: P 1  1 bar ; P 4  20 bar ; n  1.5 ; R  287 J/kg K T1  27  273  300 K ; Z  3

For perfect intercooling P2 P1 

P4 P3



P3 P2





P3 P2 P2 P1





P4 P3

P1

3

P1

1

1/3

 20  3 1   2.714 bar  1 

P22 P1

P3 P2

 P4  or P2  P1   P  1

P3 

P4

PV

1 .5

=C

4

P4

P4  P2     P P1  1

P 22  P 3  P1

P

3 2

1 V

Air Compressors

P3 

4.87

2.7142  7.36 bar 1 1.5  1

n1 1

 P4  n . z  20  1.5  300  Also T 2  T 1    P1  1   



1 3

 418.48 K

n1

 P2  n (or) T 2  T 1   P  1

0.5

 2.714  1.5  300    1   418.48 K

Mass of air handled  P 1V 1 1  10 5  20   23.23 kg/min m  RT1 287  300 Total workdone in 3 stages n1

   P4   n nZ  P1V 1   1 WZ  P n1   1  0.5

3

20   20  1.5  3 1.5  1  1  10 2  60   1  0.5 

 118.486 kW

or Indicated power I.P  118.486 kW Problem 4.35: A multi-stage air compressor is to be designed to evaluate the pressure from 1 bar to 200 bar, such that stage pressure will not exceed 5. Determine (i) Number of stages (ii) Exact stage pressure ratio (iii) Intermediate pressure.

4.88

Thermal Engineering - I

Solution: (i) Number of stages Z Assuming perfect intercooling between stages and the condition of minimum work of compression in multistage compression is PZ  1 PZ

 PZ  1  P1 

1

Z  

...(1)

Given: PZ  1 PZ

4 ...(2)

Substitute (2) in (1) 1

 PZ  1  Z 4  P1   1

1  200  Z 4  or ln 4  Z ln 200 1   Z

ln 200  4  3.82 ~ ln 4

No of stages Z  4 (ii) Exact stage pressure ratio PZ  1 PZ

 PZ  1  P1 

1

1

 Z  200  4     3.76  1  

Exact pressure  P Z  1  3.76  ratio PZ 

Air Compressors

4.89

(iii) Intermediate Pressures P 1  1 bar ; P 4  1  P 5  200 bar P4  1

P4 Also

Also



P5

200  53.19 ba r  3.76 or P 4  P4 3.76 P4 P3 P3 P2

 3.76 or P 3 

 3.76 or P2 

P4 3.76

P3 3.76



53.19  14.147 bar 3.76



14.147  3.762 bar 3.76

 The pressures are P 1  1 bar ; P 2  3.76 ba r ; P 3  14.147 bar P 4  53.19 bar ; P 5  200 bar Problem 4.36: A 3 stage air compressor supplying air for an engine has a capacity of 15 m3 of free air/min and is driven by an engine at 200 rpm. The pressure at suction to L.P cylinder is 1 bar and at delivery at H.P cylinder is 100 bar. The clearance ratio is for L.P and H.P cylinders are 0.04 and 0.07. Assuming the temperature at the end of suction in all the cylinders is 27C. Perfect intercooling, stage pressure in geometric progression and the law of compression PV1.5  C, calculate the swept volume of each cylinder. Free air conditions are 1 bar and 15C

Solution: The P  V diagram is shown in figure.

4.90

Thermal Engineering - I

We know that P4 P3



P3 P2



P2 P1

P P4

PV

P3



P3 P2



P2 P1

=C

1

4

 Z pressure ratio P3

P4

1 .5



P4 P1

 Z3

2

P2

1

1

3

 P 4  3  100  3 Z     4.6415 P  1   1

P1

1 V Fig

 P2  ZP 1  4.6415  1  4.6415 bar P 3  ZP 2  4.6415  4.6415  21.54 bar 1

vol L.P  1  kLP  kLP Z

n

 1  0.04  0.04 4.6415 

1 1.5

 0.9287

1

vol H.P   1  kHP  kHP z

n

 1  0.07  0.07 4.6415 

1 1.5

 0.8752

Swept volume for Each cylinder Volume of free air to suction condition of L.P cylinder P 1V 1 T1



V1 

P amb Vamb Tamb 1  10 5  15  300 288  1  10

5

 15.625 m 3/min

Air Compressors

4.91

Given: [P amb  1 bar ; V amb  15 m3/min ; T amb  15C ; T1  27C, P 1  1  10 5]  V 1  15.625 m 3/min

Swept capacity of L.P cylinder 

15.625 15.625   0.084 m 3 vol L.P  N 0.9287  200

Volume of free air at the suction conditions of L.P cylinder 

1  10 5  15  300

288  4.6415  10 5

 3.366 m3/min

Swept capacity of L.P cylinder 

3.366  0.0192 m 3 0.8752  200

Volume of free air reduced to suction condition of H.P cylinder 

1  10 5  15  300

288  21.54  10 5

 0.7254 m 3/min

Swept capacity of H.P cylinder 

0.7254  0.00414 m 3 0.8752  200

Problem 4.37: A single acting 2 stage air compressor with complete intercooling delivers 10.5 kg/min of air at 16 bar. The suction occurs at 1 bar and 27C. The compression and expansion follows for both stages PV1.3  C. Calculate (a) power

4.92

Thermal Engineering - I

required (b) isothermal efficiency (c) the free air delivered. (d) heat transferred in the intercooler (e) if the clearance ratio for LP and HP cylinders are 0.04 and 0.06 respectively, find swept and clearance volume for each cylinder. Assume speed of compressor as 440 r.p.m.

(JNTU - January 2014 - Set 1)

Solution:  P 1P 3 The optimum intermediate pressure P 2       4 bar 1  16

To Find Power n1

  P 3  2n  2n  1 m RT1   Power   P n1  1  1.3  1

2  1.3  10.5    16  2  1.3    1   0.287  300   1  60 0.3       49.232 kW P P 3 =16

(bar) 3 4

H .P

P 2 =4

1 -3

PV =C 2

6

1

2

5 L.P

P 1 =1 V c (L.P) V c (H .P)

1

7 V s (H .P)

v V s (L.P)

Air Compressors

4.93

To Find Isothermal Efficiency Isothermal Efficiency 

P owerisotherm al Powerpolytropic

 P3   Powerisothermal  m RT1 ln   P  1 

10.5  16   0.287  300 ln   60  1 

 41.776 kW  isothermal 

41.776  0.84855  84.86 % 49.232

To Find FAD Since the atmospheric conditions are not given, here  V a  FAD   P 1V a  m RT1  10.5  287  300 Va  1  10 5  9.0405 m 3/min FAD  9.0405 m 3/min

To Find Q in the intercooler  Q  m C p T2  T3 [T 3  T1 since we assume it is perfect intercooling] Q

10.5  1.005 T2  300 60

4.94

Thermal Engineering - I n1

 P2  n   T1 P  1 T2

n1

 P2  n So , T 2  T 1  P  1

0.3

4  300    1.3  413.103 K 1 Q

10.5  1.005 413.103  300 60

 19.892 kW

To Find Vs and V c    P 2 1n vol LP  1  k    1 P   1 1

4   1  0.04    1.3  1   0.92381 1     92.381 %    P 3 1n  vol HP  1  k    1 P   2 1

   16  1.3  1   0.88571  1  0.06    4     88.571 %

For LP Cylinder

 Va  vol LP  0.92381   Vs

 9.0405   60      0.1631 m 3/sec Vs  0.92381

Air Compressors

4.95

 N V s  Vs  60  Vs 0.1631 Vs    N   440   60   60       0.022241 m 3

Clearance volume  4% o f V s  0.04  0.022241  8.8965  10  4m 3

HP Cylinder

 Va  vo l HP  0.88571   Vs

 9.0405   60     Vs   0.170118 m 3/sec 0.88571

Swept volume V sH P 

 Vs

N Stage pressure ratio  60



0.170118  440  4   60 

 5.7995  10  3m 3

Clearance volume V cH P  0.06  V sHP  3.48  10  4m 3

4.96

Thermal Engineering - I

Note L LP  L HP D 2LP D 2HP

 Pressure ratio

V sLP VsHP

 Pressure ratio

Problem 4.38: A two stage single acting air compressor delivers air at 20 bar. The pressure and temperature of air before compression in L.P. cylinder are 1 bar and 27C. The discharge pressure of L.P. cylinder is 4.7 bar and the air is cooled to 27C. The diameter and stroke of LP cylinder are 40 cm and 50 cm respectively. The clearance volume is 4% of stroke in both cylinders. The speed of compressor is 200 r.p.m. Assuming the index of compression and expansion in both cylinders are 1.3, find the indicated power to run the compressor and the heat rejected in intercooler per minute.

(Oct.99, Madras University)

Given: y  1 for single acting; Z  2 for two stage; Delivery pressure

=

20

T1  27  273  300 K;

bar; LP

Initial cylinder

pressure delivery

P 1  1 b ar

pressure

P 2  4.7 bar; T3  27  273  300 K. D  0.4 m; L  0.5 m; k  0.04; N  200 r.p.m. ; n  1.3.

Solution: To Find vol   P 2 1n   vo l  1  k LP    1 P  1 

Air Compressors

4.97

1

   4.7  1.3  1  0.04   1  1     0.90846  90.85 %  Swept volume of LP cylinder  Vs  D 2L 4 

  0.4 2  0.5 4

 0.062832 m3

Swept volume rate of air in LP cylinder  N yZ V s  Vs  60  0.062832 

200 11 60

 0.20944 m 3/sec . . [ . y  1 for single acting; for single LP cylidner Z  1]

Actual   V a   vol  V s

volume

rate

of

air

in

LP

 0.90846  0.20944  0.1903 m 3/sec  To Find Mass Rate of Flow of Air m   P 1V a  m RT1  1  10 5  0.1903  0.220985 kg/sec m 287  300

cylinder

4.98

Thermal Engineering - I

To

Find

Indicated

Power:

[T 1  T 3

So

Perfect

Intercooling] n1

  P 3  2n   n m RT 1   1 I.P  2   P n1  1  0.3

1.3   20  2  1.3   0.220985  0.287  300    1 2  0.3  1    68.0904 kW

To Find Power: Alternate method   Power  W LP  W HP n1

  P2  n m RT1     n1   P1  n



n1

  P3  n   n mR T3   1  P  2  n1



1 

0.3

1.3   4.7  1.3   0.220985  0.287  300   1   0.3  1   0.3

1.3   20  1.3    0.220985  0.287  300   1  0.3   4.7    35.39  32.72  68.107 kW

To Find Heat Rejected in Intercooler ‘Q’ n1

0.3

 P2  n  4.7  1.3   428.76642 K ; T2  300    P1 T1  1     Q  m Cp T2  T3 T2

 0.220985  1.005 428.77  300  28.598 kW  1715.9 kJ/min

Air Compressors

4.99

Problem 4.39: A two-cylinder single-acting air compressor is to deliver 16 kg of air per minute at 7 bar from suction conditions 1 bar and 15C. Clearance may be taken as 4% of stroke volume and the index for both compression and re-expansion as 1.3. Compressor is directly coupled to a four-cylinder four-stroke petrol engine which runs at 2000 r.p.m. with a brake mean effective pressure of 5.5 bar. Assuming a stroke-bore ratio of 1.2 for both engine and compressor and a mechanical efficiency of 82% for compressor, calculate the required cylinder dimensions. (May/June 2012 - AU)

Solution Amount of air delivered per cylinder 

16  8 kg/min 12

P 1  1 bar, T1  15  273  288 K

Suction conditons:

From the gas equation, P 1V 1  V 4  mRT 1   m RT 1 8  0.287  288 3 or V1  V4  m   6.61 /min  V a P1 1  10 2 1

   P2  n vo l  1  k    1 P   1 1

 1  0.04  7 1.3  1     0.863  86.13%

4.100

Thermal Engineering - I

 Va Also, vol   Vs

 6.61  7.6744 m3/min Vs  0.8613  Vs  Vs  N  1 Vs 

If

. . [ . y  1]

7.6744  0.00384 m 3 2000

L  Length of stroke of compressor, and D  diameter of the cylinder of the compressor, then L  1.2D ...(given) 

 2 D  L  Vs 4

or

 2 0.00384  4 D  1.2 D  0.00384 or D 3  4   1.2

i.e

D  0.1596 m or 159.6 mm L  159.6  1.2  191.5 mm

Now indicated power of the compressor n1

  P2  n  n  m RT 1     P n1  1

 1  1.3  1

16 1.3 7    0.287  288    1.3 1.3  1 60 1 Brake power of the engine 

Now,

66 

  1   54.14 kW 

54.14 54.14   66 kW 0.82 mech ne P mb L e AN 60

Air Compressors

If

De 

diameter of the engine cylinder

Le 

length of the stroke of engine  1.2 D e,

ne 

number of engine cylinders

 4  5.5  10 2  1.2 D e    D 2e 4 Then, 66  60  D 3e 

i.e.,

66  60 2

5.5  10  1.2    2000

4.101

   2000 

 0.0009549 m 3

D e  0.0985 m or 98.5 mm L e  1.2  98.5  118.2 mm

Problem 4.40: A six - cylinder double acting compressor is required to compressor 40 m3/min of air at 1 bar and 30C to a pressure of 20 bar. Determine of size of motor required and cylinder dimension for the following data given: Speed of compressor  400 rpm ; k  4%, L/D  1.2 ; mech  82%, n  1.5. Assume no pressure change in suction values and the air gets heated by 20C during suction stroke.

Solution: The P  V diagram is given below n1

  P2  n n  P 1  V 1  V 4   Net workdone   P n1  1 V 1  V 4  40 m3/min 

 1 

 40  0.667 m 3/sec  V a 60 1.5  1

1.5   20  1.5  1  10 5  0.667   W  1.5  1  1 

 1 

4.102

Thermal Engineering - I

W  343054 J/s

P P 2 =P 3 3

Theoretical Power  W  343.05 kW

2

n

P V =C

(i) Motor power 343.05 343.05   418.36 kW  0.82  mec h

P 1 =P 4 Vc

(ii) Cylinder dimensions L and D

V3 = Vc

4 V s=V 1 -V 3

1 V

V1 V4 Veff=V 1 -V 4

F ig

We know 1

vol

 P2  n  Pi Ta  1kk   P P T  1  a i 

‘i’ for inside and ‘a’ for atmospheric conditions. 1

1  273  30   20 1.5    0.7  vol   1  0.04  0.04       1  273  30  20   1  vol  70%  V a  Vs  y  Z  N  vol  Va  Vs  y  Z  N  vo l

Swept volume per one cylinder  Vs 

. . [ . y  2 ; Z  6]

1 1 40    0.0119 m 3 0.7 6 2  400  2  D  L   D 2  1.2D  0.0119 4 4  D 3  0.1262 or D  0.232 m

 Diameter D   232 mm and L  1.2  232  280 mm

Air Compressors

4.103

Problem 4.41: A two stage air compressor with complete intercooling delivers air to the mains at a pressure of 30 bar, the suction conditions being 1 bar and 30C. If both cylinder have the same stroke, find the ratio of cylinder diameters, for the efficiency of compression to be a maximum. Assume the index of compression be 1.5 (JNTU - Jan/Feb 2015 - Set - 4)

Given: P 1  1 b ar ; T1  30  273  303 K ; P 3  30 bar V 1  Volume of L.P cylinder; V 2  Volume of H.P cylinder

Solution:

Now

V1 V2



 2   D 1L 4  2   D 2L 4

or

D1 D2



 

V1

V2

From the curve 1  2 PV 1.5  C

1.5

P 1V 1  P2V2

1.5

1/1.5

 P2   or  V 2  P1  V1

P P3 4

3

For maximum efficiency P 2   P 1P 3

P2 5

P2    1  30   5.48 bar



V1 V 2

 5.48 

1 1.5

2

2’

1

P1 6

V

 3.1082

V2

V1

4.104

Thermal Engineering - I n1

 P2  n T 2  T1   P  1

1.5  1

 5.48  1.5  30  273     1 

 534.2 K

Constant pressure process 2  2 [T 2  T 1  303 K] [P 2  P 2] 

V2 T2 V1

Now

V2



D1 D2







V 2

V 2

or

T 2 V1



V2 

V2

V 2 V2

 

V1 V2



T2 T2



534.2  1.763 303

 3.1082  1.763  5.4798

  5.4789  2.34

Problem 4.42: In a single acting two stage reciprocating air compressor, 5 kg of air per min is compressed from 1 bar and 27C through a pressure 9. Both the stages have the same pressure ratio, and the law of compression and expansion in both stages is PV1.5  C. If perfect intercooling is present, Calculate (i) The indicated power (ii) The cylinder swept volume. Assume that the clearance volume of both stages are 3 % of their respective swept volumes and that the compressor runs at 310 rpm.  Given: m  5 kg/min ; Ps  1 bar ; Ts  27  273  300 K ;

Pd Ps

9

Air Compressors

P2 P1

Also

Pi Ps

3; Pd



Pi

P3 P1

4.105

P

9

Pd

; n  1.5 ; V c  0.03 V s ; Pi

4

3

2

6 5

N  310 rpm PS

Solution:

7

1 V

(i) Indicated power   P3  n  m RT 1   IP  2   P n1  1

n1 2n

 1  0.5

5 1.5 2   0.287  300  9 2  1.5  1   19.04 kW 0.5 60   Indicated power  19.028 kW (ii) The cylinder swept volume For L.P cylinder   m.R.T 1 5  287  300  4.305 m3/m in  V a  V1  V 4  5 P1 1  10   P1  vol  1  k   P  2

1

n   1  

1  1  0.03  3 1.5  1   

 vol  0.967 or 96.7 % Vs 

 Va  vol  N



4.305  0.01436 m 3/cycle 0.967  310

4.106

Thermal Engineering - I

 The swept volume for L.P cylinder V s L.P  0.01436 m 3/cycle

Swept volume for H.P cylinder For high pressure stage P 2  3 bar   m  R  T3 Volume drawn in V HP  T3  5  287  300  1.435 m 3/m in V HP  5 3  10  Swept volume for H.P stage 

1.435 1.435  vol  N 0.967  310

V s H.P  0.004787 m 3

Also we can obtain V s H.P 

V s L.P

3



0.01436  0.004787 m3 3

Problem 4.43: A two stage double acting air compressor, operating at 300 rpm, takes in air at 1 bar and 27C. The size of L.P cylinder is 200  300 mm. The stroke of H.P cylinder is same as that of L.P cylinder. The clearance volume of both cylinders is 5%. The L.P cylinder discharges the air at a pressure of 5 bar. The air passes through the intercooler so that it enters the H.P cylinder at 27C and 4.8 bar. Finally it is discharged from the compressor at 20 bar. The value of n in

both

the

cylinders

is

1.5;

CP  1.005 kJ/kg K

and

R  0.287 kJ/kg K Calculate. (i) The heat rejected in the

Air Compressors

4.107

intercooler (ii) Diameter of H.P cylinder (iii) Power required to drive H.P cylinder.

Given: *1 Z2;y2 N  300 rpm : P 1  1 bar , T 1  300 K; D LP  200 mm ; L LP  300  L HP ; V c  0.05 V s ; P 2  P 3  5 bar; P 5  P 8  4.8 bar ; P 6  P 7  20 bar; n  1.5 ; C P  1.005 kJ/kg K ; R  0.287 kJ/kg K

Solution: Swept volume of L.P cylinder V s L.P  

  D 2LP  L L.P  300  2 4

  0.2 2  0.3  300  2 4

P 20 bar 5 bar 4.8 bar

 5.656 m 3/min 1 bar

1

 P2  n vol  1  k  k   P  1 5  1  0.05  0.05   1  0.9038

7

6 3

2 5

8

4

1 V

1/1.5

Vc    k  V  0.05  s  

 V1  Volume of air drawn at stages  0.9038  5.656  5.112 m 3/min

4.108

Thermal Engineering - I

Mass of air/min  P 1V 1 1  10 5  5.112    5.937 kg/min m RT 1 287  300  P2  T2  T1     P1 

n1 n

5  300   1

1.5  1 1.5

 513 K

(i) Heat rejected in the intercooler Heat rejected in the intercooler  Q  m C P T2  T1  5.937  1.005 513  300 Q  1271 kJ/m in

(ii) Diameter of the H.P cylinder Volume of air in H.P cylinder per min   m RT5 5.937  287  300   1.065 m 3/min V 5H.P  P5 4.8  10 5 Swept volume of H.P cylinder Vs 

Now

1.065 1.065  1.178 m 3/min  0.9038  vol

  D 2HP  L HP  300  2  1.178 4   D2HP  0.3  300  2  1.178 4

D 2HP  8.333  10  3 o r D HP  0.091 m  Diameter of H.P cylinder D HP  91 mm

. . [ . y  2]

Air Compressors

4.109

(iii) Power required to drive H.P cylinder n1

  P6  n  n m RT 5   W  P n1  5

 1  0.5



1.5 5.937    20  1.5  0.287  300   1   0.5 60 4.8     15.569 kW

Power required P   15.57 kW Problem 4.44: A single acting two stage compressor with perfect intercooling delivers 20 kg/min of air at 25 bar. The suction occurs at 1 bar and 27C. The compression and expansion are reversible, polytropic index n  1.5. Calculate (i) Power to drive compressor (ii) isothermal (iii) F.A.D (iv) Heat transferred in intercooler (v) Swept volume for L.P and H.P cylinders if their clearance ratio are 0.04 and 0.06 and compressor runs at 400 rpm.

Given: y1;Z2 P 1  1 bar ; P3  25 bar ; T 1  27  273  300 K ; n  1.5 ; m  20 kg/min ; k LP  0.04 ; kHP  0.06 ;

Solution:  P 1P 3   1  25 Intermediate pressure P 2      5 bar

(i) Power required to drive compressor Workdone in two stages with perfect intercooling n1

  P 3  2n  2n  1 m.R.T 1   W  P n1  1 

4.110

Thermal Engineering - I 1.5  1

2  1.5 20    20  2  1.5   0.287  300   1   60 1 1.5  1    W  111.508 kW

(ii) Isothermal Efficiency iso  We know that Isothermal workdone  P3    m.R.T1 ln    P1  

20  20   0.287  300  ln    85.978 kW 60  1 

W iso  85.978 kW

Isothermal Efficiency iso 

85.978  100  77.10 % 111.508

(iii) F.A.D Free Air Delivered (F.A.D)  m .R.T 1 20  0.287  300    17.22 m3/m in 2 P1 1  10 (iv) Heat transferred to Intercooler Q  Heat transferred to intercooler   Q   m  C P  T 2  T5  m .CP T2  T1 n1

 P2  n T2  T1    P  1 Q

. . [ . T 1  T5 ]

1.5  1

5  300   1.5 1

 513 K

20  1.005 513  300   71.355 kW 60

Air Compressors

(v) Swept volume Swept volume for L.P cylinder F.A.D  V1  V3  speed   vol L.P 1

 P2  n  vol L.P  1  k LP  k LP   P  1 1

5  1  0.04  0.04   1.5  0.9230 1

Swept volume for LP cylinder V s L.P 

17.22  0.0466 m 3 400  0.9230

Swept volume for H.P cylinder V S H.P 

F.A.D Stage pr. ratio  s peed   vol H.P

 P3 vol H.P  1  kHP  kHP  P  2  1  0.06  0.06 5 V S H.P 

1

n  

1 1.5

 0.8845

17.22  0.009734 m 3 5  400  0.8845

4.111

4.112

Thermal Engineering - I

4.17 VARIOUS TYPES OF COMPRESSORS ROTARY POSITIVE DISPLACEMENT COMPRESSORS Rotary compressors are used to compress large quantity of air at a relatively low pressure. In rotary air compressors, the air is trapped in between two sets of engaging surfaces (lobes) and the pressure of air is increased by squeezing action or back flow of air. 4.17.1 Different Types Displacement Compressor

of

Rotary

1.

Roots blower compressor

2.

Vane type blower compressor

3.

Lysholm compressor.

4.17.2 Different Types of Displacement Compressors 1.

Centrifugal compressors

2.

Axial flow compressors.

Rotary

Positive

Non-positive

Note: ‘Displacement Compressor’ means the air is compressed by being trapped in the reduced space formed by two sets of engaging surfaces.

4.18 ROOTS BLOWER Roots blower compressor consists of two rotors with lobes. These lobes are rotating in air tight casing. It has inlet and outlet port. The shape of the lobes is epicycloid or hypocycloid or involute. During the rotation, volume of air at atmospheric pressure is trapped between left hand lobe and casing. The further rotary motion of the lobes delivers the entrapped air into the receiver. So more and more flow of air into the receiver increases its pressure.

Air Compressors

4.113

4.18.1 Back Flow of Air When the rotating lobe uncovers the exit port, the high pressure from receiver flows back into the casing. The high pressure air and trapped air gets mixed. The back flow process continues till the trapped air and receiver air attains the equal pressure. Then the pressure of air trapped in the pocket (in between lobe and casing) is increased at constant volume entirely by back flow of air. Finally, the high pressure air from receiver is delivered to the requirement. P 1  Inlet pressure; P 2  Delivery pressure

  1.4 for air; V  Volume of air compressed in a space. This process is carried out four times per revolution of the driving shaft. To R e ceiver D is cha rg e

lobe lobe E xtern al driv e g ea r V

Fig .4.13

Inle t R o ots B lo w er, tw o lo be rotors

4.114

Thermal Engineering - I

For this process the PV diagram is shown in Fig. Work done  P 2  P 1 V

per

cycle

P P2 Irreversib le pressure rise of a ir due to b ack flo w from re ceiver

 Work done per revolution  4 P 2  P 1 V  If Vs is the volume per sec

P1 V S = 4V V

at P 1 and T1, then  Power  P 2  P 1 V s The ideal compression process is a reversible adiabatic (i.e., isentropic) process. Ideal power (or) Isentropic power 1

   P2     P1 Vs    P1 1  

i.e.,

 Roots efficien cy   or Ise ntropic efficiency   or Compresso r efficiency 1 

   P2   P 1 Vs    1   P 1   P2  P 1 V s

    



 1 

Isentropic power Actual pow er

 1 

1

     P 1 V s  rp  1   1   P 1 V s rp  1

P2    Pressure ra tio   where rp  P1  

Air Compressors

4.115

1

   1    rp    Roots efficiency    1  rp  1   

Roots blowers are useful supercharging of I.C. engines.

for

scavenging

and

The efficiency of roots blower decreases with the  P2  increase in pressure ratio   P  1 Problem 4.45: A roots blower compresses 0.08 m3 of air from 1 bar to 1.5 bar per revolution. Calculate compressor efficiency. (JNTU January 2014 - Set 1)

Given data: V s  0.08 m 3; P 1  1 bar; P 2  1.5 bar

Solution: Roots compressor efficiency 

Isentropic w orkdon e Actual work don e 1

  P2    P1 Vs    P 1  1  V s P 2  P 1 0.4

 1 

  1.4  100  1.5 1.4  1    0.4  150  100   0.8598  85.98%

4.116

Thermal Engineering - I

Problem 4.46: A roots blower compresses 0.06 m3 of air from 1 bar to 1.45 bar per revolution. Calculate compressor efficiency.

(January - 2014 - JNTU - Set 2)

Given Data: V s  0.06 m 3; P 1  1 bar; P2  1.45 bar;

Roots compressor efficiency 

Isentropic w orkdone Actual workdone 1

   1   r     1 r1   0.4

  1.4 1 1.4  1.45  0.4  1.45  1   0.8711  87.11% Problem 4.47: A roots blower handles free air of 0.5 m3/s at 1 bar and 27C and delivers air at a pressure of 2 bar. Determine the indicated power required to drive compressor and isentropic efficiency.

(JNTU April - May 2013 - Set 1)

Given Data:  V  0.5 m3/s; P 1  1 bar; T1 27  273  290 K P 2  2 bar;  Indicated power  P 2  P 1 V  2  1  100  0.5  50

Actual power = 50 kW

Air Compressors

   P2   P1 V   Isentropic power   P 1  1



1 

1.4 2  100  0.5    0.4 1

4.117

 1 

0.4 1.4

  1   38.33 

Isentropic power = 38.33 kW isentropic 

Isentropic power 38.33   0.7665 50 Actual power

 76.65%

4.19 VANE TYPE BLOWER OF COMPRESSOR This compressor comprises of a disk rotating eccentrically in an air tight casing. It has inlet and outlet port. The disc has many number of vanes. When the rotor rotates, the air is trapped in the pockets formed between the vanes and casing. The compression is occurred due to decreasing volume and back flow of air. First of all, the rotary motion of the vanes compresses the air. Next, when the Inle t rotating vane uncovers D ischarge the exit port, the high pressure air from Fig.4.15 receiver flows back into the pocket. This process is called back flow process.

4.118

Thermal Engineering - I

P 2 (D e live ry P re ssu re)

P2

Irre versib le p ress ure rise o f a ir d u e to b ac k flo w from re ceive r

B

Pi

i(interm e diate p re ssure)

A

1

P1 VS

Fig:4.16 P -V d iag ram fo r vane b low er

V

Fig. 4.16 shows the PV diagram. The work done per revolution with N v vanes is given by the following expression: 1

  Pi    P1 V s   W  Nv P  1  1

  1   N v P 2  P i Vi 

Problem 4.48: Compare the work inputs required for a Roots blower and a Vane type compressor having the same induced volume of 0.05 m3/rev., the inlet pressure being 1.01325 bar and the pressure ratio 1.6. For the Vane type assume that internal compression takes place through half the pressure range.

Solution: Inlet pressure, P 1  1.01325 bar P2  1.6 Pressure ratio, P1

Air Compressors

 P 2  1.6 P 1  1.6  1.01325  1.62 ba r

4.119

P P2

For the Roots blower, refer Fig. (a) P1

Work done/rev.  P 2  P 1 V s

VS Fig:(a )

V

 1.62  1.01325   10 2  0.05  3.03 kJ

Intermediate pressure P i 

1.62  1.01325  1.32 bar 2

. . [ . Internal compression takes place through half the pressure range]

For the Vane type Refer Fig. (b) Work required  Area A  Area B   Pi   P1 V s   Now, Area A   P 1  1

  1 

 1 

1.4  1



1.4    1.32  1.4  1  kJ/rev  1.01325  10 2  0.05    1.01325 1.4  1   

 1.39 kJ/rev

Area B  P2  P i V B Now, 



P1 V S  Pi VB

4.120

Thermal Engineering - I P (bar) P 2 =1 .6 2 B VB

P c =1 .32



A

P V =C

P 1 =1 .0 13 25

VS

VB V B =0.5 V S

V (m 3 )

VS Fig:(b)



P1  VB     Pi  VS  1

1

 P1    1.01325  1.4 VB  V S    0.05   P  1.32   i   0.041 m 3

Area B  1.62  1.32   100  0.041  1.23 kJ/rev Compression work  Area A  B  1.39  1.23  2.62 kJ /rev Problem 4.49: A vaned compressor handles free air of 0.6 m3/s at 1 bar and compresses to 2.3 bar. There occurs 30% reduction in volume before the back flow occurs. Determine the indicated power required and isentropic efficiency. (JNTU - April/May 2013)

Air Compressors

P 2 P2 B

I

Pi



A

P V =C

P1

VS

3 0% R e du ctio n

V B = 0.7 V S VS

Given Data:  V s  0.6 m 3/s; P 1  1 bar; P 2  2.3 bar V B  0.7 V S  P 1 V S  Pi V B   P 1 V S  P i 0.7 Vs    Vs   Pi  P1    0.7 Vs   1

1 0.7 

 1.65 Pi  1.65 bar

3

V m /3

4.121

4.122

Thermal Engineering - I

Power required in Area A 1

   Pi     P 1 Vs    P 1  1



 1 

1.4   1.65   100  0.6    0.4  1 

0.4 1.4

 1 

Power A  32.3 kW Power required in Area B  P 2  P i  V B

. . [ . V B  0.7 V S]

 2.3  1.65   100  0.7  0.6  27.3

Power B  27.3 kW Indicated power  A  B power  32.3  27.3  59.6 kW To find isentropic 1

   P2    P1 VS   Isentropic Power   P 1  1

 1  0.4



1.4   2.3   100  0.6    0.4  1 

1.4

 56.42 kW  isentropic 

Isentropic po wer 56.42  0.9467  Actual power 59.6  94.67%

 1 

Air Compressors

4.123

Problem 4.50: A rotary compressor receives air at 1 bar and 17C and delivers it at a pressure of 6 bar. Determine per H of air delivered, workdone by the compressor and heat exchanged with the jacket water when the compression is isothermal, isentropic and by the relation PV1.6  constant. (JNTU - Apr/May 2013 (Set 3)) & (January - 2014 (Set 3))

Given Data: P 1  1 bar; T1  17  273  290 K; P 2  6 bar m  1 kg; W  ?; Q  ?

Isothermal compression:  P2  W  mRT1 ln    P1  6  1  0.287  290  ln   1 W  149.13 kJ /kg

For isothermal process, U  0 Q  W  U

So,

Q  W  149.13 kJ/kg

Isentropic compression: 1

   P2    1  m RT 1   W  P 1   1



1.4 6  1  0.287  290    0.4 1

0.4 1.4

W  194.74 kJ /kg

 1 

4.124

Thermal Engineering - I

and Q  0 for reversible adiabatic process (isentropic) PV1.6  C : Polytropic compression

  P2  n  mRT1   W  P n1  1

n1 n

 1  0.6



1.6    6  1.6 1  1  0.287  290    0.6 1   

 212.62 kJ/kg Q  m C n T 2  T 1 C n  Polytropic specific heat  1.4  1.6   n   Cv     0.718   n 1  1  1.6    C n  0.2393 kJ/Kg K n1

 P2  n   P T1  1

T2

0.6

T 2  290 6 1.6  567.81 K Q  m C n T 2  T 1  1  0.2393 567.81  290  66.48 kJ /kg

Air Compressors

4.125

4.20. CENTRIFUGAL COMPRESSOR The figure shows the centrifugal compressor with double sided impeller. 



  

 

 

It consists of curved radial vanes which are attached to shaft and rotate. The impeller is a disc fitted with radial vanes. The casing is surrounding the rotating impeller. The diffuser is housed in a radial portion of the housing. Air enters the eye of impeller. Due to the centrifugal action of impeller, the air moves radially outwards with the guidance of impeller vanes. The impeller transfers energy to air by increasing its pressure. Then the air enters diverging portion called diffuser. Here, the kinetic energy is converted into pressure rise further. Nearly half the pressure rise is achieved by impeller and remaining by diffuser. A pressure ratio of 5:1 is obtained by single stage centrifugal compressor and for higher pressure ratio’s multi stage compressors are used.

Centrifugal compressor is a head or pressure producing machine. It has larger frontal area than the axial flow compressor. It is capable of producing pressure ratio of about 4 : 1 per stage.

4.126

Thermal Engineering - I

C om pre sse d a ir d elivery

D ischa rge S cro ll

A ir flow

Im p elle r

A ir flow C asing R adial van es

D iffu ser Passages

C om pre sse d a ir d elivery

Fig. 4.17 Cen trifu gal Com pressor.

Tip Van es Im p elle r e ye S haft

Fig. 4.18 mp eller (Single - Eyed) and Radial vanes o f Centrifugal Compresso r.

Air Compressors

4.127

4.20.1 Principle of Operation Fig. 4.19 shows the components of a centrifugal compressor. Fluid enters into the impeller through an accelerating nozzle and a row of inlet guide vanes (IGV). The accelerated flow in the nozzle enters the IGV and it directs the flow in the desired direction at the entry of the impeller.

3

2

Volu te

D iffu ser

C asing S hroud Im p elle r E ye Indu cer

Im p elle r

1 Dt Dh IG V D riving Sh aft N ozzle Fig:4.19

The impeller is made by one piece consisting of both the inducer section and a largely radial portion. The inducer receives the flow between the hub and tip

4.128

Thermal Engineering - I

diameters of the impeller eye and passes it on to the radial portion of the impeller blades. The tips of the blades can be shrouded to prevent leakage. The impeller discharges the flow to the diffuser through a vaneless space. Here the static pressure of the fluid rises further on account of the deceleration of the flow. The flow at the periphery of the diffuser is collected by a spiral casing known as volute which discharges it through the delivery pipe.

4.21 VELOCITY AND PRESSURE VARIATION The changes of pressure and velocity of air passing through the impeller and diffuser are shown in Fig. 4.20

P3 P re ssu re rise in d iffu ser P re ssu re rise in im pe ller

P2 P re ssu re C urve

P1 C2 Velo city d ecrea se in d iffu ser

Velo city increase in im pe ller Velo city C u rve

C3

C1 Fig:4.20 Variatio ns of pressure and velocity of air passing through im peller and diffuser

Air Compressors

4.129

4.22 STATIC TEMPERATURE AND TOTAL HEAD (OR) STAGNATION TEMPERATURE Since the velocities in centrifugal compressors are very large, total head quantities should be considered. The total head quantities include the kinetic energy of the air passing through the compressor. Consider a horizontal passage of varying area (Fig. 4.21). Applying steady flow energy equation to the system for 1 kg of air flow assume Q  0; W  0 h1 

C 21 2

 h2 

1

Q =0 2

C 22

2

W =0 Fig:4 .21

C 21 C 22  Cp T2  Cp T1  2 2 Cp T 

C2  constant 2

Temperature ‘T’ is called the static temperature T is measured by the thermometer when the thermometer is moving at the air velocity. If the moving air is brought to rest under reversible conditions, the total kinetic energy of the air will be converted into heat energy, increasing the temperature and pressure of the air. This increased temperature and pressure of the air is known as “stagnation” or “total head” temperature and pressure. The total head temperature and pressure are denoted by a suffix ‘O’.

4.130 Thermal Engineering - I

 Cp T 

C2  Cp T 0 2

where T 0 is known as total head or stagnation temperature In otherwords, T 0  T 

C2 C2  h0  h  2 2C p

To find the total head pressure    T0  1

   P  T 

P0

T 02

T 02

2

P2

T2

2

T 0 2’

2

C2 2 cP

C2 2 cP

0 2’

T 2’

2’

P1 T01 2

C1 2 cP

01

T1

1 S

Fig. 4.22 Total head and static P and T on Ts diagram

Air Compressors

4.131

where, P  S tatic pressure; T  Static temperature, P 0  To tal hea d pressure or Stagnation pressure, and T 0  To tal hea d temperature or Stagnation temperature

4.22.1 Stagnation State and Stagnation Properties The state of a fluid attained by isentropically decelerating it to zero velocity at zero elevation is referred to as the stagnation state. It is often used as a reference state. The properties of the fluid at the stagnation state are the stagnation properties of the gas. 4.22.2 Stagnation Enthalpy Stagnation enthalpy of a gas or vapour is its enthalpy when it is adiabatically decelerated to zero velocity at zero elevation. Therefore, h0  h 

C2 2

For an adiabatic energy transformation stagnation enthalpy remains constant.

process,

In a power generating turbomachine, W is positive so that  h0 is negative, i.e., the total enthalpy of the flowing fluid decreases from the inlet to the exit. In a power absorbing device (compressors), mechanical energy input occurs, so that the stagnation enthalpy of the fluid increases from the inlet to the exit. Hence W will be negative.

4.132 Thermal Engineering - I

4.23 STEADY-FLOW ENERGY EQUATION C 22 C 21  q  h2  gZ 2   W1  2 h 1  gZ 1  2 2

... (1) [per unit kg mass] This is the steady flow energy equation for a control volume or an open system. This will be rewritten for processes in various turbomachines. Most of the compressible flow turbo machines, such as turbines, compressors and blowers are assumed to be an adiabatic machines i.e., q  0. In these machines, the change in potential energy  Z is negligible as compared to changes 1 2 [C  C 22]. 2 1

in

enthalpy

 h

and

kinetic

energy

Therefore Eq (1) yields h1  

C 21 2

C 22

 h2  W 2

Energy compressor.

equation

...(2) [W = Work transfer] for

a

turbine

and

 C21   C22     h2   W 1  2   h1  2   2    h01  h02   h 0

For the turbine output work W 1  2  C p T 01  T 02 

...(3)

For the compressor input work W 1  2  C p T 02  T 01 

...(4)

Air Compressors

4.133

4.24 EULER’S EQUATION - (ENERGY TRANSFER) Energy transformation can occur in both stator and rotors of turbomachines. Energy transfer can occur only in its moving or rotating elements i.e rotors/impeller. An expression for estimating the amount of energy transfer taking place in a turbomachine is derived below. The Fig. 4.23 shows the velocity triangles at the entry and exit of a general turbomachine. All the velocity vectors are shown in same plane and are assumed to remain constant.

u2

Cw

C f2 2 C r 2 2

2

O utlet VelocityTriang le 2 C 2 E x it 

E ntry

C1 1 r2

r1

C f1

C r1 1

Cw1 u1

In tet Velocity Trian gle Fig 4.23

The angular speed of rotor is  radians per second. 

2 N 60

[N: Speed of rotor / impeller in r.p.m]

The peripheral velocity of the blades at entry & exit are

4.134 Thermal Engineering - I

1

1

2

2

Fig. 4.24

u1  u2 

D 1 N 60

D 2 N 60

in m/s

in m/s

4.24.1 Velocity components at the entry and exit of the rotor In order that the fluid enters and leaves the blade passage without shock, the relative velocities C r1 and C r2 at inlet and outlet are in the direction of the respective tangents to the blade. The

relative

velocities

make

angles

1 and 2

respectively with u1 and u 2. The absolute velocities of the fluid at inlet and outlet to the blades are obtained by a vectorial combination of C r and u, u1 and u 2 1 and  2 make angles with respectively.

Air Compressors

4.135

The components of absolute velocities parallel to the tangential directions are C w and C w and those to the 1

2

radial directions are C f and C f . 1

2

Clearly, C f and C f are associated with the flow rate 1

2

through the impeller / runner.   The three velocity vectors C , u and C r are related at

a section by a simple vector equation.    C  u  Cr

The absolute velocity C at both the entry and exit has a tangential component C w and a radial component C f.

The torque on the rotor/impeller (exerted by the impeller on the fluid) is obtained by employing Newton’s second law of motion for the change of moment of momentum. Torque  Rate of change in moment of momentum. The tangential momentum at a given station is   mass   m  tim e   

  m Cw

 and its moment is  m rV w  Torque on the impeller is given by   T  m2 r 2 C w  m 1 r 1 C w 2

1

   For a constant flow machine m 1  m 2  m   Torque T  m [C w r2  C w r1] 1 2

4.136 Thermal Engineering - I

If the value of the torque given by this equation is positive [r2 C w  r1 C w ], it applies to a head producing 2

1

tubomachines (pumps, fans, blower, compressors etc).   T c  m [C w r 2  C w r 1 ] 2

1

The work done is given by Work  Torque  Angular Velocity of the rotor For compressors, pumps, etc.  Work  W c  T c    m [Cw r2   C w r1 ] 2 1  W c  m [Cw u2  Cw u1] 2

...(1)

1

. . [ . u 1  r1  u 2  r2  ]

The above equation is known as Euler’s equation for pump and compressor (or) Euler’s work. The workdone per kg of air W c  C w2 u 2  C w1‘ u1

...(2)

 h02  h01  C p T02  T01  u2 C w2

C r1 1

1

C 1 = C f1 C w1 = 0

u1 In let Velocity diag ram (a)

2

2 Cf2 C2

Fig:Velo city D iag ram s

Cr2

O utlet Velocity diag ram (b)

Air Compressors

4.137

Using the inlet and outlet velocity triangles, we have C 2r1  u 21  C 21  2u 1 C w1  u1 C w1  C 21  C 2r1  u21/2

...(i)

C 2r2  u 22  C 22  2u 2 C w2  u2 C w2  C 22  C 2r2  u22/2

...(ii)

Inserting the values of C w2  u 2 and C w1  u 1 from the above expressions (i) and (ii) in eqn. (2), we get C 22  C21 C2r1  C r22 u 22  u 21  W  2 2 2 First term Second term Third term







...(3)

The first term represents the increase in K.E. of 1 kg of working fluid in the impeller which will be converted into the pressure energy in the ‘diffuser’. The second term represents the pressure rise in the impeller due to ‘diffusion action’ (since the relative velocity decreases from inlet to outlet). The third term represents the pressure rise in the impeller due to ‘centrifugal action’ (since the working fluid enters at a lower diameter and comes out at a higher diameter).

4.25 IMPELLER BLADE SHAPE - BACKWARD, RADIAL AND FORWARD BLADE IMPELLERS The vanes of a radial outward flow machines e.g. a centrifugal pump, fan or a compressor can be set at different outlet angles to the direction of the local periphery velocity. The impeller is then classified according to the direction of the blade angle as shown in Fig. 4.26 (Backward, Radial and Forward)

4.138 Thermal Engineering - I

4.25.1 Impeller Blade shape The exit vane shape of the centrifugal compressors are generally any one of the three configurations. (i) Backward curved (ii) Radial and (iii) Forward curved If the angle between the rotor blade tip and the tangent to the rotor at the exit is acute i.e., 2  90, the vanes are backward curved vanes. If this angle is a right angle 2  90 the blade is said to be radial. If it is greater than 90 [2  90] the blade is forward curved as shown in Fig. 4.26.

2 < 9 0



2 > 9 0

o

o

u 

2 = 90

o

(ii) FO R W A R D

(i) B A C K W A R D

u 

Fig:4.26

(iii) R A D IA L

Air Compressors

4.139

The velocity triangles at the inlet and exit of centrifugal machines are shown in Fig. 4.27. It is assumed that there are no guide vanes, therefore C 1 will be radial and 1  90 and Cw1  0. Since there is no whirl velocity at inlet, the particular condition is “zero whirl and swirl” at inlet. The Fig. 4.27(a) shows the entry and exit velocity triangles for backward curves vanes with zero whirl at inlet. Fig. 4.27(b) shows radial curved vanes in which exit vane angle 2 is 90.

C r2 = C f2 2

2

2

C w 2 =u 2

C1 =

C

f1

1  1

Fig: 4.27 (a) B ackw a rd C urve Van es w ith Z ero W h irl at Inle t

1

Fig: 4.27 (b) R a dial C urv e Vanes

Fig (a)

Fig (b)

1  90; C w1  0

2  90; C r2  C f2 and

u 1  u 2; 2  90 ,

C w2  u2

W act  C w2 u 2  C w1 u1  C w2 u2

1

4.140 Thermal Engineering - I

Fig. 4.27 (c) shows the velocity triangles for forward curved vanes [2  90] with zero whirl at

2

2

inlet.  1  90 ;

C w1  0

C f1  Cf2;

u1  u2 1

W  Cw2 u2

4.26 VELOCITY TRIANGLE AND WORK DONE BY THE CENTRIFUGAL COMPRESSOR The work done by the impeller on the air is obtained by drawing the inlet and outlet velocity triangles as shown in Fig. 4.28. Let, C1



D1



u1



Cr Cf

1

1



Absolute velocity of the entering air  r1  Diameter of the impeller at inlet    2  Mean velocity of the impeller at inlet (or) peripheral velocity at inlet Relative velocity of air to the wheel at inlet



Velocity of flow at inlet and

C2, D 2, u 2, Cr , C f  Corresponding values at the outlet 2

N 1

 

2

Speed of the impeller in rpm Blade angle at inlet

Air Compressors

4.141

u2 Cw2 2

2 C f2

C r2

C2

O utlet Velocity diag ram

o utlet 

in le t

f1

=C 0 C 1 1= Cw 1

r1

r2

1

1

u

C r1

Inlet Velocity diag ram (a)

Fig 4.28 Velocity Triangles at Inlet and Outlet

1



Angle at which air leaves the impeller

2



Blade angle at outlet

 The power by the impeller on the air per sec per m kg of air flow.   m  C w u 2  C w1 u 1  ... (4) 2  

As the air enters the impeller radially which means that, the absolute velocity of air at inlet is in the radial direction and hence fluid angle 1  90  and V w1  0. Therefore, the equation (4) becomes.

4.142 Thermal Engineering - I

 Power  m Vw u 2

[work/unit mass] ... (5)

2

 Power by the impeller on air per sec.   m Vw u2 watts 2

... (6)

4.27 IMPORTANT FORMULAE  1. Power required for m kg/s of air flow  m C w2 u2 P kW, 1000 

If the blade is u2= C w2 radial (ideal case), then the velocity diagram at the C r2 =C f2 C2 outlet of the impeller is as Fig. 4.29 shown in Fig. 4.29  As C w2  u 2, the work done per m kg/s of air flow is given by W  u22

Since the air cannot leave the impeller at a velocity greater than the impeller tip velocity, the maximum work supplied per kg of air flow per second is given by the above eqn. 

Now write the steady flow energy equation at the inlet and outlet of the impeller, assuming the heat transfer through the impeller is zero. [Q  0]

Air Compressors

4.143

SFEE h1 

C22  Z1 g  W  h 2   Z2 g  Q 2 2

C 21

Z 1  Z 2; Q  0; C 21 C 22  h1   W  h2  2 2  C 22   C 21     h1   W   h2  2   2     C 21  C 22    C p  T1    C p  T2  2C p  2C p    . . [ . h  C p T]  Cp T 02  C p T01  C p T 02  T 01 

Multiply & Divide by T01 1     P02   T2   P  T 02    2   W  C p T 01   1   cp T01  1 T   01     P 01    T1   P  1  1  

  P 02  C p T 01    P   01 

   1   C p T 01  rp0  

1 

    1    

 1 

where rp0 is the pressure ratio based on stagnation pressures.

4.144 Thermal Engineering - I

2. Width of Blades of Impeller and Diffuser  Let, m  Mass of air flowing per second, b 1  Width or height of im peller at inlet, C f1  Velocity of flow a t inlet of the impeller,

1  density of air at inlet r1  Radius of impeller at the inlet,  Then, m  1  (Volume of air flowing per second)

 m  1  2r1 b1  C f1

But as the air is entering radially, C f1  C 1   m  1 2 r1 b 1  C 1

i.e.,

b1 

 m 1 21 r1 C 1

...(1)

...(1a)

Similarly the width of impeller blade at the outlet can be found by using suffix 2 in eqn.  m  2 2r2 b2  C f2 ...(2) The width or height of the impeller blades at the outlet and height of diffuser blade at the inlet should be same theoretically. The width of height of the diffuser blades at the outlet, is given by  m  d 2 rd bd  C fd ...(3) where suffix ‘d’ represents the quantities at the outlet of the diffuser.

