Thermodynamics for Engineers [1 ed.] 1133112862, 9781133112860

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KENNETH A. KROOS | MERLE C. POTTER — u

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Thermodynamics

—

ad ti

a

Thermodynamics

Kenneth A. Kroos Villanova University

Merle C. Potter Michigan State University

wuirg GENGAGE a” Learning Australia « Brazil

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2

To my wife Kathleen. Ken

To my wife Gloria. Merle

=

SS

Nomenclature List

Preface

xvii

xix

About the Authors

XXiii

Chapter 1 Basic Concepts and Systems of Units 1.1 Introduction 5 1.1.1. 1.1.2

1.2

1.3

1.4

What is thermodynamics?

5

| How do we use thermodynamics?

5

1.1.3

How do engineers use thermodynamics?

6

1.1.4

What is the history of thermodynamics?

6

1.1.5

What is the future of thermodynamics?

1.1.6

What are the fundamental concepts and assumptions?

1.1.7.

What are the phases of matter?

Dimensions and Units 1.2.1

The SI system

1.2.2

The English system

8

9 9

11

Properties, Processes, and Equilibrium 1.3.1

Properties and state of a system

1.3.2

Density and specific volume

1.3.3.

Processes and equilibrium

Pressure

12

12

13 15

17

1.4.1.

What is pressure?

1.4.2

Absolute and gage pressure

1.4.3.

Units of pressure

1.4.4

6

17 17

18

| Pressure-measuring devices

19

vii

vill

1.5

Contents

Temperature 1.5.1

What is temperature?

1.5.2

Absolute and relative temperature scales

1.5.3.

Temperature measurement

1.6

Energy

1.7.

Summary

2.2

2.3 2.4

22 22

24

27

Chapter 2. 2.1

22

29

Phases of a Substance

39

2.1.1

Phase-change process

2.1.2.

Quality and compressed liquid calculations

2.1.3.

Superheated vapor

2.1.4

Properties using the IRC Property Calculator

2.1.5

Phase diagrams

40 44

47 49

51

Internal Energy and Enthalpy

56

2.2.1

‘Internal energy

2.2.2.

Enthalpy

2.2.3.

Internal energy and enthalpy for liquids and solids

2.2.4

Latent heat

56

57

Refrigerants

58

60

61

Ideal-Gas Law

64

2.5

Real Gas Equations of State

2.6

Internal Energy and Enthalpy of Ideal Gases

2.7

Specific Heats of Liquids and Solids

2.8

Summary

66 69

73

75

Chapter 3

The First Law for Systems

3.1

88

Work

37

Properties of Pure Substances

3.1.1.

Definition and units

3.1.2

Work due to pressure

88

3.1.3.

Other forms of work

91

3.2.

Heat Transfer

3.3.

Problem-Solving Method

88

95 98

85

Contents 3.4

‘The First Law Applied to Systems

3.5.

The First Law Applied to Various Processes The constant-volume process

3.5.2

The constant-pressure process

3.5.3.

The constant-temperature process

3.5.4

The adiabatic process

3.5.5

The polytropic process

3.7

Summary

Chapter 4

102

105

106 110

112

The First Law Applied to Control Volumes

The Conservation of Mass for Control Volumes 4.1.1

Basic information

4.1.2

The continuity equation

4.2.1

Turbines, compressors, and pumps

4.2.2

Throttling devices

4.2.3.

Mixing chambers

145

4.2.4

Heat exchangers

148

4.2.5

Nozzles and diffusers

137 139

143

151

Unsteady Flow

4.4

Devices Combined into Cycles

155

4.4.1.

The Rankine power cycle

4.4.2

The refrigeration cycle

4.4.3

The Brayton cycle

Chapter 5

158 158

160 163

165

The Second Law of Thermodynamics

5.1

Second-Law Concepts

5.2.

Statements of the Second Law of Thermodynamics 5.2.1. 5.2.2

130

131

The First Law for Control Volumes

Summary

127

130

= ae)

4.5

101

112

Cycles

4.2

99 101

3.5.1

3.6

4.1.

mn

180

Kelvin-Planck statement—heat engines Clausius statement—refrigerators

182 183

182

179

iat

5.3

Contents

Cycle Performance Parameters 5.3.1.

The heat engine

5.3.2

The refrigeration cycle

5.4

The Carnot Cycle

5.9

Summary

Chapter6

185

185 186

189

194

203

Entropy

6.1

Inequality of Clausius

205

6.2

Entropy

6.3

Entropy Change in Substances for Systems

207

209

6.3.1

Basic relationships

209

6.3.2

Entropy change of an ideal gas with constant C, and C,

6.3.3.

Entropy change of a solid, a liquid, and a reservoir

6.3.4

Entropy change of a phase-change substance

6.3.5

Entropy change of an ideal gas with variable specific heats

6.4

Entropy Changes for a Control Volume

6.5

Isentropic Efficiency

213 216 218

221

225

6.5.1

Isentropic turbine efficiency

6.5.2

Isentropic compressor efficiency

6.5.3

Pump efficiency

6.5.4

—_Isentropic efficiency of a nozzle

225

226

228 229

6.6

Exergy (Availability) and Irreversibility

6.7

Summary

Chapter 7

210

230

237

Thermodynamic Relations

7.1

The Maxwell Relations

7.2

The Clapeyron Equation

7.3

Relationships for Internal Energy, Enthalpy, Entropy, 258 and Specific Heats

7.4

The Joule-Thomson Coefficient

7.5

Real-Gas Effects

7.6

Summary

269

265

252

255

263

251

Contents

zt

Part App il licati == ons 275 Chapter 8

The Rankine Power Cycle

8.1

Energy Sustainability

8.2

The Rankine Cycle

8.3

279

279

8.2.1

Basic configuration and components

8.2.2.

Improving Rankine cycle efficiency

Modified Rankine Cycles

286

290

8.3.1.

The ideal reheat Rankine cycle

8.3.2

The ideal regenerative Rankine cycle

8.3.3.

A combined reheat-regenerative ideal Rankine cycle

8.4

Cogeneration Cycles

8.5

Losses in Power Plants

8.6

Summary

Chapter 9

290 293 296

299

302

303

Gas Power Cycles

9.1

Air-Standard Analysis

9.2

Reciprocating Engine Terminology

9.3

The Otto Cycle

315

317 320

322

9.3.1

The four-stroke Otto cycle

9.3.2

Otto cycle analysis

322

9.3.3

Two-stroke Otto cycle engine

9.3.4

The Wankel engine

323 327

328

9.4

The Diesel Cycle

9.5

Other Gas Power Cycles

9.6

279

330 334

9.5.1

The dual cycle

9.5.2

The Stirling and Ericsson cycles

The Brayton Cycle

335

337

341

9.6.1.

The Brayton cycle with regenerative heating

9.6.2

The Brayton cycle with regeneration, intercooling, and reheat

9.6.3

The turbojet engine

350

9.7

The Combined Brayton-Rankine Cycle

9.8

Summary

354

346

351

tia

Contents

Chapter 10 10.1

The Vapor Compression-Refrigeration Cycle 10.1.1

Refrigeration cycle terminology

10.1.2

The ideal refrigeration cycle

10.1.3.

An actual refrigeration cycle

10.1.4

Heat pumps

371

10.1.5

Refrigerants

373

10.3

Absorption Refrigeration

10.4

Gas Refrigeration Systems

10.5

Summary

11.2

367

370

Cascade Refrigeration Systems

Chapter 11

374

377 377

380

Mixtures and Psychrometrics

Gas Mixtures

387

389

11.1.1

Definitions and terminology

389

11.1.2

The Amagat and Dalton laws

391

Air-Vapor Mixtures and Psychrometry 11.2.1

Terminology and definitions

11.2.2

Adiabatic saturation temperature

11.2.3

Psychrometrics

396

396 399

401

11.3

Air-Conditioning Processes

11.4

Summary

Chapter 12

367

368

10.2

11.1

365

Refrigeration Cycles

405

412

Combustion

423

12.1

Introduction

12.2

Combustion Reactions

12.3

The Enthalpy of Formation and the Enthalpy of Combustion

12.4

Flame Temperature

12.5

Equilibrium Reactions

12.6

Summary

425

450

426

442

447

Contents

Chapter 13 13.1

13.2

Alternative Energy Conversion 463

13.1.1

Ethanol

13.1.2

Biodiesel

13.1.3

Algae fuel

464 467

467

Solar Energy 13.2.1

Photovoltaic cells

13.2.2

Active solar heating

13.2.3

Passive solar heating

467 468 470

13.3

Fuel Cells

13.4

Thermoelectric Generators

13.5

Geothermal Energy

13.6

Wind Energy

13.7.

Ocean and Hydroelectric Energy

470 472

473

473

13.7.1 Wave energy

474

474

13.7.2

Ocean thermal energy conversion

13.7.3.

Hydroelectric power

13.8

Osmotic Power Generation

13.9

Summary

475

477

478

479

Thermodynamics of Living Organisms

14.1.

Energy Conversion in Plants

14.2.

Energy Conversion in Animals

14.3.

The Generation of Biological Work

14.4

Temperature Regulation in Biological Systems

14.5

485 487

14.4.1.

Endothermic organisms

492

14.4.2

Ectothermic organisms

492

14.4.3.

Temperature regulation in plants

Summary

461

462

Biofuels

Chapter 14

a

494

490

493

491

483

XIV

Contents

497

The Appendices A

Conversions of Units

B

Material Properties B-I B-1E

499 501

Properties of the U.S. Standard Atmosphere

501

__ Properties of the U.S. Standard Atmosphere

502

B-2

Properties of Various Ideal Gases (at 25°C)

B-3

Critical-Point Constants

B-4

Specific Heats and Densities of Liquids and Solids

505

B-4E

Specific Heats and Densities of Liquids and Solids

505

B-5

506

_Ideal-Gas Specific Heat of Several Common Gases as a Cubic Function of Temperature

B-6

506

Ideal-Gas Specific Heats of Several Common Gases at Different Temperatures

C

507

B-6E

_Ideal-Gas Specific Heats of Several Common Gases at Different Temperatures 508

B-7

Enthalpy of Formation and Enthalpy of Vaporization

B-8

Enthalpy of Combustion and Enthalpy of Vaporization

B-9

Natural Logarithms of the Equilibrium Constant K,

oi

B-10

Constants for the van der Waals Equation of State

511

Steam Tables

509 510

513

C-1

Properties of Saturated HO—Temperature Table

C-2

Properties of Saturated H,O—Pressure Table

C-3

Superheated Steam

C-4

Compressed Liquid

C-5

Saturated Solid—Vapor

C-1E

Properties of Saturated H,O—Temperature Table

C-2E

D

504

Ideal-Gas Specific Heat of Several Common Gases as a Cubic Function of Temperature

B-5E

503

515

516

522

523

_—_—Properties of Saturated H,O—Pressure Table

C-3E

Properties of Superheated Steam

C-4E

Compressed Liquid

C-5E

Saturated Solid—Vapor

Properties of R134a

Sis

524 526

528

531

532

533

D-I

Saturated R134a—Temperature Table

D-2

Saturated R134a—Pressure Table

. 533 535

Contents D-3

D-1E

Superheated R134a

536

—_Properties of Saturated R134a—Temperature Table

D-2E

Properties of Saturated R134a—Pressure Table

D-3E

Superheated R134a Vapor

Properties of Ammonia Saturated Ammonia

E-2

Superheated Ammonia

E-1E

Saturated Ammonia

E-2E

| Superheated Ammonia

542

545 547

549 550

553

F-|

Properties of Air

F-2

Molar Properties of Nitrogen, N,

F-3

Molar Properties of Oxygen, O,

F-4

Molar Properties of Carbon Dioxide, CO,

F-5

Molar Properties of Carbon Monoxide, CO

F-6

Molar properties of hydrogen, H,

F-7

Molar Properties of Water,H,O

F-1E

Properties of Air

F-2E

Molar Properties of Nitrogen,N,

F-3E

Molar Properties of Oxygen, O,

F-4E

Molar Properties of Carbon Dioxide, CO,

F-5E

Molar Properties of Carbon Monoxide, CO

F-6E

—_Ideal-Gas Properties of Hydrogen, H,

566

F-7E

Molar Properties of Water Vapor, H,O

567

Psychrometric Charts

541

545

E-|

Ideal-Gas Tables

540

553 554 555 556

Shy

558

= 559

560

562 563 564 565

569

Figure G-1

Psychrometric chart, P = 1 atm, SI units

Figure G-1E

Psychrometric chart, P = 1 atm, English units

Compressibility Charts

571

Figure H-1

Compressibility chart, low pressures

Figure H-2

Compressibility chart, high pressures

Enthalpy Departure

569

571 572

573

Figure I-1

Enthalpy departure chart, SI units

Figure I-1E

Entropy departure chart, English units

573 574

570

Contents

J

Entropy Departure Charts Figure J-1

Entropy departure chart, SI units

Figure J-1E

Entropy departure chart, English units

Answers to Selected Problems Index

575

585

577

575 576

Constant, Acceleration, Helmholtz function Constant Air/fuel mass ratio Bulk modulus Bottom dead center Back work ratio

Specific heat, A constant

Speed of light Coefficient of performance of a heat pump Coefficient of performance of a refrigerator Constant pressure specific heat

Constant pressure specific heat Change in energy per unit time Differential length segment A representative function Force Force vector

Fuel/air mass ratio = 275

99 7©

Normal component of a force Gibbs function Gravity, Gibbs function Gravitational constant Specific enthalpy, Height, Planck’s constant

°

Enthalpy at the reference state Enthalpy of formation Enthalpy of vaporization

°

Convection heat transfer coefficient

s|a SF SS >

> x»

jee ee = 2.678 m*

Chapter 2

Properties of Pure Substances

The pressure in the container is the saturation pressure at 120°C, which is 198.5 kPa. The liquid, which is 40% of the mass, is only about 0.08% of the volume. That’s because the density of the liquid is 840 times that of the vapor. Figures 2.3 and 2.4 are not to scale.

perenne ec

wes

Sto one

volumes, the liquid phase would not be noticed. Problem 2.46 requests

that the sketches be drawn to scale.

The volume of a mixture ina container

Example

2,2

SESS

A container holds 5 kg of water at 120°C with a quality of 0.6. Determine the total volume of the water in the container. This is the same volume as in Example 2.1, but it will be calculated using the average specific volume v. Solution

The average specific volume of the saturated mixture, using Eq. 2.5 and the

values listed in Example 2.1, is

Note: The specific volume v of the water

is igs es X(V, = 2) = 0.001060 + 0.6 X (0.8919 —

~ combines the liquid and ~ the vapor phases.

0.001060) = 0.5356 m°/kg

We can now calculate the volume of the container using this average specific volume and the total mass:

V=m-v=5

X 0.5356 = 2.678m?

The pressure and quality of

amixture

2S SOTaA

A 2000-cm? container holds 0.4 kg of water at 300°C. Determine the pressure and the quality of the mixture in the quality region at state 1 indicated in Fig. 2.6. Superheated

vapor P =const

300°C Saturated mixture

(Continued)

Example

2.3

bry

Part 1

Example

Concepts and Basic Laws

2.3

(Continued) — Solution

__Comment | | If P > 8.58 MPa, the water

The pressure and temperature are coupled in the saturated mixture region. So, the pressure can be found next to the temperature in Table C-1 in the Appendix. It is

would be compressed. If

P < 8.58 MPa, the water

P, = 8.58 MPa

would be superheated.

Par yer aa

Recognizing that 2000 cm? is 2000 x 10 °m’, we find that the specific volume of the saturated mixture is

Ve

2000 «107

ne

0.4

Note: Quantities given in

a problem statement are

v1, =

assumed known to four

Ran diets oovthe

mass of 0.4 kg is assumed tobe 0.4000 kg. Probably |, three significant digits would be closer to the . truth, but assuming four

significant digits is usually

= 0.005 m°/kg

This is less than v,8 but greater than v,,f So it is a mixture of liquid and vapor. We Can now calculate the quality:

UV = Up +r x(U, a :

0.005 =.0.001404

2)

+ x,(0.02168

—

0.001404)

.«.x, = 0.1774

acceptable.

The state is relatively close to the liquid saturation point. Its quality is 1774%.

Example

2.4 — The change in volume of a compressed liquid AEMEG LSLLEE DELETE

A cylinder holds 200 perature is increased in Fig. 2.7 Determine percent change in the

kg of water at 120°C at a pressure of 20 MPa. The temto 340°C while holding the pressure constant, as shown the volume of the water at both temperatures and the volume. Use 1) Table C-1 and ii) Table C-4. P =20 MPa

\ Saturated mixture

he

Figure 2.7

\.

Chapter 2

Properties of Pure Substances

Solution

i)

Assuming the specific volume depends on temperature only, the volumes are found, using Table C-1, to be

V, = mv, = 200 X 0.00106 = 0.212 m° V, = mv, = 200 X 0.001638 = 0.3276 m> The percent change in the volumes is

Yochange =

03276 0212 0.212

x 100 = 54.5%

ii) Using the compressed liquid Table C-4 provides

V, = mv, = 200 X 0.0010496 = 0.2099 m3 V, = mv, = 200 X 0.001633 = 0.3266 m> The percent change in the volumes is

0.3266 — 0.2099 %

ch

Boveuee

=

0.2099

x 100 = 55.6%

Observe that the values using Table C-1 are quite acceptable for engineering calculations.

2.1.3 Superheated vapor When the last drop of liquid in a saturated mixture vaporizes, the state becomes a saturated vapor with a quality of x = 1. If heat continues to be added to the

— Remember: Steam, vapor,

saturated vapor, it becomes a superheated vapor for which quality has no mean-

and gas are often used

ing. In this state, the substance exists as a gas, and temperature and pressure now become independent,properties. We once visited a major power plant that, as is common, used steam as the working fluid that drove the turbines. The steam generator (boiler) had a viewport on the side with a sign that read, “superheated steam.” When we looked through the viewport, we saw absolutely nothing. A superheated vapor is invisible, as is the water vapor in the atmosphere on a nice sunny day. The properties of superheated water vapor are presented in Table C-3 in the Appendix.

_synonymously.

Interpolate to find specific volume tee

Find the specific volume of superheated steam at 323°C and 2 MPa. Assume a linear relationship between temperature and specific volume in the vicinity of the desired state. (Continued)

Example

2.5

ace

Part 1

Example

Concepts and Basic Laws

2.5

(Continued) v

A straight line

300°C

323°C

350°C

Figure 2.8 Solution

The specific volume at 300°C and 350°C is listed in Table C-3, with several entries displayed as Table 2.3. A straight-line interpolation, sketched in Fig. 2.8, provides the specific volume at 323°C. The ratios from the triangles provide 323 — 39, — 0.1255 ia 350 — 0.1386 — 0.1255

300

300

: k 1345 m’/kg BA 0323 ="0. 0.1315

There are computer programs that provide properties at any temperature and pressure, but it is important that one can interpolate for values in tables of numbers. Tabulated sets of numbers are encountered in numerous areas of engineering, and interpolation or extrapolation is often necessary; computer programs are not always available. Practice interpolating until you are fairly good at it. It is necessary when you are using tables in various subject areas and may be required during an exam! Table 2.3

T (°C) 300 350

Example

2.6

P=2MPa

v (m3/kg) 0.1255 0.1386

Double interpolation of water vapor properties re

Water exists at 1120 kPa and 470°C. Determine the specific volume of this superheated water vapor. Solution

Appendix C-3 presents the properties for superheated water vapor. Tables

exist for pressures of 1.0 MPa and 1.2 MPa, so an interpolation for P = 1.12 MPa is necessary. The tables also list properties for steam at temperatures of 400°C and 500°C, so an interpolation for T = 470°C is also necessary. This is

Chapter 2

Properties of Pure Substances

Pa

referred to as a double interpolation. The following table shows the specific volume at the four listed states from Table C-3:

P = 1.0 MPa

P = 1.2 MPa

T = 400°C:

v = 0.3066 m/kg

v = 0.2548 m/kg

P= 300°C:

v = 0.3541 m*/kg

v = 0.2946 m/kg

The interpolation for v, at P = 1.12 MPa and 400°C is performed first, although the temperature interpolation could be selected first. At T = 400°C: AO bie 12-10

248 0, 0.2548 — 0.3066

0; =_0.2755 m*/kg

Next, interpolate for v, at P = 1.12 MPa and 500°C: gre

eae

12 -— 1.0,

se 0

D>

0.2946 — 0.3541

0) = 0.3184 m*/kg

Next we interpolate between 400°C and 500°C to get the specific volume v at

Note: Three rather than

470°C and 1120 kPa: A AOD 500 — 400

| = 0275). 0.3184 — 0.2755

four significant digits are written since the straight-line interpolation

v0 = 0.306 m°*/kg

'

method loses at least one

significant digit.

(2)

[¥{ You have completed Learning fy)

2.1.4 Properties using the IRC Property Calculator In the several examples of this section, we have used the tables in Appendix C to find the properties of water. Several other substances will also be of interest in our study of thermodynamics. Rather than using the tables in the Appendix to find properties, the IRC (Industrial Refrigeration Consortium), in conjunction with the University of Wisconsin,' developed the IRC Fluid Property Calculator, displayed in Fig. 2.9, and located at http://www.irc.wisc.edu/properties where properties of interest can be found rather easily. This site will be used in some examples and solutions to problems if it is not requested that the tables in the Appendix be used, and especially if interpolation is required. It should be pointed out that, on this Web site, density is used as an input rather than specific volume, so remember

Note: You can also Google “IRC Fluid Property ~

Calculator.”

to use

aa re)

2.8 (2.8)

when converting from specific volume to density. The specific volume is one of the properties calculated, although it is called “Volume.” The tables in the appendix are generated not by using measured values of properties but by using curve fits of measured values, allowing properties to be carried 'There may be periods of time when the IRC Fluid Property Calculator, referred to as the IRC Calculator, is not operating due to computer system updates and/or changes being made in the operation of the calculator.

It takes practice to become familiar with this IRC Calculator. Check the solutions in Example 2.7 and make sure they are correct.

wis

Part 1

Concepts and Basic Laws

RC - The Web site automatically

Fluid Property Calculator

selects English units. If SI

© international Units @ English Units

units are desired, check the

sa

appropriate box on the site.

Choose a fluid.

gr

Molar Weight Triple Point Temperature:

=

|

E

ae

Notmal Boiling Point Gas Phase Dipole at NBP: Acentric Factor: Critical Ternperature: Critical Pressure: Critical Density: Gas Constant:

Enter two conditions (properties).

Click on “Calculate Properties.”

—

Temperature:

Thermal Expansion: Sound Speed: Viscosity: Thermal Conductivity: Prandtl Wumber: Helmholtz Energy: Gibbs Free Eneray:

' —

Pressure: Density: Volume: Quality: Internal Energy: Enthalpy: Entropy: Isochoric Heat Capacity: lsobaric Heat Capacity: Surface Tension:

www.irc.wisc IRC. of Courtesy

Read the values of the

properties of interest. Not

A display of the IRC Property Calculator.

all properties are shown here, and many shown are not of interest in our study.

out to many significant numbers. This is an undesirable feature for engineering students since exaggerated accuracy is implied. Several sets of curve fits are available, so numerical values from one table may not match all the significant digits of the values from another table. The values generated with the IRC Calculator may be slightly different from the values in the tables in this book. The answers should, however, be the same to within acceptable engineering accuracy, which is usually taken as about 3%. The three-digit accuracy on the Web page is more realistic than the multidigit accuracy in the tables and will hopefully prevent both students and faculty from presenting answers to five and six significant digits.” The following example will illustrate the use of the Web page, which resembles the reproduction above.

“It is suggested that one point be deducted per answer on a partial-credit exam when five or more significant digits are given in an answer. Measured properties (temperature, pressure, etc.) are not known to four significant digits unless extreme care is taken, and probably are never known to five significant

digits unless you're a scientist working at the National Bureau of Standards. An answer to five or more significant digits is not acceptable.

Chapter 2

Properties of Pure Substances

Use the IRC Calculator to find properties

zh

Example

2,7

i) Find the volume of 5 kg of water if it is at 120°C with x = 0.6. ii) A 2000-cm* container holds 0.4 kg of water at 300°C. Find the quality. ili) Determine the volume of 200 kg of water at 340°C and 20 MPa. iv) Find the specific volume of superheated steam at 470°C and 1120 kPa.

Solution i)

From the Web site www.irc.wisc.edu/properties select “International units” and “Water.” Select “Temperature” as 120°C and “Quality” as 0.6. Press “Calculate Properties” and under “Volume” find v = 0.535 m*/kg. The volume is

V = mv = 5kg X 0.535 m*/kg = 2.68 m’ “5

This compares with 2.678 m* in : Example 2.2. ; } :

ahias

ii) Since specific volume is not an input, the density is found to be

m

0.4ke

Be AE p

V

2000

ee

y, 107° m?/cm>

= 200 ke/m: g/

The number 0.535 has three significant digits so

the answer is expressed using three significant digits, not four.

2

SERRA

ters

Oe

rn eee eer

Follow the directions in (i) with T = 300°C, and find x = 0.178. This compares with x = 0.1774 in Example 2.3. iii) At 340°C and 20 000 kPa, following the directions given in (i), we find

v = 0.00157 m*/kg so that

V = mv = 200kg X 0.00157 m*/kg = 0.314 m* This compares with a value of 0.328 m’, based on a linearly interpolated

value for v, in Example 2.4, a difference of 4%. A more accurate interpolation method for v, would reduce the difference. The value using the IRC Calculator is more accurate. iv) At 470°C and 1120 kPa, following the directions given in (i), the Web site

provides v = 0.303 m*/kg, which compares with 0.306 m*/kg in Example 2.6.

(3) 2.1.5 Phase diagrams A phase diagram is a two- or three-dimensional graph that shows how a pure substance (a substance that has the same chemical composition throughout) changes state as temperature, pressure, and volume of the substance vary. If numerous experiments were run, similar to the mental experiment described earlier (see Fig. 2.3), and the results plotted on one graph, all the saturated liquid states and the saturated vapor states could be connected to form the 7-v phase diagram sketched in Fig. 2.10. The diagram is divided into three regions: the compressed liquid region, the saturated mixture region, and the superheated vapor region. A P-v diagram

—

Part 1. Concepts and Basic Laws

if

Superheated vapor

th

pe Critical

oe

Olt

5

\ \\

n

o2

i

aie: S oO

pomeee

C85 \

| ~ mixture aied Saii liquid line

P= const

Saturated — vapor line Me SS

A T-v phase diagram for a pure substance.

Saturated vapor line: The line separating the superheated vapor from the saturated mixture.

Note: The saturated

mixture region is also

referred to as the “wet region” or the “quality region.”

SSS Critical point: Where the saturated liquid line and the saturated vapor line intersect. Supercritical region: Ata

pressure and temperature above the critical point values in a state where a liquid is no longer distinguished from a vapor.

|

;

ee Chitical

\

‘point

:\

\ > \\ i ad

Saturated SN

Saturated liquid line: The solid line separating the compressed liquid from the saturated mixture.

IP

i}

Be raceme

S

|

\ \ clBN

Saiurated

|

vapor

\\

ie A 5 oo:

=

Superheated

: mixture

|Saturated liquid line

Saturated vapor line

\ >. \

Sg < See

=

&

const

=e

A P-v phase diagram for a pure substance.

created with constant-temperature experiments, similar to those that created Fig. 2.4, would appear quite similar and, of course, exhibit the same three regions; it is shown in Fig. 2.11. The region on the left represents the range of pressures and temperatures where the substance behaves as a compressed liquid. The region under the solid curve shows the range of pressures and temperatures where the substance behaves as a saturated mixture of liquid and vapor. The solid line separating the compressed liquid from the saturated mixture is called the saturated liquid line. It is on this line that the phase-change process from a liquid to a vapor begins. To the left of the line, the fluid is a compressed liquid. To the right of the line it is a saturated mixture. On the line, the fluid exists as a saturated liquid with a quality of zero. The region to the far right is the range of pressures and temperatures where the fluid exists as a superheated vapor. The line separating the superheated vapor from the saturated mixture is the saturated vapor line. On this line, the fluid exists as a saturated vapor with a quality of one. The saturated mixture region that appears as a dome between the saturated liquid line and the saturated vapor line is also referred to as the wet region or the quality region. All pure substances have phase diagrams like the ones shown in Figs. 2.9 and 2.10. The saturated liquid line and the saturated vapor line intersect at a very high pressure at a point called the critical point. For water, the pressure and temperature

at the critical point are P,, = 22.09 MPa (3203 psia) and T., = 374.1°C (705.4°F). The critical temperatures and pressures for other common substances are included in Appendix B-3. The superheat region in which the pressure and temperature are above the critical point values is often referred to as the supercritical region; the substance can no longer be distinguished as a liquid or a vapor. Very few engineering processes outside of power plants operate in the supercritical region. The cost is often too high to safely contain the working fluid. The Eddystone power plant, described at the beginning of this chapter, operates in the supercritical region, as do about 150 other power plants in the United States. If we combined the two diagrams into one and included the solid phase as well, it would appear as the three-dimensional sketch in Fig. 2.12 for a substance that contracts on freezing. The dashed lines are constant-temperature lines. Another phase diagram is the pressure-temperature diagram, where the pressure of the substance is plotted as a function of the temperature, as in Fig. 2.13. Three lines separate the three phases. The phase change occurs as a line is crossed.

Chapter 2

Properties of Pure Substances

[EE

Critical point

e€ =

Sublimation

EE

The P-v-T diagram for a substance that contracts on freezing.

Figure 213 The P-T phase diagram.

This can be observed if the P-v-T diagram of Fig. 2.12 is viewed from the right; the saturated-mixture dome would appear as the curved vaporization curve of Fig. 2.13. It may be difficult to visualize Fig. 2.13, but with some effort it is possible. The T-v or P-v diagrams are used exclusively in the examples and problems. The melting line (or fusion line) separates the solid from the liquid. The vaporization line separates the liquid from the vapor, and the sublimation line separates the solid from the vapor. The point where all three lines come together is the triple point, where all three phases coexist. (The solid-vapor region wasn’t displayed on the bottom, as is common, of the phase diagrams of Figs. 2.10 and 2.11.) We refer to a liquid changing phase to a vapor as vaporization (or boiling or evaporation) and a vapor changing phase to a liquid as condensation. Melting and freezing is commonly known in the phase change between solids and liquids. Sublimation, which occurs when a solid changes phase directly to a gas without passing through the liquid phase, is less well known and not of significant interest in our study of thermodynamics. Water vapor can become a solid directly if the temperature drops suddenly, resulting in frost. If we could generate power during that process, we’d definitely be interested in it! None of the diagrams are drawn to scale. In order to display the various regions, the diagrams are skewed much to the right so that the various regions are observable. The saturated liquid lines in Figs. 2.10 and 2.11 are actually very close to the P-axis, and the triple point in Fig. 2.13 is actually very close to the origin. A good exercise would be to draw the 7-v and P-T diagrams for water to scale; see Problem 2.46. Before extensive tables or computer software were used to determine thermodynamic properties, phase diagrams displayed the values of several properties, and numerical values were read from the diagrams. They are often used in problem solving to help visualize a process, much like the free-body diagram used in statics. Some engineers find them very useful in understanding a problem, whereas others have no interest in them. One phase diagram that may be provided when taking the Fundamentals of Engineering Exam (FE Exam) is the pressure-enthalpy diagram (enthalpy will be defined in the next section). Fig. 2.14 shows a pressure-enthalpy diagram for the

Triple point: Where all three phases coexist. Sublimation: When a solid changes phase directly to a gas without passing through the liquid phase.

None of the phase diagrams are drawn to scale in order to make the various regions more apparent.

A phase diagram can be

constructed using any two independent properties. The F-v and P-v diagrams are uSually selected. The P-h diagram shown in Fig. 2.14 is unnecessary in this text since the information is provided in tables in the Appendix.

Part 1

Concepts and Basic Laws

juogng 0 Asayinog

200 180 160 140

120

100

80

(BTU/Ib) Enthalpy $F =0

i i

emperature

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Suva® DuPont™ HFC-134a =| llPressure

—20 800

1000

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R134a. refrigerant for diagram Pressure-enthalpy

Chapter 2

Properties of Pure Substances

—

refrigerant R134a. When presented as part of a page in a textbook, it is too small to be read accurately; if a much larger diagram is available, it can be used to read values to three significant digits.

on phase diagram

Example

2.8

The properties of the following three states are provided as (TJ, P. v). Indicate where the states are on a T-v phase diagram provided in Fig. 2.15 and determine the unknown property.

:

®D (340°C, 2000 kPa, v) @ (140°C, P,0.5 m’/kg) @ (T,5000 kPa, 0.00108 m*/kg) Solution

The T-v phase diagram is sketched as shown in Fig. 2.15. By scanning Tables C in the Appendix, state 1 is found in the superheat Table C-3 at 2 MPa with 340°C between 300°C and 350°C. That’s 80% of the difference between those two temperatures. Interpolation provides

0, = (0.13857 — 0.12547) x 0.8 + 0.12547 = 0.136 m3/kg State 2 is observed to be very close to the saturated vapor state, so it is in the mixture region. The pressure is then found next to the temperature in Table C-1 to be P, = 0.3613 MPa or 361.3 kPa.

Now, for state 3. In Table C-2 at 5 MPa (between 4 MPa and 6 MPa), observe that v, is less than v, Hence, it is a compressed liquid. Its temperature can be found in Table C-1 or Table C-4. From Table C-1 the temperature is 140°C. Table C-4 would provide essentially the same temperature.

[=_=milaiiiilitttattisitessss: Obeenniic atc: you drink in a glass, be it ice water or hot tea, is compressed liquid. It’s

T ia

5 MPa

Ne /

a a

300°C

compressed liquid if it is below the boiling point.

2 MPa

4 \

a ey

f

ae

TT

361 kPa

/

f

f:

|

Qo

(4)

ey

Part 1

Concepts and Basic Laws

2.2.1 Internal energy In Chapter 1 we considered some of the basic properties of substances, such as density, specific volume, pressure, and temperature. In this section we will add

Internal energy: The amount of thermal energy contained in a substance. Absolute zero temperature: The state in which there is no motion of the molecules.

We often omit “specific” when referring to u. In the tables in the Appendix, it may be titled simply “Energy.”

Note: A joule isa relatively small energy

unit whereas the Btu is | relatively large so that PBtU=

Example

105505

2.9

new properties to this list,which will give us information about the energy level of a substance. A key step in being able to solve problems in thermodynamics is the ability to determine numerous properties of a substance. In Chapter 1 we observed that temperature is proportional to the amount of thermal energy contained in a substance. In our study of thermodynamics, with the exception of combustion, the thermal energy contained in a substance is the internal energy. Internal energy is a derived property in that it is a function of basic properties, primarily temperature. The concept of internal energy is easier to understand if you consider the effects of changing it. Molecules move according to the amount of thermal energy contained in a substance. As the motion of the molecules increases, the internal energy increases, as does the temperature. In fact, absolute zero temperature is defined as the state in which there is no motion of the molecules. For this reason,

there are no negative energy levels in a substance. In the wet region, specific internal energy u (often referred to simply as internal energy) is calculated in the same way volume was calculated by using u = Us + x(u, — Uy)

(2:9)

The units of internal energy are the same as the units for any energy term. Consider potential energy mgz; it has units in the SI system of (kg)(m/s*)(m), or using N = kg-m/s’ (see Eq. 1.7), the units are N-m, which is called a joule (J). A joule is a small amount of energy, so we mostly use units of kilojoules (kJ). All forms of energy in the SI system use this unit. In the English system, the units of potential energy are the foot-pound force (ft-lbf), whereas the unit for thermal energy is the British thermal unit, or Btu, which is defined as the amount of heat needed to raise one lbm of water from 68°F to 69°F. The standard symbol for internal energy is U and for specific internal energy the letter uw. Specific internal energy has units of kJ/kg in the SI system and Btu/Ibm in the English system.

Find the change in internal energy The temperature of 2 lbm of water, contained in a volume of 2000 in’, is in-

creased from 212°F to 600°F holding the pressure constant, as illustrated by Fig. 2.16. What is the change in internal energy i) using the equations and ii) using the IRC Calculator? Solution

i)

First, let’s determine the initial state of the water. The specific volume is v

Vy. 2 2000/127) tt 7)

2 Ibm

= 0.5787 ft?/lbm

Chapter 2

Properties of Pure Substances

At 212°F, state 1 is in the wet region since 0.016712 < 0.5787 < 26.8 (see Table C-1E). To find the internal energy at state 1, we must use the quality. It is found to be Piston

Uy

0.5787

v, + xX,(0, — 2)

001672 “x, (26.80° —-0.01672)

“x, = 0.02099

The specific internal energy is then Uy = Uy + XU

180.1 + 0.02099 x 897.5 = 198.9 Btu/lbm The pressure at state 1 is found next to the temperature in Table C-1E to

be 14.7 psia. State 2 with P, = 14.7 psia and T, = 600°F is found in the

Figure 2.16

superheat Table C-3E, where we read u, = 1218.6 Btu/lbm. The change in internal energy is calculated to be

AU = m(u, — u,) = 2lbm X (1218.6 — 198.9) Btu/Ibm = 2039 Btu

ii) Follow the steps outlined in Example 2.7 and find at 7, = 212°F and p, = 1/0, = 1/0.5787 = 1.728 Ibm/ft* [using the specific volume from part (i)], u, = 199 Btu/Ibm and P, = 14.7 psia.At T, = 600°F and 14.7 psia, we obtain u, = 1220 Btu/lbm. We have

AU = miu, — U,) = 2lbm (1220 — 199) Btu/lbm = 2042 Btu

We often refer to the saturated-mixture region as the wet region.

Remember: Density is input using the IRC

The difference between the two answers is negligible.

Calculator, not the specific

volume.

2.2.2 Enthalpy When working with equations in thermodynamics, we often encounter the combination u + Pv of properties. Because this happens so often, especially when analyzing devices involving a flowing working fluid, such as steam flowing through a turbine, we define enthalpy H to be H =U

+ PV

(2.10)

Specific enthalpy, often referred to simply as enthalpy, is then found by dividing by the mass:

wu

4°Ro

(2.11)

It simplifies calculations since three properties are combined into one. In fact, in engineering practice, enthalpy is used more often than internal energy, so a table may list enthalpy but not specific internal energy. If that is the case, Eq. 2.11 is used to calculate specific internal energy. Enthalpy is also found to be a measure of the amount of energy in a substance in some instances, especially in a flowing fluid. If high-pressure vapor or air is allowed to flow over a set of turbine blades, the flow will cause the blades to rotate and

Ss

a

Enthalpy: The combination of properties (U + PV) occurs quite

often. It is particularly useful when analyzing devices through which a fluid flows, such as

a turbine. It simplifies calculations. Units: The units on Pu are

(kN/m?)(m?/kg), which are equivalent to kJ/kg, the same as the units on u.

“

Part 1 Concepts and Basic Laws

State postulate: If two intensive, independent properties of a simple pure substance are given, all others can be determined.

create work, which will be shown to be related to the enthalpy change. Enthalpy is the amount of thermal energy contained in a substance plus the potential of that substance to do work. Enthalpy has the same units as internal energy, kJ in the SI system and Btu in the English system. Similarly, specific enthalpy has units of kJ/kg (Btu/Ibm). Solving problems in thermodynamics is like solving a mystery. You are given some clues, and from those clues you have to determine what happened and in what order it happened. The state postulate maintains that if two intensive, independent properties of a simple pure substance are given, all others can be determined. So we need two “clues” about a state in order to determine all other properties of that state. In textbooks, you are typically given the exact information needed to solve a problem. In real life, you may have too little or too much information. But what information is

not of interest, or what information is necessary that is not available? Experience is often required to make the assumptions required for a possible solution.

Example

2.10

2 kg Water

The change in enthalpy

=

The temperature of 2 kg of saturated liquid water contained in a volume of 9000 cm? is increased from 60°C to 600°C holding the pressure constant, as illustrated by Fig. 2.17 What is the change in enthalpy? Solution

The initial enthalpy of the saturated liquid water is found in Table C-1 in the Appendix at T, = 60°C as h, = h, = 251.1 kJ/kg. The pressure is noted to be P, = 19.94 kPa, which is close enough to 20 kPa. At 600°C a pressure of 0.02 MPa is not a table entry, so an interpolation is required to find h,. It is h

Figure 2.17

=

_= 0.02 — 0.01 005 = 0.01

(3705.1 — 3705.4) + 3705.1 = 3705 kJ/ke

Notice that the enthalpy is insensitive to the pressure in the superheat region and we could have observed that its value is 3705 kJ/kg without interpolating. Finally, the change in enthalpy is

AH = m(h, — h,) = 2kg X (3705 — 251.1) kJ/kg = 6908 kJ

Note: We are using water

as the working fluid to establish the relationships that will be used and

the techniques used to | determine properties of phase-change substances. In the problems, refrigerants R134a and ammonia will also be of interest.

2.2.3 Internal energy and enthalpy for liquids and solids When liquid water is compressed with a pump, as it is in most power plants, it is important to determine its state. As mentioned earlier, Tables C-4 and C-4E provide the properties of compressed water. By studying the entries in those tables, we observe that all the entries, with the exception of enthalpy, change very little with pressure when the temperature is held constant. For example, at 80°C, the percentage changes in v, u, and s (s will be introduced in Chapter 6) between 5 MPa and S50 MPa are at most 2.8%, a relatively small percentage change

and

within

engineering

accuracy.

Only h experiences

a significant

Chapter 2

Properties of Pure Substances

change between those two pressures, which is expected since h depends directly on the pressure (h = u + Pv). Consequently, we can assume that the

et

properties, except h, are dependent only on the temperature and assume that,

Ti Tae a Ne OH

for liquids, v = Up, U = Uy, and s = s, where the saturated liquid values with the f subscript come from Tables C-1 and C-1E when water is of interest. The

pressure fora liquid when the temperature is held constant,

enthalpy is then calculated using h = u + Pv. In fact, because the compressed

liquid table is not needed, it is often omitted in tables of properties for phasechange substances. The properties needed to analyze the change of phase of water between a saturated solid and a saturated vapor are found in Tables C-5 and C-5E. For ice, this represents ice changing phase directly into a vapor; an example is snow changing phase directly into vapor without melting. It’s a very slow process but real. How do clothes hung on a clothesline in the winter dry when the temperature is below freezing? First, the water freezes and then the ice changes phase into a vapor and dry clothes appear the next day. The ice does not melt; it sublimates, that is, changes phase, into a vapor. Examples will illustrate use of tables for liquids and solids. Water will be the working fluid.

Properties of

aliquid

_—&xCept use Eq. 2.11 for (erey ee

La

Example 2,11

A ARS

Determine u and h for the water in Fig. 2.18 at 60°C and 10 MPa using i) Table C-4, ii) the saturated liquid values from Table C-1, and ii) the IRC Calculator. Solution

i)

©

10 MPa

Ata temperature of 60°C and a pressure of 10 MPa, the water is a compressed liquid, so Table C-4 is used to find the desired properties. To four significant digits, they are

u = 249.4kJ/kg

and

h = 259.5 kJ/kg

ii) Using the saturated liquid values from Table properties are

C-1, we find that the

u = u,= 251.1kI/kg and h = h, = 251.1kI/kg A more accurate value for / must include the Pv term. It is

h = ug + Po, = iiall = + S

Figure 2.18

; 10000 se x 0.001017 a = Wil} kJ/kg m Baie A ei

iii) The IRC Calculator provides u = 249 kJ/kg and h = 260 kJ/kg. Obviously, when determining wu values and h values for a compressed liquid, using any of the three sources provides acceptable values, but when calculating / using Table C-1, the pressure term Pv must be included, especially if the pressure is relatively high.

Note: If P= 5 MPa, the Pu term must be included in

the calculation of h.

as Units: kPa = kN/m?

vy

Part 1

Concepts and Basic Laws

Example 2.12

Find a property change of a solid DET

of 2 Ibm of water at 2°F between

Determine the change in internal st saturated vapor and ice. Solution

The saturated ice-vapor Table C-5E is not sensitive to the pressure, so only the temperature is needed. At a temperature of 2°F, interpolation is required since 2°F is not a direct entry. Also, u;,,which is the change between ice and saturated vapor, is an entry in the table so interpolation provides, at 2°F,

u,— u,= U, a — >(11688 — 1169.5) + 1169.5 = 1169.2 Btu/Ibm

lI

Actually, we observe that u,, is insensitive to temperature, so a value of 1169 Btu/lbm could have been selected by observation. That value would be accurate enough for engineering calculations. Then

AU = mAu = 2lbm X 1169 Btu/lbm = 2338 Btu

Latent heat: The change in enthalpy during a change of phase. Heat of vaporization: The energy required to vaporize a unit mass of saturated liquid. Heat of fusion: The energy required to melt (or freeze) a unit mass of

a substance.

Heat of sublimation: The energy required to completely vaporize a unit mass of a solid.

2.2.4 Latent heat The change in enthalpy at the saturated conditions of the two phases experienced during the change of phase of a substance at constant pressure is called /atent heat. The heat of vaporization is the energy required to completely vaporize a unit mass of saturated liquid, or condense a unit mass of saturated vapor; it is equal toh, = h, — h, The energy that is required to melt or freeze a unit mass of a substance at constant pressure is the heat of fusion and is equal to h, — h;, where h,is the enthalpy of saturated liquid and h, is the enthalpy of saturated solid. Sublimation occurs when a solid changes phase directly to a vapor; the heat ofsublimation is equal to h;,, = h, — h,, The heat of fusion and the heat of sublimation are relatively insensitive to pressure or temperature changes. For ice, the heat of fusion is approximately 330 kJ/kg (140 Btu/lbm) and the heat of sublimation is about 2840 kJ/kg (1220 Btu/Ibm). The heat of vaporization of water is very dependent on temperature and is included as hy in Tables C-l, C-1E, C-2, and C-2E.