Air Compressors

4.145

If, n  Number of blades on the impeller, and t  Thickness of the blade,  m  1 2 r1  nt b1 C f1

...(4)

 m  2 2 r2  nt b2 C f2

...(5)

 m  d 2 rd  nt bd C fd

...(6)

3. Isentropic Efficiency of the Compressor The following losses occur when air flows through the impeller: (i)

(ii) (iii)

Friction between the air layers moving with relative velocities and friction between the air and flow passages. Shock at entry. Turbulence caused in air.

The above losses cause an increase in enthalpy of the air without increase of pressure therefore the actual temperature of air coming out from the compressor is more than the temperature of air if it is compressed isentropically. The actual work required for the same increase in pressure is more due to irreversibilities. The isentropic efficiency isen 

Isentropic work h02   h 01  Actual work h 02  h 01 

T 02   T01 T02  T01

...(7)

4.146 Thermal Engineering - I

T 02

T 02

2

P2

T2

2

T 0 2’

2

C2 2 cP

C2 2 cP

0 2’

T 2’

2’

P1 T01 2

C1 2 cP

01

T1

1 S

If C 1  C 2, then isen 

T2  T1 T 2  T1

...(8)

Isentropic efficiency “Isentropic efficiency” of rotary compressor is defined as the ratio of isentropic temperature rise to actual temperature rise. Isentropic efficiency  isen  

Isentropic te mperature rise Actual temperature rise

T02   T 01 ...based on stagnation temperature T 02  T01 T 2   T1

T2  T 1

...based on static temperature

Air Compressors

4.147

During compression process, work has to done on the impeller. The energy balance equation will be as follows: C p T1 

C 21 2

 C p T2 

C 22 2

W

[ sign indicates work input to the system is negative] or

C p T 01  C p T02  W or W  Cp T02  T01 

T02   T01 Isentropic efficiency, isen  T02  T01 cp T 02   T01 Isentropic work h02  h 01   or isen  A ctual work h 02  h 01 cp T02  T01 

Thus the isentropic efficiency of a rotary compressor may be defined as the ratio of isentropic compression work to actual compression work. 4. Slip factor The difference between u2  C w2 is known as slip. Slip factor s is defined as the ratio of actual whirl component C w2 and the ideal whirl component u 2 Slip Factor For an ideal impeller, the fluid is assumed to follow the blade contour, so that the blade exit angle 2 is used to find out the actual work transfer. But the fluid always leaves the vanes at an average angle which is less than the geometrical blade angle. This is because of the fact that, due to its inertia, the air trapped between the impeller vane is reluctant to move around with the impeller and this

4.148 Thermal Engineering - I

results in higher static pressure on the leading face than on the trailing face.

Cs

Theo re tical A ctu al

C2

C

r2 The air will C f2 thus tend to flow 2 2 around the edges of 2 " the vanes in the Cw2' clearance space C w2 u2 between the impeller and the casing. This Fig.4.30 is clearly a loss of efficiency and clearance must be kept as small as possible. This tendency of air not to flow with the impeller with its speed is known as slip and is denoted by  s. The greater the number of

vanes, the smaller is the ‘slip’. Therefore, it is necessary in design to assume suitable value for slip factor s. From the diagram, when 2 decreases, C w2 and W act will also decrease. We know that, W th  C w2 u2 W act  W th  W act   s C w2 u2

where  s  slip factor and is always less than 1. s 

W act C w2 u 2



C p T02  T01  C w2 u 2

4.149

Air Compressors

5. Work factor Theoretical work done per kg of air is given by C p T02  T01   u2  C w2

Therefore the actual work is obtained by multiplying u2 C w2 by a factor  w known as work factor or power input factor.  C p T02  T01   w u2 C w2

The actual work input to the air is greater than the theoretical value due to friction between the casing and the air carried around by the vanes. In order to take this into account “work factor”, w is introduced, so that the actual work done on the air becomes Work factor w 

Actual work supplied Theoretical work supplied

 W act  w C w2 u 2  C p T02  T01 

6. Pressure Co-efficient p It is defined as the ratio of isentropic work to Euler work. p 

Isentropic w ork C p T02   T01  Euler work Cw2 u2

7. Stage Work 1. Specific Work W act  C w2 u2  C w1 u1  h 0

2. The flow co-efficient at the impeller exit

u2 C w2 2

2 C f2 C2

C r2

Fig.4 .31 O u tlet Velo city d iag ram

4.150 Thermal Engineering - I

f 

C f2 u2

3. Pressure co-efficient  p or  

W act

u2 Cw2

C f2 tan 2  u2  C w2 cot 2 

u2  C w2 C f2

C w2  u2  C f2 c o t 2 W act  u2 C w2 [zero whirl at inlet   1  0; Cw1  0]  W act  u2 [u 2  Cf2 c o t 2]  u22 [1  f cot 2] W act

u22

 1  f c o t  2

The theoretical performance characteristics of different types of impellers are shown in figure. The backward curved and radial blade impellers gives stable characteristics whereas forward curved vanes gives unstable flow conditions.

Pressure co-efficient ( ) (o r) P

  p or  

ard F orw

( 2 > 9

o 0 )

o

R a dial (2 =90 ) B ack

w a rd

( 2 < 9

0 o)

Flow co -efficien t ( f ) Fig:4.32

Air Compressors

4.151

8. Stage Pressure Rise The static pressure rise in a centrifugal compressor stage occurs in the impeller, diffuser and the volute. The static pressure rise across the diffuser and volute occurs due to the energy transformation processes accompanied by a significant deceleration of the flow. The fluid is assumed to be a perfect gas and isentropic process.  W act  h 0  Cp T0   T 0  1 2  T0   2  C p T 0   1  1  T01   

We know that, 1

T0 

 P0   2 2    T01  P 0  1  

 [rp0

1 

 1]

where rP0  Stagnation pressure rise 

P0

2

P0

1

9. Enthalpy-Entropy Diagram Fig. 4.33 shows an enthalpy-entropy diagram for a centrifugal compressor stage. The flow processes are i  1  Accelerating nozzle 1  2  Impeller 2  3  Diffuser 3  4  Volute

4.152 Thermal Engineering - I

O4

O3

h 0 2 =h 0 3 =h 0 4 O2

C3 2

0 4’ 4’

h

C4 2

2

4

2

C2 2

2

3

3’

W a ct

W ideal 2 2’ Oi C1 2

2

Ci 2

O1

2

1 i 1 s Fig:4.33

Accelerating Nozzle i  1 Only energy transformation takes place, therefore, the stagnation enthalpy remains constant. C 21 h0  h0  h1  i 1 2

But the stagnation pressure loss is due to the increase in entropy and losses. [P 0  P 0 ]. i

1

Only energy transfer occurs in the impeller blade passages. Therefore the stagnation enthalpy is not constant. h 0  h 0 . The difference in enthalpy is used to drive 2 1

the impeller.

Air Compressors

4.153

In the diffuser 2  3 and volute 3  4, static pressure rises and the stagnation enthalpy remains constant from station 2 to 4. The stagnation pressure decreases progressively. i.e., h 0  h 0  h 0 2

3

4

W act  h 0  h 0  u 22 1  2 cot 2 1 4 W ideal  h 0   h0 1 4

Total-to-total efficiency t  t 



W ideal W actual



h0   h 0 1 4 h0  h0 1 4

T0   T 0 1 4 T0  T0 1 4

Static-to-static efficiency s  s 

T4  T 1 T0  T0 4

1

10. Degree of Reaction The degree of reaction is defined as Rd  

Change in static enthalpy in the impeller Ch ange in stagnation enthalpy in the stage

1 1   cot 2 2 2 2

11. Mass Flow Rate  m  1  Area of flow at inlet  C f  2  Area of flow at exit  C f  m  1  D h1 C f  2  D h2 C f

...(5.11)

4.154 Thermal Engineering - I

Problem 4.51 A centrifugal compressor handles 150 kg/min. of air. The suction pressure and temperature are 1 bar and 20C. The suction velocity is 80 m/s. After compression in the impeller, the conditions, are 1.5 bar and 70C and 220 m/s. Calculate: (i) Isentropic efficiency (ii) Power required to drive the compressor (iii) The overall efficiency of the unit. It may be assumed that K.E. of air gained in the impeller is entirely converted into pressure in the diffuser. [JNTU - Jan/Feb 2015 - Set 4]

Solution:  150  2.5 kg/s; P 1  1 bar; T1  20  273  293 K; m 60 C 1  80 m/s ; P 2  1.5 bar; T2  70  273  343 K; C 2  220 m/s

(i) Isentropic efficiency, isen.: 1

T 2

 P2     P T1  1

or

1.4  1  1.4

 1.5    1 

 1.1228

T 2  293  1.228  328.98 K

 Isentropic work done C 22  C 21  C p T2  T 1  2  1000  1.005 328.98  293  

220 2  802 2  1000

 36.16  21  57.16 kJ/Kg

Air Compressors

Actual work done  C p T 2  T 1 

4.155

2202  802 2  1000

 1.005 343  293 

2202  802 2  1000

 50.25  21  71.25 kJ/kg  isen 

Isentropic w ork 57.16   0.8022 or 80.22% Actu al work 71.25

(ii) Power required to drive the compressor, P:  P  m  Actual work done kJ/kg  2.5  71.25  178.125 kW

(iii) The overall efficiency of the unit, overall As K.E. gained in the impeller is converted into pressure, hence C 22  C 21 C p T3  T2  2  1000 1.005 T3  343 

or

2202  802 2  1000

T3  363.89 K.

The pressure of air after leaving the diffuser, P 3: 1

 P3     T2  P2  T3



or 

P3

 T3    1

1.4

 363.89  1.4  1   1.2298   343 

  T  2 P 3  1.5  1.2298  1.844 ba r

P2

4.156 Thermal Engineering - I

After isentropic compression, the delivery temperature from diffuser, T3: 1

T 3

 P3     P T1  1

1.4  1  1.4

 1.844    1 

 1.191

T 3  293  1.191  348.96 K T3  T 1 348.96  293  0.7893 or 78.93% or overall   T 3  T 1 363.89  293 Problem 4.52: A single inlet-type centrifugal compressor handles 600 kg/min. of air. The ambient air conditions are 1 bar and 20C. The compressor runs at 21000 r.p.m. with isentropic efficiency of 80 percent. The air is compressed in the compressor from 1 bar static pressure to 4 bar total pressure. The air enters the impeller eye with a velocity of 145 m/s with no prewhirl. Assuming that the ratio of whirl speed to tip speed is 0.9, calculate: (i) Rise in total temperature during compression if the change in K.E. is negligible. (ii) The tip diameter of the impeller. (iii) Power required. (iv) Eye diameter if the hub diameter is 15 cm.

Solution:  600 m  10 kg/s; P 1  1 bar, T 1  20  273  293 K; 60 N  21000 r.p.m.; isen  80% ; P 02  4 bar; C 1  145 m/s;

C w2 U2

 0.9 ; d h  15 c m  0.15 m

Air Compressors

4.157

T 02

T 02 T 02 ’

0 2’

T2

d2 d1

P2

2 2’

T 2’

1 5 cm

P1 T 01

01

T1

1

S

(i) Rise in total temperature during compression if the change in K.E. is negligible: Refer Fig. The suffix ‘0’ indicates the condition

stagnation

The stagnation temperature at inlet to the machine, C 21 1452  293   303.5 K T 01  T 1  2C p 2  1.005  1000 1

T 01  P 01   Now,   T1 P  1 



 T01    1 or P 01  P 1    T  1 

1.4

or

 303.5 P 01  1    293 1

T02 

 P 02     P T 01  01 



 1.4  1  

 1.131 bar 1.4  1

 4  1.4    1.131 

T02   303.5  1.435  435.5 K

 1.435

4.158 Thermal Engineering - I

 Isentropic rise in total temperature

 T 02   T01  435.5  303.5  132C Hence, Actual rise in total temperature 

132 132  165 C.  isen 0.8

(ii) The tip diameter of the impeller, d2 Work required by the compressor  C p  Tactual  1.005  165  165.83 kJ /kg

Work required by the compressor is also given by Euler’s equation without prewhirl as: W

But

C w2  U 2 1000 C w2 U2

kJ/kg  165.83 kJ/kg

 0.9  C w2  0.9 U 2 u22  0.9



165.83 

or

u2 

But

u2  429.25 



d2 

1000

 

 165.83  1000     429.25 m/s 0.9   d 2N 60



d2  21000 60

429.25  60   21000

 0.3904 m or 39.04 cm say 39 cm

Air Compressors

4.159

(iii) Power required, P:  P  m  165.83  10  165.83  1658.3 kW (iv) Eye diameter if hub diameter is 15 cm, d1: From continuity equation, we have   m  d 21  d 2h  C 1  1 4

But density at entry is given by, 1 

P1 RT 1



1  10 2  1.189 kg/m 3 0.287  293



10 

 2 d  0.15 2  145  1.189 4 1

or

d 21 

10  4  0.15 2  0.0964 m 2   145  1.189



d 1  0.3104 m or 31.04 cm

Problem 4.53: A centrifugal compressor running at 9000 r.p.m delivers 600 m3/min. of free air. The air is compressed from 1 bar and 20C to a pressure ratio of with an isentropic efficiency of 82%. Blades are radial at outlet of impeller and flow velocity of 62 m/s may be assumed throughout constant. The outer radius of impeller is twice the inner and the slip factor may be assumed as 0.9. The blade area co-efficient may be assumed 0.9 at inlet. Calculate: (i) Final temperature of air (ii) Theoretical power (iii) Impeller diameters at inlet and outlet (iv) Breadth of impeller at inlet (v) Impeller blade angle at inlet (vi) Diffuser blade angle at inlet

(JNTU - Apr/May 2013 - Set 4)

4.160 Thermal Engineering - I

Given: N  9000 r.p.m ; Volume of air delivered, V  600 m 3/min; P 1  1 bar, T 1  20  273  293 K; rp  4,  isen  0.82 ; C f2  62 m/s;  C f1 r2  2 r1;  s  0.9; Blade area co-efficient, k a  0.9.

Solution: (i) Final temperature of air, T2: 1

T 2

 P2     P T1  1

 4

1.4  1 1.4

 1.486

 T2  293  1.486  435.4 K

Now, isen 

T 2  T1 T2  T1

or T 2  T1 

T 2  T1 435.4  293  293   466.7 K 0.82 isen

(ii) Theoretical power:  2  PV 1  10  600 /60  Mass flow rate, m    11.89 m3/s RT 0.287  293  Power

  m C p T 2  T 1   11.89  1.005 466.7  293  2075.95 kW

(iii) Impeller diameters at inlet and outlet, d1, d2: For radial blades, work input to the compressor is given by,

Air Compressors

Work done 

s u 22 1000

 C p T 2  T 1

4.161

[ V w2  U 2]

Here T2 is the final temperature of air from the exit of compressor. 1/ 2

 1000  C p T2  T 1  u2    s  

1/2

 1000  1.005 466.7  293     0.9   d 2N Also, u2   440.4 m/s 60

or

d2  d1 

 440.4 m/s

60  440.4  0.9346 m or 93.46 cm   9000 d2 2



93.46  46.73 c m 2

(iv) Breadth of impeller at inlet, b1: Volume flow rate  2r1 b 1 C f1 k a, where k a is the blade area coefficient  b1  

Volume flow rate 2r1  C f1  ka

600/60  0.1221 m or 12.21 cm.  0.4673    0.9 62 2    2  

(v) Impeller blade angle at inlet 1 tan 1  

C f1 u1



62  0.2816 440.4 /2

1  tan  1 0.2816   15.73 

4.162 Thermal Engineering - I

(vi) Diffuser blade angle at inlet, 2: C f2 62  0.1564  tan  2   s  U 2 0.9  440.4 

2  tan  1 0.1564   8.9.

Problem 4.54: A centrifugal compressor delivers 54 kg of air per minute at pressure of 200 kPa, when compressing air from 100 kPa and 15C. If the temperature of air delivered is 97C and no heat is added to the air from external sources during compression, determine the efficiency of the compressor relative to ideal adiabatic compression and power absorbed. (JNTU - April/May - 2013 - Set 1)

Given Data  54  0.9 kg/s ; P 2  200 kPa; P 1  100 kPa ; m 60 T1  15  273  288 K; T2  97  273  370 K ; Q  0 T

Solution:

02

T 02

P2

To find isentropic efficiency T 2

T02’

1 P2  

0 2’

T2

   T1 P  1

2 2’

T 2’

1 P2  

P1

 T 2  T1   P  1

T01 T1

0.4  1.4

 200  288    100 

 351.08 K

T2  351.08 K

01 1

S

Air Compressors

Isentropic efficiency isen 



4.163

T 2  T1 T2  T1 351.08  288  0.7692 370  288  76.92%

Power absorbed for compression work Since C 1 and C 2 are not given, the power can be calculated based on static head condition.  Power  m  w ork don e in the impeller  0.9  C p T 2  T 1  0.9  1.005 370  288  74.169 kW Problem 4.55: A single sided centrifugal compressor is to deliver 18 kg/s of air when operating at a stagnation pressure ratio of 4:1 and a speed of 220 rps. The inlet stagnation conditions may be taken on 288 K and 1.0 bar. Assuming the slip factor of 0.92 and as a power input factor of 1.04 and an overall efficiency of 0.84, estimate the overall diameter of the impeller and power input. (JNTU - May - 2013 and Apr/May 2013 - Set 3)

Given Data: P 02   4; N  220 rps; T 01  288 K and P 01  1 bar m  18 kg/s; P 01 s  0.92;  w  1.04 ; isen  0.84

4.164 Thermal Engineering - I

Solution: (i) To find overall diameter of impellar D2: 1

T 02 

 P 02     P T01  01 

 4

0.4 1.4

 1.486

T 02   T01  1.486  288  1.486  427.97 K  isen 

0.84 

T02   T 01 T 02  T 01 427.97  288 T 02  T 01 

T02  T01   166.63 K

Euler work W

 w s u22 1000

 Cp T02  T01 

1.04  0.92  u 22  1.005 166.63  1000 J kJ     1000 is to convert kg in to kg   

u22  175020.6  u2  418.35 m/s We know u2   D 2 N 418.35    D 2  220 D 2  0.605 m  60.5 cm

rev    N is in s   

Air Compressors

4.165

(ii) Power input  Power  m C p T 02  T01   18  1.005 166.63   3014.34 kW Problem 4.56: A centrifugal air compressor having a pressure ratio of 5, compresses air at the rate of 10 kg/s. If the initial pressure and temperature of air is 1 bar and 20C, calculate final temperature and power required to drive the compressor. (JNTU - January - 2014 - Set 2)

Given Data: P2 P1

  5; m  10 kg/s; P 1  1 bar; T1  20  273  293 K ;

To find T2, 1 P2  

   P T1  1

T2

1

 P2    T2  T1    P1  0.4

T 2  293   5  1.4  464.06 K

Power to drive the compressor   m C p T2  T1  10  1.005 464.06  293  1719.15 kW

[assuming C 1  C2]

4.166 Thermal Engineering - I

4.28 AXIAL FLOW COMPRESSORS An axial compressor is a pressure producing machine. As the name indicates, the flow of air or fluid is in the direction of the rotor axis. It consists of a rotor with moving blades and a stator fixed to casing which serve to recover part of kinetic energy imparted to the working fluid. T his kinetic energy imparted to the fluid by means of the rotating blades is then converted into a pressure rise. Inlet guide vanes

S

D e livery vanes

Stato r (C asing )

R

S

R

S

R

S

R

Air D e livery

R o tatin g drum

Air in

D rive shaft Air D e livery S = S tator (Fixed) blad es R = Ro tor (M oving ) blade s Fixed bla de s Fig. 4.34 Axial Flow Com pressor.

M oving bla de s

   

In this axial flow compressor, air is flowing parallel to the axis of compressor. It consists of stators (fixed blades) and Rotors (rotating blades) in an alternate rows. T he rotors are fixed with rotating drum and stators are fixed to the casing. One stage of compressor means a row of stator blades with a row of rotor blades.

Air Compressors

 









T he work energy of moving transferred to air to accelerate.

4.167

blades

is

T he blades are so arranged that the space between the blades forms diffuser passage and hence air pressure is increased at the expense of velocity. T he air is then further flown through stator blades and gets diffused and its pressure is further increased. After air gets pressure rise in one stage, it will be allowed to pass second stage and pressure is continuously raised. T he annular area is normally reduced from inlet to outlet of the compressor to keep the flow velocity constant. T he stator blades serve to convert a part of kinetic energy to pressure energy and to guide air from one stage to next stage without shock.

4.28.1 Working Principles of a Compressor Stage A compressor stage consists of a rotor followed by a diffuser ring. T he first stage of a multistage or a single stage compressor may consist of a ring of inlet guide vanes (IGV) and rotor blades as shown in Fig. 4.35. Air enters axially into the inlet guide vanes where it is turned through a certain angle to impinge on the first row of rotating blades with proper angle of attack. T he first row of moving blades imparts the kinetic energy and increase in total pressure. T he static pressure increase is because of the flow area increases i.e., A2  A1 in the Fig. 4.35. T

he stator blades of the first stage decrease

4.168 Thermal Engineering - I AX IA L FLO W

IG V (In let G uid e Va nes) 1 A1

1 Stage

R o tor B lad es 2

A2

2 Stato r B lade s (D iffuser)

3 Fig:4.35

the absolute velocity to bring about a static pressure rise. T he air at a proper angle enters the second stage of rotating blades and it is repeated for the remaining stages. T able shows the variation of properties occurs in an axial flow compressor. [ increase; decrease ]

Stator

Abs.

Relative

Flow

Velocity

Velocity

width

C

Cr 

B

P 









Static Total Pressure Pressure P 0 about constant

Rotor











Air Compressors

4.169

4.28.2 Stage Velocity Triangles T he stage velocity triangles for an axial flow compressor is shown in Fig. 4.36. Air enters with an absolute velocity C 1 and angle 1 [from the axial direction] into the rotor blades. T he inlet guide vanes, guide the flow in a proper angle of attack.

4.170 Thermal Engineering - I

A general cycle includes, the entry to the rotor (1), exit from the rotor (2) and the diffuser (stator) blade exit (3) respectively. In axial compressors, the following assumptions are (a) Axial Velocity

C f1  C f2  C f  constant

(b) Blade speed

u1  u2  u  constant

(c) C 1  C 3 and 1  3 (d) In rotor, relative velocities are tangent and stator absolute velocity is tangent. C w1  C w1  C f tan  1 tan  1  Cf tan 1 

...(1)

u  Cw 1 Cf

 Cw1  u  Cf tan 1

...(2)

Cw2 tan  2  Cf C w2  Cf tan  2 tan 2 

...(3)

u  C w2 Cf

 Cw2  u  Cf tan 2

...(4)

1  4  2  3 C f tan  1  u  C f tan2  u  C f tan 1  Cf tan 2

[C f  constant] C f [tan  1  tan 2]  C f [tan  2  tan 1]

Air Compressors

4.171

tan  1  tan 2  tan  2  tan 1

(or) tan  1  tan 1  tan  2  tan 2

...(1)

4.28.3 Blade Loading, Flow Coefficients and Specific Work T his is a dimensionless quantity used for comparing different stages of various sizes and speeds. i.e., Blade Loading Co-efficient  

W actual u2

Flow co-efficient is the ratio between the axial velocity to its blade speed. i.e.,  f 

Cf u

Specific Work W act  u [Cw 2  C w1] 

C 22  C 21 C 2r1  C 2r2  2 2

4.28.4 Static Pressure Rise in a Stage T his pressure rise in a stage depends upon the blade geometry and the speed of the rotor. T he total static pressure rise in the stage = Pressure rise in rotor  Pressure rise in diffuser T

he various assumptions before deriving are

1. T he flow is assumed to be incompressible [ = constant]

4.172 Thermal Engineering - I

2. Reversible adiabatic flow takes place in the stage 3. T

he axial velocity is constant throughout the

stage Applying Bernoulli equation across the rotor blade rows, it gives P1 

 C 2r1 2

 P2 

 C 2r2 2

 P 2  P 1  C 2r1  C2r2  Protor 2 C 2r1  u  C w12  C 2f 

C 2r2  u  C w22  C 2f

 C 2r1  C 2r2  C w2  C w1 [2u  C w1  C w2]  C w2  C w1 [u  C w 1  u  cw2] tan 1 

u  C w2 u  C w1 and tan 2  Cf Cf

 u  C w1  C f tan 1  u  C w2  C f tan 2

By substituting this in the above equation, we get  Protor 

 C  Cw1 C f [tan 1  tan 2] 2 w2

Applying Bernoulli equation across the diffuser P2 

 C 22  C 23  P3  2 2

 P 3  P 2  [C 22  C 23] 2

Air Compressors

4.173

From the velocity triangle C 22  C 2w2  C2f C 21  C 2w1  C 2f



C 22  C 21  C 2w2  C 2w1

T

herefore,

P 3  P 2  Pstator 

  C2  C2  w1  2  w2

Pstage  Protor  Pstator 

  [Cw2  C w1] C f [tan 1  tan 2]  C 2w2  C 2w1 2 2



 Cw2  Cw1   Cf [C w2  C w1]  tan 1  tan 2   Cf 2  

C w2  C f tan 2 C w1  C f tan 1

Substituting this in the above equation, we get 

 2 C [tan  2  tan  1] [tan 1  tan 2  tan 2  tan  1] 2 f

From equation (1) tan  1  tan 1  tan  2  tan 2  Pstage 

 2 C tan  2  tan  1 2 tan  1  tan 1 2 f

u  C w1 C w1 , tan 1  tan 1  Cf Cf

4.174 Thermal Engineering - I

tan 1  tan 1 

C w1 Cf



u  C w1 Cf



1 u [C  u  C w1]  C f w1 Cf

 u   Pstage   C 2f [tan 2  tan  1]  C   f  Pstage   C f u [tan  2  tan  1]

...(b)

From the velocity triangles, u  tan  1  tan 1 Cf u  tan 2  tan  2 Cf

and,

Work absorbed by the stage per kg of air, W st  u C w2  C w1  Cp T02  T01 act

where C w1 and C w2 are the whirl components of absolute velocities at inlet and exit of rotating blades. Note: Here whirl component at the entrance of the compressor is not zero because air flows axially and not radially. So W st  u C f tan  2  tan  1 . .  . C w1  C f tan  1, and  W st  u C f tan 1  tan 2   C w2  C f tan  2  

Air Compressors

4.175

4.29 DEGREE OF REACTION Degree of reaction Rd is defined as the ratio of pressure rise in the rotor blades to the compressor stage. i.e., Rd 

Pressure rise in the rotor blades Pressure rise in the stage

Pressure rise in the compressor stage equals work input per stage Wact  u C w2  C w1 Pressure rise in the rotor blades is at the expense of K .E. and is C 2r1  C 2r2  2 Rd 



C 2r1  C 2r2 2u C w2  C w1

Refer inlet and outlet velocity triangles: C w2  u  C f tan 2 

C w1  u  C f tan 1 C w2  C w1  C f tan 1  tan 2



Similarly from velocity triangles, C 2r  C f2  C f tan 12 1  

C 2r  C f2  C f tan 22 2

C 2r  C 2r  C 2f tan 2 1  tan 2 2 1 2

4.176 Thermal Engineering - I

So Rd 

C 2f tan 2 1  tan 2 2 1 Cf tan 1  tan 2  2u  C f tan 1  tan 2 2 u

Degree of reaction is usually taken as 0.5,  0.5 

or But

1 Cf  tan 1  tan 2 2 u

u  tan 1  tan 2 Cf u  tan 1  tan 1  tan  2  tan 2 Cf

(from velocity triangles) 

u  tan 1  tan 2  tan  1  tan 1  tan  2  tan 2 Cf

From this  1  2;  2  1 So with 50% reaction blading, the compressors have symmetrical blades and with this type of set-up, losses in flow path are greatly reduced. In symmetrical blades, the tip clearance and fluid friction losses are minimum.

4.30 INFINITESIMAL STAGE EFFICIENCY (OR) POLYTROPIC EFFICIENCY Polytropic efficiency is the isentropic efficiency of one stage of a multi-stage compressor. T his small stage efficiency is constant for all stages of a compressor with infinite number of stages. T o see the true aerodynamic performance of a stage, a finite stage is divided into an infinite number of small stages of the same efficiencies. T his is an imaginary stage with an infinitesimal pressure rise and the effect of preheat is negligible.

Air Compressors

4.177

Fig. 4.37 shows the 2 T2 actual (1-2) and isentropic P 2’ = 1  2 compression process T2’ 2’ P 2 P of a finite stage. A small P +d dT dT’ stage is considered in P h between 1-2 whose P and pressures are P1 T1 P  dP . T he pressure 1 and temperature at the S entry of the small stage are Fig:4.37 Infinitesim al and P and T respectively. T he Com pressio n Processes actual and isentropic temperature rise are dT and dT respectively. T herefore, the efficiency of a small stage is P  

isentropic tempe rature rise dT dh   dT dh actual temperature rise dP    C P dT



dP  RT  P  R dT 1   1 dP T   dT P 

  1 dP dT 1    T P  P

T

...(1)

he value of the polytropic efficiency P is

constant and is determined by integrating the finite state compression process between 1 and 2. 2

2

dT   1 1 dP  T     P P 1 1

4.178 Thermal Engineering - I

 T2   P2  1   1 ln  ln    T P   1  P  1  ln    1  P    ln  

P2   T2    ln  P1  T1   T2   T2   ln   T T1   1

...(2)

(or) 1

 T2   P2      T1   P 1 





1 P

...(3)

T he irreversible actual compression process can be considered as equivalent to a polytropic process with an index ‘n’. T

hus equation (3) can be rewritten as 1

 P2     T1 P  1

T2



1 P

n1

 P2  n   P  1

Equating the powers, n  1    1 1   n  P

 P 

  1 n  n  1 

...(4)

4.31 FINITE STAGE EFFICIENCY T he efficiency of a finite stage can be expressed interms of the small stage efficiency. T ake static values of temperature and pressure and assuming perfect gas,

Air Compressors

 T2  1  T2  T 1 T1    st  T2  T 1  T 2  1  T  1 We know that T 2

 P2    P T1  1

1 

 rp

1 

4.179

...(5)

a nd

P 2 P 2   pressure ratio    where rp  P1 P1  

T2

1 1

 rp   T1

P

By substituting this in equation (5) 1

st 

rp    1 1

1

 rp     P  1

...(6)

T his equation is also applicable to a multi stage compressor. i.e.,

0 

R P

1 

1

   1  1       P  R P

1

Polytropic or small stage efficiency, P : 1 n  or p       n1 

Polytropic efficiency depend on exponent n and the adiabatic exponent .

4.180 Thermal Engineering - I

 P02   T02    1  Also, P ln  ln    P T   01   01  1  

 P 02  P 02  1 ln   ln   P 01 P01      P    T 02   T02  ln  ln    T T  01   01 

or

4.38 IMPORTANT FORMULAE 1. Flow coefficient f f 

Since

Cf u

u  C f tan 1  tan  1 f 

Cf C f tan 1  tan  1



1 tan 1  tan  1

Since  1  2 and 1  2, 1  f  tan 2  tan  2

2. Head or work coefficient h  It is defined as the ratio of actual work done to the kinetic energy developed by the mean peripheral velocity. T

hus,  h 

C p T u     2  2



2 C w2  C w1 u

 tan  2  tan  1  2   tan 2 tan  2 

Air Compressors

4.181

3. Deflection co-efficient def It is defined as,  def 

u C w2  C w1 u2



Cw2  Cw1 u

or  h  2 def

4. Pressure co-efficient p It is defined as the ratio of isentropic work done to kinetic energy developed by the peripheral velocity. hus,  p 

T

C p Tisen   isen  h  u2     2 

5. Pressure ratio  T2  T1    1   isen  P1  T1  P2

/  1

6. Stagnation pressure ratio   T0    1  0 isen  T 01  P 01  P 02

/  1

7. Number of stages If the work done per stage is assumed to be the same, then the number of stages N  N

T0 T0stage

...(24.112)

Since the pressure ratio per stage is the same, rpstage 

P 02 P 01



P 03 P 02



P 0 N  1 P 0N

4.182 Thermal Engineering - I

T

he overall pressure ratio rp   rp stage    N

N

ln rp

ln  rpstage   

and rpstage varies from 1.12 to 1.2.

4.32 LOSSES IN AXIAL FLOW COMPRESSOR STAGE Various losses occur while the air (or) gas flows through a compressor stage. T he total pressure loss occurs in three ways: 1. Profile losses on the surface of the blades. 2. Skin friction on the annulus walls. 3. Secondary flow losses.

2 .29 % 1 00

S A n n u lu k in F ric ti on s lo s s S e co n d a ry lo s s

4 4% 4 2%

P ro file lo ss

D es ig n

S u rge

S ta ge e fficie ncy, s tag e

90

80

70

60 0 .5

0 .7

0 .9

1 .1

1 .3

Flow C oe ffic ie nt f Fig:4.38 L osse s in com pressor stage

1 .5

1 .7

Air Compressors

4.183

T he various losses are represented on graph between stage efficiency and flow coefficient as shown in Fig. 4.38 1. Profile losses on the surface of the blades: 



Profile loss means that the total pressure loss of two dimensional rectilinear cascade arising from the friction on the surface and due to the mixing of flow particles after the blades. T hese losses experimentally.

are

normally

determined

2. Skin friction loss on the annulus walls:  

T hese losses are arising from the skin friction on the annulus walls. Empirical relations derived by Howell and Haller are available for determining drag co-efficient and skin friction losses.

3. Secondary flow losses:  



Secondary flows are produced by combined effects of curvature and boundary layer. Secondary flow is developed when the components of velocity are developed from the deflection of an initially sheared flow. Such secondary flow occurs when there is a bend, when a sheared flow passes over an aerofoil shape with finite lift or when a boundary level meets an obstacle. Secondary flow loss occurs when boundary layers are growing on the casing and hub walls of the machines are deflected by rows of blades - stator and rotor.

4.184 Thermal Engineering - I

Air Compressors

4.185

4.33 SURGING

Fig. 4.39 shows the relation between mass flow and pressure ratio for some typical performance characteristics curves at different speeds (N 1, N 2,

Pressu re ratio

It is the phenomena of excessive aerodynamic pulsation which is transmitted throughout the machine by virtue of sudden drop in delivery pressure or complete breakdown of the steady through flow.

S urge C ycle D U nstable A

B S ta b le

S urge lin e

E N1

T1 etc). T he surge N2 N3 phenomenon is explained T2 C T3 with the aid of one of the M ass flow rate curves in this figure. It is Fig:4.39 assumed that, a valve placed in the delivery line of a compressor running at constant speed, is slowly opened. When the valve is shut and the mass flow is zero, the pressure ratio will have some value A , corresponding to the pressure head produced by the action of the rotor on the air trapped between the blades. When the valve is opened, flow commences and the pressure ratio increases. At some point B , where the efficiency approaches its maximum value, the pressure ratio will reach a maximum, and any further increase in mass flow will result in a fall of pressure ratio. For mass flows greatly in excess of that corresponding to the design mass flow, the air angles will be widely different from the vane angles, breakaway of the air will occur, and the efficiency will fall off rapidly. In this hypothetical case, the pressure ratio drops to unity at C, when the valve is fully opened

4.186 Thermal Engineering - I

and all the power is absorbed in overcoming internal frictional resistance. T he region between A and B (having positive slope) is unstable. Let the compressor be operating at point D on left of B. Now the mass flow is reduced and the pressure ratio is also reduced according to the graph. If pressure of air downstream of the compressor does not fall quickly enough, the air will tend to reverse the direction and flow back in the direction of resulting pressure gradient in to the compressor. When this occurs, the pressure down stream of the compressor has also fallen, so that the compressor will now pick up again to repeat the cycle of events, which occurs at high frequency. Such a situation should be avoided by keeping the operating point to the right of B. T his is because the decrease in mass flow in this region is accompanied by increase in pressure ratio and thus stability is obtained. T he point to the left of B where surging begins is a function of the ability of flow passage downstream of the compressor to swallow the flow. Such points for various N values of would lead to a locus called the surge line  T1  as shown in figure. Surging leads to vibration of the entire machine which can ultimately lead to mechanical failure. T herefore, the operation of compressors on the left of surge line is injurious to the machine and must be avoided and that range is unstable range. T he stable range of operation of the compressor is on the right-hand side of this line.

Air Compressors

4.187

4.34 STALLING

i= - v

e +v i=

i= +

ve

e

Stalling is the separation of flow from the blade surface. A partial blockage or uneven flow in the blade passages due to the change of angle of incidence is called stalling. At low flow rates, the axial velocities are lower and the angle of incidence is increased as shown in Fig. 4.40 At large values of the incidence, the flow separation occurs on the suction side of the blades which is referred to as positive stalling and the pressure hide is negative stalling.

i=0

A

Pro pa gating

B

C

Fig:4.40

D

U n stallin g

T his is a separate phenomenon, which may contribute to surge but can exist in the nominally stable operating range. T his is the rotating stall. When there is any non-uniformity in the flow or geometry of the channels between blades, breakdown in the flow occurs in one channel, say B in figure. T his causes the air to be

4.188 Thermal Engineering - I

deflected in such a way that channel C receives fluid at a reduced angle of incidence and channel A at an increased incidence. Channel A then stalls, resulting in a reduction of incidence to channel B enabling the flow in that channel to recover. T hus the stall passes from channel to channel and the stall cells will rotate in a direction opposite to that of the rotor blades. It may lead to aerodynamically induced vibrations and high frequency noises resulting in fatigue failures in other parts of the compressor. T he blades can fail due to resonance. T his occurs when the frequency of the passage of stall cells through a blade coincides with its natural frequency. Both the efficiency and delivery pressure drop considerably on account of rotating stall.

4.35 COMPARISON BETWEEN RECIPROCATING AND CENTRIFUGAL COMPRESSORS Reciprocating compressors

Centrifugal compressors

Greater vibration problems Less vibrational problems due to the presence of since it does not have reciprocating parts which reciprocating parts. are partially balanced. Due to the presence of Due to the absence of many several sliding or bearing sliding or bearing members, members, it has lesser  mech is more.  mech. Higher initial cost.

Lower initial cost.

Pressure ratio per stage is Pressure ratio per stage is about 5 to 8. about 3 to 4.5.

Air Compressors

Reciprocating compressors

4.189

Centrifugal compressors

High delivery pressure upto Medium delivery 5000 atm. upto 400 atm.

pressure

Smaller Free Air Delivered Greater FAD. (FAD). Greater Flexibility in No flexibility in capacity and capacity and pressure range. pressure range. Higher maintenance cost.

Lower maintenance cost.

Compression efficiency is  compressor is higher, at higher, at compression ratio compression ratio less than above 2. 2. Adaptability to low speed Adaptability to high speed drive. drive. More operating needed.

attention Less operating attendance.

Always a chance of mixing No chance of mixing of air with lubricating oil. lubricating oil with air.

of

Suitable for low, medium Suitable for low and and high pressures and low medium pressures and large and medium gas volumes. gas volumes.

4.36 COMPARISON BETWEEN RECIPROCATING AND ROTARY AIR COMPRESSORS Reciprocating air compressors

Rotary air compressors

Suitable for low discharge of Suitable for handling large air at high pressure. volumes of air at low pressures.

4.190 Thermal Engineering - I

Reciprocating air compressors

Rotary air compressors

Low speed (RPM).

High speed (RPM).

Pulsating air supply.

Continuous air supply.

More cyclic vibrations occur. Less vibrations occur. Complicated system.

lubricating Simple lubrication system.

Air delivered is generally Air delivered is relatively contaminated with oil. more clean. Large compressor size for Small size the given discharge. discharge 250  300 m 3/min

Free

for

same

air 2000  3000 m 3/min FAD.

Delivery. High delivery pressure.

Low delivery pressure.

4.37 COMPARISON BETWEEN CENTRIFUGAL AND AXIAL FLOW COMPRESSORS Centrifugal compressors Radial flow

Axial flow compressors Axial flow (Parallel to the direction of axis of the machine)

Pressure ratio per stage is Low pressure ratio per stage high, about 4.5:1. T his about 1.2:1. T his is due unit is compact. to absence of centrifugal action. Less compact and less rugged. Isothermal efficiency about 80 to 82%

is With modern aerofoil blades,  iso is about 86 to 88%.

Air Compressors

Centrifugal compressors Frontal area is larger

4.191

Axial flow compressors Frontal area is smaller. Hence the axial flow compressor is more suitable for jet engines due to less drag.

More flexibility of operation Less flexibility of operation. due to adjustable prewhirl and diffuser vanes. Low starting required. Multistaging difficult.

is

torque High starting required. slightly More suitable multi-staging.

Upto 400 bar delivery Delivery pressure pressure is possible. upto 20 bar.

torque for is only

It is used in application of blowing engines in steel mills, low pressure refrigeration, big central air conditioning plants, fertiliser and industry, supercharging I.C. engines, gas pumping in long distance pipe lines etc.

Mostly used in jet engines due to higher efficiency and smaller frontal area. Also used in power plant gas turbines and steel mills.

Efficiency vs. compared to here.

curve

speed

of

both

compression

Efficiency (vs) speed curve of both compressor is compared here in Fig. 4.41.

4.192 Thermal Engineering - I

E fficien cy C entrifu ga l co m pressor

A xia l flo w com pressor

S peed Fig:4.41

Problem 4.57: The following data relate to an axial flow compressor: u  250 m/s; Cf  180 m/s; 1  40; 2  15   1 kg/m3; Calculate (i) the pressure rise (ii) the workdone per kg of air

Given Data: T

he pressure rise in rotor.

p 

 2 C [tan2 1  tan2  2] 2 f



1

2  10

5

 180 2 [tan 2 40  tan 2 15]

 0.102 bar

Workdone per kg of air W

u [C w1  C w2] 1000

Air Compressors



u  C f [tan  1  tan  2] 1000



250  180 [tan 40  tan 15] 1000

4.193

 25.702 kJ/kg Problem 4.58: An axial flow compressor with an overall isentropic efficiency of 85% draws air at 20 C and compresses it in the pressure ratio of

4 : 1. The mean blade speed and

flow velocity are constant throughout the compressor. Assuming 50% reaction blading and taking blade velocity as 180 m/s and work input factor as 0.82, calculate: (i) Flow velocity (ii) Number of stages Take 1  12, 1  42 and Cp  1.005 kJ/kgK (JNTU - January - 2014 - Set 4 & Set 3)

Given: C1

C r2

Solution: 1.4  1

 4 1.4

2 2 C f2

C2

C w2

Work input factor   0.82

1 

C w1

u1

Pressure ratio, P2  4, u  180 m/s P1

 P2    P T1  1

1 C f1

T1  20  273  293 K

T2

1

C r1

isen  85% ,

 1.486

4.194 Thermal Engineering - I

 T2  293  1.486  435.4 K

Now isen  0.85 

T2  T 1 T2  T1

435.4  293 T 2  293

 T2  460.5 K

T

heoretical work required per kg

 C p T 2  T 1  1.005 460.5  293   168.33 kJ/kg

From velocity triangles (Refer Fig.) u  tan  1  tan 1  tan 12  tan 42  0.212  0.9  1.112 Cf u 180  Flow velocity C f   161.8 m/s.  1.112 1.112

Work done per stage  u C w2  C w1  work input factor . . Now, C w2  C f tan  2  161.8 tan 42  145.7 m/s  .  2  1

and

C w1  C f tan  1  161.8 tan 12  34.4 m/s

 Work done per stage  180 145.7  34.4   0.82  10 3 kJ/kg  16.4 kJ/kg

[10  3 is to make J/kg to kJ/kg]  Number of stages 

168.33  10 16.43

i.e., Number of stages  10

Air Compressors

4.195

Problem 4.59: In an eight stage axial flow compressor, the overall stagnation pressure ratio achieved is 5 : 1 with an overall isentropic efficiency of 90 per cent. The inlet stagnation temperature and pressure are 293 K and 1 bar. The work is divided equally between the stages. The mean blade speed is 175 m/s and 50% reaction design is used. The axial velocity through the compressor is constant and is equal to 100 m/s. Calculate: (i) The blade angles (ii) The power required. (JNTU August 2014 - Set 4)

Given: N s  8 ; rp  5:1; isen  90%; T01  293 K; P 01  1 bar; u  175 m/s; Degree of reaction = 50% ; C f  100 m/s.

Solution: (i) The blade angles, 1, 1, 2, 2: T 08 07

P 08 P 07 P 06

1

C r1

1

C1

C f1

0 8’ 06

C w1

u1

P 03 P 02 P 01

C r2

03 02

2 2 C f2

C2

C w2

01 S

4.196 Thermal Engineering - I

Refer to Fig. for velocity diagrams. Since the degree of reaction is 50%, the blades are symmetrical and hence the velocity diagrams are identical. T hus 1  2 and  2  1 Let suffix N denotes the number of stages. With isentropic compression, the temperature of air leaving the compressor stage is  P 0N  T 0N  T 01   P  01 

But isen  0.9 

 T0N 

  1/

1.4  1

 293  5 1.4

 464.06 K

T 0N  T 01 T0N  T01 464.06  293 T 0N  293 464.06  293  293 0.9

 483.07 K

T

he work required by the compressor  C p T0N  T 01   C w2  C w1 u  N s

or C p T0N  T 01   C f tan  2  tan  1 u  Ns  tan  2  tan  1 



C p T 0N  T01  Cf  u  N

1.005 483.07  293  10 3  1.364 100  175  8  tan  1  tan 2  1.364

...(1)

Air Compressors

4.197

. .  .  2   1

From velocity triangles, we have 175 u  tan  1  tan 1   1.75 100 Cf

...(2)

Adding (1) and (2), we get tan 1 

1.364  1.75  1.56 2

or

1  tan  1 1.56   57.29 



2  2  57.29 

Substituting the value of tan 1 in (2), we have tan  1  1.56  1.75 tan  1  0.19 or  1  tan  1 0.19  10.76  

1  2  10.76 

(ii) The power required by compressor P:  P  m C p T0N  T01   1  1.005 483.07  293   191.02 kW Problem 4.60: In an axial flow compressor, the overall stagnation pressure ratio achieved is 4 with overall stagnation isentropic efficiency 86 percent. The inlet stagnation pressure and temperature are 1 bar and 320 K. The mean blade speed is 190 m/s. The degree of reaction is 0.5 at the mean radius with relative air angles of 10 and 30 at the rotor inlet and outlet respectively. The work done factor is 0.88. Calculate: (i) Stagnation polytropic efficiency. (ii) Number of stages.

4.198 Thermal Engineering - I

(iii) Inlet temperature and pressure. (iv) Blade height in the first stage if the hub-tip ratio is 0.45, mass flow rate 20 kg/s. [JNTU Aug 2014 Set (2) & Apr/May 2013 - Set (2)]

Given: rp 

P 0N P 01

 4;  isen  86%; P 01  1 bar; T01  320 K

u  190 m/s ; Degree of reaction, R d  0.5. Work done  factor = 0.88 ; Hub-tip ratio = 0.45; m  20 kg/s.

Solution: Refer Fig. For 50% reaction, the inlet and outlet velocity diagrams are identical. Hence 1  2  10 ; 2  1  30 . (i) Stagnation polytropic efficiency, p: T he temperature at the end of the compression stage due to isentropic expansion is

T

1

C r1

1

P 0N

C1

C f1

0N C w1

u 0N’

2 = 1 = 30

P 01

o

In le t ve locity trian gle C r2

2 2 C f2

C2

01 S

C w2 1 = 2 = 10

o

O utle t velocity trian gle

Air Compressors 1

 P 0N    T 0N  T 01    P01   0 isen  0.86 

 T0N 

1.4  1

 320 4

T 0N  T01 T0N  T 01



1.4

 475.52 K

475.52  320 T 0N  3.20

475.52  320  320  500.84 K 0.86 1 

 P0N  ln   P 01   p   T0N ln  T  01

Now

4.199



  

0.4 1.4



ln 4  500.84  ln    320 

0.396  0.8839  88.39% 0.448

(ii) Number of stages, N s: From the configuration of velocity triangles, U  tan  1  tan 1  tan 10   tan 30  0.7537 Cf 

Cf 

u 190  252.1 m/s  0.7537 0.7537

Now C w 1  C f tan  1  252.1  tan 10  44.5 m/s C w 2  C f tan  2  252.1  tan 30  145.6 m/s

Work required per stage  u C w2  C w1  work done factor

4.200 Thermal Engineering - I



190 145.6  44.5  0.88 1000

 16.9 kJ/kg

T

otal work required by the compressor

 C p T 0N  T01   1.005 500.84  320  181.74 kJ/kg  Number of stages, NS 

181.74  11 stages  10.75 ~ 16.9

(iii) Inlet temperature and pressure, T 1, P 1: T

he absolute velocity C 1 at exit from the guide

vanes and approaching to moving blades of first stage, C1 

T

Cf

cos  1



emperature, T 1  T 01 

252.1  255.99 m/s cos 10

C21 2C p

 320 

255.99 2 2  1.005  1000

 287.4 K

Assuming reversible flow through the guide vanes put ahead of the first stage, 

1.4

 T1    1  287.4  0.4   0.687 bar ; P1  1     P01  T 01   320  P1

(iv) Blade height in the first stage, 1: T

he density of air approaching to first stage, 1 

P1 RT 1



68.7  0.833 kg/m 3 0.287  287.4

From the continuity equation,

Air Compressors

4.201

 1 A1 C f  m  20 kg /s 0.833   r21 [1  0.452]  252.1  20

or r1 

    20

0.833    [1  0.452]

But

rh r1

252.1

 0.195 m o r 19.5 cm

 0.45  rh  19.5  0.45  8.8 cm

Hence height of the blade in the first stage, L  r1  rh  19.5  8.8  10.7 cm Problem 4.61: A multistage axial compressor is required for compressing air at 293 K through a pressure ratio of 5 to 1. Each stage is to be 50% reaction and the mean blade speed is 275 m/s, flow coefficient is 0.5 and stage loading factor is 0.3 are taken, for simplicity, as constant for all stages. Find the flow angles and number of stages required if the stage efficiency is 88.8%. (JNTU August 2014 - Set 1 and Apr/May 2013 - Set (3))

Given Data: Flow coefficient f 

C f1 u



1  0.5 tan 1  tan 1 1

 P 02   Stage efficiency  ln    P 01 1 n  or    0.88  p     T P olytropic    1 n 02     ln    efficiency  T 01  

4.202 Thermal Engineering - I

Ns 

ln rp P 02 P 03 P 0 N  1   ; rpstage  P 01 P 02 P0N ln rp stage

Also Ns 

1

C r1

1

 T 0 T 0stage

u In le t velocity tria ngle

P 02 P 01

 5; u  275 m/s; P 01  1 bar; T 01  293 K 1 

 P 02  ln   P 01   p   T 02 ln  T  02

  

 T02  0.4 ln  ln 5  0.4598   0.88  1.4 293    T02  ln    0.523  293  T02 293

C1

C f1

 e0.523  1.686

T02  494 K

Work required/kg of air  C p T02  T01   1.005 494  293  202 kJ /kg

C w1

Air Compressors

Stage loading factor   0.3 

W stage 

No. of stages N s  

T

Wstage u2

0.3  275 2  22.69 kJ/kg 1000

otal workdo ne/kg of air W stage

202  8.9 ~  9 stages 22.69

1 2 tan  1  tan 1  0.5 Cf u

4.203

...(i)

 0.5  C f  0.5  275  137.5 m/s

Also;    f tan 1  tan 2 0.3  0.5 tan 1  tan 2 tan 1  tan 2  0.6

...(ii)

Add (i) & (ii) tan 1  tan  1  2 tan 1  tan  1  0.6 2tan 1  2.6 tan 1  1.3 1  52.43    2

Substitute 1  52.43 in eqn (i), we get

. . [ .  1   2]

4.204 Thermal Engineering - I

2 2   1.538 tan  1  tan 1 tan 52.43  1  56.98   2 Problem 4.62: A multistage axial flow compressor delivers 18 kg/s of air. The inlet stagnation condition is 1 bar and 20C. The power consumed by the compressor is 4260 kW. Calculate (i) Delivery pressure (ii) Number of stages (iii) Overall

isentropic

efficiency

of

the

compressor.

Assume

temperature rise in the first type is 18C, the polytropic efficiency of compression is 0.9 and the stage stagnation pressure ratio is constant.

(JNTU - August 2014 - Set 3)

Given Data:  m  18 kg/s; P 01  1 bar; T 01  20  273  293 K Power = 4260 kW; T02  18  293  311 K; p  0.9 (i) Delivery pressure: P 0N  P02  1 ln   P 01    P   T02  ln   T  01   P 02   T02    p  ln  ln    P  01    1  T01   P 02 P 01

1.4  311   0.9  ln    0.1878 0.4  293 

 e0.1878  1.207

Air Compressors

P 02  1  1.207  1.207 bar

Stagnation pressure at 2 P02  1.207 bar

Power required by the compressor  Power  m C p T ON  T01 4260  18  1.005 T0N  293 T0N  528.5 K

Also, the  p is given by  1  n  p        n1  0.9  1.4 n  3.15  0.4 n1 n  n  1 3.15  3.15n  3.15 2.15 n  3.15 n  1.465

During polytropic compression, n 3.15  TON  n  1  528.5     6.411   P01  T01   293 

P 0N

P 0N  1  6.411 bar

Delivery pressure P0N  6.411 bar

4.205

4.206 Thermal Engineering - I

(ii) To find no. of stages: Ns  P 02  P   01  T

Ns



P 0N

P01

aking ln on both sides,

 P 0N   P02  Ns  ln    ln   P P  01   01   P 0N   6.411  ln   ln   P  01    1  Ns   P 02   1.207  ln   ln   P  1   01   9.87 ~  10 stages Ns  10 stages

(iii) Overall isentropic efficiency of the compressor: overall isen 1

T 0N

 P 0N     P T 01  01   overall isen 

0.4

 6.411  T 0N  293    1

T 0N  T 01 T0N  T 01



 1.4  498.22 K  

498.22  293  0.8714 528.5  293

 87.14% Problem 4.63:_An 8 stage axial flow compressor takes in air at 20C at the rate of 180 kg/min. The pressure ratio is 6 and isentropic efficiency is 0.9. Determine the power required. (JNTU - January - 2014 - Set 2)

Air Compressors

4.207

Given Data: T 1  20  273  293 K; r 

P2 P1

 6;

 180  3 kg/s;  isen  0.9 m 60

Solution: 1

T 2

 P2     P T1  1

1

 P2    T 2  T1    P  1 0.4

T 2  293  6 1.4  488.9 K T2  488.9 K

isen  0.9 

T2  T 1 T2  T1 488.9  293 T 2  293 T2  510.64 K

Power required  Power  m C p T 2  T 1  3  1.005 510.64  293  656.18 kW Problem 4.64: An axial flow compressor is to have constant axial velocity of 250 m/s and 50% degree of reaction. The mean diameter of blade ring is 45 cm and speed is 18000 rpm. The exit angles of the blades are 25C. Calculate blade angle at

4.208 Thermal Engineering - I

inlet and workdone per kg of air with the help of velocity triangles.

(JNTU - Apr/May - 2013 Set 1)

C f  250 m/s; R d  0.5; Dm  0.45 m; N  18000 rpm; 2  25;

1  2  25 for 50% reaction u

 D m  N   0.45  18000  60 60 1

C r1

 424.12 m/s

1

From velocity triangle

C w1

u

u 424.12  tan  1  tan 1  Cf 250  1.696

C1

C f1

o

1 = 2 = 12 In le t velocity tria ng le 2 2 C f2

C r2

tan 1  1.696  tan  1

C2

C w2

 1.696  tan 25 2 = 1 = 4 2

 1.23

o

O utle t velocity tria ng le

1  50.88    2

Workdone/stage  u C w2  Cw1 C w2  C f tan  2  250  tan 50.88  307.42 m/s C w1  C f tan  1  250  tan 25  116.6 m/s

Workdone 

424.12 307.42  116.6  80.4 kJ/kg 1000

Problem 4.65: An axial flow compressor is to have constant axial velocity of 150 m/s and 50% degree of reaction. The mean

Air Compressors

4.209

diameter of blade ring is 35 cm and speed is 15000 rpm. The exit angles of the blade are 27. Calculate blade angle at inlet and workdone per kg of air with the help of velocity triangles. (JNTU - April/May - 2013 - Set 4)

Given Data: C f  150 m/s; Rd  0.5; Dm  0.35 m; N  15000 rpm; 2  27;

1

C r1

1

1  2  27 for 50% Rd.

C f1 C w1

u

Solution:

o

1 = 2 = 2 7 Inle t velocity tria ng le

 D m N   0.35  15000  u 60 60  274.89 m/s

2 2 C f2

C r2

By triangle,

C2

C w2

u 274.89  tan  1  tan 1  Cf 150

C1

2 = 1 = 5 2.9 3

o

O utle t velocity tria ng le

 1.833 tan 1  1.833  tan  1  1.833  tan 27  1.323

Blade angle at inlet 1  52.93    2 Work done/stage  u [C w2  Cw1] C w2  C f tan  2  150  tan 52.93  198.52 m/s C w1  C f tan  1  150  tan 27  76.43 m/s

4.210 Thermal Engineering - I

Work done 

274.89 198.52  76.43  33.56 kJ/kg 1000 W  33.56 kJ /kg

Problem 4.66: An axial flow compressor with compression ratio as 4, draws air at 20C and delivers it at 197C. The mean blade speed and flow velocity are constant throughout the compressor. Assuming 50% reaction blading and taking blade velocity as 180 m/s, find flow velocity and the number of stages. Take work factor = 0.82, 1  12 and 1  42 and Cp  1.005 kJ/kg K.

(JNTU - January 2014 - Set 3)

Given Data: r  4; T 1  20  273  293 K

1

C r1

C1

C f1

T2  197  273  470 K;

C w1

u

R d  0.5; u  180 m/s;   0.82

o

1 = 2 = 12 In le t velocity tria ng le

1  12  and 1  42; C r2

C p  1.005 kJ/kgK

1

2 2 C f2

T heoretical work required per kg

C2

C w2 2 = 1 = 4 2

o

O utle t velocity tria ng le

 C p T2  T1  1.005 470  293  177.9 kJ /kg

From velocity triangles u  tan  1  tan 1  tan 12  tan 42 cf  0.212  0.9  1.112

Chapter - 5

Refrigeration Mechanical Refrigeration and types - Units - Air refrigeration system, Details, Principle of operations application Vapour compression refrigeration - Calculation of COP - Effect of super heating and sub cooling - Desired properties of refrigerants and Common refrigerants - Vapour absorption system - Mechanical Details, Working and Principle - Use of p-h chart for calculation.

5.1 INTRODUCTION Refrigeration is defined as the science of providing and maintaining temperature below surrounding atmosphere. Refrigeration is a method to achieve and maintain low temperature by supplying work input continuously. 5.1.1 Fundamentals of Refrigeration: Refrigeration may also be defined as the process by which the temperature of a given space or a substance is lowered below that of the atmosphere or surroundings. In simple, refrigeration means the cooling of or removal of heat from a system. The equipment employed to maintain the system at a low temperature is termed as refrigerating system and the system which is kept at lower temperature is called refrigerated system. Refrigeration can be generally produced in one of the following ways. (i) By melting of a solid (ii) By sublimation of a solid (iii) By evaporation of a liquid. Generally by evaporation of liquid called refrigerant is used in commercial refrigeration.

5.2 Thermal Engineering - I

5.1.2 Applications of Refrigeration Refrigeration has a wide applications in a person’s daily life. Some of the important one’s are listed below. (i)

For comfort purpose: Air conditioning of residential buildings, offices, cinema houses, restaurants, departmental stores, hospitals, halls etc.

(ii)

For industrial purpose: cotton mills, textile industries, liquefaction of gases, treatment of metals, machine tool industries, marketing industries etc.

(iii)

For medicine purpose: Preservation of drugs, bloods, eyes, preservation of surgical equipments, human tissues etc,.

(iv)

For preservation of food products: Preservation of foods, highly perishable foods, produce ice creams, beverages, cold water, diary products etc.

(v)

For research work: For research under low temperature application, cryogenics study, rocket, fuels, synthetic rubber and oil factory.

(vi)

For computer functioning: Maintaining temperature in computer environments.

low

5.1.3 Important elements of a refrigeration system: (i)

A low temperature thermal sink

(ii)

A means of extracting energy from the sink, raising the temperature level of this energy and delivering it to a heat receiver.

(iii)

A receiver to which heat will be transferred from the high temperature high pressure refrigerant

(iv)

Means of reducing of pressure and temperature of the refrigerant as it returns from the receiver to the sink.

Refrigeration 5.3

5.1.4 Types of Mechanical Refrigeration system The important refrigeration systems are (i) (ii)

Vapour compression refrigeration system Vapour absorption refrigeration system

(iii)

Ice refrigeration system

(iv)

Air refrigeration system

(v)

Special refrigeration systems. (a) Thermoelectric refrigeration system (b) Adsorption refrigeration system (c) Cascade refrigeration system (d) Vortex tube refrigeration system

5.1.5 Unit of Refrigeration: (Ton of Refrigeration) The unit used in the field of refrigeration is Ton of refrigeration. One ton of refrigeration is defined as heat removed from 1000 kg of water at 0C to make (1000 kg) 1 ton of ice at 0C within 24 hours. Heat removal rate is one ton of refrigeration. h fg  Latent heat of the fusion = 301.5 kJ/kg. (It can

be taken from steam table) 1 ton of refrigeration 1 TR  1 TR 

1000 kg  301.5 kJ /kg 24  3600 sec kg  kJ kJ   kW sec  kg sec

1 TR  3.4892 kW

So, 1 TR  Heat removal rate of 3.49 kW or 210 kJ/min or 50 kcal/min.

5.4 Thermal Engineering - I

(1 to n of w ater) 1 00 0 k g of o w a te r a t 0 C Q out

w a ter b eco m e ice Q out

(1 to n) 1 00 0 k g of ic e o at 0 C

Q out Fig 5 .1

Tim e duratio n : 24 ho urs

For problems, take C p w ater  C pw  4.187 kJ/kg K C p ic e  2.09 kJ/kg K 1 TR  3.5 kW Problem 5.1: 2000 kg of water at 30C is cooled into ice at  20C and 6 hours. Determine heat removal rate in kW and in TR. Take latent heat of ice as 334 kJ/kg.  Solution: (m w  mass in kg/sec) Total Q remo ved  Q R water  Q R water to ice  Q R ice

1. Q removed from water at 30 C to make it 0C  Q R w ater  m w C pw T  2000  4.187 30  0 } 6  3600  11.631 k W

2. Q removed to make water at 0C to ice at 0C hfg  Latent heat  334 kJ/kg  Q removed  m w  hfg

Refrigeration 5.5



2000  334 6  3600

W a ter a t o 30 C

 30.9259 kW

3. Q removed

to make ice from

0C to  20 C: QR ice  Q R ice  m ice C p ice Tice 

2000  2.09  0   20 6  3600

 3.8704 kW

Total heat removed  Q R water  Q R water to ice  Q R ice  11.631  30.9259  3.8704 Q R  46.4273 kW

Q out (1 ) Q re m o v e d = Q R w a te r W a ter a t o 0 C (2 )

Q out

Q re m o v e d = L ate n t he a t ice a t o ice 0 C

Q out (3 ) Q = Q re m o v e d R ic e ice o a t-2 0 C

ice a t o -2 0 C

Tonne of Refrigeration TR 

QR

3.5



46.4273  13.265 3.5

. . [ . 1 TR  3.5 kW] TR  13.265 TR Problem 5.2: 5000 kg of water at 20C is converted into ice at  30C in 12 hrs. Determine heat removal rate in kW and in TR. Latent heat of freezing of ice 335 kJ/kg.

Solution: Q R  QR water  Q R water to ice  QR ice   m w [ C p wTw  h fg  C p ice Tic e 

5.6 Thermal Engineering - I



5000 12  3600 [ 4.187 20  0  335  2.09 0   20]

 53.303 kW TR 

QR 3.5

 15.229

. . [ . 1 TR  3.5 kW]

Tonnes of refrigeration TR  15.229 TR

5.2 AIR REFRIGERATION SYSTEM An air refrigeration system is a simple refrigeration cycle where in the working fluid is air. It has relatively low coefficient of performance despite its high operating costs. Hence, the usage of air refrigeration system has been predominantly limited to aircraft refrigeration system due to its low weight and the availability of cabin air as per the necessity. The notable feature about this system is that the refrigerant remains in gaseous state throughout the cycle. It can be primarily divided into: 



Closed system - It is also known as dense air refrigeration system. In this system, the refrigerating air is contained within the components of the system at all times. It usually operates in pressures exceeding the atmospheric pressure. Open system - In this system, the air is not circulated repeatedly within the system. Cooled air from the turbine directly comes in contact with

Refrigeration 5.7

the substances to be cooled, and is released into the atmosphere. Hence, the operating pressure is limited to the pressure inside the refrigerator. Consequently, an open cycle air refrigerating system offers low COP with high operating costs. Closed cycle Can operate pressure.

at

Open cycle high Limited to pressures.

Moisture is eliminated.

atmospheric

Moisture may choke valves present in system.

the the

The size of the compressor The size of compressor and expander is reduced due directly depends upon the to use of dense air. pressure inside the refrigerator. No fog is formed due to Fog formation due to absence of moisture. moisture at the turbine. Hence a drier is needed. Based on the principle of operation, an air refrigeration system can be classified into three types. 5.2.1 AIR REFRIGERATION CYCLES Refrigeration system is working under the following 1.

Reversed carnot cycle

2.

Bell-Coleman cycle

3.

Aircraft refrigeration cycle

5.3 REVERSED CARNOT CYCLE Heat is removed from sink at low temperature to source at high temperature by supplying work input.

5.8 Thermal Engineering - I

COP-Coefficient of Performance C.O.P means the ratio of the (Refrigeration effect) to the work input.

desired

effect

C.O.P for reversed carnot cycle 

Heat removed from cold body Work input per cycle



Refrigeration effect W



Q2 W



Q2 Q1  Q2



T2 T1  T2

Reversed carnot cycle will give more C.O.P. But this cycle is not practically possible. So this cycle C.O.P is used to rate the other cycles C.O.P. Pv diagram and TS diagram of Reversed Carnot

Cycle

f (a)

Fig . 5.3

e

(b)

Air is the working fluid in the reversed carnot cycle. So it is called air refrigerator.

Refrigeration 5.9

This cycle is used to find the maximum C.O.P. for given temperatures. From the T-S diagram Heat abstracted from the cold body  A rea dcfe  T 2  cd Workdone per cycle  Area bcda  T1  T2 cd Coefficient of performance, C.O.P 



Heat extracte d from the cold body Workdone per cycle T2  cd T1  T2 cd



T2 T1  T2

A Carnot cycle can run either as a refrigerating machine, or a heat pump or as a heat engine. (i) As a refrigerating machine H o t b od y R o om T 1 S ource

C.O.P ref  



Heat extracted from cold bo dy Wo rkdone per cycle

Q 1 = H ea t sup plie d H e at p um p

T2 T1  T2

W

Fig . 5.4 Q2

(ii) As a Heat engine C.O.P heat engine 



W ork obtained/cycle Heat supplied/cycle T 1  T 2  cd T1  T2  T1 T1  ba

C o ld bo dy T 2 S ink a tm osph ere

. .  . cd  ba

5.10 Thermal Engineering - I

(iii) As a Heat Pump If the desired effect is heat supplied to hot body, then the device is called heat pump. The C.O.P of heat pump 





C.O.P 

Desired effect Work input Q 1 Heat supplied  W

Q1 Q 1  Q2 Q1

Q1  Q 2



T1 T1  T2

1

T2

T1  T2

So COP of heat pump is always greater than COP of refrigerator working on reversed carnot cycle between same temperature limits T1 and T 2 by unity (1). Problem 5.3: A/C room is to be maintained at 20C. The atmospheric temperature is at 45C. The power given to the compressor is 3 kW. Determine the ton of refrigeration.

Solution: C.O.P 



T2 T1  T2 293 318  293

 11.72

Refrigeration 5.11

Also C.O.P   11.72 

Q2

T1 = 45+273 = 318 K

W Heat remo ved Wo rk input

Q1

Q2

W = 3 KW

R ef

3

Q 2  3  11.72 Q2

 35.16 kW Q2 

T2 = 20+273 = 293 K

35.16  10.045 TR 3.5

Heat removal rate  10.045 TR Problem 5.4: A carnot refrigerator requires 2 kW per ton of refrigeration to maintain a space at  40C. Determine 1. COP of carnot refrigerator. 2. Temperature of the hot body. 3. The heat delivered and COP when this device is used as heat pump.

Solution: T2   40  273  233 K

Work input = Power required = 2 kW Refrigeration effect = 1 Ton of refrigeration = 3.5 kW C.O.P  

Refrigeration effect Work input

3.5  1.75 2

But C.O.P for carnot cycle 

T 1 = 3 66 .14 3 Q1 H e at pum p D e vice Q2

T2 T1  T2

T 2 = 2 33

W

5.12 Thermal Engineering - I

1.75  T 1  233 

233 T 1  233 233  133.14 1.75

T 1  366.143  K Q1  Q2  W Q 2  Refrigeration effect  3.5 kW W  Power required  2 kW Q 1  3.5  2  5.5 kW

C.O.P of heat pump 

Q1 W



5.5  2.75 2

Note: C.O.P of heat pump (2.75) = 1 + C.O.P of Refrigerator (1.75). Problem 5.5: Find the least power to produce 400 kg of ice per hour at  10C from feed water at 20C. Assume specific heat of ice as 2.09 kJ/kg-K and latent heat 334 kJ/kg.

Solution: Least Power means more COP. More COP means it is reversed carnot cycle. h fg  334 kJ/kg

Mass of ice produced per hour   m w  m ice  400 kg/hr

T1 2 0 + 27 3 = 2 9 3 K Q1 Ref

W

Q2 T2 -1 0 + 2 7 3 = 2 6 3 K

Refrigeration 5.13



400  0.1111 kg/sec 1  3600

C.O.P 

T2 T1  T2



263  8.767 293  263

To Find Least Power C.O.P 

Net refrigeration effect Q 2  W W

Q 2  Q removed from water at 20C to make it ice at  10C  Q 2  m w [ Q R water  QR water to ice  Q R ice ] 

400 [ C pw Tw  hfg  C p ic e Tic e ] 1  3600



400 [ 4.187  20  0  334  2.09 0   10 ] 48.74 kW 3600

Least power W 

Q2 C.O.P



48.74  5.56 kW 8.767

Relative COP Relative COP is the ratio of actual COP to the Ac tual COP carnot COP. Relative COP  Carnot COP Actual COP 

Carnot COP 

Q2

W



QR W

T2 T1  T2



Q rejected or Q remo ved Power input

5.14 Thermal Engineering - I

Problem 5.6: An A/C room is to be maintained at 20C and the atmospheric temperature is 40C. The power required to run the compressor is 5 kW. Determine capacity of the refrigerator when relative COP is 50%.

Solution: Capacity refrigeration. Carnot C OP 

of

refrigerator

T2

T1  T2



is

given

by

ton

of

293 313  293

 14.65 Actual COP  Carno t COP  Relative COP  14.65  0.5

T1 = 40+27 3=313K

 7.325

Also Ac tual COP 

Q1

Q2 R ef

W

Q 2  Actual COP  W  7.325  5

W =5K W

Q2 T2 = 20+27 3 = 293K

 36.625 kW

Capacity of refrigerator  Q 2 in TR 

36.625 kW 3.5 kW

 10.46 TR Problem 5.7: The capacity of the refrigerator working on reversed Carnot cycle is 280 tonnes when operating between  10C and 25C. Determine (i) quantity of ice produced within

Refrigeration 5.15

24 hours when water is supplied at 20C (ii) Minimum power required. Latent heat of ice is 335 kJ/kg

Solution: (i) Heat removed from water at 25C to make it ice at  10C per kg of water Q 2  [Q R Water  Q R

Water to rice

 QR ice]

 [C P T w  h fg  C P Tice ] w ice  [4.187 20  0  335  2.09 0   10] Q 2  439 kJ/kg.

Heat extracting capacity of refrigerator  280 tonnes. 1 tonne  3.5 kW  Heat extracting capacity  280  3.5  980 kW.

Quantity of ice produced in 24 hours M ice 

980  24  60  60  192874 kgs or 193 tonnes . 439

(ii) Minimum power required: T1  25  273  298 K; T2   10  273  263 K

C.O.P 

T2 T1  T2



263  7.51 298  263

C.O.P 

Heat extracting capacity Power required

Power required 

Heat extracting capacity C.O.P



980  130.49 kW 7.51

5.16 Thermal Engineering - I

Problem 5.8: A cold storage plant is required to store 20 tonnes of food. The temperature of the food when supplied is 25C; storage temperature of food is  8C. Specific heat of food above freezing point is 2.93 kJ/kgC. Specific heat of food below freezing point  1.25 kJ/kgC; Freezing point of food is  3C. Latent heat of food is 232 kJ/kg. If the cooling is achieved with in 8 hrs; find (i) Capacity of the refrigeration plant (ii) C.O.P of Carnot cycle (iii) If actual C.O.P is

1 5th

of

the Carnot C.O.P, find out the power required to run the plant.

Solution: (i) Heat removed from 1 kg of food  Q R  mf [Q R above  Q R Freeze  Q below ]  1 [CP above T  H fg  C P below T]  1 [2.93 25   3  232  1.25  3    8] Q 4  320 kJ/kg

Heat removed by the plant 

Q R  20  1000

8  60  60



320  20  1000  222.22 kJ/sec 8  3600  222.22 kW

Capacity of Refrigeration plant 

Heat removed by plant 222.25   63.49 to nnes or TR 3.5 3.5

(ii) COP of Carnot Cycle C.O.P 

T2 T 1  T2



 8  273 265  8.03 .  25  273   8  273 298  265

Refrigeration 5.17

(iii) Power required Actual COP 

1 1  COP   8.03  1.606 5 5

Actual C.O.P 

Refrigeration or Heat rem oved 222.22  W Work done/min Power

Power required W 

222.22  138.49 kJ/sec or kW 1.606

5.4 BELL-COLEMAN CYCLE If the isothermal processes in carnot cycle is replaced by constant pressure processes, then the cycle is called Bell-coleman cycle. Process 1-2 Air is compressed isentropically during first part of the stroke. During the remainder stroke, the compressed air (at high temperature) is forced into cooler at constant pressure. Process 2-3 The cooler cools the air at constant pressure. Process 3-4 The cold air is now drawn into expansion cylinder and expanded isentropically. During isentropic expansion, the air is further cooled below the refrigerator space temperature. Process 4-1 The cold air passes through the refrigerator and absorbs the heat at constant pressure.

5.18 Thermal Engineering - I

Closed cycle air-refrigerator working on Bell - Coleman cycle.

w ate r ou t

C oo le r

3

2 H ea t E xc ha ng er (co oler) 4

w ate r in Air

3 2 C om pres so r

R efrige ra tor 1 E xp an de r

C ou pling

A ir

A ir E le ctric m o to r

B ea rin g Fig. 5.5

B ea rin g

Conclusion The heat is absorbed from refrigerator and rejected into the circulating water of cooler. So the refrigerator is maintained at low temperature. The bell-coleman cycle consists of 1-2 isentropic compression process [ Pv  C ] 2-3 constant pressure cooling process [ P  C ]

t .p

process [ Pv  C ]

r.

expansion

P2

ns

isentropic

2

Co

3-4

T

3

g

pv

4-1 constant pressure heat absorbing process [ P  C ]

C

g

pv

=c P1

=c

s on

r. t .p 1

S

4 Fig. 5.6

Refrigeration 5.19

C.O.P isentropic process 

T4 T3  T4

But in actual practice, perfect isentropic process is not possible. So the process will be polytropic process. Pvn  C.

C.O.P polytropic process 

T4  n 1  n  1     [ T3  T4 ]   

BELL-COLEMAN CYCLE B ell C olem en Cycle C onsta nt pressure co olin g

P 3 Isentro pic g expansion P V =c (or) P olytro pic n ex pansion (P V =c)

2

Isentro pic co m p res sion pv g= c (or) P olytro pic n co m p res sion (P V =c)

Q out

C ons ta nt press ure hea t absorp tion

4

1 Q ab so rbe d = C p(T 1 -T 4 ) V

Fig. 5.7

Polytropic Law n1

n1

 P2  n T3  P3  n   ;   P P T4 T1  1  4 T2

n1

 P2  n   P  1

Heat absorbed from refrigerator (cold chamber) per kg of air  CpT1  T4 Heat rejected in the cooling tower per kg of air  CpT 2  T 3

5.20 Thermal Engineering - I

Work done  

C.O.P 

n [ P 2v2  P 1v1  P 3v3  P 4v4 ] n1 n R [ T 2  T 1  T 3  T4 ] n1 Hea t absorbe d W ork

Problem 5.9: A refrigerator working on Bell-coleman cycle operates between pressure limits of 1.05 bar and 8.5 bar. Air is drawn from the cold chamber (refrigerator) at 10C. Air coming out of compressor is cooled to 30C before entering the expansion cylinder. Expansion and compression follow the law Pv1.35  constant. Determine theoretical C.O.P. of the system. (JNTU - May/June 2009)

Solution: P 1  1.05 bar; P 2  8.5 bar; T1  10  273  283 K T 3  30  273  303 K ; Pv 1.3  C

Consider Polytropic Compression 1-2 n1

P

 P1  n   P T2  2 T1

35 1.

pv

=c

 486.69 K

283 0.58147

io n



ns

0.58147

pa

T1

C oo ling 2 ch am ber

ex

0.35

 1.05  1.35   0.58147   8.5  T2 

3

co

m

pr

es s io n C old 4 ch am ber 1v

Refrigeration 5.21

Consider Polytropic Expansion 3-4 n1

 P1  n   T3 P  2 T4

 0.58147

T 4  T 30.58147   3030.58147   176.185 K Q a -Heat

(absorbed) extracted from cold chamber

(refrigerator) per kg of air. Q a  C p T 1  T 4   1.005 283  176.185   107.348 kJ/kg Q r-Heat rejected in the cooling chamber per kg of air.  C p T2  T3  1.005486.69  303  184.61 kJ/kg

Since the compression and expansion are not isentropic, the difference between heat rejected and heat absorbed is not equal to work done. So work done is found as follows for polytropic process. Work done T1  283 K; T 2  486.69 K; T3  303 K; T 4  176.185 K W 

n R [ T 2  T1  T 3  T 4 ] n1 1.35  0.287 [ 486.69  283  303  176.185  ] 0.35

 85.1 kJ/kg C.O.P 

Heat absorbed 107.348   1.261 85.1 Work done

5.22 Thermal Engineering - I

Problem 5.10: A refrigerator of 6 ton capacity working on Bell-coleman cycle has an upper limit of pressure of 5 bar. The pressure and temperature at the start of the compression are 1 bar and 15C respectively. The compressed air is cooled at a constant pressure to a temperature of 40C enters the expansion cylinder. Assuming both expansion and compression processes to be isentropic with   1.4, calculate (i) C.O.P. (ii) Quantity of air in circulation per minute. (iii) Piston displacement of compressor and expander. (iv) Bore of compressor and expansion cylinders. The refrigerating unit runs at 240 r.p.m. and is double acting. Stroke length = 250 mm. (v) Power required to drive the unit. For air take   1.4 and Cp  1.005 kJ/kg K.

Solution: Capacity  6T.R  6  3.5  21 kW  Refrigerating effe ct produced by the refrigeration. P 2  P 3  5 bar; P 1  P 4  1 bar T 1  15  273  288 K; T3  40  273  313 K

Isentropic Compression Process 1-2 1

0.4

 P2    5  1.4     P T1 1  1 T2

P 2= P 3= 5b ar 2 C o oling cha m be r  P v =c Co



P v =c

so

on

re s

nsi

mp

pa

r

T2  1.58382  T 1

3

Ex

 1.58382

P (b ar)

cyl in d er

 1.58382  288

P 1= P 4= 1 bar

 456.14 K T2  456.14 K

4 C o ld ch am b er(refrige rator)1 v(m 3 ) volum e

Refrigeration 5.23

Isentropic Expansion Process 3-4 1

1

 P4     T3 P  3 T4

0.4  1.4

1   5

 P1     P  2

 0.6314

T 4  0.6314  T 3  0.6314  313  197.624 K

To Find C.O.P Since both compression and expansion are isentropic, C.O.P isentropic 

T4

T3  T4



197.624  1.7128 313  197.624 

 (ii) To Find Mass of Air in Circulation: m a

R.E = Refrigerating Effect R.E per kg of air  C pT 1  T 4  1.005 288  197.624   90.828 kJ/kg

R.E. produced by the refrigerator  C apacity of refrigerator  6  3.5  21 kW

 Capacity of the refrigerator  m a  R.E. per kg of air  where m a  Mass of air in circulation.  21 kJ Capacity of refrigerator  ma  R.E. per kg of air 90.828 sec  kJ/kg  m a  0.2312 kg/sec

5.24 Thermal Engineering - I

(iii) To Find Piston Displacement of Compressor  V   V  Volume of air in circulation in m 3/sec   Volume corresponding to point 1 i.e. V 1.   P 1V 1  maRT1   m RT 1 V1  P1 

[P 1 in kPa ; 1 bar  1  10 2 kPa]

0.2312  0.287  288 1  10 2

 0.1911 m 3/sec

(iv) To Find Dia. of Compressor Cylinder  V1 for double V s  Swept volume per stroke V s N 2 60 acting compressor. N  R.P.M. V s  Swept volume per stroke 

0.1911  60 m 3  sec  0.023887 m 3 sec 2  240

V s  0.023887 m 3

Also, V s 

  d 2c  L 4

where L  Stroke length d c  dia. of compressor cylinder

Refrigeration 5.25

V s  0.023887 

 2 d  0.25 4 c

d 2c  0.12165 d c  0.3488 m

To Find Diameter of Expander Cylinder  V 4  Volume of air in circulation in m 3/sec .  = Volume corresponding to point 4 is V4.   P 4V 4  maRT4   m aR T4 0.2312  0.287  197.624  V4  P4 1  10 2  0.13113 m 3/sec V s  Swept volume per stroke 

acting expander. 

0.13113  60 2  240

 0.01639 m 3/sec Vs  0.01639 

  d2e  L 4   d2e  0.25 4

where de  dia. of expander cylinder. d 2e  0.08347 d e  0.2889 m

 V4 2  N /60

for double

5.26 Thermal Engineering - I

(v) To Find the Power Required to Run the Unit C.O.P  1.7128 

R.E W 6  3.5 W

where W  Power required W

6  3.5  12.26 kW 1.7128

Power required

 12.26 k W

Problem 5.11: An air refrigerator working on Bell-coleman cycle takes air into the compressor at 1 bar and  7C and it is compressed isentropically to 5.5 bar and it is further cooled to 18C at the same pressure. It is then expanded into the refrigerating chamber. Find the C.O.P. of the system if (a) the expansion is isentropic, (b) the expansion follows the law Pv1.25  constant. Take   1.4 and Cp  1.005 kJ/kg K for air.

Solution: T 1   7  273  266 K T 3  18  273  291 K

(a) If both Isentropic

the

Compression

are

1.

Pv

4

Co

pa ns io n

 1.6275

=c

0.4

 5.5  1.4    1 

Ex

1

 P2     T1 P  1

Expansion

P P 2 =P 3 3 C ooling 2 5.5bar cham be r

Pv  Constant T2

and

P 1 =P 4 1 bar

4

m

pr

es s io n C old cham be r 1 V

Refrigeration 5.27

T 2  1.6275  266  432.93 K T 2  432.93 K 1  

 P4   T3  P 3 

T4

1

 P1     P  2

0.4

 1  1.4  0.6144    5.5 

T 4  0.6144  T 3  0.6144  291  178.797 K

C.O.P 

(or) C.O.P 

T4 T3  T4 T1

T2  T1





178.797  1.594 291  178.797 

266  1.594 432.93  266

So, C.O.P.  1.594 (b) If the Compression is Isentropic and Expansion is Polytropic n1

T4

 P4  n   T3 P  3

P

0.25  1.25

 1    5.5 

2

3 

 0.7111

T4  0.7111  291  207 K

Work done   n     RT2  T 1    RT 3  T 4     n 1 1       1.4   0.287  433  266    0.4   1.25  0.287 291  207    0.25  

P v =c

P v 1.25=c 4

O

4

1 v

5.28 Thermal Engineering - I

from the Cold Chamber: Process 4-1

Qabsorbed

Q absorbed  C pT 1  T 4  1.005266  207  59.3 kJ/kg C.O.P 

Q absorbed W



59.3  1.259 47.11

C.O.P  1.259

So, Problem

5.12: An air refrigerator working on Bell-coleman

cycle has a pressure limits of 1 bar and 4 bar. The temperature of air entering the compressor is 15C and entering the expansion cylinder is 30C. The compression follows the law Pv1.35  C. The expansion follows the law Pv1.25  C. Take for air Cp  1.005 kJ/kg K; Cv  0.718 kJ/kg K;

R  0.287 kJ/kg K.

Find (i) C.O.P (ii) If mass flow rate of air is 0.417 kg/sec, find the refrigeration capacity of the system.

Solution: P 1  1 bar; P 2  4 bar ; T 1  15  273  288 K ; T 3  30  273  303 K ;

To Find T 2 n1  1

 P2  n1   P T1  1 T2

0.35

 4  1.35  1.4325   1

Refrigeration 5.29

T2  1.4325  288  412.56 K T2  412.56 K

To Find T 4 n2  1

n2  1

 P4  n2   T3  P 3  T4

 P1  n2   P  2

0.25

 1  1.25   4

 0.75785 T 4  0.75785  303  229.631 K T 4  229.631 K

Heat Absorbed from the Refrigerator (Cold Chamber) per kg of air (Refrigeration Effect/kg of air) R.E./kg of air  C pT 1  T 4  1.005 288  229.631   58.661 kJ/kg   n2   n1 RT2  T1    RT 3  T 4  Work done   n  n    2 1   1 1  1.35    0.287 412.56  288  0.35    1.25    0.287 303  229.631    0.25   137.89  105.285  32.6055 kJ /kg C.O.P 

R.E/kg of air 58.661   1.799 W 32.6055

Refrigeration Capacity

 R.E in kW = R.E/kg of air  m

5.30 Thermal Engineering - I

 where m  Mass of air in circulation in kg/sec.

R.E in kW  58.661  0.417  24.46 kW R.E in TR  Capacity

24.46  6.989 ~  7 3.5

 7 tons of Refrigeration

5.13: The

Problem

capacity

of refrigerator

working

on

Bell-coleman cycle is 50 tons of refrigeration. The temperature of air entering the compressor is 8C. and the temperature before entering into expander is 27C. Find (a) Actual C.O.P of cycle, (b) Power required to run the compressor. The mass of the air circulated in the refrigerator is 1.67 kg/sec. The law of compression and expansion is Pv1.3  C. Assume   1.4; Cp  1.005 kJ/kg K for air.

Solution: To Find T 4 Heat removed from the cold chamber per kg of air = R.E/kg of air. R.E in kW  50  3.5  175 kW  R.E in kW  m  R.E/kg of air 175  1.67  R.E/kg of air R.E/kg of air  104.79 kJ/kg

But R.E/kg of air  C pT1  T 4 104.79  1.005 281  T 4 T 4  176.731 K

[T 1  8  273  281 K] [T3  27  273  300 K]

Refrigeration 5.31

To Find T 2 n1

 P3  n   T4 P  4

n1

 P2  n   P  1

T3

n

1.3

 T3  n  1  300  0.3    9.905   T P1  176.731   4 P2

n1

P 3

2

 P2  n   P T1  1

T2

1.3

P v =c

0.3

C old 1 4 cham ber v

 9.905  1.3  1.6975 T2  1.6975  281  476.996 K

To Find Power W  Work done 



n R [ T2  T1  T3  T4 ] n1 1.3  0.287 [ 476.996  281 0.3  300  176.731  ]

 90.45 kJ/kg of air

 Power  m  W  1.67  90.45  151 kW To Find C.O.P C.O.P 

R.E in kW 175   1.159 W in kW 151

C.O.P 

R.E/kg o f air 104.79   1.159 90.45 W /kg of air

(or)

5.32 Thermal Engineering - I

Problem 5.14: An air refrigerator working on reversed joule cycle works between pressure of 1 bar and 8 bar. The temperature of the air entering the compressor is 7C and after compression the air is cooled to 27C before entering the expansion cylinder. Expansion and compression follow the law PV1.25  constant. Determine theoritical COP of machine. Take Cp  1 kJ/kgK, Cv  0.7 kJ/kgK

(JNTU - Dec 2014)

Solution: P 1  1 bar, T 1  7 C  280 K P

P 2  8 bar, T 3  27 C  300 K

3

2

For polytropic compression n1

 P1  n   P T2  2 T1

4

V

0.25

280  1  1.25    T2  424.40 K T2 8

For polytropic expansion n1

 P4  n   T3 P  3 T4

1

n1

 P1  n   P  2

0.25

 1  1.25    T4  197.926 K 300  8   Heat Absorbe d     or , Q a  C p T 1  T 4    R. E/kg of air  T4

1 424.40  197.926   226.474 kJ/kg

Refrigeration 5.33

Work Done, w  

n R [T2  T1  T3  T4] n1 1.25  0.287 [924.40  280  300  197.926 ] 0.25

 60.73 kJ /kg COP 

Qa

w



226.474 60.73

COP theo  3.7287 Problem 5.15: In a Bell - Coleman refrigerator, air is taken in at 1 bar and a temperature of  8C. The compression ratio is 4. The expansion and compression follow the law PV1.2  constant. The air is cooled at the upper pressure to 25C. Find the MEP of the cycle and the COP.

(JNTU Dec 2014)

Given: Compression ratio

V1 V2

 4; PV 1.2  C

T 1   8 C  265 K; P 1  1 bar; T3  25 C  298 K

To find The MEP of the cycle and COP Solution: P 1 V n1  P 2 V n2

 V1    P1  V2  P2

n

P 2  1  4 1.2  5.27 bar

For polytropic compression

5.34 Thermal Engineering - I n1

 P2  n   T1 P  1 T2

0.2

T 2  265  5.27  1.2 T 2  349.58 K

For polytropic expansion n1

P 3

2

T4

 1  1.2  298  5.27 

io n

io n

0.2

ss

ns

re

pa

mp

ex

co

 P1  n   T3 P  2 T4

1

4

T 4  225.74 K

V2

V VS

V1

Q a, Heat Absorbed  C p T 1  T 4  1.005 265  225.74  39.45 kJ/kg

WorkDone, W  

n R [T2  T1]  T3  T4] n1 1.2  0.287 [349.58  265  298  225.74 ] 0.12

 21.22 kJ /kg COP 

He at Absorbed 39.45   1.86 Work done 21.22

Mean effective pressure 

Work Done 21.22  Stroke volume V 1  V 2

Refrigeration 5.35

For unit mass, m  1 P 1 V 1  m RT1 V1  V1 V2

RT 1 P1



0.287  265 1  10 2

 0.76 m 3/kg

 4 compression ratio

V 2  0.76 /4  0.19 m 3/kg

Mean Effective pressure P m 

21.22  37.23 kNm  2 0.76  0.19

 COP  1.86 MEP  37.23 kNm 2 Problem 5.16: A dense air refrigerator is used for absorbing 2000 kJ/min of heat. The pressure limits for compressor and expander are 16 bar and 4 bar. Compressor sucks in air at 40C and the temperature of air entering expander cylinder is 20C. Mechanical efficiency of the system is 82%. If the compressor and expander are double acting. Find (i) Power required to run the system. (ii) Bore of compressor and expander. (iii) Ice tonnage at 0C/24 hours. Speed of compressor is 250 rpm and stroke of compressor and expander is 15 cm. Assume isentropic compression and expansion.

(JNTU-June/July 2014)

Solution: T 1  40  C  313 K, T 3  20 C  293 K P 1  4 ba r; P 2  16 bar

5.36 Thermal Engineering - I

mech  0.82 ; y  2

P 1 6 bar

2 co

4 bar

io n

io n

1

 P2     P T1  1

ss

ns

(1-2) Isentropic compression T2

re

pa

mp

ex

L  0.15 m

3

1

4 V

0.4

 16  1.4 T2     313  465.11 K  4  (3-4) Isentropic expansion T4 T3

1  

 16    4 

0.4

 4  1.4 T4     293  197.17 K  16 

Refrigerating effect/kg  C p T1  T4  1.005 313  197.17  116.40 kJ/kg  Mass of air/minute, m 

R.E 2000  17.18 kg/min  R.E/kg 116.40

Compressor work W com p  

 mR T2  T1 1 1.4  17.18  0.287 465.11  313 0.4

 2625 kJ /m in

Refrigeration 5.37

 mR T3  T4 Expander work W exp  1 

1.4  17.18  0.287 293  197.17  0.4

 1653.76 kJ /m in

Network done

W  2625  1653.76  971.23 kJ/min

Power required

P

Actual power



971.23  16.18 kW 60 P 16.18   19.73 kW 0.82 m ech

Bore of compressor   m RT 1 17.18  0.287  313  V1  P1 4  10 2  3.85 m 3/min V 1 per stroke  V1  7.7  10  3 

3.85  7.7  10  3 m3 2  250 . . [ . 2 for double acting]  2 d l 4 c

 2 15 d  4 c 100

dc  0.255 m  255 mm

Bore of expander  m RT 4 17.18  0.287  197.17   V4  P4 4  10 2  2.43 m 3/min

5.38 Thermal Engineering - I

V 4 per stroke  V4  4.86  10  3 

2.43  4.86  10 3 m 3 2  250  2 d l 4 e

 2 15 d  4 e 100

de  0.203 m  203 mm

To find Ice tonnage Heat to be removed from water at 0 C to ice at 0 C Refrigeration effect in kJ/24 hours  R.E kJ/min  60  24  2880000 kJ/day

Ice produced



RE / day C p of ice



2880000 335

 8597.01 kg  8.597 tonnes

Refrigeration 5.39

5.5 VAPOUR COMPRESSION REFRIGERATION A simple vapour compression refrigeration system is shown in Fig. 5.8. M aintained at H igh tem perature

Atm osphere SO URCE 3

2

C ON D EN SER

Q1

Throttling D evice 1

C om pressor

1

3

4

2

R e frig erato r

W ork input

EVAP OR ATO R

SINK

W

W = h 2 - h1 Q 2 = R .E = h1 - h4

4 Space to be cooled

SO URCE

M aintained at low tem perature Fig. 5.8

SINK

The vapour compression cycle consists of following processes. Process 1-2 Isentropic compression The vapour at low temperature and pressure (state 1) enters the compressor. Then it is compressed isentropically to state 2. Work input  W  h 2  h 1 in kJ/kg

5.40 Thermal Engineering - I

Process 2-3 Condensation Compressed vapour (high pressure and temperature) is passing through condenser where it is condensed into liquid. While condensing into liquid, it rejects heat to the atmosphere. But the temperature and pressure remains same. Q rejected in condenser  h2  h3 in kJ/kg.

Process 3-4 Throttling The high temperature and high pressure liquid is passing through throttling device (Expansion valve or capillary tube) and throttled down to low pressure and low temperature liquid. Here h 3  h 4. Process 4-1 Evaporation The low pressure and low temperature liquid is now passing through evaporator. In the evaporator, the liquid absorbs heat (Extracts heat) from the room or space to be cooled and the liquid becomes vapour. The low pressure and low temperature vapour again goes to compressor and the cycle repeats. This process gives the desired effect. i.e., Refrigerating effect is obtained by this process. So, R.E  h 1  h 4 in kJ/kg

 Q rejected in kW  m h2  h3 Q extracted in kW  R.E in kW   m h1  h4

2

3

n) a n s io ( E x p o t t li n g Th r

R.E h 1  h 4  W h2  h1

4

Isen tro pic co m pressio n

C.O.P 

T C o nsta nt pre ssure he at rejection o r C ond ensa tion

1 s

Fig. 5.9

Refrigeration 5.41

 Power W in kW  m h 2  h 1  where m  Mass flow rate of refrigerant in kg/sec.

The Following Formulae can be used to Find the Compressor Dimensions V s  Swept Volume or Stroke Volume

 mv1  2  d L for single acting N 4  vol  no . of cylinders 60  m v1  for double acting N  vol  2  no. of cylinders 60 where v1  specific volume m 3/kg N  r.p.m    v1  vol  Volumetric efficiency  1  K   1  v   2

where K  Clearance ratio 

vc vs

vc  Clearance volume; vs  Swept volume

If K is not given, K  0, then vol  1. d  dia. of cylinder o r bo re L  stroke length

5.42 Thermal Engineering - I

5.5.1 Different Conditions of the Vapour 1. Beginning of compression is dry and end of compression is super heated

F ig. 5.10

2. End of compression is dry and Beginning of compression is wet

Fig. 5.11

Refrigeration 5.43

3. End of compression is wet and Beginning of compression is also wet. P P2= P3

h 3 =h 4

T

2

s= c

h= c

P 1= P 4

P =C

f

1

4

2

P =C

3

s= c

3

4

1 g

f g

S

h F ig. 5.12

4. Beginning of compression is wet; End is superheated. p

t 3

4 f

2

f

1

2O

3

4

2

1 g

g h Fig. 5.13

s

5.44 Thermal Engineering - I

5. Beginning of compression is superheated; [Also end of compression is superheated] 2

p

t 3

P3 = P2 = P2

2

O

2

3

2

O

O

1 1 1 g

4

4

O

f

h

Advantages system

1

O

s

Fig. 5.14

of

vapour

compression

(i) It has refrigeration.

smaller

size

for

Refrigeration

given

capacity

of

(ii) It has less running cost. (iii) It can be employed over a large range of temperatures. Disadvantages (i) Leakage cannot be detected easily. (ii) Mechanical efficiency is low. Problem 5.17: A refrigerator using Ammonia works between the temperatures  10C and 25C. The gas is dry at the end of compression and there is no undercooling of liquid. Calculate the theoretical C.O.P of the cycle.

Refrigeration 5.45

The properties of ammonia are given below: Temperature

Liquid heat

Latent heat

Liquid entropy

C

hf kJ/kg

hfg kJ/kg

sf kJ/kg K

25 (298 K)

100.5

1231

0.348

 10 (263 K)

 33.58

1348.2

 0.1382

Note: sfg 

h fg T

Solution: P

T

3

3

2

2

f 4

T 1= 2 6 3

1

4

h

Theoretical C.O.P 

g S

R .E W

where R.E  Refrigerating effect  h1  h4 W  h2  h1

To Find h2 h 2  h g at 298 K h g  h f  hfg  100.5  1231  1331.5 kJ/kg h 2  1331.5 kJ/kg

1

5.46 Thermal Engineering - I

To Find h1 h 1  h f  x1h fg at 263 K from table.   33.58  x 11348.2 

After finding x1, we can calculate h 1 s2  sg at 298 K hfg 1231  0.348  sg  sf   4.479 298 T s2  4.479 kJ/kg K s2  s1

. . [ . Isentropic compression]

hfg s1  4.479  sf  x 1 at 263 K T 4.479   0.1382  x1 

1348.2 263

x 1  0.9007 So, h 1   33.58  0.9007 1348.2   1180.71 h 1  1180.71 kJ/kg

To Find h4 h4  h3 h 3  h f at 298 K  100.5 kJ/kg  h4 h 4  100.5 kJ/kg

. . [ . Throttling process]

Refrigeration 5.47

C.O.P 

R.E h1  h4  W h2  h1 

1180.71  100.5 1331.5  1180.71

 7.164 C.O.P  7.164 Problem 5.18: The pressure in the evaporator of an ammonia refrigerator is 1.902 bar and the pressure in the condenser is 12.37 bar. The refrigerant is in dry saturated condition at the entry of the condenser. Calculate the refrigerating effect per unit mass of refrigerant and the COP. [Apr/May 2008 -AU] Similar type (JNTU May/June 2009)

Solution:

Fig.