Example

2.13

What's the energy needed to melt an ice cube? PORE DONESEIOENADEL

SE

The ice cube in Fig. 2.19 is 1 in. on a side. How much energy is needed to melt the ice cube if its initial temperature is 32°F? Solution First, let’s determine the mass of the cube of ice. In Table C-5E, using v; at 32°F, we find

“Vy he

eee O04 ibm

set fe oo

=

Chapter 2

Properties of Pure Substances

wil

Ice cube

Now, what is the heat of fusion for ice, that is, what is the energy to melt the ice? It is the difference between the enthalpy of water h; at 32°F and the enthalpy of ice h, at 32°F. The enthalpy of water h, is found in Table C-1E to be 0.01 Btu/lbm. The enthalpy of ice h,; is found in Table C-S5E to be — 143.34 Btu/lbm.The difference is hy —

h, = 0.01 —

eee (—143.34)

= 143.3 Btu/lbm

Suggestion: Alternately, the value of 140 Btu/lbm

The energy needed to melt the ice is found to be mh, — h,) = 0.0331 lbm x 143.3 Btu/lbm = 4.74 Btu

given in the text material

could have been used with

less than 2% error.

(5)

The working fluid considered in the previous two sections has been water. Refrigerants are another class of working fluids that change phase. We have no interest in the solid phase, only the liquid and the vapor phases. Refrigerant 134a is

acommon refrigerant for automobile air conditioners, household refrig-

erators, and air conditioners. It is marketed under the names R134a, HFC-134a, Genetron134a, and Suval34a; it is the chemical compound Tetrafluoroethane

and has the chemical formula CH,FCF,. R134a is a more environmentally friendly replacement for Freon, the refrigerant R12 (CCI,F,), which was found to have harmful effects on the ozone layer of the atmosphere. It is now illegal

in the United States to use or design equipment that uses R12. The property

tables for R134a are in Appendix D and are organized in the same manner as) the steam tables, with the exception that the compressed liquid table and the solid-vapor table are not of interest. R134a has also been found to have negative environmental effects, so in a few years HFO-1234yf (Tetrafluoropropene) will be the refrigerant replacing R134a in automotive air conditioners. An advantage in using HFO-1234yf is that it is much

Note: Freon was used for anumber of years

in airconditioners and

ae by R134a.

Part 1

Concepts and Basic Laws

SRE Obes auon: engineer can make coolout. of heat! Google

“absorption refrigeration”

jess likely to catch fire in the event of an accident. Europe has required its use since 2011. General Motors plans to begin marketing cars using HFO-1234yf, possibly in 2013, although there is significant opposition by the air-conditioning sector, so that date may be postponed. Its performance as a refrigerant is similar to that of R134a.

It tends to have a higher vapor density and a lower saturated vapor enthalpy.

The first refrigeration system utilized ammonia as the refrigerant. The system

eS

Ammonia poses ahigh

hazard, soitisnot health used in residential and MERE

Tiida coolications OR END SEO OED,

Example

2.14

was invented by Dr. John Gorrie in Apalachicola, Florida, to help patients recover from yellow fever in a local hospital. In large commercial applications, ammonia 1s often used as the refrigerant, especially in absorption-refrigeration systems in which cooling is generated using waste heat from a power plant. This is especially popular

ny college campuses that have their own power plants. Appendix E provides the

: a ‘ : properties needed to solve problems that utilize ammonia as the refrigerant. The superheated properties are presented using a different format that may be encountered in other property tables. It’s wise to become familiar with various forms of tables.

The enthalpy change of R134a A piston maintains a constant pressure of 500 kPa, as shown in Fig. 2.20. Determine the temperature and the enthalpy change required to completely vaporize the 2 kg of saturated liquid R134a. i) Use the tables and 11) the IRC Calculator. Solution

i)

Remember: The numbers in the tables are curve fits,

and expressing an answer

to three significant digits is

Bee eceep lacie

From Table D-2, the temperature at 500 kPa is 15.74°C, which stays constant until vaporization is complete. The enthalpy change required to completely vaporize (boil) a kilogram of R134a at 500 kPa is h, = 184.74 kJ/kg. For 2 kg, the enthalpy change is 2 X 184.74 = 369 kJ.

ii) Use the IRC Calculator: Select International Units, R134a, Pressure, and

Quality. At 500 kPa and x = 0, the temperature is 15.7°C and h, = 73.4 kJ/kg.

At 500 kPa and x = 1,h, = 259 kJ/kg. For 2 kg, the enthalpy change for complete vaporizations is

Note: To an engineer, this is essentially the same answer asin (i).

AH = mAh I] 2kg X (259 — 73.4) kI/ke = 371 kJ 2 kg R134a

Chapter 2

Properties of Pure Substances

The enthalpy change of ammonia

ea

Example 2.15

Determine the temperature and the enthalpy change required to completely vaporize the 2 kg of saturated liquid ammonia shown in Fig. 2.21 at a constant pressure of 500 kPa. i) Use the tables, and ii) the IRC Calculator. These are the same conditions as those of Example 2.14. Solution

i)

From Table E-1, the temperature at 500 kPa is 4°C, which stays until vaporization is complete. The enthalpy change required to completely vaporize a of ammonia at 500 kPa (4°C) is hy, = 1248 kJ/kg. For 2 kg, the change is 2 X 1248 = 2500 kJ. ii) Use the IRC Calculator: Select International Units, Ammonia, and Quality. At 500 kPa and x = 0, the temperature is 4.14°C

constant

comment.

26

At 4°C the pressure Is very

kilogram enthalpy

Se enn er sects

Pressure, and h, =

Wis deed ho ordec ance

200 kJ/kg. At 500 kPa and x = 1, h, = 1450 kJ/kg. For 2 kg, the enthalpy

the pressure is so close to

change for complete vaporization is

500 kPa.

AH = mAh

= 2ke-x

(1450 — 200) kJ/kg = 2500 kJ

2 kg ammonia

Double interpolation of R134a properties Refrigerant R134a exists at 73 psia and 175°F. Determine the specific volume and the specific enthalpy. Use the tables.

Solution Appendix D-3E lists the properties for superheated R134a. Tables exist for pressures of 70 psia and 80 psia, so interpolation for 73 psia is necessary. The tables also list properties for R134a at temperatures of 160°F and 180°F, so another interpolation for 175°F is also necessary, resulting in a double interpolation. (Continued)

Example 2.16

—

Part 1. Concepts and Basic Laws

Example

2.16

(Continued) SQUEEZE

The following table shows Appendix D-3E:

the properties at the four listed states from

P = 70 psia

T = 160°F

v = h = v= h =

T =180°F:

0.8752 133.82 0.9103 138.62

P = 80 psia

ft*/Ibm Btu/Ibm ft*/lbm Btu/Ibm

v h v h

= = = =

0.7584 133.41 0.7898 138.25

ft°/lbm Btu/lbm ft*/lbm Btu/Ibm

First, at T = 160°F, interpolate for a pressure of 73 psia to find v, and h;:

0, = 8192

Units: All units on the left

are ft?/Ibm, and all units on

O7584

the right are psia.

=

= 0:8752>—

ht, =

133.82

133.41 — 133.82

jem

i)

“0, = 0.8402 ft?/Ibm

80 =/0

=

43

40)

80 — 70

oh, = 133.7 Btu/lbm

Then, at T = 180°F, interpolate for a pressure of 73 psia to find v, and h,:

v, - 0.9103 73 — 70 = 0.7898 — 0.9103 80 — 70

2p,

hy-= 138.62 2 13825. = 138.02.

hy = 138.5 Btu/Ibm

= 70 80. 90

ae

: = 08142 fE/Ib ®2 /Abm

Next, interpolate to find the properties v and h at 175°F: Note: Three rather than four significant digits

|

are written since the

v — 0.8402

0.8742 — 0.8402

straight-line interpolation

:

;

a : =

method loses at least one significant digit.

1385

; =-1337

175 2 160

180 — 160 bs

a

= 13k > 30-70

“v = 0.866 ft?/Ibm Pe “Ah =

135 Btu/Ibm

This is a rather tedious procedure. Engineering design requires the use of numerous tables, and interpolation is often required. Double interpolation is not encountered very often in the home problems in this text, but it may be needed when working a problem on the FE Exam, an exam required for becoming a professional engineer. Programmable calculators and computers are not allowed on that exam. The IRC Calculator described in Section 2.1.4 is a much quicker way to obtain properties. The software does all the interpolation for you. It provides v = 0.863 ft*/lbm, h = 138 Btu/lbm, which is more accurate than the double in-

terpolation results. However, both are acceptable for engineering calculations.

™ You have completed L«

gum

«2:4 Ideal-GasLaw

This law isbasedon ions; eee

— The ideal-gas law is undoubtedly a familiar law from courses in chemistry and Hinds : ie ; : physics. It is an empirical equation of state based on experimental observa-

i

beso bs Le

|

STOO

II

(6)

:

ES.

tions of the behavior of gasses at relatively low pressure. An equation of state

Chapter 2

Properties of Pure Substances

ate

is an algebraic equation that relates two or more thermodynamic properties. The ideal-gas law relates pressure, temperature, and specific volume in a very simple algebraic equation. The most common form of the ideal-gas law is

Pv = RT

(2.12)

Air: R = 0.287 ki/kg-K = 53.34 ft-lbf/lbm-°R

where P must be the absolute pressure of the gas, v is the specific volume, R is the gas constant, and T is the absolute temperature. The ideal-gas law is also called the ideal-gas equation of state, or the ideal-gas equation. The gas constant is different for each gas and is related to the universal gas constant R,, which is the same for all gases, by

R=—

(2.13)

where M is the molar mass of the gas, tabulated in Table B-2 in the Appendix for a number of gases. The molar mass of oxygen is 32, meaning that there are 32 kg

Nitrogen: R = 0.297 ki/kg:K R = 55.15 ft-lbf/lom-°R

= 8.314 ki/kmol-K 1545 ft-lbf/lbmol-°R 1.986 Btu/lbmol-°R

10.73 psia-ft?/lbmol-°R

in one kmol, or 32 Ibm in one Ibmol.

Other forms of the ideal-gas law that are used are PV = mRT

P = pRT

PV = NR,T

(2.14)

where N is the number of moles (kmol if using SI units and Ibmol if using English units).

Pressure of anidealgas STW

Example 2,47

STEIN

Ten kilograms of oxygen exist at 20°C in the 1.2-m* container of Fig. 2.22. Determine the pressure in the container.

Solution

:

E

The pressure and temperature are given so that the ideal-gas law of Eq. 2.14 (the first version) provides

Note: In thermodynamics,

B oressure is absolute, unless otherwise stated.

— mRT

V =

poke

200

.

LE 12m:

.

Se

+-273)

@

K

= 635 kPa ey

The temperature must be absolute, and the pressure determined is also absolute. If R were used as 260 J/kg:K, the pressure would be in Pa rather than kPa.

Units: Use kJ = kN-m when writing the units on Rk.

“

Part 1

Example

Concepts and Basic Laws

2.18

The volume of an ideal gas ETE DTCELLED

ADE n NE

Ten pounds mass of helium exist in a container at 110 psia and 60°F. What is the volume of the container?

Solution The gas constant for helium is found in Table B-2 to be 386 ft-lbf/lbm-°R. The ideal gas law of Eq. 2.14 is used to find the volume: mRT

VS

ft-lbf

10 lbm xX 386 ibmoR

=

a

HO

1

in

Compressibility factor: A factor Z that accounts for real gas behavior.

Generalized compressibility chart: It provides the factor Z for all gases.

x (60 + 460)°R

=

= 127 ft

ft

The ideal-gas law can be modified to better model real gases by the use of a compressibility factor. The standard symbol for the compressibility factor is the letter Z and is defined by ere

(2.15)

For an ideal gas, Z = 1. A graph of the compressibility factor for nitrogen as a function of the pressure and temperature is shown in Fig. 2.23. A generalized |

| |

|

110k |

Critical

| | |

NEN

a)

point

|

Pressute (MPa)

Compressibility factor for nitrogen.

=

‘

Chapter 2

Properties of Pure Substances

compressibility chart that provides the factor Z for all gases is included as Figs. H-1 and H-2 in Appendix H. The generalized charts use reduced temperature Tp, reduced pressure Pp, and pseudo-reduced specific volume v rp» defined by P

P jek

a

Se

Tr

=

jk i.

v oR Po

—

=>S

br Observe: The reduced properties are nondimensional properties.

Gls)

where the critical-point constants P., and T., are found in Table B-3. The quantity Up is called a pseudo-reduced specific volume because it is not defined as v/v... There are more sophisticated real gas equations of state that also relate pressure, temperature, and specific volume but use algebraic equations that require more than one constant.A relatively simple equation is the van der Waals equation,

P= hyPOane aa

(2.17)

Le Van der Waals equation: An equation of state that accounts for real gas behavior.

where the constants a and b are related to the critical-point values found in Table B-3 by

ees

* 64P

_

Fe

cr

(2.18) cr

The numerical values of the constants a and b are presented in Table B-10. Another real gas equation of state is the Beattie-Bridgeman equation, which offers more precise modeling of gases:

ie dl D

or

b

A

a)

ar

“)

(2.19)

10)

Pa

where 7 is the molar specific volume defined by 0 = vM. (A bar over a variable signifies a per mole basis.) The gas constants Aj, By, a, b, and c are included in Table 2.3 for four gases. In general, the more gas constants in the model, the better the fit between the state equation and the actual relationship of the properties. Another real gas equation of state that should be mentioned is the virial equation of state. It is expressed as the series =

see

a)

0

OG

=

Be 3

Units: The units on a are

Pa-m®/kg’ and on b the units are m*/kg.

Molar specific volume: tt is defined by the relationship D = 0M, where M is the

molar mass (see Table B-3).

(2.20)

A

oO

The constants a(7), b(T), and so on must be given if this virial state equation is used. It is not used in this text and is presented for information purposes only. The following example compares the pressure determined by the van der Waals and the Beattie-Bridgeman models to that determined by the ideal-gas law.

Table 2.3.

The Constants in the Beattie-Bridgeman Equation

Gas

Air Nitrogen Oxygen Carbon Dioxide

Ay

By

a

b

131.84 136.23 151.09 507.28

0.04611 0.05046 0.04624 0.10476

0.01931 0.02617 0.02562 0.07132

0.00110 0.00691 0.00421 0.07235

—SSE LL

c

4.34 4.20 4.80 6.60

x x x x

—————————eeee

Note: The constants in Table 2.3 have units » requiring P to be in kPa,

10° 10° 10* 10°

~ Din m?/kmol, Tin K, and R,, in kJ/kmol-K. wee

rs

Part 1

Concepts and Basic Laws

Example 2.19

The pressure using three equations of state Forty kilograms of nitrogen exist at 600°C in the 2-m° container of Fig. 2.24. What is the pressure in the container? Use i) the ideal-gas law, ii) the van der Waals equation, and iii) the Beattie-Bridgeman equation.

Solution i)

Figure 2.24

The gas constant for nitrogen is found in Table B-2 to be 0.297 kJ/kg-K. The ideal-gas law of Eq. 2.14 is used to find the pressure: ARI.

40kg

X 0.297 kJ/kg:-K

x (600 + 273)°R

= 5190 kPa

p=

ees

2

ii) The two constants in the van der Waals equation are calculated (or look

them up in Table B-10) to be

DRT 64P.

a=

=

RT b=—=

27 207 10627 = 174.8 Pa-m®°/ke* 64 x 3.39 x 10° 297 X 1262

8P.

8 x 3.39 x 10°

= 0.001382 m3/k

/kg

The van der Waals equation, with v = 2/40 = 0.05 m*/kg, gives see

iS

Ses

@ =

1p)

vu

297-873

oa Observe: The Z-factor of Fig. H-2 in the Appendix

would be

greater than 1.

itis en Fea on

accurate value of Z from

Ay = 136.23, By = 0.05046, a = 0.2617, b = — 0.00691, c = 4.2 x 10° :

:

:

gives

eee

RE =

ol

S314

anarad Results:

pee

The Beattie-Bridgeman equation, with 0 = Mv = 28 x 0.05 = 1.4 m*/kmol,

v

sean

foe = 52610"

iii) The Beattie-Bridgeman equation constants from Table 2.3 are

the chart.

-

174.8

G00

= O05

ive

873

136.23

AD SOTO"

(1

14 X873°

b

A

0

v

v

ju + 0.0s046( 1+ a)

|

0.2617

aa 1.42 (

14

)= 5316 kPa

Ideal-gas equation: P =

Wane noe ie 5.26 MPa Beattie-Bridgeman: P = 5.32 MPa

Compared to the Beattie-Bridgeman equation, which is considered the most accurate, the ideal-gas law has an error of —2.4%. An ideal-gas assumption is good at relatively low pressures; 5 MPa is relatively low. Because pressures exceeding 5 MPa are not encountered that often in our study, the ideal-gas law is most often used.

¥ You have completed

Learni g

Ou con .

|

(7)

Chapter 2

Properties of Pure Substances

Pa

2.6 Internal Energy and Enthalpy Bas, The properties of internal energy and enthalpy for an ideal gas are calculated using properties called specific heats. The specific heat of a substance is the amount Specific heat: The amount of energy needed to raise a unit mass of that substance one degree. Textbooks in _ of energy needed to raise other fields may use the term heat capacity rather than specific heats. a unit mass of a substance

For a simple compressible system, only two independent variables are nec-

—_°"€ degree.

essary to establish the state of the system. Consequently, we can consider specific internal energy to be a function of temperature and specific volume; that is, u = u(T,v).Using the chain rule from calculus, we express the differential in

terms of partial derivatives as

P P Tee (=) av + (%) WA oT

v

v

(2.21)

fi

eee Constant-volume specific

heat: C, = (du/a7),

Since u, v, and T are all properties, each partial derivative is also a property; the first one on the right-hand side is defined to be the constant-volume specific heat C,; that is,

Ee =

Ou ae

(2.22) Thermometer

One of the classical experiments of thermodynamics, performed by Joule in 1843, is illustrated in Fig. 2.25. A high-pressure volume with an ideal gas is connected to an evacuated volume, as shown. After equilibrium is attained, the valve is opened. Even though the pressure and volume of the ideal gas have changed markedly, the temperature of the water does not change. If there was no change in the water temperature, there was no energy flow to the water, so the internal energy of the air must not have changed. Because the pressure and specific volume of the air certainly changed, Joule concluded that the internal energy of an ideal gas does not depend on pressure or volume so that, for an ideal gas,

High-pressure air

eee

eo Evacuated — Ban

A sketch of the experimental setup used (=) = 0

(2.23)

by Joule.

dv)

For a gas that behaves as an ideal gas, Eqs. 2.21 and 2.22 allow us to write du = C,dT

(2.24)

Similarly, if we consider enthalpy to depend on the two properties T and P, that is, h = h(T, P), we can express an infinitesimal change in enthalpy as J} dh = (#) dT + (=) dP

a

np

(2.25)

ee Constahinressure epeciiit

We define the constant-pressure specific heat to be

heat: C, = (ah/aT),

=

oh

C, A, ee

7 (2.26)

ii

=Part1

Concepts and Basic Laws and, since h = u + Pu = u + RT, assuming an ideal gas, we observe that h is also only a function of temperature; hence, dh/dP|, = 0. With the definition of C,, we see that

dh

gt

(2.27)

Comment

The specific heats are properties and do not

require constant volume or constant-pressure processes when used. That is simply what they have been named.

The changes in internal energy and enthalpy needed to change from state 1 to state 2 can now be found for an ideal gas by integrating Eqs. 2.24 and 2.27: T,

Au = Us.— U, = |car T,

(2.28)

T,

Ah I =

There are five methods used to find Au and Ah.

=

II

Gy aX I

Like most physical properties, C, and C, vary with the temperature of the substance. However, if the temperature change for a process is not large, they can be assumed to be constants. Constant values for C, and C, for temperatures around 300 K are shown in Table B-2 in both sets of units. If a large change in temperatures is encountered, or if the temperatures are relatively high, there are five methods to find Au and Ah for an ideal gas. 1.

Use the ideal-gas tables included as Appendix F, each of which lists u(7) and h(T). This method is relatively simple, and it is the most accurate.

2.

Assume constant specific heats from Table B-2 and perform the integrations in Egs. 2.28 to obtain

u, — u, = CP, h,—

— T;)

(2.29)

hy = C,(T5 oat)

Using this method, the temperature should not exceed about 500°C (900°F), which is the case for many of the problems we will encounter. It is used, however, for comparison purposes when analyzing the performance of power cycles in engines with much higher temperatures. 3.

Table B-6 lists the values of C, and C, over a wide temperature range. The procedure for using them is to calculate the average temperature for the process and use the specific heat at that average temperature. Then use Eqs. 2.29 to find the desired quantity.

4.

Use an equation for the specific heat C, that is a direct function of the temperature. Table 2.4 lists the coefficients of an equation for C,, that is a cubic function of temperature:

Cae Pp The units on C,, are

sometimes listed as kJ/kg:K and sometimes as kJ/kg-°C. These are equivalent since the size of 1°C is the same size as 1 K. Likewise, we use

Btu/Ibm-°R or Btu/lbm-°F.

lable 2.4__C, (in W/kg:K)as.4 Gas

Air Oxygen Nitrogen

a

0.9703 0.7965 1.0317

lieteC

ee lle

(2.30)

Function of Absolute Temperature Cc

by

6:8. 10 ATG X10 =5.6

d

166.6 1077 > 6.85 10-8 = 2241 0a ee Ate ae 2,88.5010" = 02 x10. 4

ee

Chapter 2

Properties of Pure Substances

where C, is in kJ/kg-K and T is in kelvins. The constants a, b,c, and d have dimensions in order to make Eq. 2.30 dimensionally homogeneous. Molar values to calculate cerfor numerous gases are included in Table B-5. 5.

Use the IRC Cine

for air as described in Example 2.7.

Let’s find the relationship among C.,, C,, and the gas constant R that exists for an ideal gas. Return to the definition h =u + Pv in Eq. 2.11 and write the differential form as dh = du + RdT

(2.31)

using d(Pv) = d(RT) = RdT, since R is a constant. Now, substitute the expressions for dh from Eq. 2.27 and du from Eq. 2.24 and divide by dT and obtain

Coa Car

(2.32)

Now we know why only C, is listed in many tables. If C,, is known, C, can be easily determined using Eq. 2.32. Since C, is most often used, it is the one listed. If the definition of h on a per mole basis was used, we would find

CA Ca:

(2.33)

Often the functional form similar to Eq. 2.30 is given on a per mole basis so that the units on @ are kJ/kmol-K.

Before we leave this section, let’s define another property that is related to the specific heats, the ratio of specific heats k: G 2) i

Té

mS : os The ratio of specific

(2.34)

i

heats: It is defined as

k= GJC,.

It occurs quite often when working with ideal gases, so it is usually listed in tables. It also allows, with Eq. 2.32, to express C and C,, as

ia R ee el Ge

kR

eS Omar HE gear

2.35 (2.35)

These relationships will.be used to simplify expressions.

Estimate C,ofsteam

Example 2.20

eR Estimate C, of steam at 6 MPa and 600°C. Use i) a forward-difference method, fe a backward-difference method, and iii) a central-difference method.

Solution i)

A forward-difference method is based on the entries at a forward state and the desired state—in this case, at the same pressure as required by Eq. 2.26. We use the definition

Ah

(3894.2 — 3658.4) kJ/kg

hil ee we £5 (900-— G00)

= 2.358 kI/kg-°C (Continued)

NE

eae

Part 1

Example

Concepts and Basic Laws

2.20

(Continued) ENDED

ii) A backward-difference method is based on the entries at the desired state and a backward state. We have

C= Ah NEP Se

_ 3658.4 ~ 3540.6 _ 2.356 kI/kg-°C 600 — 550

iii) A central-difference method is based on the entries at states on each side of the desired state at an equal temperature difference. The method gives

CS

Ah

=

77

3894.2 — 3422.2

100

500

= 2.36

kJ/kg-°C tke

To three significant digits, each method gives C, = 2.36 kJ/kg-°C. If C, were more sensitive to temperature change, as in Fig. 2.26, the difference would be significantly larger and the central-difference method would obviously be the most accurate. The desired C, is the slope of the curve at 600°C. h

Forward difference

Backward

difference

Central

|

difference

Example

2.21

Find AH of air using five methods SE LDLLIED

OSS Suggestion: To interpolate, observe that from 660 K to 680 K, Ah = 21.4 kJ/kg

Estimate the change in the enthalpy of 5 kg of air if the temperature changes from 27°C to 400°C. i) Use the air table, Table F-1. ii) Assume the constant specific heat from Table B-2. «86 ili) Use an average specific heat found in Table B-6. iv) Use Eq. 2.28. v) Use the IRC Calculator. P

so that’s a little more than

Solution

1 kJ/kg per degree. So add

The temperature 27°C is equal to 300 K, and 400°C is 673 K. Remember:

os to 670.47 and

Loe eu Three significant digits are

Enthalpy does not depend on pressure for an ideal gas.

i) In Table F-1 we find h, = 300 kJ/kg as a direct entry and interpolate to find h, = 684 kJ/kg, There results

:

fine. When you interpolate,

think alittle first.

AH = m(h, — h,) = 5kg X (684 — 300) kJ/kg = 1920 kJ

Chapter 2

Properties of Pure Substances

ii) In Table B-2, we find C, = 1.003 kJ/kg-K. This provides

A

mC (1,

1.)

kJ =) Keg x 7003 ke-K x (673 =— 300) K == 1870kJ iii) The specific heat for an average temperature of (673 + 300)/2 = 486 K, found in Table B-6, is 1.026 kJ/ kg-K (this is estimated since the variation is slight, and we are not interested in extreme accuracy). We find

AH = mC(T, — T,) kJ x = 5kg g X 1.026—— ~ 1910 kJ kek * (673 a~ 300)K= iv) The coefficients in the cubic equation used to estimate C, are found in Table 2.4. They are used in Eq. 2.28 and integrated to give T,

T>

ae m|T, C,aT = m|1 (a + bT + cI’ + dT*)dT Die = 5 kg|0.9703(673 — 300) + 6.8 x 10° ee

+1.66 x 1077

673°

Ss

3

0

P

ee

ee

2

Aen

4

4

ke

= 1950'kJ v) Using the IRC Calculator selecting “Dry air” at 27°C (select P, = 100 kPa since it doesn’t matter), h, = 300 kJ/kg, and at 400°C we find h, = 685 kJ/kg. Hence,

AH = m(h, — h,) = Skg X (685 — 300) kJ/kg = 1925 kJ

The answer from the air table in (i) is considered the most accurate. The errors in the other parts are found to be (ii) —2.6%, (iii) —0.5%, (iv) 1.6%, and

(v) 0.26%. All answers are within engineering accuracy. The answer in (ii) is probably the one most often used, providing AT < ~ 500 K and T, or T, is near standard temperature.

2.7 Specific Heats of Liquids The volume of a liquid or a solid changes very little as the temperature and pressure change, even though a very small change can lead to large stresses in a solid under certain conditions and to significant movement in a liquid, such as the expansion of a liquid in a thermometer. In a later chapter, we will show —that is, one for which the specific that C, = C, for an incompressible substance volume does not change during a process. In general, we make the assumption

Part 1 Concepts and Basic Laws iit Say ON a Ra ae

Liquids and solids: We usually make the assumption that C, = C,.

We most often assume

that C, = C, for a liquid and a solid. Often the subscript is simply dropped, and C represents the specific heat for liquids and solids. In most thermodynamic books, C, will be used. For water we use C, = 4.18 kJ/kg: ‘K (1.0 Btu/Ibm-°R). For ice C, == 1 + 0.00697 kJ/kg-°C (0.47 + 0.0017 Btu/Ibm-°F), with T measured ina “(or °F). The specific heat does not vary significantly with pressure, except for very special situations, and will be considered independent of pressure in this introductory text. The specific heats of some common liquids and solids are presented in Table B-4 in the Appendix. For most applications to solids and liquids, Eq. 2.29 provides the internal energy change as

the constant specific heats

i

found in Table B-4 as Se

1, Shy

fy

CO

(2.36)

unless there’s a large pressure change with liquids, in which case the Pv term must be included.

Example 2.22

Find AU of ice SSIES

Five pounds of ice exist at O°F in the open container of Fig. 2.27 Estimate the change in the internal energy of the ice if the temperature changes from 0°F to 180°F. Solution

SS

O

C]

O Di

Integrate C, = 0.47 + 0.0017 Btu/lbm-'F to find the internal energy change of

the ice between 0°F and 32°F. Assuming C,, = C.,, AU = mAu

CI =

22

(0.47 + 0.0017)dT= s( 047 Des

VAR

0.001 ae

xe ‘)= 77.8 Btu

The ice is now at 32°F at which state it melts, requiring the heat of fusion, 140 Btu/Ibm: AU melt = ™

Figure 2.27

X (heat of fusion) =

5lbm X 140 ele = 700 Btu Ibm

If the temperature of the water is raised from 32°F to 180°F the internal energy change is AO

CGwater Ai Pte Suliman 1

=

x (180 — 32)°F = 740 Btu

The overall change AU = 778 + 700 + 740 = 1518 Btu. AH = AU tor this problem since pressure was held constant and the change in volume was considered insignificant. In general, A(Pv) = 0 for a solid or a liquid. oo

/% You have completed Learning Outcom

(8)

Chapter 2

Properties of Pure Substances

In this chapter we considered the different phases in which a pure substance exists and introduced properties of fluids used in power generation and refrigeration systems. The equations of state for an ideal gas and state equations for real gases were presented. Enthalpy and internal energy for an ideal gas, a solid, and a liquid were determined with the use of tables and specific heats. Numerous additional terms were defined: Compressed liquid: The temperature is below the boiling point. Compressibility factor: A factor Z that accounts for real gas behavior. Dependent property: One that cannot vary if the other property is held constant. Fluid: A substance that flows when subjected to a shear stress, no matter how small that stress may be. Fusion:

The process of a liquid becoming a solid.

Heat of fusion: The energy required to melt a unit mass of a substance. Heat of sublimation: of a solid.

The energy required to completely vaporize a unit mass

Heat of vaporization: The energy required to completely vaporize a unit mass of saturated liquid. Law: The summary of the results of repeated observation. Quality: The mass of the vapor divided by the total mass of a mixture. Saturated liquid: The state in which a substance is just starting to vaporize (boil). Saturated mixture: A mixture consisting of both liquid and vapor. Saturated vapor: The state in which a mixture is just completely vaporized.

Specific heat: The amount of energy needed to raise a unit mass of a substance one degree. State postulate: /f two intensive, independent properties of a simple pure substance are given, all others, can be determined.

Sublimation:

The phase change from a solid directly into a gas.

Vapor: The gas phase of a substance, usually when it is near condensation. Vaporization:

The phase change from a liquid to a gas.

Working fluid: A fluid to which we add or subtract energy while it is most often undergoing a cycle.

Several important equations were presented: LE

Quality: x = ye

Specific volume of a mixture: v = 0, + x(V, — U,) Enthalpy: ) = u + Pv

Part 1

Concepts and Basic Laws

Enthalpy change: h, — h, = C,(7T, — T;) Ideal-gas law: Pv = RT

Internal energy change: U, — u, = C,(T, — Gas constant:

T))

R= —

M

Compressibility factor:

Po

Z = ——

IRAE

Reduced properties: P, = —, cr

; 1albe van der Waals equation: P = —

Oe

ae,

5

0

Constant-volume specific heat: C,, = (=)

oh Constant-pressure specific heat: C,, = () P

Specific heat relationship: C, = C, + R

Cc Ratio of specific heats: k = Ei; 0

FE Exam Practice Questions 2.1

2.2

What two properties are dependent during the phase-change process?

2.3

The height of Mount Everest is nearly 9000 m. Water would boil at that height at a

(A) (B)

Pressure and specific volume Temperature and specific volume

temperature nearest: (A) 65°C

(C)

Specific volume and enthalpy

(B)

70°C

(D)

Temperature and pressure

(C)

75°C

(D)

80°C

When molecules break away from a solid and become a vapor, the process is referred to as:

2.4

(A)

Melting

When you take a very hot shower, the water is: (A) Subcooled

(B)

Sublimation

(B)

A mixture

(C)

Vaporization

(C)

Saturated water

(D)

Evaporation

(D)

Superheated

Chapter 2 2.5

Water is maintained at aconstant temperature of 200°C, while the specific volume changes

2.9

from 0.002 m*/kg to 0.2 m°/kg. Select the

A 3-L cylinder holds 2 L of steam and 1 L of liquid at 200°C. The quality is nearest:

(A) 18%

diagram that best represents this process.

(B)

(D) :

1

e

2.10

1

(B)

(C) (D)

é 2.11

1 2 2

;

oe ee

200°C. Its quality is nearest: (A) 0.175

(B) 0.211 2.7

A 1-m-diameter bike tire is pumped up to 500 kPa. It is assumed to have a circular cross section with a diameter of 2 cm. The mass of 20°C air in the tire is nearest:

(B)

ame

0.287

(D) 0.314 The volume occupied by 20 kg of water at 300°C and 10 MPa is nearest:

0.0082 kg

(C) 0.018 kg (D) 0.028 kg

(D) (C) Two kilograms of water occupy 80 000 cm? at

(C)

3836 kJ/kg 3840 kJ/kg

(A) 0.0059 kg 1

2.6

3828 kJ/kg

(B) 3832 kJ/kg

t

(A)

3.2%

Superheated steam exists at 6.5 MPa and 675°C. The enthalpy is nearest:

(A)

i e

2.2%

(C) 2.6%

‘

:

Properties of Pure Substances

2.12

The cylinder shown in Fig. 2.28 contains 0.02 kg of air. The weight of the frictionless piston is

Aveaes < ‘ee AAS yi LORS (©) (D) 3000N

(A) 0.028 m? (B) 0.088 m? (C)

0.22 nie

Piston

(D) 0.36 m? 2.8

Superheated steam is contained in a rigid volume at 400°C and 4 MPa. Energy is released from the steam, resulting in a decreased pressure. The pressure that will cause the steam to just begin to condense is nearest: (A)

2.4 MPa

ae a : "es is (D) 3.0 MPa

1

Figure 2.28

Part 1

Concepts and Basic Laws

2.13 The mass of air contained in a rigid 20-L volume at 3 MPa with a temperature of 130 K

2.18 What is the pressure in kPa inside a container if it holds water that is just beginning to boil at a) 140°C, b) 200°C, and c) 320°C?

is nearest:

(A)

2.7 kg

(B)

2.1kg

(C)

1.8kg

(D)

15kg

€ @ Phases of a Substance

Instructor: This isn't multiple choice. Each part with a lower-case italic letter

2.14 Infiltration in a house in the winter where

should be assigned as a separate

—

problem.

T = —25°C occurs even when there is no wind

due to the difference in the air density between the inside and outside of a house. For an inside

temperature of 22°C, the density difference is nearest:

(A) (B) (C) (D)

0.207 0.224 0.241 0.288

kg/m? kg/m? kg/m? kg/m?

2.15 Estimate the value of the C . of steam at 400°C and 4 MPa:

(A) (B) (C) (D)

2.6kg 2.4kg 22kg 2.0kg

2.19 Estimate the temperature of the water in a container if the water is on the verge of boiling at a pressure of a) 12 psia, b) 28 psia, and c) 50 psia. 2.20 A group of engineers wish to know their elevation when they reach the apex on their mountain climb. They forgot their elevation-measuring device, but they have a thermometer and their thermo book. They build a campfire and boil a pot of water. The water is noted to boil at 90°C. What is their elevation? 2.21

A mountain climber reaches an a) 12,000 ft, b) 20,000 ft, and c) on the mountain of Fig. 2.29. At temperature will water just boil selected elevation?

2.22

A container holds 10 kg of liquid water and 7 kg of water vapor at 140°C. Calculate the quality of this saturated mixture.

2.16 Estimate the percent error if the C, value is used from Table B-2 to find the change in enthalpy of steam that has a temperature change from 200°C to 600°C at 200 kPa:

(A)

—4%

(B)

—6%

(C)

—8%

(D)

—10%

2.17 The enthalpy change of 10 kg of ice initially at —20°C when heated to 200°C at atmospheric pressure is nearest:

(A) (B) (C) (D)

9900 kJ 9100 kJ 8300kJ 6400 kJ

elevation of 32,000 ft what at the

3 A container holds 1 ft* of liquid water and 5 ft? of water vapor at 200°F. Calculate the quality of this saturated mixture.

Chapter 2 2.24 The 4 kg of saturated liquid water in Fig. 2.30 are completely vaporized at a constant pressure of 200 kPa. Determine the

Properties of P

ances

2.29 The atmospheric pressure on the surface of Mars is 0.0003 atmospheres. At what temperature would water boil on Mars?

volume of the water at state 1 and at state 2. State 1

State 2

com Roman/Shutterstock Vorotylin

Pressure cooker

Liquid

Vapor

Figure 2.30

2.30 Use the Internet to find three brands of kitchen pressure cookers. What is the maximum rated pressure for each of these? What would be the temperature at which water would boil at these pressures? 2.31 A 5-m* container holds water that has a quality 4

2.25 Saturated liquid water is contained in a 3 ; volume of 400 cm”. Heat is added at a constant pressure of 400 kPa until the water is completely vaporized. What is the final volume? 2.26 A vessel contains 10 kg of water with a quality of 0.85 at a pressure of a) 140 kPa, b) 200 kPa, and c) 2 MPa. Determine the volume of liquid and the volume of vapor. What is the temperature of the mixture? f eat held i 2.27 Ten kilograms of saturated mixture are held in the container of Fig. 2.31 at 40°C and a quality of 0.6. What is the volume of liquid, the volume of vapor, and the pressure of the mixture?

ee

oO

/

eee

te)

ee

te)

Wh

.

are the pressure and the mass of water in the container?

2

2.32

A constant-pressure, piston-cylinder device contains 0.5 kg of saturated liquid water at 50°C. Heat is added to the water until it becomes a saturated vapor. What is the initial volume of the water? What is the final volume of the water?

2.33

A rigid container holds water at the critical point. The container is cooled until the pressure reaches a) 2 MPa, b) 1200 kPa, and c) 400 kPa. Determine the final quality of the water.

2.34 Water with a quality of 0.1 is contained in the rigid volume of Fig. 2.32 at 200 kPa. It is heated until the temperature reaches a) 140°C, b) 180°C, and c) 210°C. Calculate the quality of

the water and the pressure at the final state.

Figure 2.31 2.28

We want to boil water in a cylinder that maintains a constant pressure of a) 70 psia, b) 150 psia, and c) 400 psia. To what temperature do we need to heat the water to make this happen?

i

Part 1

Concepts and Basic Laws

2.35 Four kilograms of water at 25°C are heated at constant pressure, as in Fig. 2.31, until it is saturated liquid at 120°C. Approximate its change in volume. 2.36 Two kilograms of water are contained in a piston-cylinder arrangement at 20°C and 5 MPa. It is heated at constant pressure until the temperature reaches 200°C. Calculate its change in volume using i) Table C-4 and ii) Table C-1. What do you conclude?

2.43 A mass of 0.02 kg of saturated water vapor is contained in the cylinder of Fig. 2.23. The spring with spring constant K = 60 kN/m just touches the top of the frictionless 160-kg piston. Heat is added until the spring compresses a) 6 cm, b) 10 cm, and c) 15 cm. Estimate the final temperature of the steam. (Don’t forget the atmospheric pressure acting on the top of the piston.)

Instructor: All parts with choices (i, ii, ’ etc.) are intended to be solved.

2.37 Two kilograms of saturated liquid water at 200 kPa are heated until the pressure and temperature reach 2 MPa and 400°C, respectively. Calculate the change in volume. 2.38 A container holds 10 kg of water at 130°C with a quality of 0.4. What is the pressure in the container? What is the volume of the container? What is the volume of the liquid phase? 2.39 A 70-ft? container holds 10 Ibm of water at i) 50 psia, 11) 62 psia, and iii) 100 psia. What is the state of the water?

Note: It is assumed that the IRC Calculator has been introduced and is now available to be used for property determination. Your instructor may, ~

however, require that the tables in the Appendix be used.

2.40 Saturated water vapor is held in a container at 200°C. What is the pressure in the container? If the volume of the container is increased by a) 50%, b) 100%, and c) 200%, holding the temperature constant, estimate the final pressure.

2.41

A 10-ft* container holds water at 40 psia and 400°F. How much water is in the container?

2.42

A container holds 8 kg of water at a pressure of 5 MPa. Calculate the volume of the container if the temperature is a) —5°C, b) 20°C, c) 400°C, and d) 800°C.

-~——

20cm

———>|

Figure 2.33 2.44 A mass of 0.02 kg of steam with x, = 0.8 is contained in the cylinder of Fig. 2.33. The spring with spring constant K = 60 kN/m just touches the top of the frictionless 160-kg piston. Heat is added until the temperature reaches 800°C. Estimate the final pressure of the steam. A trial-and-error procedure may be necessary. 2.45 The spring of Problem 2.43 is 2 cm above the top of the piston. All other quantities are the same as stated in the problem. Calculate the final temperature for part (a). 2.46

Use the steam tables in the Appendix and make a plot to scale of a) a T-v diagram showing a constant-pressure line at 400 kPa, and b) a P-v diagram showing a constant-temperature line at 200°C. Include a point in the subcooled region and a point in the superheated region. Observe the distortions of the various diagrams in this text. The distortions are necessary to show the various regions, as the diagrams of parts (a) and (b) should show.

Chapter 2

€ B Internal Energy and Enthalpy 2.47 Water at a pressure of 250 psia has a quality of 0.4. Determine the temperature, the average internal energy, and the average enthalpy for this mixture. Use the steam tables. 2.48 Superheated water exists at 1.86 MPa and 420°C. Determine its specific enthalpy using:

Properties of Pure Substances

ci

2.53 Determine the change in internal energy and enthalpy of water at 190°F and 500 psia if the temperature increases to 500°F while the pressure in Fig. 2.34 is held constant. Use a) the steam tables and b) the IRC Calculator.

i) The steam tables in the appendix (a double interpolation is needed)

ii) iii)

The IRC Calculator

The equation h = u + Pv by using u and v from the IRC Calculator

Are all the answers acceptable?

The word “specific” is often omitted when it is obvious that the specific property is desired since the mass

is not given. In fact, in the tables, the specific internal energy may be referred to simply as “energy” and the specific enthalpy as “enthalpy.”

2.49 Ten pounds of saturated water vapor at 300°F are cooled at constant temperature to a saturated liquid. What is the change in pressure? The change in the total internal energy? The change in total enthalpy? Does AH = AU + A(PV)? a) Use the steam tables and b) the IRC Calculator. 2.50

2.51

Saturated water vapor is compressed from 200°C to 2 MPa and 600°C. Determine the change in specific volume and enthalpy. a) Use the steam tables and b) the IRC Calculator.

Four kilograms of water at 20°C are heated at constant pressure of 200 kPa to 400°C. Determine the change in density, internal

energy, and enthalpy. a) Use the steam tables and b) the IRC Calculator.

2.52 Twenty kilograms of water are heated at constant pressure from 300°C and 0.6 MPa to 400°C. Calculate the change in enthalpy. a) Use the steam tables and b) the IRC Calculator.

Figure 2.34 2.54

How much energy is needed to increase the temperature of 10 kg of ice from —20°C to 20°C at atmospheric pressure? Estimate the amount of energy required to sublimate the 4 kg of water contained in the frozen clothes in a northern state which are at =A) GaSec Big. 7.35.

€ 8 Refrigerants 2.56

A 0.05-m° container contains 50 kg of R134a. If the pressure is a) 2.5 MPa and b) 240 kPa, what is the state of the refrigerant?

A 10-ft* container contains 100 Ibm of R134a at —20°F. What is the state of the refrigerant? A constant-pressure, piston-cylinder device contains 100 kg R134a at —12°C and 200 kPa. If the refrigerant is heated at constant pressure until the temperature reaches 60°C, determine:

i) The change in volume ii) The change in total internal energy

iii) The change in total enthalpy

—

Part 1. Concepts and Basic Laws

pares) A flow of saturated liquid R134a at 80°F passes through a throttle where the pressure drops to 10 psia. If the enthalpy remains constant as the refrigerant passes through the throttle of Fig. 2.36, determine the temperature and the quality of the flow exiting the valve. Throttle

+

80 F

R134a

10 psia

>

|

Figure 2.36 2.60 Ammonia is compressed from —40°C and a quality of 0.95 to 400 kPa and 40°C. Calculate the change in specific enthalpy of the refrigerant.

2.63 The atmosphere of Mars is composed of carbon dioxide at 1 kPa and —60°C. Determine the specific volume of the atmosphere. 2.64 A 5-m?-rigid container holds 0.2 kg of air at 300 K. How much air at 300 K must be added to the container to bring the pressure to 500 kPa? 2.65 Ten pounds of air at 14.7 psia and 80°F are compressed so that the final volume is one-fourth the initial volume and the pressure is 40 psia. Determine the final temperature of the air.

2.66 A 4-m? volume contains 2 kg of an unknown gas at 400 kPa and 112°C. What gas do you think occupies the volume? 2.67

€ B Ideal-Gas Law

An enclosed football stadium is maintained at

100 kPa and 20°C. If its volume is 2 X 10° m’,

estimate the weight of the air.