P 1  P 4  1.902 bar, P2  P3  12.37 bar

Refrigerating Effect  h1  h4 kJ /kg Work input

W  h2  h1

5.48 Thermal Engineering - I

COP 

RE h1  h4  W h2  h1

To find h 1, h 2, h 4 from Refrigeration Table, we can take following temp C

P in bar

hf

hg

sf

sg

 20C

1.902

89.78

1420.02

0.3684

5.6244

253 K

P1  P4

32C

12.37

332.71

1469.94

1.2350

4.9624

305 K

P2  P3

To find h 2  h 2  h g at P 2  12.37 bar  1469.94 kJ/kg  h2

To find h 1  s2  sg at P 2  12.37 bar  4.9624 kJ/kgK s2  s1  4.9624 kJ/kg K

Also

s1  sf  x1 sg  sf at P 1  1.902 bar

4.9624  0.3684  x1 5.6244  0.3684 x 1  0.874 h 1  h f  x1 h g  h f at P 1  1.902 bar  89.78  0.874 20.02  89.78  h1  1252.40 kJ/kg

To find h 3 and h 4 h 3  h f at high pressure P 3  P 2  12.37 bar  332.71 kJ/kg

Refrigeration 5.49

h3  332.71 kJ/kg

During throttling process, h3  h4 h4  332.71 kJ/kg

Refrigerating Effect RE  h 1  h 4  1252.4  332.71 RE  919.69 kJ/kg COP 

RE h1  h4 919.69    4.227 W h2  h1 1469.94  1252.4

COP  4.227 Problem 5.19: Find the theoretical C.O.P for CO2 refrigerator working between the temperature limit range of 25C and 5 C. The dryness fraction of the CO2 gas during the suction stroke is 0.6. Temp.

Liquid hf

Vapour sf

hg

Latent heat sg

hfg kJ/kg

kJ/kg kJ/kg K kJ/kg kJ/kg K 25C 298K

81.23

0.251

202.7

0.6297

121.423

 5C 268K

 7.54

 0.042

237

0.842

245.4

Solution: Entropy at (1) = Entropy at (2) [... Isentropic compression] s1  s2 s 1  s f  x1

h fg T

at 268 K

s1   0.042  0.6 

245.4 268

5.50 Thermal Engineering - I

 0.5074  s2

hfg s 2  s f  x2 at 298 K T 0.5074  0.251  x2 

121.423 298 T

x 2  0.6293

3

2

298K

To Find h1 h 1  h f  x1h fg at 268 K   7.54  0.6  245.4  139.7 kJ/kg

To Find h2 h 2  h f  x2h fg at 298 K  81.23  0.6293  121.423  157.64 kJ/kg

To Find h4 h 4  h3 h 3  h f at 298 K  81.23  h4 h 4  81.23 kJ/kg

To Find C.O.P C.O.P 

R.E h1  h4  W h2  h1

f

h 3 =h 4 s 1 =s 2

g

268K 4

1

s

Refrigeration 5.51



139.7  81.23 157.64  139.7

 3.26 C.O.P .  3.26 Problem

5.20: A

vapour compression refrigerator works

between the pressure limits of 60 bar and 25 bar. The working fluid is just dry at the end of compression and there is no under cooling of the liquid before the expansion valve. Determine (i) COP of the cycle and (ii) capacity of the refrigerator if the fluid is at the rate of 5 kg/min. Pressure

Saturation temperature

Enthalpy (kJ/kg)

Entropy (kJ/kg K)

hf

hg

sf

sg

(bar)

(K)

Liquid

Vapour

Liquid

Vapour

60

295

151.96

293.29

0.554

1.0332

25

261

56.32

322.58

0.226

1.2464

(Nov/Dec 2011 - AU)

Solution: P 2  P 3  60 bar P1  P4  25 bar h1  hf1  x1 hfg s1  s2 sf1  x1 sfg  sg 1 2 0.226  x1  1.2464  0.226   1.0332 x1  0.791 h 1  h f1  x1 h fg 1  56.32  0.791  322.58  56.32

5.52 Thermal Engineering - I

h1  266.93 kJ/kg h 2  293.29 kJ/kg, h4  h3 h3  hf

S 3 S 4 (S 1 =S 2 )

h 3  151.96 kJ/kg  h 4  151.96 kJ/kg COP 

h1  h4 h2  h1

266.93  151.96  4.66 COP  293.29  266.93

(h f3 =h 4 )

Capacity of the refrigerator Heat extracted (or) refrigerating effect.  5  m h1  hf2  266.93  151.96    9.58 60 COP 

9.58  2.74 TR 3.49

Problem 5.21: A refrigerant plant using CO2 as a refrigerant works between 298 K and 268 K. The dryness fraction of CO2 is 0.8 at entry of compressor. Find out the ice formed per month if the relative efficiency is 50%. Take that ice is formed at 0C from water at 10C. The quantity of CO2 circulated is 6 kg/min. Assume Cp for water as 4.187 kJ/kg.K and latent heat of fusion of ice as 335 kJ/kg.K Properties of CO2 are given below

Refrigeration 5.53 Temperature

Liquid Heat

Latent Heat

Entropy of

K

kJ/kg

kJ/kg

Liquid kJ/kg.K

298

81.25

135

0.2513

268

 7.53

245.8

 0.04187

T

Enthalpy of point 1 3

cond

2

1

  7.53  0.8  245.8 h1  189.1 kJ/kg

com p

h 1  h f1  x1 h fg

exp 4 evop

s1  s2

1 S

sf1  x 1 sfg  sf2  x2 sfg 1

S 1 =S 2 2

x2  ?  0.04187 

0.8  245.8 135  0.25134  x2  268 298

0.6918  0.2513  x 2 0.4533  x2  0.97

Enthalpy at point 2 h 2  h f2  x2 h fg  81.25  0.97  135 2 h2  212.2 kJ/kg COP 

Refrigerant effect hf  h f3 189.1  81.25   workdone h2  h1 212.2  189.1

Theoretical COP 

107.85  4.668 23.1

5.54 Thermal Engineering - I

Actual COP  0.5  4.668  2.33 Q removed  RE 

107.85  6  10.785 60

Actual cooling affect  5.3925 kW Heat removed to form each kg of ice  10  4.187  335  377 kJ

Ice formed/hr



5.3925  3600  51.49 kg 377

Ice formed/day



51.49  24  1.23 tonnes 1000

Ice formed per month  1.23  30  33.08 tonnes Problem 5.22: A water cooler using F12 as refrigerant works between temperature limit of 26C and 2C. The vapour leaves the evaporator dry and saturated. Find the C.O.P of the refrigerator. The properties of F12 are given below. Liquid

Temperature

Vapour

C

hf

sf

Cp

hg

sg

Cp

26C 299 K

443.9

4.274

0.997

585

4.75

0.674

2C 275 K

420.6

4.2

0.942

574.5

4.754

0.62

Solution: To Find h2 Entropy at 1 = Entropy at 2 s 1  s2 s1  sg at 275 K  4.754 kJ/kg K

Refrigeration 5.55

s2  4.754 kJ/kg K T2 s2  s2  C p ln T2

[where s2  sg at 299 K  4.75 kJ/kg K ] 4.754  4.75  0.674 ln ln

T2 299 T2

299

T2 299

 5.93  10  3 3  e5.93  10

T2  300.78 K h2  h2  CPT 2  T 2 [Here h   h at 299 K  585 ] 2 g  585  0.674  299 h2  586.2 kJ/kg

To Find h1 h 1  h g at 275 K h 1  574.5 kJ/kg

To find h 4 h4  h3 h 3  h f at 299 K  443.9 kJ/kg  h4 So, h 4  443.9 kJ/kg

To Find C.O.P. C.O.P 

R.E h1  h4 574.5  443.9   11.163  W h2  h1 586.2  574.5

5.56 Thermal Engineering - I

5.6 PH CHART A Pressure Enthalpy chart is a graphical representation of a number of thermodynamics properties of a given refrigerant. It typically contains of two major axes namely Absolute pressure and specific Enthalpy. For a given refrigerant, based on its enthalpy and pressure, properties like temperature, entropy and specific volume can be directly deduced from the chart. Thus, if any of the two properties of a refrigerant is known, the remaining properties can be directly obtained from the chart.

B C onst te mp lin e

G

if ic S p e c in e s eL m lu Vo

es lin py ro nt tE

E

A

En thalpy Fig:5.15

Co

ns

C o nst P ressure line

S up er heated reg ion

u r lin e D ry s a tu r a

te d V apo

C onst E nthalp y line

n t D ry ne

s s fr a c tio n lin e

W et R e gion

C o n s ta

S a tu ra

te d liq u id

lin e

C o nsta nt te m perature line

p li n e t te m

Su b C oo led liq uid regio n

C

C ons

Pressure

F

D

H

Refrigeration 5.57

Entities of PH chart AB

–

Saturated Liquid line

CD

–

Dry saturated vapour line

Area ABCD

–

Wet region (Two phase region)

Area EFBA

–

Liquid region

Area CDHG

–

Super heated region

The use of PH chart in solving problems will be illustrated through an example given below. Problem 5.23: An ideal refrigerating cycle operates between 1 bar and 7 bar. Dry saturated vapour enters the compressor and F-12

at

7

bar

and

15C

enters

the

expansion

valve.

Compression is isentropic. Determine the power input in kW for 40 tons refrigerating capacity.

(JNTU - Dec 2014)

Solution: Power required 

3.5 T 3.5T  R.E h1  h4

To determine the refrigeration effect, it is imperative to know h 1 and h4. Finding h 1 



Since the vapour is dry and saturated at 1, locate the 1 bar pressure point on the dry saturated vapour line. Mark this point as 1. From 1, drop a perpendicular line to the enthalpy axis. This point on the enthalpy axis gives h1  175 kJ/kg.

5.58 Thermal Engineering - I

7 ba r o

15 C

1 ba r

Finding h 4 

 

 

Since the compression is isentropic, s1  s2. Hence, from 1, trace the constant entropy line until it reaches the constant pressure 7 bar line. Mark this point as 2. Drop a perpendicular line from 2 to the enthalpy axis to get h 2. Process 2-3 is of constant pressure. Hence, draw a horizontal along the 7 bar pressure from 2 until it reaches T 3  15 C. Mark this point as 3. Since process 2-3 is the condensing process, point 3 should be either on the saturated liquid curve or in the sub-cooled region. Drop perpendicular line from 3 to the enthalpy axis to get h 3. Since the process 3-4 is expansion with constant enthalpy, draw vertically downward line to cut horizontal line produced from 1 at 4, h 3  h 4.  h3  h4  50 kJ/kg

Refrigeration 5.59

Power required P 



3.5  T h1  h4 3.5  40  1.12 kW 175  50

Problem 5.24: A refrigerating unit is working between 40C and  10C. The load on the unit is 5 tons. Assume the refrigerant is NH3 which is dry and saturated vapour leaving the evaporator and the compression is isentropic. Find – COP of the system – Power required to run the system If the temperature of the refrigerant required in the evaporator is  20C, find the change in COP of the system and power required.

(JNTU - Dec 2014)

Solution: COP 

Refrigeration Effect h 1  h 4  Work Done h2  h 1

Case - 1 T 1   10 C, T 2  40  C Locating h 1 on PH chart Since, the fluid is dry saturated vapour, locate.  10C on the saturated vapour curve. Drop a perpendicular to the specific enthalpy axis and this point corresponds to h1  1435 kJ/kg Locating h 2 on PH chart Since h the compression is isentropic, s1  s2. trace a constant entropy curve in the vapour region from T1 till it meets up with the 40C constant temperature line.

5.60 Thermal Engineering - I

o

15 C

o

-1 0 C

Mark it as 2. From this point 2, drop a perpendicular to the specific enthalpy axis meeting at the point h2  1525 kJ/kg Locating h 3 and h 4 Since process 2-3 is constant pressure, draw a horizontal from 2 along constant pressure line, till it reaches the saturated liquid curve at 3. From this point, drop perpendicular to X axis which gives the h3 value. h3  225 kJ/kg . . . h3  h 4 as per the graph, h4  225 kJ/kg COP 

h1  h4 h2  h1



1435  225  13.44 1525  1435

  Power required  m h2  h1; Also R.F in kW  m h1  h4

 3.5  5 3.5T m   0.0144 kg/s h1  h4 1435  225 Power required  0.0144 1525  1435   1.296 kW

Refrigeration 5.61

Case 2 - T 1   20C, T 2  40 C h 1, h2, h 3, h4 for the corresponding temperatures and pressure conditions on the PH using the

Locate

same method. h 1  1420 kJ/kg, h 2  1535 kJ/kg , h 3  h 4  210 kJ/kg COP 

h 1  h 4 1420  210  10.52 h 2  h 1 1535  1420

 3.5  5 3.5T  m  0.0144 kg/s h 1  h 4 1420  210  Power required  m h 2  h 1  0.0144 1535  1420   1.66 kW Problem 5.25: The temperature range in a Freon-12 plant is  6C to 27C. The compression is isentropic and there is no under cooling of the liquid. Find the COP assuming that the refrigerant (i) after compression is dry and saturated (ii) leaving the evaporator is dry and saturated. The properties of F-12 and given in the table. tC

hf

hg

sf

sg

Cp

27 (300 K)

445

585

4.28

4.75

0.714

– 6 (267 K)

413

571

4.17

4.76

0.641

(i) After compression is dry and saturated T1   6 C   6  273  267 K

5.62 Thermal Engineering - I

T2  27 C T

 27  273  300 K COP 

3

3 00 K

h 1  hf3 h2  h1

h1  hf1  x1 hfg 1

2 67 K

2

g

f 4

 hf1  x1 hg1  hf1

S

h 2  h g at 300 K

S f3

s1  sf1  x1 sfg1  4.17  x1 4.76  4.17  s2  sg2 Dry saturated at 300 K s2  sg  4.75 kJ/kg K s1  s2  4.17  x1 0.59  4.75 x 1  0.98 h 1  h f1  x1 h fg1  413  0.98 571  413 h1  568.3 kJ/kg h 2  h g at 300 K;

h2  585 kJ/kg

h 3  h f at 300 K  445 kJ/kg

and h3  h4  445 kJ/kg COP 

h1  h4 568.3  445   7.38 h2  h1 585  568.3

COP  7.38

1

(S 1 = S 2)

s1  s2 h 3  h 4  h f3

Refrigeration 5.63

(ii) Leaving the Evaporator is dry and saturated

2

T 3 C o nd

3 00 K

2’

Comp

E xp 2 67 K f

4

1

E vap

g S (S 1 = S 2 )

h 1  hg at 267 K  571 kJ/kg h2  hg a t 300 K  585 kJ /kg h 3  h f  at 300 K  445 kJ/kg  h4 h 2  h 2  C p  degree of su per heat h 2  585  0.714   T 2  T 2   

...(1)

s1  sg at 267 k  4.76  s2 s2  sg at 300 K  4.75 kJ/kgK  T2  Using s2  s2  C p ln    T2  T2 4.76  4.75  0.714  ln 300 T2  304.2 K

(sub in (1))

h 2  h2  C p  T2  T2  585  0.714  304.2  300  h 2  587.99 kJ/kg COP 

h1  h4 571  445   7.41 h 2  h 1 587.99  571

COP  7.41

5.64 Thermal Engineering - I

Problem 5.26: The Freon 12 compressor of 15 cm dia., 15 cm stroke has 4 single acting

cylinders running at 970 r.p.m. It

is working between  10C and 30C find (i) Cooling capacity. (ii) Power required by electrical motor during the compression. Volumetric efficiency 75%; Mechanical efficiency 95%. Also draw the T.S. diagram. The vapour enters the compressor as dry saturated. P

(FAQ) T2

T o 38 C

T2

30 C 3

2

3

2

o

2O

2O T2

4

1

T 2 Oo 30 C

o

-10 C 4

h

1

s

Solution: Freon 12 (R12) d  Dia.  15 c m  0.15 m L  stroke length  0.15 m

Single Acting No. of cylinders = 4;  vol  0.75 ; mech  0.95 ; N  970 r.p.m.

From P-h chart, we can find the following enthalpy values. h1  185 k J/kg; h 2  210 kJ/kg ;

Refrigeration 5.65

h 3  68 kJ/kg; T 2  38 C  311 K; h 4  h 3  68 kJ/kg C.O.P 

185  68 R.E h1  h4    4.68 W h2  h1 210  185

To Find Cooling Capacity (R.E in kW) v1  vg for  10C  0.0766464 m 3/kg

Swept volume Vs 

 2  d L   0.152  0.15  2.65072  10  3m 3 4 4

But Vs 

 m v1

. . [ . 1 for single acting ]. N vol    1  no. of cylinders  60   m 0.0766464  3 2.65072  10   970  0.75   14  60   m  1.67731 kg/sec  Cooling capacity  R.E in kW  m h1  h4 Q 4  1 

 1.67731185  68  196.24 kW R.E  196.24 kW

To Find Power   Power  m  W  m h 2  h 1  1.67731 210  185  41.93 kW

5.66 Thermal Engineering - I

(or) C.O.P 

R.E in kW W in kW

So Power  W in kW  

R.E in kW C.O.P 196.24  41.93 kW 4.68

mech  95%

But

Actual power required to run the compressor P 41.93   44.136 kW 0.95 mech



Actual power required to run the compressor  44.136 kW

5.7 SUBCOOLING OR UNDERCOOLING In the condenser, the refrigerant is cooled (condensed) at constant pressure. If the refrigerant is cooled below its saturation temperature, then it is called sub cooling or under cooling. O

3

O

p= c

3

2

2

O

O

s=c

O

2

O

O

O

h4

O

3

h 3=

4

3 p= c

T3 T3

h 3= h 4

P

D egree of subcooling = (T 3 -T 3 ) T 2

s=c

D egree of subcooling = (T 3 -T 3 )

p= c 1

p= c

1

4 h

s F ig. 5.16

Refrigeration 5.67

In the problem, ‘subcooled by 10C ’ means the degree of subcooling i.e., T 3  T3  10C “Subcooled to 10C ” means the final temperature of subcooling. i.e., T3  10C . 5.7.1 Effect of Subcooling (i) (ii)

Refrigeration effect increases. Work input remains same.

(iii)

C.O.P increases [C.O.P  R.E/W ; when R.E increases C.O.P will increase

(iv)

Condenser heat rejection capacity increases.

Subcooling is also defined as the process of cooling the liquid refrigerant below the condensing temperature for a given pressure. Problem 5.27: A refrigerating plant operates in vapour compression cycle. The refrigerant is R12 and the saturation temperature

in

the

condenser

and

evaporator

are

40C and  5C respectively. The vapour enters the compressor as saturated vapour and it is subcooled to 20C in the condenser. Calculate 1. C.O.P., 2. compressor displacement. Assume vol  100%, Power  1 kW.

Solution: From the P-h chart for R12, we can find enthalpy values.

(FAQ) P 3

3

O

P= C

4

h 4  h 3  58 kJ/kg

2

O

h 3= h 4

h 2  208 kJ/kg

O

s=C

h 1  185 kJ/kg

2

P= C 1 h

5.68 Thermal Engineering - I

To Find C.O.P R.E h 1  h 4  C.O.P  W h2  h1

T

D e gre e of subcoolin g = (T 3 -T 3 )= 20 o C

2 T 3 =4 0 C T 3 

P=C

3



%

3

2

O

O

O

h3=

185  58  5.521  208  185

O

s=c

O

h4

O

T 1 =T 4 =-5 C

C.O.P  5.521

P=C

1

4 s

To Find Compressor Displacement: m 3/sec Q 2  1  Power in kW  1 k W  Power in kW  m  Work input/kg of refrigerant

 m

1 1  h2  h1 208  185  0.0434 kg/sec

v1  vg at  5C  0.0649629 m 3/kg

Assume No. of cylinder = 1, and single acting Swept volume Vs 

 mv1  N  vol   11  60 



 m v1  N  1 11  60 

N    Compressor displacement in m 3/sec   Vs   mv1 60    0.0434  0.0649629  2.819  10  3m 3/sec

Refrigeration 5.69

5.8 SUPER HEATING When the liquid refrigerant is passing through evaporator, it absorbs heat from the space to be cooled. Thus this liquid refrigerant becomes vapour. If this liquid refrigerant absorbs more heat, then it becomes super heated vapour i.e. it is heated above the saturation temperature. So the process of heating the vapour above the saturation temperature in the evaporator is called as super heating. In the problem, ‘Super heated by 20C, means the degree of super heating i.e., T 1  T 1  20C ‘Super heated to 20C means T 1 i.e. temperature of vapour leaving the evaporator (or) temperature of vapour entering the compressor is 20C . 5.8.1 Effect of Super heating The effect of super heating is to increase the refrigerating effect, but this increase in refrigerating effect is at the cost of increased in work input between the upper pressure limits. So, (i) (ii)

R.E is increased. Work input is increased.

[But the increase in work input is more as compared to increase in refrigerating effect; so the overall effect of super heating will reduce the C.O.P.] (iii)

The C.O.P. is reduced.

(iv)

Specific volume is increased.

5.70 Thermal Engineering - I Su per heating P

2

T 2O

3 3

P 3 = P 2 = P2

2O

O

2

T1 - T1 = D egree of Super h eat 4 1O

1

O

1O f

4

1

g D egree of Super h eat h Fig. 5.17

(v)

s

Since the compressor is totally free from liquid refrigerant, corrosion and erosion is minimum. So the life of the compressor is increased.

Problem 5.28: The evaporator and condenser temperature of 20 ton capacity refrigerator are  28C and 23C respectively. The R22 is subcooled by 3C before it enters the expansion valve and is super heated by 8C before leaving the evaporator. A six cylinder single acting compressor with stroke equal to bore running at 1000 r.p.m. is used. Determine 1. Refrigerating effect/kg, 2. Mass flow rate of refrigerant in kg/sec, 3. Bore and stroke of the compressor, 4. Power, 5. C.O.P. 6. Heat removed through condensor.

(FAQ)

Solution: Locate the points 1, 1, 2, 2, 3, 3 and 4 in the P  h chart for R22 by the following procedure.

Refrigeration 5.71 D egre e o f S ubc ooling T 3 - T 3 = 3 OC O

O

O

23 C

20 C

P

O

O

2

h3 = h4

S=

C

C

T hrottling

O

O

3

2

23 C C ond en sation

23

3

O

D egre e of S upe rheat

1

1

O

O 0

-2 O

8

f

O

C

g

h 615k J/kg

446k J/kg

O

T 1 -T 1 = 8 C

C

-2

4

E vap o ratio n O -2 8 C

664k J/kg

1.

Locate 1 at the intersection of  28C and ‘g’ line. (saturated vapour line).

2.

Locate 1 by extending 1 horizontally upto  20C . .  . Degree of super heating  8C So ,T1   28  8   20C

3.

Mark 2 at 23C in g line.

4.

From 1, trace a line through constant entropy line and from 2 draw horizontal line. Both lines intersect at 2.

5.

From 2 draw horizontal lines in left direction, cut the ‘f’ line and mark as 3. Extend this line upto . . temperature 20C [ . Degree of subcooling  3C, So ,

T3  T3  23  T3  3 .

T3  23  3  20C ] and mark as 3

So,

5.72 Thermal Engineering - I

6.

From 3 draw vertical line to cut upto temperature  28C. This point is 4.

Now we can take enthalpy values from all the points. These are given below. h 1  240 kJ/kg; h2  290 k J/kg; h 4  h 3  70 kJ/kg; v1  0.14 m 3/kg.

(i) Refrigerating effect: R.E R.E  h 1  h 4  240  70  170 kJ/kg

(ii) Mass flow rate of refrigerant. R.E in kW = Cooling capacity of refrigerator  20  3.5  70 kW  R.E in kW  m  R.E in kJ/kg  70  0.4117 kg /sec m 170

(iii) Compressor Dimensions Vs 

 2 d L 4  m v1

Also V s   vo l  N/ 60  1  6 . . [ . 1 for Single acting and 6

for 6 cylinders]  0.4117  0.14  d2  d  4 1  1000/60  1  6  5.7988  10 4 d 3  7.3424  10  4 d  0.0902 m L  d  0.0902 m

. . [ . d  L;  vol  1 ]

Refrigeration 5.73

(iv) Power

 Power  m  W  0.4117  h 2  h 1  0.4117  290  615  20.56 kW

(v) C.O.P C.O.P 

240  70 R.E h1  h4   W h2  h1 290  240  3.4

(vi) Heat removed in the Condenser   m h2  h 3  0.4112 290  70  90.46 kW Problem 5.29: In a theoretical single stage ammonia vapour compression

refrigeration

system,

the

liquid

leaves

the

condenser at 12 bar, 20C. Evaporator pressure is 1.2 bar. The vapour leaves the evaporator at  20C. Speed is 200 r.p.m. Single acting with L/D  1.5  vol is 80%. Determine (i) C.O.P (ii) Power required per ton of refrigeration. (iii) Compression cylinder dimensions. (Apr-99 - Madras University)

Solution 1.

~ 1.2 bar mark 1. At  20 C and 1 bar 

2.

From 1 draw line through constant entropy line and it cuts the constant pressure line 12 bar line at 2.

5.74 Thermal Engineering - I

D e gree of Su bcooling O T 3 - T 3 =9 C

29 C

lea ving the condenser

O

O

O

20 C

P

O

O

2

O

O

h3 = h 4

s=

C

C

O

C

50

O

Throttling

2

31

3

29 C (12 bar) C o ndensation

=1 T2

3

O

8

O

270kJ/kg

C

g

C

-2

O 0

f

3.

1

-2

Evap oration 1 bar (=1.2 b ar)

4

lea ving the evaporator

1

h 1425kJ/kg 1760kJ/kg

Draw horizontal line from 2 and cut 20C line at 3.

4.

From 3 draw vertical line to cut 1 bar line at 4.

5.

Take the enthalpy value from all the points. h1  1425 kJ/kg h2  1760 kJ/kg and T2  150C h 3  h4  270 kJ/kg

Degree of super heat  T1  T1   20   28  8C Degree of subcooling  T3  T3  29  20  9C C.O.P C.O.P 

R.E h1  h4  W h2  h1 

1425  270  3.4477 1760  1425

Refrigeration 5.75

Power Required/Ton of Refrigeration 1 T R  3.5 kW  R.E in kW C.O.P  3.4477 

R.E in kW R.E in kW  Power W in kW 3.5 po w er

Power  1.0151 kW  Mass Flow Rate of Refrigerant: m  R.E in kW  R.E in kJ/kg  m  m

3.5  3.0303  10  3kg/sec 1425  270

Compressor Cylinder Dimensions vo l  0.8 ; Single ac ting; No. o f cylinder  1; N  200 r.p.m. ; L/D  1.5

 Swept volume V s   D2  L  4

 m v1  vol  

First of all find v1 v1 v1



T1 T 1

v1  vg at  28 C  0.7397 T 1   20  273  253 K T1   28  273  245 K v1  0.7397 

253  0.7638 m 3/kg 245

N 11 60

5.76 Thermal Engineering - I

 Now Vs   D2  1.5D  4



 mv1 vol  

N 11 60

3.0303  10  3  0.7638 200 11 0.8   60

  1.5D 3  8.679  10  4 4 D 3  7.3711  10  4 D  0.09033 m L  1.5D  0.13549 m Problem 5.30: In an vapour compression refrigeration system R22

has

condensing

and

evaporating

temperature

as

40C and  20C respectively. The refrigerant is super heated to 0C in the evaporator and subcooled by 10C in the condenser. The bore and stroke is 100 mm each, clearance ratio 5%, speed 600 r.p.m. Double acting, 2 cylinder. Determine (i) Refrigeration effect in ton of refrigeration. (ii) Power required.

Solution T sat 1  Evaporator temperature   20C  253 K T sat 2  Condenser temperature  40C  313 K

Refrigerant is super heated to 0C i.e.T1  273 K Refrigerant is subcooled by 10C i.e. t3  t3  10 t3  tsat   40C 3 t3  40  10  30C

Refrigeration 5.77

D e gree of S u bc oo lin g O T3 - T3 =10 C O

O

O

30 C

40 C

P

3O

2 O

O

h3

O

-2 0 C

=

h4

s=

C

C

T hrottlin g

2O 40

O P =C 40 C C o nd en sa tio n

3

1O

1

88 k J/kg

C

g

C

0

O

f

O

-2

4

0

E v ap ora tio n -20 O C p=C

h 2 5 2 k J /k g 3 1 0 k J /k g

From p-h chart, h 1  252 kJ/kg; h2  310 kJ/kg , t2  104C h 4  h 3  88 kJ/kg.

To Find Mass Flow Rate of Refrigerant Swept volume V s 

   D 2  L   0.1 2  0.1 4 4 V s  7.853982  10  4m 3

Also Vs 

 m v1 vol  N/60  2  2

. . [ . Double acting and two cylinder]

So, first of all, we have to find v1 and  vol To Find v1 v1 v1



T1 T 1

5.78 Thermal Engineering - I

T 1  273 K ; v1  vg at  20C  0.0930 m 3/kg T1   20  273  253 K v1  0.093 

273  0.1003 253

v1  0.1003 m 3/kg

To Find v2 v2 v2



T2 T 2

v2  vg at 40C  0.0148 T 2  104  273  377 K T 2  40  273  313 K T2 v2  v2  T2 v2  0.0148 

377  0.0178 m 3/kg 313

v2  0.0178 m 3/kg

To Find vol vo l  Volumetric efficiency  v1  1K 1 v  2    0.1003 1  1  0.05  0.0178    vo l  0.7682 or 76.82%

Refrigeration 5.79

Swept volume Vs  7.853982  10  4 

 m v1 vol  N/60  2  2

 7.853982  10  4  0.7682  600  2  2 m 0.1003  60  0.2406 kg/sec

To Find R.E in TR  R.E in kW  m h1  h4  0.2406 252  88  39.46 kW R.E in TR 

39.46  11.28 TR 3.5

R.E  11.28 TR

Power Required  Power  m  W  0.2406  h 2  h 1  0.24063  310  252  13.95 Power  13.95 kW Problem

5.31: A

vapour

compression

refrigeration

uses

methyl chloride and operates between  10C and 45C. At entry to the compressor, the refrigerant is dry saturated and after compression it acquires a temperature of 60C. Find the C.O.P.

5.80 Thermal Engineering - I

Saturation

Enthalpy in kJ/kg

Temperature

Liquid

Entropy in kJ/kg K

Vapour hg

Liquid

 10C

45.4

460.7

0.183

1.7619

45C

133.0

483.6

0.485

1.587

hf

Vapour sg

sf

(FAQ)

Solution h 1  h g at  10C  460.7 kJ/kg

To Find h2 h 2  h 2  C pt2  t2

where h2  h g at 45C  483.6 kJ/kg t2  45C ; t2  60 C

To find h2, we have to find C p.

P

ts a t 2=

T

o 45 C

2 O

3

45 C 2

O

2

2

O

ts a

t2 =

45

o

C

3

O

-1 0 C 0 Co

g

f

t1 =

g

ts a

f

1

4 -1

1

4

h

s

Refrigeration 5.81

To Find C p s2  s1 [Isentropic compression]

Here s2  sg at 45 C

s1  sg at  10C  1.7619 kJ/kg K

 s2  1.587

So, s2  1.7619 kJ/kg K

T2  45  273  318 K

T2 But s2  s2  C p ln T2

T2  60  273  333 K

1.7619  1.587  Cp ln

333 318

1.7619  1.587   3.7946 kJ/kg K 333    ln 318   

Cp 

h 2  h 2  CpT 2  T 2  483.6  3.7946 60  45  540.519 kJ/kg

To Find h4 h 4  h3 and h3  hf at 45C  133 h4  133 kJ/kg

So, C.O.P 

h1  h4 460.7  133  h2  h1 540.519  460.7

 4.1055 Problem 5.32: A

vapour

compression

refrigeration

plant

works between pressure limits of 5.3 bar and 2.1 bar. The vapour is super heated at the end of compression with temperature being 37C. The vapour is super heated by 5C before entering the compressor. If the specific heat of super heated vapour is 0.63 kJ/kg K, find the coefficient of performance (C.O.P).

(FAQ)

5.82 Thermal Engineering - I Pressure (bar)

Saturation Temperature

Liquid heat

Latent heat

kJ/kg hf

kJ/kg hfg

5.3

15.5

56.15

144.9

201.05

2.1

 14.0

25.12

158.7

183.82

T

P 3 7OC P 2 =C

3

2O

O

37 C

2

O

1 5.5 C 3

2 o

15

hg kJ/kg

.5

2O

C

O

-9 C

1

O

g

1O

f

1

4

4 Co

f

4

5 C

O

-14 C

1O

g s

-1

P 1 =C

h

Solution: Super heated by 5C i.e. Degree of super heat  t1  t1  5C.

So, t1   14  5   9C Find h 1, h2, h4 to find C.O.P. To Find h1 h 1  h 1  C pt1  t1

where h1  h g at  14 C  183.82 kJ/kg C p  0.63 kJ/kg K h 1  183.82  0.63 5  186.97 kJ/kg h1  186.97 kJ/kg

. . [ . hg  hf  h fg ]

Refrigeration 5.83

To Find h2 h 2  h 2  C pt2  t2

where h2  h g at 15.5C  201.05 kJ/kg t2  37 C; t2  15.5C h 2  201.05  0.63 37  15.5  214.595 k J/kg h2  214.595 kJ/kg

To Find h4 h 4  h 3 (Throttling process) h 3  h f at 15.5C  56.15 k J/kg h 4  56.15 kJ/kg C.O.P 



h1  h4 h2  h1



Refrigeration effect Work input

186.97  56.15  4.73585656 214.595  186.97

C.O.P  4.736 Problem 5.33: A CO2 refrigeration system works between 5.25 bar and 21.2 bar. The refrigerant leaves the compressor at 32C with total heat 246.2 kJ/kg. Determine the theoretical C.O.P of the plant. The properties of CO2 are Pressure Sat. bar Temp.

Enthalpy kJ /kg

Entropy kJ /kg K

Liquid hf

Vapour hg

Liquid sf

Vapour sg

C 56.25

18.5

52.75

214.37

0.165

0.7244

21.2

 18

 37.68

234.69

 0.1507

0.9170

(FAQ)

5.84 Thermal Engineering - I

Solution: Total heat of vapour leaving the compressor = 246.2 kJ/kg i.e., h2  246.2 kJ/kg Inlet to the compressor - it may be dry saturated or super heated or wet. We have to find the condition of vapour entering the compressor. Then only we can draw P-h diagram and T-S diagram. h 2  h 2  C pt2  t2

where h2  h g at 18.5 C  214.37 kJ/kg t2  32C; t2  18.5C 246.2  214.37  Cp32  18.5 Cp 

246.2  214.37  2.357777 13.5

C p  2.35778 kJ/kg K T2 s2  s2  C p ln T 2

where s2  sg at 18.5  C  0.7244 kJ/kg K T 2  32  273  305 K ; T 2  291.5 K s2  0.7244  2.35778 ln  0.83114 kJ/kg K s2  0.83114 kJ/kg K s1  s2  0.83114

305 291.5

Chapter - 6

Actual Cycles and Their Analysis Introduction, Comparision of air standard and actual cycles, Time loss factor, Heat loss factor, Exhaust Blowdown - Loss due to Gas exchange processes, Volumetric efficiency. Loss due to Rubbing friction, Actual and fuel - air cycles of CI engine.

6.1 INTRODUCTION Air standard cycle is a cycle which uses air as the working medium and assumes that the working medium behaves as a perfect working substance. Air standard cycle also prevents all the heat losses that could occur in an engine and pictures it as an imaginary perfect engine. The steps involved in air standard cycles are simple and idealistic, which is why They are also called as Ideal cycles. The following assumptions are made to consider a process as an air standard cycle. 1.

The working substance should be a perfect gas with standard specific heats.

2.

The heat transfer should be simple and no chemical reactions should occur.

3.

It must be a reversible process.

4.

The heat losses are assumed to be zero.

The otto cycle and diesel cycle of an IC engine are said to be air standard cycles but their real working processes greatly vary with respect to their physical properties. A normal engine operation is subjected to various heat losses, change in heat input, change in working medium, etc.

6.2 Thermal Engineering - I

The cycle which accounts to the every possible losses in real time environments may be referred to as an actual cycle. The actual cycle efficiency is much lesser than the air-standard efficiency because of the various losses occurring in an actual engine. If the losses due to variable specific heats due to varying temperatures are neglected from an air standard cycle, then it is called as Fuel-air cycle. In fuel air cycle, the working medium is the mixture of air and fuel vapour or atomized liquid fuel. Each cycle varies with each other in many aspects.

6.2 COMPARISON OF AIR-STANDARD AND ACTUAL CYCLES As discussed earlier, the actual cycle is different from air standard cycle due to various losses that occur in an engine. So, the losses must be taken into account while studying the difference between both the cycles. The major losses occurring in an heat engine are:(i) (ii)

Dissociation losses. Loss due to incomplete combustion.

(iii)

Time loss.

(iv)

Heat loss.

(v) (vi) (vii) (viii)

Loss due to temperature.

variation

of

specific

heats

with

Loss due to exhaust blowdown. Loss due to rubbing friction. Loss due to blowby gases.

The above losses are used to differentiate actual cycles from air-standard cycles and also fuel-air cycles.

Actual Cycles and Their Analysis 6.3

Each of the losses results in the decrease of thermal efficiency and power output of the actual engine. So, understanding them clearly would help us to improve the performance of an actual engine. Along with these losses, there are some factors that also amount to the difference between these cycles. They are, (a) The working medium In case of a fuel-air cycle, the working medium would be air-fuel mixture, whereas for air standard cycle, it is pure air. (b) The nature of working substance In case of air-standard cycle, the working substance remains unchanged from the start to end whereas in other cycles the chemical composition of the working substance may change. (c) The variation of temperature, pressure and composition of fresh charge indicates the difference between the cycles. (d) Heat transfer is a property which differentiates actual cycles and air-standard cycles because, in air-standard cycles there is no heat transfer between the components. In an actual engine, there is a significant amount of heat transfer to and from the working substance as well as the cylinder walls and various other components. Hence the efficiency of the working cycle changes accordingly. (e) Gas leakage, fluid friction, engine design, rpm, etc. are all responsible for defining the losses in an actual engine cycle.

6.4 Thermal Engineering - I

All the above factors could be used to compare actual cycles, air standard cycles and fuel-air cycles. Efficiencies of each cycles also vary with one another. For example, the efficiency of actual cycle will always be lower than air-standard cycle because of the mentioned losses that occur in an actual engine. All the major losses are discussed briefly in the coming sections.

6.3 TIME LOSS Under actual cycle, it is almost impossible to obtain a homogeneous mixture of air and fuel as there are contaminants like carbon deposits, fuel charge from previous cycles etc. Thus, a part of the fuel in the combustion chamber reacts better than its counterpart when heat is applied. Hence, the time taken for the charge to burn completely varies for every cycle. Moreover, the time taken to completely burn a charge also varies due to the following factors:      

Homogeneity of the air-fuel mixture. Flame speed. Configuration of the combustion chamber. Turbulence in the air-fuel mixture. Distance needed to travel by the flame. Air-fuel ratio.

A homogeneous mixture burns steadily and uniformly within a preset time. However, it depends upon the mixing before the inlet. As little time is available at the inlet for proper mixing of air and fuel, there is always a compromise of the homogeneity of the mixture.

Actual Cycles and Their Analysis 6.5

Flame speed depends upon the air-fuel mixture, type of fuel used and the available volume. It is easier for a flame to burn through a dense charge of air and fuel rather than a small amount of charge. Flame speed is also higher at richer air-fuel ratios. Configuration of the combustion chamber deals with the geometry and the space available for the propagation of flame. A cylinder with smaller stroke and larger bore burns the fuel more quickly than its alternate. Turbulence of the mixture depends upon the inlet valve and its position. A turbulent mixture burns more rapidly than a less agitated mixture. By positioning the inlet valve suitably, several motions of air-fuel mixture within the cylinder can be obtained. A turbulent mixture burns faster than a non-turbulent mixture. The air-fuel ratio is also responsible for the time required to complete the burning process. So, the correct proportion of air-fuel ratio must be provided in order to achieve proper burning within a given interval of time. The fuel’s chemical structure and also its ignition temperature must be taken into account to determine the time required for it to burn completely during combustion. Any variation in air-fuel ratio could result in a time loss due to the time consumed by the burning process. Maximum power depends on the peak pressure of the cycle. If the spark is too early in the cycle, work is done against the expansion of gases. Thus the energy is lost. If the spark occurs after TDC, the peak pressure may not be achieved easily, consequently leading to reduced power and efficiency.

6.6 Thermal Engineering - I

These above factors are the reasons for power loss due to time factor in an internal combustion engine. However, in an ideal cycle, the heat addition is assumed to be instantaneous. Thus, with the actual losses, the total work done by a cycle is depicted by a graph of Volume vs Pressure as shown in the Fig. 6.1.  

  

 



,

   



  



 

  







  'F\O 'F



, 3   



Fig:6.1

6.4 HEAT LOSS FACTORS The losses due to heat transfer play a major role in the efficiency of an IC engine. To determine the total heat loss of an engine, it is necessary to understand the distribution of heat among various components of an engine. During the combustion process, an enormous amount of heat is generated. This heat is used to drive the piston. However, a fraction of this heat is dissipated into

Actual Cycles and Their Analysis 6.7

various components such as cylinder walls, piston rings etc. through conduction or convection process. The engine as a whole radiates a negligible amount of heat per cycle. Moreover, a part of the generated heat is quenched instantaneously by the coolants flowing through the jackets. A fraction of the heat is also transferred to the lubricating oil. Thus, the heat generated through the combustion process is transferred to several processes as shown below. Workdone piston

to

drive

the

Convection and conduction heat transfer to internal components Heat Generated

Heat transfer cooling systems

through

Heat lost at the exhaust stroke Heat lost due to unburnt fuel It may be noted that a sizeable fraction of the heat produced is repelled out of the engine during the exhaust stroke even before it gets converted into work. This is due to the constraints in the time required to convert all the heat into work.

6.8 Thermal Engineering - I

Thus, even if the losses due to heat are minimized, only one-fifth of the total heat energy produced is converted into quantifiable work. On a larger picture, heat losses however contribute to only 12% - 15% of the total losses of an IC engine.

6.5 EXHAUST BLOWDOWN Exhaust blowdown is the process in which some of the combustion gases escape through the exhaust port if the exhaust valve is opened before Bottom dead center. This is due to the self pressure created by the piston at the end of the combustion stroke and also it depends upon the compression ratio of the engine. The basic operation of the exhaust stroke is to pump out all the exhaust gases through the exhaust valve. Since the piston has to work against the self pressure created, extra power is consumed to pump all the gases. This results in loss of power and this phenomenon is called as exhaust blowdown loss. It can be avoided, if the exhaust valve opens at or after the BDC, (or) the best way is to open the exhaust valve between 40  70 before BDC which inturn

P ressu re (b ar)

E xpa n sion stro ke

Id ea l diag ram

E a rly e xh a ust v a lv e o p en in g 7 .0

O ptim um exha us t va lve op en in g

3 .5 E xha u st Valve O pe n in g a t b ottom d ea d ce ntre

Vo lu m e

Fig:6.2

Actual Cycles and Their Analysis 6.9

reduces the exhaust pressure of the cylinder into half (from 7 bar to 3.5 bar). The Fig.6.2 explains the exhaust blowdown process.

6.6 LOSS DUE TO GAS EXCHANGE PROCESS The phenomenon of gas exchange process is the admission of fresh charge during the inlet stroke and the removal of exhaust gases at the end of the combustion stroke. In other words, the difference between the workdone in removing the exhaust gases, and the workdone in admitting fresh charge is called as Pumping work. Also, the loss occurring due to this pumping work is called as pumping loss. This loss is due to the difference of pressure between lower inlet pressure (Pi) and the highest exhaust pressure (Pe). P

Ex

Co

Pe

E xha u st

mp

re s

pa

sio

ns

io n

n

P a tm Pi

P u m ping w o rk V Fig:6.3 G as Exch an ge pro cesses

The Fig.6.3 shows the gas exchange process in SI engine. The shaded area indicates the pumping work of pressure from lower value to higher value; from suction to exhaust. The pumping loss increases with throttling,

6.10 Thermal Engineering - I

because throttling reduces the suction pressure and also it increases with speed. The volumetric efficiency of the engine is greatly affected by the losses occurring due to gas exchange processes which will be explained briefly in the next article.

6.7 VOLUMETRIC EFFICIENCY The volumetric efficiency is one of the important parameters in deciding the performance of an IC engine. It is defined as the ratio of volume of air intake during the suction stroke to the swept volume of the piston. It is also defined as the ratio of mass of charge sucked into the cylinder during suction stroke to the mass of air corresponding to the swept volume of engine at atmospheric pressure and temperature. A number of factors affect the volumetric efficiency of an engine. Some of them are given here: [Note - For further efficiency refer Chapter 4].

explanation

on

volumetric

6.7.1 The amount of air-fuel ratio When a fresh charge is sucked inside the cylinder, the temperature of the charge is increased by transfer of heat from the cylinder walls and the hot exhaust gases. It is known that for a constant pressure and volume, the temperature is inversely proportional to the mass. As a result, the mass of the fresh charge decreases which inturn reduces the volumetric efficiency. But the volumetric efficiency could be increased by the high pressure from the incoming fresh charge because high pressure results in increased mass of charge.

Actual Cycles and Their Analysis 6.11

6.7.2 The design of intake and exhaust manifold and port design The intake and exhaust manifold should be designed in such a way that the intake manifold must allow the maximum fresh charge to come in and the exhaust manifold must remove all the burnt gases to escape. So, there is no limitation in designing the intake and exhaust manifold if they carryout the above process correctly. But, if the intake and exhaust process does not take place properly, it will affect the volumetric efficiency and so, the design should be modified accordingly. 6.7.3 Compression ratio Compression ratio is defined as the degree to which the fuel mixture is compressed before ignition takes place. The compression ratio does not have much effect on the volumetric efficiency because it does not change the displacement volume. But, there may be a little effect, due to the amount of residual gas in the cylinder. 6.7.4 The intake and exhaust valve timings Valve timings in IC engines is the time at which the inlet and exhaust valves are set to operate. A finite period of time is needed for the valves to operate at a smooth rate which is provided by a cam operation. A cam and follower mechanism is used to achieve a proper opening and closing of the valves, and also to determine the timing of the valve. The inlet valve timing and exhaust valve timing affect the volumetric efficiency. Usually, the inlet port is opened at some degrees before the TDC in order to allow maximum possible charge to come in. Many S.I engines are designed to open at 10 before the TDC and the charge is sucked in

6.12 Thermal Engineering - I

until the piston reaches the BDC. Now, the intake valve should be closed when the piston reaches BDC, but modern SI engines tend to close at a few degrees (say 10) after the BDC for low speed engine and after say 60 for high speed engine. After reaching BDC, the piston starts the compression stroke by moving up. So if the intake valve is still open beyond BDC, the incoming charges are pushed out by the piston which tend to move up and reach BDC. Hence, the charge is pushed out by the piston and some workdone in doing so results in the reduction of Volumetric efficiency. TD C

Inta ke closes

E xhau st closes

BDC

E xhau st op en s

Fig:6.4 (a)

Inta ke closes

o

20

o

60

o

55

Po w er

o

10

IV O

25

o

C om pressio n

o

EVO

5

10

Inta ke o pe ns

o

Pow er

10

TD C E xhau st close s Valve o verlap

IV O

C om pression

E VO

Inta ke o pe ns

o

E xhau st op en s BDC Fig:6.4 (b)

However, it is different for the high speed engines because the high speed engines bring high pressurized charges inside the combustion chamber, which acting against the moving piston and produce a ram effect. This ram effect could be used as an advantage in order to suck more of the high pressured fresh charges in. So, the valve

Actual Cycles and Their Analysis 6.13

time for the high speed engines could be delayed until after BDC in order to make use of the ram effect. Now, coming to the exhaust valve timings, the exhaust valves tend to open slightly before the piston reaches BDC. For a low speed engine, the valve opens at 25 before the BDC and at 55 before BDC for high speed engines. Actually, this process of opening the valve before BDC helps in removing the burnt gases easily without consumption of much work from the expansion stroke. And like inlet valves the exhaust valves of modern engines tend to close at some degrees after the TDC. In a way it is advantageous because it helps to carry out all the exhaust gases from the cylinder, but there may be a chance for valve overlap. Valve overlap occurs when both the inlet and exhaust valves are opened at the same time. Due to this effect, scavenging takes place (ie) removing out the fresh inlet charge through the exhaust. Also, it may result in loss due to blowdown effect which may affect the volumetric efficiency. [NOTE - The valve taming diagrams for both 4 - stroke and 2 - stroke engines are explained in detail in the earlier sections, refer Pg No. 1.27 to 1.34.] Other miscellaneous factors that also affect the volumetric efficiency are: (i) (ii)

Stroke to Bore ratio. Fuel type, fraction of fuel vapourised in the intake system and fuel enthalpy of vapourisation.

(iii)

Engine speed.

(iv)

Ratio of inlet and exhaust manifold.

6.14 Thermal Engineering - I

(v)

The temperature inside the cylinder.

6.8 LOSS DUE TO RUBBING FRICTION The internal components of an IC engine experience an enormous amount of thermal and compressive stresses. The components are usually designed to withstand these stresses, simultaneously providing a better efficiency. However, as discussed in the earlier sections, several losses deteriorate the efficiency. Friction is one of the primary losses in an engine. Some common factors which influence the frictional forces in an engine are:     

Surface finish of piston and cylinder Lubricant properties Engine power Heat dissipation Other auxillary engine components

6.8.1 Surface finish The prime components like piston, piston rings, cylinder walls etc. are expected to maintain a highly accurate surface finish, with tolerances in the order of few microns. Also, the materials used should have low frictional coefficient inorder to reduce the opposing frictional force. An unsatisfactory surface finish may lead to increased wear and opposing frictional force, thereby loading the engine. This load, which may seem to be negligible, could cause significant losses in the overall efficiency of the engine. 6.8.2 Lubricant properties The lubricant used should possess high viscosity and should have the capability to remain stable under varying thermal conditions. Usually, the viscosity of a substance

Actual Cycles and Their Analysis 6.15

varies inversely with the applied temperature. As the temperature increases, the viscosity decreases thus the lubricating effect also decreases. Due to this reduced lubricating effect, the frictional coefficient increases. This leads to significant frictional losses in the system. 6.8.3 Engine power The engine power is a function of engine speed and mean effective pressure. As the engine speed increases, the frictional resistance between the piston rings and the cylinder wall multiplies. An increase in mean effective pressure also contributes to a small increase in frictional resistance. Moreover, as the speed and pressure increase, the heat developed could affect the proper functioning of the lubricants. Thus, engine power plays a major role in the frictional components of an engine. 6.8.4 Heat Dissipation A proper cooling system ensures proper heat dissipation from the engine. This reduces the temperature of the lubricant present in the engine. Thus the lubricant functions properly and accounts to reduced frictional forces. 6.8.5 Other miscellaneous components Auxiliary components like super-chargers, cooling pumps, radiators, etc. could offer a small frictional resistance in the system. Bearings are the most commonly present components of any mechanical system. Though a bearing offers frictional coefficient only of the order of micron   mm  10  3, the collective resistances of a number of

bearings could cause a small amount of rise of frictional forces in the system.

6.16 Thermal Engineering - I

Thus the collective frictional resistances of a number of auxiliary components could directly affect the efficiency of the engine.

6.9 BLOWBY LOSSES Blowby may be defined as the phenomenon by which the air-fuel mixture reaches the crank case below the piston. At high pressures, the air-fuel mixture is susceptible to enter the crank-case through the cracks and crevices in the piston or the cylinder head. Moreover, blowby could also occur if there is improper sealing between the cylinder wall and the piston ring. Losses due to blowby are more likely to increase when the engine has operated for a long time. The various thermal and compressive stresses on the engine could easily cause the piston and cylinder to develop cracks. Mentioned below are some of the factors which could cause a blowby loss in an IC engine.       

Improper sealing by the piston rings. Cracks developed on the cylinder walls. Oversized bore of the piston. Dents and other physical modifications of the pistons. Improper assembly of piston rings. More end gaps in the piston rings. Late opening of exhaust valves.

6.9.1 Effects of blowby loss on efficiency 

Since a part of the air-fuel mixture escapes to the crankcase, the gas pressure on the piston reduces.

Actual Cycles and Their Analysis 6.17



Thus the mean effective pressure also reduces and this leads to decreased efficiency. A high blowby could cause a considerable pressure rise within the crankcase. This rise in pressure could create resistive force on the rotation of the crank, leading to reduced efficiency.

6.9.2 Crankcase ventilation Crankcase ventilation may be defined as the process by which the pressure developed inside the crankcase is reduced by external means. As discussed in the previous sections, the essential cause of the increased pressure in the crankcase is due to blowby gases from the combustion chamber. Crankcase ventilation is achieved by employing of a PCV valve. The valve is typically a pressure controlled flow valve with a spring loaded mechanism. $LU

$LU &OHDQHU

&RPEXVWLRQ &KDPEHU

&DUEXUHWRU

3&9 &UDQNFDVH

Fig:6.5

6.18 Thermal Engineering - I

The working and the arrangement of a PCV is shown in Fig.6.6. During idling process, a rich mixture of fuel is needed at the inlet of the combustion chamber. However, the blowby mixture at the crankcase predominantly consists of air and less amount of fuel. Due to high manifold pressure at the carburetor, the PCV adjusts accordingly and a small amount of blowby gases enter the inlet port.

1

Idle Sp eed - H igh M anifold Vacu um - Low F lo w

2

H ig h S p ee d - Low M a nifold Va cuu m - M a x F low

3

B acklin e - Pre ssure F rom M anifold - N o Flow

Fig:6.6 PCV W orking

Actual Cycles and Their Analysis 6.19

At normal or cruising speeds, the movable end of the spring shifts to the left due to reduced pressure. As it shifts to the right, a sizeable amount of blowby gases pass through the PCV and enter the inlet of the combustion chamber. At high speeds, the throttling valve is completely open and the suction pressure is increased. At this instant, the entire mixture of air and fuel from the crankcase is forced inside the combustion chamber through the PCV. However, when the quantity of the blowby gases exceed a preset limiting value, the remaining gases are directed towards the air cleaner and then to the carburetor respectively. When a backfire occurs, due to reverse pressure, the PCV completely seals the crankcase from the inlet, thereby preventing the backflow of gases into the crankcase. Thus, the crankcase ventilation process is achieved.

6.10 ACTUAL AND FUEL-AIR CYCLES OF CI ENGINES

P re ssure (ba r)

In reality, a diesel cycle is more efficient than an otto cycle. Incomplete combustion is the main difference between an actual cycle and a fuel-air cycle. As the 80 Fu el-air c ycle combustions are assumed to be instantaneous, the 60 fuel-air cycle are assumed A ctu al cy cle 40 to have complete combustion. 20 In an fuel air process, the allowances are made for the amount and type of fuel. These are bound to

0 0

4

8 12 V cy l / V c Fig:6.7

16

6.20 Thermal Engineering - I

reduce the cycle’s efficiency. In an actual cycle, the allowances are also made for heat transfer, combustion time and other miscellaneous losses as explained previously. Thus the efficiency of the cycle further reduces. To completely understand the exact working of a CI engine, several computer aided models are being developed. It is to be noted that in diesel engines, the ratio between actual efficiency and air-fuel efficiency is around 0.85.

Refrigeration 5.85

But sg at  18 C  0.9170 Since s10.83114   sg0.9170 , the vapour entering the compressor is wet. P

T

2 O

3

2

O

2

2

3

O

O

-18 C

P 1=C 4 f

f

1 g h

4

1

sg s g = 0.9170

18.5 C P 2 =C

g s

Note If s1  sg, then vapour entering compressor is dry saturated. If s1  sg, then vapour entering compressor is super heated. If s1  sg, then vapour entering compressor is wet. Since the vapour at entry of compressor is wet, we have to find the dryness fraction ‘x’, to find h 1. . . . h1  hf  x 1hfg at  18C s1  sf  x 1sfg at  18C

5.86 Thermal Engineering - I

0.83114   0.1507  x10.9170   0.1507 . . [ . sfg  sg  sf ] x1  0.919584 h 1   37.68  0.919584 234.69  37.68   212.78714 kJ/kg h 4  h3

where h3  hf at 18.5C  52.75 kJ/kg So , h 4  52.75 kJ/kg

Now C.O.P  

h1  h4 h2  h1 212.78714  52.75  4.78969 246.2  212.78714

Problem 5.34: In a 15 TR NH3 refrigerant plant, the condenser temperature is 25C and evaporating temperature  10C. The refrigerant NH3 is subcooled by 5 C. The vapour leaving the evaporator is 0.97 dry. Find (i) the C.O.P (ii) power required.

(FAQ)

Solution: From NH 3 refrigeration table, take the following properties Enthalpy (kJ/kg) Temp.

Pressure

hf

25C

10.027 bar

298.90

 10C

2.908

135.37

hg

hfg

Entropy (kJ/kgK) sf

sg

1465.84 1166.94

1.1242

5.0391

1433.05 1297.68

0.5443

5.4770

Refrigeration 5.87

To Find C.O.P C.O.P 

h1  h4 h2  h1

Inlet to compressor is wet vapour, Outlet of compressor may be dry saturated or super heated, which should be found out. If s2  sg at 25C, then vapour leaving compressor is super heated. If s2  sg at 25C, then vapour leaving compressor is dry saturated. If s2  sg at 25C, then vapour leaving compressor is wet. s2  s1 [Isentropic compression] s1  sf  x1sfg at  10C  0.5443  0.97 5.4770  0.5443  s1  5.3290 kJ /kg K  s2 sg at 25C  5.0391 kJ/kg K s25.3290   sg at 25C 5.0391 , heated vapour at the end of compression.

Since

Find h1, h2 and h4 To Find h1 h1  hf  x 1hfg at  10C  135.37  0.971297.68  h1  1394.119 kJ/kg

it

is

super

5.88 Thermal Engineering - I D e gre e of sub co o lin g

T

2

3

o

25 C

2

O

o

Before that, we have to find T2

5C

To Find h2

s2  s2  C p

vap

ln

T2

o

20 C

h3= h4 O

o

-1 0 C

C p  4.84 kJ/kg K

[It can be taken from HMT table] T 2  25  273  298 K s2  s1  5.3290 kJ /kgK

T2   5.3290  5.0391  4.84  ln  298   T2 298 T2

298

O

T 2

where s2  sg at 25 C  5.0391

ln

3

 0.059

 0.059

T2  316.394 K; t2  43.39C h2  h2  C p t2  t2 vap

where h2  h g at 25 C  1465.84 kJ/kg h 2  1465.84  4.84 43.39  25 h2  1554.847 kJ/kg

4

1 s

Refrigeration 5.89

To Find h4 h 4  h 3 h3  h f at 20C  275.16 kJ/kg  h4 [From Refrigeration Table]

C.O.P 

h1  h4 1394.119  275.16   6.9618 h2  h1 1554.847  1394.119

Also, C.O.P 

Refrigeration effect in kW R.E Power W

R.E  15 T R  15  3.5  52.5 kW Power 

52.5  7.5411 kW 6.9618

Po wer  7.5411 kW Problem 5.35: An NH3 refrigeration produces 30 tons of ice from and at 0C in a day of 24 hrs. The temperature range in the compressor is from 25C to  15C. The vapour is dry saturated at the end of compression. Assume actual C.O.P is 60% of theoretical C.O.P. Calculate the power required to drive the compressor. Assume latent heat of ice as 335 kJ/kg. For properties, refer table or charts.

(FAQ)

Solution: Steps: 1. Find theoretical C.O.P and actual C.O.P. 2. Find R.E in kW 3. Find power using actual C.O.P. From Refrigeration table

5.90 Thermal Engineering - I

Temp

hf

hg

h fg

sf

sg

25C

298.9

1465.84

1160.94

1.1242

5.0391

 15C

112.39

1426.58

1314.19

0.4572

5.5497

T

p 3

2 T 2 =2 98

3

2

(2 5 OC )

O

25 C f 4

1

T 1 =2 58 O

-1 5 C

(-1 5 OC )

4

h

Find h1, h2 and h4 h 2  h g at 25C  1465.84 kJ/kg h 4  h 3 and h3  hf at 25C  298.90 kJ/kg So, h 4  298.90 kJ/kg

To Find h1 s1  s2  sg at 25C  5.0391 kJ/kg K s1  sf  x1sfg at  15C 5.0391  0.4572  x1 5.5497  0.4572  x 1  0.8997 h 1  h f  x1h fg at  15C h 1  112.39  0.8997 1314.19   1294.766 kJ/kg

1

g S

Refrigeration 5.91

C.O.P 

h1  h4 1294.766  298.9   5.8212 h2  h1 1465.84  1294.766

Theoretical C.O.P  5.8212 Actual C.O.P

 0.6  5.8212  3.4927

Actual C.O.P 

R.E in kW Power in kW

To Find R.E in kW Heat removed from water at 0C to produce 30 tons of ice at 0C in 24 hours = R.E. in kW = Q total 

30  1000   335  116.32 k W 24  3600 

 kg 

kJ  kW kg  sec

Power Power required to drive the compressor 

R.E in kW 116.32  Actual C.O.P 4927

 33.30 kW

Power  33.30 kW Problem 5.36: An ice plant using NH3 as refrigerant works between  15C and 35C and produces 10 tons of ice per day from water supplied at 0C. The ice temperature is  5C. Assuming simple saturated cycle and using the following properties of NH3, determine (i) The capacity of the refrigeration system required.

5.92 Thermal Engineering - I

(ii) The discharge temperature. (iii) The diameter and stroke of the compressor cylinder if its speed is limited to 1250 r.p.m. Take L/D as 1.2 and volumetric efficiency of the compressor as 0.75. (iv) The power of the motor required to run the compressor if the isentropic efficiency is 85% and mechanical efficiency of 95%. (v) The theoretical and actual C.O.P. Saturation Pressure Temp.

(Dec 2014 JNTU)

Specific enthalpy

Specific entropy

hf

hg

sf

Specific volume sg

vg

C

bar

kJ/kg

kJ/kg

 15

2.36

112.3

1426

0.457

5.549

0.509

35

13.5

347.5

1471

1.282

4.930

0.096

kJ/kg-K kJ/kg-K

Given: T 2  T3  35 C  308 K

h 3  h f at 35  C

T 1   15C  258 K

 347.5 kJ/kg  h 4

m3/kg

Refrigeration 5.93

Solution: From table h1  hg a t  15  C  1426 kJ/kg

[Saturated dry vapour]

s1  sg at  15 C  5.549 kJ/kgK s1  s2

[Isentropic compression]

h 2  hg at 35 C  1471 kJ/kg s2  sg at 35 C  4.93 kJ/kg  T2  s2  s2  cp ln   T   2   T2  5.549  4.930  4.8 ln    308   T 2  350 K

[C p  4.8 kJ/kgK, from table] Discharge temperature, T2  350 K h 2  h 2  C p T 2  T 2  1471  4.8 350  308  h 2  1672 kJ /kg  3.5T Mass flowrate m  h1  h4  m

3.5  10  0.032 kg/s 1426  347.5

L/ D  1.2 L  1.2D

5.94 Thermal Engineering - I

  2 N  m vg1 D L v  60 4

From refrigeration table page 26 vg 

0.5303  0.4896 2

 0.509 m3/kg  3 1250 D  1.2  0.75   0.032  0.509 60 4  D  0.103 m  10.3 cm L  12.4 cm

 Power of motor P  m h2  h1  0.032 1672  1426  7.87 kW

Actual power 

P mech   isen

7.87  9.74 kW 0.85  0.95  Total Actual Power Required = 9.74 kW COP theo 



h 1  h4 h2  h1

1426  347.5  4.38 1672  1426

COP act  4.38  0.95  4.161

Refrigeration 5.95

Problem 5.37: A vapour compression system of 5 ton capacity operates

at

40C

condenser

and

 16C

evaporator

temperatures. Vapour is superheated at the entry to the compressor by 5C. Determine the COP and power requirement. Use the following properties of the refrigerant. (JNTU - Dec 2014)

At tsat  40 C P sat  1.0166 MPa  ; h f  256.41 kJ/kg hg  419.43 kJ/kg; sg  1.711 kJ/kgK, C p  1.145 kJ/kgK At tsat   16 C P sat  0.1572 MPa ; hg  389.02 kJ/kg sg  1.7379 kJ/kgK ; C p  0.831 kJ/kgK T1   16 C  257 K T 2  40 C  313 K

Degree of superheat t1  t1  5 C COP 

o

40 C

o

-11 C o -1 6 C

h1   h 4 h 2  h 1

h1  h 1  Cp [Degree of superheat] h 1  h g at  16 C h1  389.02  0.831  5  393.17 kJ/kg h 2  h2  Cp T 2  T2

To find T2

[s1  sg at  16 C  1.7379 kJ/kgK] s1  s2 [Isentropic compression]

5.96 Thermal Engineering - I

s1  s1  C p ln

T 1 T1

 262   1.7379  0.831  ln    1.754 kJ/kgK  257   T2  s2  s2  Cp ln    T 2   T2  1.754  1.711  1.145  ln    313   T2  ln    0.0376  313   T2  324.98 K h2  h2  C p T2  T2 h 2  hg at 40C h2  419.43  1.145 324.98  313  433.14 kJ/kg h3  hf at 40 C  256.41 kJ/kg  h 4  COP 

h 1  h4 393.17  256.41   3.42 h 2  h 1 433.14  393.17

 Also, R.E in kW  m h 1  h4  m

3.5  5 3.5T   0.127 kg/s h 1  h4 393.17  256.141

 Power Required  m h2  h1 kW

Power

 0.127 433.14  393.17  5.11 kW

Refrigeration 5.97

Problem

5.38: In

an

Ammonia

Vapour

compression

refrigerator, the temperature of refrigerator is  10C. Vapour is condensed in a condenser at 30C. Find the theoretical cop of the cycle when the vapour at the end of the compression is 0.9 dry.

(JNTU - Dec 2013)

T 1   10 C  263 K ; T2  30 C  303 K x 2  0.9

From Refrigeration table - Khurmi - Pg 26 For  10 C

For 30C

h f1  135.37 kJ/kg

hf2  323.08 kJ/kg;

h fg1  1297.68 kJ/kg

hfg2  1145.79 kJ/kg

sf1  0.5443 kJ/kgK

sf2  1.2037 kJ/kgK; sg2  4.9842 kJ/kgK

sg1  5.477 kJ/kgK COP 

RE h1  h4  W h2  h1

h2  hf2  x2 hfg2  323.08  0.9  1145.79  1354.291 kJ/kg

5.98 Thermal Engineering - I

s2  sf2  x2 sfg2  sf2  x2 sg2  sf2  1.2037  0.9 4.9842  1.2037   4.6061 kJ/kgK s2  s1 [Isentropic compression] s1  sf1  x1 sfg1 4.6061  0.544  x1 5.4770  0.544 4.6061  0.544  4.93x 1 x 1  0.8239 h 1  h f1  x1 h fg1  135.37  0.8239  1297.68  1204.528 kJ/kg h3  hf 2

[After condensing at 30 C ]

 323.08 kJ/kg

Also h 3  h 4  h 4  323.08 kJ/kg  COP 



h1  h4 h2  h1 1204.528  323.08 1354.291  1204.528

COP  5.8856

Refrigeration 5.99

5.9 PERFORMANCE CALCULATION AFFECTING PERFORMANCE OF

A

FACTORS VAPOUR

COMPRESSION SYSTEM The factors which affect the performance of a vapour compression system are (i) Effect of suction pressure (Fig. 5.18) The effect of reduction in suction pressure is shown in the P-H diagram Fig. 5.18. C.O.P of original cycle h1  h4  h2  h1

P (Pressu re )

P2

P1

C.O.P of cycle with reduction in suction pressure C.O.P 



2 2O

3

4

1 1O

4O

H (E ntha lp y) Fig. 5.18

h 1  h4 h 2  h 1

h1  h4  h1  h1 h 2  h1  h 1  h 1  h2  h 2

This shows that the refrigeration effect is reduced and work required is increased. The net effect reduces the refrigerating capacity of the system and the C.O.P. (ii) Effect of delivery pressure (Fig. 5.19) The effect of increase in delivery pressure is shown in the Fig. 5.19. Original C.O.P 

h1  h4 h2  h1

5.100 Thermal Engineering - I The C.O.P of increas ed  h1  h4  delivery Pressure  h2  h1

This shows clearly that the refrigeration effect is reduced and work required is increased as in previous case. The net effect reduces the refrigerating capacity of the system and hence C.O.P.

P ( P re ssu re ) 3 P2

3

P1

4



O

O

2

4

O

Fig. 5.19

1 H (E nthalp y)

Note: (i)

As the discharge temperature required in the summer is more as compared with winter, the same machine will give less refrigerating effect (load capacity decrease) at a higher cost.

(ii)

The increase in discharge pressure is necessary for high condensing temperatures and decrease in suction pressure is necessary to maintain low temperature in the evaporator.

(iii) Effect of superheating and subcooling The effect of superheating and subcooling have been already discussed in the previous section. (iv) Effect of suction or vapourising temperature and condenser temperature The performance of a vapour compression system varies considerably with both vapourising and condensing temperatures. Of these, vapourising temperature has a greater effect. It is seen that the capacity and performance of the refrigerating system improve as the vapourising temperature increases and the condensing temperature decreases.

Refrigeration 5.101

5.9.1 Calculations in refrigeration system.

a

vapour

compression

(i) Refrigeration effect: It is the amount of heat absorbed by the refrigerant in its travel through the evaporator. R.E  h 1  h 4 kJ/kg.

(ii) Mass of refrigerant circulated: Mass of refrigerant circulated  One tonne of refrigeration m Refrigeration Effect  m

3.5 in kg/sec  tonnes . h1  h4

(iii) Theoretical piston displacement: Theoretical piston displacement  m  vg1  Ma ss of refrigerant 

Sp. volume of refrigerant gas at its e ntrance of compres sor

(iv) Power (Theoretical) required): (a) For isentropic compression Work of compression  h 2  h 1 Power required

 m h2  h1 kW

(b) For polytropic process P vn  C Work of compression 

n P v  P 1v1 in Nm/kg n1 2 2

5.102 Thermal Engineering - I

Power required

m

n P v  P 1v1 in Watts n1 2 2 where m : m ass of refrigera nt circulated in kg/s

(v) Heat rejected to compressor cooling water: Heat rejected to compressor cooling water   n  P 2v2  P1v1  h 2  h 1   n1

(vi) Heat removed through condenser Heat removed through condenser  m h2  h3 in kJ/s or kW.  m : mass of refrigerant circulated in kg/s.

5.10 VAPOUR ABSORPTION SYSTEM The vapour absorption system differs from the compression system in a way that it uses heat energy instead of mechanical energy to make a change in the condition necessary to complete the refrigeration cycle. The heat energy for this purpose may be obtained from a gas burner, kerosene lamp or electric heater. The system uses a minimum number of moving parts. The only moving parts used by smaller units are valves and controls but larger units are circulating pumps and fans also. Due to the absence of moving parts such units are quiet in operation and may be used for both commercial and domestic installations. The working of an absorption machine depends upon the use of two substances which have a great affinity for each other and which can be easily separated by the application of heat. The principal combinations are sulphuric acid and water or ammonia and water etc.

Refrigeration 5.103

5.10.1 Working principle of vapour absorption A simple vapour absorption system is shown in the Fig. 5.20. If a compressor in a vapour compression system were replaced with a generator absorber assembly, the result would be a simple vapour absorption system. CO NDEN SOR A M M O N IA VA P O U R C O O L IN G W AT E R

G E N E R ATO R STRONG S O LU T IO N PU MP

W E A K S O L U T IO N R E C E IV E R

C O O L IN G A B S O R B E R W AT E R

EVA POR ATOR

E X PA N S IO N VA LV E Fig . 5.20 A S IM P L E VA P O U R A B S O R P TIO N S Y S TE M

A low pressure refrigerant vapour (Ammonia) coming from the evaporator is absorbed in the absorber by the weak solution of refrigerant in water. Absorption of ammonia lowers the pressure in the absorber, which in turn draws more ammonia vapour from the evaporator. Cooling arrangement evolved in absorber. This increases ammonia absorption capacity of water. The pump draws strong solution from the absorber, builds up a pressure upto 10 bar and forces the strong solution in the generator. In the generator, strong solution of ammonia is heated by some external source such as a gas or steam. In the heating process, the ammonia vapour is driven out of the solution leaving behind the generator a weak solution. The weak

5.104 Thermal Engineering - I

solution flows back to the absorber through a control valve which maintains the pressure differential between the high and low sides of the system. From the generator the refrigerant vapour goes to the condenser where it is condensed. Then the high pressure liquid ammonia is passed through a throttle valve to the evaporator where it absorbs its latent heat thus producing cooling effect. 5.10.2 Practical vapour absorption system The simple vapour absorption system can provide refrigeration but its operating efficiency is very low. In order to make it more practical it is fitted with accessories like heat exchanger, an analyser and a rectifier. The practical vapour absorption system is shown in the Fig. 5.21. AM M O N IA VA PO U R

S trong solution

C ooling w ater R EC TIFIE R

Analyser R eceiver W eak solution H eat exchanger

Pum p

Expansion valve Fig. 5.21

Refrigeration 5.105

(i) Heat Exchanger Heat exchanger is located between the generator and absorber. The strong solution pumped from the absorber to the generator is heated. The weak solution from generator to absorber is cooled. This is done in the heat exchanger. The heat exchanger therefore reduces both the cost of heating the generator and the cost of cooling the absorber. (ii) Analyser Analyser is a direct contact heat exchanger consisting of a series of trays mounted above the generator. Its function is to remove partly some of the unwanted water particles associated with ammonia vapour going to the condenser. The water vapour if allowed to enter the condenser may enter the expansion valve, get freezed and hence chock the pipeline. (iii) Rectifier The final reduction (elimination) of the percentage of water vapour occurs in the rectifier. It is a water cooled heat exchanger which condenses water vapour (and some ammonia) and returns it to the generator. The net refrigerating effect of the machine is the heat extracted in the evaporator. The total energy supplied for operating the machine is the sum of the workdone by the liquid pump and the heat supplied in the generator. C.O.P of machine 

Heat extracted from the evaporator W ork done b y pu mp  Heat supplied in generator

5.106 Thermal Engineering - I

5.11 REFRIGERANT A refrigerant is defined as the substance which absorbs heat through expansion or vapourization and loses heat through condensation in the refrigeration system. Usually refrigenants consist of those working mediums which pass through the cycle of evaporation, recovery, compression, condensation and recovery. These substances absorb heat from a body at low temperatures and reject them at a place of higher temperature at the cost of some mechanical work. Refrigerants are classified into 



Primary refrigerants - These are heat carriers which directly take part in the refrigeration process and absorb the latent heat of the substances, thereby cooling them Eg NH 3, CO2, SO2, CH3Cl, Freon groups etc. Secondary refrigerants - These substances are initially cooled down by primary refrigerants and then used for refrigeration purposes. Eg - Ice, solid CO 2 etc.

5.11.1 Characteristics of good refrigerants 1.

The refrigerant should not be poisonous.

2.

It should not be corrosive.

3.

It should not be explosive.

4.

It should not be inflammable.

5.

It should operate under low pressure (i.e. Its boiling point should be low).

6.

It should not be toxic.

Refrigeration 5.107

7.

The difference between vapourising pressure and condensing pressure of refrigerant should be minimum.

8.

The standard evaporator temperature  15 C and condenser temperature  30 C is being preferred for refrigerant.

9.

It should be cheap and abundantly available.

10.

It should be eco-friendly.

11.

The refrigerant should have higher latent heat to increase the refrigeration effect.

12.

It should have lesser specific heat to decrease the work input.

13.

It should have higher critical point.

5.11.2 Refrigerants Number ASHRAE - American Society of Heating, Refrigerating and Air-conditioning Engineers have given numbers as follows. Refrigerant No.

Name

Chemical Formula

R-11

Trichloromono fluoro meth ane

CCl3F

R-12

Dichloro di fluoro me thane

CCl2F2

R-22

Mon ochlorodifluoro methane

CHC lF2

R-500

Aze otropic mixture of 73.8%

R12

and 26.2% R-502

R-717

Azeotropic mixture of 48.8%

R-152(a) R22

and 51.2%

R-115

Ammonia

NH 3

5.108 Thermal Engineering - I

5.12 AMMONIA - WATER ABSORPTION SYSTEM 5.12.1 Absorption Refrigerator)

Refrigeration:

(Electrolux

Steps 1.

Ammonia enters the condenser as vapour.

2.

It is condensed into liquid in the condenser.

3.

The liquid ammonia enters the expansion valve as a liquid.

4.

It partially flashes into vapour in the expansion valve.

5.

Remaining liquid ammonia is further vaporized as it absorbs heat in the evaporator.

Refrigeration 5.109

Note: Upto this point, (i.e. the part of the flow diagram to the left of line AA in Fig. 5.22) is same as that of vapour compression cycle. 6.

Ammonia vapour from the evaporator is dissolved in the water in the absorber. Now the heat is removed from the absorber to keep its temperature constant. [This process is called exothermic process which is a condensation process of a pure substance. The absorber temperature is maintained as low as possible.]

7.

The strong ammonia-water solution is then pumped to the generator which is at the condenser pressure.

8.

Heat is added to the solution in the generator to drive much of the ammonia out of solution. [This process is called endothermic process which is an evaporation process of pure substance.]

9.

Ammonia vapour (only ammonia-not water) goes to condenser and the weak ammonia-water solution left in the generator passes through (return line) a valve back to the absorber.

10.

Then the cycle is repeated and the evaporator removes heat from the space to be cooled.

5.12.2 Lithium Bromide Absorption System Lithium Bromide absorption system is a type of refrigeration system which uses water as refrigerant. Lithium Bromide is used to transfer the heat inside the system by acting as an absorbent. Fig.5.23 shows the working principle of this system.

5.110 Thermal Engineering - I

Process cycle   



The water to be chilled is sent through tubes inside the system. The refrigerant water cools the tube and receives enough heat from it to produce vapour particles. These vapour particles are mixed with a strong solution of a spray of Lithium Bromide, thereby diluting it. This process is called Absorption and it takes place inside the absorption chamber. The diluted Li-Br solution is then pumped to generator through a heat exchanger.

C onden ser

C oolin g Wa ter O ut

S trong LiB r

C oolin g Wa ter

H eating C o il

G en erator Wa ter Tube

E xpan ding Valve S trong L iB r

D iluted L iB r E vapo rato r

~

H eat E xchanger

A bsorb er

D iluted L i B r C oolin g W ate r Fig. 5.23 L ithiu m Bro mide Vapo ur A bsorption S yste m - w o rkin g

Refrigeration 5.111









The generator is either heated with a heating coil or a superheated steam. Thus the water vapour in the diluted solution evaporates, and condenses inside the condenser, leaving behind a strong Li-Br solution. This strong solution is taken back to the absorber to be sprayed in to the water vapour. This process is cyclic. The condensed water in the condenser, reaches the evaporator through an expansion valve and makes up for the water vapour produced. An external supply of cooling water is given to the condenser through the absorber, to aid the heat transfer.

5.13 GAS LIQUEFACTION SYSTEM 5.13.1 Hampson-Linde Gas Liquefaction System After multistage compression, the gas is cooled from state 2 to state 3 at constant pressure in an after-cooler. The gas is further cooled to state 4 by a regenerative heat exchanger. After expansion through a throttle valve, the fluid is in the liquid-vapour mixtures state 5 and is mechanically separated into liquid (state a) and vapour (state 6). The liquid is drawn off as the desired product and the vapour flows through the regenerative heat exchanger to cool the gas flowing toward the throttle valve. The area under line 6-7 on the T-S diagram equals that under 3-4.

5.112 Thermal Engineering - I

C on stan t-p ress ure co olin g in a fte rco oler

2 W in

Q ou t A fterco oler

Expa nsio n va lve

3

7 H e at E xcha ng er

Q

4

G a s from o utside sys te m is m ix ed w ith g as in sta te 7 to fo rm g as in sta te 1 th at e nte rs th e m ultista ge co m p res so r.

Q o ut

M ultista ge com p ressor

1

0 G as

T

A d ditio na l co o lin g in re pre se nta tive h ea t e xch a ng er= h ea t a d de d to ta ke the flu id fro m sta te 6 to 7

2

M u ltistag e co m p res sio n 0

6 3

1

5

7

S e pe ra tor a

L iqu id

Irre versib le exp a nsion throu gh valve

4

Gas liquefaction: H am pson - Linde system

6 a

Fig. 5.24

5 F lu id is m ech a nically se p ara ted in to va po r in state 6 an d liq uid in sta te a S

The gas at state 7 is mixed with an amount of gas from outside equal to the amount of liquid removed and this mixture in state 1 enters the compressor. 5.13.2 Claude System for Liquefying Gases From state 1 to state Hamson-Linde system are same.

4,

both

Claude

and

After the gas is cooled to state 4 by the compressor after cooler followed by a regenerative heat exchanger, most of it is expanded through an engine and then it is mixed with vapour from the separator and flows back toward the compressor through a heat exchanger which precooles appreciably the small fraction of the flow that is directed toward the throttle valve instead of the engine. In the separator, the liquid is separated from the vapour-liquid mixture. Simplified flow and T-S diagrams are shown in Fig. 5.25.

C om pressor

Refrigeration 5.113 Q out 2

3 Q

regenerative heat exchanger

M ultistage com pressor

10 9

4

8

6 7 S eparator

1

Q out 0 G as W E ngine is used to obtain w ork from the system 2 T

Q 5

W in

e 3 a

Liquid

G as liquefaction : C laude system Fig. 5.25

5.13.3 Advantages absorption system:

4

and

5 a

6

0 1 10

9 W ork is obtained 8 e from expansion 7 process s

Limitations

of

vapour

Advantages (i)

Compression of liquid requires lesser power.

(ii)

Low grade energy (fire wood heat) is sufficient to run the system.

(iii)

Friction loss is minimum due to the absence of compressor. Therefore mechanical efficiency is high.

(iv)

Noise and vibration is minimum.

Limitations (i) (ii)

It occupies more space and more volume. Initial investment is more.

(iii)

It is not suitable for smaller capacity.

(iv)

It is more complicated for fabricating the plant.

5.114 Thermal Engineering - I

5.14 COMPARISION BETWEEN VAPOUR COMPRESSION AND VAPOUR ABSORPTION SYSTEM Vapour Compression (i)

Compressor is used.

(ii) It is suitable capacity.

for

Vapour Absorption No compressor is used; Absorber pump, generator and throttling valves are used.

low It is suitable for higher capacity.

(iii) Only high grade energy Low grade heat energy is mechanical energy and sufficient. electrical energy are needed. (iv) C.O.P is less. (v)

C.O.P is more.

Initial investment is less. Initial more.

investment

is

(vi) It occupies less volume It occupies more volume and space. and space. (vii) Leakage cannot detected easily.

be Leakage can be easily detected due to presence of water.

(viii) Mechanical efficiency is Mechanical efficiency is low. more.

5.15 APPLICATION OF CRYOGENIC Cryogenic is the study of materials at very low temperature upto  200C (i) Application of Liquid Oxygen (a)

Liquid oxygen is used in fire extinguisher.

Refrigeration 5.115

(b)

It is used for artificial respiration in medical field.

(c)

It is used for deep sea diving and swimming.

(d)

It is used for climbing mountains and treckking.

(e)

It is used for oxy-acetylene welding.

(ii) Application of Liquid Nitrogen (a)

It is used in hydrogenation. (for manufacturing vegetable oil and vanaspathi)

(b)

It is used for manufacturing fertilizers. (like NH 3, ammonia acid)

(c)

It is used for storage in low temperature.

(iii) Application of Carbon-di-oxide (a)

It is used for fire extinguisher.

(b)

It is used for storage purpose.

(c)

It is used for special type of welding.

(iv) Application of Inert Gases (a)

Inert gases are filled up in space exploration balloons.

(b)

Advertisement boards are filled up with inert gases.

(c)

TIG - Tungsten Inert Gas welding. MIG - Metal Inert Gas Welding - These are used for welding.

Chapter - 6

Actual Cycles and Their Analysis Introduction, Comparision of air standard and actual cycles, Time loss factor, Heat loss factor, Exhaust Blowdown - Loss due to Gas exchange processes, Volumetric efficiency. Loss due to Rubbing friction, Actual and fuel - air cycles of CI engine.

6.1 INTRODUCTION Air standard cycle is a cycle which uses air as the working medium and assumes that the working medium behaves as a perfect working substance. Air standard cycle also prevents all the heat losses that could occur in an engine and pictures it as an imaginary perfect engine. The steps involved in air standard cycles are simple and idealistic, which is why They are also called as Ideal cycles. The following assumptions are made to consider a process as an air standard cycle. 1.

The working substance should be a perfect gas with standard specific heats.

2.

The heat transfer should be simple and no chemical reactions should occur.

3.

It must be a reversible process.

4.

The heat losses are assumed to be zero.

The otto cycle and diesel cycle of an IC engine are said to be air standard cycles but their real working processes greatly vary with respect to their physical properties. A normal engine operation is subjected to various heat losses, change in heat input, change in working medium, etc.

6.2 Thermal Engineering - I

The cycle which accounts to the every possible losses in real time environments may be referred to as an actual cycle. The actual cycle efficiency is much lesser than the air-standard efficiency because of the various losses occurring in an actual engine. If the losses due to variable specific heats due to varying temperatures are neglected from an air standard cycle, then it is called as Fuel-air cycle. In fuel air cycle, the working medium is the mixture of air and fuel vapour or atomized liquid fuel. Each cycle varies with each other in many aspects.

6.2 COMPARISON OF AIR-STANDARD AND ACTUAL CYCLES As discussed earlier, the actual cycle is different from air standard cycle due to various losses that occur in an engine. So, the losses must be taken into account while studying the difference between both the cycles. The major losses occurring in an heat engine are:(i) (ii)

Dissociation losses. Loss due to incomplete combustion.

(iii)

Time loss.

(iv)

Heat loss.

(v) (vi) (vii) (viii)

Loss due to temperature.

variation

of

specific

heats

with

Loss due to exhaust blowdown. Loss due to rubbing friction. Loss due to blowby gases.

The above losses are used to differentiate actual cycles from air-standard cycles and also fuel-air cycles.

Actual Cycles and Their Analysis 6.3

Each of the losses results in the decrease of thermal efficiency and power output of the actual engine. So, understanding them clearly would help us to improve the performance of an actual engine. Along with these losses, there are some factors that also amount to the difference between these cycles. They are, (a) The working medium In case of a fuel-air cycle, the working medium would be air-fuel mixture, whereas for air standard cycle, it is pure air. (b) The nature of working substance In case of air-standard cycle, the working substance remains unchanged from the start to end whereas in other cycles the chemical composition of the working substance may change. (c) The variation of temperature, pressure and composition of fresh charge indicates the difference between the cycles. (d) Heat transfer is a property which differentiates actual cycles and air-standard cycles because, in air-standard cycles there is no heat transfer between the components. In an actual engine, there is a significant amount of heat transfer to and from the working substance as well as the cylinder walls and various other components. Hence the efficiency of the working cycle changes accordingly. (e) Gas leakage, fluid friction, engine design, rpm, etc. are all responsible for defining the losses in an actual engine cycle.

6.4 Thermal Engineering - I

All the above factors could be used to compare actual cycles, air standard cycles and fuel-air cycles. Efficiencies of each cycles also vary with one another. For example, the efficiency of actual cycle will always be lower than air-standard cycle because of the mentioned losses that occur in an actual engine. All the major losses are discussed briefly in the coming sections.

6.3 TIME LOSS Under actual cycle, it is almost impossible to obtain a homogeneous mixture of air and fuel as there are contaminants like carbon deposits, fuel charge from previous cycles etc. Thus, a part of the fuel in the combustion chamber reacts better than its counterpart when heat is applied. Hence, the time taken for the charge to burn completely varies for every cycle. Moreover, the time taken to completely burn a charge also varies due to the following factors:      

Homogeneity of the air-fuel mixture. Flame speed. Configuration of the combustion chamber. Turbulence in the air-fuel mixture. Distance needed to travel by the flame. Air-fuel ratio.

A homogeneous mixture burns steadily and uniformly within a preset time. However, it depends upon the mixing before the inlet. As little time is available at the inlet for proper mixing of air and fuel, there is always a compromise of the homogeneity of the mixture.

Actual Cycles and Their Analysis 6.5

Flame speed depends upon the air-fuel mixture, type of fuel used and the available volume. It is easier for a flame to burn through a dense charge of air and fuel rather than a small amount of charge. Flame speed is also higher at richer air-fuel ratios. Configuration of the combustion chamber deals with the geometry and the space available for the propagation of flame. A cylinder with smaller stroke and larger bore burns the fuel more quickly than its alternate. Turbulence of the mixture depends upon the inlet valve and its position. A turbulent mixture burns more rapidly than a less agitated mixture. By positioning the inlet valve suitably, several motions of air-fuel mixture within the cylinder can be obtained. A turbulent mixture burns faster than a non-turbulent mixture. The air-fuel ratio is also responsible for the time required to complete the burning process. So, the correct proportion of air-fuel ratio must be provided in order to achieve proper burning within a given interval of time. The fuel’s chemical structure and also its ignition temperature must be taken into account to determine the time required for it to burn completely during combustion. Any variation in air-fuel ratio could result in a time loss due to the time consumed by the burning process. Maximum power depends on the peak pressure of the cycle. If the spark is too early in the cycle, work is done against the expansion of gases. Thus the energy is lost. If the spark occurs after TDC, the peak pressure may not be achieved easily, consequently leading to reduced power and efficiency.

6.6 Thermal Engineering - I

These above factors are the reasons for power loss due to time factor in an internal combustion engine. However, in an ideal cycle, the heat addition is assumed to be instantaneous. Thus, with the actual losses, the total work done by a cycle is depicted by a graph of Volume vs Pressure as shown in the Fig. 6.1.  

  

 



,

   



  



 

  







  'F\O 'F



, 3   



Fig:6.1

6.4 HEAT LOSS FACTORS The losses due to heat transfer play a major role in the efficiency of an IC engine. To determine the total heat loss of an engine, it is necessary to understand the distribution of heat among various components of an engine. During the combustion process, an enormous amount of heat is generated. This heat is used to drive the piston. However, a fraction of this heat is dissipated into

Actual Cycles and Their Analysis 6.7

various components such as cylinder walls, piston rings etc. through conduction or convection process. The engine as a whole radiates a negligible amount of heat per cycle. Moreover, a part of the generated heat is quenched instantaneously by the coolants flowing through the jackets. A fraction of the heat is also transferred to the lubricating oil. Thus, the heat generated through the combustion process is transferred to several processes as shown below. Workdone piston

to

drive

the

Convection and conduction heat transfer to internal components Heat Generated

Heat transfer cooling systems

through

Heat lost at the exhaust stroke Heat lost due to unburnt fuel It may be noted that a sizeable fraction of the heat produced is repelled out of the engine during the exhaust stroke even before it gets converted into work. This is due to the constraints in the time required to convert all the heat into work.

6.8 Thermal Engineering - I

Thus, even if the losses due to heat are minimized, only one-fifth of the total heat energy produced is converted into quantifiable work. On a larger picture, heat losses however contribute to only 12% - 15% of the total losses of an IC engine.

6.5 EXHAUST BLOWDOWN Exhaust blowdown is the process in which some of the combustion gases escape through the exhaust port if the exhaust valve is opened before Bottom dead center. This is due to the self pressure created by the piston at the end of the combustion stroke and also it depends upon the compression ratio of the engine. The basic operation of the exhaust stroke is to pump out all the exhaust gases through the exhaust valve. Since the piston has to work against the self pressure created, extra power is consumed to pump all the gases. This results in loss of power and this phenomenon is called as exhaust blowdown loss. It can be avoided, if the exhaust valve opens at or after the BDC, (or) the best way is to open the exhaust valve between 40  70 before BDC which inturn

P ressu re (b ar)

E xpa n sion stro ke

Id ea l diag ram

E a rly e xh a ust v a lv e o p en in g 7 .0

O ptim um exha us t va lve op en in g

3 .5 E xha u st Valve O pe n in g a t b ottom d ea d ce ntre

Vo lu m e

Fig:6.2

Actual Cycles and Their Analysis 6.9

reduces the exhaust pressure of the cylinder into half (from 7 bar to 3.5 bar). The Fig.6.2 explains the exhaust blowdown process.

6.6 LOSS DUE TO GAS EXCHANGE PROCESS The phenomenon of gas exchange process is the admission of fresh charge during the inlet stroke and the removal of exhaust gases at the end of the combustion stroke. In other words, the difference between the workdone in removing the exhaust gases, and the workdone in admitting fresh charge is called as Pumping work. Also, the loss occurring due to this pumping work is called as pumping loss. This loss is due to the difference of pressure between lower inlet pressure (Pi) and the highest exhaust pressure (Pe). P

Ex

Co

Pe

E xha u st

mp

re s

pa

sio

ns

io n

n

P a tm Pi

P u m ping w o rk V Fig:6.3 G as Exch an ge pro cesses

The Fig.6.3 shows the gas exchange process in SI engine. The shaded area indicates the pumping work of pressure from lower value to higher value; from suction to exhaust. The pumping loss increases with throttling,

6.10 Thermal Engineering - I

because throttling reduces the suction pressure and also it increases with speed. The volumetric efficiency of the engine is greatly affected by the losses occurring due to gas exchange processes which will be explained briefly in the next article.

6.7 VOLUMETRIC EFFICIENCY The volumetric efficiency is one of the important parameters in deciding the performance of an IC engine. It is defined as the ratio of volume of air intake during the suction stroke to the swept volume of the piston. It is also defined as the ratio of mass of charge sucked into the cylinder during suction stroke to the mass of air corresponding to the swept volume of engine at atmospheric pressure and temperature. A number of factors affect the volumetric efficiency of an engine. Some of them are given here: [Note - For further efficiency refer Chapter 4].

explanation

on

volumetric

6.7.1 The amount of air-fuel ratio When a fresh charge is sucked inside the cylinder, the temperature of the charge is increased by transfer of heat from the cylinder walls and the hot exhaust gases. It is known that for a constant pressure and volume, the temperature is inversely proportional to the mass. As a result, the mass of the fresh charge decreases which inturn reduces the volumetric efficiency. But the volumetric efficiency could be increased by the high pressure from the incoming fresh charge because high pressure results in increased mass of charge.

Actual Cycles and Their Analysis 6.11

6.7.2 The design of intake and exhaust manifold and port design The intake and exhaust manifold should be designed in such a way that the intake manifold must allow the maximum fresh charge to come in and the exhaust manifold must remove all the burnt gases to escape. So, there is no limitation in designing the intake and exhaust manifold if they carryout the above process correctly. But, if the intake and exhaust process does not take place properly, it will affect the volumetric efficiency and so, the design should be modified accordingly. 6.7.3 Compression ratio Compression ratio is defined as the degree to which the fuel mixture is compressed before ignition takes place. The compression ratio does not have much effect on the volumetric efficiency because it does not change the displacement volume. But, there may be a little effect, due to the amount of residual gas in the cylinder. 6.7.4 The intake and exhaust valve timings Valve timings in IC engines is the time at which the inlet and exhaust valves are set to operate. A finite period of time is needed for the valves to operate at a smooth rate which is provided by a cam operation. A cam and follower mechanism is used to achieve a proper opening and closing of the valves, and also to determine the timing of the valve. The inlet valve timing and exhaust valve timing affect the volumetric efficiency. Usually, the inlet port is opened at some degrees before the TDC in order to allow maximum possible charge to come in. Many S.I engines are designed to open at 10 before the TDC and the charge is sucked in

6.12 Thermal Engineering - I

until the piston reaches the BDC. Now, the intake valve should be closed when the piston reaches BDC, but modern SI engines tend to close at a few degrees (say 10) after the BDC for low speed engine and after say 60 for high speed engine. After reaching BDC, the piston starts the compression stroke by moving up. So if the intake valve is still open beyond BDC, the incoming charges are pushed out by the piston which tend to move up and reach BDC. Hence, the charge is pushed out by the piston and some workdone in doing so results in the reduction of Volumetric efficiency. TD C

Inta ke closes

E xhau st closes

BDC

E xhau st op en s

Fig:6.4 (a)

Inta ke closes

o

20

o

60

o

55

Po w er

o

10

IV O

25

o

C om pressio n

o

EVO

5

10

Inta ke o pe ns

o

Pow er

10

TD C E xhau st close s Valve o verlap

IV O

C om pression

E VO

Inta ke o pe ns

o

E xhau st op en s BDC Fig:6.4 (b)

However, it is different for the high speed engines because the high speed engines bring high pressurized charges inside the combustion chamber, which acting against the moving piston and produce a ram effect. This ram effect could be used as an advantage in order to suck more of the high pressured fresh charges in. So, the valve

Actual Cycles and Their Analysis 6.13

time for the high speed engines could be delayed until after BDC in order to make use of the ram effect. Now, coming to the exhaust valve timings, the exhaust valves tend to open slightly before the piston reaches BDC. For a low speed engine, the valve opens at 25 before the BDC and at 55 before BDC for high speed engines. Actually, this process of opening the valve before BDC helps in removing the burnt gases easily without consumption of much work from the expansion stroke. And like inlet valves the exhaust valves of modern engines tend to close at some degrees after the TDC. In a way it is advantageous because it helps to carry out all the exhaust gases from the cylinder, but there may be a chance for valve overlap. Valve overlap occurs when both the inlet and exhaust valves are opened at the same time. Due to this effect, scavenging takes place (ie) removing out the fresh inlet charge through the exhaust. Also, it may result in loss due to blowdown effect which may affect the volumetric efficiency. [NOTE - The valve taming diagrams for both 4 - stroke and 2 - stroke engines are explained in detail in the earlier sections, refer Pg No. 1.27 to 1.34.] Other miscellaneous factors that also affect the volumetric efficiency are: (i) (ii)

Stroke to Bore ratio. Fuel type, fraction of fuel vapourised in the intake system and fuel enthalpy of vapourisation.

(iii)

Engine speed.

(iv)

Ratio of inlet and exhaust manifold.

6.14 Thermal Engineering - I

(v)

The temperature inside the cylinder.

6.8 LOSS DUE TO RUBBING FRICTION The internal components of an IC engine experience an enormous amount of thermal and compressive stresses. The components are usually designed to withstand these stresses, simultaneously providing a better efficiency. However, as discussed in the earlier sections, several losses deteriorate the efficiency. Friction is one of the primary losses in an engine. Some common factors which influence the frictional forces in an engine are:     

Surface finish of piston and cylinder Lubricant properties Engine power Heat dissipation Other auxillary engine components

6.8.1 Surface finish The prime components like piston, piston rings, cylinder walls etc. are expected to maintain a highly accurate surface finish, with tolerances in the order of few microns. Also, the materials used should have low frictional coefficient inorder to reduce the opposing frictional force. An unsatisfactory surface finish may lead to increased wear and opposing frictional force, thereby loading the engine. This load, which may seem to be negligible, could cause significant losses in the overall efficiency of the engine. 6.8.2 Lubricant properties The lubricant used should possess high viscosity and should have the capability to remain stable under varying thermal conditions. Usually, the viscosity of a substance

Actual Cycles and Their Analysis 6.15

varies inversely with the applied temperature. As the temperature increases, the viscosity decreases thus the lubricating effect also decreases. Due to this reduced lubricating effect, the frictional coefficient increases. This leads to significant frictional losses in the system. 6.8.3 Engine power The engine power is a function of engine speed and mean effective pressure. As the engine speed increases, the frictional resistance between the piston rings and the cylinder wall multiplies. An increase in mean effective pressure also contributes to a small increase in frictional resistance. Moreover, as the speed and pressure increase, the heat developed could affect the proper functioning of the lubricants. Thus, engine power plays a major role in the frictional components of an engine. 6.8.4 Heat Dissipation A proper cooling system ensures proper heat dissipation from the engine. This reduces the temperature of the lubricant present in the engine. Thus the lubricant functions properly and accounts to reduced frictional forces. 6.8.5 Other miscellaneous components Auxiliary components like super-chargers, cooling pumps, radiators, etc. could offer a small frictional resistance in the system. Bearings are the most commonly present components of any mechanical system. Though a bearing offers frictional coefficient only of the order of micron   mm  10  3, the collective resistances of a number of

bearings could cause a small amount of rise of frictional forces in the system.