2.61 A spherical balloon contains 0.1 kg of air at 120 kPa and 20°C. If the temperature of the air is increased to 35°C and the pressure remains

2.68 Fill in the missing properties in the following table if the air occupies a 10-m* volume.

the same, determine the initial and final

volume of the sphere. 2.62 A spherical 24-in.-diameter balloon contains helium at 72°F. What is the mass of helium in the balloon assuming the pressure is approximately 14.7 psia? Approximate the diameter of the balloon at an elevation of 5000 ft, as shown in Fig. 2.37 (Assume the pressure difference across the balloon surface is negligible.)

P

T

v

p

Ww

(kPa)

(°C)

(m*/kg)

(kg/m*)

—(N)

i) 200m

200

li)

400

iii) 600 iv)

0.08

40 —40

20

2.69 Fill in the missing properties in the following table if the air occupies a 30-ft® volume.

Pansy:

5000 ft

Figure 2.37

v

(psia)

(°F)

(ft'/lbm)

i) ii)

20

400 600

12

iii)

100

>|

iv)

p (Ibm/ft*)

, —60

Ww (bf)

80 1.6

Chapter 2 2.70 The air in a house is maintained at 72°F. The air outside the house is at —20°F. What is the difference in the density of the air between

the inside and the outside of the house?

Properties of Pure Substances

ca

2.74 Ten kilograms of air occupies 1 m? of space at —40°C. Calculate and compare the pressure using the:

i) Ideal-gas law

There are always small passages where air can infiltrate into or from a house even if extreme

ii) Van der Waals equation

care is taken to seal all those passages. Explain

iii) Beattie-Bridgeman equation

what the air particles do even if the outside air

iv) Z-factor (see Appendix H)

is perfectly motionless. See Fig. 2.38.

v) IRC Calculator 2.75 Estimate the pressure of nitrogen at —80°F and v = 0.6 ft*/lbm using the: i) Ideal-gas law ii) Van der Waals equation

iil) Z-factor (see Appendix H) 2.76

The pressure and temperature are measured in

a tank containing air to be 60 kPa and —80°C, respectively. Determine the specific volume

Figure 2.38 2.71

The tires on an automobile are inflated to

using the: i) Ideal-gas law

35 psi in New York, where the temperature is

ii) Van der Waals equation

—10°F. The auto is driven to Arizona, where

ili) Z-factor (see Appendix H)

the temperature on the blacktop is 160°F. Estimate the pressure in the tires assuming the volume does not change.

iv) IRC Calculator

2.72 The golfer in Fig. 2.39 on a very humid day states that the ball doesn’t travel as far because

€ 8 Internal Energy and Enthalpy of ideal

Gasses

of the heavy air. If the air were less humid at

the same pressure and location, would the ball have traveled further? Explain, referring to the ideal-gas law and the fact that the drag F,, on the golf ball is proportional to the density of

theain

= pAV/2)

© B Real-Gas Equations of State 2.73

Nitrogen exists at 4 MPa and 130 K.

Approximate the specific volume using i) the ideal-gas law and ii) the compressibility factor from Fig. 2.23.

2-77 Calculate the internal energy change and the enthalpy change if the temperature changes from 20°C to 450°C for each of the following, assumed to an ideal gas with constant specific

heats: a)

Air

b)

Nitrogen

c)

Hydrogen

d)

Propane

e)

Steam

Note: Use specific heat values from

.

Table B-2 unless otherwise stated.

‘i

rk

Part 1

Concepts and Basic Laws

2.78 Steam is assumed to be an ideal gas. Determine the enthalpy change if state 1 is saturated vapor at 30°C and state 2 is at 150°C and 200 kPa. i) Assume constant specific heats. ii) Use the steam tables. iii) Use the IRC Calculator.

2.79 Nitrogen is heated from 20°C to 500°C. Calculate the change in the internal energy and enthalpy. Use: i) Constant specific heats from Table B-2

ii) The ideal-gas table in Appendix F iii) The average specific heat from Table B-5

iv) Equation 2.30 2.80 Twenty pounds of air at 25°F is heated to 1000°F. Determine the change in internal energy and enthalpy. Use:

i) Constant specific heats from Table B-2

ii) The ideal-gas table in Appendix F iii) The specific heat at the average temperature from Table B-6 iv) Equation 2.30

2.83 Estimate the specific heat C, of steam at:

a)

40 psia and 600°F

b)

140 psia and 600°F

c)

600 psia and 600°F

€ B Specific Heats of Liquids and Solids 2.84 One hundred pounds of water at atmospheric pressure are heated from 60°F to 200°F. What is the enthalpy change? The internal energy change? Why is the difference between the internal energy change and the enthalpy change so small? 2.85 If the enthalpy of 5 kg of mercury at 25°C is raised 200 kJ, determine the final temperature of the mercury.

2.86 The temperature of 10 kg of ice at —20°C is raised to 50°C, displayed by Fig. 2.40. Determine the enthalpy change.

a) b)

Assume C,;.. = 2.1 kJ/kg°C. Usere pice = 2.1 + 0.0069T kJ/kg-°C.

v) The IRC Calculator 2.81 Estimate the specific heat C, of steam at 1.6 MPa and 400°C using i) backward differences, ii) forward differences, iii) central differences, and iv) Table B-2. Which is the most accurate? (b)

2.82 Estimate the specific heat C,, of steam at:

a)

0.2 MPa and 400°C

b)

1.8 MPa and 400°C

c)

7MPa and 400°C

Figure 2.40

Outline 3.1 Work 3.1.1 3.1.2 3.1.3.

88 Definition and units 88 Work due to pressure 88 Other forms of work 91

3.2 Heat Transfer

95

3.3 Problem-Solving Method

98

3.4 The First Law Applied to Systems

99

3.5 The First Law Applied to Various Processes 101 3.5.1 3.5.2 3.5.3 3.5.4 3.5.5

The The The The The

3.6 Cycles

constant-volume process 101 constant-pressure process 102 constant-temperature process 105 adiabatic process 106 polytropic process 110

112

3.7 Summary

112

FE Exam Practice Questions Problems

116

114

The following variables are introduced in this chapter: R r S i t V WwW

Electrical resistivity, R-factor Radius Entropy Torque Time Voltage Work

C ds F h, i K k

A constant Differential length segment Force vector Convection heat transfer coefficient Current Spring constant Thermal conductivity

L

Thickness

W

Work rate

n

A constant

w

Work per unit mass

Rate of heat transfer Specific heat transfer

o op a

Q

O qd

Heat transfer

=

Emissivity Stefan—Boltzmann constant A special property Shear stress

Learning Outcomes

We have interest primarily in the conservation of mass and the ‘st law.

First law of thermodynamics: An accounting of the flow of energy into and out of a system and its accumulation.

System: A volume with fixed mass. The volume can change, as it does in a cylinder when the air is compressed by a piston.

41

Solve work integrals

1

Identify heat transfer mechanisms

1)

Apply the first law to systems

J

identify and analyze common thermodynamic processes

Vi any of the problems we solve in engineering involve the application of one of the three laws: the conservation of mass, Newton's second law,' and the first law of thermodynamics. This chapter focuses on the conservation of mass and the first law of thermodynamics, with minor emphasis on Newton's second law. The first law of thermodynamics is simply an accounting of the energy that flows into and out of a system and its accumulation. The general approach to solving problems applying the two laws of interest can be summed up with Input — Output = Accumulation

(3.1)

The system that is being analyzed must be carefully defined. A system consists of a fixed mass of substance that is contained in a volume. Mass is not allowed to enter or leave the volume. Even though the mass of the system is fixed, the shape of the volume that contains this mass may vary; the volume in a piston-cylinder arrangement decreases as the piston compresses the air. The boundary of the system we are analyzing will be defined by the boundary of the mass of the system. In Chapter 4, the substance will enter and/or leave the vdlume, referred to as a control volume. 1

.

.

A fourth law, the conservation of angular momentum, is a result of Newton’s second law.

Motivational Example—tThe piston-cylinder processes in an automobile engine A good example of a thermodynamic system is the air in a cylinder being compressed by a piston in an automobile engine. The engine cycle consists of five distinct processes. First, air is drawn into the cylinder through the inlet valve by moving the piston down, as shown in Fig. 3.1. Next, the inlet valve closes and the piston moves upward, compressing the air during the compression stroke thereby increasing its pressure and temperature. When the piston nearly reaches its maximum extension into the cylinder, a spark plug ignites

: é

Mess The five strokes in a piston engine: Intake Compression

Ignition/Combustion Power

Exhaust

the injected fuel, with the result that the pressure and temperature of the air are greatly increased. Next, this high-pressure air exerts a large force on the

Process 0-1: Intake stroke

Process 3-4: Power (expansion) stroke

Piston-engine processes.

:|

TDC—Top dead center

DC

Process 2-3: Ignition and combustion

Process 1-2: Compression stroke

EE BDC—Bottom

dead center

BDC

Process 4-1: Exhaust

Process1-0: Exhaust stroke

(Continued)

The piston-engine process can be seen in motion on the Internet on the site at: howstuffworks.com /engine1.htm

Part 1

Concepts and Basic Laws

Motivational Example (Continued)

|

piston during the power stroke. As the cycle continues, the exit valve is opened and the piston moves upward, forcing the products of combustion from the cylinder. The system is now ready to repeat the series of processes. The compression stroke, ignition, and power stroke are processes that involve a fixed amount of mass, a system. Once the valves are closed, the mass of air in the cylinder is fixed. During the intake and exhaust strokes, the system is not identified.

Note: The mass of fuel injected is negligible compared to the mass ©

of air.

System

3 1

Work

3.1.1

Definition and units

‘ boundary r

q

= |‘F-ds

(3.3)

Ss

This is a vector expression in which the integral of the scalar product of the force Comment | vector F and the differential path element ds, a vector tangential to the path, is integrated over the path from the initial state to the final state. The work done is a Work not a property and pt Biever iswaiter a5 Wor We pé himeiionsh inction; hence th the su sci scripts on Wand Win icate a process from state 1 to state 2. is EE So Since it depends on the path, its differential is not exact, so it is represented by

|

Chapter 3

The First Law for Systems

dW. Work is not a property and is never written as W, or W, but always as W,., or simply as W. In thermodynamics, the primary force used to do work is a pressure force. Pressure forces always act normal to the surface in contact with the fluid; for example, in Fig. 3.3 the pressure force is moving the piston a small displacement ds. The resulting displacement is often perpendicular to the surface in which case the dot product, using F = PA, takes the form

Ww

|‘PAds

(3.4)

Sy

Recognize that Ads = dV, the infinitesimal increase in volume due

a pressure force doing work on a moving

to the displacement ds. The expression for the work due to pressure

boundary.

acting on a boundary is an important and oft-used relationship, in

both differential and integral form,

comment?) The force F = PA is in the

é6W = PdV

W,,=

e 7”

PdV

W,,= 2

1

ba

| Pdv F ee:

(3.5)

direction of ds, so the

Aa

eg Gore

differential work is simply Fds = PAds = PdV.

If the pressure is a known function for each position of the boundary in Fig. 3.3, the integration could be performed and the work determined. Notice how the path can be very different between state 1 and state 2, resulting in less work (the area under the curve) in Fig. 3.4a than in Fig. 3.4b. If the pressure is constant between states 1 and 2, the expression for the work would be

Wi 5 = P(0x — 0;)

if P = const

(3.6)

The sign convention: Work done by the system on the surroundings is considered positive, and work done by the surroundings on the system is negative. One additional observation should be made concerning the work defined by the integral of Eq. 3.5. It is assumed that the pressure and volume are known at each state between the initial and final states; that is, a quasi-equilibrium process is assumed between state 1 and state 2 of Fig. 3.4. So, Eq. 3.5 represents a quasiequilibrium work mode. Nonequilibrium work modes involve, for example, friction or electrical resistance heaters. If a nonequilibrium work mode is present, the work from state 1 to state 2 is not given by Eq. 3.5. If the information given does not allow one to determine the type of process, a quasi-equilibrium process will be assumed.

p

6)

Y

®

=

Cy

PCV

=

(a)

(b)

Work is the cross-hatched area from state 1 to state 2

ev

Remember:

Work done on

a system is negative.

es

Part 1

Example

Concepts and Basic Laws

3.1

Work needed to condense steam 2ST

SOE

DLA

Ten kilograms of steam at 200 kPa and 400°C are condensed in the cylinder of Fig. 3.5 at constant pressure until the quality is 50%. Determine the work required between the two states.

Superheated steam

Solution

200 kPa 400°C

The specific volumes at states 1 and 2 are found using Tables C-2 and C-3 in the Appendix. They are

State 1

v, = 1.5493 m*/kg UV, = Ve + X(U, — U,) 0.00106 + 0.5 X (0.8857 — 0.00106) = 0.4434 m°*/kg The work during this condensation process, assumed to be a quasi-equilibrium process, is then, using Eq. 3.6,

W,., = Pm(v, — 2)

= 200kN/m? x 10 kg x (0.4434 — 1.5493) m°/kg = —2212kJ

State 2

recognizing that kN-m = kJ. Note that the work is negative since the volume is decreasing. A negative work means that work is being done on the system, the steam.

Example

3.2

The work needed toraise a piston One hundred grams of water at 50°C are contained in a 220-mm-diameter cylinder at state 1, as shown in Fig. 3.6. Energy is added until the temperature reaches 150°C in the steam at state 2. If the frictionless piston has a mass of 387 kg, find the work done by the steam on the piston.

Solution Pressure: The absolute pressure Is 99.9 + 100 = 199.9 ~ 200 kPa. Assume

Pom = 100 kPa, unless

The pressure in the cylinder, due to the piston resting on top of the water is L=

mg

387kg X 9.81 m/s*

Ae

0

= 99 870 Pa gage

otherwise stated.

Weight

Piston

j One can visualize atmospheric pressure acting on top of the piston or simply add 100 kPa to give absolute pressure, as required.

State 1

Wr

Restaias

or ~ 200 kPa abs

Chapter 3

The First Law for Systems

wie

The work done by the steam to lift the piston is due to the force PA and is given by mPAv (see Eq. 3.6). State 1 is compressed liquid: at 50°C,

: taePee

Table C-1 gives 0, =-0.0102

the piston. The air above

m°/kg. State 2 is superheat; from Table C-3,

V, = 0.9596 m*/kg at 0.20 MPa and 150°C. The work is

ae

se

the piston does negative

work on the piston due to Wi

Ae Pm(v, aes v,)

kN ie =

the atmospheric pressure.

100 m3 oe: 20: (i on)|kag X (0.9596 — 0.01012) =%

= 18.99 kJ Because we used kPa for pressure, the work is in kJ.

Work needed to increaseavolume

Example

3.3

SS

Air at 694 psia is contained in a cylinder at state 1, as shown in Figure 3.7 A piston is moved such that the volume changes according to PV = 10°, with P in psfa and V in ft®. The piston moves until the volume doubles. Find the work done by the air on the piston.

®

Solution

The work done due to the motion of the piston is given by the integral of Eq. 3.5, so the volumes must be known since they are the limits of integration. They are V

V,=—P,

694 Ibf/in? x 144 in?/t = 1.00ft? and V,=2f0

Note that 10° is a constant that has units of ft-lbf, the same units as on PV. The work is then

V>

2 j |PdV=

W,, = V,

10° ev

5 = 10°(1In2—In1)

= 69,300 ft-lbf

1

The work is positive, so it is doing work on the surroundings, that is, the piston, as expected since the volume increases.

3.1.3 Other forms of work There are four other forms of work that we will occasionally encounter in our study. They are the work required to stretch a linear spring, to rotate a shaft, to turn a paddle wheel, and to cause an electrical current to flow through a resistor. Other forms of work exist, such as work due to electrical and magnetic fields and the work needed to stretch a liquid film, but they will not be included. Let’s analyze each of the four.

igure:

Part 1

Concepts and Basic Laws

Unstretched

j

=F

iS F= Kx

Stretched

Figure 3.8 A linear spring being stretched.

Comment

We have used x for displacement. The work here is written as a positive quantity. If a spring acts on a piston, the appropriate

The stresses acting on an area element of a rotating shaft.

A linear spring: The force needed to stretch a linear spring, shown in Fig. 3.8, from its unstretched position is F=

Kx

(3.7)

where K is the spring constant with units of N/m. The work required to stretch the spring from the stretched position x, to position x, is

sign will be used.

W=

Power:

The rate of

doing work. Shear force: Shear stress

times the area upon which it acts.

2g

[ Xp

1

xy

a

=

°R |rwtdA =

A

|ret(2ardr) = 27w|

J0

Se J0

trdr

(3.9)

he DE 4

i. we “a

(3.8)

A rotating shaft: Next, consider the rate of doing work, called power, that is transmitted by a rotating shaft. The differential sear force due to the shearing stress 7 that acts on the infinitesimal area dA, displayed in Fig. 3.9, is dF = 7dA. Recognizing that the velocity with which this differential force dA moves is ro and that force times velocity is the rate of doing work, symbolized by W, we see that

: 1

5

where we have written W rather than W,.,, as is often done. If the spring is the system, the work done on the spring has crossed the boundary of the spring and energy is stored in the spring.

W= fe

>

|Fd = K |xdx = —K(x5 — x;)

dA = rdédr

/,

Center of circle

Actually, a double integration could have been performed using the area element shown in Fig. 3.10. First, the integration would be from 6 = 0 to @ = 27, then from r=Otor=R. The torque T transmitted by the shaft is found by integrating the infinitesimal torque rdF = r7dA over the area:

T= aaa Units: [wT] = (rad/s)(N-m) =N-m/s = J/s =W

°R ioe |r7dA = | r(2ardr) = 27} trdr A J0 J0

(3.10)

Comparing Eq. 3.9 with Eq. 3.10, we observe that the power transmitted by a shaft is (3.11) er If the work over a time increment Aris desired, the power is multiplied by the time, giving W = oTAt The power or work delivered by a shaft can now he calculated.

(3.12)

Chapter 3

The First Law for Systems

[EE

V

Figure 3.12 A paddle wheel doing work on a fluid.

Figure 3.13

An electrical resistance heating a fluid.

A nonequilibrium process.

A paddle wheel: One situation where a rotating shaft contributes to a thermodynamic system is when a paddle wheel is stirring a fluid, as in Fig. 3.11. The work can be found either by using Eq. 3.12 or by using the weight times the distance the weight falls. This is an example of a nonequilibrium’ work mode. The viscous effects in the fluid dissipate this negative work into stored energy by raising the temperature of the fluid.

An electrical resistance heater: Another example of a nonequilibrium work mode is the electrical resistance heater in a volume of fluid, as shown by Fig. 3.12. This provides the same effect to the fluid as does that of the paddle wheel. It is considered negative work since it is energy that crosses the boundary of the fluid, resulting in an increase in the temperature of the fluid. The power transferred to the fluid by the resistance R is given by 2 Power = Vi = i (3.13) where the usual variable for the electric potential (voltage) is V,i is the current, and R is the resistance. The power

Vi is measured

Note: Don’t confuse voltage V with volume V. The context will make the meaning obvious

in watts, related to English

units by 1 W = 3.412 Btu/hr = 0.7374 ft-lbf/s. The work transferred by the resistor

is power multiplied by the time, that is, 7RAt. If a P-V diagram were sketched after a period of time elapsed of the setup of Fig. 3.12, the actual process from state 1 to state 2 would not be known, so a dashed line connecting:the states could be used, as in Fig. 3.13.

The work from a spring Air at 150 kPa is contained in an 80-cm-diameter cylinder by a frictionless piston, initially 60 cm from the bottom of the cylinder. A spring with K = 80 kN/m is brought into contact with the piston so that no force exists in the spring, as shown in Fig. 3.14. Energy is then added to the air until the volume doubles. Calculate the work done by the air on the piston.

Solution Since V, is double V,, the height of air in the cylinder must double so the piston is raised 60 cm, thereby compressing the spring 60 cm. The work can be (Continued)

An equilibrium work mode can be reversed and the energy recovered. A paddle wheel does not provide an equilibrium work mode, since if it could be reversed, it would not result in a raising of the weight and a lowering of the temperature of the fluid.

Example

3.4

a

Part 1 Concepts and Basic Laws

Example

3.4

(Continued) separated into two parts: the work W, needed to raise the piston 60 cm at constant pressure, and the work W, needed to compress the spring 60 cm. The total is

Units: Never use cm for

distance or cm? for area in an equation. Either Pa or kPa can be used for pressure, but

make sure the same units are used on each term in an equation.

Example

3.9

1

W=w, + Ws = PAh + 5Kx° kN S507ne Xia

2 1 xX 0A) mma xX 0.0m+ 5

The 400-pound weight implies a force, i.e., 400 Ibf; however a 400-pound weight has a mass of 400 lbm anywhere near the Earth’s surface (where

ee

0.6° De) m° eres= 59.6kJ

Rather than using Eq. 3.6 to find the work to raise the piston, we used the relationship W = F X distance. Because both the pressure and the spring constant were expressed using kPa, the work is in kJ.

Work provided by a paddle wheel DDE

p Comment |

kN

IOA EE

A four-hundred-pound weight attached to a paddle wheel via a pulley, as in Fig. 3.15, drops 4 ft. Estimate the Btu’s added to the water in the tank.

g = 32.2 ft/s’). So, either 400 Ibf or 400 Ibm could be used. On the Earth, gravity varies between

32.1 ft/s? (on the highest mountain) and 32.3 ft/s? (in the lowest ocean trench). so assuming sea-level gravity is acceptable for all problems unless you're in a space station!

Solution The energy added to the water by the rotation of the paddle wheel is W=F

xX h= 400lbf X 4ft = 1600 ft-lbf

Work is usually measured in ft-lbf when using English units, but it can also be measured in Btu’s. Using Table A in the Appendix, the energy added to the water by the paddle wheel can also be expressed as

1600 ft-lbf

Ff IbiBtu This would be a negative nonequilibrium work mode relative to the fluid since it represents energy added to the fluid. Sees)

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ngy I

(1)

Chapter 3

The First Law for Systems

3.2 Tr Heatansf = er

Heat transfer: The

Heat transfer is another mechanism used to transfer energy to or from a system. Heat transfer, indicated by Q, is the transfer of energy due to a temperature difference between a system and its surroundings. The rate of heat transfer is designated Q. If the heat transfer for a process is zero, it is referred to as an adiabatic process. An insulated container would provide for zero heat transfer. Heat transfer is accomplished by one or more of three physical mechanisms. These mechanisms will be briefly described but will not be analyzed in detail in this text. This would require a full semester course called Heat Transfer. Conduction heat transfer is the transfer of thermal energy by intermolecular action. If the molecules on the boundary of the surroundings are more active, and have a higher kinetic energy than the molecules that make up the system boundary, a transfer of energy takes place from the faster molecules to the slower ones. We know that temperature is related to the speed of the molecules; consequently, in order for energy to be transferred from the surroundings to a system, the temperature of the boundary of the surroundings must be higher than the temperature of the system boundary. Fourier’s law of heat conduction, for the one-dimensional plane wall of Fig. 3.16, takes the form

ATER L = A— R Q = kA—

transfer of energy due to a temperature difference between a system and its surroundings. Adiabatic process: A process for which the heat transfer is zero. Conduction: The transfer of thermal energy by intermolecular action.

(3.14) A plane wall

where k is thermal conductivity with units of W/m-K (Btu/s-ft-°F), A is the wall area through which the heat flows, AT is the temperature difference across the wall, and L is the thickness of the wall. Sometimes the R-factor in Eq. 3.14, the resitivity, is provided; it is R = L/k. It should be noted that the R-factors can be added for multilayered surfaces. Convection heat transfer is the transfer of thermal energy by conduction and by the bulk flow of a fluid, called advection. This means

Note: R-factors can be added for

~ mulilayered surfaces:

oR

= Ry oR Pe

that a fluid that

contains thermal energy will transfer energy primarily by conduction when the fluid motion is very low, and then primarily by advection when the fluid motion is relatively high. Consider a hot plate sitting in air. Initially, the air will absorb heat from the plate by conduction from the hot plate to the cooler air. As the air absorbs thermal energy, it expands and becomes buoyant. The air then rises from the plate, carrying thermal energy away from the plate, while cooler air takes its place, and the process continues with the air speeds increasing as the plate heats up. If air is blown over the plate in some manner, it is forced convection; otherwise; with no forced air, it is referred to as free convection. Newton's

law of cooling provides an equation that allows convective heat transfer to be

Convection: The transfer of thermal energy primarily due to the motion of a fluid.

Note: First, convection occurs primarily due to

~ conduction and then "primarily due to advection as the air begins to rise.

calculated from an area A; it is

O-1AG = 1)

(3.15)

where h, is the convective heat transfer coefficient with units of W/m?-K (Btu/s-

ft?-°R), T, is the surface temperature, and 7, is the bulk fluid temperature. Convection is a complex combination of heat transfer and fluid motion and can be difficult to model. It is studied in detail as part of a heat transfer course. For con-

vection, the R-factor would be R = 1/h,.

n

Forced convection: When air is forced over a surface.

Newton's law of cooling: The equation that allows convective heat transfer to be calculated.

ce

Part 1

Concepts and Basic Laws

Radiation: The transfer of thermal energy by electromagnetic radiation. Stefan-Boltzmann law: The equation that allows radiation heat transfer to be calculated.

Radiation heat transfer is the transfer of thermal energy by electromagnetic radiation. Any substance with a temperature above absolute zero emits electromagnetic energy. The frequency of the radiation emitted is a function of the temperature of the substance. Radiation is unique in that it can be transferred through a perfect vacuum as well as through gases. The human body has a normal temperature of 310K. At this temperature, the frequency of radiation is primarily in the infrared zone, which is why infrared night vision goggles are so effective in locating people. Radiation is very unique in that the energy output varies with absolute temperature to the fourth power. This type of relationship does not occur elsewhere in nature. To calculate the heat transfer from a solid surface to the surroundings, the Stefan—Boltzmann law provides

O=c0A(Te

125) SUIT

(3.16)

where ¢isthe emissivity and cis the Stefan—Boltzmann constant (5.67 X 10 *J/s-m*-K* or 0.1714 x 10°* Btu/hr-ft’-°R*). The temperature T, of the surface and T,,,, of

Black body: A body that emits and absorbs the maximum amount of radiation

the surroundings must be absolute temperatures. The emissivity is between 0 and 1 where e = 1 is for a black body that emits and absorbs the maximum amount of radiation. Radiation does not transfer significant amounts of heat until high temperatures are reached. Before the development of rockets and jet engines in the 1940s, radiation was mainly a subject reserved for the astronomical modeling of stars. In this text we will not analyze the details of the heat transfer process. A quantity of heat will simply be transferred to or from a system without specifying how it is done. We will, however, provide several examples and problems illustrating the simplified heat transfer equations listed above. They are the type of problems included on national exams such as the Fundamentals of Engineering Exam and the GRE/Engineering Exam. The sign conventions are very important to the analyses of processes. Heat transferred to a system is positive, and heat removed from a system is negative. The unit of heat transfer in the SI system is the joule (J). In this text we will use the kilojoule (kJ) since the joule is quite small. The unit of heat in the English system is the British thermal unit (Btu). Heat transfer and work have several traits in common.

1.

Both work and heat transfer are boundary phenomena. Work and heat represent energy that enters or leaves a system through the system boundary. A system does not contain work or heat.

2.

Both heat transfer and work are path functions. The amount of energy transferred is not just a function of the initial and final states but of the path the process takes to get from one state to the other. Hence, the differential of

Heat transferred to a system is positive, and heat

removed from a system is negative.

heat transfer is inexact and is written 6Q.

3. 4.

Both heat transfer and work are time-dependent phenomena. The processes take place over a period of time and are not instantaneous. Both heat transfer and work are processes, not state functions. As processes,

heat transfer and work cause a change of state. Heat transfer is written as Q,.,, never Q, or Q,. Often it is simply Q when the process is obvious.

The amount of energy transferred by heat transfer can be defined in specific terms by dividing the amount of energy transferred by the mass of the system. The symbol for specific heat transfer is g, defined by d=

Ga m

(3.17)

Chapter 3

The First Law for Systems

In the SI system it is common to assign units of kJ/kg to q and kN-m/kg to w, even though they are equivalent units; in the English system it is common to assign Btu/lbm to q and ft-lbf/lbm to w.

Conduction heat transfer

Example

3.6

Estimate the rate of heat transfer through a wall that measures 8 ft by 12 ft

composed of two identical wood layers with R = 0.6 ft*-hr-°F/Btu and an insu-

insulation

lation layer with R = 5 ft’-hr-°F/Btu, as displayed in Fig. 3.17 The inside temperature is 70°F, and the outside temperature is 0°F. Neglect the convective heat transfer due to the air layers on the inside and outside. (The convective layers on a glass pane cannot be neglected.) Solution

Equation 3.14 provides the heat transfer through the wall. Using R rather than

Oak

_ k (we can add R-factors) gives

a ANE AAT Pee total CIR woodet Re:

Wood

_ 8 x 12) f? x (70 - 0) °F = 1084 Btw/hr ~ (2 X 0.6 + 5)ft?-hr-°F/Btu

Radiation heat transfer

Routesealz

Example

3.7

NSE LEAD

The walls of the electric oven of Fig. 3.18 are maintained at 2000°C. An artist inserts a 10-cm-diameter spherical mass of glass at 20°C into the center of the oven. Estimate the initial maximum rate of heat transfer to the glass sphere. Solution

The heat transfer rate due to radiation is given by Eq. 3.16. The maximum rate of heat transfer occurs with a black body for which e = 1. That rate is

Q BOAT i

Ta)

Pc 567 K 10

J/s

x (4a X 0.05") m? x (2273° — 293°) K*

m’:K*

47 500 J/s or 47.5 kJ/s

We used the surface area of a sphere as 47°. Temperature must be in kelvins

Figure-3.18

and the area in m’.

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Outcome©

(2)

ae

Part 1

Concepts and Basic Laws

_ ing Method Problem-Solv 3.3

Note: Most often it is assumed that g=

2 ft/s? 9.81 m/s? = 32.

s

14.7 psia

BS

Patm =

100 kPa =

A course in thermodynamics is by nature a problem-solving course. You are being taught the skills necessary to solve practical problems in thermodynamics. In high school this type of problem was called a “word” problem. A story is told about the physical system and what happens to it. Usually, but not always, a specific answer is requested at the end of the story. It is up to the reader to extract the necessary information and develop a solution that will yield the answer. You will be taking many courses like this in your engineering career, and it would serve your career well to develop sound, systematic tools for setting up and solving problems. In this text, emphasis will be on solving problems. The language of thermodynamics is unique, and special attention must be paid to the wording and interpretation of problems. Read the problem carefully. All problems will contain exactly enough information to solve the problem. Some of the necessary information will be given directly or can be found in the Appendix. Other information must be obtained from the description of the processes used. A systematic procedure for setting up and solving problems is highly recommended. An eight-step, problemsolving method is listed below and will be demonstrated in solving example problems. If an example is solved without adhering to each of these steps, the reader can insert the desired information in the margin next to the example’s solution. The eight steps are as follows. 1.

Read the problem carefully and identify all the information given in the problem. Treat the problem statement as though every word holds a clue to solving the problem.

2.

Sketch a figure showing the system or the control volume. Engineers often think in visual terms. Try and picture a problem. Drawing a sketch of the physical problem will often give insight into understanding and solving the problem.

3.

List the given information for all states. Know what type of substance you are analyzing. Convert all data to useful units. Check the problem for special processes.

State any assumptions you make about the problem. This is important if you are making simplifying assumptions. If you simplify a problem too much, the problem you finally solve may not be the problem originally stated. 6.

Determine the required states and/or properties.

7.

Apply the appropriate equations. In this course it will most often include the first law of thermodynamics, the energy equation. Be sure to use the correct sign conventions on your terms, and always be aware that the units on every term in an equation must be the same.

8.

Check your answer thoroughly. Does it appear to be too large or too small? Are the units on your answer the proper units? If the units are incorrect, you definitely made a mistake somewhere.

These steps are not followed literally, maybe occasionally, but they are in one’s

mind while solving all problems. They become part of the thought process as a problem is being solved.

Chapter 3

The First Law for Systems

ey

3.4 The First Law Applied The first law of thermodynamics, usually simply called the first law or the energy equation, focuses primarily on thermodynamics. We use it to analyze thermodynamic processes and thermodynamic cycles, which are made up of several sequential processes. In this chapter the first law will be applied to a fixed mass of a substance, a system. Change in the energy level of a system can be caused by heat transfer or work. The standard form of the first law applied to a system undergoing a process is

Q-

Remember: The sign convention is that heat transferred from the surroundings to the system is positive, whereas work done by the system on the surroundings is positive, which is why the negative sign exists in Eq. 3.19. In

(3.18)

W=AE

some areas of study, the

The terms on the left-hand side of the equation represent the energy transferred across the boundary of a system. They are the heat QO transferred to the system and the work W done by the system. They represent the two ways in which energy is transferred into and out of a system. Energy can also be transported into and out of a volume by a flowing fluid, but that will be considered in the next chapter. In this chapter, no fluid will cross the boundary of

same convention is used for both, resulting in a plus sign, although the negative sign used in thermodynamics is more common.

a system.

The work and heat transfer across the boundary of a system results in a change in the energy of the system, represented by Eq. 3.18. The change in the system’s energy AF is the sum of three terms:

OQ -W-=AU + AKE + APE

(3.19) ®

where AU is the change in the internal energy of the system, AKE is the change in the kinetic energy of the system, and APE is the change in the potential energy of the system. The energy Q and W transferred across the system’s boundary represents the amount of energy stored or released from the system. The systems of interest in the remainder of this chapter are stationary, so the changes in potential and kinetic energy are zero. This reduces Eq. 3.19 to

a

-2

x

Se

is he

® ea ie

Wa.4

Q-W=AU

(3.20)

If a system undergoes a cycle so that the system returns to its initial state, such as the cycle sketched in Fig. 3.19, which consists of four processes, the internal energy change is zero for the cycle. Then Eq. 3.20 takes the form OFF

a

Wee

@ (6)

A cycle composed of four processes.

(3.21)

1* law for a cycle:

where OQ... and W,,, are the net heat transfer and the net work for the entire cycle. This form of the first law will be used quite often when considering cycles. We can write the first law in terms of specific properties by dividing Eq. 3.20 by the mass of the system to give, for a system undergoing a process, q—- w=

Au

(3.22)

Qnet me W,net

ey

Part 1)

Concepts and Basic Laws In this equation, q is the heat transfer per unit mass and w is the work done per unit mass. All terms in this equation have units of kJ/kg (Btu/Ibm). The first law is also used in differential form as

dq — dw = du

(3.23)

where the differentials of g and w are inexact but the differential of u is

exact since u is a property; q and w are not properties.

Example

3.8

An application of the first law LLL

NDE

DIED

One hundred kilojoules of heat is transferred to a system while the system does 200 kJ of work on the surroundings. Calculate the change in the energy possessed by the system for this process. A system is any fixed quantity of mass, like that sketched in Fig. 3.20. Solution W

Q

Since heat is added to the system, Q = 100 kJ. The work is done by the system on the surroundings so W = 200 kJ. When we apply Eq. 3.20, the change in the system’s energy is

O-W=AE 100kJ — (+200kJ) = AE

=.AE = —100kJ

The system loses 100 kJ of energy.

Example

3.9

A first-law problem — ELLEN

Water

Pee

a

The covered pot of Fig. 3.21 contains 2 kg of liquid water at 20°C. How much heat will the stovetop have to transfer to the water to heat the water to 50°C? Use more than one method to find Q.

Q

For many problems, in fact most of the problems in a first course in thermodynamics, the kinetic and potential energy changes are either zero or negligible, as in this example.

Solution

At 20°C and 50°C and atmospheric pressure, the water is a subcooled liquid, SO we ignore the pressure and use the temperature. Table C-1 provides the following properties:

At T, = 20°C: u, = 83.9 kJ/kg At T, = 50°C: u, = 209.3 kI/kg The top of the pot is covered, so the heat loss from the top of the water will be neglected. With W = 0, the first law reduces to the following:

O—

AU] in, —u,) = 2kg X (209.3 — 83.9) kJ/kg = 250.8 kJ

Chapter 3

The First Law for Systems

et

A positive Q indicates that heat was indeed transferred to the water. |

Or, the IRC Calculator at 20°C provides u, = 83.9 kJ/kg, and at 50°C it

provides u, = 209 kJ/kg, which gives

Q = AU= m(u, — uw)

= 2kg X (209 — 83.9) kI/ke = 250kJ Or, we could have used the specific heat C,, of water from Table B-4 to be 4.18 kJ/kg and found, using Eq. 2.36,

OF, AL

mC, eke

dT Ti;

find Ah. If the temperatures are relatively large, say

.

=

Ci)

=f)

Remember: Any of the five methods presented in Section 2.6 can be used to

(af C, = const)

(3.33)

:

above 500°C, it’s best to use the ideal-gas tables, the

For large temperatures, he ideal-gas tables in the Appendix F or the average C,, value from Table B-6 could be used, especially if accuracy is important rather than

average C, from Table 8-6 (or the IRC Calculator for

simple comparisons. ws If any nonequilibrium work mode is present, such as an electricresistance heater, that work must be included as an additional term in the above energy equations.

air only).

A constant-pressure process OE MATILDA

A vertical piston-cylinder device contains 20 kg of saturated liquid water. The weight of the piston is such that a constant pressure of 200 kPa is maintained in the cylinder. Heat is added to the water by a resistance heater until the quality is 0.3, as shown at state 2 in Fig. 3.24. A 12-V battery supplies a current of 2 A for 200 minutes. Determine the initial and final volumes of the water in the cylinder, the work done, and the heat transfer required for this process. (Continued)

Example 3.11

en

Part 1

Concepts and Basic Laws

Example 3.11

(Continued) = const

qT,

P

P, = = P

0, \* |= or 02

Pv"

k

(3.45)

= const

(3.46)

Equations 3.44, 3.45, and 3.46 are for an ideal gas undergoing an adiabatic (q = 0), quasi-equilibrium process. No nonequilibrium work modes are allowed; they cannot simply be added. Most often an engine is designed with a predetermined compression ratio r, represented in Fig. 3.26, which is defined to be the maximum volume divided by the minimum volume of the gas in the cylinder: P Roa

V,

==

Ko

(3.47)

! \

ne

If the working fluid is not an ideal gas, we return to Eq. 3.39 and observe that the sum of the two differential properties represents a third differential property, which we shall call dy. This is analogous to our definition of enthalpy in Eq. 2.11. This allows us to write

!

P; ae |

V4/V9 =F

SS

® ~ bee

c

dis = du

+ Pdv

(3.48) The pressure-volume diagram

Since dy = 0, p, = ,. We do not give the property y a formal name, but we do recognize that it is constant whenever the property,

for a quasi-equilibrium adiabatic compression process.

itt

Part 1

Concepts and Basic Laws denoted by s, the entropy, is constant. Hence, it is not necessary to create a new function and list it in the tables since s is listed in the various tables, as

Note: w= const fora quasi-equilibrium adiabatic process:

we shall discover in Chapter 6 (it is not necessary or helpful to define entropy at this point in our study). Consequently, an adiabatic, quasi-equilibrium pro-

=

cess requires &, = w, or S, = $j. Comment: Entropy will be defined in Chapter 6.

Example 3.13

&

An adiabatic (Q = 0) process for air Air at 20°C and 100 kPa is compressed in an insulated cylinder from a volume of 400 cm*, as shown in Fig. 3.27 For a compression ratio of 8 to 1, find

the required work. Assume constant specific heats and a quasi-equilibrium process. Insulated

100 kPa 20°C Air State 1

State 2

Figure 3.27

Solution Air is an ideal gas with a gas constant R = 0.287 kJ/ke-K.

Given:

State 1: T, = 20°C = 293K,

P,=100kPa,

V,=400cm?

State 2: V, = V,/8 (a compression ratio is a volume ratio) For a compression of an ideal gas in a cylinder, the variable ris used as the compression ratio. It is

The first law states that —W = mu, — u,) = mC{T,

—

T,)

We must determine m and T,. The mass of the system can be determined from state 1: m

_ PV,

RE, 2

100 x (400 % 107°)

0287,

03

= 4.757 X 10°-*ke

Chapter 3

The First Law for Systems

wily

The final temperature is found using Eq. 3.44 to be 0

k-1

ares a1 es 2 0,

= 293 x g!4-1 = 673K

Absolute temperature must be used, and k = 1.4 was found in Table B-2 in the Appendix. The work is now calculated to be Winn

PADS

Cl, 27, )

KAO

oOke x HT gaa

7293) K = 129,65

The specific heat C, was found in Table B-2 to be 717 J/kg-K. If 0.717 kJ/kg-K was used for C,,, the answer would be 0.1296 kJ.

An adiabatic process forsteam mone

Steam at 400°C and 4 MPa expands in an insulated cylinder to 600 kPa, as sketched in Fig. 3.28. Estimate the work output if the cylinder’s initial volume is 8000 cm®. Assume a quasi-equilibrium process, as was done in the preceding four examples. Insulated

P, = 600 kPa uperheat

State 1

State 2

Figure 3.28 Solution State 1 is in the superheat region, so Table C-3 is used.

Given: State 1: T, = 400°C, P,=4MPa,

V, = 8000 cm”

State 2: P, = 600 kPa The work output for the The process is adiabatic since the cylinder is insulated. quasi-equilibrium process, with Q = 0, is given by

W = m(u, — U)) (Continued)

Example

3.14

kite

Part 1

Concepts and Basic Laws

Example 3.14

(Continued) To find the work, we must know m, u,, and u,. The mass is found by using v7; =

|

0.07341 m*/kg from Table C-3:

aoe Vy _ 8000 x ae-6 mn 3 VY 0.07341 m°/kg Comment

If an entry is close, an interpolation is not needed. An answer to three significant digits is acceptable.

0.109 kg

Also, from Table C-3, we find u, = 2919.9 kJ/kg..To find u, we must use y, = ws, (see the discussion below Eq. 3.48) or s, = s, = 6.769 kJ/kg-K (we aren't concerned about the units since we are equating the numerical values). At P, = 600 kPa = 0.6 MPa, we search for the state where s, = 6.769. It is very close to the first entry in Table C-3 where P, = 0.6 MPa (600 kPa), so we use u, = 25674 kJ/kg. The work is estimated to be W = m(u, — 4)

= 0.109 kg(2919.9 — 2567.4) kJ/kg = 38.4 kJ

3.5.5

The polytropic process

In a polytropic process, the pressure and volume vary according to the relation

SSE

Polytropic process: A process in which the pressure and volume vary according to Pu" = CG where C is a constant.

Po" =C

(3.49)

where n and C are constants. This process is motivated by the adiabatic process equation (3.46) for an ideal gas with constant specific heats, which can be writ-

ten as P,v; = Pv or Po“ = C. To obtain the relationships among temperature, pressure, and specific volume, simply replace k with n in Eqs. 3.44, 3.45, and 3.46. An expression for the work for a quasi-equilibrium process is found by performing the integration:

oe

|‘Pdo = c]"orae = a

Note: For an ideal gas with constant

specific heats.

0;

& =p

Koes

xe ee)

(3.50)

ae il

But C = Pu" = P,v! = P,v3. Inserting these into Eq. 3.49, we get: |—

NRai) meted Fe) 62

PV 5 oa Ve WV eee

rn

(3.51)

=n

For an ideal gas, the ideal-gas law allows this to be expressed as

ep aaa) len

(3.52) :

The work integral for a polytropic procegs can be evaluated if you know either

the pressure and volume changes for the process or the temperature change for an ideal gas. But we’re reminded that these equations for the work are for quasiequilibrium processes only.

Chapter 3

The First Law for Systems

aT

Assume that an ideal gas is the working fluid. Then, we observe for a quasiequilibrium polytropic process: He)

constant-pressure process

n=

constant-volume process

n=1

isothermal process

n=k

adiabatic process

If n is not one of these numbers, it is simply a polytropic process. Also, observe that the expressions for the work are not valid if n = 1, that is, for an isothermal process. For an isothermal process, the work is found using Eq. 3.36 or 3.37

Air at a pressure of 200 kPa and a volume of 2 m? is allowed to expand in a polytropic process until the volume of the gas is 5 m’, as shown in Fig. 3.29. _ Determine the work done and the heat transfer during this process if

Poe = CC P (kPa)

200

@ @

\ \

Note: If Po'® = C, then = Pv'° = Csince mis constant for a process and P

hee

V=mo.

!

v(m?)

2

5

Solution Given:

State 1: P, = 200kPa, V,= 2m’, C = P,V;° = 200 X 2'° = 565.7

State 2: V,=5m° The pressure at state 2, using the polytropic equation, is

P, =

565.7 ys

2

us 565 = 0.002 m3/s

In all three examples above, the flows were steady flows. In Example 4.1 the density was constant, in Example 4.2 the conditions were specified and determined by the steam tables, and the inlet density and exit density were different. Example 4.3 involved air that was assumed to be incompressible. If the flow involves a compressible substance, as in flow through a nozzle, a process equation may be needed. In addition, the first law of thermodynamics, developed for a control volume in Section 4.2, may be required.