6.16 Thermal Engineering - I

Thus the collective frictional resistances of a number of auxiliary components could directly affect the efficiency of the engine.

6.9 BLOWBY LOSSES Blowby may be defined as the phenomenon by which the air-fuel mixture reaches the crank case below the piston. At high pressures, the air-fuel mixture is susceptible to enter the crank-case through the cracks and crevices in the piston or the cylinder head. Moreover, blowby could also occur if there is improper sealing between the cylinder wall and the piston ring. Losses due to blowby are more likely to increase when the engine has operated for a long time. The various thermal and compressive stresses on the engine could easily cause the piston and cylinder to develop cracks. Mentioned below are some of the factors which could cause a blowby loss in an IC engine.       

Improper sealing by the piston rings. Cracks developed on the cylinder walls. Oversized bore of the piston. Dents and other physical modifications of the pistons. Improper assembly of piston rings. More end gaps in the piston rings. Late opening of exhaust valves.

6.9.1 Effects of blowby loss on efficiency 

Since a part of the air-fuel mixture escapes to the crankcase, the gas pressure on the piston reduces.

Actual Cycles and Their Analysis 6.17



Thus the mean effective pressure also reduces and this leads to decreased efficiency. A high blowby could cause a considerable pressure rise within the crankcase. This rise in pressure could create resistive force on the rotation of the crank, leading to reduced efficiency.

6.9.2 Crankcase ventilation Crankcase ventilation may be defined as the process by which the pressure developed inside the crankcase is reduced by external means. As discussed in the previous sections, the essential cause of the increased pressure in the crankcase is due to blowby gases from the combustion chamber. Crankcase ventilation is achieved by employing of a PCV valve. The valve is typically a pressure controlled flow valve with a spring loaded mechanism. $LU

$LU &OHDQHU

&RPEXVWLRQ &KDPEHU

&DUEXUHWRU

3&9 &UDQNFDVH

Fig:6.5

6.18 Thermal Engineering - I

The working and the arrangement of a PCV is shown in Fig.6.6. During idling process, a rich mixture of fuel is needed at the inlet of the combustion chamber. However, the blowby mixture at the crankcase predominantly consists of air and less amount of fuel. Due to high manifold pressure at the carburetor, the PCV adjusts accordingly and a small amount of blowby gases enter the inlet port.

1

Idle Sp eed - H igh M anifold Vacu um - Low F lo w

2

H ig h S p ee d - Low M a nifold Va cuu m - M a x F low

3

B acklin e - Pre ssure F rom M anifold - N o Flow

Fig:6.6 PCV W orking

Actual Cycles and Their Analysis 6.19

At normal or cruising speeds, the movable end of the spring shifts to the left due to reduced pressure. As it shifts to the right, a sizeable amount of blowby gases pass through the PCV and enter the inlet of the combustion chamber. At high speeds, the throttling valve is completely open and the suction pressure is increased. At this instant, the entire mixture of air and fuel from the crankcase is forced inside the combustion chamber through the PCV. However, when the quantity of the blowby gases exceed a preset limiting value, the remaining gases are directed towards the air cleaner and then to the carburetor respectively. When a backfire occurs, due to reverse pressure, the PCV completely seals the crankcase from the inlet, thereby preventing the backflow of gases into the crankcase. Thus, the crankcase ventilation process is achieved.

6.10 ACTUAL AND FUEL-AIR CYCLES OF CI ENGINES

P re ssure (ba r)

In reality, a diesel cycle is more efficient than an otto cycle. Incomplete combustion is the main difference between an actual cycle and a fuel-air cycle. As the 80 Fu el-air c ycle combustions are assumed to be instantaneous, the 60 fuel-air cycle are assumed A ctu al cy cle 40 to have complete combustion. 20 In an fuel air process, the allowances are made for the amount and type of fuel. These are bound to

0 0

4

8 12 V cy l / V c Fig:6.7

16

6.20 Thermal Engineering - I

reduce the cycle’s efficiency. In an actual cycle, the allowances are also made for heat transfer, combustion time and other miscellaneous losses as explained previously. Thus the efficiency of the cycle further reduces. To completely understand the exact working of a CI engine, several computer aided models are being developed. It is to be noted that in diesel engines, the ratio between actual efficiency and air-fuel efficiency is around 0.85.

Supplementary 7.1

Supplementary 7.1 AXIAL FANS AND PROPELLERS A fan usually consists of a single rotor with or without a stator element, and causes only a small pressure rise of the flowing fluid. The total pressure developed by fans is of the order of a few mm of water gauge. The example of axial fans are ceiling, table and ventilation fans. 7.1.1 Axial Fan Stage An axial fan stage consists of a rotor made up of a number of blades fitted to the hub, when it is rotated by an electric motor. A flow is estabilshed through the rotor and it causes an increase in the stagnation pressure of air or gas across it. A cylindrical casing which encloses the rotor receives the flow through a converging passage (nozzle) and discharges it through a diverging passage (diffuser) as shown in Fig. 7.1 (a). (i) W stage  u C w3  C w2  h0stage (ii) Mass flow rate  m   A Cf 

Where

 D 2  d 2 C f 4

D  tip diameter and d  h ub dia meter

(iii) Power required to drive the fan   P  m C P  T0st  m u [C w3  C w2]

7.2 Thermal Engineering - I

(iv) Stage pressure rise P 0stage   h 0stage   Wact 

P 0  W act

 0

P 0   u2 [1   tan 3]

(v) Stage pressure co-efficient  

(vi) Stage reaction Rd 

(vii) Fan efficiency

P 0stage 

u2 2

 Protor P 0stage



Static pressure rise in the rotor Stagnation pressure rise in the stage



Isentropic work Actual work input



where

Pst

P 0stage  [u C w3  C w2]

 Isentropic work

u C w3  C w2  Actual work   m P 0st v P0stage   Power developed P    0 0  m3   Where v  volume rate of flow    sec 

Supplementary 7.3

7.1.2 Types of Axial Fan Stages (a) Stage without guide vanes [Fig. 7.1 (a) & (b)] (b) Stage with upstream guide vanes [UGV] [Fig. 7.1 (c) & (d)] (c) Stage with downstream guide vanes [DGV] (d) Stage with upstream and downstream guide vanes N o zzle R o tor

2

w2

D iffuser

C f2 = C w 2

u

Hub

Inle t W3

O utlet

3  C f3

C3

Cw3 C a sing

u

Fig. 7.1 (a)

Fig. 7.1 (b)

C 1 =C f1

U .G .V R o tor

IG V 2

2

W2 C2 u

O utle t

Inle t

W3

Cw2

3 C 3 =C f3

u

Fig. 7.1 (c)

Fig. 7.1 (d)

C f2

7.4 Thermal Engineering - I

Problem 7.1: An axial fan takes in 2.5 m3/sec at 102 kPa and 315 K and delivers it at 75 cm of water head and 325 K. Determine the mass flow rate through the fan and power required to drive the fan and static fan efficiency.

Given data  v  2.5 m 3/sec

[FAQ]

2

P 1  102 kPa

T

T1  315 K

P  750 mm W.H 1

T2  325 K S

Solution: 10336 mm of W.H  101.325 kPa  750 mm of W.H 

101.325  750 10336

 7.3523 kPa  P 2  P 1  P  109.3523 kPa 1

T 2

 P2     P T1  1

 109.3523   T2     315 102    321.3268 K  u

DN  21.9911 m /sec 60

Supplementary 7.5



P1 RT1

 1.14645 kg/m 3

 m   A C f  1.16478 kg/sec



Cf u

 0.2501

u  3.99838  2  75.958 tan 2  Cf P 0st   u 2 [1   tan 3]  1.14645 [21.9911] 2 [1  0.2501  tan 10]  529.9827 N/m 2 101.325  10 3 N/m 2  10336 mm of W.H P 0 

10336  529.9827 101.325  10 3

P 0  54.0626 mm of W.H

The ideal power required to drive the fan  m P 0  538.4576 Watts   0 

ideal power actual power

0 

538.4576  44.871% 1200

Static pressure rise in the stage Pst 

1  u2 [1   2 tan 2 3] 2

7.6 Thermal Engineering - I

Protor  276.6773 N/m 2 Protor  0.522  Rd  Pstage

Result   0.2501

(a) Flow co-efficient (b) Rotor blade angle at inlet

2  75.958 

(c) Static pressure rise P 0stage  54.0626 mm of W.G (d)

0  44.871 %

(e) Degree of reaction

R d  52.2%

Problem 7.2: In an axial flow fan the rotor and IGV blades are symmetrical and arranged for 50% reaction with blade angles at inlet and exit are 65 and 12 respectively. The mean diameter of the blade is 620 mm and height of the blade is 15 cm. The static properties of air at inlet are 101 kPa and 310 K. The speed of rotor is 1000 rpm. Determine stage pressure co-efficient, power required for a fan efficiency of 82% and drive efficiency 85%.

Given data R d  0.5

2  65 

3  12 

D m  0.62

h  0.15 cm

P 1  101 kPa

T 1  310 K

N  1000 rpm

f  0.82

d  0.85

Um 

 Dm N 60

 32.4631 m/sec

Supplementary 7.7

A   Dm  h  0.2921681 m 2 

101  10 3 P   1.135214 kg/m 3 RT 287  310

In a 50% reaction stage  2  3 ;

2   3,

 2  12  ;

2  65,

C w2 ; tan  2  Cf

Cw2  Cf  tan  2

tan 2 

u  Cw2 Cf

 Cf 

;

u  Cw2 tan 2

...(1)

...(2)

Equating (1) and (2) C w2 tan 2





u  C w2 tan 2 C w2 u  C w 2



tan  2  0.0991167 tan 2

C w2  3.2117637  0.0991167 C w 2  C w2  2.927475 m /sec  C f  13.7726 m/sec C w3 tan  3  ; Cf  Cw3  29.5354 m /sec

7.8 Thermal Engineering - I

 Wact  u C w3  C w 2  863.7768 J/kg  m W act P  m   A Cf mech  4.642 kW  4.568 kg/sec P stage

=W C3

 W act

2

 Pstage  804.0686 N/m 2 U 2 3 

2

Cf

=W C2

 1.3442

Cw2

2P

3



=W

U 2

2

2

P

C



u

3

2 2 Cf

=W C3

For efficiency f 

Cw3 u

Result 1. Stage pressure co-efficient   1.3442 2. Power required to drive the fan P  4.642 kW

Supplementary 7.9

7.2 CENTRIFUGAL FANS AND BLOWERS A blowers consists of one or more stages of compression, with its rotor mounted on a common shaft. The air is compressed in a series of successive stages and is often led through a diffuses located near the exit. The overall pressure rise may range from 1 to 2.5 atm with shaft speeds up to 30000 rpm. Blowers are used in ventilation, power stations, workshops, etc. 7.2.1 Centrifugal Blower The centrifugal blower consists of s rotor or impeller which rotates causing air-flow by centrifugal action. The air

Volu te C a sing

Im p elle r

A ir in S haft

G uide Vane s (D iffuser)

Fig. 7.2 Centrifugal Blower

7.10 Thermal Engineering - I

usually enters the impeller at the axis and leaves at the tip in a direction determined by the angle of the impeller blades as shown in Fig. 7.2 Upon leaving the tip, the air flows through a volute chamber, some times provided with a vaned diffuser casing. The diffuses casing utilizes part of the K.E. of the outflowing fluid and raises its static pressure. The volute chamber collects all the fluid at constant velocity and leads it to a diverging discharge pipe which may again provide more of diffusive action. 7.2.2 Types of Centrifugal Fans There are three types of impellers used in centrifugal fans. 1. Backward Swept Blades For low pressure and lower flow rates 2  90, refer Fig.(a) [same as compressor] C w2 u2

1

1  Rd  2

2. Radial Blades The swirl at the entry is zero,  C w1  0 used for medium pressure and medium flow rates. 2  90 and  Rd 

C w2 u2 1 2

1

[refer Fig]

Supplementary 7.11

3. Forward Swept Blades 2  90 and C w2  u2 Rd 

1 2

It is used for high pressure and high flow rates. Refer Fig. (c) 7.2.3 Centrifugal Fan Stage Parameters    1. Mass flow rate m  1v1  2v2 2. Area of cross section A1  d 1b1 and A 2  d2b 2 [ 1- impeller inlet and 2 - Exit] 3. u1 

d 1N 60

and u2 

d 2N 60

4. Stage work W st  u 2 C w2  u 1 C w1. For zero whirl at C w1  0,  Wst  u2 C w2 since the velocity triangles are same as centrifugal compressors.

inlet

Therefore, W st  u 22 [1  2c o t 2]  h0

...(5.3)

5. Power required to drive the fan   P  m h0st  mu2C w2 6. Stage pressure rise P 0st  h0st  P 0st  P 0  P 0   P 2  P 1  C 22  C 21 2 1 2 P 0st  Protor 

 2 [C  C 21] 2 2

7. Stage pressure co-efficient  st 

...(5.15) P 0st u 22 2

...(5.16)

7.12 Thermal Engineering - I

8. Degree of reaction Rd 

Protor P 0stage

1



9. Stage efficiency 

C w2 2 u2

1  1   2 c o t 2  2

P 0  W act



...(5.17)

...(5.11)

P 0st  C w2 u 2

10. 0  mech  st

...(5.18) ...(5.19)

Problem 7.3 A centrifugal blower takes in 180 m3/min of air at pressure 101.3 kPa and temperature 315 K and delivers it at 750 mm of water head. Taking the efficiency of blower as 80% and mechanical efficiency 82%, determine the power required to drive the blower and exit condition of air. [FAQ]

Given data  180  3 m 3/sec ; v 60

P 1  101.3 kPa ;

T 1  315 K

st  0.8 ;

mech  0.82 ;

P  750 mm of water head

Solution: 1 

P1 RT 1



101.3  10 3  1.12051 kg/m 3 287  315

10336 mm of W.H  101.325 kPa  750 mm of W.H 

101.325  750 10336

Supplementary 7.13

P  7.352336 kPa  P 2  P 1  P  108.652 kPa  Ideal power  v  P P i  22057.008 Watts  Actual power required 

Pi st   mech



22057 0.8  0.82

 33.62348 kW P 7.35 Ideal work supplied W i    6.56159 kJ/kg 1.12   Actual work supplied 

Wi st



6.56  8.201997 kJ/kg 0.8

W act  C P T 2  T 1 8.2  1.005 T2  315  T2  323.1611 K

Result 1. The condition of air at exit P 2  108.652 kPa and T2  323.1611 K 2. Power required to drive the blower P  33.62348 kW Problem 7.4 A centrifugal fan has the following data:

Inner diameter of the impeller

:

18 cm

Outer diameter of the impeller

:

20 cm

Speed

:

1450 rpm

The relative and absolute velocities respectively are:

7.14 Thermal Engineering - I

Air entry 20 m/sec and 21 m/sec At exit 17 m/sec and 25 m/sec Flow rate is 0.5 kg/sec. Motor efficiency is 78% Determine: (a) The Stage pressure rise (b) Degree of reaction and (c) The power required to drive the fan. Take density of air as 1.25 kg/m3

(FAQ)

Given data D 1  0.18 m ;

D 2  0.2 m ;

N  1450 rpm

C r  20 m /sec

C 1  21 m /sec  m  0.5 kg/sec

C r  17 m /sec

C 2  25 m /sec

  0.78

  1.25 kg/m 3 Solution: u1 

D 1N

u2 

60

 13.6659 m/sec

D 2N 60

 15.184369 m/sec

1 2 u  u21  21.9 J/kg 2 2 1 2 C  C 2r2  55.5 J/kg 2 r1 1 2 C  C 21  92 J/kg 2 2

From Euler’s equation for a stage work

Supplementary 7.15

W st 

1 2 1 1 u2  u21  C 2r1  C 2r2  C 22  C 21 2 2 2

 169.4 J/kg

Total pressure rise across the stage is P 0st    W st  211.75 N/m 2 101325 N/m 2  10336 mm of water head  211.75 N/m 2 

10336  211.75 101325

 21.6 mm of water head

Static pressure rise in the rotor Protor 

 2 [u  u21  cr21  C 2r2] 2 2

 1.25 [21.9  55.5]  96.75 N /m 2

Degree of reaction 



Protor P 0stage 96.75  0.4569 211.75

 m W st Power required to drive the fan P    108.5897 Watts

7.16 Thermal Engineering - I

Results 1. Stage pressure rise P 0  21.6 mm of W.H 2. Degree of reaction R d  0.4569 3. Power required to drive the fan P  108.5897 Watts Problem 7.5 A centrifugal fan has  of 80%. Its impeller diameter is 1 m runs at 720 rpm. The impeller tip angle is backward curved to 51 tangent to the wheel. The density of the air is 1.25 kg/m3 and mass flow rate is 3 kg/sec. The impeller width at the exit is 10 cm. Determine the power required, pressure co-efficient, stage reaction, pressure head developed and flow co-efficient at exit. Assume zero whirl at inlet and mech is 82%.

Given data f  0.8 ;   1.25 kg/m 3 ;

D2  1 m ;  m  3 kg/sec

mech  0.82

,

2  51 b 2  0.1 m

Solution: u2 

 D2 N 60

 37.69911 m/sec

A 2   D 2 b2    1  0.1  0.314159 m 2  m   A C f2  m  7.6394 m/sec  C f2    A2

Assuming constant radial velocity,

Supplementary 7.17

 C f1  C f2  7.6394 m/sec

Refer velocity diagram, C f2 tan 2  u  C w2

C2

C r2 C f2

u  C w2 

C f2

tan 2

2

2

 6.18626

Cw2

 Cw2  31.51284 m /sec

u2

W act  C w2 u 2  1188.0062 J/kg f 

P  W act

 P   W act   f  1188.0062 N/m2

Power required 

 m Wact m ech

 4346.3643 Watts  4.34636 kW

Pressure co-efficient  st 

P

u22

 1.33744

2 Degree of reaction R d  1 

Flow co-efficient 2 

C f2 u2

Cw2 2 u2

 0.582

 0.20264

7.18 Thermal Engineering - I

Result 1.

Power required

P  4.34636 kW

2.

Pressure co-efficient

  1.33744

3.

Stage reaction

R d  0.582

4.

Pressure head developed

P  1188.0062 N/m 2

5.

Flow co-efficient at exit

 2  0.20264

Supplementary 7.19

7.3 WANKEL ENGINE A Wankel engine is a type of internal combustion engine wherein the applied gas pressure creates a rotational moment on the rotor. Unlike a reciprocating engine, a Wankel engine does not have any parts undergoing linear motion. Instead, the applied pressure is used to create an eccentric rotary motion of the crank present in it. For these reasons, a Wankel engine is often known as a rotary engine. Note: A 6 - cylinder radially placed engine is also known as a rotary engine and it is often employed in aircraft.

Fuel

Fuel injector

In jecte d F u e l S ide sea l A pex seal C o olan t jackets

Inle t A ir

E xhau st gas

C o rne r seal

S park p lu ge

Fig. 7.3 Wankel Engine

7.20 Thermal Engineering - I

WORKING PRINCIPLE A schematic diagram of the Wankel engine is shown in the figure. It consists of a crank which is eccentrically mounted onto the centre shaft which is often referred to as the E-Shaft or the Eccentric shaft. The crank is triangular in shape and the inner periphery of the outlet casing is in the form of an epi-trochoid shape as shown in the figure. This geometric orientation ensures the contact between the vertices of the crank and the inner periphery of the casing at all times.

S u c tio n Fu el

A

C o m pre sse d Fu el

In le t A E x h a u st

B C C 1

2

( i ) Intake

( ii ) Com pression

C B

C

B 3

C o m bu stio n of Fuel

A 4 B u r nt Fu el

( iii ) Pow er

( iv )Exhaust

Fig. 7.4 Working of Wankel engine

Supplementary 7.21

The suction stroke starts when the intake port opens up and the air-fuel mixture occupies the upper portion between AB and the periphery as shown in the Fig 7.4 (i). As the rotor rotates, the sucked air is compressed against the inner periphery of the casings. This rotation corresponds to the compression stroke of a reciprocating engine. After another partial rotation, the fuel is compressed against the spark plugs as shown in the Fig 7.4 (iii). At this time, the spark plug produces a spark and the fuel ignites. This causes the fuel to expand and the gas pressure further rotates the rotor in the same direction. To prevent rotation in the opposite direction, the spark is initiated only after a major part of the rotor passes the plugs. At this time, the exhaust valves are opened and the burnt gases are expelled out of the system at shown in Fig. 7.4 (iv). The above mentioned cycle applies to one of the 3 sides of the rotor. However, all the 3 sides of the rotor perform the same function for a given position. Thus, in a rotary engine, the suction, compression, power and exhaust strokes are always happening at all times. Advantages of Wankel Engines 



There are two power strokes for every three strokes in a Wankel engine when compared to one power stroke for every four strokes in a reciprocating engine. The number of components in a Wankel engine is lesser than the number of components in a reciprocating engine.

7.22 Thermal Engineering - I

 

The power to weight ratio is lesser than that of a reciprocating engine. The engine is immune to knock and other phenomenon which could lead to its seizure.

Disadvantages of Wankel Engine    

 

The amount of unburnt hydrocarbons emitting out of a rotary engine is quite high. The fuel economy is reciprocating engine.

less

than

that

of

a

Upon failure, the components are difficult to replace. The edges of the rotor time more susceptible to failure and wear which lead to reduce the efficiency. The design of the inner periphery and the rotor is quite difficult. Since the power produced is quite high a top-notch cooling system is required.

Zenith Carburetor Zenith carburetor, often called as ‘British carburetor’ is employed in many famous cars, largely in cars manufactured in England. Construction It consists of a float chamber where the fuel from the fuel tank is supplied through a pipe. When the fuel drains in the float chamber, the needle valve moves up allowing the fuel to flow into the chamber. The flow of fuel into the float chamber is faster than the consumption of fuel by the engine. When the fuel chamber receives its needed quality

Supplementary 7.23

T h ro ttle Va lv e

Ven tu r i

Id lin g jet

M a in J e t

Id le A d justing K n ob A A ir f or Id lin g & S lo w R u n n in g B To g gle B le v e r

C o m pe n sa tin g jet Tub e

A ir

F loa t

F loa t C h am b er

C h ok e

C o m pe n sa tin g jet

Fuel

of fuel, the float rises up, there by the needle moves down stopping the fuel flow into the chamber. Three jets namely auxiliary jet main jet and idle jet are provided. The main jet is directly connected with the float chamber while the auxiliary jet or compensating jet fetches fuel from the reservoir / auxiliary chamber. An idle jet is where the fuel flows during idle and slow speed conditions. An orifice connects the float chamber and auxiliary chamber, while both auxiliary jet and main jet are opened up in the venturi. The air is supplied through a passage to the carburetor. The throtle valve is located at the end of the carburetor and connected to the engine suction pipe. The opening and closing of this valve controls the quantity of air-fuel mixture supplied to the engine suction manifold.

7.24 Thermal Engineering - I

Working Initially the choke is used for starting. During idling the throttle valve is closed. When engine suction is applied to the idling jet, the fuel is supplied. The required quantity of air for idling enters through the holes A and B and the mixture then passes out as shown in Fig. 7.5. For the idle adjustment a separate knob is provided. It is connected with the threaded extension tube that controls the opening B, to control the quality of mixture. When the throttle is opened a little, a small quantity of air flows via venturi. Both main jet and slow running jet supplies the mixture at this condition. With further more throttling the suction is applied to the main jet and compensating jet and the compensating jet ensures the correct air-fuel ratio being supplied at various speed.

Short Questions and Answers QA.1

SHORT QUESTIONS AND ANSWERS Chapter 1

I.C. Engines 1.1. Name the types of injection system in CI engines. 1.

Air injection system

2.

Airless (or) solid (or) mechanical injection system (a) Individual pump system (b) Common rail system (c) Distributor system.

1.2. Why compression ratio of petrol engines is low while diesel engines have high compression ratio? In petrol engine, the compressed air-fuel mixture gets ignited by sparks produced by spark plug. But in case of diesel engine, the compressed air should attain very high temperature in order to ignite injected diesel. Hence the air should be compressed to very high pressure. So the compression ratio is more in diesel engine than that of petrol engine. 1.3. What are heat engines? Classify. A type of engine or machine which derives heat energy from combustion of fuel or any other source and converts this energy into mechanical work is termed as Heat engines. They are classified as (i) External combustion engines (ii) Internal combustion engines (IC. Engines).

QA.2 Thermal Engineering - I

1.4. Classify I.C engines according to the cycle of combustion. (i) Otto cycle engine (ii) Diesel cycle engine (iii) Dual combustion Engine. 1.5. Classify I.C engines arrangement of cylinders.

according

to

the

(i) Horizontal engines (ii) Vertical engines (iii) Radial engines (iv) V-type engines 1.6. Classify I.C engine according to the method of Ignition. (i) Spark Ignition engines (S.I engines) (ii) Compression Ignition engines (C.I engines) 1.7. What do Engines?

you

mean

by

scavenging

in

I.C

(Anna Univ. Apr’ 2003)

The process of removing burnt exhaust gases from the combustion chamber of engine cylinder is known as scavenging. Scavenging helps in reducing the dilution of fresh charge of mixture in the I.C. engines. Basically there are 3 ways of Scavenging (i) Cross flow scavenging (ii) Backflow or loop scavenging (iii) Uniflow scavenging. 1.8. What is the purpose of Thermostat in an engine cooling system? Whenever engine is started from cold, large amount of heat is required to attain its correct temperature (working). A thermostat prevents the flow of water below a certain temperature, from the engine to the radiator. It

Short Questions and Answers QA.3

is a bellow type thermostat which maintains water at desired temperature. 1.9. Explain exhaust blow down in case of IC engines. (Anna Univ. Dec 2003) When the exhaust port is opened, the gases existing in the cylinder at the end of expansion stroke discharge spontaneously into the exhaust manifold and the pressure of the main cylinder drops to a value lower than that existing in the scavenge air manifold. This process is called blow down 1.10. List out advantages of electronics ignitions system over the conventional systems. (Anna Univ. - Dec 2003)

1.

In conventional type contact breaker interrupts high inductive current during its operation which causes excessive wear and requires frequent services which is not present in electronic system.

2.

Misfiring takes places in conventional system which does not occur in electronic system.

3.

The electronic systems are more efficient than conventional system even at low speeds.

4.

Slow opening of breaker points and cranking speed results in poor starting ability in conventional systems.

1.11. What is the function of push rod and rocker arm? (Anna Univ. - Apr’ 2004) The push rod and rocker arm are required to push the valve against the spring pressure. The roller arm rotates about the rocker arm shaft under the force exerted

QA.4 Thermal Engineering - I

by the push rod. A clearance is kept between the rocker arm and the valve stem and can be adjusted by screw adjuster. 1.12. What are the basic requirements of a fuel injection system of a diesel engine? (Anna Univ. - Apr’ 2004)

1.

Fuel should be introduced into the combustion chamber within a precisely defined period of cycle.

2.

Metered amount of fuel must be injected per cycle very accurately.

3.

The rate of injection should be such that it results in desired heat release pattern.

4.

Quantity of fuel metered should vary with the changing speed and load requirements.

5.

The spray pattern must be in the form of fine droplets.

6.

Height and size of injection system must be minimum.

1.13. What are the types of Ignition systems? (i) (ii) (iii)

Battery coil Ignition system Magneto Ignition system Electronic Ignition system

1.14. List the types of cooling systems in I.C. Engines. (i) Thermo-syphon cooling (ii) Forced or pump cooling (iii) Thermostatic cooling (iv) Pressurised cooling (v) Evaporative cooling.

Short Questions and Answers QA.5

1.15. List types of Lubrication System. (i) Wet sump lubrication system (ii) Dry sump lubrication system (iii) Mist lubrication system. 1.16. What is the use of fuel system? The fuel system is used for the following reasons.   

To store fuel in the fuel tank To supply fuel to the required amount and proper condition To indicate to the driver the fuel level in the fuel tank.

1.17. What is Gravity fuel system? In gravity system, the fuel tank is placed above the carburetor. The fuel flows from the tank to the carburetor due to the gravitational force. Thus the system does not have fuel pump. This system is cheap and simple one. The fuel tank is directly connected to the carburetor. Motor cycles and scooters use this system. 1.18. Explain about Carburetor. Carburetor is a device which is used for atomizing and vapourizing the fuel (petrol) and mixing it with the air in varying proportions, to suit the changing operating conditions of the engine. Atomization is the breaking up the liquid fuel (petrol) into very small particles so that it is properly mixed with the air. But vaporization is the change of state of the fuel from liquid to vapour. Carburetor performs both the process i.e., atomization of the fuel and vaporization of the fuel.

QA.6 Thermal Engineering - I

1.19. What is stoichiometric air fuel ratio? Oxygen is very much necessary to burn the fuel. This oxygen is taken from atmospheric air. The proper proportion of air and fuel mixture should be obtained for complete combustion of fuel. For complete combustion, the Air-Fuel ratio should be approximately 15:1 by weight. This is known as chemically correct or stoichiometric air fuel ratio. 1.20. What are various compensation in carburetors? A simple carburetor can not supply different air-fuel ratio according to the speeds and loads of the engine. To supply correct airfuel ratio to meet the existing condition is known as the compensation in carburetor. The various compensations in carburetor are given below. 1. Auxiliary (or) extra air valve compensation 2. Restricted air bleed compensation 3. Compensating jet compensation 4. Economiser needle in metering jet. 1.21. What are the types of carburetors? There are three important types of carburetor 1. Zenith carburetor 2. Solex Carburetor 3. Amal Carburetor

Short Questions and Answers QA.7

1.22. What are the fuel injection systems in C.I engines? There are two methods of fuel injection in compression ignition engines. 1. air injection, 2. airless or solid or mechanical injection. 1.23. What is the use of spark plug? Spark plug is used in SI engines (Petrol engines) to produce electric spark to ignite the compressed air fuel mixture inside the engine cylinder.

Term ina l C en tral e le ctrod e P orce la in in su la to r

M e ta l S crew

S pa rk g a p or G ro u nd e le ctro de A ir g a p Fig 1.22 Spark plug

1.24. What are requirements of a Good fuel? Requirements of a Good fuel: 1.

Good fuel should have high calorific value.

2.

It should have low ignition temperature.

3.

It should burn freely with high efficiency.

4.

It should not produce any harmful gases.

5.

It should produce very less smoke.

6.

It should be economical.

7.

It should be easily stored and transported.

QA.8 Thermal Engineering - I

Chapter 2

Combustion in SI and CI Engine 2.1. What are the harmful effects of detonation?   

A loud pulsating noise vibration of the engine.

occurs

resulting

in

An increase in the heat lost to the surface of the combustion chamber. An increase in carbon deposits.

2.2. What is the importance of delay period in CI engine combustion phenomenon? If the delay period is more, more fuel (diesel) will be injected inside the cylinder and more will be the pressure rise. This causes Diesel knock. Some delay period is needed to disperse and atomize the fuel in the air for complete combustion. So we have to keep (maintain) the delay period as short as possible. 2.3. How CI engine classified?

combustion

chambers

are

1.

Non-turbulent type - Open combustion chamber.

2.

Turbulent type - (i) Turbulent chamber (ii) Precombustion chamber.

3.

Energy cell.

2.4. What is pre-ignition? How can it be detected? A very high temperature carbon deposits formed inside the combustion chambers ignites the air fuel mixture much before normal ignition occurred by spark plug. The

Short Questions and Answers QA.9

ignition occurred pre-ignition.

by

hot

carbon

deposits

is

called

The standard test for pre-ignition is to shut-off ignition. If the engine still fires, it is assumed that pre-ignition was taking place when ignition was on. Sudden loss of power with no evidence of mechanical malfunctioning is fairly good evidence of pre-ignition. 2.5. What is knocking in diesel fuels? How can it be prevented? If the delay period is prolonged, a large amount of diesel will be injected in the chamber. Combustion of large amount of fuel may cause high pressure rise and this high pressure rise cause knocking. Knocking can be prevented     

by reducing the delay period. by raising the compression ratio. by increasing the turbulence of compressed air. by adjusting the fuel injector. by supercharging.

2.6. Why for the same power CI engines are larger and heavier than SI engine. In CI engine, the air is compressed to a very high pressure i.e the compression ratio in CI engine (diesel cycle) is more. To withstand the high pressure, the CI engines are larger and heavier. 2.7. How are the CI fuels rated? Explain the same (or) Define Cetane Number. (Anna Univ. - Apr’ 2003) CI fuels are rated by its cetane number.

QA.10 Thermal Engineering - I

Cetane number is a number to rate its ability to auto ignite quickly when it is injected into the high pressure, high temperature air in the cylinder. of

Cetane number of a fuel is the percentage by volume cetane in a mixture of cetane and   methyl

naphthalene C 10 H7CH 3 that has the same performance in the standard test engine as that of fuel. 2.8. List the factors that affect the ignition lag. 1.

Temperature and pressure in chamber at the time of injection.

the

combustion

2.

Air fuel ratio.

3.

Turbulence of air.

4.

Pressure of residual gases.

5.

Rate of fuel injection.

6.

The extent of atomization, vaporization and fineness of fuel spray.

2.9. What are desirable qualities of SI engine fuels to inhibit detonation? 1.

Too lean or too rich air-fuel mixture will inhibit detonation, (preferably too rich mixture).

2.

Use of high octane number fuel can eliminate detonation.

2.10. How are the SI fuels rated? The SI engine fuels are rated by its octane number. Octane number is the number to rate the SI engine fuels according to its detonating tendency. If the fuel has the tendency to denotate less, then it has high octane number and vice versa.

Short Questions and Answers QA.11

2.11. What are the requirements of CI combustion chamber for diesel engines? The requirement of CI combustion chamber is to provide proper mixing of fuel and air in a short duration. For this purpose, an organised air movement called as ‘swirl’ is provided to produce high relative velocity between the fuel droplets and air. The combustion chamber should withstand high temperature and pressure. 2.12. What are the requirements of a good combustion chamber for SI engines? 1.

The high power output with minimum octane requirement.

2.

High thermal efficiency.

3.

Smooth engine operation.

2.13. What do you understand by octane number 100? Highest fuel octane number is 100. This type of fuels will not have tendency to detonate. 2.14. Explain about the octane number. Iso-octane is a high rating fuel (i.e. detonation is less). Normal heptane is a low rating fuel (i.e detonation is more). Iso-octane and normal heptane are mixed together and this sample mixture is used for running a test engine. The sample mixture called as Primary Reference Fuel (PRF). The octane number of the fuel is the percentage of Iso-octane in this sample mixture (Primary Reference Fuel (PRF)) which detonates or knocks as similar way as the fuel under the same condition.

QA.12 Thermal Engineering - I

2.15. What is auto-ignition and why it occurs in SI engines? (What is detonating zone?) When sparking occurs, the combustion of fuel nearby spark plug commences. The flame travels through combustion chamber with high speed. The high pressure and high temperature gases produced by this ignition compress the fresh charge in front of the moving flames. Hence the temperature and pressure of the fresh charge is increased beyond the limit and a spontaneous ignition (i.e. auto-ignition) takes place in far away from spark plug. This zone far away from spark plug where auto-ignition takes place is the zone called ‘detonating zone’. 2.16. Knocking in SI ____________ Compression ratio.

engine

increases

with

2.17. Name the index to measure the ignition quality of petrol and diesel engine fuel. 1.

Octane number for petrol.

2.

Cetane number for diesel.

2.18. Name two methods to find the indicated power of IC engines. 1.

By indicator diagram - with the help of PV diagram drawn by an indicator.

2.

By measuring B.P and losses. (losses = Friction power) I.P  B.P  losses

2.19. What are the desirable combustion chamber? (i) Short combustion time

characteristics

of

Short Questions and Answers QA.13

(ii) Short ratio of path to bore (iii) Absence of hot surfaces (iv) High velocity at inlet valve (v) Adequate cooling. 2.20. What are the phases of C.I engine combustion? (i) Ignition lag or delay (ii) Period of uncontrolled combustion (iii) Period of uncontrolled combustion. 2.21. What are stratified engines? Engines having different A/F ratio at different locations within combustion chamber are called stratified engines. 2.22. Write the Dulong’s H.C.V.

formula for calculating

O2   H.C.V  33800 C  144000  H 2    9270 S KJ/kg 8  

where C , H2, O 2 and S are the fractions of carbon, hydrogen, oxygen and sulphur contained in 1 kg of fuel. 2.23. What are the factors affecting flame speed? 1. Turbulence 2. Engine speed 3. Engine size 4. Compression ratio 5. Inlet temperature and pressure 6. Fuel-Air ratio 7. Engine output

QA.14 Thermal Engineering - I

2.24. What is meant by abnormal combustion? The abnormal combustion deviates from the normal behavior resulting in loss of performance and physical damage to the engine. There are two types of Abnormal combustion. 1. Pre-ignition 2. Knocking (or) Detonation (or) Pinking 2.25. What are the effects of knocking in SI Engine? 1. Noise and Roughness 2. Mechanical Damage 3. Carbon deposits 4. Increase in heat transfer 5. Decrease in power output and efficiency 6. Pre-Ignition 2.26. Name the anti-knock additives. The widely used antiknock agents are:  

Tetraethyl lead [TEL] CH3CH 24 Pb Methylcyclopentadienyl manganese (MMT) CH3C 5H 4MnCO3



Ferrocene Fe C5H52



Iron pentacarbonyl



Toluene



Iso octane

tricarbonyl

Short Questions and Answers QA.15

2.27. What are the factors affecting delay period 1.

Temperature and pressure in chamber at the time of injection.

the

combustion

2.

Air-fuel ratio.

3.

Turbulence of air.

4.

Presence of residual gases.

5.

Rate of fuel injection.

6.

The extent of atomization and vaporization and fineness of fuel spray.

2.28. What are the factors that affect delay period in diesel engine? (i) Compression ratio (ii) Intake temperature (iii) Intake pressure (iv) Engine size (v) Fuel temperature (vi) Injection time (vii) Output

QA.16 Thermal Engineering - I

Chapter 3

Testing and Performance of Engines 3.1. Exhaust gas composition analysed by ____________. 1.

of

IC

engine

is

Flame ionisation detector (FID). Extra: Other instruments are as follows.

2.

Spectroscopic analyzers (a) Dispersive analysers (b) Non-dispersive infra-red analysers (NDIR)

3.

Gas chromatography.

3.2. Name two methods to find the indicated power of IC engines. 1.

By indicator diagram - with the help of P V diagram drawn by an indicator.

2.

By measuring B.P and losses. (losses = Friction power) I.P  B.P  losses

3.3. Differentiate between brake power and indicated power of I.C Engines. (Anna Univ. - May 2003) Brake power (i)

Power available at the (i) crankshaft is called Brake power (B.P)

Indicated power Power developed at the engine is called indicated power (IP)

(ii) B.P = I.P – Friction (ii) I.P = B.P + Friction power power (iii) B.P is obtained by (iii) I.P is obtained Brake Dynamometer Indicator diagram.

by

Short Questions and Answers QA.17

3.4. Differentiate between SFC and TFC in engine performance. (Anna Univ. - Dec. 2003) Specific Fuel Consumption (S.F.C) (i)

Total Fuel Consumption (T.F.C)

S.F.C is the ratio of T.F.C is the ratio of fuel amount of fuel consumed consumed in kg/hr to the in kg/hr to the brake indicated power developed power developed in kW in kW of the engine Fue l co nsumed in kg/hr of the engine. T.F.C  I.P in kW F ue l used in kg/hr S.F.C  B.P in kW

3.5. What is Morse test?

(Anna Univ. - Apr 2004)

Morse Test is a test for multicylinder engine to measure the indicated power without employing any elaborate equipments - consider a three cylinder engine 1, 2, 3. With I1, I2, I3 be indicated power and F 1 F 2 F 3 be friction power. Total I.P  I1  I2  I3; Total friction power FP  F 1  F 2  F 3. Total Brake power when all are running B  I1  I2  I3  F 1  F 2  F 3.

When cylinder 1 is cut, Brake power is B 1  0  I2  I3  F 1  F 2  F 3  I.P of cylinder 1  B  B 1

Similarly I.P of cylinder 2  B  B 2

QA.18 Thermal Engineering - I

IP of cylinder 3  B  B 3. 3.6. Define firing order. The order in which the spark plugs of a multicylinder engine are ignited or order in which the cylinder are fired is called firing order. For (Eg) for 3 cylinder engine 1, 3, 2. 4 cylinder engine 1, 4, 3, 2 or 1, 3, 4, 2. 3.7. What is the apparatus to carry out flue gas analysis. Orsat’s apparatus. 3.8. Write the composition of air by mass. Nitrogen N 2  77%; Oxygen O 2  23% 3.9. Write the composition of air by volume. Nitrogen N 2  79% ; Oxygen O 2  21% 3.10. Define L.C.V. The Lower Calorific Value is the heat liberated by 1 kg of fuel after subtracting heat used to vapourize the steam formed from hydrogen. L.C.V  HCV  9 H 2  2466  KJ/kg 3.11. Name the apparatus used for finding calorific values for fuels. Bomb Calorimeter is used for determining calorific value for solid and liquid fuel. Junker’s gas calorimeter is used for determining calorific value for gaseous fuel.

Short Questions and Answers QA.19

3.12. Mass of carbon in 1 kg of flue gas is given by ________. 3 3     11 CO2  7 CO   

3.13. What is an element? The smallest quantity of a substance, which can exist by itself in a chemically recognizable form. 3.14. What is molecular weight of Oxygen? Atomic weight of oxygen  16 Molecular weight of oxygen O 2  32 3.15. Name the molecule which has the minimum molecular weight. Hydrogen  for hydrogen H 2  1  1  2 3.16. 1 kg of CO requires

4 kg of oxygen and produces 7

________ . Ans:

11 kg of CO 2 7

3.17. 1 kg of C requires

4 kg of oxygen and produces 3

________.  7 Ans:  kg of carbon monoxide  3  

QA.20 Thermal Engineering - I

3.18. What is Indicator diagram? An indicator diagram is a P-V diagram traced by the indicator which is attached to the piston. The P-V diagram represents the work done by the engine in one cycle. The power developed inside the engine cylinder is known as indicated power. This is measured by indicator diagram.

P

a d= A rea re pre se n tin g w o rkd on e in on e cy cle E q uiva le nt 1 re ctan gle o f a rea = a d

2 V Fig. 3.1

3.19. How the measurement of cylinder pressure in done? Cylinder pressures can be easily obtained using several analytical equations relating to the temperature and the volume of the working fluid. Indicator diagrams are one of the most common tools needed to compute the pressure of a cycle. However, several electronic and mechanical components are being used to verify the pressure at each stage manually. 3.20. What is meant by pressure transducer? A transducer is a device capable of converting one form of energy into another. A pressure transducer typically converts the pressure exerted on an object into noticeable output like displacement, electrical signals etc. Since the cylinder of an engine is subjected to various thermal and shear stresses, the pressure transducer must be capable of withstanding all the stresses.

Short Questions and Answers QA.21

3.21. What is meant by Friction Power (FP)? The Power available in the engine flywheel (crankshaft end) is less than the power developed inside the engine. i.e. The BP is less than the IP. Because, there is a loss of power due to friction between the moving parts. The Power lost in this way is known as friction power. So, F.P  I.P  B.P The difference between the indicated power and brake power is known as friction power. 3.22. What is Relative Efficiency or Efficiency Ratio? The ratio of the indicated thermal efficiency or the brake thermal efficiency to the air standard efficiency is known as relative efficiency or efficiency ratio. Relative efficiency,  relative 

Indicated o r Brake thermal efficiency Air standard efficiency

3.23. How Dynamometer is classified? Dynamometer can be classified into two types (i) Absorption dynamometers (ii) transmission dynamometer (i) Absorption dynamometer These types of dynamometers are used to measure and absorb the power output of the engine to which they are coupled. The power absorbed is usually released as heat (or) by other forms of energy. Eg., Hydraulic, Eddy current dynamometer etc.

QA.22 Thermal Engineering - I

(ii) Transmission Dynamometer Transmission dynamometer is also called as Torque meter. The purpose of these meters is to simply sense the torque. It doesnt supply (or) receive any energy. Torque meters employ normal measuring units like a strain gauge to directly determine the torque acting on a shaft. 3.24. What is the method for measurement of air consumption?

M easu re me nt of A ir C onsum p tio n th rou gh the orifice cham b er m eth od

Orifice chamber method is used in laboratory for measuring the consumption of air. The arrangement of this system is shown in Fig. 3.12. 3.25. Explain about exhaust gas emission. Emissions can be defined as the unburnt fuel and other by-products exiting the combustion chamber. Based on the visibility, emissions can be broadly classified into  

Visible emissions Invisible emissions

Short Questions and Answers QA.23

Some common emissions found are: (i) Carbon dioxide (ii) Water vapour (iii) Unburnt hydrocarbons (iv) Oxides of nitrogen (v) Aldehydes (vi) Carbon monoxide (vii) Smoke (viii) Particulate matter Based on the type of emission, it’s effect on the environment can vary.

QA.24 Thermal Engineering - I

Chapter 4

Air Compressors 4.1. How are air compressors classified? 1.

Reciprocating compressors and rotary compressors.

2.

Single stage and multistage compressors.

3.

Single acting and double acting compressors.

4.2. Define isothermal efficiency of air compressors. It is defined as the ratio of isothermal work to polytropic work.  isothermal 

W iso thermal W polytropic



I.Pisothermal I.Ppolytropic

4.3. Define mean effective pressure. How it is related to indicated power of an IC engine? Mean effective pressure is defined as hypothetical pressure which is considered to be acting on the piston throughout the power stroke. If it is based on I.P, it is called Indicated mean effective pressure. I.P 

pm ALN or N/2  n 60

in kW

where I.P  Indicated power in kW pm  mean effective pressure in KPa; A  Area of piston

in m 2; L  Length of stroke in m; N  (for 2 stroke) in r.p.m; N/2  (for 4 stroke) in r.p.m; n  No. of cylinders.

Short Questions and Answers QA.25

4.4. What is the meaning of free air delivered (FAD)? Free air delivered is defined as the actual volume rate of air reduced to atmospheric condition and usually expressed in m 3/min. 4.5. Explain how flow of air is controlled in a reciprocating compressor cylinder? The flow of air is controlled.    

By centrifugal governor. By maintaining the speed of motor constant. By providing air pocket adjacent to the cylinder. By making the suction valve open for part of the compression stroke.

4.6. Mention the compressed air. 

 

important

applications

of

Compressed air is used to operate pneumatic circuits such as pneumatic drills, hammers, hoist, air brakes, pile drivers and blast furnaces. It is used for supercharging the IC engines. It is used for cleaning, inflating and injecting purposes.

4.7. What factors limit the delivery pressure in a reciprocating compressor? 1.

The size of the cylinder will be too large for too high pressure.

2.

Due to compression, there is a rise in temperature of air. So the delivery pressure is limited so that rise in temperature of air is not going beyond the limit and size of cylinder is not too large.

QA.26 Thermal Engineering - I

4.8. Mention two main advantages of multistage compression over a single stage compression for the same pressure ratio. 1.

The work done per kg of air is reduced in multistage compression with intercooler as compared to single stage compression for the same delivery pressure.

2.

It improves the volumetric efficiency for the given pressure ratio.

4.9. Name the methods adopted for increasing isothermal efficiency of reciprocating air compressors. By perfect intercooling, isothermal is increased. The following methods are employed to achieve nearly isothermal compression. 1.

Spray injection

2.

Water jacketing

3.

Intercooling

4.

External fins

5.

By suitable choice of cylinder dimensions.

4.10. What is meant by perfect intercooling? In two stage compressors, If the temperature of air leaving the intercooler T3 is equal to the original inlet temperature T1, then the intercooling is known as perfect intercooling. By having perfect intercooling, we can approach the isothermal process. So the isothermal efficiency will be increased by perfect intercooling.

Short Questions and Answers QA.27

4.11. Why clearance is necessary and what is its effect on the performance of reciprocating compressors?   



Clearance volume is necessary so that the piston will not hit the cylinder end at the end of the stroke. It is necessary to provide some space for valve movements. Clearance makes the volume of air taken in per stroke less than the swept volume. So the size of compressor is increased. So the power to drive the compressor is increased. The maximum compression pressure controlled by the clearance volume.

is

also

4.12. What is the part played by the intercooler in multistage air compressor? Explain perfect intercooling. The intercooler is placed in between L.P cylinder and H.P cylinder. The temperature of air leaving L.P cylinder is high T2 and is cooled in the intercooler to T3 i.e the original temperature of inlet air T 1[ i.e T 3  T 1 ]. By cooling the air to its original temperature the process is approaching isothermal process, so that the power required to drive the compressor will be minimum. If T3  T1, then it is perfect intercooling. If T3  T1, then it is imperfect intercooling. 4.13. Why multistage compression reciprocating air compressors?

is

required

in

If we use single stage compression for getting high pressure air, the following problems will arise.

QA.28 Thermal Engineering - I

1.

The size of the cylinder will be too large.

2.

Due to compression, there is a very high rise in temperature of air.

3.

It is not possible to reject the heat from the air in the available duration.

4.

The high temperature of air at the end of compression will heat up the cylinder head and burn the lubricating oil.