{ You have completed Learning Out

(2)

Chapter 4

The First Law Applied to Control Volumes

137

4.2 The First Law for Control

Mee VOUINOS 2 Ft

i

The major difference between the first law applied to control volumes and the first law applied to systems is that the equation is now a rate equation. It still represents conservation of energy, except now we consider rate of heat transfer and work rate across the boundary of the control volume. All terms in the equation will now have units of energy per unit time: kJ/s or kW, or Btu/s. Consider the control volume sketched in Fig. 4.12. To arrive at the desired energy equation, consider the rate of

For a control volume, we consider rate of heat transfer and work rate across the boundary of the control volume.

heat transfer Q, work rateW, and the rate of change of the kinetic

energy (review Eq. 1.22), the internal energy, and the potential energy. The work rate term W is composed of two parts: the work rate required to move the fluid into and out of the control volume due to the pressure forces, and the work rate W, that crosses the surface usually due to a shaft crossing the boundary. For a steady flow with one inlet and one outlet, the conservation of energy demands that The control volume used to derive the

ee

ee

is eee release ua ae

control surface

: Ona

.

the c.v.

ike

energy equation.

the c.v.

ealigl scale ot m(SVi all, = #2) = m( 3V3feline ty #2.)

(4.15)

The mass flux into, the mass flux out, and the work rate (a force times velocity) are, respectively,

MS Work rate = Force X Velocity

= PAX V = PAV

m, = pA\V,

My = pA;

(4.16)

This work due to a pressure force is occasionally called flow work.

LW = P,A,V, — P,A\V, + Ws where W, is the shaft work that crosses the boundary primarily through a rotating shaft. Substitute Eqs. 4.16 into Eq. 4.15 and arrive at

OF

2A

PAW,

: — We

1 pAv(3v3 +H, 2

+ ¢2,]

1

= aAw(sv} + u, + gz.)

(4.17)

If the flow is not steady,

the term dE_,/dt is added

Additional terms can be added to this equation to account for additional inlets and/or exits. Since there is one inlet and one exit in this steady flow, continuity

provides m = p,A,V, = p,A,V, and Eq. 4.17 takes the form ; Pp es

5

2

Vi ti, SS

Uy

=

Uu;

P, 1 toe oo ]p; Sb

p>

to the right-hand side of Eq. 4.17. It accounts for the change in energy in the control volume. The energy

E., would include internal le 2 SK

g(

| 1)

(4.18)

energy U, kinetic energy KE, and potential energy PE.

VEX

=Part1

Concepts and Basic Laws Enthalpy (h = u + P/p) can be introduced to obtain

Assumptions for the energy equation:

; Ove

1. Steady flow. 2. One inlet and one outlet.

3. Uniform flow at the inlet and outlet (the velocity, pressure, and enthalpy are uniform across the areas). 4. Heat into and work out of the control volume

are positive.

Control surface: It includes the areas where the fluid enters and exits the control volume.

Note: The shaft work term could include a resistance heater or a paddle wheel

in the control volume, in which case the term would be negative.

1

rahySU higes Ae: ~ Vi) er eeteatzy

(4.19)

a.

We refer to this equation as the energy equation or the first law. The left-hand side of Eq. 4.19 is the sum of the net rate of heat transfer into the control volume minus the net rate at which work is done by the control volume, most often by a rotating shaft. Similar to the energy consideration in Chapter 3, the boundary phenomena of the heat transfer rate minus the work rate represents a net rate of energy transfered to the control volume across the control surface. Several points must be considered in the analysis of most problems in which the first law is used. As a first step, it is very important to identify the control volume selected. Those parts of the control surface should be identified where the fluid enters and exits the control volume. The control surface should be chosen sufficiently far from an abrupt area change (an entrance, an exit, a valve, or a sudden contraction) that the velocity and pressure can be approximated by uniform distributions. It is also necessary to specify the process by which the flow variables change from the inlet to the outlet. If the fluid can be treated as an ideal gas with constant specific heats, then simplified equations may be used; if not, one of the methods listed in Section 2.6 must be applied. If the fluid is a low-speed gas or a liquid, a constant density process is often assumed. In most devices of interest in thermodynamics, the kinetic and potential energy changes are negligible. For such devices, illustrated in Fig. 4.13, the steadyflow energy equation takes the simplified form

Q — W, = m(h, — h,)

(4.20)

Devices for which Eq. 4.20 provides acceptable results include those for which the fluid is compressible such as turbines, compressors, and heat exchangers. The kinetic energy term must be retained in nozzles and diffusers. The gravity term is most often, if not always, neglected; it is very important in the analysis of a dam used to generate power, but that subject is covered in a fluid mechanics course. If a device utilizes an incompressible fluid (constant density) with negligible kinetic and potential energy changes, with u, = u, (negligible temperature change is assumed), return to Eq. 4.18 and a simplified form of the energy equation results: Figure 4.13 The steady flow of energy

*

QO =

into and out of a device.

;

W,

=

fk

a

n(2—1)

(4.21)

This equation is used to find the minimum power required to drive, for example, a water pump, a device that exists in every electrical power plant, or to determine the maximum power generated by a hydroturbine. The heat transfer term is typically neglected. The next several subsections present the common devices encountered in thermodynamics that are modeled as control volumes. v

™ You have completed Learning

C

(3)

Chapter 4

The First Law Applied to Control Volumes

[EE

4.2.1 Turbines, compressors, and pumps A pump is a device that transfers energy to a liquid with the result that the pressure of the liquid is increased. Power is supplied to the pump by a motor. Compressors and blowers also fall into this category but have the primary purpose of increasing the pressure in a gas. A turbine, on the other hand, is a device in which work is done by the fluid on a set of rotating blades. As a result, there is a pressure drop from the inlet to the outlet and mechanical power leaves the turbine. In some situations, heat may be transferred from any of these devices to the surroundings, but often the heat transfer is negligible. In addition, the kinetic and potential energy changes are negligible. For adiabatic devices (oi== 0), for which the kinetic and potential energy changes can be neglected, operating in a steady-state mode, the energy equation (see Eq. 4.20) takes the form

—-w,=h,-h,

or

-W,=m(h,—h,)

If an ideal gas, such as air, is the fluid in a compressor or a turbine, as in the

engine of a commercial aircraft, the process from the inlet to the outlet can be assumed to be an adiabatic quasi-equilibrium process (no heat transfer and no losses). Thus, assuming constant specific heats, an approximation to the unknown temperature is

P,\&-D/k (4.23)

This is a very important equation since the compression or expansion of a gas is a common process in several devices. The specific heats of choice in this chapter will be found in Table B-2 in the Appendix, which is common practice, although a more accurate answer would be found using those found in Table B-6 (an average temperature would be assumed for problems where the temperatures are not given). Variable specific heats for ideal gases, which will provide for more accurate calculations, will be considered in Chapter 6. The constant specific heats used in this chapter allow quick comparisons of various design changes; they are also used on national exams such as the Fundamentals of Engineering Examination. For liquids that are incompressible, such as water flowing through a pump, neglecting heat transfer and losses, the power required by the pump is W, = — W, so Eq. 4.21 becomes

ae

e/a

W, = th{ For a hydroturbine W, = W,,so

with

the

same

Wee i

—

assumptions,

P, — P; - :)

(4.24) the

power

Pumps, compressors, and blowers: Devices that increase the pressure in a fluid. PS -

eee

ee
Pid)

. S . ke-K . = 0488 x tO x (464 — 293)K roar = 68.4kW

This represents the least power required since an ideal process was assumed. Losses in the form of heat transfer from the compressor, and friction and separation as the air moves over the blades could be accounted for by using an efficiency for the device. The losses would increase the required power. Efficiency will be introduced in Chapter 6.

Example

4.5

A turbine generating power using steam LEP DISA A

Sieam enters the turbine of Fig. 4.15 at 4000 kPa and 500°C and leaves as a saturated vapor at 80 kPa. For an inlet velocity of 200 m/s, calculate the turbine power output. Neglect any heat transfer and kinetic energy change. (This process does not allow Eq. 4.23 to be applied since it is not an ideal gas. Also, there may be significant losses around the turbine blades.)

Chapter 4

The First Law Applied to Control Volumes

ity

4 MPa 500°C

—— >

5 cm dia.

Z

Wr 80 kPa x

=

be

Solution Given:

State 1

State 2

P, = 4000 kPa

P, = 80 kPa

T, = 500°C Steam tables:

ie Sik

h, = 3445 kJ/kg

h, = 2666 kJ/kg

The energy equation, in the form of Eq. 4.22, is se W, = mh, — h) The mass flux m is found as follows:

i m= p,AV, = Boe =

0.08643

Units: Always use kg, m,

x am X 0.0257 X 200 = 4.544 kg/s

and s and the units will work out. It is not necessary to check the units in every problem, just the ones that include terms with unusual units. But check them occasionally to be sure.

The maximum power output is then

: W,=

kg kJ Read eee X (2666 — 3445) a = 3540kJ/s

or

3540kW

Example

A pump increasing the pressure of water PE

4.6

RELESIETIOS

Determine the maximum pressure increase provided by the pump that requires 10 hp. Figure 4.16 shows a velocity of 4 m/s at the 6-cm-diameter inlet; there is a 10-cm-diameter exit. Do not neglect the kinetic energy change.

Note: If the kinetic energy change is to be ignored,

"the velocity V, would not —

be of interest so continuity would not be used.

vy

We Figure 4.16

Note: Using the density of

Solution

The energy equation (4.18) is used. By neglecting the heat transfer and assuming no increase in internal energy (100% efficiency), we establish the maximum pressure (Continued)

water as 1000 kg/m? (62.4 lbm/ft?) is acceptable unless the temperature exceeds about 100°C, as in Example 4.2.

ey

Part 1

Concepts and Basic Laws

Example 4.6

(Continued) SITUS

STE.

rise. Neglecting the potential energy change but retaining the kinetic energy change, the energy equation takes the form Units on AP/p: This is encountered quite often so make sure it is understood:

N/m? kg/m

Nem J Kg = kg

=

:

We

=

Poe

m( 2

ae

eae ez:

')

2 The velocity V, is given, and V, is found from the continuity equation as follows:

AV, =AVy

aX

003? X 4=a7%

005 X Vy Y=

14 mss

The mass flow rate needed in the energy equation, using p = 1000 kg/m°, is Units on V7/2:

kg rn = pA,V, = 1000 x (a x 0.03°) m> x 4— = 11.31 kg/s m

me Nem mJ ea N-s?/m

J kg

Recognizing that the pump work is negative, we find that the energy equation is W —((—10) hp AG 6 ip Engineers, in general, accept

to just two or three, at best.

The numbers in the tables are not measured values but are simply curve fits. So, a kinetic energy change of under 3% is usually ignored.

Example

4.7

kg}

AP

1.44" — 47]

J

2

ko

s |1000

..AP = 667 000 Pa

accuracy within 3%, or

so. Most problems have dimensions or material properties that are not known to four significant digits, and probably known

leo

or a maximum pressure rise of 667 kPa. The pump efficiency (to be introduced in Chapter 6) would decrease this pressure. Check the units in this equation. If

AP is in kPa, then W, would be in kW, and a factor of 1000 would be needed in the denominator of the kinetic energy term to convert from J/kg to kJ/kg. Note: In this example the kinetic energy terms account for about 1% of the energy transfer. This is very typical, and in most applications the inlet and exit areas are not given. But even if they are, as in this example, kinetic energy changes can usually be ignored in a pump or turbine.

Compressing a refrigerant Refrigerant 134a enters the adiabatic compressor of Fig. 4.17 at 15 psia and 20°F. It leaves the compressor at 140 psia and 150°F. If the mass flow rate of the refrigerant is 2 Ibm/s, determine the power required by the compressor to accomplish this. Neglect heat transfer and the change in kinetic energy. (There could be losses, so we do not assume a quasi-equilibrium process.) R134a:

State 1 (entrance)

Py = 15 psia T, = 20°F

superheated vapor

State 2 (exit)

» P,= 140 psia T, = 150°F

superheated vapor

Chapter 4

The First Law Applied to Control Volumes

ee

R-134a 20°F, 15 psia

Figure 4.64

Vo

——_ >

T> = 30°C

Part 1

176

Concepts and Basic Laws

4.62 Steam enters a nozzle at 4 MPa and 500°C at a velocity of 100 m/s and leaves the nozzle at 1 MPa and 250°C. The entrance area of the nozzle is 0.01 m?. Determine the mass flow rate

of steam through the nozzle, the exit velocity of the steam, and the exit diameter.

4.63 Steam enters a nozzle at 20 psia and 400°F at a velocity of 200 fps and leaves the nozzle at 14.7 psia and 360°F. The entrance area of

4.67 The air in a laboratory reservoir is maintained at 20°C. It flows out through a convergingdiverging nozzle into the laboratory maintained at 100 kPa, through a shape like that sketched in Fig. 4.66. Estimate the maximum velocity at the exit assuming an adiabatic, quasi-equilibrium flow with constant specific heats and a reservoir pressure of a) 600 kPa, b) 800 kPa, and c) 1200 kPa.

the nozzle is 0.1 ft”. Determine the mass flow rate of steam through the nozzle and the exit velocity of the steam.

4.64 Air at 140 kPa and 100°C enters the nozzle of Fig. 4.65. It leaves at 100 kPa with a velocity of 400 m/s. Determine the ratio of the exit area to the inlet area. Assume an adiabatic quasiequilibrium flow so that Eq. 3.45 applies. Figure 4.66

V; —

|400 m/s

€ B Unsteady Flow Ae

Figure 4.65 4.65

Refrigerant 134a enters a diffuser at 800 kPa, a temperature of 50°C, and a flow velocity of 160 m/s. It leaves the diffuser at 1 MPa and

60°C. The inlet area of the diffuser is 0.006 m?.

4.68 A valve separates the 10-m° insulated, evacuated tank of Fig. 4.67 from a pipe line that contains a gas at 2000 kPa and 300°C. The gas in the pipe is held at constant conditions after the valve is opened. Determine the final temperature and mass of the gas in the tank after the tank is completely filled if the gas in the pipe line is a) air, b) steam, and c) nitrogen.

Determine the mass flow rate of the refrigerant and the exit area.

Pipe line

2000 kPa, 300°C

4.66 Air enters an adiabatic diffuser at 500 kPa and 30°C with a velocity of 20 m/s. The diffuser has an inlet diameter of 8 cm. Air leaves the diffuser at 800 kPa and 80°C. If the exit diameter of the diffuser is 20 cm, determine:

i)

the mass flow rate

ii)

the exit velocity

ii)

the exit volumetric flow rate

iv)

the heat transfer to or from the diffuser

Figure 4.67

Chapter 4

The First Law Applied to Control Volumes

4.69 The evacuated tank in Problem 4.68 is not insulated. After a long period of time, the temperature is measured to be 347°C. Determine the heat transfer from the tank if the gas in the pipe line is a) air, b) steam, and c) nitrogen.

4.70 An insulated 4-m? volume contains a vacuum of air at 20°C. A valve is opened and atmospheric air at 100 kPa and 25°C rushes into the volume. After the air has stopped flowing in, estimate the temperature in the volume and the mass of air that entered if the initial pressure in the volume is a) 0 kPa, b) 10 kPa, and c) 25 kPa. 4.71

An air line at 100 psi and 70°F is used to fill

the 2800-in® tire of a vehicle that contains air at only 5 psi and 70°F. Assuming negligible heat transfer from the rigid tire during filling, estimate the temperature of the air in the tire when the pressure reaches 35 psi. Also, estimate the added mass of air. Assume that the conditions in the air line remain constant and that P.,,,, = 15 psia.

4.73 After a long period of time, the volume of Problem 4.72 a will reach the temperature of the atmosphere. Estimate the final pressure in the volume if the initial pressure was a) 2 MPa, b) 1200 kPa, and c) 800 kPa.

4.74 A 10-ft-diameter insulated spherical balloon contains air at 180 psia and 70°F. A special valve allows the air to leave at a constant rate of 0.4 Ibm/s. Calculate the temperature and pressure after a) 8 minutes and b) 15 minutes. Estimate the time needed for the temperature to reach —20°F;

€ B Cycles 4.75 The Rankine cycle of Fig. 4.69 operates between 20 kPa and 4 MPa with a maximum temperature of 400°C. For a mass flow rate of 10 kg/s, determine the adiabatic turbine power output, the pump horsepower required, and the cycle efficiency (the pump power is negligible in the efficiency calculation) if the quality at the turbine exit is a) 1.0, b) 0.92, and c) 0.85. The boiler and condenser are treated as heat exchangers.

4.72 The insulated 4-m? pressurized rigid volume of Fig. 4.68 contains air at 20°C. A valve is opened, and air discharges to the atmosphere at 100 kPa and 25°C. After the air has stopped flowing and the valve is closed, estimate the temperature in the volume if the initial pressure P. in the volume is a) 2 MPa, b) 1200 kPa, and c) 800 kPa.

—

Qe

Pump

4-m3 Volume ee

Qc

eo CS Pee

Figure 4.69 4.76

Figure 4.68

177

The Rankine cycle of Fig. 4.69 operates between 2 psia and 800 psia with a maximum temperature of 800°F. For a mass flux of 20 Ibm/s, determine the adiabatic turbine horsepower output, the pump horsepower required, and the cycle efficiency if the quality at the turbine exit is a) 1.0, b) 0.92, and c) 0.85. The boiler and

condenser are treated as heat exchangers.

178

Part 1

Concepts and Basic Laws

4.77 The refrigeration cycle of Fig. 4.70 operates with R134a. It is powered by an input of 30 kW to the compressor. Determine the cooling rate and the COP. The inlet pressure to the compressor is a) 120 kPa, b) 140 kPa, and c) 180 kPa. Q Cond

eonseiee!

4.79 Air at 3.8 kg/s enters the Brayton cycle of Fig. 4.72 (such a cycle could be used in a large truck engine). For a combustor outlet temperature of a) 650°C, b) 850°C, and c) 960°C, determine:

i)

(The The heat emit

iv)

The cycle efficiency

1 MPa, 60°C Compressor|

Throttle

520 kPa

:

Compressor

Figure 4.70 4.78 The refrigeration cycle of Fig. 4.71 operates with ammonia. If it is powered by an input of 50 hp to the compressor, determine the cooling rate and the COP. The inlet pressure to the adiabatic compressor is a) 16 psia, b) 25 psia, and c) 40 psia. Q Cond

;

180 psia, 160°F

Compressor

Evaporator

‘ Q Evap

Figure 4.71

Combest:

2)

QeEvap

Throttle

BWR (see Example 4.20) heat rate emitted (add a fictitious exchanger between states 4 and 1 to the heat)

Assume an adiabatic quasi-equilibrium compressor and turbine.

Evaporator

Condenser

The power output

ii) iii)

20°C

m = 3.8 kg/s

Figure 4.72 4.80 Air at 8 lbm/s enters the Brayton cycle of Fig. 4.73 (such a cycle could be used in a large truck engine). For a combustor outlet temperature of a) 1400°F, b) 1600°F, and c) 1800°F, determine:

i)

The power output

ii)

The BWR (see Example 4.20)

ili)

The heat rate emitted (add a fictitious heat exchanger between states 4 and 1 to emit the heat)

iv)

The cycle efficiency

Assume an adiabatic quasi-equilibrium compressor and turbine.

70°F m=

Figure 4.73

8 lbm/s

Outline 5.1 Second-Law Concepts

180

5.2 Statements of the Second Law

oe

of Thermodynamics 5.2.1 5.2.2

~The

Second

5.3 Cycle Performance Parameters 5.3.1 5.3.2

. Law

of

.

_ Thermodynamics

182

Kelvin-Planck statement—heat engines Clausius statement—refrigerators 183

:

Theheatengine 185 The refrigeration cycle

5.4 The Carnot Cycle 5.5 Summary

186

189

194

FE Exam Practice Questions Problems

185

197

196

182

The following variables are introduced in this chapter:

Q,

Heat transfer from a high-temperature

T,,

reservoir Heat transfer from a low-temperature reservoir

T,

Q,

Temperature of a high-temperature reservoir Temperature of a low-temperature reservoir

mE

Learning Outcomes 1

The concepts of reversibility and irreversibility

Gi

Statements of the second law

(1

General cycle performance parameters

Cy

The Carnot cycle performance parameters

[:this chapter, we will examine the second law of thermodynamics and study its implications. Unlike the first law of thermodynamics, which is a fundamental tool for problem solving, the second law informs us of not what can be done but of what cannot be done in thermodynamic processes and cycles. The second law of thermodynamics also puts very real limits on the efficiency of devices and cycles, limits that are often surprisingly low. The second law has led to many discussions and statements. Here are two interesting statements:

1.

2.

“The tendency for entropy to increase in isolated systems is expressed by the second law of thermodynamics—perhaps the most pessimistic and amoral formulation in all human thought,” by Gregory Hill and Kerry Thornley, Principia Discordia (1965).

“There are almost as many formulations of the second law as there have been discussions of it,” by Philosopher/Physicist P. W. Bridgman (1941).

Unlike the philosophers, we will make a technical statement of the second law, one that can be used not only to make decisions on’ proposed designs but also to decide whether improvements should be made in existing devices and cycles.

2.1 Second-Law Concepts. Reservoir: A body that able to Brcvideheat oF absorb heat withouta. change in its temperature.

aserie a oN 1S ee - need of a ae of energy, but it must also Pp or energy disposal. Such soyrces and disposals are termed reservoirs. | source of thermal energy is a heat source, and a thermal energy disposal site is a heat sink. They will usually be depicted as shown in Fig. 5.1. A reservoir is a sufficiently large body that it is able to provide heat or absorb heat without a

Chapter 5

The Second Law of Thermodynamics

change in its temperature. The ocean, a lake, a river, and the atmosphere are the most common heat sinks that serve as reservoirs where heat is dumped. A reservoir need not be large; a furnace that supplies heat at a constant temperature could

én

Heat source

be considered a reservoir that serves as a heat source, as sketched in the schematic

of Fig. 5.1. An ice bath that stays at 0°C could be a heat sink. The important property of a reservoir is its constant temperature. In the case of an industrial furnace or the coal fire in a power plant, the reservoir’s temperature could be very high. Between a heat source and a heat sink, a process takes place. A reversible process is one that, once completed, can be reversed and returned to its original state, leaving no change in either the system or the surroundings. Every system can be returned to its original state with some type of process, but every real process will require that the surroundings provide more energy than the surroundings received during the initial process. An example of a process that approaches a reversible process, which does not utilize heat reservoirs, is a spring that requires the surroundings to do work to stretch it; when released, it returns to its original shape, providing essentially the same amount of work to the surroundings. A frictionless piston compressing the air in the cylinder of an engine is essentially the same process as the spring and will be considered reversible; during its expansion, it gives back the energy required during the compression. All thermodynamic processes that utilize heat reservoirs are irreversible due to physical aspects of the processes that allow transformations only in one direction. Unlike the spring, heat flows in only one direction: from hot to cold. Several common effects that introduce irreversibilities follow. 1.

Heat transfer— Heat transfer across a finite temperature difference is not reversible. Heat will naturally flow from a hotter region to a colder region, as in Fig. 5.2, but will never naturally flow from a colder region B to a hotter region A. The only way for heat transfer to flow from a colder region to a hotter region is to use a refrigeration cycle; it will not happen naturally. The surroundings must provide work to operate the cycle, as in the familiar refrigerator. The irreversibility of heat transfer across large temperature differences leads to the substantial losses in most thermodynamic processes, especially in vehicle engines and power plants.

2.

Friction— When two surfaces rub against one another, some of the energy in one or both of the surfaces is turned into heat energy (see Fig. 5.3). A good example is when you rub the palms of your hands together to warm them. The friction products heat. This process is not reversible. The two surfaces will not return to their original positions by returning the heat.

3.

Mixing—If you mix two substances together, such as salt and pepper, or the red water and blue water in Fig. 5.4, they will not separate back into the original substances by themselves even with significant effort from the surroundings. The mixing process is not reversible. Velocity

Body A

Body B

Figure 5.1 A heat source and a heat sink.

Reversible process: One that can be reversed and returned to its original state, leaving no change in either the system or the surroundings. The quasiequilibrium process did not involve the surroundings.

Red water

_ Blue water

a

F

Hot

pW

Heat flows from a hot to a cold

body.

Friction between two surfaces.

Mixing red water and blue water in a container.

182

Part 1

Concepts and Basic Laws

Partition

Evacuated

4. Unrestrained expansion—The unrestrained expansion of a gas is not reversible. If you rupture or instantly remove the partition of Fig. 5.5, the gas will suddenly flow from the high-pressure region into the evacuated region. It will not, without energy supplied from the surroundings, return to its original state.

These four physical phenomena result in irreversible processes. No known process is free of all of these effects. All four of these phenomena demand that a process occurs in a particular “direction.” For example, heat only flows with no outside effort, from hot to cold; friction does not help propel an object; rotating a blade in the opposite direction will not Gas suddenly flows from the highreorganize water that has been mixed; and the two gases will not return pressure region to the evacuated to their separate regions without outside help. region when the partition is ruptured. Some processes used in industry come close to being reversible —so close that we will consider the processes to be reversible in our solutions to problems. The compression and expansion processes in the cylinder of an engine are examples since friction can be essentially eliminated. The turbines described in There is no known Chapter 4 can also be designed to be nearly reversible. Note that most, if not all, reversible process. of the processes that are nearly reversible do not involve heat reservoirs. All processes are inherently irreversible, especially those with heat transfer across large temperature differences. There is no such thing as a perpetual motion machine that operates with reversible processes, even though some clever entrepreneur may attempt to sell the idea to investors! region

[Y/ You have completed Learning 0

(1)

5.2 Statements of the Second Law

Energy source

Work

An impossible device that operates on a cycle.

This text uses two statements of the second law of thermodynamics: the KelvinPlanck statement, which applies to heat engines, and the Clausius statement, which applies to refrigeration systems. Both are statements of observation and are not derived from other physical laws. Each also indicates the direction in which a process will occur.

9.2.1 Kelvin-Planck statement—heat engines The Kelvin-Planck statement of the second law of thermodynamics (see Fig. 5.6) states: It is impossible to construct a device that operates on a cycle whose sole effect is to take energy from a heat source and convert it into work.

Heat engine: A device that accepts heat from a reservoir and supplies work while operating on a cycle.

The device that accepts heat from a reservoir and supplies work while operating on a cycle is called a heat engine. During its operation, it must reject heat to the surroundings. Consider the operation of the simple heat engine in Fig. 5.7 Heat OF As transferred to the engine from the heat source at temperature T,,, work W is

Chapter 5

The Second Law of Thermodynamics

produced, and heat Q, is transferred from the engine to the heat sink maintained at temperature T,. The Kelvin—Planck statement of the second law states that the heat transfer Q, to the heat sink cannot be zero. Some heat must be

Heat source

[MEE Ty

rejected by the engine. Essentially, W cannot equal Q y Lhe first law of thermodynamics applied to a boundary around the engine would demand that One

On

(5.1)

The second law will actually provide the minimum size of the discarded heat Q,, as we shall determine in Section 5.4. Actually, Q, must often be quite large, which is why inventors are encouraged to invent engines that violate the second law. Heat engines will be analyzed in detail in Chapters 8 and 9.

Figure 5.7 A heat engine accepts heat from a heat source and rejects heat to a heat sink.

5.2.2 Clausius statement—refrigerators The Clausius statement of the second law of thermodynamics (see Fig. 5.8) states: It is impossible to construct a device that operates on a cycle whose sole effect is the transfer of heat from a low-temperature reservoir to a high-temperature reservoir. Heat does not naturally flow from a cooler region to a hotter region; in fact, it does the reverse. In order to move heat from a cooler region to a hotter region, we must force it to do so by providing work from the surroundings. The cycle that does this is called a refrigeration cycle used by refrigerators, air conditioners, freezers, and heat pumps— generically referred to as refrigerators. The energy requirements of a refrigerator are shown in Fig. 5.9. The refrigerator removes heat Q, from a low-temperature reservoir maintained at temperature 7, and rejects heat to the high-temperature reservoir maintained at temperature 7,,. In order to operate the refrigerator, work W is required from the surroundings. The low-temperature region could be a freezer, and the

High-temperature reservoir

High-temperature

ip

region

Figure 5.8 An impossible device.

A refrigerator requires work to move heat from a cool region to a hotter region.

Refrigeration cycle: A cycle that moves heat from a cooler region to a hotter region.

Part 1

Concepts and Basic Laws high-temperature region would be the hot coils on the back or bottom of the freezer. The Clausius statement of the second law states that heat will not travel from the low-temperature reservoir to the high-temperature reservoir without the addition of work. In layperson’s terms, you can’t cool your house in the summer unless you spend money running an air conditioner. The first law applied to the refrigeration cycle is Wei

(5.2)

Equation (5.2) is the same as Eq. 5.1 because the refrigeration cycle is simply the reverse of a heat engine cycle. The difference is that the heat engine creates work from purchased heat, whereas the refrigerator operates on purchased work. The refrigeration cycle will be studied in detail in Chapter 10.

Example

9.1

The Clausius and Kelvin-Planck statements | PIRI

EON

Show that the Clausius statement of the second law is equivalent to the KelvinPlanck statement. Solution

To show that the two statements are equivalent, consider the refrigerator on the left of Fig. 5.10a. This refrigerator violates the Clausius statement of the second law since it transfers heat from a low-temperature reservoir to a

Chapter 5

The Second Law of Thermodynamics

high-temperature reservoir with no work input. Assume that the engine on the right, which does not violate the Kelvin-Planck statement of the second law,

rejects the same amount of heat Q, used by the refrigerator. Now, place a boundary around the engine and the refrigerator, as displayed in Fig. 5.10b, which shows that the heat rejected by the engine is used to supply the refrigerator. Since Q, = Q,,, as required by the first law applied to the refrigerator, and Q,,, = Q, + W,as required by the first law applied to the engine (Qu,, 1s the heat from the high-temperature reservoir to the engine),

the result is a net heat transfer of (Q,,; — Q,) from the high-temperature reservoir directly into the “device.” This device is composed of the refrigerator plus the engine inside the dotted boundary. This net heat transfer (O ue ~ Q1) is converted directly into work W. Heat that is converted directly into work by a device operating on a cycle is a violation of the Kelvin-Planck statement of the second law. Hence, it is concluded that the Clausius statement and the Kelvin-Planck statement are equivalent.

[™ You have completed Learning Outcome =

The performance parameter for a thermodynamic cycle is defined as Output Performance parameter = =a Energy input

(5.3)

The required input is the energy (heat or work) that must be supplied to the cycle to make it operational. The output is the energy transferred by the cycle. In the case of a heat engine, it is the work transferred to the surroundings by the engine. In the case of an air conditioner, it is the heat accepted from the low-temperature region. In the case of a heat pump, it is the heat transferred to the high-temperature region. In the next sections, we will analyze performance parameters for the above cycles and determine how well these cycles function.

The required input is _ typically the energy that must be purchased, although work can be created using “green” energy such as wind, hydro, or solar. This sustainable energy !s gaining considerable interest, but it

still must be “purchased.”

Efficiency: Output

se

Required input

5.3.1 The heat engine The performance parameter for a heat engine is the cycle efficiency », defined by Output

if) =

_W

rE = Requiredinput Q,,

(5.4)

The required input need to make a heat engine work is the heat transfer Q,, from a high-temperature reservoir. The output of an engine is the work produced W. The Kelvin-Planck statement of the second law states that the heat rejected to the

The heat rejected by an engine or a power plant cannot be zero. We shall see in Section 5.4 that the rejected heat must usually be quite large.

ae

Part 1

Concepts and Basic Laws low-temperature reservoir cannot be zero. If we insert the left-hand side of Eq. 5.1

The maximum efficiency of most engines is much less than 100%. A power plant has an efficiency of around 40%, and that of an automobile engine is around 20%.

Example

5.2

for the work, we obtain

apiliean Qy

Q1

(5.5)

pines

Q1

Since Q, cannot be zero by the second law, the engine efficiency must be less than unity (7 < 1),so that a heat engine can never have an efficiency of 100%. In fact, we will discover that the efficiency of most heat engines is considerably less than 100%. Proposed engines with efficiencies close to 100% undoubtedly violate the second law. It should be pointed out that rates of work W and heat transfer QO, and QOyi can also be used in Eq. 5.5.

The efficiency of a heat engine | The heat engine of Fig. 5.11 receives 600 kJ of heat from a heat source. It rejects 400 kJ of heat to the environment. Calculate i) the work done by the heat engine and ii) the engine efficiency.

Quy = 600 kJ Sar

WwW

Q, = 400 kJ

Solution i) The work done by the engine is found by applying the first law: Woe

On OF

= 600 kJ — 400 kJ = 200 kJ ii) Equation (5.5) allows the efficiency of the engine to be calculated. It is WwW a] = eS O,,

200 kJ 600KI

0333,

S08

35.59a

This would represent a typical engine.

5.3.2 The refrigeration cycle ss

Refrigerator: A refrigeration cycle that cools a space.

Note: An air conditioner and a freezer are included

in the definition of a

refrigerator. —

The performance parameter for a refrigeration cycle is called the coefficient of performance, or simply the COP. A refrigerator operates on a cycle with the objective of cooling a space. It has a coefficient of performance of

Output COP, * =~~ Required = Oy,W input

(5.6)

The required input to make a refrigerator operate is the work input W to the compressor. The output is the heat removed Q, from the low-temperature region, for example, the inside of a refrigerator. If we substitute Eq. 5.2 for the compressor work, we obtain

Cr COP, =

W

eae O7

=

OQ;

Qu

(5.7)

Chapter5

The Second Law of Thermodynamics

The coefficient of performance has no defined upper limit as does the efficiency of a heat engine; that’s why it is not referred to as “efficiency.” If the work required by a refrigerator approaches zero, then the COP, for the system could theoretically approach infinity. When the second law is applied to the refrigeration cycle, we will discover that it has a limit that is closer to 10 than to infinity. ACOP, < 1 for arefrigeration cycle is an indicator of poor performance, although in certain situations, a COP oul is acceptable, as in a commercial airplane or on a university campus, where absorption refrigeration (to be presented in Chapter 10) may be used. Heat transfer rates Q, and Q,,, rather than Q, and Q,,, are often used

in the equations for COP. Figure 5.12 shows the basic components of a central air-conditioning system for a building. The compressor takes in refrigerant, the fluid used ina refrigeration cycle, at low pressure and low temperature and compresses it to a higher-temperature superheated vapor. As the pressure of the refrigerant increases, so does the temperature. The hot refrigerant exits the compressor at about 50°C (120°F). The hot refrigerant is then piped outside, where it passes through a heat exchanger, the condenser. The high-pressure refrigerant exchanges heat with the outside air where it is condensed to a saturated liquid. The refrigerant then passes through a valve where the sudden drop in pressure creates a sudden drop in the refrigerant temperature. The mixture is now at about —30°C (—20°F). This cold refrigerant then passes through an evaporator where it exchanges heat with the air in the house. The refrigerant then returns to the compressor. The processes in this cycle will be studied in detail in Chapter 10. A heat pump is a device that operates on a refrigeration cycle in order to heat a building, for example, the house of Fig. 5.13. The hot refrigerant from the compressor is diverted into the heat exchanger used in the house for air conditioning. Heat is exchanged between the hot refrigerant and the air in the house. The refrigerant then flows through the valve, where the pressure drops and the refrigerant becomes very cold. Next the cold refrigerant passes through the heat exchanger outside the house, where it absorbs heat from the outside atmosphere. This happens because the refrigerant must be much colder than the outside air, so heat pumps are not common in locations where the outside temperature is less than about 20°F. The refrigerant is then sent back to the compressor. This is actually accomplished using clever valving and piping with the same system used for the air conditioner. The performance parameter for a heat pump is also called the coefficient of performance. The output is now the heat delivered to the house Q,,,80 the coefficient of performance of a heat pump Is

COP,

= os

Qy Ww

=

On QO, —

1

= Q,

er

ray

Compressor

Evaporator

Figure 5.12 Basic components of an air conditioner. Comment

A simple valve suddenly reduces the pressure in the refrigerant with a simultaneous reduction in temperature.

Ww

Evaporator

Expansion valve

Compressor

Condenser

A heat pump uses the same refrigerant and components used by the air conditioner of Fig. 5.12.

(5.8) Heat pump: A device that heats a space by operating on a refrigeration cycle.

A comparison of Eq. 5.7 with Eq. 5.8 would show that

There is also no limit on COP,,p. In Section 5.4 very large —usually less than 10 for an ideal heat It should be pointed out that the COP of ever, listed. Manufacturers list the EER (energy

eee

Condenser

‘(ee Q.

+ 1

the coefficient of performance of a refrigerator has no defined upper limit.

Expansion

Qi COP,» = COP,

187

(5.9)

we will find that the COP is not pump. an air conditioner is seldom, if efficiency ratio), which is Q,/W

Manufacturers use an EER which is Btu/W-h, so that

EER = 3412 x GOR

3

=Part1

Concepts and Basic Laws with units of Btu’s for Q, and watt-hours for W. There are 3.412 Btu/W-hr, so EER

Observation: A SEER

EER: the Btu’s divided by the watt-hours for the entire cooling season. The government mandates a minimum SEER of 12 for all air conditioners, although 13 is being considered.

of 13 is a COP of 3.8,

considerably less than infinity, which is the upper limit for the COP.

Example

= 3.412 COP. Also listed by the manufacturer is the SEER, the seasonal

5.3

The performance of an air conditioner The compressor in the schematic of an air conditioner in Fig. 5.14 requires 5 hp. It rejects 12 kJ/s of heat to the environment. Calculate i) the cooling rate, in tons, and ii) the performance of the air conditioner.

Solution

i) The cooling rate is found by applying the first law to the air conditioner as the “refrigerator” box in Fig. 5.9: Note:

Aton isa

customary unit for a measure of the rate of heat transfer in a refrigeration cycle. It is the rate of heat transfer

O, ma OF =

ora oe

needed to melt a ton of —

ice at O°C in one day.

W

ww) = ((12kW — Shp: x 0.746—— bp)

dS

1 ton asea

——

ii) Equation (5.6) allows the coefficient of performance to be calculated. It is

8.27 kW COP, = 1 = i = 222 Ww 5X0746kW —— Unit:

1 ton = 3.517 kw

Example

5.4

The heat pump A refrigeration cycle operates as a heat pump requiring 5 hp by the compressor, as shown in Fig. 5.14. It supplies 12 kJ/s of heat to a space. Calculate i) the rate at which heat is absorbed from the environment and ii) the performance of the heat pump.

Chapter 5

The Second Law of Thermodynamics

Solution

i) The rate at which heat is absorbed from the environment is found by apply-

ing the first law to the heat pump as a unit (the “refrigerator” box in Fig. 5.9): QO, aa Oo; oe Ww

=

12kW

—- Shp

kW Xx wary

= 8.27 kJ/s

Observe that the energy equation is identical to that used for the air conditioner of Example 5.3. ii) Equation (5.8) allows the coefficient of performance to be calculated. It is a

COPyp = Qu _ Be I ete = WwW De GAG Wem

RSs

39)

Observe: The COPs from Examples 5.3 and 5.4 satisfy Eq. 5.9.

The definition of the coefficient of performance changed compared to that of the air conditioner of Example 5.3 since the desired output changed

from Q, to Oy.

(3)

Now that the efficiency of a heat engine has been defined, we consider an engine that operates with quasi-equilibrium processes—

‘ a

processes that have no friction, no sudden expansion, and no heat

transfer across finite temperature differences. Such an engine does not exist, but it does define the most efficient engine possible since it has no irreversibilities. It is referred to as a Carnot engine. Consider our fictitious engine to operate using a piston-cylinder arrangement with the four quasi-equilibrium processes of Fig. 5.15; an ideal gas with constant specific heats will be assumed. The processes are as follows. 152:

2-3:

GQ. =

;

:

a NS

T, = const

@

ty, processes that make up the Carnot gas cycle.

An adiabatic compression. The piston compresses a gaS frOM 2 low temperature 7, to a high temperature T, with the cylinder completely insulated

ae

An isothermal expansion. Heat is transferred from a high-temperature reservoir to the gas at the high temperature 7, = 7, = T,,.The reservoir

guys Carnot engine: An ideal engine that operates in a

cycle with no irreversibilities. _|t can operate with any

fluid, but an ideal temperature is a fraction of a degree above the gas temperature, so both —_Working gas will be assumed for

are approximated to be at 77,

334:

which simplified equations

The gas is expanded from a high temperature 7; _ are available. An adiabatic expansion. to a low temperature 7,, with the cylinder completely insulated.

Part 1

Concepts and Basic Laws 41:

Note: It is the two heat transfer processes that are assumed to occur across an infinitesimal temperature difference that make the Carnot cycle a fictitious

cycle. For heat transfer to occur quickly, alarge temperature difference is — necessary.

An isothermal compression. low-temperature reservoir at temperature is a fraction of a both are approximated to be

Heat is transferred from the gas to a the low temperature 7, = T; = T,.The gas degree above the reservoir temperature, so

at 77.

This Carnot engine operates with ideal processes that are reversible with no friction, so they cannot be improved upon; the cycle must possess the maximum possible efficiency. The two processes that make up the cycle have been analyzed in Chapter 3. The processes from state 1 to state 2 and from state 3 to state 4 are adiabatic pro-

cesses for which Q,, = Q,., = 0.The isothermal process from state 2 to state 3 was analyzed in Section 3.5.3, with the result that the heat transfer Q, ,= Q,, is given for the ideal gas by Eq. 3.35 and Eq. 3.36 to be The two processes used in the Carnot gas cycle are the adiabatic process and the isothermal heat transfer process.

V; QO, = mRT,In— (5.10) V, The heat transfer Q, = —Q,., for the isothermal process from state 4 to state 1 is given by

OO; = mk

Note: We are assuming an ideal gas with constant

specific heats. The objective of Problem 5.32 is to show that the working fluid does

©

not affect the efficiency of | the Carnot cycle. So, the ideal gas is an acceptable working fluid.

positive in the equations.

(5.11)

The heat transfers Q,, and Q, are expressed to be positive so that Eq. 5.5 for the cycle efficiency is acceptable (we recognize that Q, ,from state 4 to state 1 is negative since heat is leaving the gas). Equation (5.5) for the cycle efficiency then provides

aT

Cm

1

?

Oy a

mRT, In V,/V,

Maman

Suc The heat transfers Q,, and Q, are defined to be

Ven

ln V,

Ly a Vags V; Sone : Te nV V2

(5.12)

But during the adiabatic quasi-equilibrium processes, Eq. 3.44 is used to give

i2

E H=

iepeals

V\ eo (2)

IE

VN as sje 18“= (2)

i

(iy

V;

(5.13)

so that

Vive V; =

Vea

V;

Or

a

=

Wa

(5.14)

Consequently, the two In terms in Eq. 5.12 cancel out, providing us with the surprisingly simple expression for the efficiency of the Carnot cycle:

i

T,

(5.15)

Compare Eq. 5.5 with Eq. 5.15. We have replaced Q,/Q,, with T,/T,,,; that is, The ratio Q,/Q,, is replaced with T//T,, in this ideal Carnot cycle.

Qn _ es Ty eee

Op,

5.16 ae

The efficiency of this ideal reversible cycle, the Carnot cycle, is dependent only on the absolute temperatures of the two heat reservoirs. This is, in fact, true of all

reversible cycles; they all have the same efficiency. The efficiency does not depend

Chapter 5

The Second Law of Thermodynamics

on the working fluid (the conclusion of Problem 5.32), and it is impossible for any cycle operating between the same two reservoirs to have an efficiency greater than the Carnot efficiency. The Carnot cycle can be reversed resulting in a refrigerator or a heat pump, depending on the objective of the cycle. If the objective is to cool, we use the

COP,, defined by Eq. 5.7 We substitute T,,/T, for Q,,/Q,, and we obtain the

maximum possible COP for a refrigerator:

COR

eee Caer

er

If the Carnot power cycle of Fig. 5.15 is reversed, the Carnot refrigeration cycle results.

(5.17)

lh seared es

rT, If the objective is to heat, we substitute T,/T,, for Q,/Q,, in Eq. 5.8 to obtain the maximum possible COP for a heat pump:

CORES =

1

=

ie H

(5.18)

It is true that the Carnot cycles considered above are fictitious cycles that cannot be built, let alone tested in the laboratory. Why then are they defined? It is important to know the maximum possible performance that an engine or a refrigerator can have when operating between two temperature reservoirs. If an engine has an efficiency of 30% and the Carnot efficiency operating between the same temperatures

All Carnot cycles are fictitious. They serve as limits for real cycles.

is 33%, it would not be advisable to devote substantial effort to increase the perfor-

mance of that engine. Also, if an inventor proposes an engine that operates with an efficiency of 60% but the Carnot engine, operating between the same high and low temperatures, has an efficiency of 55%, the engine will not operate as proposed.

Can the Carnot efficiency be exceeded? Show that the efficiency of a Carnot engine cannot be exceeded.

Solution

.

A very clever engine is proposed that has an efficiency that exceeds the Carnot efficiency. The engine accepts Q,, from the high-temperature reservoir, and a Carnot refrigerator rejecs the same amount of heat to that reservoir, as shown in Fig. 5.16.

The work W,, produced by the engine is greater than the work W required by the Carnot refrigerator since the proposed engine is more efficient than the refrigerator.

(Continued)

Example

5.5

er

Part 1 Concepts and Basic Laws

Example

5.9

(Continued) Now, consider the engine and the refrigerator to operate as a single device, as identified inside the dotted boundary of Fig. 5.17 The heat rejected by the refrigerator is used to drive the engine. The heat Q, from the low-temperature reservoir is greater than the heat rejected Q, , by the engine since W,, > W. Consequently, the single device inside the dotted boundary accepts heat (Q, — Q, ,) from a single reservoir and produces net work (W,, — W); this is a violation of the Kelvin-Planck statement of the second law. It is concluded that no engine operating on a cycle can have an efficiency greater than the Carnot efficiency.

d oie aeS Bropeeer| engine is possible.

Example

oS

5.6 _ The efficiency of an engine The heat engine of Fig. 5.18 receives 600 kJ of heat from a heat source at 600°C. It rejects 300 kJ of heat to an environment at 10°C. Calculate i) the work done by the heat engine, ii) the actual engine efficiency, and iii) the maximum possible engine efficiency.

Ty = 600°C

Solution w_

For the two reservoirs:

Q,, = 600 kJ

T,, = 600°C = 873 K

QO, = 300kI

TA

ee es

i) The work done by the engine is found by applying the first law: Figure 5.18

WO

0.

= 600kJ — 300kJ = 300 kJ ii) Equation (5.5) allows the actual efficiency of the engine to be calculated. It is n=

W Q,

300KJ 600kJ

= 0.5 or 50%

Chapter 5

The Second Law of Thermodynamics

iii) The maximum possible efficiency, the Carnot efficiency, is found, using Eg.5.15,t0 be i Neca

ene

ig =

]-—-

283K ——_ 873K

=

RB 0.676-or-

569 67.6%

The actual engine efficiency can never be greater than the Carnot efficiency. If it is greater, it is considered a violation of the 2"? law.

Is this invention possible?

itty

Observation: When a difference of temperatures is desired, it doesn’t matter if absolute or relative temperatures are used, but when a temperature appears alone, as in the numerator of Eq. 5.17, or as a ratio, invariably absolute units must be used.

Example

5.7

MNES

An inventor claims to have developed an automobile engine that burns biofuels at 600°F with an engine efficiency of 50%, as demonstrated by Fig. 5.19. If the local air temperature is 60°F, can this claim be believed?

Heat

source

7, = GOOF

Solution

The upper and lower absolute temperatures for this engine are

T,, = 600°F = 1060°R

1 = 60°F = 520°R

The Carnot efficiency of the engine, using Eq. 5.15, is T, ifC.armo

t

Se

i! Beet Ti

Se |

520°R 2054 71060°R

as

51%

An engine efficiency of 50% is possible according to the second law but is not very likely. Real effects, especially heat transfer across a finite temperature difference, would lower the actual efficiency by a considerable amount.

The COP of an air conditioner An air conditioner maintains a house temperature of 23°C while the outside temperature is 40°C. If the refrigeration cycle removes 8 kJ/s from the house and the compressor does 2 hp of work on the refrigerant, as shown in Fig. 5.20, determine i) the heat rejected to the outside environment, ii) the actual coefficient of performance for the air conditioner, and iii) the maximum possible coefficient of performance for this air conditioner. (Continued)

Example

5.8

gry

Part 1

Example

Concepts and Basic Laws

5.8

(Continued)

Solution The given information is

T, = 23°C = 296K

T,, = 40°C = 313K

QO, = 8kJ/s

W = 2hp = 1.492kW

i) The first law provides the heat rejected to the high-temperature reservoir:

O,=W+

oO,

1.492 + 8 = 9.49kJ/s ii) The actual COP for this air conditioner, using Eq. 5.6, is

Se

Ae i 2

ili) To obtain the maximum possible COP, the Carnot COP, we use Eq. 5.17:

T,

Ee P

—

296

ie

ra

BEC

= 17.41

This must be greater than the actual COP, which it is.

(4)

In this chapter, the second law was intyoduced along with the concepts of reversibility and irreversibility. The primary contributors to irreversibility were identified: friction, heat transfer across a finite temperature difference, mixing,

and unrestrained expansion. The Kelvin-Planck and Clausius statements of the

Chapter 5

The Second Law of Thermodynamics

Mar

second law were then presented. The performance of cycles was investigated with the most efficient cycle possible, the Carnot cycle, given special attention. It was analyzed for a four-process, ideal-gas cycle with its maximum efficiency expressed in terms of temperatures. The following additional terms were defined in this chapter: Carnot cycle: An ideal cycle that operates with no losses. It can be an engine, a refrigerator, or a heat pump. Clausius statement: /¢ is impossible to construct a device that operates on a cycle whose sole effect is the transfer of heat from a low-temperature reservoir to a high-temperature reservoir. . EER (energy efficiency ratio): Q,/W with units of Biu’s for Q, and watt-hours for W.

Ay’

Heat engine: A cycle that takes energy from a heat source and converts a portion of it to work. Heat pump: A device operating on a refrigeration cycle that heats a space. Kelvin-Planck statement: /t is impossible to construct a device that operates on a cycle whose sole effect is to take energy from a heat source and convert it to work.

Performance parameter:

The ratio of the output of a device to the required input.

Refrigeration cycle: A cycle that moves heat from a cool region to a hotter region. Refrigerator: A device operating on a refrigeration cycle to cool a space. Reservoir: A body that it is able to provide heat or absorb heat without a change in its temperature.

Reversible process: One that can be reversed and returned to its original state leaving no change in either the system or the surroundings. No process is reversible. SEER (the seasonal EER): season.

The Btu’s divided by watt-hours for the entire cooling

Several important equations were presented:

Performance:

Neg ot

Relation between COP, and COP,,,:

COP;

F

Heat to temperature ratio for Carnot cycle:

Efficiency of a Carnot engine: COP of a Carnot refrigerator: COP of a Carnot heat pump:

COP — — COP,

desired effect

purchased energy

+ I

QO; ia Ty

On aha E is

Comment

“Carnot” can be replaced by “reversible.”

eh

Part 1

Concepts and Basic Laws

Questions 5.1

The fire in a coal furnace in a house in

A refrigerator requires 5 hp to provide 24 000 kJ/h of cooling. The COP of the refrigerator is nearest:

the country

(A)

18

(B)

2.0

Which one of the following could not be

5.5

considered a reservoir?

(A)

(B) A bathtub filled with ice water (C) The combustion process in an automobile engine

(D) 5.2

5.3

The water in a river

Which of the following processes could not be considered a reversible process?

(A)

The process of a piston compressing the air in a cylinder

(B) (C) (D)

Inflating a balloon

60)

(C) 2.4 (D) 2.8 If the ideal heat engine of Fig. 5.22 rejects 50 kJ of heat, estimate the energy required to operate the engine.

(A) (B) (C) (D)

The stretching of a rubber band

110kJ 100kJ 90kJ 80kJ

The freezing of water into ice ina refrigerator

Ty = 200°C

Which statement comes closest to being a Kelvin-Planck second-law statement?

(A) You can’t get something for nothing. (B) You can’t convert heat completely into work transmitted by a rotating shaft.

(C) (D) 5.4

You can’t use heat to cool a space. You can’t move heat from a cold space to a hot space.

The heat engine of Fig. 5.21 operates between the two reservoirs shown, producing 100 kW of power and 50 kJ/s of rejected heat. The engine’s efficiency is nearest:

(A) (B) (C) (D)

hy!

50%

An inventor proposes an engine that utilizes the boiling water of a hot spring and suggests it can operate with an efficiency of 22% in an area where a local stream has an average temperature of 12°C. Such a proposal is:

67%

(A)

Not possible

80%

(B)

Very likely possible

(C)

Possible but improbable

(D)

Possible many days of the year

33%

5.8

A Carnot engine expands 0.2 kg of air at 40°C

from 30 cm? to 300 cm? during the isothermal heat addition process in the gas cycle of Fig. 5.15. The heat required is nearest:

(A) (B) (C)

24kJ 31kJ 36kJ

(TY)

AlLkT

Chapter 5 3.9

The heat pump of Fig. 5.23 requires 5 hp and rejects 17 kJ/s of heat to the high-temperature reservoir. The COP is nearest: (A)

4.98

(B)

4.56

(C)

3.82

(D)

3.56

Qy= 17 kd/s

The Second Law of Thermodynamics

197

€ B The Heat Engine 3.12 A telemarketer advertises an emergency electric generator powered by a gasoline engine. The gasoline-powered engine supposedly produces 500 ft-Ibf/s of mechanical power while the electric generator produces i) 600 W, ii) 800 W, and iii) 1000 W of electric power. Which of these generators are possible? 5.13 The heat engine of Fig. 5.24 accepts 30 kJ/s of heat from a high-temperature reservoir and produces a) 10 hp, b) 15 hp, and c) 20 hp. Determine the rejected heat and the efficiency of the engine.

Qy = 30 kd/s

W

Figure 5.23

5.10 If the high-temperature reservoir in Problem 5.9 is at 160°C, the low-temperature reservoir for a Carnot heat pump is at:

(A) (B) (C) (D)

65°C 55°C 45°C 35°C

€ @ Statements of the Second Law 5.11

Show that a violation of the Kelvin-Planck

statement of the second law implies a violation of the Clausius statement of the second law. This requires an argument similar to that used in Example 5.1.

5.14 A power plant produces 20 MW of power by burning trash and rejects 14 x 10’ kJ each hour from a cooling tower. Calculate the efficiency of the power plant and the amount of heat supplied by the trash burner every hour. 5.15 A gas turbine engine is used to propel a ship. It has a thermodynamic efficiency of 0.75. How much heat must be supplied to this engine for it to produce a) 4000 hp, b) 200 000 Btu/min, and c) 3 300 000 ft-lbf/s? Remember:

Parts with lower-case italic

letters are separate problems and may be assigned separately.

Key

Part 1

Concepts and Basic Laws

5.16 An automobile engine produces 200 hp at an efficiency of 28%. What rate of heat transfer is required from the gasoline? How many kilograms of gasoline would be needed per hour if there are 50 MJ/kg of gasoline? 5.17

A power plant burns coal to deliver 80 MJ/s of heat to the steam. The plant turbines produce 15 000 hp. What is the rate of heat rejection from this plant, and what is its thermodynamic efficiency?

5.23 The power plant of Fig. 5.25 provides a) 200 GJ/hr, b) 3000 MJ/min, and c) 60 000 kJ/s from the boiler to the steam that flows through the plant. It rejects 120 GJ/hr to the environment, and the pump requires the power indicated. Calculate the horsepower output and the cycle efficiency. Is it necessary to include the pump power?

5.18 A heat engine is required to produce a) 120 hp, b) 70,000 ft-lbf/s, and c) 100 Btu/s by accepting 500 000 Btu/hr from a heat source. How much heat is dumped to the low-temperature reservoir, and what is the efficiency of the engine? 5.19 An automobile engine uses gasoline at a rate

of 1 ft*/hr. Gasoline has a density of 45 lbm/ft* and a heating value of 21 000 Btu/Ibm. If the engine has a thermodynamic efficiency of a) 30%, b) 25%, and c) 20%, what is the horsepower produced by the engine?

An automobile has a gas mileage of 30 miles/gal when traveling at 60 mph on a flat road with no head wind. Gas contains about 21,000 Btu/Ibm

with a density of 45 lbm/ft*. Estimate the efficiency of the engine if it produces a) 17 hp, b) 14 hp, and c) 12 hp under those conditions.

The engine of an automobile requires 20 hp to travel at 100 km/hr. If the engine’s efficiency is 18%, determine the rate of fuel consumption, in L/km, assuming an energy content of 45 MJ/kg and a density of

720 kg/m?.

Q¢= 120 Gu/hr

€ B The Refrigeration Cycle 5.24 A refrigerator is advertised that will remove 10 kW from the refrigerator while dumping 15 kW of heat into the kitchen. How much horsepower is the compressor using to run this system? ae The refrigeration cycle of Fig. 5.26 is intended to (5.25 \ ~ cool a space by supplying a) 10 kJ/s, b) 15 kJ/s, and c) 20 kJ/s of cooling. Determine the rate of heat transfer that is reyected from the condenser and the performance of the cycle.

A power plant burns 50 000 kg of coal each hour. Coal contains about 26 MJ/kg of energy. If the efficiency of the plant is 34%, estimate the heat transfer each day to the nearby river used to receive the rejected

energy. p

K and it is

A central air-conditioning system is reversed to operate as a heat pump in the winter. If the inside temperature is maintained at 68°F when the outside temperature is —10°F and the

desired to remove 0.02 J of energy from the specimen, what is the minimum work requirement if the laboratory at 18°C is the high-temperature reservoir?

house loses heat at a rate of 72,000 kJ/hr, what

is the minimum power required to operate the compressor? Note: A kJ is nearly equal to a Btu.

An experiment is proposed to reduce the temperature of a specimen to nearly absolute

a hal wai

Morph Creat

Outline 6.1 Inequality of Clausius 6.2 Entropy

205

207

6.3 Entropy Change in Substances for Systems 209 6.3.1 6.3.2

Basic relationships 209 Entropy change of an ideal gas with constant

Cand:G,-

210

6.3.3

Entropy change of a solid, a liquid, and

6.3.4

areservoir 213 Entropy change of a phase-change

6.3.5

substance 216 Entropy change of an ideal gas with variable specific heats 218

6.4 Entropy Changes for a Control Volume 221 6.5 Isentropic Efficiency .

;

6.5.1 6.5.2 6.5.3 6.5.4

225

Isentropic turbine efficiency 225 Isentropic compressor efficiency 226 Pump efficiency 228 Isentropic efficiency of anozzle 229

6.6 Exergy (Availability) and Irreversibility 230 6.7 Summary FE Exam

Problems

237 Practice Questions

241

239

/ Nomenclature | The following variables are introduced in this chapter:

otek =u: gQoO i=}

Irreversibility Irreversibility per unit mass Relative pressure Entropy Entropy generated Entropy of surroundings Entropy production

Cryo AE CCA [e}

An

Specific entropy Specific entropy per mol An entropy function

T> v,

Dead state temperature Relative specific volume

W, Mh

Actual work Second-law efficiency

WV wv AS,

Exergy (or availability) Exergy per unit mass Generated entropy

AS

Net entropy change

AS...

Entropy change in the universe

a

Learning Outcomes ‘J

Understand the inequality of Clausius for engineering cycles

1

Understand the concept of entropy and how it is used in thermodynamics

J

Evaluate chan ges in entropy for thermodynamic processes

“J

Use the conce pt of entropy to develop performance

parameters for common engineering devices 1

Use availability and exergy to solve second-law problems

n Chapter 5 the second law of thermodynamics was observed to limit the :performance of hea t engines and refrigerators. In this chapter we will quantify the second law and make it a tool for the analysis of cycles using the property of entropy. Changes in entropy for a system will be evaluated, and cycle performance will be analyzed. Entropy is a diffic ult property to define. Like energy, the effects of entropy can be observed, but it is not a directly measurable property such as pressure or temperature. In this ch apter we will define entropy and calculate the value of this property for various su bstances and the change in entropy for several processes of interest. This property will help us determine if the second law of thermodynamics is Satisfied or if it is violated. Entropy will also be used in the definition of performance parameters for some common engineering devices. )

Motivational Example—High-Efficiency Electric Motor A research group from Tokai University under the direction of Professor Hideki Kimura has developed a 24-volt DC motor that has an energy efficiency in excess of 96%. The motor, shown in Fig. 6.1, was developed to be used in electric vehicles such as the one shown in Fig. 6.2. The motor was developed by studying the causes of energy loss in motors and then making improvements to reduce these losses. The two major areas of energy loss were friction in the bearings and wind friction over the rotating core. Electric motors are very efficient; the inefficiency in electrical energy production is in the processing of the steam in power plants where efficiencies of about 33% have been realized since the 1950s. New technologies have raised efficiencies, but added regulations have managed to decrease efficiencies with the result that power-plant efficiencies have remained relatively constant. Google “electricity-generating efficiency” to read more about this subject.

> = &

Zz

2 Bez= ©

Core of a high-efficiency DC motor.

University Tokai

A streamlined electric car.

The study of entropy begins by developing the Clausius inequality, expressed as

dQ7 = 0 p

(6.1)

where the circle through the integral symbol indicates integration around a cycle. The German physicist J. Clausius (1822-1888) postulated, based on experiments, that the ratio of the differential heat transfer 6Q to the absolute temperature T at which that heat transfer took place integrated around a cycle is conserved for reversible cycles and negative for irreversible cycles (for all real cycles). Let us interrupt this analysis for a short discussion of a reversible process. The Clausius inequality is applicable to a system that undergoes a reversible or an irreversible cycle. For example, the air in a cylinder could be compressed with a piston with negligible friction between the piston and the cylinder wall, or it could be compressed the same distance, with the piston being forced to move due to extreme friction between the piston and the cylinder wall. The air, the system, would undergo the same process in both cases; the process with friction could be considered a quasi-equilibrium process’ for the system, the air, but irreversible when both the system and surroundings are considered. The friction would make the process, as a whole, irreversible. Generally speaking, both the system and surroundings are included when a process is declared reversible. However, in this discussion of entropy, it is the system as it passes through a cycle that is being considered in the Clausius inequality. If all processes in the cycle are quasi-equilibrium processes, the “equal” sign is used. If even one process involves heat transfer with the system boundary at a different temperature than the system, a sudden expansion, internal heating (a resistor), or mixing, the process is a

nonequilibrium process and the “less than” sign is used for the cycle. 'Some authors refer to this process as being internally reversible.

Clausius inequality:

a Reversible process: The system and surroundings possess no irreversibilities. A reversible cycle possesses all reversible processes. Irreversible process: The system or surroundings experience irreversibilities.

VA

=Part1

Concepts and Basic Laws To show that the Clausius inequality is true, consider }5Q/T

for the heat

engine of Fig. 6.3 that operates on a cycle. Since T), and 7, are constant during

each respective heat transfer process, the integral becomes 5Q

pi

tt Oy

OQ,

ee

(6.2)

Te

For a Carnot heat engine, the integral in Eq. 6.2 will equal zero since it has been

shown (see Eq. 5.16) that O,,/O, = T/T, or.tewritten, Q,,/T;, = Q,/T,.Con-

sequently, for the reversible cycle, A heat engine.

pie= 0

(reversible cycle)

(6.3)

Now, consider a heat engine operating on an irreversible cycle between the same two reservoirs with an equal amount of energy supplied to each, that is,

z

(Qi rev = (Quix Since W,., > W;,,. the first law demands that (Q,),. < (Qy irr

Remember: The first law for the heat engine requires OF, Fe

OF =

80 Qn

W.

< 0

je a! ei

2)

Figure 6.4 A two-process cycle.

Example

Hence, for the irreversible cycle,

6.1

(irreversible cycle)

(6.4)

and the Clausius inequality is shown to be true. Equation (6.3) implies that 5Q/T for a reversible process is an exact differential since the cyclic integral is zero, an observation to be used in the next section. To show that this is true, consider the two-process cycle of Fig. 6.4. A perfect differential df integrated between states 1 and 2 would givef, — f,. Then, integrated between states 2 and 1, to complete the cycle, the result would be f, — f,. The cyclic integral would be the sum, which is zero. If an inexact differential, such as the work dW (or heat 5Q), were integrated around a complete cycle, the result would not be zero but the net work (or net heat transfer) for the cycle, that is, the area under the P-v diagram of Fig. 6.4. It is concluded that 5Q/T is an exact differential. The above analysis holds true for both heat engine cycles and refrigeration cycles.

Check the Clausius inequalit The schematic

of a steam

power

plant operates

on the cycle shown

in

Fig. 6.5 with its P-v diagram in Fig. 6.6. Heat transfer takes place in the boiler and condenser only. (The boiler and condenser are heat exchangers, so it is assumed that they operate at constant pressure.) Check the Clausius inequality for this cycle if the properties of the steam at the four states are as follows:

State 1: P,= 10 kPa,x,=0.01

State 2: P, = 800 kPa,x, = 0

State 3: P, = 800 kPa,x, = 1

State 4: P, = 10 kPa,x, = 0.9

Chapter6

Entropy

PAZ

Figure 6.6 Solution

The enthalpies and temperatures are found in the steam tables (the IRC Calculator can also be used) and are listed or calculated as follows:

h, = 216 kJ/kg 3 = 2770 kJ/kg, T, = 170°C

rT Kiko: 4 = 2340 kJ/kg, T, = 45.8°C

Using the energy equation (4.16), the specific heat transfer in the boiler is

dy = hy — hy = 2770 — 721 = 2049 kI/kg The heat transfer emitted by the condenser is

dy, = Ng — hy = 2340 — 216 = 2124kI/kg The boiler temperature is T,, = 170°C = 443 K, and the low temperature is T, = 45.8°C = 318.8 K. Because the high and low temperatures are constant, we can apply Eq. 6.3 to verify the Clausius inequality:

6q lier

nu 4 _ 2049 — 2124 | ae £1 > AAG 3188

2.04 kI/kg-K

The value of the cyclic integral is less than zero, which it must be. If the integral was positive, the cycle would not be possible since it would violate the second law, as expressed by Eq. 6.1.

(1)

; , : ; ; _ : . Spe is : a [he Clausius inequality has identified 6Q/Tas an exact differential for a reversiblee

process (refer to the discussion referencing Fig. 6.4). Hence, it represents the differential of a property of the system. This property, called entropy, will be used as an expression of the second law as it applies to physical systems. It was first

Entropy: A property defined by dS = 6Q/T|....

} SS

ev

Aeriseinest bakin thermodynamics this one suffices

208

Part 1

Concepts and Basic Laws presented by Clausius in 1865 and, for our purposes, it is defined by its differential to be

dS

awe 8Q/T. The effect of the irreversibility (e.g., heat transfer across a finite temperature difference) is to increase the change in entropy for a process.

Chapter 6 = Entropy 2.

Consider an isolated system, one that does not interact with the surroundings so that there is no heat transfer. Equation (6.9) would then be expressed as

iS Bet)

(6.10)

The second law demands that the entropy of an isolated system remains constant for a reversible process or increases for a real process. 3.

Since the universe is an isolated system, its entropy must always increase for every real process, as demanded by Eq. 6.10. This can be stated for any process as

Ree univ

AS

AS surree 0

(6.11)

The entropy change of the universe, AS,,,., is also referred to as AS,,.,, the net entropy increase, or AS,.,,, the entropy generated. This has led to various statements of the second law, all of which refer to a generation of entropy as the result of every real process.

Specific entropy, often referred to simply as entropy, is defined as the amount of entropy per unit mass; the standard symbol is a lower-case s, defined by s = S/m. This can often be confusing since upper-case S and lower-case s look quite similar, especially when hand written during the solution to a problem. Entropy S has units of kJ/K (Btu/°R), and specific entropy s has units of kJ/kg-K (Btu/Ibm-°R). A more qualitative definition of entropy is that it is the measure of the amount of disorder in a substance. In a solid, the molecules are held in a tight crystalline structure. Solids have a very low level of entropy. The molecules in a solid move very little from their set position. A solid existing at absolute zero temperature has zero entropy. In a liquid, the molecules are able to move freely, with intermolecular forces keeping them together. Liquids have a higher level of entropy compared to a solid at the same temperature. In a gas, the molecules move independently and randomly with a relatively high level of entropy.

[™ You have completed Learning Outcome

The entropy of every real process increases in an isolated system.

A result of the second law: Every real process results in a net entropy increase, represented by

AS univ ini = ance net

esOaeae

in the universe.

Suggestion: Be careful when writing S and s. Make the S quite large and the s quite small so that they are obviously different.

Entropy is also considered to be a measure of the amount of disorder in a substance.

Observation: Since the _ actual value of entropy is not of interest in a liquid or a gas, a zero value for

entropy is assigned at an arbitrary temperature.

(2)

6.3 Entropy Change in Substances

BerOTas Stem See sin 6.3.1 Basic relationships The incremental heat transfer for a reversible process (no friction, sudden expansion, electrical resistance heaters, or paddle wheels) can be related to the differential entropy change by Eq. 6.4 as

5Q = TdS

209

(G02)

et

Part 1. Concepts and Basic Laws or

6g. = Ts

(6.13)

When integrated, there results D,

The area under a 7-S diagram between two states represents the heat transfer between those two states for a reversible process. The area under a T-S diagram for a complete cycle represents Q net W, “net for a cycle composed of reversible processes.

Os

|TS,

(Ol

5p

2

|Tds

1

(6.14)

1

which shows that the area under a 7-S diagram between two states represents the heat transfer between those two states for the reversible process. If the process is irreversible, the area does not represent the heat transfer, much like work is not represented by the area under a P-v diagram for an irreversible process. Consequently, for a closed cycle of reversible processes, the area under a T-S diagram represents Q.,., for the cycle, which, according to the first law (see Eq. 3.21), is also equal to W,,... If the Carnot cycle of Fig. 5.15 were to be sketched on a 7-S diagram, it would appear as shown in Fig. 6.8. The rectangular area represents Quer = Wher hence, the 7-S diagram is often the diagram of choice for cycles since the adiabatic process, a vertical line on the 7-S diagram, is so common. To find an expression for the entropy change for a process, we return to the differential form of the first law, expressed by Eq. 3.23, MGs = Wp -se IG

(6.15)

where we used 6g = Tds (from Eq. 6.13) and d6w = Pdv (from Eq. 3.5). In terms of enthalpy, h = u + Pv, Eq. 6.15 can be written as Ss

Figure 6.8

Tds = du + d(Pv) — vdP dh — vdP

The Carnot cycle.

Calculus: d(Pv) = Pdv + vdP so that Pdv = d(Pu) — vdP

Entropy is dependent on two other properties for a simple substance.

The differential relationships of Eqs. 6.15 and 6.16 can be used for reversible or irreversible processes since they relate properties.

(6.16)

It should be emphasized that entropy is a property that, for a simple substance, is dependent on two other properties of the substance, usually temperature and pressure, the two measureable properties. The differential relationships of Eqs. 6.15 and 6.16 can be used for reversible or irreversible processes since they simply relate properties. The entropy change over a process is often of interest since it is so closely related to the second law. It will be of particular interest for the adiabatic reversible process, since the entropy change for such a process is zero; it is an isentropic process, as defined by Eq. 6.6. In the next several subsections, entropy changes will be related to various processes for systems.

6.3.2 Entropy change of an ideal gas with

constant C, and C, To relate the entropy change for an ideal gas with constant specific heats to temperature and specific volume, divide Eq. 6.15 by the absolute temperature and use Eq. 2.22 and the ideal-gas law to find Ok=

du T

+: Pdo T =

Cal | Rado a

5

(6.17)

Chapter6 Entropy

eit

Assuming constant specific heat C,, this equation can be integrated for a process to obtain the change in entropy:

IE

peel

V1

eo

wv

ne

eae

6.18

"O,

(6.

)

If the first law were used in terms of enthalpy as Eq. 6.16, the entropy change after dividing by T and integrating, assuming constant C,, would be

2—5=

Tee cL | aa

(fe

pete of I

ea= 1p

pa

>

dP Ree

de

P. 2 BOTS

d (6.19)

' Either Eq. 6.18 or Eq. 6.19 can be used to find the entropy change in an ideal gas with constant specific heats. Which form is used depends on the information given. It should be emphasized that the above relationships can be used for either reversible (quasi-equilibrium) of irreversible processes since they are simply equations relating properties for an ideal gas. For an isentropic process, that is, one for which As = 0, Eq. 6.18 and Eq. 6.19 and the ideal-gas law result in the three relationships By Me (2)

ha

V>

iS IE

(By Ps

Py (2) Xo, Pee

Aah (Cee

a ideal gas must be constant for these equations to be used.

The specific heats of the

'ee aS - used

These are the same equations as we found for the quasi-equilibrium adiabatic process of Section 3.5.4, which they should be since the same assumptions apply. These equations will be used extensively in the remainder of our study because it is most common to assume constant specific heats for an ideal gas. It does introduce significant error if the temperatures are relatively high, but the results are of interest in comparisons and do give quick estimates of unknown properties. The following subsection presents the equations for ideal gases with variable specific heats.

The entropy change in an idealgas (iL

Ten kilograms of air at 600 kPa and 40°C are heated in the rigid container of Fig. 6.9 to 250°C. Assuming constant specific heats, calculate i) the heat transfer and ii) the change in entropy for this process.

Solution From the example statement, we know that the air has the following properties:

P, = 600 kPa, T, = 40°C = 313 K, T, = 250°C = 523K (Continued)

Example

6.2

ci

Part 1 Concepts and Basic Laws

Example

6.2

(Continued)

Figure 6.9

i) The heat transfer can be calculated from the given information. The first law provides

Q

mau = mCi,

— J)

10kg X 0.717kJ/kg-K X (250 — 40) = 1506kJ ii) To find the entropy change, Eq. 6.18 is used since v, = v, (the container is rigid). The entropy change is Since the mass is given, we have found AS. If the mass had not been given, we

would have found As.

AS = mCi in

T,

0,

+ mR lp —

T,

O71

523 = 10 kg| (0.717 kJ/ke-K) LD as + 0.287 lat | = 3.68 kJ/K We have used C, = 0.717 kJ/kg-K and R = 0.287 kJ/kg-K from Table B-2 in the

Example

6.3

An isentropic process of an ideal gas Air at 100 kPa and 20°C is compressed in a cylinder to a pressure of 1800 kPa. Assuming an isentropic process with constant specific heats, calculate i) the final temperature, ii) the initial and final specific volumes, and iii) the work required. The process is shown on 7-s and P-v diagrams in Fig. 6.10.

Figure 6.10

Chapter6 Solution From the example statement, we know that the air has the following properties:

P,=100kPa,

7, = 20°C = 293K, P, = 1800kPa

i) The temperature at the end of the compression process is found, using Eq. 6.20 with k = 1.4, to be P,\(&-D/k

T,=T Us 2) 2

1800

= 293(100

\°4/14

=

669K

or

p 396°C

ii) The initial and final specific volumes are found to be see 1

RT, 0,25) Rp;

RT,

Vv, = Se

XK 293

0.841

100

02875

m3/k

O71 MKS

< 669

eee

;

= 0.1067 m°/kg

iii) To find the work, the first law is used. Since Q = 0, it is

—w

=u,

—

uw, = C7,

—

T,) = 0.717396

— 20) = 270kI/ke

We have used C, = 0.717 kJ/kg-K from Table B-2. Observe that the work is negative, which was expected since work must be input in order to compress a system. The work per unit mass was calculated since neither the mass nor the volume of the air was specified.

6.3.3 Entropy change of a solid, a liquid, and a reservoir For a solid or a liquid for which the specific heats can be assumed constant, as is

usually done, the first law, Eq. 6.15, is put in the form du

ds = —7

(6.21 )

since solids and liquids are assumed to be incompressible so that dv = 0. Letting du = CdT, Eq. 6.21 can be integrated to provide

g, = 4. = ke op Tein T,

dT

If,

(6.22)

For most solids and liquids, the specific heat C = C, = C, = const, so the integration was easily performed. For some solids or liquids, it may be necessary to integrate using a given C(7). In tables, the specific heat C is most often listed as C,.

Entropy

([P4AE

ay

Part 1

Concepts and Basic Laws The entropy change of a substance, whether a solid, a liquid, or a gas, which composes a reservoir (a source of energy at constant temperature) is often of interest. Since the temperature of a reservoir is constant, its entropy change is

) 1 Q 6Q == sea 1p ip iE:

AS =

(6.23)

Several examples now illustrate the above equations for entropy change of a system.

Example

6.4

Entropy change of the universe REEL

DE

Air in the 2-m?-rigid container of Fig. 6.11 cools from 200°C and 400 kPa to 100°C. The surroundings are at 20°C. Determine the entropy change of the universe. Q

Air

initially at

200°C, 400 kPa

Solution The mass of air in the container is

m

EePy RE

400 0287

= 5,893 ke

ae

Using Eq. 6.18, we find the entropy change in the air to be

AS,

T, y = mC, ln + mk in T, Vy

kJ 505 ko —— 2 x 0717 ok

in

373

1.004 kJ/K

where we have used v, = v, since the volume is rigid. The heat transfer from the air is found using the first law:

Q., —W

= AU =me,

— T))

: g Xx 0717 ta“L973 — 473)K )K = = 5.893 kg The heat transfer is negative since it is leaving the system.

KS —422.5k

Chapter6 = Entropy

eta

The entropy change of the surroundings (a large reservoir), which are at 20°C, is positive since heat is added to the surroundings using Eq. 6.23; it is equal to

a We

ee ee

a

The entropy change of the universe is found to be OS cae

ae AS. ir

e

BS

ae

= —1.004 + 1.442 = 0.438 kJ/K The net entropy change is positive as demanded by the second law.

Entropy change of ice and water Two pounds of ice at 20°F and 12 lbm of water at 60°F are mixed in an insulated container, as shown in Fig. 6.12. Estimate 1) the final temperature of the mixture and ii) the entropy change.

Solution i) The first law is applied to the mixture to determine the final temperature. If all the ice is assumed to melt so that the final temperature is above 32°F, we have

WG ©(32)

220)

Al + CT

32) =,w ~ pw (60. T)

Insulated container

(Continued)

Example

6.5

Val

=6Part1 © Concepts and Basic Laws

Example

6.5 where T is the final temperature of the mixture. The subscript ‘7’ refers to ice and ‘w’ to water. Substitute the known quantities into the above equation, using units on C,, of Btu/Ibm-°F, and obtain

It takes 140 Btu's to melt one

21049

«12 + 140 41.0 x7

pound of ice.

— 32)| = 12x 10 x 60 GR = 35.0°F

C, for water and ice was found in Table B-4E. The latent heat Ah, of ice was listed in Section 2.2.4 as 140 Btu/lbm. ii) The entropy change of the ice and the water is found to be

Units: The units on AS are the same as the units on (mC,), which are lbm(Btu/lbm-°R) = Btu/°R.

Ae

ml

In Cy,

492 0 te AS See

Coun

495.2 2

a MgC

ao

495.2 520

The entropy of saturated water at 32°F is found in Table C-1E, and that of ice at 32°F is found in Table C-SE. This provides

492 AS netne,= 2|0.49 In =~ +[0 Mg fl —(—0.292)]

+1.01 495.2 |+12 X 1.0)

495.2 n= eg

= 0.0554 Btu/°R It is positive, as it must be.

6.3.4 Entropy change of a phase-change substance The entropy of a pure substance is calculated in the same manner as internal energy and enthalpy. Values of specific entropy are listed in tables in the Appendix for steam, R134a, and ammonia. Follow the same steps taken when calculating the properties of steam in Chapter 2. The specific entropy for a saturated mixture is calculated using wn

The entropy is arbitrarily set equal to zero at 0°C (32°F) for saturated water.

The saturated temperature table can be used for the entropy of a compressed liquid.

II S, + x(S, =

= Sp + XSp

Sy)

(6.24)

Observe that the entropy of saturated water at 0°C (other substances use a different temperature) is arbitrarily set equal to zero, as were u,and h,. This is of no consequence since it is only the changes in those properties that te of interest in our study. For compressed liquid water, either Table C-4 or the saturated value at the given temperature can be used. The superheated values for steam are listed in Table C-3. Similar tables for R134a and ammonia are included in the Appendix. The IRC Calculator can also be used.

Chapter 6 = Entropy

217

Critical

T

point

v = const P=const

A T-s diagram for steam.

A T-s diagram is often sketched, especially when the working fluid is one of the three substances listed above. For water, it is relatively symmetric and resembles Fig. 6.13 when drawn to scale.

The entropy change insteam SELES

Steam is contained in a rigid volume at 1400 kPa and 300°C. Heat is transferred from the volume until the pressure is 60 kPa. Determine i) the entropy change and ii) the heat transfer required. i1i) Also, sketch the process on a T-s diagram.

Solution From the example statement, we know properties:

P, =1400kPa,

that the steam has the following

7, = 300°C, P, = 60kPa

i) Forarigid volume,v, = v,.The steam table C-3 provides v, = 0.1823 m’/kg. Table C-2 then allows the quality to be calculated:

V, = 0, + x,(v, — %)

0.1823 => 0.00103 + x,(2.732 — 0.00103)

WX) = 0.0664

The entropy at state 2 is then Sp = Spt XS, = 1.1455 + 0.0664 X 6.387 = 1.57 kJ/kg:K ii) The heat transfer can be calculated from

Oates,

ey

[359.8 + 0.0664 x (2489.6 — 359.8)] — 2785.2 = —2284 kJ/kg The minus sign means heat leaves the volume. (Continued)

Example

6.6

Part 1. Concepts and Basic Laws

Example 6.6

(Continued) ZELDA

LEIA EDEZIAD

iii) The process is sketched on the T-s diagram in Fig. 6.14. The specific volume must be held fixed at 0.1823 m*/kg.

Figure 6.14

6.3.5 Entropy change of an ideal gas with variable specific heats If the temperatures are sufficiently high, the assumption of constant specific heats results in significant error. To account for variable specific heats, return to Eq. 6.16 using’ dh = C,dT and v/T = R/P:

ds =

dh

vdP

=P

G

=

ae”

Ti

R

dP

P

6.25

(62>)

Integrate this equation, recognizing that R is a constant and C, = C(t): Note: C, depends only on

temperature.

aa

Ss

=

T,

| Ge

BIE

Te

P

Rin

P,

(6.26)

The integral in this equation is a function of the temperature only. It can be evaluated from the gas tables in Appendix F using a function s° (some authors prefer rather than s°) that is defined so that SoS le}

[ Oo

fe}

p

(6.27)

dT

Using this equivalence, we see that Eq. 6.26 becomes |

Comment

This relationship is most often used when considering variable specific heats.

|

Sy

S) SS

Se ‘

le2

P

(6.28)

1

This more exact expression for entropy change is used only when requested. Equations (6.18) and (6.19) are often used for comparison purposes, even though they are not accurate when either of the temperatures is relatively high.

Chapter6

Entropy

PAL

For some gases listed in Appendix F, the entropy function is presented on a per mol basis, so the change in entropy takes the form

De

ae

aaa

P,

(6.29)

Py

For an isentropic process, Eqs. 6.20 cannot be used if the specific heats are not constant. We can, however, use Eq. 6.28 to obtain, for As = 0, Py as eS.

2

s°y/R

it~ fT)

=

eR

o

ex HUE) a

oe

6.30

This leads to the definition of the relative pressure P., defined by

(Ses

(6.31)

which is included as an entry in the air Table F-1. In terms of P., Eq. 6.30 takes the form

Pie P,

Observe: The relative pressure P_ is nondimensional, as was the reduced pressure P, of Chapter 2.

6.32




The minimum work input assumes no heat transfer, no losses, and no changes in

kinetic and potential energy.

Example 6.13 _ Efficiencyof apump—

SSL

SIDELINE

The pump of Fig. 6.24 has an efficiency of 0.75. If the pressure of 8 kg/s of water is increased from 120 kPa to 1200 kPa, determine the horsepower required by the pump.

= 408 m/s where C,, for nitrogen was found in Table B-2. This is a supersonic flow (V > 300 m/s), so the nozzles of Fig. 6.26 must be combined to form a converging diverging nozzle similar to the one shown in Fig. 6.27 that would provide a supersonic flow. This type of nozzle is observed on rockets to put satellites into orbit. Vy

———>V>

Figure 6.27

(4)

[{ You have completed Learning Outcom e

=e Exergy: The maximum reversible work possible during a process that brings a system into equilibrium

; with the surroundings, that SS tate dasa ciaie.

pee Be caaiibnneesye@] win te surroundings, takenas 25°C and 100 kPa, unless

otherwise stated.

6.6 Exergy (Availability) TI

;

,

;

a heat engine

(B) (C)

igure 635

All processes in heat engines and refrigerators

(D)

6.3

All cycles involving heat engines and refrigerators

a

\

Assume ideal conditions for each component of the refrigeration cycle of Fig. 6.33 (with 7-s

i

diagram in Fig. 6.34) and find 5q/T. R134a is x : @ 120 KPa @ Ne

the refrigerant. (Remember, q,, is negative and q,, is positive.)

(A) —0.062 kJ/kg-K (B) —0.085 kJ/kg-K

(C) —0.095 kJ/kg-K (D) —0.16 kJ/kg-K

—

y | Figure 6.34

Ss

[EE

eH 6.4

Part 1.

Concepts and Basic Laws

heats, is nearest:

The air in the constant-pressure cylinder of Fig. 6.37 is initially at 400°C and 120 kPa. It is cooled until its temperature is 40°C. The surroundings are at 20°C. The entropy generated during this process is nearest:

(A)

0.0123 kJ/K

(A)

0.463 kJ/kg-K

(B)

0.0972 kJ/K

(B)

0.591 kJ/kg-K

(C)

0.563 kJ/K

(C)

0.766 kJ/kg-K

(D)

3.11 kJ/K

(D) 1.229 kJ/kg-K

Air is heated from 20°C to 800°C at constant pressure of 200 kPa in a cylinder with an initial

6.8

volume of 4000 cm*. The entropy change, assuming an ideal gas with constant specific

6.5

6.6

A piston compresses air in a cylinder from 20°C and 100 kPa to 2 MPa. The final temperature, assuming an ideal gas with constant specific heats, is approximately:

(A)

690°C

(B)

540°C

(C)

490°C

(D)

420°C

Figure 6.37

The 200°C heat reservoir of Fig. 6.35 gains 600 kJ of heat during a process. Its change in entropy is nearest: (A)

0.863 kJ/K

(B)

1.02 kJ/K

Two kilograms of steam enter a turbine each second at 600°C and 4 MPa and leave as saturated vapor at 20 kPa while 100 kJ/s of heat leave the turbine. The surroundings are at 25°C. The entropy production is nearest:

(C)

127kJ/K

(A)

1.42 kJ/K-s

(D)

1.59 kJ/K

(B)

1.08 kJ/K-s

(C)

0.63 kJ/K:s

(D)

0.34 kJ/K:s

6.9

Reservoir

6.10 Q

6.7

The 20-kg slab of copper (C = 0.395 kJ/kg-K) in Fig. 6.36 is heated from 100°C to 200°C. Its entropy change is nearest

(A) (B) (C) (D)

0.63 kKJ/K 1.88 kJ/K 3.27kI/K 5.48 kJ/K Copper

Q Figure 6.36

Steam at 500°C and 4 MPa enters a turbine with an isentropic efficiency of 86%. The exit pressure is 100 kPa. The exiting temperature of the steam is nearest: (A)

95°C

(B)

107°C

(C)

118°C

(D)

126°C

Chapter6 6.11 Saturated Ri34a at 120 kPa enters an adiabatic compressor with an efficiency of 80%. The exit pressure is 1.6 MPa, as shown in the T-s diagram of Fig. 6.38. The exiting temperature of the refrigerant from the compressor is

Entropy

(4X

6.12 The exergy of the steam at state 1 is nearest: (A) 1340 kJ/kg

nearest:

eS (C)

EUG ase 3030 kI/kg

(D)

4400 kJ/kg

(A)

104°C

(B)

99°C

nearest:

(C)

95°C

(A)

6.13 The second-law efficiency of the turbine is

(D) 89°C

68%

ae (C)

78%

(D)

82%

6.14 The isentropic efficiency of the turbine is nearest:

(A) (B) (C) (D) .

:

. ; € B Questions

6.15 The irreversibility of the process is nearest:

6.12-6.15

Steam enters a turbine (see Figures 6.39 and 6.40) at 4 MPa and 500°C and leaves as saturated steam at 40 kPa. Assume standard conditions for the dead state.

®

)

(A)

529 kJ/kg

(B) (C)

389 kJ/kg 319 kJ/kg

(D)

229 kJ/kg

€ B Inequality of Clausius 6.16

Wr wie

80% 75% 70% 67%

The 7-s diagram of a Carnot refrigeration cycle utilizing R134a is shown in Fig. 6.41. Heat transfer takes place in the condenser and evaporator at constant temperature. Verify the Clausius inequality using x, = 1 and x, = 0. T.

Figure 6.39

1 - 2: Compressor

as

15 psia

Figure 6.41

Figure 6.40

Dae

242

Part 1

Concepts and Basic Laws

6.17 The 7-s diagram of a Carnot steam power cycle is shown in Fig. 6.42. Heat transfer takes place in the boiler and condenser only. Assume an isentropic turbine and pump. Determine the Clausius inequality numerical result for this

6.22

cycle if x, = 0 and x, = 1 and Pz, is a) 2 MPa, b) 4 MPa, and c) 6 MPa.

Four kilograms of an ideal gas are initially at 25°C and 150 kPa in the rigid container of Fig. 6.43. Heat is added until the pressure reaches 900 kPa. Determine the entropy change of gas if it is a) air, b) nitrogen, c) carbon dioxide, and d) hydrogen. Assume constant specific heats from Table B-6. Q

2 - 3: boiler 4-1: condenser

é

20kPa

@) NG Figure 6.43

Figure 6.42

Remember: Each part with a lower-case italic letter is a separate problem.

€ B Entropy Change in an Ideal Gas

with Constant C, and C, 6.18

A piston moves in a cylinder at constant temperature while heat transfer occurs. If the pressure doubles, determine entropy change in the air assuming constant specific heats from Table B-2.

6.19

A piston moves in a cylinder at constant pressure while heat is added. If the volume doubles, determine entropy change in the air assuming constant specific heats from Table B-2.

6.20 Air at 200°C and 300 kPa is compressed to 400°C and 800 kPa. Calculate the change in entropy and the specific volume for this process. Assume constant specific heats from Table B-2. 6.21

Nitrogen at 500 psia and 200°F is allowed to expand in an insulated control volume until the temperature is 30°F and the specific volume is increased by 90%. Calculate the change in entropy for this process. What is the final pressure? What work is required? Assume constant specific heats from Table B-2E.

6.23

An ideal gas is initially at 120 kPa and 20°C when contained in a 5-m° volume. If it experiences an entropy increase of 4 kJ/K as the temperature is raised to 100°C, what is its final volume if the gas is a) air, b) nitrogen, c) carbon dioxide, and d) hydrogen? Assume constant specific heats from Table B-2.