4.14. Which type of compression reciprocating compressors?

is

the

best

in

Isothermal compression. Because isothermal compression requires less power to drive the compressor. 4.15. Show the work saved by shading area in P V diagram of 2 stage with intercooling compared to single stage compressor. D elivery P pressure H P

4

W ork saved

5

Perfect intercooling

Isotherm al process P2

3

2

LP P1

1 1-2-5 = Single stage com pression 1-2-3-4 = Tw o stage com pression w ith perfect intercooling V

Short Questions and Answers QA.29

4.16. What is the effect on compressor capacity and power consumption at high altitude compared to sea level? At high altitude, the atmospheric pressure is less. So the high power is required to attain the required pressure ratio. The compressor capacity to deliver free air/min is reduced. 4.17. How do you classify the compressor? I. Positive displacement Compressors (ii) Reciprocating (ii) Rotary (a) Roots blover (b) Screw type (c) Vane type II. Dynamic Compressor (i) Centrifugal (ii) Axial 4.18. Write the expression for workdone clearance for air compressor. n1

  P2  n  n Work done W   m.R. T1    P n1  1

without

 1 

QA.30 Thermal Engineering - I

4.19. Write the expression for clearance for air compressor.

workdone n1

  P2  n n Workdone W   P1 V1  V4    P n1  1

4.20. Define volumetric compressor?

efficiency

of

with

 1 

an

air

Volumetric efficiency of an air compressor is the ratio of free air delivered to the displacement of the compressor (or) Ratio of effective swept volume to the swept volume. Volumetric efficiency  vol 

F.A.D Effective s wept volu me or Displacement volume swept volu me

4.21. Define clearance ratio of an air compressor? The ratio between clearance volume to swept volume is called clearance ratio. Ratio K 

Clearance volume V C  Swept volume VS

4.22. Give the expression for volumetric efficiency in terms of clearance Ratio.  V1  Vol  1  k  k   V  2

Also, Vol amb  

P 1  Tamb   V2   1kk  V T1  Tamb   1

Short Questions and Answers QA.31

4.23. Draw the actual PV diagram for compressor P P2

D elivery pressure

A tm osphe ric pressure P2 V

Intake depre ssio n

4.24. Write an expression for workdone for a two stage compressor with and without intercooling. Workdone without Intercooling n1

  P4  n n P1 V 1   W  P n1  1

n1

  P5  n  n 1 P4 V 4    P  4  n1

 1 

Workdone with perfect intercooling n1

  P2  n n P1 V 1   W  P n1  1

n1

 P3  n   P  2

 2. 

4.25. Write the condition for minimum work in an air compressor? The condition of minimum workdone is  P 1 P3 P2   P 1 : Inlet pressure P 3 : Exit pressure (delivery)

QA.32 Thermal Engineering - I

P 2 : Intermediate pressure.

4.26. Give the expression for minimum work in two stage air compressor? n1

   P 3  2n 2n 1. P 1 V1   Workdone W min   P n1   1

4.27. Give the expression for workdone for multistage compressor? n1

   P z  1  Zn Zn 1 P1 V1   Workdone W   P1 n1    Z  No. of stages.

4.28. Give the expression for heat rejected and isentropic change with perfect intercooling.  n   Heat rejected W   C p  C v    T 2  T 1 per kg n1  of air.  Cp  Cv   P3  Isentropy change S 2  S1    ln  . P 2    1

4.29. Give the methods to increase efficiency of air compressor? (i) Spray injection (ii) Water jacketing (iii) Intercooling (iv) External fins.

isothermal

Short Questions and Answers QA.33

4.30. Define F.A.D? Free air delivered (F.A.D) is defined as the actual volume delivered at the stated pressure reduced to intake temperature and pressure. 4.31. Explain the compressor.

working

principle

of

rotary

In rotary compressor the air is entrapped between two sets of engaging surfaces and the pressure rise is either by back flow of air (roots blower) or by both squeezing action and backflow of air (vane type). 4.32. What are the merits of rotary compressor over reciprocating compressor? 1.

Rotary compressors are suitable for large discharge at low discharge pressure. The maximum free air discharge may be as high as 3000 m 3/min and maximum delivery pressure is 10 kgf/cm 2 (10 bar)

2.

Air supply is continuous in rotary compressors and therefore no receiver is required. But receiver is an essential requirement of a reciprocating compressor.

3.

The rotary compressors are small in size for the same discharge compared with reciprocating compressors.

4.

There is no balancing problem in rotary compressor whereas it is one of the major problems with reciprocating compressors.

4.33. What are stagnation properties. The state of a fluid attained by isentropically decelerating it to zero velocity at zero elevation is referred to as the stagnation state. The properties of the fluid at

QA.34 Thermal Engineering - I

the stagnation state are the stagnation properties of the gas. 4.34. What is meant by stagnation enthalpy? Stagnation enthalpy of a gas or vapour is its enthalpy when it is adiabatically decelerated to zero velocity at zero elevation. Therefore, Stagnation enthalpy h 0  h 

C2 2

4.35 Explain the classification of impellers based on blade shape.

2 < 9 0



2 > 9 0

o

o

u 

2 = 90

o

(ii) FO R W A R D

(i) B A C K W A R D

u 

Fig:4.26

(iii) R A D IA L

Short Questions and Answers QA.35

The exit vane shape of the centrifugal compressors are generally any one of the three configurations. (i) Backward curved (ii) Radial and (iii) Forward curved If the angle between the rotor blade tip and the tangent to the rotor at the exit is acute i.e., 2  90, the vanes are backward curved vanes. If this angle is a right angle 2  90 the blade is said to be radial. If it is greater than 90 [2  90] the blade is forward curved. 4.36. State the loses which occur at the impeller of a centrifugal compressor. The following losses occur when air flows through the impeller: (i)

(ii) (iii)

Friction between the air layers moving with relative velocities and friction between the air and flow passages. Shock at entry. Turbulence caused in air.

4.37. Define isentropic compressor.

efficiency

for

a

rotary

Isentropic efficiency “Isentropic efficiency” of rotary compressor is defined as the ratio of isentropic temperature rise to actual temperature rise. Isentropic efficiency 

Isentropic temperature rise Actual tem perature rise

It is also defined as the ratio of isentropic compression work to actual compression work.

QA.36 Thermal Engineering - I

4.38. What is meant by slip factor. The tendency of air to flow around the edges of the vanes in the clearance space between the impeller and the casing is known as slip. Slip factor s is defined as the ratio of actual whirl component C w2 and the ideal whirl component u 2 4.39. Explain the term work factor. The actual work input to the air is greater than the theoretical value due to friction between the casing and the air carried around by the vanes. In order to take this into account “work factor”, w is introduced, so that the actual work done on the air becomes Work factor w 

Actual work supplied Theoretical work supplied

 W act  w C w2 u 2  C p T02  T01 

4.40. Define the term pressure coefficient. It is defined as the ratio of isentropic work to Euler work. p 

Isentropic w ork C p T02   T01  Cw2 u 2 Euler work

4.41. Write about axial flow compressors. An axial compressor is a pressure producing machine where the flow of air or fluid is in the direction of the rotor axis. It consists of a rotor with moving blades and a stator fixed to casing which serve to recover part of kinetic energy imparted to the working fluid. This kinetic energy imparted

Short Questions and Answers QA.37

to the fluid by means of the rotating blades is then converted into a pressure rise. 4.42. What is heat compressor.

coefficient

of

an

axial

flow

Head or work coefficient h  It is defined as the ratio of actual work done to the kinetic energy developed by the mean peripheral velocity. Thus,  h 

Cp  T  u2 



   2 

 tan  2  tan  1  2 C w2  C w1 2  u  tan 2 tan  2 

4.43. What are the basic losses in an axial flow compressor. 1. Profile losses on the surface of the blades. 2. Skin friction on the annulus walls. 3. Secondary flow losses. 2 .29 % 1 00

S A n n u lu k in F ric ti on s lo s s S e co n d a ry lo s s

4 4% 4 2%

P ro file lo ss

D es ig n

S u rge

S ta ge e fficie ncy, s tag e

90

80

70

60 0 .5

0 .7

0 .9

1 .1

1 .3

Flow C oe ffic ie nt f Fig:4.38 L osse s in com pressor stage

1 .5

1 .7

QA.38 Thermal Engineering - I

4.44. Explain secondary flow losses in an axial flow compressor. 





Secondary flows are produced by combined effects of curvature and boundary layer. Secondary flow is developed when the components of velocity are developed from the deflection of an initially sheared flow. Such secondary flow occurs when there is a bend, when a sheared flow passes over an aerofoil shape with finite lift or when a boundary level meets an obstacle. Secondary flow loss occurs when boundary layers are growing on the casing and hub walls of the machines are deflected by rows of blades - stator and rotor.

It is the phenomena of excessive aerodynamic pulsation which is transmitted throughout the machine by virtue of sudden drop in delivery pressure or complete breakdown of the steady through flow.

Pressu re ratio

4.45. Explain the term surging.

S urge C ycle D U nstable

S urge lin e

B S ta b le

A

E N1

N3 T3

N2 T2

T1 C

M ass flow rate Fig:4.39

4.46. Explain the term stalling. Stalling is the separation of flow from the blade surface. A partial blockage or uneven flow in the blade passages due to the change of angle of incidence is called

Short Questions and Answers QA.39

stalling. At low flow rates, the axial velocities are lower and the angle of incidence called stalling. At low flow rates, the axial velocities are lower and the angle of incidence is increased. 4.47. State the differences between reciprocating and centrifugal compressors. Reciprocating compressors

Centrifugal compressors

Greater vibration problems Less vibrational problems due to the presence of since it does not have reciprocating parts which reciprocating parts. are partially balanced. Due to the presence of Due to the absence of many several sliding or bearing sliding or bearing members, members, it has lesser  mech is more.  mech. Higher initial cost.

Lower initial cost.

Pressure ratio per stage is Pressure ratio per stage is about 5 to 8. about 3 to 4.5. High delivery pressure upto Medium delivery 5000 atm. upto 400 atm.

pressure

Smaller Free Air Delivered Greater FAD. (FAD). Greater Flexibility in No flexibility in capacity and capacity and pressure range. pressure range. Higher maintenance cost.

Lower maintenance cost.

Compression efficiency is  compressor is higher, at higher, at compression ratio compression ratio less than above 2. 2.

QA.40 Thermal Engineering - I

Reciprocating compressors

Centrifugal compressors

Adaptability to low speed Adaptability to high speed drive. drive. More operating needed.

attention Less operating attendance.

Always a chance of mixing No chance of mixing of air with lubricating oil. lubricating oil with air.

of

Suitable for low, medium Suitable for low and and high pressures and low medium pressures and large and medium gas volumes. gas volumes. 4.48. State the differences between reciprocating and rotary compressors. Reciprocating air compressors

Rotary air compressors

Suitable for low discharge of Suitable for handling large air at high pressure. volumes of air at low pressures. Low speed (RPM).

High speed (RPM).

Pulsating air supply.

Continuous air supply.

More cyclic vibrations occur. Less vibrations occur. Complicated system.

lubricating Simple lubrication system.

Air delivered is generally Air delivered is relatively contaminated with oil. more clean. Large compressor size for Small size the given discharge. discharge

for

same

Short Questions and Answers QA.41

Reciprocating air compressors 250  300 m 3/min

Free

Rotary air compressors air 2000  3000 m 3/min FAD.

Delivery. High delivery pressure.

Low delivery pressure.

4.49 What are the differences between centrifugal and axial flow compressors. Centrifugal compressors Radial flow

Axial flow compressors Axial flow (Parallel to the direction of axis of the machine)

Pressure ratio per stage is Low pressure ratio per stage high, about 4.5:1. This unit about 1.2:1. This is due to is compact. absence of centrifugal action. Less compact and less rugged. Isothermal efficiency about 80 to 82% Frontal area is larger

is With modern aerofoil blades,  iso is about 86 to 88%. Frontal area is smaller. Hence the axial flow compressor is more suitable for jet engines due to less drag.

More flexibility of operation Less flexibility of operation. due to adjustable prewhirl and diffuser vanes. Low starting required.

torque High starting required.

torque

QA.42 Thermal Engineering - I

Centrifugal compressors Multistaging difficult.

is

Axial flow compressors

slightly More suitable multi-staging.

Upto 400 bar delivery Delivery pressure pressure is possible. upto 20 bar. It is used in application of blowing engines in steel mills, low pressure refrigeration, big central air conditioning plants, fertiliser and industry, supercharging I.C. engines, gas pumping in long distance pipe lines etc.

for is only

Mostly used in jet engines due to higher efficiency and smaller frontal area. Also used in power plant gas turbines and steel mills.

Short Questions and Answers QA.43

Chapter 5

Refrigeration 5.1. What is commonly used unit of refrigeration? The ‘Ton of Refrigeration’ is commonly used unit of refrigeration. One ton of refrigeration is defined as the heat removed from 1000 kg of water at 0C to make 1000 kg of ice at 0C within 24 hours.  3.5 kW Simply 1 ton of refrigeration  3.4892 kW~

5.2. What are the desirable properties of an ideal refrigerant? The refrigerant should not be poisonous, corrosive, explosive, inflammable and toxic. It should operate under low pressure. 5.3. Why air cycle refrigeration is more popular in air craft air conditioning? The air cycle refrigeration is more popular in air craft air conditioning because of its lower equipment weight. This cycle utilizes the portion of the cabin air according to the supercharger’s capacity. 5.4. What are the merits electrolux refrigerator?

and

demerits

of

an

Merits 1.

No compressor or any reciprocating components are used.

2.

No leakage of the refrigerants.

QA.44 Thermal Engineering - I

3.

No mechanical troubles.

4.

No noise-smooth operation because of the absence of moving parts.

Demerits 1.

Once the system is down, we can not repair it. The complete system should be replaced.

2.

Charging of refrigerant is difficult.

5.5. How the second law of thermodynamics is satisfied in refrigerator and heat pump? S pace to be heated

A tm o sph ere T 1 H ot H eat flow ing d ire ction

Q1

R ef

H ot b ody

T1

W =Q 1-Q 2

H eat flow ing d ire ction

Q1

H eat p um p

Refrigerator Q2 T2 C old bod y S pa ce to be coo le d

W =Q 1-Q 2 Heat pu mp

Q2

L ow Tem p era tu re

T2 C old atm o sph ere

According to second law, heat cannot flow of itself from low temperature body to high temperature body. Some external work W net  must be expended to achieve this. The refrigerator and heat pump satisfy the second law. 5.6. Why throttle valve is used in place of expansion cylinder for vapour compression refrigeration machine? We want the refrigerants pressure to be lowered. At the same time, its enthalpy should be constant. To achieve

Short Questions and Answers QA.45

this, we use throttle valve. During throttling process, the pressure is reduced and the enthalpy remains constant. 5.7. What is undercooling? Sketch the process in TS diagram of refrigerator?

S up er he ated co ndition S aturated liq uid

S aturated va pour

In the condenser, the refrigerant is cooled (condensed) at constant pressure. If the refrigerant is cooled below its saturation temperature, then it is called subcooling or undercooling. The objective of subcooling is to increase the refrigerating effect. So subcooling increases the C.O.P without supplying extra work. 5.8. Why should the refrigerant temperature inside the evaporator be at a lower temperature than the refrigerator cabinet temperature? Because, the heat from the cabinet (at high temperature) will flow to evaporate the refrigerant (at low temperature). Hence the cabinet gets cooled.

QA.46 Thermal Engineering - I

5.9. In a refrigeration system, where do we have saturated and superheated conditions of refrigerant? D egree of sub cooling = (T 3 -T 3 ) T 2 O

T3

3

P =C 2

3

O

O

O

h 3=

s= C

T3

O

h4 P =C

1

4

s

According to given diagram,   

At the end of condensation (3), we have saturated liquid refrigerant. At the end of compression superheated vapour refrigerant.

(2),

we

have

Also, at the beginning of compression, we have saturated vapour refrigerant.

5.10. What is net refrigerating effect of refrigerant? Net refrigerating effect is the total heat removed from the refrigerant in the evaporator. Net refrigerating effect  C.O.P  work done where C.O.P = Coefficient of performance. 5.11. Name the refrigerant normally used in vapour absorption refrigeration system. Ammonia.

Short Questions and Answers QA.47

5.12. Name the various components used in simple vapour absorption system? 1. Absorber, 2. Generator, 4. Expansion valve, 5. Evaporator.

3. Condenser,

5.13. State the condition of the refrigerant at the end of compression in vapour compression system.  

Usually superheated vapour (or) Dry saturated vapour.

5.14. Name the different components used in vapour compression system. 1. Compressor, 4. Evaporator.

2. Condenser,

3. Throttling

device,

5.15. Mention important applications of cryogenics.    

It is used for upto  200C

storage

in

low

temperature.

It is used for fire extinguisher. It is used for oxy-acetelene welding, TIG and MIG welding. It is used for manufacturing fertilizers.

5.16. Define Refrigerant? Refrigerant is defined as the substance which absorbs heat through expansion or vapourization and loses it through condensation in the refrigeration system. 5.17. What type of condensers are in common use for vapour compression refrigeration system? Water cooled condensers. This type of condensers (heat exchanger) are used to provide heat transfer surface

QA.48 Thermal Engineering - I

through which heat passes from the hot refrigerant vapour to the condensing medium i.e. cooling water. 5.18. What part of absorption refrigerator is heated and what function does the heating perform? 



The ‘generator’ is heated by burner. The heat is added to drive NH 3 from the strong solution to condenser. The weak solution will be sent to (return back) absorber.

5.19. What is open air and dense air refrigeration system? Which is better? In an open air refrigeration system, the air is directly taken to the space to be cooled. Since atmospheric air is handled, volume of air handled is large. So size of the compressor and expander will be large. Also the moisture in the air form a frost of air and clog the passage. In a closed or dense air refrigeration, the air is not having direct contact with the space to be cooled. The air is taken through pipe line. Dense air refrigeration is better. 5.20. What is the effect of superheating of refrigerant in vapour compression refrigeration on the performance? Superheating increases the refrigerating effect. But superheating requires more work input. The increase in refrigerating effect is less than the increase in work input. Hence the overall effect of superheating is to reduce the C.O.P. Due to superheating, no moisture content is present in the refrigerant. So no corrosion in the machine parts.

Short Questions and Answers QA.49

Chapter - 6

Actual Cycles and Their Analysis 6.1. Write a note on air standard cycles. Air standard cycle is a cycle which uses air as the working medium and assumes that the working medium behaves as a perfect working substance. Air standard cycle also prevents all the heat losses that could occur in an engine and pictures it as an imaginary perfect engine. 6.2. What are the assumptions made in the working of an air standard cycle. 1.

The working substance should be a perfect gas with standard specific heats.

2.

The heat transfer should be simple and no chemical reactions should occur.

3.

It must be a reversible process.

4.

The heat losses are assumed to be zero.

5.

The working medium at the end of the cycle must be the same as it was at the beginning of the cycle.

6.3. What is meant by an actual cycle? The cycle which accounts to the every possible losses in real time environments may be referred to as an actual cycle. 6.4. What is the difference between fuel-air cycle and air standard cycle. If the losses due to variable specific heats due to varying temperatures are neglected from an air standard cycle, then it is called as Fuel-air cycle. In fuel air cycle,

QA.50 Thermal Engineering - I

the working medium is the mixture of air and fuel vapour or atomized liquid fuel. 6.5. What are the major losses occuring in a heat engine? (i)

Dissociation losses.

(ii)

Loss due to incomplete combustion.

(iii)

Time loss.

(iv)

Heat loss.

(v)

Loss due to temperature.

(vi)

variation

of

specific

heats

with

Loss due to exhaust blowdown.

(vii)

Loss due to rubbing friction.

(viii)

Loss due to blowby gases.

6.6. What are the factors which affect the efficiency due to time loss in a heat engine.      

Homogeneity of the air-fuel mixture. Flame speed. Configuration of the combustion chamber. Turbulence in the air-fuel mixture. Distance needed to travel by the flame. Air-fuel ratio.

Short Questions and Answers QA.51

6.7. How is the heat produced in an transferred to various components? Workdone piston

to

engine,

drive

the

Convection and conduction heat transfer to internal components Heat Generated

Heat transfer cooling systems

through

Heat lost at the exhaust stroke Heat lost due to unburnt fuel 6.8. What is meant by exhaust blowdown. Exhaust blowdown is the process in which some of the combustion gases escape through the exhaust port if the exhaust valve is opened before Bottom dead center. This is due to the self pressure created by the piston at the end of the combustion stroke and also it depends upon the compression ratio of the engine. 6.9. Explain the terms pumping work and pumping loss. The difference between the workdone in removing the exhaust gases, and the workdone in admitting fresh charge is called as Pumping work. Also, the loss occurring due to this pumping work is called as pumping loss. This loss is due to the difference of pressure between lowest inlet pressure and the highest exhaust pressure.

QA.52 Thermal Engineering - I

6.10. Define volumetric efficiency. It is defined as the ratio of volume of air intake during the suction stroke to the swept volume of the piston. It is also defined as the ratio of mass of charge sucked into the cylinder during suction stroke to the mass of air corresponding to the swept volume of engine at atmospheric pressure and temperature. 6.11. What are the factors which influence frictional forces acting on an engine? 

   

the

Surface finish of piston and cylinder Lubricant properties Engine power Heat dissipation Other auxillary engine components

6.12. What is meant by blowby?. Blowby may be defined as the phenomenon by which the air-fuel mixture reaches the crank case below the piston. At high pressures, the air-fuel mixture is susceptible to enter the crank-case through the cracks and crevices in the piston or the cylinder head. 6.13. What are the factors which could cause blowby losses in an engine? 

   

Improper sealing by the piston rings. Cracks developed in the cylinder walls. Oversized bore of the piston. Dents and other physical modifications of the pistons. Improper assembly of piston rings.

Short Questions and Answers QA.53

 

More end gaps in the piston rings. Late opening of exhaust valves.

6.14. What are the effects of blowby losses on the efficiency of an engine. 



Since a part of the air-fuel mixture escapes to the crankcase, the gas pressure on the piston reduces. Thus the mean effective pressure also reduces and this leads to decreased efficiency. A high blowby could cause a considerable pressure rise within the crankcase. This rise in pressure could create resistive force on the rotation of the crank, leading to reduced efficiency.

6.15. Define crankcase ventilation. Crankcase ventilation may be defined as the process by which the pressure developed inside the crankcase is reduced by external means.

Solved Question Papers SQP.1

Code No:R22033 R10

SET -1

II B.Tech II Semester Supplementary Examination Jan/Feb-2015 THERMAL ENGINEERING -I (Com. to ME, AME) Time: 3 hours

Maximum Marks: 75 Answer any Five Questions

All Questions carry Equal Marks 1. (a) Briefly explain major losses and differences in actual engine cycle and air-standard cycle. Refer Dr.S.R. Book Page No.6.2 (b) Explain following: (i) Exhaust blow down loss factor (ii) Loss due to gas exchange process. Refer Dr.S.R. Book Page No.6.8 2. (a) With a neat sketch explain the working of four stroke diesel engine. Refer Dr.S.R. Book Page No.1.6 (b) Explain briefly about important qualities of SI engine fuel. Refer Dr.S.R. Book Page No.1.81 3. (a) Explain about the mixture requirements at different loads and speeds for a IC engine. Refer Dr.S.R. Book Page No.1.40 (b) Explain the phenomenon of knock in SI engines. Refer Dr.S.R. Book Page No.2.10

SQP.2 Thermal Engineering - I

4. (a) Explain about the stages of combustion in CI engine. Refer Dr.S.R. Book Page No.2.29 (b) Explain different types of combustion for SI engine. Refer Dr.S.R. Book Page No.2.26 5. (a) Explain the procedure to estimate the friction power of a multi cylinder engine by using Morse test. Refer Dr.S.R. Book Page No.3.21 (b) Explain Retardation method to evaluate the friction power of an engine. Refer Dr.S.R. Book Page No.3.6 - Brake power methods 6. A three stage compressor is used to compress hydrogen from 1.04 bar to 35 bar. The compression in all stages follows

the

law

PV 1.3  C .

The

temperature

of

hydrogen at inlet is 288 K . Neglecting clearance and assuming perfect inter cooling, find (i) indicated power required to deliver 14 m 3 of hydrogen per minute measured at the inlet conditions. (ii) Intermediate pressure. Take R  4125 J/Kg K . Refer Dr.S.R. Book Page No. 4.84 - Problem 4.33 7. (a) Explain the construction, working of Vane type compressor. Refer Dr.S.R. Book Page No.4.117 (b) Explain about effect of impeller blade shape on performance of centrifugal compressor. Refer Dr.S.R. Book Page No.4.138

Solved Question Papers SQP.3

8. What is degree of reaction? Draw and explain the velocity of an axial flow Compressor when the degree of reaction is 0.5. Refer Dr.S.R. Book Page No.4.175 and Fig. 4.36 ...........................................................................

Code No:R22033 R10

SET -2

II B.Tech II Semester Supplementary Examination Jan/Feb-2015 THERMAL ENGINEERING -I (Com. to ME, AME) Time: 3 hours

Maximum Marks: 75 Answer any Five Questions

All Questions carry Equal Marks 1. (a) Compare the actual and fuel-air cycles of a gasoline engine. Refer Dr.S.R. Book Page No.6.19 (b) Explain with a neat diagram the typical fuel feed system of a CI engine. Refer Dr.S.R. Book Page No.1.53 2. (a) With a neat sketch explain the working of Wankel engine. Refer Dr.S.R. Book Supplementary (b) With a neat sketch explain the working of four stroke petrol engine. Refer Dr.S.R. Book Page No.1.4

SQP.4 Thermal Engineering - I

3. (a) Explain the factors influencing the flame speed in an SI engine. Refer Dr.S.R. Book Page No.2.5 (b) Explain the stages of combustion in SI engines. Refer Dr.S.R. Book Page No.2.1 4. (a) Explain about the factors that affect the delay period in CI engine. Refer Dr.S.R. Book Page No.2.7 (b) With a neat diagram explain about different direct ignition chambers. Refer Dr.S.R. Book Page No.2.42 5. (a) Explain the heat balance diagram of a typical CI engine. Refer Dr.S.R. Book Page No.3.52 (b) In a test of four cylinders, four stroke petrol engine of 75 mm bore and 100 mm stroke, the following results were obtained at full throttle at a constant-speed and with a fixed setting of the fuel supply of 0.082 kg /min . BP with all cylinder working  15.24 kW BP with Cylinder number 1 cut off  10.45 kW BP with Cylinder number 2 cut off  10.38 kW BP with Cylinder number 3 cut off  10.23 kW BP with Cylinder number 4 cut off  10.45 kW Estimate the indicated power of the engine under this condition. If the Calorific Value of the fuel is 44000 kJ/kg ,

Solved Question Papers SQP.5

find the indicated thermal efficiency of the engine. Compare this with the air standard efficiency. The clearance volume of one cylinder being 155 CC . Refer Dr.S.R. Book Page No.3.28 6. (a) Derive the equation for shaft work for single stage air compressor with clearance (i) When law of compression followed is isothermal. (ii) When law of compression followed is PV n  C Refer Dr.S.R. Book Page No.4.16 (b) A single acting compressor has zero clearance, stroke of 200 mm and piston diameter 150 mm . When the compressor is operating at 250 rpm and compressing air from 10 N/cm 2 , 25 C to 40 N /cm 2, Find:(i) The volume of air handled (ii) The ideal power required Refer Dr.S.R. Book Page No.4.26 7. (a) In two stage compressor, prove that the work done on 1 kg of air is minimum with perfect inter cooling when the intermediate pressure is the geometric mean P dP of the suction and delivery pressure or P i    s

Refer Dr.S.R. Book Page No.4.70, Section: 4.16.3 (b) Explain the construction, working of Roots blower and derive the expression for roots efficiency. Refer Dr.S.R. Book Page No.4.112 8. (a) Show that for axial flow compressors if the degree of reaction is 50% the compressors have symmetrical blades. Refer Dr.S.R. Book Page No.4.175

SQP.6 Thermal Engineering - I

(b) Explain about surging, chocking and stalling in a axial flow compressor. Refer Dr.S.R. Book Page No.4.184 ...........................................................................

Code No:R22033 R10

SET -3

II B.Tech II Semester Supplementary Examination Jan/Feb-2015 THERMAL ENGINEERING -I (Com. to ME, AME) Time: 3 hours

Maximum Marks: 75 Answer any Five Questions

All Questions carry Equal Marks 1. (a) What are the merits and demerits of two stroke IC engines over the four stroke IC Engines. Refer Dr.S.R. Book Page No.1.20 (b) Explain the following: (i) time loss factor (ii) heat loss factor Refer Dr.S.R. Book Page No.6.4 2. (a) Explain briefly about valve timing diagram of four stroke petrol engine. Refer Dr.S.R. Book Page No.1.26 (b) Explain briefly about important qualities of CI engine fuel. Refer Dr.S.R. Book Page No.2.50

Solved Question Papers SQP.7

3. (a) With a neat sketch explain the working of a simple carburetor. Refer Dr.S.R. Book Page No.1.38 (b) Explain about flame front propagation in a SI engine Refer Dr.S.R. Book Page No.2.4 4. (a) Explain how Rating of CI engine fuels is done. Refer Dr.S.R. Book Page No.2.52 (b) Compare the Knock in SI engine with CI engine. Refer Dr.S.R. Book Page No.2.38 5. (a) Explain the analytical method to evaluate the engine performance. Refer Dr.S.R. Book Page No.3.1 (b) A four stroke four cylinder gasoline engine has a bore of 60 mm and a stroke of 100 mm . On test it develops a torque of 66.5 Nm, when running at 3000 rpm . If the clearance volume in each cylinder is 60 cc the relative efficiency with respect to brake thermal efficiency is 0.5 and the calorific value of the fuel is 42 MJ/kg , determine the fuel consumption in kg/ hr and the brake mean effective pressure. Refer Dr.S.R. Book Page No.3.19 6. (a) Explain the factor that affect the efficiency of a reciprocating compressor. Refer Dr.S.R. Book Page No.4.20

volumetric

SQP.8 Thermal Engineering - I

(b) A three stage reciprocating air compressor compresses the air from 1 bar 17 to 35 bar. The law of compression is PV 1.25  C and is same for all the stages of compression. Assuming perfect intercooling and neglecting the clearance and valve resistance, find the minimum power required to compress 15 m 3/min of free air. Also find the intermediate pressure. Refer Dr.S.R. Book Page No.4.84 7. (a) With a neat diagram explain the working of a centrifugal compressor. Refer Dr.S.R. Book Page No.4.125 (b) With help of velocity diagram derive expression for work done for a centrifugal compressor. Refer Dr.S.R. Book Page No.4.140 8. (a) With a neat diagram explain the construction and working of an axial flow compressor. Refer Dr.S.R. Book Page No.4.166 (b) Make comparison centrifugal compressors.

between

reciprocating

and

Refer Dr.S.R. Book Page No.4.187 ...........................................................................

Solved Question Papers SQP.9

Code No:R22033 R10

SET -4

II B.Tech II Semester Supplementary Examination Jan/Feb-2015 THERMAL ENGINEERING -I (Com. to ME, AME) Time: 3 hours

Maximum Marks: 75 Answer any Five Questions

All Questions carry Equal Marks 1. (a) What are the merits and demerits of four stroke petrol engine over the four stroke Diesel Engines. Refer Dr.S.R. Book Page No.1.21 (b) Define volumetric efficiency and discuss about various factors affecting the volumetric efficiency. Refer Dr.S.R. Book Page No.6.10 2. (a) Explain briefly about port timing diagram of two stroke petrol engine. Refer Dr.S.R. Book Page No.1.29 (b) Explain the working of a simple carburetor with a neat sketch. Refer Dr.S.R. Book Page No.1.38 3. (a) Explain how Rating of SI engine fuels is done. Refer Dr.S.R. Book Page No.2.21 (b) Explain about the effect of engine variables on knock. Refer Dr.S.R. Book Page No.2.14

SQP.10 Thermal Engineering - I

4. Explain about the factor that affect the delay period in CI engine. Refer Dr.S.R. Book Page No.2.32 5. With a neat diagram explain the working of forced circulation cooling system. Refer Dr.S.R. Book Page No.1.76 6. (a) Derive the equation for shaft work for single stage air compressor with clearance. Refer Dr.S.R. Book Page No.4.16 (b) A two stage air compressor with complete intercooling delivers air to the mains at a pressure of 30 bar, the suction condition being 1 bar and 27 C . If both cylinders have same stroke, find the ratio of cylinder diameters, for the efficiency of compressor to be maximum. Assume the index of compression to be 1.3. Refer Dr.S.R. Book Page No.4.103 7. (a) A centrifugal compressor handles 150 kg/min of air. The suction pressure and temperature are 1 bar and 20 C . The suction velocity is 80 m/s. After compression in impeller the conditions are 1.5 bar and 70 C and 220 m/s . Determine (i) Isentropic efficiency. (ii) Power required to drive the compressor. Refer Dr.S.R. Book Page No.4.154 (b) With a neat sketch explain the working of axial flow compressor. Refer Dr.S.R. Book Page No.4.166

Solved Question Papers SQP.11

8. (a) With the help of velocity diagram derive the expression for work done in a stage of a axial flow compressor. Refer Dr.S.R. Book Page No.4.169 (b) Make comparison centrifugal compressors.

between

axial

flow

and

Refer Dr.S.R. Book Page No.4.189 ...........................................................................

Code No:R22033 R10

SET -1

II B.Tech II Semester Supplementary Examination August-2014 THERMAL ENGINEERING -I (Com. to ME, AME) Time: 3 hours

Maximum Marks: 75 Answer any Five Questions

All Questions carry Equal Marks 1. (a) Explain the reasons for the difference between the air-standard cycles and the actual cycles of Internal Combustion Engines. Refer Dr.S.R. Book Page No.6.2 (b) What is the main loss in a Diesel engine? By means of a P-V diagram indicate the actual and fuel air cycle for a two stroke Diesel engine. Refer Dr.S.R. Book Page No.6.19

SQP.12 Thermal Engineering - I

2. (a) Explain the operations of “High tension Magneto ignition system” with a sketch. Refer Dr.S.R. Book Page No.1.63 (b) Explain the property “Volatility”. Refer Dr.S.R. Book Page No.2.51 3. (a) Explain the phenomenon of knock in S.I. engine with sketches Refer Dr.S.R. Book Page No.2.10 (b) Describe the following combustion chambers of S.I. engine with line diagrams: (i) T-head type (ii) F-head type (ii) L-head type Refer Dr.S.R. Book Page No.2.26 4. (a) What is the ignition delay period in C.I. Engine combustion? Explain the physical and chemical delay periods. Refer Dr.S.R. Book Page No.2.28 (b) Table the important characteristics that tend to reduce knock in C.I. engines. Refer Dr.S.R. Book Page No.2.35 5. During the trail of a single cylinder 4-stroke oil engine, the following results were obtained: cylinder diameter 20 cm , stroke 40 cm , mep 6 bar, torque 407 Nm , speed 250 rpm , oil consumption 4 kg/h , calorific value 43 MJ/kg , cooling water flow rate 4.5 kg/min , air used

Solved Question Papers SQP.13

per kg of fuel 30 kg , rise in cooling water temperature 45 C , temperature of exhaust gases 420 C , room temperature 20 C , mean specific heat of exhaust gas 1 kJ/kg K. Find the indicated power, brake power and draw the heat balance sheet for the test in kJ/h . Refer Dr.S,R.Book Page No.3.59 6. A two stage air compressor with perfect intercooling takes in air at 1 bar pressure and 27 C . The law of compression in both the stages is PV 1.3  constant. The compressed air is delivered at 9 bar from the H.P. cylinder to an air receiver. Calculate per kg of air (i) the minimum work done, and (ii) The heat rejected to intercooler. Refer Dr.S.R. Book Page No.4.82 7. Explain the principle of working of centrifugal compressor with the help of h-s diagram. Refer Dr.S.R. Book Page No.4.127 8. A

multistage axial compressor is required for compressing air at 293 K through a pressor ratio of 5 to 1. Each stage is to be 50% reaction and the mean blade speed is 275 m/s , flow coefficient is 0.5 and stage loading factor is 0.3 are taken for simplicity as constant for all stages. Find the flow angles and the number of stages required if the stage efficiency is 88.8%. Refer Dr.S.R. Book Page No.4.200

...........................................................................

SQP.14 Thermal Engineering - I

Code No:R22033 R10

SET -2

II B.Tech II Semester Supplementary Examination August-2014 THERMAL ENGINEERING -I (Com. to ME, AME) Time: 3 hours

Maximum Marks: 75 Answer any Five Questions

All Questions carry Equal Marks 1. Discuss the effect of “the timing of the intake and exhaust valves” on the volumetric efficiency. Refer Dr.S.R. Book Page No.6.12 2. Explain the working of a single cylinder Jerk pump type fuel injection system with a sketch. Refer Dr.S.R. Book Page No.1.53 3. (a) Describe how the following parameters influence the flame speed in an S.I engine: (i) Turbulence (ii) Compression ratio (iii) Fuel-air ratio (iv) Engine output (v) Engine size. Refer Dr.S.R. Book Page No.2.5 (b) Define Octane number and explain how the S.I Engine fuels are rated. Refer Dr.S.R. Book Page No.2.21

Solved Question Papers SQP.15

4. Compare the phenomenon of knock in S.I and C.I Engines, by means of P diagrams. Refer Dr.S.R. Book Page No.2.38 5. A two stage single acting reciprocating air compressor draws in air at a pressure of 1 bar and 17  C and compresses it to a pressure of 60 bar. After compression in the L.P. cylinder, the air is cooled at constant pressure of 8 bar to a temperature of 37  C . The low pressure cylinder has a diameter 150 mm and both the cylinders have 200 mm stroke. If the law of compression is pv 1.35  C, find the power of the compressor, when it runs at 200 rpm . Take R  287 J/kg K . Refer Dr.S.R. Book Page No.4.80 6. Draw the velocity diagrams for a centrifugal compressor at impeller inlet, impeller outlet, diffuser inlet and outlet and explain the various terms. Refer Dr.S.R. Book Page No.4.140 7. In an axial flow compressor, the overall stagnation pressure ratio is 4 with an overall stagnation isentropic efficiency of 86%. The inlet stagnation temperature and pressure are 320 K and 1 bar. The mean blade speed is 190 m/s. The degree of reaction is 0.5 at the mean radius with relative air angles of 30 and 10 at rotor inlet and outlet respectively. The work done factor is 0.88. Calculate the stagnation polytropic efficiency, number of stages, inlet pressure and temperature. Refer Dr.S.R. Book Page No.4.196 ...........................................................................

SQP.16 Thermal Engineering - I

Code No:R22033 R10

SET -3

II B.Tech II Semester Supplementary Examination August-2014 THERMAL ENGINEERING -I (Com. to ME, AME) Time: 3 hours

Maximum Marks: 75 Answer any Five Questions

All Questions carry Equal Marks 1. (a) Explain the loss due to rubbing friction of an I.C.Engine. How will it vary with the engine load. Refer Dr.S.R. Book Page No.6.14 (b) With the help of a P-V diagram for an S.I.Engine, explain the consequences of the finite time of combustion. Refer Dr.S.R. Book Page No.6.4 2. (a) Sketch and explain the working principle of “pressure cooling system”. Refer Dr.S.R. Book Page No.1.76 (b) Explain the principle of working of Wankel engine. Refer Dr.S.R. Book Supplementary 3. (a) By means of P diagram, explain the stages of combustion in an S.I engine. Refer Dr.S.R. Book Page No.2.1 (b) Describe few anti-knock additives for S.I Engine. Refer Dr.S.R. Book Page No.2.23

Solved Question Papers SQP.17

4. Define Cetane number and explain how the C.I Engine fuels are rated. Refer Dr.S.R. Book Page No.2.52 5. In a test of oil engine, under full load, the following results were obtained: ip 33 kW , brake power 27 kW , oil consumption 8 kg/h , calorific value 43 MJ/kg , cooling water flow rate 7 kg/min , rate of flow of water through gas calorimeter 12 kg/h, rise in cooling water temperature 60 C , final temperature of exhaust gases 80 C , room temperature 17 C , air-fuel ratio on mass

basis 20, rise in water temperature through exhaust gas calorimeter 40 C , mean specific heat of exhaust gas 1 kJ/kg K . Draw the heat balance sheet and find thermal and mechanical efficiencies. Refer Dr.S.R. Book Page No.3.61 6. A two-stage single acting reciprocating compressor takes in air at the rate of 0.2 m 3/s. The intake pressure and temperature of air are 0.1 MPa and 16 C . The air is compressed to a final pressure of 0.7 MPa . The intermediate pressure is ideal and intercooling is perfect. The compression index in both the stages is 1.25 and the compressor runs at 600 r.p.m . Neglecting clearance, determine (i) The intermediate pressure, (ii) The total volume of each cylinder, (iii) the power required to drive the compressor, and (iv) the rate of heat rejection in the C p  1.005 kJ/kg K and intercooler. Take R  287 J/kg K .

Refer Dr.S.R. Book Page No.4.78

SQP.18 Thermal Engineering - I

7. (a) Explain the significance of the following dimensionless parameters of centrifugal compressor: (i) Flow coefficient (ii) Head coefficient (iii) Pressure coefficient. Refer Dr.S.R. Book Page No.4.149 (b) What are rotary compressors? Where do vane type compressors find application? Refer Dr.S.R. Book Page No.4.112 8. A multi stage axial flow compressor delivers 18 kg/s of air. The inlet stagnation condition is 1 bar and 20 C . The power consumed by the compressor is 4260 kW . Calculate: (i) delivery pressure (ii) number of stages and (iii) overall isentropic efficiency of the compressor. Assume temperature rise in the first stage is 18  C , the polytropic efficiency of compression is 0.9 and the stage stagnation pressure ratio is constant. Refer Dr.S.R. Book Page No.4.203 ...........................................................................

Solved Question Papers SQP.19

Code No: R22033 II B.Tech II Semester Regular Examinations, August - 2014 (Set - 4) THERMAL ENGINEERING - I (Com. to ME, AME) Time: 3 hours

Max.Marks:75

Answer any FIVE Questions All Questions carry Equal Marks 1. Define volumetric efficiency and name the variables that affect the volumetric efficiency. Refer Dr. S.R Book Pg.No.6.10 2. (a) Explain the main metering and idling system of a carburetor for cruising and full throttle operations. Refer Dr. S.R Book Pg.No.1.40 (b) Sketch and explain the working of a gear type lubricating pump. Refer Dr. S.R Book Pg.No.1.70 3. (a) Describe the following two general objectives of combustion chamber design for S.I.Engine: (i) Smooth Engine operation, (ii) High power output & thermal efficiency. (b) Explain the factors influencing flame speed in combustion of S.I.engines. Refer Dr. S.R Book Pg.No.2.5 4. (a) By means of P- diagram, explain the stages of combustion in a C.I.Engine. Refer Dr. S.R Book Pg.No.2.29

SQP.20 Thermal Engineering - I

(b) Differentiate between combustion induced swirl.

compression

and

Refer Dr. S.R Book Pg.No.2.40 5. A four stroke cycle gas engine has a bore of 20 cm and a stroke of 40 cm. The compression ratio is 6. In a test on the engine the imep is 5 bar, the air to gas ratio is 6:1 and the calorific value of the gas is 12 MJ /m 3

at NTP. At the beginning of the compression stroke, the temperature is 77 C and pressure 0.98 bar. Neglecting residual gases, find the indicated power and thermal efficiency of the engine at 250 rpm. Refer Dr. S.R Book Pg.No.3.76 6. An air compressor takes in air at 0.98 bar and 20 C and compresses it according to the law pv 1.2  C . It is then delivered to a receiver at constant pressure of 9.8 bar. Determine: (i) the temperature at the end of compression;(ii) the work done per kg of air(iii) the heat transferred during the compression; and (iv) the work during delivery. Take R  287 J/kgK and   1.4 Refer Dr. S.R Book Pg.No.4.36 7. (a) What is slip factor of centrifugal compressor? With the help of a velocity triangle diagram, show the slip on the tip of radial bladed centrifugal impeller. Refer Dr. S.R Book Pg.No.4.147 (b) Derive an expression for the work supplied in a stage of a centrifugal compressor Refer Dr. S.R Book Pg.No.4.143

Solved Question Papers SQP.21

8. In an 8 stage axial flow compressor, the overall stagnation pressure ratio is 5:1 with an overall isentropic efficiency of 90%. The stagnation temperature and pressure at inlet are 20 C and 1 bar. The work is divided equally between the stages. The mean blade shape is 175 m/s and 50% reaction design is used. The axial velocity through the compressor is constant and equal to 100 m/s. Calculate the power required and blade angles. Refer Dr. S.R Book Pg.No.4.194 ...........................................................................

Code No: R22033 II B.Tech II Semester Regular Examinations, January - 2014 (Set - 1) THERMAL ENGINEERING - I (Com. to ME, AME) Time: 3 hours

Max.Marks:75

Answer any FIVE Questions All Questions carry Equal Marks 1. (a) Compare the actual and fuel air cycles of a gasoline engine. Refer Dr. S.R Book Pg.No.6.19 (b) Explain the ideal and actual port timing diagrams of a 2-stroke S.I.engine Refer Dr. S.R Book Pg.No.1.29 2. (a) With neat sketches explain the working principle of simple carburetor. Refer Dr. S.R Book Pg.No.1.38

SQP.22 Thermal Engineering - I

(b) Discuss the important qualities of an S.I and C.I engine fuel. Refer Dr. S.R Book Pg.No.2.18 3. (a) Describe the phenomenon of pre-ignition in S.I. engines and discuss its effect on the performance. Refer Dr. S.R Book Pg.No.2.8 (b) What are the various types of combustion chambers used in S.I. engines? Explain them briefly. Refer Dr. S.R Book Pg.No.2.26 4. (a) Explain the various factors that influence the delay period in a C.I.engine. Refer Dr. S.R Book Pg.No.2.29 (b) Explain the stages of combustion in C.I.engine. Refer Dr. S.R Book Pg.No.2.28 5. The following data were recorded during a test on a 4-stroke cycle gas engine. Area of indicator diagram  90 cm 2 Length of indicator diagram = 7 cm Spring scale = 0.3 bar/mm Diameter of piston = 20 cm Length of stroke = 25 cm Speed = 300 rpm Determine (i) Indicated mean effective pressure(ii) Indicated power Refer Dr. S.R Book Pg.No.3.75

Solved Question Papers SQP.23

6. Explain the following as referred to air compressors. (i) Isothermal efficiency(ii) Volumetric efficiency Refer Dr. S.R Book Pg.No.4.18 Prove the following relationship    P1    vol  1  k   P   1     2 

Refer Dr. S.R Book Pg.No. 4.19 7. (a) What do you mean by multistage compression? State its advantages and disadvantages. Refer Dr. S.R Book Pg.No.4.5 (b) A single acting two stage compressor with complete intercooling delivers 10.5 Kg/min of air at 16 bar. The suction occurs at 1 bar and 25C . The compression and expansion process are reversible, polytrophic index = 1.3. Calculate (i) Power required to drive the compressor(ii) Isothermal efficiency (iii) Volumetric efficiency of both the compressors; If the clearance ratios for LP and HP cylinder are 0.04 and 0.06 respectively. Refer Dr. S.R Book Pg.No.4.91 8. (a) Describe with a neat sketch the working of roots blower. Refer Dr. S.R Book Pg.No.4.112 (b) A Roots blower compresses 0.08 m 3 of air from 1.0 bar to 1.5 bar per revolution, calculate compressor efficiency. Refer Dr. S.R Book Pg.No.4.115 ...........................................................................

SQP.24 Thermal Engineering - I

Code No: R22033 II B.Tech II Semester Regular Examinations, January - 2014 (Set - 2) THERMAL ENGINEERING - I (Com. to ME, AME) Time: 3 hours

Max.Marks:75

Answer any FIVE Questions All Questions carry Equal Marks 1. (a) Draw the neat sketch of fuel pump for C.I.engine and explain. Refer Dr. S.R Book Pg.No.1.53 (b) Explain ideal and actual port timing diagrams of 2-stroke S.I.engine. Refer Dr. S.R Book Pg.No.1.29 2. (a) What is the difference between air standard cycle and fuel air cycle analysis? Explain significance of fuel air cycle. Refer Dr. S.R Book Pg.No.6.2 (b) What is meant by crank case ventilation? Explain the details. Refer Dr. S.R Book Pg.No.6.17 3. (a) Briefly explain the stages of combustion in S.I.engines elaborating the flame front propagation. Refer Dr. S.R Book Pg.No.2.1 (b) Explain with a neat sketch the battery ignition system Refer Dr. S.R Book Pg.No.1.60

Solved Question Papers SQP.25

4. (a) Explain the phenomenon of knock in C.I.engines. Refer Dr. S.R Book Pg.No.2.36 (b) Explain the working of Ricardo swirl chamber. Refer Dr. S.R Book Pg.No.2.41 5. The following data was recorded during testing of a 4-stroke cycle gas engine. Diameter = 10 cm ; Stroke = 15 cm; Speed = 1600 rpm Area of the positive loop of the indicator diagram  5.75 cm2 ;

Area of the negative loop of the indicator

diagram  0.25 cm2 ; Length of the indicator diagram = 55mm; Spring constant = 3.5 bar/cm; Find the indicated power of the engine. Refer Dr. S.R Book Pg.No.3.75 6. (a) Compare centrifugal and axial flow compressors. Refer Dr. S.R Book Pg.No.4.189 (b) A roots blower compresses 0.06 m 3 of air from 1.0 bar to 1.45 bar per revolution. Calculate compressor efficiency. Refer Dr. S.R Book Pg.No.4.116 7. (a) Explain the working of an axial flow compressor with a neat sketch. Refer Dr. S.R Book Pg.No.4.166 (b) An 8 stage axial flow compressor takes in air at 20 C at the rate of 180 kg/min. The pressure ratio is 6 and isentropic efficiency is 0.9. Determine the power required. Refer Dr. S.R Book Pg.No.4.205

SQP.26 Thermal Engineering - I

8. (a) Explain with a neat sketch the working of a centrifugal compressor. Refer Dr. S.R Book Pg.No.4.125 (b) A centrifugal air compressor having a pressure ratio of 5, compresses air at the rate of 10 kg/s. If the initial pressure and temperature of air is 1 bar and 20 C , calculate final temperature and power required to drive the compressor. Refer Dr. S.R Book Pg.No.4.165 ...........................................................................