6.24 The ideal gas a) air, b) nitrogen, c) carbon dioxide, or d) argon is compressed from 125°C and 100 kPa to 500 kPa in an isentropic process, as displayed in Fig. 6.44. Calculate the final temperature. How much work occurs during this process? Assume constant specific heats from Table B-2.

Ideal gas initially at 400 K, 100 kPa

Figure 6.44

Chapter 6

Entropy

243

6.25 The ideal gas a) air, b) nitrogen, c) carbon dioxide, or d) argon is allowed to triple its volume in an isentropic process. If the final temperature is 20°F, what is the initial temperature and what work is produced? Assume constant specific heats from Table B-2E.

6.30 Four hundred kilojoules of heat are transferred to 2 kg of air in a piston/cylinder arrangement that is maintained at constant pressure. Calculate the final temperature and the entropy change if the initial temperature is a) 100°C, b) 200°C, and c) 400°C. Assume constant specific heats from Table B-2.

6.26 Two thousand kilojoules of heat are added to 10 kg of air contained in a rigid container. Calculate the final temperature and the entropy change of the air if the initial temperature and pressure are, respectively, a) 200°C and 200 kPa, b) 100°C and 400 kPa, and c) 400°C and 100 kPa. Assume constant specific heats from Table B-2.

6.31 An electric heater provides 800 kJ of heat to nitrogen in a rigid 2-m* volume. If the initial temperature is 100°C, determine the entropy

6.27 The frictionless piston/cylinder arrangement of Fig. 6.45 contains 2 kg of nitrogen at 40°C and 200 kPa. If the pressure is held constant, calculate the final temperature and the entropy change if a paddle wheel inserted in

increase if the initial pressure is a) 100 kPa,

b) 400 kPa, and c) 1000 kPa. Assume constant specific heats from Table B-2. 6.32 A frictionless piston compresses 0.2 kg of air in the insulated cylinder of Fig. 6.46 from the initial conditions shown. Estimate the final temperature and the work required if the final pressure is a) 800 kPa, b) 1400 kPa, and c) 2 MPa. Assume constant specific heats from Table B-2.

the insulated cylinder adds a) 100 kJ of work,

b) 200 kJ of work, and c) 400 kJ of work. Assume constant specific heats from Table B-2.

T, = 40°C P, = 200 kPa Ty = 20°C

P1 = 100° kPa AIR 2kg Nitrogen

Figure 6.46 Figure 6.45 6.28

Heat is added to a volume that contains 2 kg of air, initially at 80°C and 200 kPa. Determine the work, the heat transfer, and the entropy change if the temperature is held constant

while the volume is expanded to a) 2 m’,

b) 3 m°, and c) 4 m’. Assume constant specific heats from Table B-2.

6.29 Air is compressed in an isentropic process from 20°C and 100 kPa until the final volume is a) six times the initial volume, b) eight times the initial volume, and c) ten times the initial volume,. Calculate the final pressure and temperature assuming constant specific heats from Table B-2.

The gases that closely resemble air after combustion in an automobile cylinder are expanded such that the volume increases by a factor of 8. If the initial pressure is 275 psia, estimate the final temperature and the work produced if the initial temperature is a) 800°F, b) 1100°F, and c) 1400°F. Assume constant specific heats from Table B-2E. Assume a frictionless process.

sity

Part 1

Concepts and Basic Laws

6.34 A torque of 50 N-m is required to rotate the paddle wheel shown in Fig. 6.47 at 60 rad/s. If it rotates for 2 minutes during which time 200 kJ of heat is transferred to the air from a reservoir, calculate the final temperature in this rigid volume and the entropy increase in the air. Assume constant specific heats from Table B-2.

© B Entropy Change of a Solid, a Liquid, and a Reservoir 6.36 The engine in Fig. 6.49 delivers 50 kJ by operating between two reservoirs at 900°C and 40°C. Its efficiency is 40%. Determine the entropy change of each reservoir.

Ty = 900°C Piston

T, = 40°C P, = 200 kPa

Qh Engine

Figure 6.49

Figure 6.47 6.35

Two kilograms of air at 40°C and 200 kPa are contained in half of the insulated volume shown in Fig. 6.48. The partition is suddenly ruptured, and the air fills the entire volume. Determine the final pressure and the entropy change for this process.

Perfect

6.37 A 4-lbm copper slab at 200°F is brought into contact with a 10-lbm aluminum slab at 60°F. Insulation allows the two slabs to arrive at the same temperature with no heat transfer to the surroundings. Determine the entropy generated by this process. Assume constant specific heats from Table B-4E. Two kilograms of ice at —10°C are mixed in the insulated container of Fig. 6.50 with 12 kg of water at 25°C. Estimate the final temperature of the mixture and the net entropy change.

vacuum

Figure 6.48

Figure 6.50 vy

Chapter6 Entropy 6.39

Twenty kilograms of copper at 80°C are submerged in a container that holds water at 24°C. Two hundred kilojoules of heat is transferred from the water to the 25°C surroundings. Calculate the eventual temperature of the copper, the entropy change of the water plus the copper, and the entropy change of the universe if the mass of water is a) 6 kg, b) 10 kg, and c) 15 kg. Use average C values from Table B-4.

ety

6.44 Two kilograms of steam are contained in a 40-L rigid container originally at 120 kPa. If 2 MJ of heat are transferred from a 400°C reservoir, estimate the entropy generated. 6.45

Superheated steam is in a 10-ft* rigid container at 200 psia and 800°F. The container is cooled until the pressure reaches 20 psia. Determine the necessary heat transfer and the entropy change of the universe if the heat is transferred to a 70°F atmosphere.

6.46

Steam at 200°C is compressed reversibly in an insulated cylinder from 100 kPa to a) 500 kPa, b) 800 kPa, and c) 1200 kPa. Calculate the final temperature and the work requirement.

€ B Entropy Change of a Phase-Change Substance 6.40

Steam at 4000 kPa and 400°C is allowed to cool to a saturated vapor at 4000 kPa. Calculate the ratio of the initial specific volume to the final specific volume and the change in specific entropy for this process.

6.47 Determine the work output steam, contained in a 0.4-m° and 8000 kPa, are expanded until the pressure is a) 1200 and c) 400 kPa.

6.41

Refrigerant 134a undergoes a process from 400 kPa and 400°C to a pressure of a) 400 kPa, b) 800 kPa, and c) 1200 kPa ina rigid container. Calculate the heat transfer and the change in specific entropy.

6.48 The temperature of 5 kg of water changes from 40°C to 400°C, while the pressure remains constant at 200 kPa. Calculate the required heat transfer and the entropy increase of the universe if the heat comes from a 600°C reservoir.

6.42 Ten pounds of water at 14.7 psia and 40°F are heated to a) 280°F, b) 320°F, and c) 500°F in a constant-pressure process. Calculate the change in entropy for this process. 6.43

Four kilograms of saturated steam are being condensed at a constant pressure of 120 kPa in the cylinder of Fig. 6.51 by transferring heat to the 25°C surroundings. Calculate the entropy generated by this process if a) x, = 0.6, b)ix, —0.2, and'c)T, = 40°C.

when 4 kg of cylinder at 600°C isentropically kPa, b) 800 kPa,

€@ B Entropy Change of an Ideal Gas with Variable Specific Heats 6.49

Air is compressed from 400 K and 100 kPa in an isentropic process. Assuming constant specific heats from Table B-2 and variable specific heats, calculate the final temperature if the final pressure is a) 500 kPa, b) 1000 kPa,

and c) 2000 kPa.

6.50 Air is allowed to quadruple its volume in an isentropic process. If the final temperature of the air is 17°C and we assume variable specific heats, what is the initial temperature? What

would it be if constant specific heats (from Table B-2) were assumed?

4kg

P,=120 kPa

—

Part 1

Concepts and Basic Laws

6.51 The farmer in Fig. 6.52 inserts nitrogen from a pressurized tank to fertilize a field. An isentropic process can be assumed as the gas leaves the tank. Estimate the temperature of the nitrogen at the point where it enters the soil if the pressure in the tank is a) 8 MPa and b) 14 MPa. Make any necessary assumptions and assume constant specific heats and then variable specific heats. (This would be the same temperature should a hose burst and the nitrogen spray across the farmer.)

6.56 Water enters a boiler at 6 MPa and 60°C and leaves at 400°C. Calculate the rate of heat transfer if 2000 kg of water pass through the boiler each minute. What is the entropy generated if the heat that stokes the boiler comes from a flame maintained at 2000°C? 6.57 Steam at 20 kPa and a quality of 0.9 enters the condenser of Fig. 6.54 and leaves as saturated liquid. Calculate the heat transfer and entropy change. If the heat is transferred to a 10° @illakes

what entropy is generated due to this process?

Figure 6.52 6.52 Rework the following problems assuming variable specific heats:

a) Prob.6.22(a) c) Prob. 6.23(a) e) Prob.6.25(a)

—_b) Prob. 6.22(b) d) Prob. 6.24(a) ff) Prob. 6.26(b)

Figure 6.54 6.58 Two kilograms of steam at 6 MPa and 600°C enter an insulated turbine each second. If the losses as the steam moves over the blades in the turbine are neglected, estimate the

g) Prob. 6.27(a)

€ B@ Entropy Changes for a Control Volume 6.53

R134a enters the valve of Fig. 6.53 at 1.2 MPa and x = 0. It leaves the valve at 120 kPa. Determine the entropy change. 1200 kPa

120 kPa

maximum horsepower output of the turbine if the exit pressure is 20 kPa.

6.59 Superheated steam enters an insulated turbine at 20 MPa and 600°C at a mass flow rate of 1000 kg/min. If the steam leaves the turbine at a) 10 kPa, b) 40 kPa, and c) 80 kPa asa saturated vapor, what are the power output and the rate of change of entropy of the steam? 6.60

Superheated steam flows into an insulated turbine at 4 MPa and 500°C. Calculate the maximum horsepower output if the exit pressure a) P, = 10 kPa, b) P, = 60 kPa,

c) T, = 50°C, andd) T, = 80°C. The mass flux 6.54

Ammonia enters a throttle as a saturated liquid at 600 kPa. It exits the throttle to a pressure of

120 kPa. Calculate the exit temperature of the refrigerant and the change in specific entropy for this process. Use the tables. 6.55

Refrigerant 134a enters a heat exchanger at 120 psia and a temperature of 160°F and leaves at the same pressure with a quality of 0.6. The mass flow rate is 20 Ibm/s. What are the heat transfer and rate of change of entropy for the refrigerant?

is 4000 kg per minute. 6.61

Saturated R134a vapor is compressed from 15 psia using an insulated compressor. Determine the minimum horsepower needed to compress 4000 Ibm each hour if the compressor outlet pressure is

a) 200

psia, b) 300 psia, and c) 400 psia.

Saturated liquid leaves the condenser.

Chapter 6 6.62

Saturated ammonia at 120 kPa is compressed to 1200 kPa by the insulated compressor of Fig. 6.55. Neglect any losses and determine the horsepower requirement if 40 kg passes through the compressor each minute. What is the entropy change of the universe due to this compression process if the surroundings are at 20°C? 120 KPa =

1

10 Weomp —

1200 KPa

Entropy

247

6.66 Water at atmospheric pressure and 180°F enters a mixing chamber at a flow rate of 80 gpm. Cold water enters the mixing chamber at 40°F at a flow rate of 120 gpm. Calculate the temperature of the water leaving the mixing chamber and the rate of change of entropy for this insulated chamber. 6.67 The boiler in a power plant heats water to a temperature near 600°C before it enters the turbine. The water is preheated by mixing it with superheated steam extracted from the turbine before it enters the boiler, resulting in less fuel used by the boiler. The insulated preheater is shown in Fig. 6.57 Determine the entropy production if all pressures are at 600 kPa and

a) m, =2kg/s of steam at 250°C

m, = 20 kg/s of water at 45°C 6.63

Two kilograms per second of superheated

b) m, m, c) m, m,

steam leave a steam generator and enter a turbine at 6 MPa and 500°C. The turbine produces 2500 hp by expanding the steam to a pressure of 20 kPa with x = 0.9. Calculate the rate of entropy production by the turbine if the surroundings are at 25°C.

= = = =

2.4 kg/s of steam 24kg/s of water 2.8 kg/s of steam 30 kg/s of water

at 200°C at 40°C at 180°C at 50°C

M3

6.64 Fifteen thousand pounds of superheated steam enter a turbine each hour at 1000 psia and 1000°F and leave the turbine at 5 psia as saturated vapor producing 2000 hp. Calculate the rate of entropy change in the steam and in the universe if the surroundings are at 70°F. 6.65

The insulated turbine in the Rankine cycle in Fig. 6.56, operating with water, produces 3000 kW of power. Determine the entropy change across each device.

Qy Figure 6.57

€ B Isentropic Efficiency 6.68

P, = 18 kPa xX4= 1

Steam at 600 kPa and 350°C enters an adiabatic turbine at a flow rate of 20 kg/s. It leaves the turbine at a pressure of 40 kPa as a saturated vapor. Calculate the power produced by this turbine and the isentropic efficiency of the turbine.

Part 1 6.69

Concepts and Basic Laws

Air enters an adiabatic gas turbine at 600 kPa and 400°C. It leaves the turbine at atmospheric pressure and a temperature of 140°C. Calculate the specific work output and the isentropic efficiency of this turbine. Assume constant specific heats.

6.70 For the adiabatic steam turbine of Fig. 6.58, calculate the isentropic efficiency if a) x, = 0.9, b) x, = 1.0,c) Ty =90°C, and d) T, = 60°C.

6.75 Saturated R134a vapor is to be compressed from 20 psia to 400 psia in a compressor rated at 86% efficiency. Determine the horsepower requirement if the mass flow rate is 400 Ibm each hour. Water enters a pump at 80 kPa and leaves the 6.76 pump at 6 MPa. The mass flow rate through the pump is 90 kg/min. If the pump requires 15 hp, calculate the pump efficiency. 6.77. The pump of Fig. 6.60 is used in a power plant to increase the low pressure of saturated liquid water exiting a condenser to a high pressure entering a boiler. Estimate the efficiency of the pump if it accepts 2400 kg of water each minute at 10 kPa and discharges it at 8MPa while consuming 4000 kWh of energy over a 10-hour period.

:

P.=10kPa

Out

Figure 6.58 6.71

An insulated steam turbine accepts 35 lbm/s at 2000 psia and 1000°F and discharges it at 2 psia. Determine the power produced and the efficiency if the exit quality is a) x, = 0.82, b) x, = 0.90, and c) x, = 1.0.

6.72

An insulated steam turbine produces 8000 hp by accepting 30 000 kg of steam at 10 MPa each hour and discharging the steam at 20 kPa with a quality of 85%. Determine the inlet temperature and the turbine efficiency.

6.73

An air compressor has an isentropic efficiency of 0.88. If air at 100 kPa and 25°C and a mass flow rate of 20 000 kg/hr is compressed to 500 kPa, what work must be supplied to the compressor? Assume constant specific heats.

6.74

Saturated ammonia vapor is compressed as shown in Fig. 6.59. Calculate the horsepower input if the compressor’s adiabatic efficiency is a) 80%, b) 90%, and c) 100%.

10 kPa

Figure 6.60

6.78 Determine the horsepower requirement of a pump that increases the pressure of water from 20 kPa to 5 MPa if it is 88% efficient and pumps 340 000 gallons of water each day. 6.79

a pressure of 140 kPa. If the exit pressure is 100 kPa and the nozzle is 96% efficient,

estimate the exiting velocity. 6.80

120 kPa x=1

Aur enters a nozzle at 106 m/s and 80°C with

a pressure of 120 kPa. If the exit pressure is 100 kPa and the nozzle is 92% efficient, estimate the exiting velocity and temperature.

20 kg/min

W,Comp 2 MPa

Figure 6.59

Air enters a nozzle at 20 m/s and 25°C with

i=

Chapter 6 = Entropy

€ B Exergy (Availability) and Irreversibility 6.81 A steam generator produces superheated steam at a rate of 140 kg/min, a temperature of 500°C, and a pressure of 5 MPa. What is the maximum amount of power that could be obtained using this steam? Use the standard dead state. 6.82 Steam enters a turbine at 10 MPa and 600°C and exits at 20 kPa with a quality of 95%. Determine the irreversibility of this process.

249

6.83 What is the exergy of air at 300°C and 600 kPa if the dead state is 20°C and 101 kPa? a) Assume constant specific heats and b) use Table F-1.

6.84 R134a enters a compressor at 30 psia and 10°F at a flow rate of 40 gpm, and leaves with a temperature of 300°F and a pressure of 160 psia. What is the rate of change of exergy for this device?

Outline 7.1 The Maxwell Relations 7.2 The Clapeyron Equation

252 255

7.3 Relationships for Internal Energy, Enthalpy, Entropy, and Specific Heats 258

Thermodynamic

7.4 The Joule-Thomson Coefficient

|

7.5 Real-Gas Effects

ae

Relations

—

3

58 sans 263

FE-Style Questions Problems

271

265

270

263

Nomenclature | The following variables are introduced in this chapter:

a B & M N

Bulk modulus Gibbs function

A

The star superscript indicates a state where the ideal-gas law PV = mRT applies Volume expansivity Joule~Thomson coefficient

T*

Helmholtz function

B #4,

general function A general function

Learning Outcomes (J

Use the Maxwell relations to develop relations among thermodynamic properties

(J

Apply the Clapeyron equation

Q)

1

Develop and apply the Joule-Thomson coefficient 5

Study the real-gas effects on thermodynamic properties

Find general relationships for internal energy, enthalpy, and entropy ~ :n this chapter, we will relate thermodynamic properties that cannot be measured, such as enthalpy, internal energy, entropy, and specific heats, to properties that can be measured: temperature, pressure, mass, and volume. In previous chapters, tables have been used to determine properties of interest. How were all those properties determined? This chapter will provide those relationships that were used to determine properties over the wide range of temperatures and pressures encountered in the various devices analyzed in our study. In addition, the relationships developed in this chapter will allow us to perform more accurate calculations of properties than those used in previous chapters.

From Chapter 6 we developed the following differential relations for specific internal energy and specific enthalpy. Equations 6.15 and 6.16 are restated as du = Tds — The Helmholtz and Gibbs functions are independent properties defined in terms of other independent properties and are found to be useful in our study, similar to our use of enthalpy. They all have the units of energy and represent energy in certain situations.

Pdv

.

dh = Tds + vdP

(7.1)

(7.2)

Two additional independent properties are now defined for a simple thermodynamic system. They are the Helmholtz function a and the Gibbs function g: a=u

gf

—

Ts

(7.3)

ls

(7.4)

In differential form, these become, with help from Eqs. 7.1 and 7.2, da —

Pau.

dg "00

ssa i

Pasa1

(7.5)

(7.6)

Chapter 7 Thermodynamic Relations From calculus the differential of a function z that depends on the two variables x and y is written as Oz

d

ae dx iG = @}

OZ

ae

& Ee. )ay

(7.7) °

where the subscript y on the first partial derivative means that the variable y is held fixed while z is differentiated with respect to x; on the second partial derivative the variable x is held fixed while z is differentiated with respect to y. Equation (77) can be written in the form

dz = Mdx + Nady

(7.8)

where M = (dz/dx), and N = (dz/ay),. The partial derivatives of the functions M and N are

(2M), _ afte) _ fe _ a dy

() Ox

dy\dx/,

/,

> /,

dydx

adaxdy

(=) ise: Ox\dy/,

Remember:

N-m = J

251.0 SK If the copper was assumed to be incompressible (dv = 0 and then B= 0), Eq. 741 would indicate AS = 0 since dT = 0 for this example. The entropy change is due to the small change in the volume of the copper.

(3)

[{ You have completed Learning Out

For a throttle, like an orifice or a valve, the enthalpy remains constant as the fluid flows through the throttle. Setting dh = 0 in Eq. 739 then provides 0 0=CAaT+|o(2) Jap < whys

(7.51)

Divide this equation by dP and solve for d7/dP to obtain

ai

dP)

eC,

[7(=)

o|

dl y>

(7.52)

We define the Joule-Thomson coefficient 4, as the rate of change of temperature with respect to pressure for an isenthalpic (constant enthalpy) process, that is,

Isenthalpic: A constant enthalpy process. It’s also called an isoenthalpic

ae ss

a ANP,

-

7

{7(2) (ES

ol

Jax

WES)

process.

Part 1

Concepts and Basic Laws This coefficient can be used to approximate the temperature change across a throttle for various refrigerants. For an ideal gas, enthalpy is a function of temperature, and thus temperature is a function of enthalpy. For an isenthalpic process, as described in Eq. 753, temperature will also remain constant, giving a Joule-Thomson coefficient of zero for an ideal gas. For nonideal gases, Eq. 7.53 can also provide an expression for C,, in terms of measurable properties, 7; P, and v.

Example

7.8 — Joule-Thomson coefficients — EETEDD EDEL ELLALALLA

Estimate the Joule-Thomson coefficient for R134a at 500 kPa and x (where h, = 73.4 kJ/kg). Then estimate the exiting temperature from the throttle of Fig. 73 if the pressure drops from 1 MPa to 100 kPa and the entering temperature is 39°C. The IRC Calculator is used in this example (the properties are slightly different from those in Appendix D). 39°C 1000 kPa

100 kPa

:

coer The IRC Calculator gives slightly different values

from those in Appendix D.

sae

se

Throttle

:

Solution Use forward differences by using a pressure at 500 kPa and 400 kPa (the refrigerant enters the throttle at x = 0, so a lower pressure is used in the quality region). The enthalpy is held constant at 73 400 J/kg as the refrigerant passes through the throttle. The IRC Calculator at 73 400 J/kg and 500 kPa gives T = 15.7°C. At h = 73 400 J/kg and 400 kPa, we find T = 8.93°C. Using the definition of the Joule~-Thomson coefficient results in

Reel h

The finite difference curacy, though, is not Note that as the perature experiences

| eee

,

_ 8.93 — 15.7 400 — 500

= 0.068°C/kPa

does not allow for more accuracy in the answer. More acdemanded here; we are estimating the values. pressure decreases across a throttle for R134a, the tema significant drop, which is the essential feature of a re-

frigerant. So, using the , calculated, we find %

p, = 0.068 =

dey Oe Ig = Oe. P,— P, 100 — 1000

Chapter 7 Thermodynamic Relations

(P£&

This exiting temperature is not accurate since yw, varies significantly from 100 kPa to 1000 kPa while holding h fixed. But it does give a rough estimate since “4, was calculated at an average pressure of 500 kPa. It is obvious why R134a is a good refrigerant. By simply passing it through a throttle, the temperature dropped about 60°C. If an ideal gas is passed through a throttle, the temperature does not change since enthalpy depends only on temperature for an ideal gas. So, if enthalpy remains constant, so does temperature for an ideal gas, but not for a refrigerant.

(4)

Gases at extreme conditions cannot always be assumed to be an ideal gas. Often, a low temperature combined with a high pressure results in nonideal-gas behavior (as Problems 740(b) and 743(c) would illustrate). Such temperature and pressure combinations are not usually encountered in applications of interest in an introductory course. This section provides the equations needed to make calculations given such extreme conditions. The properties of 7 P and v were found in Chapter 2 using the Z-factor or an equation of state, such as the van der Waals equation. The differential changes in u, h, and s in nonideal gases for which an equation of state is not known can be determined using relations developed in Section 73. Expressions for Ah, Au, and As will now be developed. The relation for the differential enthalpy change in a gas was presented in Eq. 734. It is reproduced as

dip

=

Cal ++)\0

—

oe)

T| ah —] |dP

7.54 (7.54)

This equation can be integrated from state 1 to state 2 to give the change in

enthalpy as

7 oh

Z "

[ear —

Ile_

Ov

ea

ANE

(7.55)

P

To tackle these integrations directly is a difficult if not impossible task. So, we identify a path that results in more integrations but those that are possible to perform without too much trouble. The change in enthalpy is independent of the path between states 1 and 2 so we select a different path with three segments, as shown in Fig. 74 by the dashed lines. This alternate path involves two new states that exist at a very low pressure where the ideal-gas law can be applied with

acceptable accuracy. The states will be designated as 1° and 2’ so that T7,, = 7, and

,then from 1 to2 ,and T,, = T,.To move from 1 to 2,follow the path from | tol finally from 2° to 2. The enthalpy change can then be written as

ey

yy)

(i, = Thy)

(7.56)

Part 1

Concepts and Basic Laws

Figure 7.4 The path from state 1 to state 2 replaced by the alternate path 13 1732732.

The enthalpy change (h, — h,) is for an ideal gas and is simply

h, — h, = | Cas *

(7.57)

T,

T,

since, for an ideal gas, the quantity in brackets in Eq. 755 is zero. The specific heat is assumed to be constant or the ideal-gas Tables F are used. The other two enthalpy changes are at constant temperatures 7, and T,, respectively. An isothermal enthalpy change from some general state to a state of an ideal gas (for example, from state 2 to state 2° in Fig. 74) is given by Eq. 755 to be Recall: The critical

temperature 7_, and pressure P.. can be found in Table B-3.

rv fle-2(@)Jo P

j@

Hint: Recognize that, using

v = ZRTIP 720| _RP az], RZ Atle Poet, P

(7.58)

since the first integral is zero if T = const. To account for nonideal-gas behavior, the compressibility factor Z, defined by Eq. 2.15, is introduced along with the reduced pressure Pp = P/P,, and reduced temperature T, = T/T... The above enthalpy difference can, after some manipulation, be written as

p= The Le

Pp

=

;

dP

aT, | (=| pene ie

P.

IT

(7.59)

IPhy Pr

R

If both sides of this equation are divided by the molar mass M, Eq. 759 takes the form

hh T

Hil

Pr

Pees \ es = RT, | {——) —ie u ale

h

(7.60)

oP

The difference (h” — h)/T., is the enthalpy departure and can be obtained using Appendix I. Using Figure I-1 and Eq. 756, we can obtain the enthalpy difference in a real gas. An example will illustrate.

Chapter 7

Thermodynamic Relations

The internal energy change is found using the definition of enthalpy to give

h, — h, — R(Z,T, — Z,T))

(7.61)

The entropy change for a gas is found by integrating the expression provided in Eq. 7.42, giving T,

Ss) — 5, = |C7aT

a (e) dP

(7.62)

JP

T,

An alternate path is taken as was done in Fig. 7.4. The difference 5 — 5 is the entropy departure and is presented in Appendix J. The entropy change between state 1 and state 2 is then found from SS

(Sy

) =i (s5 am )

Gms

3)

(7.63)

Entropy departure: The difference between the ideal-gas entropy at a sufficiently low pressure and the entropy at that same temperature.

There is one difference in this calculation when compared to the enthalpy difference calculation: The ideal-gas entropy change s, — s, represents the ideal-gas difference between states 1° and 2’, so the pressure is not constant. The following example illustrates.

Property changes in a real gas LLL

INO LEELA

Carbon dioxide undergoes a process from state 1 at 320 K and 200 kPa to state 2 at 600 K and 8000 kPa. Estimate the change in enthalpy, internal energy, and entropy, i) assuming’ the carbon dioxide is an ideal gas, using constant specific heats from Table B-2, ii) using the ideal-gas tables in Appendix F, and iii) using the equations of this section.

Solution i) The ideal-gas equations assuming constant specific heats result in

Ah=h,

—h, = C,AT = 0.842 x (600 — 320) =

236 kJ/kg

Au = C,AT = 0.653 X (600 — 320) = 183 kJ/kg NSS

Sy

(s

Cue

=

P,

ae

600 8000 = 0,168 kI/ke-K s 0.842Ina0: 355 — 0.1889In= = /kg (Continued)

Example

7.9

Part 1

Example

Concepts and Basic Laws

7.9

(Continued)

SZ

ii) The ideal-gas Table F-4, which allows for variableie alate heat, provides

hy — h,

i

22280 — 10186

ee

oe

.@

44

= 275 kI/kg

17 es (2S

eae

As = boe = Ring ae

orESSE

ae ie

— 0.0877 kI/kg:K

iii) Using the enthalpy departure chart of Fig. I-1 in the appendix, we need Remember: fh = Mh Pay

P,

=

0.2 = = 0.027

te

P,

8 SS;

i

Po

7.39

=

oe

Po Accuracy: Two digits is all

that can be obtained from

30

gis

ae

=

600

= 1.97

304.2

me

+=

r

0.05

0.05 x 304.2

«hi, — hy = ————"

cr

The terms

he =

(hi, — h’) and (53 — s}) accurate than those in

Ie

=

ie

:

et

are for an ideal aS Those from Part (il) are more

=U.

Mme tied08)

Using values from the chart, the enthalpy differences are

would allow for three digits.

Note:

320 iceeag ge

T.,

:

the charts. Larger charts

1 are

:

es é

= 0.35kI/kg

44

h

FA

== 24

xX 3042

sh, -h => :

:

a4

= 17k

/ 2

The property differences between state | and state 2 are found, using Eqs. 7.56,

hy = hy = (hy Si Lf

Un

is

hy) + he, = hh) = th, te 2S

4)

O33 = 258 kJ/kg

NZ

Za)

= 258° — 0.1889 x (0.96 x 600 —-0.99 x 320) = 209KI/ke Sp

Sy SS

i ie:

5) ni (s5

WOSTe =

cm)

(s;

i (-4) =

)

—0.0877 kJ/kg-K

where the entropy departure (5 — 5°) = M(s — s') was found using Table J-1.

The Z-values used to find vu, — u, are found in Appendix H.

Chapter 7 Thermodynamic Relations :

The high pressure (above about 10 MPa) at state 2 would make the ideal-

gas calculations unacceptable as approximations. If the pressure at state 2 was Aires _ d

less than about 4 MPa, which it usually is, the ideal-gas Table F would provide acceptable approximations. See Problem 7.40a. Bees

(P

Comment

ites eno high

ee

at

oi a

and 220 K) that real-gas effects are particularly significant (see Problem 7.406). In our study, such

a combination is seldom encountered.

(5)

In this chapter, relationships of thermodynamic properties were developed that could not be measured (enthalpy, internal energy, entropy, and specific heats) in terms of properties that can be measured: temperature, pressure, weight, and volume. In previous chapters, tables were used to determine properties of interest, but how were the entries in those tables determined if they were not measured? This chapter provided the relationships used to determine those properties over the wide range of temperatures and pressures that were encountered in the various devices analyzed in our study of thermodynamics. The following additional terms were defined in this chapter: Clapeyron equation: Allows the determination of the enthalpy of vaporization h,, from measurable properties P, T; and v. Enthalpy departure: The difference between the ideal-gas enthalpy at a sufficiently low pressure and the enthalpy at that same temperature divided by the criticalpoint temperature. Entropy departure: The difference between the entropy at a sufficiently low pressure and the entropy at that same temperature. Helmholtz and Gibbs functions: Properties that are defined in terms of other known properties. Joule-Thomson coefficient: The rate of change of pressure with respect to temperature for a constant enthalpy process. Maxwell relations: Relationships between entropy and measurable properties.

Several important equations were presented: Helmholtz function:

a = u—Ts

Gibbs function:

e— hI

The differential:

+ Ndy dz = (=) he ee (=) dy = Mdx

0

NZ

C

ER

=Part1 = Concepts and Basic Laws

Maxwell relations:

Additional relations:

Clapeyron equation:

Clapeyron—Clausius equation:

Differential internal energy:

Differential enthalpy:

Differential entropy:

Joule—Thomson coefficient:

Enthalpy departure:

Real-gas enthalpy change:

I

TIN

LP DIEPPE

PTL

h, — h, = (h, — hy) + (hy — hh) + (ap)

ILIV

PTEDL

A

RE

FE-Style Questions* 7.1

Estimate the partial derivative (ds/dP)., of steam at 400°C and 2.5 MPa: (A) —0.0003 mi/kg-K

(B) (C) (D)

—0.00025 m*/kg-K —0.0002 m*/ke-K —0.00015 m*/kg-K

7.2

The change in the Helmholtz function at 400°C in steam as the pressure changes from 2500

kPa to 3500 kPa is nearest:

(A) 224 kJ/kg (B) 154 kJ/kg (C) 122 kJ/kg (D)

“The FE exam does not include questions on topics from this chapter.

108 kJ/k

Chapter 7 7.3

Estimate the saturation temperature of R134a at 40 kPa using the Clapeyron equation and values from Table D-2:

(A) (B) (Cy) (D) 7.4

—44°C —46°C 426 —50°C

Estimate the Joule-Thomson coefficients for

ammonia at 1400 kPa and h = 1570 kJ/kg:

(A) (B) (C) (D) TES

0.0190°C/kPa 0.0165°C/kPa 0.0145°C/kPa 0.0120°C/kPa

If the Joule-Thomson coefficient for a refrigerant at 1000 kPa is 0.08 °C/kPa, approximate the exiting temperature if the

if

7.6

using the ideal-gas law directly.

7.10

Apply Eqs. 77 and 710 directly to Eqs. 71 and 72 and derive the relations of Eqs. 711 and 712.

711

Apply Eqs. 77 and 7.10 directly to Eqs. 75 and 7.6 and derive the relations of Eqs. 713 and 7.14.

7.12

Apply Eq. 77 directly to Eqs. 71 and 72 and derive the relations of Eqs. 715 and 7.16.

7.13

An ideal gas undergoes a process from state 1 to state 2. For the following data, approximate the change in specific entropy for this process using Eq. 7.13.

IG

1 2 7.14

21 kJ/kg 19 kJ/kg 17kJ/ke 15 kJ/kg

€ B The Maxwell Relations

Equations Use the differential form for dv (see Eq. 77) assuming v = R7/P and find the change in the specific volume Av of air if the pressure and temperature change from 200 kPa and 100°C

T (°C)

P (kPa)

v (m*/kg)

200 180

100 150

0.8 {2

An ideal gas undergoes a process from state 1 to state 2. For the data shown below, approximate the change in specific entropy for this process using Eq. 7.11.

State

21 kJ/kg 29kJ/ke 41 kJ/kg 64kJ/ke

The change in internal energy for the conditions of Problem 76 is nearest:

(A) (B) (C) (D)

7.8

State

Estimate the change in the enthalpy of air if it is heated isothermally from 10°C and 100 kPa to 10 MPa:

(A) (B) (C) (D)

Use the differential form for dP (see Eq. 77) assuming P = RT/v and find the change in the pressure of air if the specific volume and temperature change from 0.5 m*/kg and 80°C

to 0.52 m*/kg and 70°C. Compare with AP

exiting pressure is 120 kPa:

—60°C —70°C Pane —90°C

271

to 210 kPa and 120°C. Compare with Av using the ideal-gas law directly.

entering pressure and temperature are 2000 kPa and 60°C, respectively, and the

(A) (B) (Os (D)

Thermodynamic Relations

1 2

7 (°F)

P (psia)

v (ft?/lbm)

150 180

50 1

4.5 By?

Verify Eq. 712, which is (07/dP), = (00/08)p, using the properties of steam at 1 MPa and 400°C.

7.16

R134a exists at 500 kPa and 100°C. Verify

Eq. 7.14, which is (00/0T)p = —(ds/dP);, using a) the values in Table D-3 and b) the IRC Calculator.

7.17

Verify Eq. 714, which is (0v/0T)p = — (0s/0P),, using the properties of R134a at 300°F and 160 psia.

7.18

Verify, using ammonia at 200 kPa and 40°C, that (dh/ds), = T is indeed true using the values in Table E-2.

7.19

Verify, using steam at 800 kPa and 400°C, that (ah/ds) p = T is indeed true using a) the values in Table C-3 and b) the IRC Calculator.

272 7.20

Part 1

Concepts and Basic Laws

Verify, using steam at 800 psia and 800°F, that

7.29

Find a relationship for C, — C, for a gas for which the van der Waals equation of state

(du/ds), = T is indeed true. Use the IRC

P = RT/(v — b) — a/v’ is applicable. Then

Calculator.

let a = b = Oand show that C, — C,= R.

€ B The Clapeyron Equation 7.21

7.22

7.30

Show that saturated water vapor at 20 kPa can be treated as an ideal gas (compare v = RT/P with v, from Table C-2). Could saturated water vapor at 100 kPa be treated as an ideal gas?

P = RT/(v — b) — a/v’. Assume that the specific heat C, depends on temperature only. Then compare the results with those assuming constant specific heat with C, = 10 kJ/kg-K.

Using the pressure, temperature, and specific volume found in Table C-1, determine hig for

steam at 300°C using the Clapeyron equation. Also determine s,, Calculate the errors assuming the values obtained in Table C-1 are accurate. 7.23

7.31

Using the pressure, temperature, and specific

7.24

The steam table C-2E has a low pressure of 1 psia, unlike Table C-2, which has a much lower pressure of 0.611 kPa. A need is expressed for the saturation temperature of steam below 1 psia; suppose that only Table C-2E is available. Estimate the saturation temperature of steam at a) 0.1 psia, b) 0.2 psia, and c) 0.6 psia using information from Table C-2E.

p = 2700 kg/m’*. What is the contribution due to the pressure increase?

7.32 7.33 7.34

7.26

Assume an ideal gas with constant specific heats for which Pv = RT and determine expressions for As using Eqs. 741 and 7.42.

7.27

Find the expression for the change in internal energy Au for a gas obeying the equation of state equation) assuming constant specific heats.

Find the expression for the change in enthalpy Ah for a gas obeying the equation of state P = RT/(v — b) — a/v’ (the van der Waals equation) assuming constant specific heats.

Estimate B and B for R134a at 800 kPa and Gs Use

€ B The Joule-Thomson Coefficient Estimate the Joule-Thomson coefficient for R134a at 350 kPa and x = 0. Then predict the exiting temperature from the valve of Fig. 75 if the entering pressure drops to 100 kPa from a) 400 kPa and 8°C, b) 600 kPa and 21°C, and c) 800 kPa and 31°C. Use the IRC Calculator. Py

100 kPa

ios

R134a

Figure 7.5

P = RT/(v — b) — alv’ (the van der Waals 7.28

Estimate 6B and B for water at 1000 psia and 100°F and then find the difference C, — C,, O°C and then find the difference C a the IRC Calculator.

and C, Determine an expression for the quantities in the brackets of Eqs. 733 and 739 for an ideal gas for which Pu = RT:°

Estimate 8 and B for water at 10 MPa and

40°C and then find the difference C, — C,,

€ @ Relationships for Au, Ah, As, c; 7.25

A 50-kg block of aluminum experiences a pressure change from 100 kPa to 50 MPa, while the temperature increases from 50°C to 100°C. Estimate its change in entropy. Use

B=7X10°K™,C, = 0.9 kI/kg-°C, and

volume found in Table D-1, determine hy

for R134a at 60°F by applying the Clapeyron equation. Also determine s,. Calculate the errors assuming the values obtained in Table D-1 are accurate.

Air undergoes an isothermal process at 100°C from 200 kPa to 6000 kPa. Calculate the enthalpy and entropy changes using the van der Waals equation of state

7.36

Estimate the Joule-Thomson coefficient for R134a at 1000 kPa and x = 0; then predict the exiting temperature from a valve if the pressure drops to 100 kPa from 2000 kPa and the entrance temperature of 67°C. Use the IRC Calculator.

Chapter 7 IES

Estimate the Joule-Thomson coefficient for air entering the valve of Fig. 76 at 1200 kPa and 400°C and show that the temperature does not significantly change when the pressure drops to 100 kPa.

400°C P,

7.43

Thermodynamic Relations

273

Calculate the change in the entropy of nitrogen using both the ideal-gas table and the entropy departure chart if its state changes from: a) 260 K and 900 kPa to 800 K and 4 MPa

b) 350K and 1200 kPa to 700 K and 8 MPa c) 500K and 1800 kPa to 220 K and 10 MPa

7.44

100 kPa

Air undergoes a process from state 1 at 400 K and 200 kPa to state 2 at 900 K and 12 MPa. Estimate the change in enthalpy and entropy:

i) Assuming the air is an ideal gas with constant specific heats from Table B-2

—_> Air

Figure 7.6

7.38

ii) Assuming the air is an ideal-gas with constant specific heats interpolated from Table B-6 at 650 K

Estimate the Joule-Thomson coefficient of steam at 4 MPa and 400°C, and then estimate

Using the ideal-gas tables in Appendix F-1

the value for C,, at that state using Eq. 753. Compare with the value found by using

Using the equations of Section 75

C, = (h/0T)>. 7.39

v) Using the IRC Calculator

Estimate the Joule-Thomson coefficient of steam at 800 psia and 800°F, and then estimate

7.45

the value for C,, at that state using Eq. 753. Compare with the value found by using

C, = (dh/dT)p.

i) Assuming the air is an ideal-gas with constant specific heats from Table B-2 ii) Using the ideal-gas tables in the Appendix

© B Real-Gas Effects 7.40

Calculate the change in enthalpy of air using both the ideal-gas table and the enthalpy departure chart if its state changes from:

a) 260 K and 900 kPa to 820 K and 4 MPa

b) 360 K and 900 kPa to 220 K and 8 MPa c) 300 K and 200 kPa to 600 K and 10 MPa

7.41

Calculate the change in enthalpy of carbon dioxide using both the ideal-gas table and the enthalpy departure chart if its state changes from:

Air undergoes a process from state 1 at 220 K and 200 kPa to state 2 at 300 K and 10 MPa. Estimate the change in enthalpy and entropy:

Using the equations of Section 75 Using the IRC Calculator

7.46

Nitrogen undergoes a process from state 1 at 220 K and 2 MPa to state 2 at 280 K and 20 MPa. Estimate the change in enthalpy and entropy: Note: These extreme conditions are not indicative of typical problems but are meant to illustrate nonideal-gas behavior.

a) 320K and 800 Pa to 300 K and 6 MPa

b) 400 K and 2 MPa to 400 K and 12 MPa 7.42

c) 500K and 400 kPa to 900 K and 10 MPa

i) Assuming the nitrogen is an ideal gas with constant specific heats from Table B-2

Calculate the change in entropy of air using both the ideal-gas table and the entropy departure chart if its state changes from:

ii) Assuming the nitrogen is an ideal gas with constant specific heats interpolated from Table B-6 at 250 K

a) 520°R and 180 psia to 1600°R and 1000 psia

b) 600°R and 250 psia to 1800°R and 3000 psia c)

400°R and 100 psia to 440°R and 4000 psia

iii) Using the ideal-gas tables in the Appendix Using the equations of Section 75

274 7.47

Part 1

Concepts and Basic Laws

Carbon dioxide undergoes a process from state 1 at 640°R and 900 psia to state 2 at 900°R and 3000 psia. Estimate the change in enthalpy, internal energy, and entropy: i) Assuming the carbon dioxide is an idealgas with constant specific heats from Table B-2E

7.49

Calculate the minimum power required to compress 2 kg/s of nitrogen from 100 kPa and 300 K to 8 MPa assuming an adiabatic process. i) Use the ideal-gas table and ii) account for real-gas behavior.

7.50

A rigid volume contains air at 300 K and 400 kPa. If the temperature is raised to 1200 K, determine the final pressure and the heat transfer i) assuming ideal-gas behavior and ii) accounting for real-gas effects.

ii) Using the ideal-gas tables in the Appendix

ii) Using the equations of Section 75 7.48

Determine the maximum power produced by the adiabatic turbine of Fig. 77 that accepts 4 kg/s of air at 10 MPa and 1100 K and exhausts the air to the atmosphere at 100 kPa. i) Use the ideal-gas table and ii) account for real-gas behavior.

P;=10MPa T, = 1100K

Nitrogen enters a compressor at 27°C and 1.5 MPa and exits at 480 K and 15 MPa. If the mass flux is 2 kg/s, determine the enthalpy change, the entropy change, and the power required if the heat loss is 20 kJ/s. 7,52

Air is compressed isothermally from 20 psia and 100°F to 1200 psia, as shown in Fig. 78.

Calculate the enthalpy change, the entropy change, and the work required if the heat loss is 80 Btu/Ibm. 1200 psia

Compressor

Py = 100 kPa Figure 7.7

Figure 7.8

We

The Rankine Power Cycle Gas Power Cycles Refrigeration Cycles Mixtures and Psychrometrics Combustion

Outline 8.1 Energy Sustainability 8.2 The Rankine Cycle 8.2.1 8.2.2

279

279

Basic configuration and components 249 Improving Rankine cycle efficiency 286

8.3 Modified Rankine Cycles 8.3.1 8.3.2 8.3.3.

290

The ideal reheat Rankine cycle 290 The ideal regenerative Rankine cycle 293 A combined reheat-regenerative ideal Rankine cycle 296

8.4 Cogeneration Cycles 8.5 Losses in Power Plants 8.6 Summary

299 302

303

FE Exam Practice Questions (afternoon

M.E. discipline exam) Problems

306

305

Only one new variable is introduced in this chapter: 5

Utilization factor

SSS

SSS

ee

ea

Learning Outcomes 1

Analyze a simple Rankine cycle Modify a Rankine cycle with reheat and preheat Understand the effectiveness of cogeneration

Determine the overall efficiency of a power plant Ott

Te chapter will analyze power plants. These plants use what is traditionally called a vapor Rankine cycle. The term vapor is used to indicate that the working fluid, water, undergoes phase changes between the liquid and gas phases. Phase change allows us to transfer large amounts of heat to or from the water. In Chapter 9, power cycles that use a gas as the working fluid will be considered. The equations and properties of the previous chapters will be utilized, so very few additional equations and quantities will be required. Basically, what has already been presented will be applied to the vapor cycle that produces power.

Motivational Example—The Cooling Tower

Note: Asketch ofa — cooling tower is shown in Fig. 11.21 onp. 411.

No power generating plant can operate at 100% efficiency. Some of the heat produced by the steam generator must be rejected to a low-temperature reservoir. In major power generating stations the device that accomplishes this job is a heat exchanger called a condenser. Ocean water, lake water, or river water is often used to cool the condensing steam. When an adequate source of water is not available, a cooling tower is a device that uses air to condense steam as a method of rejecting heat. Smaller cooling towers are also used as condensers in large air conditioning and refrigeration systems. Cooling towers have been used to cool steam in major power plants since the early 1900's. In a dry tube cooling tower, the steam travels down pipes along the inside of the cooling tower. Cool air travels up the inside of the tower and transfers heat from the steam pipes. In a wet or evaporative cooling tower, the fluid to be cooled flows in the open, down the inside of the cooling tower. Again, air traveling up the inside of the cooling tower cools the fluid by convection and evaporation. The air moves up the cooling tower either by natural draft or by forced draft from fans. These cooling towers are so often used in nuclear power plants that they have become a symbol for nuclear plants even though they do not contain the reactor.

Chapter 8

The Rankine Power Cycle

The concept of sustainability applies to any system in which natural resources are used or impacted by the system. Any system that uses resources that cannot be replenished is ultimately unsustainable. At some point in time, the resources required in the operation of many of the systems in use today will disappear. Most of the systems that produce energy are unsustainable. They use fossil fuels like coal or oil for which there is a limited supply. Examples of energy supplies that are sustainable include solar energy, geothermal energy, hydropower, wind energy, and possibly the thermal energy and wave energy in the oceans. Sustainability also influences the by-products developed in an engineering system. The by-products of an engineering process that damage the environment in which the system operates may also produce an unsustainable system. One example is the use of fossil fuels to produce electrical power. Even if there was an unlimited supply of coal and oil, a major product of combustion is carbon dioxide, which plants use to create oxygen in the atmosphere. Excessive carbon dioxide production, coupled witha decrease in the number of forests on our planet, has created a buildup of carbon dioxide in the atmosphere. This may be leading to a pattern of global warming that may cause significant damage to many ecosystems. Energy sustainability is essential to the future of humankind. Without energy, there is no economy. Problems with the sustainability of water resources and food production can be directly tied to the availability of energy. One of the major tasks for engineers in the next several decades will be to redirect our energy dependence from fossil fuels to alternative methods of energy conversion that are more sustainable. Chapter 13 will address several of these alternatives.

Any system that uses resources that cannot be replenished is unsustainable.

By-products of a process that damage the environment may also produce an unsustainable

system. ce

i

In a nuclear power plant, the steam flows over hot rods of fis-

sile uranium or plutonium. These rods do not get as hot as burning coal, so the steam temperature produced in a nuclear reactor is less than that developed in a coal-fired plant. As a result, the second law

ie

is

Conclusion: Energy sustainability is essential to the future of humankind.

Smokestack

t

eerie:

Superheated

coal, fuel oil, nuclear fission, geothermal heat sources, or even trash.

The function of the boiler is to heat subcooled liquid water to superheated steam. In a coal-fired power plant, the coal is ground into a fine powder which is then fed down the center of the steam generator, where it burns at a very high temperature. The combustion temperature of coal increases with the amount of surface area available to burn. Pulverizing the coal creates this large surface area due to the small size of the coal particles. Water is pumped through pipes that cover the inside surface of the steam generator. As the water travels in these pipes, it absorbs heat from the burning coal in the center of the chamber.

i

Boiler (steam generator): A heat exchanger used to transfer heat from the fuel to the working fluid.

8.2.1 Basic configuration and components The Rankine cycle is used to produce power; it has four major components, which will be described in some detail in the following paragraphs. A boiler, or steam generator, is a heat exchanger used to transfer large amounts of heat to the water, the working fluid. Figure 8.1 shows a sketch of a simple boiler. The heat for a boiler can be produced from

279

t |titt

TT aan

Fire grate

ce 7

od

Figure 8.1 A sketch of a boiler.

steam out

ry

Part 2 Applications

Per pound, trash contains

about 50% of the energy content of coal. However, serious emission problems accompany trash burning. Do a Google search on “waste energy” and discover what can be done.

Figure 8.2 The blades on a steam turbine rotor. [Permission from Tesla-Turbines, photo by Christian Kuhna.]

Pollution: The boiler accounts for most of the

pollution of CO, and sulfur in a coal-fired plant. Steam turbine: The device that creates mechanical power from the superheated steam that leaves the boiler. Condenser:

A heat

exchanger that condenses the steam from the turbine into a saturated liquid, the condensate.

Heat rejected by the condenser can be used to heat.and cool buildings and/or used in industrial processes.

indicates that the efficiency of a nuclear plant will be less than the efficiency of a coal-fired plant due to the lower boiler (reservoir) temperature. A well-designed coal-fired power plant will have an overall cycle efficiency of 40 to 42%.A nuclear

power plant will have cycle efficiencies on the order of 32%. The heating process in the boiler, a heat exchanger, is assumed to be a constant-pressure process. A major concern in the design of a power plant is the impact on the environment of operating the plant. Much of the pollution created by a power plant comes from the operation of the boiler. In a coal-fired power plant the main stack gas is carbon dioxide. Plants are now being designed which have facilities to remove the carbon dioxide and contain it for industrial use. Coal-fired boilers also emit sulfur into the atmosphere, the amount depending on the type of coal used. Nuclear reactors do not emit products of combustion; their main environmental problem is how to dispose of spent nuclear fuel rods. Currently, they are stored at the reactor site. Inventories of spent fuel rods are building up and pose a significant health risk if the storage facilities are not well maintained. The 2011 disaster at the Fukushima-Daichi Nuclear Plant in Japan is a clear example of the need to find a safe and sustainable solution to this problem. A steam turbine is used to create mechanical power from the superheated © incamerastock/Alamy steam that flows from the boiler via a rotating shaft transmitting torque. The steam enters the turbine as a high-temperature, high-pressure superheated vapor and exits the turbine as a much lower energy steam, which can be superheated steam or a saturated mixture. If a mixture exits the turbine, it should not have a quality less than 0.90! in order to prevent water droplets from damaging the turbine blades. In a steam turbine, high-pressure steam impacts the blades of a rotor (see Fig. 8.2). The shaft of the turbine provides rotational mechanical power, which is the product of the torque produced by the turbine and the rotational speed of the turbine shaft. This mechanical power is used to drive an electric generator. The mechanical engineer loses control of the power after it leaves the turbine to the electrical engineer, who assumes control as the power enters the generator and passes to the electrical grid. So, our study of thermodynamics does not consider the power after it leaves the turbine, which operates ideally as an isentropic device. The steam that exits the turbine travels through a heat exchanger, called a condenser. At the condenser, heat is removed from the steam from the turbine sufficient to condense it into a saturated liquid so that it can be efficiently pumped back to a high pressure. There are a variety of techniques used to accomplish this that are dependent on the cooling water resources available near the plant. A water-cooled condenser is sketched in Fig. 8.3. Ideally, large amounts of water from a river, lake, or ocean are used to condense the steam exiting the turbine. If water is not readily available for cooling, air can be used in a cooling tower facility to condense the steam. In special circumstances, much of the heat rejected by the condenser can be used for heating and cooling’ buildings that are adjacent to a power plant, as has been done at many universities, cities, and industrial complexes. The operation of the condensers poses environmental concerns. When large amounts of heat are dumped into the environment, it is called thermal pollution. If hot water from a condenser is deposited into a river or lake, it will have an effect

on the ecology of that body of water. Many types of fish cannot live at elevated 'A condensing turbine operates with an exiting quality less than 1.0. We will assume the exiting steam has a quality of 1.0 or is in the superheat region as a design criterion for an operating turbine. ~ Cooling with heat uses a process called absorption refrigeration, which will be analyzed in Chapter 10.

Chapter 8 Cooling water out

Cooling

281

The Rankine Power Cycle

Steam in from turbine

Condensate

to pump

water in

Figure 8.3 A water-cooled condenser in which steam enters and condensate leaves.

eee

temperatures and will leave the area around a power plant, whereas other fish will be attracted to the warmer water. Some types of waterfowl will use the warm areas around a power plant as a winter sanctuary, an unnatural phenomenon that may or may not be welcomed. Also, vapors emitted by cooling towers, used to condense the steam, can cause fog and icing on local roads during cold weather, in addition to possible medical problems. The liquid condensate leaving the condenser is at a low pressure, usually below atmospheric pressure. To obtain the high pressure required by the turbine, massive feedwater pumps, like the one shown in Fig. 8.4, are used to pump the water back to the boiler, thereby completing the cycle. Power produced by the plant must be used to run these pumps, although compared to the power produced by the turbine it is so small that it can actually be ignored when considering the efficiency of the power cycle. A schematic of how the components of the simple Rankine cycle are arranged is shown in Fig. 8.5, and its’ 7-s diagram is sketched in Fig. 8.6. State 1 is a saturated liquid that leaves the condenser and enters the pump. State 2 is the high-pressure

Suggestion: Let's just skip the condenser and use a compressor to raise the pressure back to the high pressure required by the turbine. Why dump all the heat into the environment? Answer this question at the end of this chapter. Feedwater pump: The device used to produce the high pressure required by the turbine. Observe:

The boiler and

condenser are constantpressure heat exchangers.

T

©) \ \ \ \ \ \

TAffiliates, rat.net.nz

Pump

|/

Condenser

Turbine \ x

9®

| Na

Ss

Based of on

Figure 8.4 Feedwater pump.

Figure 8.5 The simple Rankine cycle includes a pump, boiler, and turbine.

Figure 8.6 The Fs diagram for the simple Rankine cycle.

282

Part 2 Applications liquid water that leaves the pump and enters the boiler. State 3 is the superheated steam leaving the boiler and entering the turbine. A saturated mixture at state 4, or possibly a superheated steam at state 4’, leaves the turbine and enters the condenser.

Ideal Rankine cycle: It provides an upper limit of the cycle efficiency since there are no losses in all the components, especially the turbine.

Rankine cycle efficiency: The net work output divided by the energy input, considering only the devices that constitute the Rankine cycle.

The components that make up the Rankine power cycle were introduced in Section 4.2. The basic Rankine cycle may have been studied in Section 4.3.2: it is often skipped in an introductory course. In this chapter it is analyzed in greater detail, including modifications and efficiencies. If losses are assumed negligible in all the components and connecting pipes, it is called the ideal Rankine cycle. It provides an upper limit of the cycle efficiency. Losses are often included in the turbine, as shown by the dashed line in Fig. 8.6, in which case it is not an ideal cycle. Losses in the other components are relatively small

and often ignored. The Rankine cycle efficiency is defined as the net work output divided by the energy input to the steam, considering only the four devices that constitute the Rankine cycle. The net work output includes the work provided by the turbine minus the work required by the pump (the pump work is often neglected since it is relatively small, less than 1% of the turbine power). The energy input is the heat transfer received by the working fluid in the steam generator. The cycle’s thermodynamic efficiency, sometimes referred to simply as the cycle efficiency, is

duis

Plant efficiency: The ratio of the energy output from the entire power plant divided by the energy in the fuel that enters the power plant. It is always less than the efficiency of the Rankine cycle.

Example

8.1

=

:

(8.1)

where Wand W, are the turbine and pump work, respectively, and Q, is the heat input by the boiler. The plant efficiency is the ratio of the energy output from the entire power plant divided by the energy in the fuel that enters the power plant. It would include the losses in the combustion chamber where the energy contained by a fuel is converted to the heat that enters the steam in the boiler pipes (losses of about 10%), and the losses that exist between the output of the turbine and the electrical energy that leaves the power plant (losses of about 8%). Also, about 6% of the electrical power is needed to operate the auxiliary equipment that is needed to operate the power plant. These losses do not enter the efficiency calculation for the Rankine cycle. ;

An ideal Rankine cycle An ideal Rankine cycle operates between the four states shown in Fig. 8.7 Analyze the cycle and determine the cycle efficiency. The mass flow rate of the steam is 25 kg/s. The cycle is referred to as “ideal” since all the components are assumed to operate with no losses. No losses in the turbine, pump, and connect-

ing pipes are considered.

:

“We use both the terms stear generator and boiler for the same heat exchanger. This is done because both terms are in use and it’s a good idea to become familiar with them. We tend to use boiler most often.

Chapter 8

The Rankine Power Cycle

Qc

Q Pump

ee

Ss

Figure 8.7

Figure 8.8

State 1: P, = P,, x,=0

State 2: P, = P,

State 3: T, = 600°C, P,= 6 MPa

State 4: P,=20kPa

Solution

First, a 7-s diagram is often helpful. It should always be sketched, as in Fig. 8.8. Compressed liquid exits the pump, and superheated steam exits the boiler. Let’s analyze the turbine first, for no particular reason. It accepts superheated steam at 6 MPa and 600°C, and, in the ideal cycle, the steam undergoes an isentropic process from state 3 to state 4; that is,s,= s;. This allows state 4 to be identified since the pressure P, = 20 kPa. From Table C-3 at 6 MPa and 600°C, s, = 7.1677 kJ/kg:K, allowing us to locate state 4. Since 5,

NNN

oS

NNA FNS

.

Exhaust jet —>

NAN

NN = NSS Compressor

Hot exhaust strearn

Bypass air

Combustion chamber

Compressor

Ss

Propeller

77)

D

Turbofan

2

D

Outline 9.1 Air-Standard Analysis zs

: oS

:

=

9.2 Reciprocating Engine Terminology

9.3 The Otto Cycle

Gas

-

Power & , :

9.3.1 9.3.2 9.3.3. 9.3.4

The four-stroke Otto cycle 322 Otto cycle analysis 323 Two-stroke Otto cycle engine 327 The Wankel engine 328

330

9.5 Other Gas Power Cycles 9.5.1 9.5.2

‘

9.6.2

9.6.3.

——

334

The dual cycle 335 The Stirling and Ericsson cycles

9.6 The Brayton Cycle 9.6.1

320

322

9.4 The Diesel Cycle

: Cycles

317

337

341

The Brayton cycle with regenerative heating 346 The Brayton cycle with regeneration, intercooling, and reheat 348 Theturbojetengine 350

9.7 The Combined Brayton-Rankine Cycle 9.8 Summary

351

354

FE Exam Practice Questions (afternoon M.E. discipline exam) 356 Problems

358

The following variables are introduced in this chapter: Bottom dead center BDC BWR _ Back-work ratio

ip

Compression ratio

r

Cutoff ratio

r

Pressure ratio

MEP Mean effective pressure TDC _ Top dead center

Se

ee

Learning Outcomes Describe gas power cycles Analyze the Otto cycle

Analyze the diesei eyile Analyze the dual, Stirling, and Ericsson cycles Analyze the Brayton cycle er Combine the Brayton and Rankine cycles ee CO

Motivational Example—tThe VariableCompression-Ratio Engine

A oA A B aye A four-bar linkage used to vary the compression ratio.

The compression ratio of a car engine is the ratio of the largest volume occupied by the air inside one of the cylinders to the smallest volume of the air in that cylinder. Its value for a gas engine is usually around 8 and for a diesel engine around 18. The efficiency of an engine is a direct function of this compression ratio. It would be desirable to vary the compression ratio in a car engine to allow it to perform better under different operating conditions, such as uphill driving or high-speed driving. The first practical variablecompression-ratio engine was developed by Saab in 2000. This engine has a split engine block that is hinged on one side. As the two halves of the engine block rotate away from each other, the compression ratio gets smaller. A hydraulic actuator is used to move the blocks. Paul Corina developed a variable-compression-ratio engine where the two halves of the engine block slide relative to each other. This allows for better sealing of the engine block and the alignment of all moving parts. A four-bar linkage (to be studied in a future design course for mechanical engineering students) is used in Fig. 9.1 to vary the compression ratio, with the highest ratio on the right. These

engines represent a significant development in engine design by allowing the performance of the engine to vary with the demands placed on the engine. In this chapter we will study power cycles that use an ideal gas, most often air, as the working fluid. For this reason they are called gas power cycles. These engines do not have the advantage of using phase change to transfer large amounts of heat. Instead, a gas-powered engine moves large quantities of air to produce power. The engines of primary interest in our study are called internal combustion engines since the fuel burns inside the engines. Engines that operate on the Rankine power cycle presented in the last chapter are external combustion engines since the combustion of fuel takes place outside the power-producing turbine. Another example of an external combustion engine is the old steam locomotive. Combustion took place in a furnace where wood or coal was used to create steam. The steam was then delivered to a reciprocating-piston device that converted the thermal energy in the steam to mechanical energy. External combustion engines that operate on a gas do not have sufficient application to be of interest in our study, although a short description will be presented of two such cycles. The power cycles to be analyzed in this chapter mostly operate as open cycles in that the working fluid does not return to its original state and location. A good example of an open power cycle is that of the automobile engine. Air at ambient conditions enters the engine, and products of combustion at high temperature leave the engine.

Gas power cycles: Power cycles that use an ideal gas as the working fluid. Internal combustion engine: The fuel burns inside the engine.

External combustion engine: The combustion of fuel takes place outside the power-producing device. Open cycle: The working fluid does not return to its original state and location.

The small amount of fuel that mixes with the air can

from the environment will enter the engine prior to compression and be vented from the engine immediately after the mechanical power is produced, as happens

in the engines

of cars, trucks, locomotives,

and tanks. The

rapid

succession of processes in these open cycles will be simplified by a series of assumptions used in the air-standard analysis, which makes use of the following assumptions: 1. The working fluid is air that will be treated as an ideal gas. 2. The compression and expansion processes in the cycle will be assumed to be adiabatic and reversible, resulting in two isentropic processes. 3.

4.

The combustion process in the engine will be modeled as a simple heat transfer process, with heat transferred to the air from an external source equal to that which the combustion process would supply. The process will depend on the type of engine being analyzed: a gas engine, a diesel engine, or a gas turbine engine. The heat rejection in the exhaust will be accomplished by transferring heat to the surroundings with a process that returns the air to the initial state while doing no work to complete the cycle.

The cold-air standard adds another assumption to the air-standard assumptions: The constant specific heats listed in Table B-2, which are at 27°C, will be assumed for all processes throughout a cycle.

be ignored when analyzing the processes in these cycles. The fuel simply introduces heat into the combustion process.

Air-standard assumptions: . Air is an ideal gas.

pay

2. Compression and power

processes are isentropic. 3. A heat transfer process replaces combustion. 4. Exhaust is a process that

transfers heat while doing no work.

Cold-air standard: It adds to the air-standard analysis (assumptions 1 through 4), requiring that the specific heats be constant as given in Table B-2.

ti

Part 2 Applications Some of these assumptions simplify the analyses compared to using the idealgas tables or including real-gas effects; even though they do introduce some error, they provide estimates of the power output and efficiency of actual cycles. They also allow quick estimates of the effects of modifications made to cycles of interest. The cycles presented in this chapter utilize processes and devices introduced in Chapters 3 and 4.

Example

9.1 — Airstandard analysis PLL

S LELLD LE

All three processes in the fictitious cycle of Fig. 9.2 are assumed to be quasiequilibrium processes. The pressure and temperature of the air at the beginning of the compression stroke, from state 1 to state 2, are 100 kPa and 25°C. Assume a piston-cylinder arrangement with a volume ratio of 4 to 1. Determine:

i) ii) iii) iv)

The The The The

maximum temperature and pressure net heat transfer net work output thermal efficiency of the cycle

-—

P®@

©)

Solution

With a piston-cylinder arrangement, the air is contained in the cylinder and treated as a system. So, the first law applied to various processes in Section 3.5 is applicable. i) At state 1, P, = 100 kPa and T, = 25°C = 298 K. The compression stroke is an isentropic process (it’s a quasi-equilibrium adiabatic process) for which we use Eqs. 3.44 and 3.46. They give the temperature at state 2 and the maximum pressure in the cycle at state 2, using k = 1.4: v,

T,

II

es

\k-1

) (D e\2

v,

= 298 x 40? = 510K ‘

\k

Dae (=) = 100 x 4'4 = 696kPa = P,

Chapter 9

Gas Power Cycles

The highest temperature in the cycle occurs at state 3. It is found for this constant-pressure process, using the ideal-gas law, to be (6)

(eee sD a = 519 x 4 = 2075K

or 1800°C

2

ii) The heat transfer for the constant-pressure process is

Qe>

A=

CUT, = 7)

= 1.0 X (2075 — 519) = 1556 kJ/kg The heat transfer for the constant-volume process (work is zero) is Gj,

Au = CAE I

GT)

0.717 x (298 — 2075) = —1274kJ/kg

The net heat transfer is

'Note: The ideal-gas law

oe Gre

as las

v=

RT/P is used.

43.

= 1556 — 1275 = 282 kJ/kg iii) Work occurs for the first two processes, and the net work is calculated as follows:

|}Comment = The first law for a cycle IS Wret = Inet (See Eq. 3.21).

Deen

ate

Cll

a

thy)

pe)

X AG19

298)

es

(about 2%) between

= —158 kJ/kg

Wee

O28)

10, --0,) = 696(

ex 20T 696

20.2875 x 2519 ) 696

= 447 kJ/kg Wey = Wiz + Weg + W,, = —158 + 447 = 289 kJ/kg iv) Assuming q,, = 4>3 and q,,, = 93. the thermal efficiency is

Wyer _ 289

Sg

1556

= 0.186

or

The small difference

18.6%

This is not a cycle used in any particular engine, but it is a cycle, and if you can find a use for it, please do! A reminder is in order to emphasize the difference between a quasiequilibrium process and a reversible process. The working fluid, the air in the above quasi-equilibrium processes, experiences heat transfer due to processes in which the temperature changes slowly (thermodynamically). The hightemperature reservoir would have to be at 2075 K or above for the heat transfer process from state 2 to state 3 to occur. Such a process would be highly irreversible since it would be across a large temperature difference. The efficiency of 92% of a Carnot cycle between the high and low temperatures would require all processes to be reversible. Some texts may refer to the processes being internally reversible but externally irreversible.

Wet ANG Jner iS NOt significant.

Part 2 Applications

9.2 ASI fetes

Engine

Many unique terms and symbols are used in describing internal combustion engines that use reciprocating pistons in cylinders. The following are some basic definitions. Refer to Fig. 9.3. TDC

BOG

Saas

Top dead center (TDC): The position of the piston in the cylinder that provides the minimum volume. Bottom

dead center

(BDC): The

position of the piston in the

cylinder that provides the maximum volume. Bore: The inside diameter of the cylinder, which is assumed to be the outside diameter of the piston. Stroke: The distance of travel of the piston from BDC to TDC. Displacement: A volume equal to the stroke times the crosssectional area of the piston, whichis V_. — V,,,. in Fig. 9.3: Displacement

= Stroke

X 7

Bore?

(9.1)

The total engine displacement is this value times the number of cylinders in the engine. The total displacement is often used to describe the power level of a particular engine. Clearance volume: The volume between the cylinder head and the top of the piston when the piston is at TDC.

Piston-engine terminology.

Compression ratio: The ratio of the maximum volume (piston at BDC) in the cylinder to the minimum volume (piston at TDC) in the cylinder. A higher compression ratio produces more engine power. The compression ratio r is r=

Vine _ V, =

Vepc

O%

=

V in

(9.2)

)

Intake valve: Allows the air or air—fuel mixture to enter the cylinder. Exhaust valve: Allows the combustion products to exit the cylinder. Spark ignition: Occurs when a spark plug ignites the air-fuel mixture. Compression ignition: Occurs when the high SON TORU: during compression ignites the air-fuel mixture. 2

icomuient | Mean effective pressure (MEP):

The higher the MEP. the better the performance for an engine with the same

displacement. mire 8 5 Be oe Orit kN —

= su =

kPa

Refers to the ratio of the net work done dur-

ing the cycle to the displacement. If the MEP acted on the piston during the power stroke, it would produce the same work produced during the actual stroke; that is, the rectangular area under the MEP

dashed line in Fig. 93

is equal to the area under the P-V diagram encompassed by the solid lines. The MEP is used to rate the effectiveness of an engine compared to one with the same displacement: the higher the MEP, the better the performance. The MEP (kPa or psia) can be expressed as MEP

=

cycle

max

Weycle

min

V7

7)

Chapter 9

Gas Power Cycles

Several of the preceding terms were presented for information purposes only. The compression ratio r and the MEP are of particular interest. An analysis of the Otto cycle will utilize these terms.

Reciprocating engine terminology

Example

9.2

The dimensions of a piston cylinder arrangement are displayed in Fig. 9.4. The work for one cycle is 400 J. Determine the displacement, the clearance volume, the compression ratio, and the MEP.

Figure 9.4

Solution The displacement is the volume displaced while the piston moves from TDC to BDC. It is Displacement = stroke X wR? = 12.

7 X 3? = 339.3 cm?

The clearance volume is the volume in the cylinder when the piston is at TDC: Clemence =H

mw « ox

Lt = 28.3 cm

The compression ratio, a ratio of great importance in engine parameters, is found to be

i ——

POS a

:

Vers

oe =e

aoe:

a

The MEP, using W,,,,, = 0.400 kJ, is MEP

=

Wescie

Vo]

Vv

3

0.400 kKN-m

T;, a regenerator would actually have a negative effect. If T,/T, = 0.25, the pressure ratio where the efficiencies are equal is 1132. The following example will include a regenerator in the ideal Brayton cycle of Example 9.8 so that the effect of the regenerator can be assessed.

we

cycle

ifs [F(R PY) ty.0 aftect of r, on cycle efficiency for the ideal Brayton cycle and the ideal regenerative Brayton cycle.

The ideal regenerative Brayton cycle Air enters the compressor of a gas turbine engine at 300 K and 100 kPa where it is compressed to 700 kPa with a maximum temperature of 1000 K, as in Example 9.8. A regenerator is inserted into the cycle, as sketched in Fig. 9.28. Determine the thermal efficiency and the BWR, assuming ideal processes with constant specific heats. out

©

Regenerator

Products of combustion

Figure 9.28 (Continued)

Example 9.10

Part 2 Applications

Example 9.10

(Continued) PRESETS

Es!

Solution The 7-s diagram for the ideal regenerative Brayton cycle is identical to that of the ideal Brayton cycle, with the exception that two additional states are included. The minimum and maximum temperatures are not affected. The efficiency of the regenerative cycle is 10 oS

=

thse

=

i (k-1)/k ie

300 =X

7000

PO

= AT

a47.7%

This represents a 12% increase in cycle efficiency compared to the cycle of Example 9.8. To find the BWR, the temperatures must be known. The temperatures at states 2 and 5 are P.\k-1/k

.

(oe| r(3) 1se

i Ae) P,

k-1/k

= Oa

ee) = oe

1 \0.2857 1000 x (+) = 574K

The turbine and compressor work are then Weg

hy

a Chi

Ts)

= 1.0 x (1000 — 574) = 426 kJ/ke Weomp Gs (hy oe h,) ae Ci(T,

ip i)

= 1.0 X (523 — 300) = 223 kJ/kg so the BWR of this ideal regenerative Brayton cycle is

BWR = ——

Weom

Wi

= —

223

426

= 0523

or 52.3%

The regenerator increased the efficiency but had no effect on the back-work ratio. The effect of the regenerator was to reduce the energy input to the combustor.

Sang

eee |

Intercooler: A device used to reduce compressor work input.

9.6.2 The Brayton cycle with regeneration, intercooling, and reheat In addition to the regenerator, two other devices can be added to the Brayton cycle to increase its efficiency. An intercooler allows compression to occur in two stages, with the intercooler located between the stages. It reduces the compressor work and the temperature at the compressor outlet, thereby controlling the maximum

Chapter 9

Air

Intercooler

Qp

Gas Power Cycles

Reheater

s

Figure 9.30

Figure 9.29 The regenerative Brayton cycle with intercooling and reheating.

The T-s diagram for the ideal Brayton cycle of

Fig. 9.28.

temperature in the cycle. Also, a reheater allows the turbine to produce power in two stages, resulting in increased turbine output. The pressures at which the intercooler and the reheater are inserted in the ideal cycle are selected such that P

P

P

2 4 and | Roe a aes

=6 (oe

P

= — paraater: A device used to increase the turbine work output.

8

a

9.38 cae

with reference to Figs. 9.29 and 9.30. It is assumed in the ideal cycle that there

is no pressure drop across a heat exchanger so P, = P; = P, and P, = P,, = P,. Figure 9.30 also indicates that 7, = T, = T,) and T; = T, = T,. The high turbine temperatures 7, and 7, are also equal. Intercooling and reheating are effective only if a regenerator is used in the Brayton cycle. If a regenerator is not in the cycle, intercooling and reheating will actually decrease the ideal cycle efficiency. Several stages in both the compressor and turbine may be used in the larger gas turbines.

The ideal regenerative Brayton cycle with intercooling and reheating Sr.

Insert an intercooler and a reheater in the regenerative Brayton cycle of Example 9.10 and calculate the efficiency, assuming ideal components and constant specific heats. Air enters the compressor at 300 K and 100 kPa; it is compressed to a maximum pressure of 700 kPa, and the maximum cycle temperature is 1000 K, as shown in Fig. 9.31.

Solution Recognizing that P, = P,, using Eq. 9.38 the intermediate compressor pressure is found to be

P2 = P,P, =100 X 700

«.P, = 265kPa (Continued)

Example 9.11

pity

Part 2 Applications

Example 9.11

(Continued) Ona ; 0)

Combustor

Regenerator

Air out

Air Intercooler 300 K 100 kPa

Q rR

Reheater

The isentropic compressor provides se (k-1)/k i =

7(3)

265 \02857 === SiO)

x

(=)

=

396K

1

From the maximum temperature of 7, = 1000 K, the temperature reheater inlet, assuming an isentropic turbine, is qr, =

de (k-1)/k r(Z) 00m

at the

265 \0.2857 (2) = HSS

6

The temperatures are all known, so the thermal efficiency is found to be Wret 7)

E>}

Donrb

—

in.total

(ee

din

ag IR

ace

CL

:

Weomp

=

ie

rl

242 +242. = 96. = 96

242 + 242

ee

yee 3 2

Ser

et)

= 0.603

or

60.3%

This represents a 26% increase in efficiency over the ideal regenerative Brayton cycle and a 41% increase over the ideal Brayton ‘cycle. These numbers would change somewhat if losses were considered and the ideal-gas tables were used

9.6.3 The turbojet engine Perhaps the primary use of the gas turbine that operates on the Brayton cycle has been in aircraft propulsion. Turbofan and turbojet engines were first developed for use in large commercial aircraft and in military jet planes since a gas turbine has a relatively large power to weight ratio. The turbofan engine is primarily used in aircraft designed to fly at subsonic speeds and the turbojet engine is used in aircraft designed to fly at supersonic speeds; however, both can fly in both

Chapter 9

Diffuser

Compressor

Burner section

Turbine

Nozzle

Gas Power Cycles

[EE

Based Cengel Boles and on

Figure 9.32 A schematic of a turbojet engine.

speed ranges. The basic configuration of the turbojet engine is shown in Fig. 9.32. A diffuser’ slows down the incoming air flow from state 1 to state 2. The compressor sends pressurized air at state 3 into the burner section (the combustor) where fuel is injected into the air; the fuel burns at constant pressure. Products of combustion leave the burner at state 4 and travel through the gas turbine to state 5 and accelerate through the diverging supersonic nozzle to state 6, providing the thrust required to propel the aircraft. The primary function of the turbine is to produce the power needed to drive the compressor. The turbine also produces auxiliary power for the aircraft. The primary purpose of the turbojet engine is to provide the thrust developed by the diverging nozzle. A turboprop engine uses the turbine power to drive a propeller. These engines are used mainly in smaller commuter aircraft. A gear system is used to change the high-speed turbine rotation into a lower-speed, high-torque power that drives the propellers. The analysis of the air as it flows through a turbojet engine described above involves the air flow through a diffuser and a nozzle as well as the compressor and turbine and is the subject of a course in aerodynamics. It was described here to indicate that a jet engine operates on the Brayton cycle. 4

(5)

9.7 The Combined BraytonThe high-temperature exhaust gases from a gas turbine operating on the Brayton cycle can be used as the heat source for a power plant that operates on the Rankine cycle. This combined-cycle power plant is an excellent example of cogeneration; rather than discard the energy-laden exhaust gases from a gas turbine engine, they can be used to generate additional power by generating steam in the boiler of Fig. 9.33. The key element in this system is the gas-to-steam heat exchanger (the boiler for the 'In a supersonic flow, a diffuser has a decreasing area, whereas a nozzle has an increasing area.

eT

Part 2 Applications

;

@ on Exhaust gases

turbine

Heat exchanger

Onn

The combined Brayton-Rankine cycle.

The T-s diagram for the ideal combined Brayton-Rankine cycle.

Rankine cycle) used to transfer heat from the products of combustion leaving the gas turbine to the water that leaves the pump. The 7-s diagram is displayed in Fig. 9.34. This combined-cycle power plant is more sustainable than either of the plants operating separately. Since Brayton cycle engines can use a low heating-value fuel in the external combustion chamber, they are compatible with the use of biofuels that could be made from used vegetable oil or cooking oil. This is especially useful in remote locations where small Brayton cycle engines can be located. Most combinedcycle power plants, however, burn natural gas in the combustor of the Brayton cycle. Conventional coal-fired power plants range in size to over 2000 MW, while combined cycle Brayton—Rankine power plants approach the 1500 MW size. The efficiency of actual combined-cycle power plants may now exceed 60%, an unheard of efficiency for a coal-fired plant, so it has become a very attractive form of power production. To make it even more attractive,.the rejected heat can be used for the various uses mentioned in Section 8.4 on cogeneration. An example will illustrate an ideal combined Brayton—Rankine power cycle.

Example

9,42.

The ideal combined Brayton-Rankine cycle Air enters the compressor of the Brayton cycle of Fig. 9.35 at 300 K and 100 kPa, where it is compressed to 600 kPa. It exits the combustor at 1200 K. The exhaust gases exit the heat exchanger at 350 K. The pressure in the Rankine cycle is raised by the pump from 20 kPa to 4 MPa. The heat exchanger provides the

Chapter 9

Gas Power Cycles

1200 K

__turbine —

Steam | turbine -

Condenser

Qout

steam turbine with steam at 400°C. Determine the efficiency of the combined cycle if the steam turbine power output is 20 MW. Solution

As usual, the pump work will be neglected in the analysis of the Rankine cycle. Refer to Figs. 9.34 and 9.35 for the appropriate state numbers. The enthalpy at the pump outlet ish, = h, = 251 kJ/kg. At the inlet to the steam turbine

h, = 3210 kJ/kg and s, = 6.77 kJ/kg-K. The isentropic turbine allows the properties at state 4 to be determined using s, = s, and P, = 20 kPa. We find , = 2230 kJ/kg with x, = 0.839. The mass flow rate passing through the steam turbine is then calculated, using the specified power output: Wor oe mh,

=

Ny)

20000 = m,(3210

—

2230)

“Ms = 20.4 ke/s

Now, focus on the Brayton cycle. The temperature at the outlet of the compressor is 600 1 P_\(k-1/k 6 = — = 300 * |—— = 501K if es : ce

The temperature that exits the gas turbine and enters the heat exchanger (the steam boiler) 1s ee P,\&-D/k

= m= = Tp)

T.

(— =1200200 x(So

=

=I

(Continued)

Note: The IRC Calculator » has been used.

Eee

6 Part2

«Applications

Example 9.12

(Continued) SPINNER

EG

OAT

The heat that is transferred from the air in the Brayton cycle is used to superheat the water in the Rankine cycle. This is expressed as follows: m,C,(Ts

of Ty) rae m,(h;

-

hy)

m, X 1.0 X (719 — 350) = 20.4 x (3210°— 251)

“.m, = 164kg/s

The output of the gas turbine is found to be Wer ae m,C,(T, =

ds)

= 164 X 1.0 X (1200 — 719) = 78900 kW The gas compressor requires hecomp

a

nC,

ci Ts)

164 xX 1.0 X (501 — 300) = 32900 kW The combustor must provide Cin a m,C,(T;

al)

164 x 1.0 x (1200 — 501) = 114600 kJ/s The combined cycle efficiency can now be determined. It is

ge

Waet

Wor

as Wer

Oe

aa

W

mp

Q.,

20000

+ 78900

—

32900

‘

Compare this efficiency with that of the Brayton cycle of Example 9.8 of 42.6% and that of the Rankine cycle of Example 8.1 of 38%. Even though a direct comparison cannot be made because of the different pressure ratios and high temperatures, the efficiency of 526% is much higher than the individual efficiencies of the Brayton and Rankine cycles. The addition of reheaters, regenerators, and intercoolers would increase the efficiency well above the 576% value.

The three basic cycles of most importanée presented in this chapter were the Otto cycle, the diesel cycle, and the Brayton cycle. Engines that produce power that use a gas, usually air, as the working fluid invariably operate using one of these three basic cycles. The ideal cycles were considered initially in which isentropic processes were used for the compressions and expansions in all cycles. The dual cycle modified the

Chapter 9

Gas Power Cycles

diesel cycle. The very efficient Stirling and Ericsson cycles, which have minimal applications, were also presented. The Brayton cycle was then introduced showing the magnified effects of losses in the compressors and turbines. The Brayton cycle was then modified to include regenerators, intercoolers, and reheaters. Finally, the

combined Brayton—Rankine cycle with its relatively high efficiency was presented. The following additional terms were introduced or emphasized in this chapter:

Air-standard analysis: Air is an ideal gas, compression and power processes are isentropic, a heat transfer process replaces combustion, and exhaust is a heat transfer process. Back-work ratio: The ratio of the work required to drive the compressor to the total work produced by the turbine. Bottom dead center: The position of the piston where the volume in the cylinder is a maximum.

Bore: The diameter of the piston, which is assumed to be the inside diameter of the cylinder. Clearance volume: The volume between the cylinder head and the top of the piston when the piston is at TDC. Cold-air standard: Air enters a cycle at standard conditions with constant specific heats assumed throughout the cycle. Compression ignition: The high temperature after compression ignites the air-fuel mixture. Compression ratio: The ratio of the maximum volume in the cylinder to the minimum volume in the cylinder.

Cutoff ratio: The ratio of the volume at the end of combustion to the volume at the start of combustion. Displacement: A volume equal to the stroke times the cross-sectional area of the piston.

Engine knock: The sound of the flame front hitting the oncoming piston. Exhaust valve: Allows the.combustion products to exit the cylinder. Intake valye: Allows the air or air—fuel mixture to enter the cylinder. Intercooler: Reduces the compressor work and the temperature at the compressor outlet.

Mean effective pressure: The ratio of the net work done during the cycle to the change in volume (the displacement) of the cycle. Regenerative Brayton cycle: The air is preheated before it enters the combustor.

Regenerator: A heat exchanger that transfers heat from one fluid to another fluid; most often it preheats air. Reheater: Increases the turbine output.

Spark ignition: A spark plug ignites the air—fuel mixture. Stroke: The total linear distance of travel of the piston. Throw: The throw ona crankshaft is the moment arm that creates the torque. Top dead center: a minimum.

The position of the piston where the volume in the cylinder is

ee

Part 2 Applications Several important equations were introduced in this chapter: Compression ratio: Mean effective pressure: Otto cycle efficiency:

Diesel cycle efficiency:

(ee A

=

TERK a

a Dual cycle efficiency:

ee

rex

1 ° Brayton cycle efficiency:

UP

=

Back-work ratio:

BWR

=

ge

= Oia

1+

yi

—k)/k

Comp Turb

: 5 Regenerative Brayton cycle efficiency:

Ty Gen/je iets F

Combined cycle efficiency:

(afternoon M.E. discipline exam) Problems 9.19.3 refer to Fig. 9.36. Assume air with constant specific heats. 9.1

The ideal Otto cycle operates with a compression ratio of 8 and inlet conditions of 20°C and 100 kPa. The high temperature is 1200°C. The work output of the cycle is nearest:

(A) (B) (C) (D) 9.2.

325 350 400 425

kJ/kg kJ/kg kJ/kg kJ/kg

Figure 9.36

The MEP of the Otto cycle of Problem 9.1 is nearest:

(A) (B) (C) (C)

370kPa 410kPa 440 kPa 475 kPa

t

ae &, Caeat)

Chapter 9 9.3

The efficiency of the Otto cycle of Problem 9.1 is nearest:

Gas Power Cycles

Problems 9.7-9.10 refer to Fig. 9.38. Assume air with constant specific heats and an adiabatic compressor.

(A) 37% 10)

(B)

42%

(C) (D)

51% 56%

[EEX

Fuel

Problems 9.4-9.6 refer to Fig. 9.37. Assume air with constant specific heats. Air

Products of combustion

Figure 9.38 9.7

The ideal Brayton cycle operates with a pressure ratio of 6 and inlet conditions of 20°C and 100 kPa. The high temperature is 1800°C. The heat input to the cycle is nearest :

(A)

1210 kJ/kg

(B) 1380 kJ/kg Figure 9.37 9.4

9.5

9.6

(C)

1420 kJ/kg

The ideal diesel cycle operates with a compression ratio of 20 and inlet conditions of | 9.8

my) aie kiikg The efficiency of the Brayton cycle of

20°C and 100 kPa. If the cutoff ratio is 2, the

Problem 9.7 is nearest:

high temperature in the cycle is nearest:

(A)

(A) 1710°C

(B) 45%

40%

(B)

1670°C

(C)

50%

(C)

1580°C

(D)

55%

(D)

1430°C

9.9

If both the compressor and the turbine of

The work output of the diesel cycle of Problem

Problem 9.7 are 80% efficient, the cycle

9.4 is nearest:

efficiency reduces to:

(A)

560 kJ/kg

(A)

27%

(B)

490 kJ/kg

(C)

450 kJ/kg

(B) (C)

31% 35%

(D)

420 kJ/kg

(D)

39%

The efficiency of the diesel cycle of Problem

9.10

If an ideal regenerator is added to the ideal

9.4 is nearest:

Brayton cycle of Problem 9.7, the) efficiency of the regenerative Brayton cycle is nearest:

Veet (D) 65%

(C) 55%

nee

(A) 41%

oi

(D)

:

60%

ey

Part 2 Applications

© @ Air-Standard Analysis

€ 8 Reciprocating Engine Terminology

9.11 A piston-cylinder arrangement operates on an air-standard cycle composed of the following four processes:

9.13 An engine cylinder has a diameter of 6 cm. The piston has a clearance of 5 mm and a stroke of 14 cm. Calculate the compression ratio for this cylinder. What is the engine displacement if this is an eight-cylinder engine?

(1-2)

(2-3)

An isentropic-compression process that compresses air from 100 kPa and 20°C to a pressure of 800 kPa A constant-volume combustion process

that increases the pressure to 6.4 MPa and the temperature to 1800°C

(3-4) An isentropic process that reduces the

9.14 An engine cylinder has a diameter of 4 in. The piston has a clearance of 0.6 in. and a stroke of 6 in. Calculate the compression ratio for this cylinder. What is the engine displacement if this is a six-cylinder engine?

pressure to 800 kPa (4-1)

Constant specific heats: When constant specific heats are assumed, they are found in Table B-2 or Table B-2E, unless otherwise specified.

Aconstant-volume process that re-

duces the pressure back to 100 kPa to complete the cycle Sketch the cycle on P-v and T-s diagrams. Using the ideal-gas relationships developed in Chapter 3 assuming constant specific heats, calculate the work output and heat input for this cycle. What is its efficiency?

9.12 A piston-cylinder arrangement operates on an air-standard cycle composed of the following four processes: (1-2)

An isentropic-compression process that takes air from 14.7 psia and 50°F to a pressure of 300 psia

(2-3)

A constant-pressure combustion process that occurs at 300 psia and increases the temperature to 3000°F

(3-4)

An isentropic process that reduces the pressure to 100 psia

(4-1)

A constant-volume process that further reduces the pressure to 14.7 psia to complete the cycle

€ B The Otto Cycle 9.15 The Otto cycle of Fig. 9.39 has a compression ratio of 8 and a maximum temperature of 2000°C. What is the efficiency of this cycle? Determine the heat transfer in and the net work of the cycle if the air enters the cylinder at a) 25°C, b) 120°C, and c) 200°C. Assume the cold air-standard applies. T

Sketch the cycle on P-v and T-s diagrams. Using the ideal-gas relationships developed in Chapter 3 assuming constant specific heats, calculate the net work output and heat input for this cycle. What is its efficiency?

Figure 9.39 9.16

Note:

If neither the volume nor the

mass is given, specific quantities are calculated.

'

|

An Otto cycle has an efficiency of 70% anda maximum temperature of 3500°F. What is the compression ratio of this cycle? Determine the heat transfer in and the net work of the cycle if the air enters the cylinder at a) 60°F, b) 150°F, and c) 250°F. Assume the cold air-standard applies.

Chapter 9 9.17 Heat is added to the air in an Otto cycle in the amount of 1200 kJ/kg. The compression ratio for this engine is 10. Air enters the cycle at 100 kPa and 20°C. Determine, using a cold airstandard analysis,

Gas Power Cycles

€ B The Diesel Cycle 9.23 The diesel cycle of Fig. 9.40 has a compression ratio of 16. Calculate the thermal efficiency of the cycle for cutoff ratios of 2, 2.5, and 3.0.

i) The temperature and pressure of the air at the end of the combustion process The cycle efficiency The net work output The mean effective pressure Remember: All parts of problems with i), ti), ili), etc., are to be worked.

'

Figure 9.40 The combustion products are assumed to have the properties of air, a reasonable assumption.