Code No: R22033 II B.Tech II Semester Regular Examinations, January - 2014 (Set - 3) THERMAL ENGINEERING - I (Com. to ME, AME) Time: 3 hours

Max.Marks:75

Answer any FIVE Questions All Questions carry Equal Marks 1. Define volumetric efficiency and discuss the effect of various factors effecting the volumetric efficiency. Refer Dr. S.R Book Pg.No.6.10 2. (a) Explain pressure cooling system with a sketch. Refer Dr. S.R Book Pg.No.1.76 (b) With neat sketch, explain the working of simple carburetor. Refer Dr. S.R Book Pg.No.1.38

Solved Question Papers SQP.27

3.(a) Explain the various factors that influence the flame speed of an S.I.engine. Refer Dr. S.R Book Pg.No.2.5 (b) What is delay period and what are the factors affecting the delay period? Refer Dr. S.R Book Pg.No.2.29 4. (a) What are the different methods used in C.I.engines to create turbulence in the mixture. Refer Dr. S.R Book Pg.No.2.42 (b) Discuss the advantages and disadvantages of the two types of combustion chambers of C.I.engines. Refer Dr. S.R Book Pg.No.2.49 5. (a) What is the purpose of engine testing? Name the various measurements which are to be taken in a test of I.C.engine. Refer Dr. S.R Book Pg.No.3.6 (b) Find the engine dimensions of a 2-cylinder, 2-stroke, I.C.engine from the following data: Engine speed = 4000 rpm, Volume efficiency = 0.77, Mechanical efficiency = 0.75, Fuel consumption 10 Lit/hr (g = 0.73), Air fuel ratio = 18:1, Piston speed = 600 rpm, Indicated mean effective pressure = 5bar. Find also brake power, Take R of gas mixture as 281 J/kg K at S.T.P. Refer Dr. S.R Book Pg.No.3.90 Problem 3.23

SQP.28 Thermal Engineering - I

6. (a) Enumerate the applications of compressed air. State how reciprocating air compressors classified. Refer Dr. S.R Book Pg.No.4.1 (b) Prove that the volumetric efficiency of a single stage compressor is given by   P1    vol  1  k   P   2 

1/n

 1 

Refer Dr. S.R Book Pg.No.4.18 7. (a) Derive an expression for efficiency of Roots blower in terms of pressure ratio and ratio of specific heats. Refer Dr. S.R Book Pg.No.4.113 (b) A rotary compressor receives air at 1 bar and 17  C and delivers it at a pressure of 6 bar. Determine per kg of air delivered, work done by the compressor and heat exchanges with the jacket water when the compression is isothermal, isentropic by the relation PV constant. Refer Dr. S.R Book Pg.No.4.123 8. (a) Differentiate between centrifugal compressor and axial flow compressor. Refer Dr. S.R Book Pg.No.4.189 (b) An axial flow compressor, with compression ratio as 4, draws air at 20  C and delivers it at 197  C . The mean blade speed and flow velocity are constant throughout the compressor. Assuming 50% reaction blading and taking blade velocity as 180 m/s, find flow velocity and the number of stages. Take work factor = 0.82,   12,   42 and C p  1.005 kJ/kgK.

Refer Dr. S.R Book Pg.No.4.209 ...........................................................................

Solved Question Papers SQP.29

Code No: R22033 II B.Tech II Semester Regular Examinations, January - 2014 (Set - 4) THERMAL ENGINEERING - I (Com. to ME, AME) Time: 3 hours

Max.Marks:75

Answer any FIVE Questions All Questions carry Equal Marks 1. Explain exhaust blow down factor Refer Dr. S.R Book Pg.No.6.8 2. (a) Mention the various parameters which affect the engine heat transfer and explain their effect. Refer Dr. S.R Book Pg.No.6.6 (b) Clearly explain wet sump lubrication system with a sketch. Refer Dr. S.R Book Pg.No.1.70 3. (a) What is meant by abnormal combustion? Explain the phenomenon of knock in S.I.engine. Refer Dr. S.R Book Pg.No.2.10 (b) What is ignition lag in S.I.engine and how does it affect performance. Refer Dr. S.R Book Pg.No.2.2 4. (a) Explain with figure various types of combustion chambers used in C.I.engine. Refer Dr. S.R Book Pg.No.2.42

SQP.30 Thermal Engineering - I

(b) Bring out clearly the process of combustion C.I.engine and also explain various stages of combustion. Refer Dr. S.R Book Pg.No.2.28 5. The following data refer to an oil engine working on Otto 4-stroke cycle Brake pressure = 14.7 kW, Suction pressure = 0.9 bar, Mechanical efficiency = 80%, Index of compression curve = 1.35, Index of expansion curve = 1.3, Maximum explosion pressure = 24 bar, Engine speed = 1000 rpm, Compression ratio = 5 Ratio of stroke bore = 1.5. Find the diameter and stroke of the piston. Refer Dr. S.R Book Pg.No.3.78 6. (a) Explain the effect of intercooling in a multi stage reciprocating compressor. Refer Dr. S.R Book Pg.No.4.67 (b) Determine the size of the cylinder for a double acting air compressor of 40 kW indicated power, in which air is drawn at 1 bar and 15 C and compressed according to the law PV 1.2  constant to 6 bar. The compressor runs at 100 rpm with average piston speed of 152.5 m/min. Neglect clearance. Refer Dr. S.R Book Pg.No.4.42 7. (a) What is a slip factor and pressure coefficient? Refer Dr. S.R Book Pg.No.4.147

Solved Question Papers SQP.31

(b) The following data relate to a performance test of single acting 14 cm  10 cm reciprocating compressor. Suction pressure = 1 bar, Suction temperature  20 C , Discharge pressure = 6 bar, Discharge temperature  180 C , Speed of compressor = 1200 rpm, Shaft power = 6.25 kW, Mass of air delivered = 1.7 kg/min Calculate the following (i) the actual volumetric efficiency (ii) Indicated power (iii) Isentropic efficiency (iv) Mechanical efficiency (v) Overall isothermal efficiency Refer Dr. S.R Book Pg.No.4.38 8. (a) What do you mean by surging & choking? Refer Dr. S.R Book Pg.No.4.184 (b) An axial flow compressor with an overall isentropic efficiency of 85% draws air at 20 C and compresses it in the pressure ratio of 4:1. The mean blade speed and flow velocity are constant throughout the compressor. Assuming 50% reaction blading and taking blade velocity as 180 m/s, and work input factor as 0.82. Calculate: (i) Flow velocity and (ii) The number of stages. Take 1  12, 1  42  and Cp  1.005 kJ/kgK . Refer Dr. S.R Book Pg.No.4.192 ...........................................................................

SQP.32 Thermal Engineering - I

Code No: R22033 - JNTU-K II B.Tech II Semester Regular Examinations, April/May - 2013 (Set - 1) THERMAL ENGINEERING - I (Com. to ME, AME) Time: 3 hours

Max.Marks:75

Answer any FIVE Questions All Questions carry Equal Marks 1. (a) Illustrate the constructional details of an I.C.engines? Explain briefly about the important components and its materials? Refer Dr. S.R Book Pg.No.1.13 (b) Discuss briefly the loss due to gas exchange process? Refer Dr. S.R Book Pg.No.6.9 2. (a) How are S.I. and C.I. engine fuels rated ? Refer Dr. S.R Book Pg.No.2.18 (b) With a neat sketch explain the magneto ignition system? Refer Dr. S.R Book Pg.No.1.63 3. What is the difference between physical delay and chemical delay? Explain its importance. Refer Dr. S.R Book Pg.No.2.31 4. (a) Explain abnormal combustion and what is delay period? Refer Dr. S.R Book Pg.No.2.8

Solved Question Papers SQP.33

(b) Explain open type and divide type combustion chamber with neat sketch. Refer Dr. S.R Book Pg.No.2.42 5. In a test of four-cylinder, four-stroke petrol engine 75 mm bore and 100 mm stroke, the following results were obtained at full throttle at a constant speed and with a fixed setting of the fuel supply of 0.082 kg/min. bp with all cylinders working = 15.24 kW bp with cylinder number 1 cut-off = 10.45 kW bp with cylinder number 2 cut-off = 10.38 kW bp with cylinder number 3 cut-off = 10.23 kW bp with cylinder number 4 cut-off = 10.45 kW Estimate the indicated power of the engine under these conditions. If the calorific value of the fuel is 44 MJ/kg, find the indicated thermal efficiency of the engine. Compare this with the air-standard efficiency, the clearance volume of one cylinder being 115 cc. Refer Dr. S.R Book Pg.No.3.28 6. (a) Derive the equation for work required for a single stage reciprocating air compressor. Refer Dr. S.R Book Pg.No.4.7 (b) A single stage reciprocating air compressor is required to compress 60 m 3 of air from 1 bar to 8 bar at the workdone by the compressor if the 22C . Find compressor is (i) Isothermal (ii) adiabatic (iii) Polytropic with an index of 1.25. Refer Dr. S.R Book Pg.No.4.12

SQP.34 Thermal Engineering - I

7. (a) Explain the working principle of Roots blower with suitable diagrams. Refer Dr. S.R Book Pg.No.4.112 (b) A centrifugal compressor delivers 54 kg of air per minute at pressure of 200 kPa, when compressing air from 100 kPa and 15 C . If the temperature of air delivered is 97 C , and no heat is added to the air from the external sources during compression, determine the efficiency of the compressor relative to ideal adiabatic compression and power absorbed. Refer Dr. S.R Book Pg.No.4.162 8. (a) Define the degree of reaction and derive its equation for the symmetrical blades of an axial flow air compressor. Refer Dr. S.R Book Pg.No.4.175 (b) Show that the degree of reaction is 50% for the symmetrical blade axial flow air compressor. Refer Dr. S.R Book Pg.No.4.175 ...........................................................................

Code No: R22033 II B.Tech II Semester Regular Examinations, April/May - 2013 (Set - 2) THERMAL ENGINEERING - I (Com. to ME, AME) Time: 3 hours

Max.Marks:75

Answer any FIVE Questions All Questions carry Equal Marks 1. (a) What are the major differences between S.I.engine and C.I.engine? Explain them with suitable examples. Refer Dr. S.R Book Pg.No.1.21

Solved Question Papers SQP.35

(b) Briefly discuss pumping and rubbing friction losses. Refer Dr. S.R Book Pg.No.6.9, 6.14 2. (a) What is meant by crankcase ventilation? Explain Refer Dr. S.R Book Pg.No. 6.17 (b) Describe the essential parts of a modern carburetor. Refer Dr. S.R Book Pg.No.1.38 3. (a) What is ignition delay in combustion of S.I.Engine? What are different parameters influencing the ignition delay? Refer Dr. S.R Book Pg.No.2.1 (b) Explain the working of fuel injector with a neat sketch Refer Dr. S.R Book Pg.No.1.56 4. (a) Explain detail what is diesel knock Refer Dr. S.R Book Pg.No.2.35 (b) Explain combustion process in C.I.Engine? Refer Dr. S.R Book Pg.No.2.28 5. In a test on two stroke oil engine, the following results were obtained: speed = 350 rev/min;Net brake load = 600 N; Mean effective pressure = 2.66 bar; Fuel consumption = 3.2 kg/h; cooling water used = 495 kg/h; Temperatures of jacket water at inlet and outlet  30C and 50C ; Exhaust gases per kg of fuel = 32 kg; Temperature of exhaust gases  432 C ; specific heat of exhaust gases = 1.005 kJ/kg K; Inlet air temperature  32 C . Draw up a heat balance for the engine if its cylinder diameter = 205 mm and stroke

SQP.36 Thermal Engineering - I

= 275 mm; brake drum diameter = 1.0 m; calorific value of fuel = 40870 kJ/kg. Refer Dr. S.R Book Pg.No.3.61 6. (a) Define the volumetric efficiency and derive its expression for the single stage reciprocating air compressor with clearance volume Refer Dr. S.R Book Pg.No.4.16 (b) A single stage single acting reciprocating compressor with 0.3 m bore and 0.4 m stroke runs at rpm. The suction pressure is 1 bar at 300 K and delivery pressure is 5 bar. Find the power required to it, if the compression is isothermal, adiabatic,

air 400 the run and

compression pv 1.3  C . Also find the isothermal efficiency. Refer Dr. S.R Book Pg.No.4.14 7. (a) Draw the velocity triangles for the centrifugal compressor and derive the equation for the estimation of power required to compress the air. Refer Dr. S.R Book Pg.No.4.140 (b) Define the term slip factor and power input factor with respect to the centrifugal compressor. Explain them. Refer Dr. S.R Book Pg.No.4.147 8. (a) With a suitable sketch and velocity diagrams, explain the working principle of simple axial flow air compressor. Refer Dr. S.R Book Pg.No.4.166 (b) What is meant by a stage of axial flow air compressor? and explain in detail about the stage velocity triangles. Refer Dr. S.R Book Pg.No.4.169 ...........................................................................

Solved Question Papers SQP.37

Code No: R22033 II B.Tech II Semester Regular Examinations, April/May - 2013 (Set - 3) THERMAL ENGINEERING - I (Com. to ME, AME) Time: 3 hours

Max.Marks:75

Answer any FIVE Questions All Questions carry Equal Marks 1. (a) Explain the working principle of four stroke I.C.Engine along with the valve timing diagram. Refer Dr. S.R Book Pg.No.1.6, 1.26 (b) Briefly explain (i) Time loss factor (ii) Heat loss factor Refer Dr. S.R Book Pg.No.6.4, 6.6 2. Explain the principle of Wankle Engine Refer Dr. S.R Book Supplementary 3. Describe the phenomenon of knocking in C.I.Engine and how it is different from S.I.engine detonation. Refer Dr. S.R Book Pg.No.2.35, 2.38 4. (a) Explain detail what is detonation? Refer Dr. S.R Book Pg.No.2.10 (b) Explain flame front propagation with suitable sketch. Refer Dr. S.R Book Pg.No.2.4 5. A four stroke petrol engine with a compression ratio of 6.5 to 1 and total piston displacement of

SQP.38 Thermal Engineering - I

5.2  10  3 m 3 develops 100 kW brake power and

consumes 33 kg of petrol per hour of calorific value 44300 kJ/kg at 3000 rpm. Find: (i) Brake mean effective pressure (ii) Brake thermal efficiency (iii) Air standard efficiency   1.4; and (iv) Air-fuel ratio by mass. Assume a volumetric efficiency of 80%. One kg of petrol vapour occupies 0.26 m 3 at 1.013 bar and 15 C . Take R for air 287 J/kg K. Refer Dr. S.R Book Pg.No.3.86 6. (a) Derive an expression for the optimum intercooler pressure for two stage reciprocating air compressors with perfect intercooling Refer Dr. S.R Book Pg.No.4.69 (b) A single stage single acting reciprocating compressor with 0.3 m bore and 0.4 m stroke runs at rpm. The suction pressure is 1 bar at 300 K and delivery pressure is 5 bar. Find the power required to it, if the compression is isothermal, adiabatic

air 400 the run and

compression follow pv 1.3  c. Also find the isothermal efficiency. Refer Dr. S.R Book Pg.No.4.14 7. (a) Derive an expression for the efficiency of roots blower in terms of pressure ratio and ratio of specific heats based on p-v and T-s diagrams. Refer Dr. S.R Book Pg.No.4.113

Solved Question Papers SQP.39

(b) A rotary air compressor receives air at a pressure of 1 bar and 17 C and delivers at a pressure of 6 bar. Determine work done by the compressor per kg of air delivered, if the process is (i) isothermal, (ii) adiabatic and (iii) polytrophic with the index as 1.3. Refer Dr. S.R Book Pg.No.4.123 8. (a) Draw the schematic diagram of axial flow air compressor and explain its working along with velocity triangles. Refer Dr. S.R Book Pg.No.4.166 (b) Derive the work input requirement for an axial flow air compressor and explain the salient points. Refer Dr. S.R Book Pg.No.4.171 ...........................................................................

Code No: R22033 II B.Tech II Semester Regular Examinations, April/May - 2013 (Set - 4) THERMAL ENGINEERING - I (Com. to ME, AME) Time: 3 hours

Max.Marks:75

Answer any FIVE Questions All Questions carry Equal Marks 1. (a) Differentiate between Magneto ignition system with battery coil ignition system. Refer Dr. S.R Book Pg.No.1.65

SQP.40 Thermal Engineering - I

(b) Why lubrication is necessary in I.C.Engine components? Explain different methods of lubrication system. Refer Dr. S.R Book Pg.No.1.69 2. (a) Explain the fuel supply system in S.I.Engine Refer Dr. S.R Book Pg.No.1.34 (b) Explain the working of Zenith carburetor with neat sketch. Refer Dr. S.R Book Supplementary 3. (a) What are different stages of combustion S.I.Engine? Explain with p   diagram.

in

Refer Dr. S.R Book Pg.No.2.1 (b) Explain the influence of different operating parameters on ignition delay during combustion process in S.I.Engine. Refer Dr. S.R Book Pg.No.2.3 4. (a) Explain need of air movement combustion induced turbulence in a C.I.engine with a neat sketch. Refer Dr. S.R Book Pg.No.2.42 (b) Explain, what are the reasons for abnormal combustion of C.I engine. Refer Dr. S.R Book Pg.No.2.8 5. A nine-cylinder petrol engine of bore 150 mm and stroke 200 mm has a compression ratio 6:1 and develops 360 kW at 2000 rpm when running on a mixture of 20% rich. The fuel used has a calorific value of 43 MJ/kg and contains 85.3% carbon and 14.7% hydrogen.

Solved Question Papers SQP.41

Assuming volumetric efficiency of 70% at 17C and mechanical efficiency of 90%, find the indicated thermal efficiency of the engine. Refer Dr. S.R Book Pg.No.3.70 6. (a) Derive the expressions for the reversible work of compression if the compression processes are (i) adiabatic, (ii) polytrophic and (iii) isothermal. Refer Dr. S.R Book Pg.No.4.5 (b) Differentiate between positive compressors and dynamic compressors.

displacement

Refer Dr. S.R Book Pg.No.4.187 7. (a) Explain the working principle of Vane sealed compressor. Refer Dr. S.R Book Pg.No.4.117 (b) What are different parameters that influence the performance of the centrifugal compressors? Explain. Refer Dr. S.R Book Pg.No.4.128 8. Define the degree of reaction and derive its equation for the symmetrical blades of an axial flow air compressor. Refer Dr. S.R Book Pg.No.4.175 ...........................................................................

SQP.42 Thermal Engineering - I

Code No: R09220303 JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY, HYDERABAD II B.Tech II Semester Regular Examinations, May - 2013 Applied Thermodynamics - I (Common to ME, AME) Time: 3 hours

Max.Marks:75

Answer any FIVE Questions All Questions carry Equal Marks 1. Discuss the optimum opening position of exhaust valve to minimize the heat loss from the combustion chamber. Explain the conditions involved in it. ...(15) Refer Dr. S.R Book Pg.No.6.6 2. What are different antiknock additives to be added to reduce detonation in S.I.Engine? and Explain ill effects on engine performance. Refer Dr. S.R Book Pg.No.2.23 3. Differentiate between overhead valve and under head valve combustion chambers used S.I.Engine. Explain with the help of line diagrams. ... ([15]) Refer Dr. S.R Book Pg.No.2.26 4. (a) What is ignition delay in combustion of C.I.Engine? How to differentiate between physical and chemical delay? Explain their influence on combustion performance. Refer Dr. S.R Book Pg.No.2.29

Solved Question Papers SQP.43

(b) How to create turbulence in C.I.Engine combustion chamber in order to get better mixing air fuel? Explain. ... ([15]) Refer Dr. S.R Book Pg.No.2.40 5. (a) Derive an expression for the calculation of indicated mean effective pressure. Refer Dr. S.R Book Pg.No.3.2 (b) A four-stroke cycle automobile engine is tested while running at 3600 rpm. Inlet air temperature is 15C and the pressure is 101.36 kN/m 2. The engine has eight in-line cylinders with a total piston displacement of 4000 cc. The air fuel ratio is 15 and the bsfc is 0.3 kg/kWh . Dynamometer readings show a power output of 89 kW. Find the volumetric efficiency. ... ([15]) Refer Dr. S.R Book Pg.No.3.84 6. (a) Derive the equation for maximum discharge in case of multi stage compression of reciprocating air compressor with perfect inter cooler. Refer Dr. S.R Book Pg.No.4.67 (b) A single acting two stage reciprocating air compressor with complete inter cooling delivers 8 kg/min at 16 bar pressure. Assume an intake condition of 1 bar and 15C and that the compression and expansion processes are polytropic with 1.35. Calculate (i) power required, (ii) the iso thermal efficiency. ... ([15]) Refer Dr. S.R Book Pg.No.4.77 7. (a) What are the required components for Vane type compressor? Explain them. Refer Dr. S.R Book Pg.No.4.117

SQP.44 Thermal Engineering - I

(b) A single sided centrifugal compressor is to deliver 18 kg/s of air when operating at a stagnation pressure ratio 4:1 and a speed of 220 rps. The inlet stagnation conditions may be taken as 288 K and 1.0 bar. Assuming the slip factor of 0.92 and as a power input factor of 1.04 and an overall isentropic efficiency of 0.84. Estimate the overall diameter of the impeller. ... ([15]) Refer Dr. S.R Book Pg.No.4.163 8. (a) Derive the work input requirement for an axial flow air compressor and explain the salient points. Refer Dr. S.R Book Pg.No.4.166 (b) What is meant by workdone factor? Derive its equation for the axial flow compressor. ... ([15]) Refer Dr. S.R Book Pg.No.4.180 ...........................................................................

Code No: 9A03402 B.Tech II Year II Semester (R09) Regular & Supplementary Examinations, April/May - 2013 (Set -1) THERMAL ENGINEERING - I (Mechanical Engineering) Time: 3 hours

Max.Marks:75

Answer any FIVE Questions All Questions carry Equal Marks 1. (a) Explain in detail the effect of exhaust valve opening time on blow down in case of gasoline engine. Refer Dr. S.R Book Pg.No.6.8

Solved Question Papers SQP.45

(b) Discuss briefly the loss due to gas exchange process in gasoline engine. Refer Dr. S.R Book Pg.No.6.9 2. (a) Briefly explain any two methods of fuel injection systems generally employed in C.I.engines. Refer Dr. S.R Book Pg.No.1.53 (b) Compare Otto and diesel engines. Refer Dr. S.R Book Pg.No.1.21 3. Briefly explain the effect of different factors on knock in S.I. engines. Refer Dr. S.R Book Pg.No.2.10 4. What are the factors affecting the delay period? Explain in detail. Refer Dr. S.R Book Pg.No.2.31 5. (a) Define indicated mean effective pressure related to I.C.engine and how it can be measured? Refer Dr. S.R Book Pg.No.3.2 (b) A two cylinder four stroke engine runs at 240 rpm developing a torque of 5 kN-m. The bore and stroke of cylinder are 30 cm and 60 cm respectively. Engine runs with gaseous fuel having calorific value of 16.8 MJ/m 3. The gas and air mixture is supplied in proportion of 1:7 by volume. The volumetric efficiency is 0.85. Determine (i) The brake power. (ii) The mean piston speed in m/s. (iii) The brake mean effective pressure. (iv) The brake thermal efficiency. Refer Dr. S.R Book Pg.No.3.81

SQP.46 Thermal Engineering - I

6. (a) A single stage double acting air compressor running at 300 rpm, delivers 15 m3 of free air per minute at 700 kPa and 200C . If the clearance volume is 8% of swept volume and if the index of compression and expansion are same, find the clearance swept volume of piston and volumetric efficiency. Initial air conditions are 10kP and 15C . Refer Dr. S.R Book Pg.No.4.34 (b) Draw the T-S diagram for the multi stage compression and show the work saving during the compression in comparison with single stage reciprocating air compression. Refer Dr. S.R Book Pg.No.4.67 7. (a) With the help of neat sketch, explain the working of vane type blower. Also show the compression process in P-V diagram. Refer Dr. S.R Book Pg.No.4.119 (b) A Vaned compressor handles free air of 0.6 m 3/s at 1 bar and compresses to 2.3 bar. There occurs 30% reduction in volume before the back flow occurs. Determine the indicated power required and isentropic efficiency. Refer Dr. S.R Book Pg.No.4.120 8. (a) Derive expression for polytropic efficiency in terms of entry and delivery pressures, temperature and ratio of specific heats. Refer Dr. S.R Book Pg.No.4.176

Solved Question Papers SQP.47

(b) An axial flow compressor is to have constant axial velocity of 250 m/s and 50% degree of reaction. The mean diameter of blade ring is 45 cm and speed is 18000 r.p.m. The exit angles of the blade are 25. Calculate blade angle at inlet and work done per kg of air with the help of velocity triangles. Refer Dr. S.R Book Pg.No.4.206 ...........................................................................

Code No: 9A03402 B.Tech II Year II Semester (R09) Regular & Supplementary Examinations, April/May - 2013 (Set - 2) THERMAL ENGINEERING - I (Mechanical Engineering) Time: 3 hours

Max.Marks:75

Answer any FIVE Questions All Questions carry Equal Marks 1. (a) Define volumetric efficiency of engine and discuss the effect of various factors that affect the volumetric efficiency. Refer Dr. S.R Book Pg.No.6.10 (b) Compare the actual and fuel-air cycles of IC engine Refer Dr. S.R Book Pg.No.6.19 2. (a) Sketch the battery ignition system and explain its working. Refer Dr. S.R Book Pg.No.1.60

SQP.48 Thermal Engineering - I

(b) Explain the various types of cooling systems employed in IC engines. Refer Dr. S.R Book Pg.No.1.74 3. Explain the phenomenon of knocking in S.I.engine. What are the effects of knocking? Refer Dr. S.R Book Pg.No.2.10 4. (a) Explain with neat sketch the working principle of combustion chamber used in C.I. engine. Refer Dr. S.R Book Pg.No.2.42 5. (a) A four stroke cylinder diesel engine running at 300 rpm produces 250 kW of brake power. The cylinder dimensions are 30 cm bore and 0.6 cm stroke. Fuel consumption rate is 1 kg/min while air fuel ratio is 10. The average indicated mean effective pressure is 0.8 MPa. Determine indicated power, mechanical efficiency, brake thermal efficiency and volumetric efficiency of engine. The calorific value of fuel is 43 MJ/kg. The ambient conditions are 1.013 bar, 27C. Refer Dr. S.R Book Pg.No.3.83 (b) How do you measure air consumption rate of I.C engine in the laboratory? Refer Dr. S.R Book Pg.No.3.31 6. A two stage air compressor compresses air from 1 bar 20C to 42 bar. If the law of compression is pV 1.35  constant and intercooling is perfect. Find per

kg of air (i) the work done in compression. (ii) the mass of cooling water necessary for abstracting the

Solved Question Papers SQP.49

heat in the intercooler, if the temperature rise of the cooling water is 25C . Refer Dr. S.R Book Pg.No. 4.76 7. (a) A roots blower handles free air of 0.5 m 3/s at 1 bar and 27C and delivers air at pressure of 2 bar. Determine the indicated power required to drive compressor and isentropic efficiency. Refer Dr. S.R Book Pg.No.4.116 (b) With the help of neat sketch, explain the working of roots blower. Also show the compression process in p-V diagram. Refer Dr. S.R Book Pg.No.4.112 8. In an axial flow compressor, the overall stagnation pressure ratio achieved is 4 with overall stagnation isentropic efficiency 86%. The inlet stagnation pressure and temperature are 1 bar and 320 K. The mean blade speed is 190 m/s. The degree of reaction is 0.5 at the mean radius with relative air angles of 10and30  respectively. The work done factor is 0.9. Calculate: (i) Stagnation polytropic efficiency. (ii) Number of stages. (iii) Inlet temperature and pressure. (iv) Blade height in the first stage if the hub-tip ratio is 0.4, mass flow rate is 20 kg/sec. Refer Dr. S.R Book Pg.No.4.196 ...........................................................................

SQP.50 Thermal Engineering - I

Code No: 9A03402 B.Tech II Year II Semester (R09) Regular & Supplementary Examinations, April/May - 2013 (Set - 3) THERMAL ENGINEERING - I (Mechanical Engineering) Time: 3 hours

Max.Marks:75

Answer any FIVE Questions All Questions carry Equal Marks 1. (a) What is the use of air-standard cycle? List some air standard cycles. Refer Dr. S.R Book Pg.No.6.1 (b) Discuss the effect of exhaust valve opening time on blow down. Refer Dr. S.R Book Pg.No.6.8 2. What is IC engine? Classify I.C. engines. Refer Dr. S.R Book Pg.No.1.12 3. What are the factors which affect the ignition lag, flame propagation in S.I. engines? Refer Dr. S.R Book Pg.No.2.4 4. Explain any three important phases of combustion in C.I.engine. Refer Dr. S.R Book Pg.No.2.28 5. (a) Define the terms: brake power, piston speed, brake mean effective pressure and brake thermal efficiency. Refer Dr. S.R Book Pg.No.3.6

Solved Question Papers SQP.51

(b) A two cylinder four stroke engine runs at 240 r.p.m developing a torque of 5.16 kN-m. The bore and stroke of cylinder are 30 cm and 58.5 cm respectively. Engine runs with gaseous fuel having calorific value of 16.8 MJ/m 3. The gas and air mixture is supplied in proportion of 1:7 by volume. The volumetric efficiency is 0.85. Determine (i) The brake power (ii) The mean piston speed in m/s. (iii) The brake mean effective pressure (iv) The brake thermal efficiency.

Refer Dr. S.R Book Pg.No.3.81 6. (a) With the help of neat sketch explain the working principle of single stage single acing reciprocating air compressor. Refer Dr. S.R Book Pg.No.4.5 (b) A single stage, double acting compressor has a free air delivery of 14 m 3/min, measured at 1.013 bar and 15C . The pressure and temperature in the cylinder during induction are 0.95 bar 15C . The delivery pressure is 7 bar and index of compression and expansion is 1.3. The clearance volume is 5% of the swept volume. Calculate: (i) indicated power required. (ii) Volumetric efficiency. Refer Dr. S.R Book Pg.No.4.41 7. (a) Define and explain the terms: pressure coefficient and adiabatic coefficient of a centrifugal compressor. Refer Dr. S.R Book Pg.No.4.149 (b) A single sided centrifugal compressor is to deliver 14 kg/s of air when operating at a pressure ratio of 4:1 and a speed of 200 rev/s. The inlet stagnation conditions are 288 K and 1.0 bar. The slip factor and power input factor may be taken as 0.9 and 1.04 respectively. The

SQP.52 Thermal Engineering - I

overall isentropic efficiency is 0.80. Determine the overall diameter of the impeller. Refer Dr. S.R Book Pg.No.4.163 8. (a) Define polytropic efficiency of an compressor and write the expression.

axial

flow

Refer Dr. S.R Book Pg.No.4.176 (b) A multi stage axial compressor is required for compressing air at 300 K, through a pressure ratio of 4.5 to 1. Each stage is to be a 50% reaction and mean blade speed of 300 m/s, flow coefficient 0.5, and the stage loading factor 0.3 are taken, for simplicity, as constant for all stages. Determine the flow angles and the total number of stages required if the polytropic efficiency is 85%. Take C p  1.005 kJ/kgK and   1.4 for air. Refer Dr. S.R Book Pg.No. 4.200 ...........................................................................

Code No: 9A03402 B.Tech II Year II Semester (R09) Regular & Supplementary Examinations, April/May - 2013 (Set - 4) THERMAL ENGINEERING - I (Mechanical Engineering) Time: 3 hours

Max.Marks:75

Answer any FIVE Questions All Questions carry Equal Marks 1. (a) Discuss the effect of exhaust valve opening time on blow down. Refer Dr. S.R Book Pg.No.6.8

Solved Question Papers SQP.53

(b) What are the factors that affect the volumetric efficiency of an engine? Refer Dr. S.R Book Pg.No.6.10 2. (a) What are the various components to be lubricated in an engine and explain how it is accomplished? Refer Dr. S.R Book Pg.No.1.69 (b) What are the various characteristics efficient cooling system in an engine?

of

an

Refer Dr. S.R Book Pg.No.1.73 3. (a) What do you understand from abnormal combustion in S.I.engines? Refer Dr. S.R Book Pg.No.2.8 (b) Explain S.I.engine

different

stages

of

combustion

in

Refer Dr. S.R Book Pg.No.2.1 4. (a) What are the factors that affect the combustion process in C.I.engines? Refer Dr. S.R Book Pg.No.2.28 (b) What is knocking in diesel engine? Explain in detail. Refer Dr. S.R Book Pg.No.2.35 5. (a) A gas engine having a cylinder 250 mm bore and 450 mm stroke has a volumetric efficiency of 80%. Air-gas ratio equals 9:1, calorific value of fuel 21000 kJ/m3 at NTP. Calculate the heat supplied to

the engine per working cycle. If the compression ratio

SQP.54 Thermal Engineering - I

is 5:1. What is the heat value of the mixture per working stroke per m 3 of total cylinder volume? Refer Dr. S.R Book Pg.No.3.64 (b) Explain the performance test of an IC engine. Why is it carried out? Refer Dr. S.R Book Pg.No.3.21 6. (a) Derive an expression for minimum work required for two stage reciprocating air compressor with perfect inter-cooling and neglect clearance volume. Refer Dr. S.R Book Pg.No.4.66 (b) A single stage single acting air compressor delivers 14 m 3 of free air from 1 bar to 7 bar. The speed of the compressor is 300 r.p.m. Assuming the compression and expansion is pv 1.35  constant and clearance is 5% of the swept volume, find the diameter and stroke of the compressor. Take stroke length is 1.5 times the bore diameter. Refer Dr. S.R Book Pg.No.4.29 7. (a) What is power input factor and slip factor? Refer Dr. S.R Book Pg.No.4.147 (b) A centrifugal compressor running at 9000 r.p.m delivers 600 m 3/min of free air. The air is compressed from 1 bar and 20C to a pressure ratio of 4 with an isentropic efficiency of 0.82. Blades are radial at outlet of impeller and the flow velocity of 62 m/s may be assumes throughout constant. The outer radius of the impeller is twice the inner and the slip factor may be assumed as 0.9. The blade area

Solved Question Papers SQP.55

coefficient may be assumed as 0.9 at the inlet. Calculate (i) Final temperature of air. (ii) Theoretical power. (iii) impeller diameters at inlet and outlet. (iv) Breadth of the impeller at inlet. (v) impeller blade angle at inlet. (vi Diffuse blade angle at inlet. Refer Dr. S.R Book Pg.No.4.159 8. (a) Define degree of reaction and write the expression for degree of reaction. Refer Dr. S.R Book Pg.No.4.175 (b) An axial flow compressor is to have constant axial velocity of 150 m/s and 50% degree of reaction. The mean diameter of blade ring is 35 cm and speed is 15000 rpm. The exit angles of the blade are 27. Calculate blade angle at inlet and work done per kg of air with the help of velocity triangles. Refer Dr. S.R Book Pg.No.4.207 ...........................................................................

1. What are the methods employed to achieve near isothermal compression for high speed compressors. 2. In detail explain about the performance curves of axial flow compressors. 3. (a) Explain procedure to estimate the friction power of an engine by using Willan’s line method. (b) By means of sankey diagram explain the energy flow through an engine. 4. Draw the typical distillation curve for Diesel fuel

SQP.56 Thermal Engineering - I

5. Describe the working cycle of the screw compressor, with a sketch. 6. What is optimum spark advance? Show the effect of spark advance on the power output by means of the P-V diagram of an S.I engine. 7. Explain the principle of direct injection of gasoline into the cylinder of an S.I. Engine with a sketch. 8. Discuss the effect of compression ratio on maximum air temperature for C.I Engine combustion by means of a graph. 9. Explain the principle of operation of “Lanova air cell combustion chamber” with a sketch and state its advantages. 10. With the help of a P-V diagram for S.I.Engine, discuss the effect of 0 and 35 spark advances. 11. What is the purpose of engine testing? 12. Why do designer go for multi cylinder engine for heavy loads and name some multi cylinder engine types. 13. Derive an expression for the efficiency of Otto cycle and comment on the effect of compression ratio on the efficiency with respect of ratio of specific heats by means of suitable graphs. 14. What is the need for air movement in S.I.Engine combustion chamber? Explain. 15. Explain the working principle of chamber with the suitable diagram.

pre-combustion

16. Describe the multi-point fuel injection system with neat sketch.

Solved Question Papers SQP.57

17. Differentiate between uncontrolled combustion and controlled combustion in S.I.Engine. 18. What is meant by low degree of reaction and high degree of reaction? How you differentiate these two? 19. How do the specific heats vary with temperature? Explain the reasons for variation of specific heat and also discuss the influence of this on thermal efficiency of the cycle. 20. Find the percentage change in thermal efficiency of Otto cycle having a compression ratio of 12 and specific heat at constant pressure increases by 1% [15] 21. How to achieve rich fuel air mixture in S.I.Engine? Under what operating conditions rich mixture is required? 22. What are power producing and power absorbing machines? List a few of them. 23. Compare external combustion and internal combustion engines.

SQP.58 Thermal Engineering - I

II Year - II SEMESTER T

P

C

31

0

3

(Kakinada) THERMAL ENGINEERING - I UNIT - I Objectives: To make the student learn and understand the reasons and effects of various losses that occur in the actual engine operation. Actual Cycles and their Analysis:Introduction, Comparison of Air Standard and Actual Cycles. Time Loss Factor, Heat Loss Factor, Exhaust Blowdown-Loss due to Gas exchange process, Volumetric Efficiency. Loss due to Rubbing Friction, Actual and Fuel-Air Cycles of CI Engines. UNIT - II Objectives: To familiarize the student with the various engine systems along with their function and necessity. I.C. ENGINES: Classification - Working principles, Valve and Port Timing Diagrams, - Engine systems - Fuel, Carburetor, Fuel Injection System, Ignition, Cooling and Lubrication, principle of wankle engine, principles of supercharging and turbocharging. UNIT - III Objectives: To learn about normal combustion phenomenon and knocking in S.I and C.I Engines and to find the several engine operating parameters that affect the smooth engine operation.

Solved Question Papers SQP.59

Combustion in S.I. Engines: Normal combustion and abnormal combustion - Importance of flame speed and effect of engine variables - Type of Abnormal combustion, pre-ignition and knocking (explanation of ) - Fuel requirements and fuel rating, anti knock additives combustion chamber - requirements, types. Combustion in C.I.Engines: Four stages of combustion Delay period and its importance - Effect of engine variables - Diesel Knock - Need for air movement, suction, compression and combustion induced turbulence - open and divided combustion chambers and nozzles used - fuel requirements and fuel rating. UNIT - IV Objectives: To make the student learn to perform testing on S.I and C.I Engines for the calculations of performance and emission parameters. Measurement, Testing and Performance:Parameters of performance - measurement of cylinder pressure, fuel consumption, air intake, exhaust gas composition, Brake power - Determination of frictional losses and indicated power - Performance test - Heat balance sheet and chart. UNIT - V Objectives: To make students learn about different types of compressors and to calculate power and efficiency of reciprocating compressors. COMPRESSORS - Classification - positive displacement and roto dynamic machinery - Power producing and power absorbing machines, fan, blower and compressor - positive displacement and dynamic types - reciprocating and rotary types.

SQP.60 Thermal Engineering - I

Reciprocating: Principle of operation, work required, Isothermal efficiency volumetric efficiency and effect of clearance, stage compression, undercooling, saving of work, minimum work condition for stage compression. UNIT - VI Objectives: To make students learn mechanical details, and to calculate power and efficiency of rotary compressors Rotary (Positive displacement type): Roots Blower, vane sealed compressor, Lysholm compressor - mechanical details and principle of working - efficiency considerations. Dynamic Compressors: Centrifugal compressors: Mechanical details and principle of operation - velocity and pressure variation. Energy transfer - impeller blade shape-losses, slip factor, power input factor, pressure coefficient and adiabatic coefficient - velocity diagrams power. Axial Flow Compressors: Mechanical details and principle of operation - velocity triangles and energy transfer per stage degree of reaction, workdone factor - isentropic efficiency -pressure rise calculations - Polytropic efficiency.

Solved Question Papers SQP.61

JNTUA COLLEGE OF ENGINEERING (AUTONOMOUS): ANANTAPUR II Year B.Tech.M.E.II-Sem T

P

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3

0

3

THERMAL ENGINEERING - I UNIT - I I.C. ENGINES: Definition of Engine and Heat Engine, I.C. Engine Classification - Parts of I.C. Engines, Working of I.C. Engines, Two Stroke & Four Stroke I.C. Engines SI & CI Engines, Valve and Port Timing Diagrams. UNIT - II Fuel System: S.I. Engine: Fuel Supply Systems, carburetor types Air Filters, Mechanical and Electrical Fuel Pump Filters - Gasoline Injection Systems. Cooling & Lubrication Systems: Cooling Requirements, Air Cooling, Liquid Cooling, Thermo Siphon, Water And Forced Circulation System, Lubrication Systems - Flash, Pressurized and Mist Lubrication. Ignition System:Function Of An Ignition System, Battery coil Ignition System, Magneto Coil Ignition System, Electronic Ignition System using Contact Breaker, Electronic Ignition using Contact Triggers - Spark Advance And Retard Mechanism. UNIT - III Fuels and Combustion: S.I. Engine: Normal Combustion and Abnormal Combustion - Importance of Flame Speed and Effect of

SQP.62 Thermal Engineering - I

Engine Variables - Type of Abnormal Combustion, Pre-Ignition and Knocking (Explanation) Fuel Requirements and Fuel Rating, Anti Knock Additives, Combustion Chambers. C.I. Engines: Stages of Combustion - Delay Period And Its Importance - Effect Of Engine Variables - Diesel Knock Combustion Chambers (DI And IDI), Fuel Requirements And Fuel Rating. UNIT - IV Testing and Performance: Parameters of Performance Measurement of Cylinder Pressure, Fuel Consumption, Air Intake, Exhaust Gas Composition, Brake Power Determination of Frictional Losses And Indicated Power Performance Test - Heat Balance Sheet and Chart. UNIT - V Air Compressors: Reciprocating Compressors, Effect of Clearance volume in Compressors, Volumetric Efficiency, Single Stage and Multi Stage Compressors, Effect of Intercooling and Pressure Drop in Multi-Stage Compressors, Problems Related to Reciprocating Compressors, Working principles of Roots blower, Vane type Blower, Centrifugal Compressor - Axial Flow Compressors.

Index In.1

INDEX Capacitance Discharge Ignition System, 1.65 Abnormal Combustion, 2.8 Carbon monoxide (CO), 3.43 Absorption Dynamometer, 3.13 Carburetor, 1.37 Actual Cycles, 6.2 Centrifugal Compressor, 4.125 Air Fuel Ratio, 1.40 Cetane Number (CN), 2.52 Air Consumption, 3.31 Characteristics of good Air Refrigeration System, 5.6 refrigerants, 5.106 Air Refrigeration Cycles, 5.7 Chemical delay, 2.31 Air standard Cycles, 6.2 Choke Valve, 1.39 Aldehydes, 3.47 CI Engine, 1.6 Ammonia - Water Absorption Claude System for Liquefying System, 5.108 Gases, 5.112 Anti-knock Agents, 2.23 Clearance Ratio ‘k’, 4.21 Anti-Knock Additives, 2.23 Combustion Stoichiometry, 1.88 Application of Cryogenic, 5.114 Combustion chamber, 2.4 Application of Liquid Oxygen, Comparison method, 3.47 5.114 Complete, 4.65 Application of Liquid Nitrogen, Composition factor, 2.17 5.115 Compression ratio, 2.4, 2.6, Application of Carbon-di-oxide, 2.33, 611 5.115 Connecting Rod, 1.17 Application of Inert Gases, 5.115 Construction, 3.39 Automatic volumetric flow Crank, 1.17 meter, 3.37 Crankcase ventilation, 6.17 Crankshaft, 1.18 B Cylinder, 1.15 Back Flow of Air, 4.113 Cylindrical chamber, 2.43 Bath-tub form, 2.26 D Bell-Coleman Cycle, 5.17 Blade Loading, 4.171 Deflection co-efficient def, 4.181 Blowby Losses, 6.16 Degree of Reaction, 4.153 Bomb Calorimeter:, 1.83 Degree of Reaction, 4.175 Brake Power (BP), 3.6, 4.22 Density factors, 2.14 Brake Thermal Efficiency Detonation, 2.9 Brake, 3.11 Diesel Knock, 2.35 Burette method, 3.35 Divided combustion chamber type, 2.45

A

C

Calorific Value of Fuels:, 1.82

In.2 Thermal Engineering - I Double Acting Air Compressor, 4.4 Dry Sump System, 1.72 Dulong’s formula, 1.83 Dynamometer, 3.12

E Eddy Current Dynamometer, 3.15 Effect of Super heating, 5.69 Effect of Subcooling, 5.67 Efficiency Ratio, 3.11 Electrolux Refrigerator, 5.108 Energy-cell chamber, 2.47 Energy Transfer, 4.133 Engine size, 2.6, 2.33 Engine power, 6.15 Engine speed, 2.3, 2.5 Euler’s Equation, 4.133 Exhaust Blowdown, 6.8 Exhaust Gas Composition, 3.41

Fuel Injector, 1.56 Fuel temperature, 2.33 Fundamentals of Refrigeration:, 5.1

G Gas Exchange Process, 6.9 Gas Liquefaction System, 5.111 Gaseous fuel, 1.78, 1.80 Governor, 1.19 Gravimetric fuel flow measurement, 3.38 Gudgeon pin , 1.17

H

Hampson-Linde Gas Liquefaction System, 5.111 Heat Dissipation, 6.15 Heat Loss Factors, 6.6 Head or work coefficient h, 4.180 Heat Balance Sheet, 3.52 Hemispherical chamber, 2.43 F Higher Calorific value:, 1.82, F-head combustion chamber, 2.28 2.20 Finite Stage Efficiency, 4.178 Hydraulic Dynamometer, 3.13 Flame Front Propagation, 2.4 Flash Point and Fire Point, 1.81 I Float and Float chamber, 1.38 I-Head combustion chamber, 2.26 Flow Coefficients, 4.171 I.C. Engines, 1.1 Flow coefficient f, 4.180 Ignition Systems, 1.57 Fly wheel, 1.19 Ignition Delay period, 2.29 Free Air Delivered (FAD), 4.22 Ignition timing, 2.4 Friction Power (FP), 3.9 Impeller Blade Shape, 4.137 Fuel choice, 2.4 Imperfect Intercooling, 4.65 Fuel Pump, 1.36 Incomplete, 4.65 Fuel Consumption, 3.34 Indicated Power (IP), 3.2 Fuel Rating for CI Engine, 2.52 Indicated Thermal Efficiency Fuel-Air ratio, 2.6 indicated, 3.10 Fuel Properties, 1.81 Indicated Power (I.P), 4.22 Fuel Pump (C.I. Engine), 1.53 Indicator Diagram, 3.1 Fuel Injection System, 1.55 Fuel Supply System, 1.34

Index In.3 Infinitesimal Stage Efficiency, 4.176 Intake pressure (Super charging), 2.33 Intake temperature, 2.33

Octane Number (ON), 2.21 Open combustion chamber, 2.42 Overhead valve, 2.26 Oxides of Nitrogen, 3.41

P

J

Perfect Intercooling, 4.65 Performance Calculation, 5.99 Period of Rapid Combustion, K 2.32 Knocking Petroil System, 1.70 Knocking, 2.9, 2.35 Petrol Engine, 1.4 PH Chart, 5.56 L Physical delay, 2.30 L-head combustion chamber, 2.27 Pinking, 2.9 Liquid fuels:, 1.78 Piston speed, 4.23 Lithium Bromide Absorption Piston, 1.16 System, 5.109 Polytropic Efficiency, 4.176 Lower Calorific Value (LCV), Port Timing Diagrams, 1.24 1.82 Pour Point of Fuel, 1.81 Lower Calorific Value (LCV), Pre combustion chamber, 2.46 2.20 Pre-ignition, 2.8 Lubricant properties, 6.14 Pre-Ignition, 2.14 Lubrication System, 1.69 Pressure System (Forced M Lubrication), 1.71 Pressure ratio, 4.181 Magneto Ignition System, 1.63 Mean Effective Pressure m.e.p, Pressure co-efficient p, 4.181 Pressure Co-efficient p, 4.149 4.22 Mean Effective Pressure Pm, R 3.2 Reciprocating Air Compressors, Mechanical Efficiency mech, 4.5 3.10 Refrigerant, 5.106 Minimum Workdone, 4.11 Refrigerants Number, 5.107 Multi Stage Compressor, 4.5 Relative Efficiency, 3.11 Multistage Compressor, 4.71 Remedies, 2.25 N Reversed Carnot Cycle , 5.7 Roots Blower, 4.112 Number of stages, 4.181 Juncker’s Gas Calorimeter, 1.85

O Obscuration method, 3.48 Octane value, 2.17

S Semipressure System, 1.72 Shallow depth chamber, 2.43

In.4 Thermal Engineering - I Simple Carburetor, 1.38 Single Stage Compressor, 4.4, 4.69 Single Acting Reciprocating Air Compressor, 4.3 Slip factor, 4.147 Smoke, 3.47 Solid fuels, 1.78, 1.79 Solved Problems 5.70-5.95 Spark Plug, 1.57 Specific Work, 4.171 Specific Fuel Consumption (S.F.C), 3.9 Splash System, 1.70 Stage Work, 4.149 Stage Pressure Rise, 4.151 Stage Velocity Triangles, 4.169 Stagnation Enthalpy, 4.131 Stagnation pressure ratio, 4.181 Stagnation Temperature, 4.129 Stalling, 4.186 Static Temperature, 4.129 Steady-flow Energy Equation, 4.132 Subcooling or Undercooling, 5.66 Sulphur Content in the Fuel, 1.81 Super Heating, 5.69 Surface finish, 6.14 Surging, 4.184 Swept Volume Vs, 4.23

T T-Head combustion chamber, 2.27 Thermal Efficiency, 3.10 Time factors, 2.16 Time Loss, 6.4 Ton of Refrigeration, 5.3 Toroidal chamber, 2.43 Total Head Temperature, 4.129

Transistorized Assisted Contact (TAC) Ignition System, 1.67 Turbulence, 2.5 Turbulent combustion chamber, 2.45 Two Stage Compression, 4.64

U Unburnt hydrocarbons, 3.45 Uncontrolled Combustion, 2.32 Unit of Refrigeration, 5.3

V Valve Timing Diagrams, 1.24 Vane Type Blower of Compressor, 4.117 Vapour Absorption System, 5.102 Vapour Compression Refrigeration, 5.39 Velocity Triangle, 4.140 Velocity And Pressure Variation, 4.128 Venturi and Throttle valve, 1.39 Viscosity of Fuel, 1.81 Visible Emissions, 3.47 Volatility, 1.81 Volume Rate, 4.21 Volume, 4.21 Volumetric Efficiency vol, 4.20, 6.10 Volumetric type, 3.35 Volumetric Efficiency, 3.11

W Wedge form, 2.26 Work factor, 4.149