9.18 Rework Problem 9.17 except that the compression ratio is a) 6 and b) 8. 9.19 Using the air tables in Appendix F-1, find the net work, the efficiency, and the MEP for the

9.24 A diesel cycle is designed to have a thermal efficiency of 70%. What must the compression ratio be if the cutoff ratio is 2 and 3?

9.25 Heat is added in the amount of 800 kJ/kg of air to a diesel cycle with a compression ratio of 19 and a cutoff ratio of 1.8. Air enters the engine at 100 kPa and 27°C. Determine, using a coldair analysis: i) The temperature and pressure of the products of combustion at the end of the combustion process

cycle of a) Problem 9.17 and b) Problem 9.18a. Remember: Parts with lower-case italic ' letters are separate problems.

9.20 Heat is added to the air in an Otto cycle in the amount of 500 Btu/lbm. The compression ratio for this engine is 10. Air enters the cycle at 14.7 psia and 65°F. Determine, using a cold-air analysis,

ii) The thermal efficiency of the engine iii) The specific work output of the cycle iv) The mean effective pressure 9.26

respectively, a) 16 and 2 and b) 20 and 1.75. 9.27

Using the air tables in Appendix F-1, determine the maximum cycle temperature, the net work, the efficiency, and the MEP for the cycle of a) Problem 9.25, and b) Problem 9.26a.

9.28

Heat is added in the amount of 300 Btu/lbm of air to a diesel cycle, with a compression ratio of 22 and a cutoff ratio of 2. Air enters the engine at 14.7 psia and 77°F. Determine, using a coldair analysis:

i) The temperature and pressure at the end of the combustion process

ii) The thermal efficiency of the cycle

iil) The work output of the cycle in ft-lbf iv) The mean effective pressure 9.21

Rework Problem 9.20 except that the

9.22

compression ratio is a) 6 and b) 8. Using the air tables in Appendix F-1E, determine the work, the efficiency, and the MEP for the

cycle of a) Problem 9.20 and b) Problem 9.2 1a.

Rework Problem 9.25 except that the compression ratio and cutoff ratio are,

i) The temperature and pressure of the products of combustion at the end of the combustion process The specific work output of the cycle

The mean effective pressure The thermal efficiency of the cycle

we

Part 2 Applications ee

9.29 Using the air tables in Appendix F-1E, calculate the maximum temperature, the net work, the cycle efficiency, and the MEP for the cycle described in Problem 9.28.

€ @ Other Gas Power Cycles

Isothermal

The dual cycle of Fig. 9.41 has a compression ratio of 20, a cutoff ratio of 3, and a pressure

ratio of 2. Calculate its thermal efficiency, the heat input, and the net work output if the inlet conditions are 100 kPa and 10°C. Assume a cold-air analysis. Jin

@

@

@

ae

Yin

out

compressor

9.30 Derive the expression for the efficiency of the ideal dual cycle given by Eq. 9.23. O31

UC

lsothermal turbine

Figure 9.42 9.34 The P-v diagram of an air-standard Ericsson cycle, shown in Fig. 9.43, operates with a minimum pressure of 60 kPa and a maximum pressure of 780 kPa. Heat is rejected at a constant temperature of 400°C, and heat is added at 1200°C. Using constant specific heats for air, calculate i) the thermal efficiency of the cycle, ii) the net work, iii) the heat addition, and iv) the heat rejection.

‘\

1.2 kPa Ge

OS

® © ¢—\ \

ne ©

\

\

S

\1200°C NY

i

®

No

800°C ~~ a

Qout 60 kPa

ns ee Se

®

ee

Figure 9.41

Vv

9.32 Air enters a dual cycle at 100 kPa and 20°C. Using a cold-air analysis, determine the thermal efficiency, the required specific heat input, and the net work output of this engine if:

De

9.33

Oe

2 UCT

2

b)

r=20,r,= 2,and r, =)

OC)

ii Sere 1.8 andr

ped

An air-standard Stirling cycle shown in Fig. 9.42 operates with a minimum pressure of 75 kPa, a compression ratio of 15, and a high temperature of 1000°C. The heat is rejected at 40°C. Using constant specific heats for air, calculate 1) the net work produced, ii) the heat addition, and iii) the heat rejection.

Figure 9.43

€ 3 The Brayton Cycle 9.35

Calculate the thermal efficiency of an ideal Brayton cycle operating with air if the pressure ratio is 1) 6, 1i) 8, and iii) 10.

9.36 What pressure ratio is needed for an ideal Brayton cycle operating with air to have an efficiency of 1) 50%, ii) 60%, and iii) 70%? 9.37 Air enters the gas turbine cycle of Fig. 9.44 at 25°C and 100 kPa with a flow rate of 1.2 kg/s. It is compressed so that the temperature at state 2 is 400°C. The temperature entering the turbine is 1400°C. Assuming constant specific heats for the air and ideal conditions in all components, calculate:

i) The thermal efficiency ii) The heat added in the combustor

Chapter 9 iii) The net power produced by the cycle

Gas Power Cycles

specific heats for the air and ideal conditions in all components, calculate:

iv) The back-work ratio

i) The thermal efficiency ii) The heat added in the combustor ili) The work produced by the turbine

iv) The back-work ratio Wret

100 kPa Ze

Heat | exchanger Qout

Figure 9.44

9.38

Rework Problem 9.37 except that the compressor exit temperature is a) 600°C and b) 800°C

9.39

Determine the thermal efficiency of the Brayton cycle of Problem 9.37 if the efficiency of the compressor and the efficiency of the turbine are both a) 85%, b) 80%, and c) 70%. Maintain the same temperatures. Remember:

The specific heats are found in Table B-2.

9.42 For the cycle of Problem 9.41, find the cycle thermal efficiency if the compressor and turbine have respective isentropic efficiencies of a) 85% and 80%, and b) 80% and 75%. Determine the percentage decrease in the thermal efficiency of the cycle from that of the ideal cycle. 9.43 Air enters the ideal regenerative Brayton cycle of Fig. 9.46 at 25°C and 100 kPa. The compressor raises the pressure to 500 kPa. The products of combustion leave the combustion chamber at 1200°C. Assuming constant specific heats and an air mass flow rate of 6 kg/s and an ideal cycle, calculate:

i) The cycle efficiency

ii) iii)

The power produced by the turbine The power used by the compressor

iv) The back-work ratio

B

v) The rate of heat exchange in the regenerator

9.40

Assume that the efficiencies of the compressor and turbine of the ideal Brayton cycle shown in Fig. 9.45 are the same. What efficiency would cause the net work output to be zero for a high temperature of 1000°C?

Os,

©

Regenerator

Products of combustion

Combustor

Heat exchanger

600 kPa Compersso

100 kPa 25°C

©)

one

25°C

Figure 9.46 9.44

Heat exchanger

Figure 9.45 9.41

®] 100 kPa

Air enters a gas turbine cycle at 20°F and 14 psia. It is compressed to 98 psia. The high temperature is 1800°F. Assuming constant

Rework Problem 9.43 except that the pressure ratio and maximum temperature are, respectively, a) 4 and 1200°C, b) 5 and 1400°C, and c) 6 and 1600°C. For the cycle of Problem 9.43, find thermal efficiency using respective compressor efficiencies of a) 92% b) 87% and 80%, and c) 83% and

the cycle turbine and and 85%, 75%.

ea

Part 2 Applications iii) The mass flux of the steam iv) The power output of the ideal Rankine

9.46 Air enters the regenerative Brayton cycle of Fig. 9.47 cycle at 14 psia and 20°F. The pressure ratio is 4 and the maximum temperature is 1800°F. Assuming an ideal cycle with constant specific heats, calculate the cycle efficiency and the back-work ratio. Compare with the results of Problem 9.41 if that problem was previously assigned. Jout

6)

Regenerator

Products of combustion

cycle v) The efficiency of the ideal Rankine cycle vi) The combined cycle efficiency

Combustor

|Combustor

mae

er

kPa Gas

comma

turbine ©

(5) 100 kPa Air 10°C

consi

325°C

Air in

(@Q) Heat exchanger

Figure 9.47

9.47

9.48

Repeat Problem 9.46 using turbine and compressor efficiencies of 0.85 and 75%, respectively. Insert two intercoolers in the regenerative ideal Brayton cycles of a) Problem 9.43, b) Problem 9.44b, and c) Problem 9.44c. Calculate the cycle efficiency assuming ideal components and constant specific heats. Refer to Figs. 9.28 and 9.29.

€ B The Combined

Brayton-Rankine Cycle 9.49 The 7-s diagram of a combined ideal Brayton— Rankine cycle is shown in Fig. 9.48. Air enters the isentropic compressor at 10°C and 100 kPa at state 5, with a mass flux of 8 kg/s. The pressure ratio for the compressor is 5. The combustor heats the air to 950°C. The air leaves the gas turbine and enters the “boiler” of an ideal Rankine cycle that operates between 10 kPa and 4 MPa. The air leaves the boiler at 325°C, and

the steam leaves the boiler at 400°C. Assuming constant specific heats for the air, calculate:

i) The net power output of the ideal-gas turbine

ii) The efficiency of the ideal-gas turbine cycle

Steam turbine

Qout

Figure 9.48 9.50

The basic layout of a combined Brayton— Rankine cycle is shown in Fig. 9.49. Air enters the isentropic compressor at 10°C and 100 kPa, with a mass flux of 30 kg/s. The pressure ratio for this compressor is 6. The exhaust gases leave the combustion chamber and enter the isentropic gas turbine at 900°C. The ideal Rankine cycle operates between 40 kPa and 4 MPa. Steam at 4 kg/s enters the isentropic steam turbine at 400°C. Assuming constant specific heats for the air, calculate:

i) The net work output of the ideal-gas turbine engine The efficiency of the ideal-gas turbine cycle The heat exchanged in the heat exchanger The work output of the ideal Rankine cycle

Chapter 9 v) The efficiency of the ideal Rankine cycle vi) The combined cycle efficiency

compressor pressure ratio is 6. Exhaust leaves the gas turbine at 2000°F and leaves the heat exchanger at 600°F. The mass flux of air through the Brayton cycle is 40 lbm/s. Steam leaves the Rankine side of the heat exchanger at 800 psia and 900°F and the steam turbine at 4 psia. Assuming constant specific heats for the

On

Combust 6 r

@ 1200°C

air, calculate:

“Gas

Compressor

i) The net horsepower produced by the gas turbine

__turbine

6) 100 kPa Air

Gas Power Cycles

ii) The mass flux of the steam through the Rankine cycle

10°C

iii) The horsepower produced by the steam turbine

@) Heat exchanger

iv) The combined cycle efficiency

“Steam : turbine

|Combustor |

Samp

Qout

6) 15 psia

Figure 9.49 9.51

Air

Repeat Problem 9.50 with the only changes being the maximum gas temperature and

BOE

600°F @) Heat exchanger

the maximum steam temperature, which are,

respectively, a) 1000°C and 450°C and b) 1400°C and 500°C.

9.52 Repeat Problem 9.50 with the following information: : a)

A gas compressor efficiency of 80%, a gas turbine efficiency of 86%, a steam turbine efficiency of 90%

b)

A gas compressor efficiency of 82%, a gas turbine efficiency of 88%, a steam

Steam turbine

Wst

Qout

Figure 9.50

=

turbine efficiency of 92% Ignore the water pump power, as usual.

9:53 Air enters the ideal Brayton—Rankine combined cycle of Fig. 9.50 at 75°F and an atmospheric pressure of 15 psia. The

9.54

Repeat Problem 9.53 except change the maximum gas temperature and the maximum steam temperature, respectively, to a) 1800°F and 800°F and b) 2400°F and 1000°F.

ere

a

ow “iF We

see ~

[=

*

| The Vapor Compression-Refrigeration Cycle 367 10.1.1

Refrigeration cycle terminology 367 2 The ideal refrigeration cycle 368 .3 An actual refrigeration cycle 370 .4 Heat pumps

371

5 Refrigerants

373

2 Cascade Refrigeration Systems Absorption Refrigeration Gas Refrigeration Systems » Summary

374

377 377

380

FE Exam Practice Questions (afternoon M.E. discipline exam) 381

Problems

382

No new nomenclature is introduced in this chapter.

J ——

| Learning =

—]}§,T

Outcomes

Understand the basic cycles and components used in refrigeration

Analyze refrigeration systems Analyze the cascade refrigeration system Introduce absorption refrigeration alain le Wires © Analyze gas refrigeration

Motivational Example—Evaporative Coolers Wall or car window

Fabric

Cooler moist air

An open box with wet fabric hanging and air blowing through

An evaporative cooler. Water flows down the fabric layers.

A serious problem in Third World countries and in remote locations is how to provide cooling without the availability of electricity to run a conventional refrigerator. One solution is the evaporative cooler sketched in Fig.10.1. These coolers have been in use for many decades by covering old-style wooden canteens with fabric. A small amount of water would seep from inside the canteen through the wood body and be absorbed by the fabric on the exterior of the canteen. The water in the fabric would evaporate slowly, and the heat of evaporation would keep the canteen cool. A modern version of this idea uses an inside waterproof container that holds food or medicine and an outside container made of a porous material. The space between the containers is filled with a material like fiber or sand that can absorb water. In use, the sun heats the outer container, causing heat to be transferred to the filler material. Water evaporates from the filler material and travels through the outside container. The heat of evaporation removed from the system will keep the inside container at about 6 € 43p):

Chapter 10

Refrigeration Cycles

10.1 The Vapor Compression10.1.1 Refrigeration cycle terminology Refrigerators, air conditioners,

The vapor compression-refrigeration cycle resembles a reverse Rankine cycle. __ freezers, and heat pumps

Use is made of the phase-change characteristics of the working fluid, the refriger- _ ©Perate on refrigeration ant, to exchange large quantities of heat. Refrigeration cycles, on which refrigerators, air conditioners, freezers, and heat pumps operate, consist of the following

four components, as shown in Fig. 10.2. An ordinary refrigerator with the components identified is sketched in Fig. 10.3.

1.

A compressor is used to raise the pressure of the refrigerant. The compressor is usually powered by an electric motor that provides the input power that drives the refrigeration system.

2. - A condenser, a heat exchanger that removes heat from the refrigerant, transfers heat from the refrigerant to the outside environment, usually as a refrigerant-to-air heat exchanger. In a home central air-conditioning system, this is the large unit that sits outside the house. 3. A throttle is then used to suddenly reduce the pressure and temperature of the refrigerant, making it very cold. This is usually a fixed throttle, though an adjustable throttling valve can be used. 4.

Finally, the cycle is completed with an evaporator, another heat exchanger that transfers heat from the volume being cooled to the refrigerant. In a refrigerator or freezer, this heat exchanger would be built into the inside wall

to transfer heat from the interior. The compression process is ideally an isentropic process. Since it is accomplished with a compressor, the refrigerant entering and leaving this device must Evaporator

Q out (to surroundings)

Condenser

Compresseor

Q,, (from refrigerated space)

Figure 10.2 The basic refrigeration cycle.

Figure 10.3 A sketch of an ordinary refrigerator (the hot coils are on the back).

poids: eRe

Ga he

eS

Part 2

T

Applications be in the gaseous state since droplets on the compressor blades cannot be tolerated. Both the condenser and evaporator are heat exchangers that operate as constant-pressure processes. The throttle is a constantenthalpy process. Figure 10.4 shows a 7-s diagram for the ideal refrigeration cycle. In the ideal refrigeration cycle, the refrigerant enters the compressor at state 1 in Figs. 10.2 and 10.4 as a saturated vapor (x = 1) and is compressed to state 2 using an isentropic process. The compressor is

Condenser

powered by an electric motor and accounts for the purchased energy in the definition of the cycle’s performance. At state 2 the refrigerant exits the compressor and enters the Figure 10.4 condenser as a superheated vapor. The temperature of the refrigerThe 7-s diagram for the ideal refrigeration ant from state 2 to state 3 must be sufficiently high that the condenser cycle. transfers heat to the surroundings. If the device is a refrigerator, the refrigerant must be significantly hotter than the air in the kitchen; the condenser is made up of the coils on the back of a refrigerator. If the device is an air condiIn the ideal refrigeration cycle, the compressor inlet tioner, the refrigerant must be hotter than the outside air. is saturated vapor (x = 1) At state 3 in the ideal cycle at the condenser exit, the refrigerant is a saturated and the condenser exit is liquid (x = 0). Because the condenser is a heat exchanger, the pressure at state 3 saturated liquid (x = 0). will be the same as the pressure of state 2 in this ideal cycle. The purpose of the condenser is to cool the refrigerant so that it undergoes a phase change from a superheated vapor to a saturated liquid. A refrigerant must have a large temperature drop across a throttle while the enthalpy remains constant.

The evaporator performs the desired effect for a refrigerator.

From state 3 to state 4, the refrigerant is forced through a throttle, a valve, or a

set of capillary tubes (very small diameter tubes). There is a large drop in the pressure and the temperature, and some of the refrigerant vaporizes while the enthalpy remains constant, as described in Section 4.2.3. State 4 is a mixture of liquid and vapor. The type of refrigerant determines the temperature drop across the throttle. The refrigerant enters the evaporator at state 4. It absorbs heat from the volume to be cooled, so it must be at a temperature significantly below the temperature of the cooled space. The cycle is completed, and the refrigerant again enters the compressor at state | as a saturated vapor (x = 1) in the ideal cycle.

(1) 10.1.2

2 aaa | Coefficient of performance: The measure of the performance of a refrigeration system. It's often greater than 1 so it is not called efficiency, but it has the same definition.

The ideal refrigeration cycle

The use of refrigerant tables is required for the analysis of refrigeration cycles because of the phase-change processes that take place. The IRC Calculator can be used for ammonia and R134a and several additional refrigerants presently being used. The tables for R134a and ammonia have been included in the appendix of this book; R134a is the selected refrigerant that will be used extensively in the examples and problems. The performance parameter for refrigeration cycle is the coefficient of performance, the COP. (It was also presented by Eq. 4.46.) The desired output of a refrigerator or air conditioner is the rate of heat removal Q,,by the evaporator from the region being kept cold. This heat transfer is called the cooling load. The necessary input to force the refrigerant through the cycle is the input power W,,, to the compressor. The COP is most often greater than 1, so it is not called

Chapter 10

Refrigeration Cycles

efficiency, but it has the same definition. For a refrigerator or an air conditioner, it is

COP,

_ desired effect _

O,,

inputenergy

Ww.in

An example will illustrate refrigeration cycle.

the calculations

required

(10.1) to analyze

the ideal

The ideal refrigeration cycle Pt

EEE

Example

10.1

EOI

Refrigerant R134a enters the compressor ofan ideal refrigeration cycle as a saturated vapor at

|r

@

120 kPa and enters the condenser at 900 kPa,

as shown in Fig. 10.5. Determine the lowest temperature, the cooling load, the necessary compressor horsepower, and the coefficient of performance for this ideal refrigeration cycle for a refrigerant mass flow rate of 0.15 kg/s.

Solution

Figure 10.5

The IRC Calculator will be used to obtain the refrigerant properties. If, for some reason, it is not available, Appendix D can be used. The properties of interest at the four states are listed below. The enthalpies are of primary interest for these control volumes. State 1 (saturated vapor): P, = P,= 120 kPa,x = 1,

SSS Remark: The IRC Calculator provides values that are slightly different from those in Appendix D.

h, = 237 kJ/kg, s, = 0.948 kJ/kg-K State 2 (superheated vapor): P, = 900 kPa,s, = 0.948 kJ/kg-K, h, = 279 kJ/kg

State 3 (saturated liquid): P, = P, = 900 kPa, x, = 0, N,= 4102 kiikg, T= 355°C

The temperature dropped substantially across the throttle from 35.5°C to —22.3°C, a necessary

State 4 (quality region): P, = P, = 120 kPa, T, = =22.3°C,

property of a refrigerant.

h, =h, = 102 kJ/kg Using the above enthalpies, we can determine the quantities of interest. The heat transferred from the condenser is

oy. = rah, ~ hy)

Note: The condenser heat

transfer is negative since it leaves the refrigerant, but

we state the heat transfer from the condenser as a positive number.

= 0.15 kg/s X (279 — 102) kJ/kg = 26.6kJ/s The cooling load provided by the evaporator is

QO,, = m(h, — h,) = 0.15(237 — 102) = 20.2 kJ/s The compressor work requirement is

W,, = m(h, — h,) = 0.15 X (279 — 237) = 63kW The coefficient of performance for this ideal cycle is

COP, =

Rea =

or 84hp

Note: The IRC Calculator uses only three significant digits, which is more reasonable than the five or six digits found in tables. One digit subtracts off so only two digits remain

Part 2 Applications

10.1.3 An actual refrigeration cycle An actual refrigeration cycle deviates from the ideal cycle in several ways. Several

deviations are intentional, while others cannot be avoided. The refrigerant enters the compressor with x > 1, which avoids any condensation on the compressor blades; an actual cycle will be designed so that the refrigerant enters slightly superheated. Also, optimum operation requires a compressed liquid to enter the throttle, so state 3 in Fig. 10.4 is designed to be several degrees in the compressed liquid region. Losses occur in the compressor, causing an entropy increase, but the compressor may also lose heat, causing an entropy decrease. A list of the deviations from the ideal cycle follow:

s The refrigerant enters the compressor slightly superheated. s The refrigerant exits the throttle slightly subcooled. a Frictional effects and heat loss exist in the compressor.

= Pressure drops due to friction in the connecting pipes between components. Sufficient heat may cause an entropy decrease across the compressor.

=

Heat transfer occurs from the connecting pipes.

= Pressure drops in the coils inside the condenser and evaporator. An actual cycle with eight states is shown with the various components in Fig. 10.6. The 7-s diagram in Fig. 10.7 shows four actual states, assuming short pipes between components with negligible losses. An example illustrates an actual cycle.

Ohsvs

iS

Figure 10.6

Figure 10.7

The actual states of a refrigeration cycle.

Example

10.2

Anactual refrigeration cycle NAPS

The pipes between components are assumed short, so only four states are needed since losses between components are neglected in

the given information.

The Fs diagram for an actual refrigeration cycle with short pipe sections between components.

SEATS

ees

The ideal refrigeration cycle of Example 10.1 experiences the following real effects: ¢ The refrigerant leaves the evaporator and enters the compressor at 120 kPa and —18°C. ° ‘The refrigerant leaves the compressor and enters the condenser at 920 kPa and

DAG

Chapter 10

Refrigeration Cycles

e The refrigerant leaves the condenser and enters the throttle at 900 kPa and 34°C.

e The refrigerant enters the evaporator at 44°C and 125 kPa. 3

=2\3

Determine the lowest temperature, the cOObAS loud and the coefficient of performance for this refrigeration cycle for a refrigerant mass flow rate of 0.15 kg/s. Solution

The IRC Property Calculator will be used to obtain the refrigerant properties, if it is available on the Web. Appendix D could also be used, but intricate interpolations would be required. The properties of interest at the various states identified in Figs. 10.6 and 10.7 are as follows.

States 8 and 1 (compressor inlet): P, = P, = 120 kPa, T= “18°C. so h, = 240 ky/ke States 2 and 3 (condenser inlet): P, = P, = 920 kPa, T= 32, -s0 hy

286 Kiiks

States 4 and 5 (throttle inlet): P, = P, = 900 kPa, T= 34°C so. = 99.4 ki/ke States 6 and 7 (evaporator inlet): P, = P, = 125 kPa, T,= —214C, so

h,= 99-4 ki/kg

The lowest temperature in the cycle is at the evaporator inlet, where T, = —21.4°C. The temperature would increase slightly to —18°C as the refrigerant passed through the evaporator. Using the above enthalpies, we can determine the additional quantities of interest. The cooling load is

O,, = m(h, — Ng) = 0.15(240 — 99.4) = 21.1 kIIs Observe that the real effects have actually increased the cooling load, a rather : interesting result. To find the COP, the work input must be known. The compressor power requirement is

W,, = m(h, — h,) = 0.15 X (286 — 240) = 6.9kW The coefficient of performance for this actual cycle has decreased about 4% because of the increased compressor power requirement. It is QO

COP,

==

in

in

11

are, of, say 85%, were included,

=> :

= 3.06

6.9

Laeeae

the CoP would decrease significantly more.

10.1.4 Heat pumps A heat pump is a refrigeration system in which the condenser is located inside

the space to be heated and the evaporator is located outside the space so that the heat rejected by the condenser warms the space. Motels often use such a system; it cools the room when it’s too warm and heats it when it’s too cool. Valves are used to reverse the flow of refrigerant through the heat exchangers and throttle. The

heat exchanger that was the condenser on the air conditioner is now the evaporator. The air-conditioner condenser is now the evaporator for the heat pump.

ciescalion Fein ERI whicT tne condenser is located inside the space to be heated.

Part 2 Applications Heat pumps work well in temperate regions where outside temperatures do

foe ee ouinp the condenser performs the desired effect. eer ys! ey A

ites

arena

not rise above about 100°F (40°C) or drop below about 20°F (—7°C), typically the midwestern United States. These systems are not advised for extreme temperatures, hot and cold. If the heat pump cannot supply enough heat in winter using the vapor cycle, an internal electric heater is used to maintain the desired temperature in the heated space. The electric heater is quite expensive to operate, so some systems use a separate natural gas backup heater, which is much more economical. In regions with hot climates (e.g., Florida and Arizona), only air conditioners are needed, with

_ Heat pumps are used — small] electric resistance heaters to provide occasional heating.

ee ae

:os

The COP of a heat pump can be expressed as :

cold. If they are used in

COPyp = purchased energy me W.

nae Oe by electric sy is supplied

resistance heaters during the colder weather; that’s too expensive.

Ow

desired effect

northern states, heating

(10.2)

in

It is related to the COP, as follows (let Q,,,/Q;, = y for simplicity in the algebra):

Cone

De.

De s Cee

=.

We

Oe

1 Site.

eT

Qsut

Q;,

=

=

1

1

Qn

Co

Wi,

ne

Os

1

Q.,

10.3 (10.3)

1 ee

he , Bs, eee Sy

OF

-_

V

—

il

Oe

- COR, = GOR.

Example

10.3

ai

Aheat pump AEST

ES

A refrigeration cycle, using R134a, operates as shown in Fig. 10.8. The mass flow rate of refrigerant is 0.3 kg/s. The power input to the compressor, which is 85% efficient, is 20 kW. Determine the coefficient of performance for this system operating in both the cooling mode and heating mode. Qcond

1000 kPa

Condenser

Compressor

@

Weomp= 20 kW Ki

| 200 kPa Q Evap

Figure 10.8

tokten 85%

Chapter 10

Refrigeration Cycles

[EZE

Solution The known properties at the four states are:

State 1: P, = 200kPa,x,=1: State 2: P, = 1000 kPa State 3: P,= 1000 kPa,x,= 0:

h,= ho = 244 Klike Fo = —10.1-C h,= 107 kJ/kg

State 4: P, = 200 kPa: h, = h, = 107 kJ/kg

The enthalpy leaving the compressor is found using the power input as follows: WwW

comp

oat

rni(y — hy)

_ 0.3 X (hy = 244)

Neomp

0.85

“Ah, = 301 kJ/kg

Using the above enthalpies, the rate of heat transfer from the condenser and evaporator is, respectively, Oe

=

mh,

FE, hs)

= 0.3 X (301 — Ons

a

mh,

107) = S58kJ/s

see h,)

= 0.3(244 —

107) = 41.1 kJ/s

The energy that is input to the refrigerant is Wi,

= nW,...comp = 0.85 X 20 = 17kIJ/s

Because the compressor is 85% efficient, only

17 kJ/s of the 20 kJ/s input power actually enters the refrigerant.

When operating in the cooling mode, the COP is

in

Note: COP. = COP] +1 mS

10.1.5 Refrigerants A refrigerant is the working fluid used in a refrigeration cycle, a cycle used to either cool (e.g., an air conditioner) or heat (a heat pump) a space. A refrigerant must experience a substantial temperature drop when it passes through a throttle, a device that provides a sudden reduction in pressure. Review Section 2.3, where properties of refrigerants were introduced, and Section 4.2.2, where throttles were analyzed. In the early days of refrigeration, the main chemicals used as refrigerants were ammonia, sulfur dioxide, and carbon dioxide. Ammonia was used the most and is

still in use in some refrigeration systems today; however, ammonia is very toxic and was first replaced with Freon refrigerants. Refrigerants R12 and R22 were used for several decades under various brand names. In the 1970s, it was concluded that

EE Refrigerant: The working fluid used in a refrigeration cycle. It must experience a large temperature drop when it passes through a throttle.

EYER

=6Part2

Applications Freon refrigerants released into the environment negatively affected the ozone

layer. This prompted the development of more environmentally benign refrigeree ants. Refrigerant R134a was developed to replace the Freons in commercial and fluorocarbons into the domestic refrigeration and air-conditioning systems. It has been in use since the atmosphere damages the early 1990s and is currently the only refrigerant allowed for domestic use. R134a ozone layer. aactag >= es vrserserernve™"""““has a negligible effect on the ozone layer. Refrigerant HFO-1234yf has been proposed as a replacement for R134a in automotive air conditioners because of its low flammability. It is being resisted by air-conditioning specialists because of the high cost with little return, according to the specialists (read comments on the Internet.) The use of refrigerants in air conditioning and refrigeration systems may be COntributing to a pattern of global warming that can cause significant damage HFO-1234yf may replace R134a, which replaced R12. to many ecosystems. Traditionally, refrigerants were vented into the atmosphere SSE aa ala al when the systems were serviced. It was determined that the release of these fluorocarbons into the atmosphere contributed to the damage of the ozone layer, which protects plants from dangerous solar radiation. Two engineering solutions were developed for this problem. First, refrigerants are now collected and destroyed, as required by law, when refrigeration and air-conditioning systems are serviced. The second solution is to switch from the refrigerants that damage the environment to refrigerants such as HFO-1234yf that are safer and more environmentally friendly.

{ You have completed Lear

: B

(

(2)

We 1atariy 5 cascada aye or a cascade system. They

The simple refrigeration system of Fig. 10.2 is adequate for most applications. For large industrial air-conditioning or refrigeration needs, however, the simple system WOuld place too much load on a large high pressure-ratio compressor. The efficien-

have the same meaning.

cy of a large compressor to power large refrigeration requirements would be rela-

ge

a

tively low. One alternative configuration is the cascade system, in which a cascade refrigerator is actually two or more separate refrigeration cycles working in series. Figure 10.9 shows the layout of a two-stage cascade system. A three-stage system would add another cycle below the bottom cycle. The two cycles of the two-stage system are joined by a heat exchanger that serves as the condenser for cycle #1 and the evaporator for cycle #2. Since the flows in the two systems never mix, the mass flow rates and even the type of refrigerant can be different. In addition to allowing Pes Comment. | multiple compressors to operate under more acceptable conditions resulting in deA cascade system creased work input, another advantage of a cascade system is to allow a lower temcan achieve the low —_—_perature in the evaporator of cycle #1. If a very low temperature is desired, several

ae

ieee

a

ees

ies fee g......

stages may be needed. Increased refrigeration also occurs in the cascade system, as

Ulustrated in Fig. 10.10.A single-stage system with the same low temperature would

result in state 9 exiting the compressor and state 10 exiting the throttle. The optimal value for the intermediate pressure P, is given by Note: If the refrigerant is different in each cycle, two different Fs diagrams will

be required.

=vV By

Pn Pe

eae

where P,, and P, are the high and low pressures shown in Fig. 10.10. The same refrigerant is assumed in both cycles.

Chapter 10

Refrigeration Cycles

Ons (to surroundings) Condenser |

T

Decreased work input

.©

AWWW @

Heat exchanger

VV K Throttle

#1

Increased refrigeration

Os, (from cooled space)

Figure 10.9

Figure 10.10

A two-stage refrigeration system.

The 7-s diagram for the ideal two-stage cycle. The shaded areas show the decreased work and increased refrigeration when compared to a single-stage system.

To determine the relationship between the mass flow rates in the two cycles, apply the first law to the heat exchanger. It is

WU,

= UE) Sali A=

109)

(10.5)

where m, is the mass flux in low-pressure cycle #1 and m, is the mass flux in highpressure cycle #2. The ratio of mass fluxes is then

ss

10.6

ae

Typically, the low-pressure mass flux 77, is known since it is cycle #1 that determines _ Historically, in the United é

;

:

:

ae

e

:

:

the required refrigeration. If X tons of refrigeration is desired, then, in the SI system, cng mi 3.52X = m,(h, hy)

(10.7)

This equation allows the mass flux in cycle #1 to be determined. The overall coefficient of performance would be determined by using Or

divided by the total compressor work input.

States. refrigeration has

pean reacimganrone

a rather unusual unit. It was considered to be the cooling effect of one ton of

Cee aueNON

dine

1 ton0 = 3.52 kW = 12,000

Btu/hr

A two-stage cascade refrigeration system PAE

The ideal refrigeration cycle that operates between high and low pressures of 20 psia and 120 psia is to be replaced with a two-stage system. Determine the cooling load for a refrigerant mass flow rate of 0.15 Ibm/s and the COP. (Continued)

Example

10.4

Ey

=6Part2 =Applications

Example 10.4

(Continued) IDL

ELLIE

LOAD

Solution First, the intermediate pressure must be determined. It is

P, = VP,P, = V20 x 120 = 49 psia If tables are used, 50 psia is sufficiently close. The properties of interest at the various states identified in Fig. 10.9 are as follows: State 1: P, = 20 psia, x, = 1,h, = 103 Btu/ Ibm, s, = 0.226 Btu/Ibm-°R

State 2: P, = 49 psia, s, = 0.226 Btu/Ibm-°R, A, = 111 Btu/lbm State 3: P, = 49 psia, x, = 0,h, = 24.6 Btu/Ibm

State 4: P, = 20 psia,h, = 24.6 Btu/lbm, 7, = 24°F State 5: P. = 49 psia, x, = 1,h, = 109 Btu/lbm, s, = 0.222 Btu/Ibm-°R

State 6: P,= 120 psia,s, = 0.222 Btu/Ibm-°R, h,= 117 Btu/Ibm State 7: P, = 120 psia,x,= 0, h, = 41.8 Btu/lbm

State 8: P,= 49 psia, h, = 41.8 Btu/Ibm Sa Units:

1 ton = 12, 000 Btu/hr

Using the above enthalpies, we can determine the quantities of interest. The cooling load is

Q,, = m(h, — h,) = 0.15(103 — 24.6) = 11.8 Btu/s

or

3.54 tons

This would compare with 9.18 Btu/s for a single-stage cycle, a 28% increase in cooling load. To find the COP, the compressor power must be known. It is the sum of the power required by each cycle. The mass flux m, is found, using Eq. 10.6, to be m, =m, 1, =m,

Ro

X ie —

Ms

= 015

Tit

246

No 48 = 0, 0.193 lbm/s xX ———

The compressor power can now be calculated. It is We

m,(h, SA

OLD

as my(he ie)

(L115—"

103) 40.993"

X01)

109) = 24 eae

The COP is then found to be

cop,

Be

=~

On am

=

ips iMoen

=

4. pp

This would compare with 3.82 using an ideal single-stage system, a 13% increase. This example illustrates that a two-stage system increases the cooling load and decreases the compressor power for the same pressure difference across the compressor. It was not intended to demonstrate the lower temperature that can be attained using a multistage system. The problems at the end of the chapter will do that.

[{ You have completed Learnin y

ol le

(3)

Chapter 10

Refrigeration Cycles

Absorption refrigeration is a refrigeration system that uses heat to power the system. It is an attractive alternative to compression refrigeration if a source of

relatively cheap heat is available, such as the heat from the condenser of a power :

qEEEEEEEEEEEEn é :

Absorption refrigeration: A refrigeration system that uses

plant or even from a solar collector. The higher the temperature of the heat source, heat to power the system. the more efficient is the absorption refrigeration unit. A pump is used in some __ Refrigerators that run on systems, so electricity may be required, but that is minimal when compared to the _ propane use such a system. power to operate a compressor. So, electricity from a small dam on a stream and bottled propane can operate an air-conditioning unit or a refrigerator, or both, at a remote cabin in the mountains. Or, the “free” heat from the power plant on a university campus can be used to cool all the buildings on campus, providing huge savings when compared to using electricity to drive compressors. Figure 10.11 shows the basic configuration of an absorption refrigeration system. The most common refrigerant used in such a system is ammonia. Theammonia in the generator exists as a mixture with water. Heat Q, is supplied to the generator by using, for example, a natural gas flame unless there Condenser Generator is a source of relatively inexpensive heat. The ammonia is heated to a superheated vapor and separates from the water allowing the generator to perform the same function as a compressor. The water is drained directly to the absorber. The ammonia then flows through _ a condenser and a throttle the same way it would in a conventional refrigerator, providing cooling Q, in the evaporator. The refrigerant then flows into an absorber, where heat is removed from the refrigerant. In the absorber the ammonia recombines with the water. The mixture is then pumped back to the generator. Because of the toxicity of ammonia, a lithium bromide—water combination is often used. Absorption refrigerators are commonly used in camping facilities to provide refrigeration when sufficient electricity to operate a compressor is not available. Alternative sources of heat for absorption refrigeration systems would include solar energy, geothermal energy, or waste heat from a power plant. Absorption refrigerators have great potential as energy sustainable devices because of the low electrical . energy requirement. Detailed problems will not be included in this

book. This subject is introduced to inform the reader of its existence.

An absorption refrigeration system.

g When a supply of compressed air is available, it can be used to power a gas refrigeration system. The most common example of this application is the opencycle air-conditioning system used in aircraft. The gas refrigeration system 1s essentially a reverse Brayton cycle, as shown in Fig. 10.12. An airplane moving at high speed can bleed air from the engines at high pressure and divert this flow into a compressor. The air leaving from the compressor is at high temperature

(4)

EYE

§=Part2

Applications

me Heat

_exchanger

Qout

ee

Compressor

Cool air

| Weomp

}

Warm air

out

Figure 10.12

:

Cool air out

i ae

The open air-refrigeration cycle.

The closed air-refrigeration cycle.

and pressure. Heat is removed from the air in a high-temperature heat exchanger, and the cooler air then flows through a turbine. The function of the turbine is to reduce the air temperature and help power the compressor. Cool air from the turbine is then directed into the airplane cabin. The closed system of Fig. 10.13 includes a low-temperature heat exchanger that shows the heat extracted from the cooled space. Its 7-s diagram is shown in Fig. 10.14 for the ideal cycle. If we consider the air-conditioning system shown in Fig. 10.13, the cooling load is the heat entering the heat exchanger between states 4 and 1. It is expressed as On z mh,

—

hy)

(10.8)

where m is the mass flow rate of the air. The isentropic compressor and turbine work rates are

Figure 10.14 The 7T-s diagram for an ideal air-refrigeration cycle.

Example

10.5

Woomp = (ht, — hy)

(10.9)

Weary = rh, — Ay)

(10.10)

Because of the low temperature at the turbine exit, the specific heats are usually not assumed constant. Either the ideal-gas tables are used, or use could be made of the IRC Calculator. For a gas refrigeration system, the performance decreases as the compressor pressure ratio increases. An example will illustrate the low temperature that can exist at the turbine exit.

Anideal-gas refrigeration cycle Air from an aircraft jet engine’ is diverted into an air-conditioning system at 12 psia and 20°F at a flow rate of 20 ft*/s. The compressor has a rated pressure ratio of 3. The inlet temperature to the turbine is 100°F. Determine the cooling

load, the net work input, and the coefficient of performance for this ideal cycle.

'At an elevation where commercial aircraft usually fly, the outside temperature is about —50°F.

Chapter 10

Refrigeration Cycles

Solution

Air is an ideal gas, and the IRC Calculator will be used to determine the properties, or the air tables in Appendix F-1 could be used, or constant specific heats could be assumed. The IRC Calculator provides: State 1: P, = 12 psia,

Te oa

h, = 115 Btu/lbm,

s, = 1.63 Btu/Ibm-°R

State 2: P, = 3P, = 36 psia, s, = s, = 1.63 Btu/Ibm-°-R h, = 160 Btu/Ibm State 3: P,= P,= 36 psia, h, = 134 Btu/lbm,

T,=100°F s, = 1.59 Btu/lbm-°R

State 4: P,= 12 psia,

S,4 = 5, = 159 Btu/lbm-"R

h, = 98.8 Btu/lbm, T,= —45.9°F

The mass flux is found, using p, = 0.0675 Ibm/ft* (from the IRC Calculator or using p, = P,/RT, = 12 X 144/53.3 Xx 480),to be

eee

m=

pA

V), =a

(OG)

Sue

ft’

eNO

Onna S

S

The flow rate of 20 ft3/s is

not constant. The mass flux is constant.

The cooling load O., can now be calculated. It is 0, = m(h, — hy)

= 1.35lbm/s x (115 — 98.8) Btu/lbm = 21.9 Btu/s The net work input is

Woo — W,turb net —= Weomp comp = m(h, —+h,) — m(h; — hy) =

1.35

X

[(160 SAID

yi (134

98.8) | = 13.2 Btu/s

The COP is found to be COP

Oe

a

15

en 2B

-O6 1.66

net

Note the low temperature at state 4 of —45.9°F. The low turbine outlet temperature is a primary feature of the gas refrigeration cycle. The relatively low COP means the gas cycle requires significantly more energy input than a comparable vapor cycle, so it is used in special situations such as on aircraft.

(5)

Part 2

Applications

This chapter presented the thermodynamic cycles that form the basis for refrigeration systems. Included were refrigerators, heat pumps, and air conditioners. A cascade system that allows for extremely low temperatures was also presented. Absorption refrigeration systems that operate using heat as the energy input and ammonia as the refrigerant were described; absorption refrigeration is used on university campuses as well as in refrigerators that run on natural gas. Finally, systems that operate with a gas, such as those used in commercial aircraft that use air, were analyzed. Terms that were used in this chapter include:

Absorption refrigeration: A refrigeration system that uses heat as a source of energy to power the system. Cascade refrigeration: Two or more separate refrigeration cycles working in series. Coefficient of performance: The measure of the performance of a refrigeration system, the ratio of the desired output to the power input. Compressor: A device used to raise the pressure of the refrigerant. Condenser: A heat exchanger that removes heat from the refrigerant until it’s a saturated liquid. Cooling load: The heat transfer from the space to be cooled. Evaporator: A heat exchanger that transfers heat from the volume being cooled to the refrigerant. Gas refrigeration: A reverse Brayton cycle.

Ideal cycle: A cycle in which the compression and expansion processes are isentropic processes.

Refrigerant: The working fluid used in a refrigeration cycle.

Throttle: A device used to suddenly reduce the pressure and hence the temperature of the refrigerant. Ton: A measure of refrigeration equal to 3.52 kW = 12,000 Btu/hr. Three important equations were utilized:

Coefficient of performance, refrigerator:

© COP, =

O.., in

Coefficient of performance, heat pump:

Intermediate cascade pressure:

COP,

Q

= ===

Chapter 10

PROBIOU Sincere

Refrigeration Cycles

ee

eee

FE Exam Practice Questions (afternoon M.E.

discipline exam) 10.1 The device that is responsible for cooling a space with an air conditioner is the: (A) Evaporator

(A) 120 kJ/kg

(B)

Regenerator

(B) 139 kJ/kg

(C)

Condenser

(C) 146 kJ/kg

(D) Absorber

10.2 The device that is responsible for heating a space with a heat pump is the: (A) Evaporator

'

10.5 If the cycle of Problem 10.3 were used to heat a space, the heating load would be nearest:

(B)

Regenerator

(C)

Condenser

(D) Absorber

- 10.3 The 7-s diagram in Fig. 10.15 indicates an ideal refrigeration cycle operating between 120 kPa and 1200 kPa. If R134a is the refrigerant, the compression work is nearest:

(A) (B) (C) (D)

42 kJ/kg 47 kJ/kg S6kI/kg 67 kJ/kg

(D) 165 kJ/kg 10.6 The compressor of the cycle of Problem 10.3 is 85% efficient, and if that is the only loss considered, the percentage drop in the COP, would be nearest:

(A) 8% (B)

11%

(C)

13%

(D) 15% 10.7 The ideal gas air-conditioning cycle, shown in Fig. 10.16, operates with air as the refrigerant. The compressor inlets air at 80 kPa and 0°C and operates with a pressure ratio of 7 If the temperature of the air entering the turbine is 100°C, the lowest temperature in the cycle is nearest:

(Ay (B) (Cy (D)

= 326 —46°C =5C -67°C

10.4 The cooling load of the refrigeration cycle of Problem 10.3 is nearest:

(A) (B) (C) (D)

118 kJ/kg 98 kJ/kg 83kI/kg 67 kJ/kg

Figure 10.16

Part 2 Applications 10.8 The cooling load for the gas air-conditioning cycle of Problem 10.7 is nearest:

(A) (B) (C) (D)

10.11

59 kJ/kg 66 kJ/kg 77kI/kg 83 kJ/kg

i) The power input to the compressor ii) The heat rejection from the ese

10.9 The COP of the gas air-conditioning cycle of Problem 10.7 is nearest:

iii) The coefficient of performance for the system

(A) 0.29

iv) The volume rate of flow entering the compressor

(B) 1.05 (C) 1.34

10.12

(D) 1.98

€ B The Basic Vapor Refrigeration Cycle 10.10

An ideal refrigeration cycle uses ammonia as the working fluid. The refrigerant enters the compressor at 120 kPa and the throttle at 1200 kPa. If it produces 12 tons of refrigeration, calculate:

i) The power input to the compressor ii) The heat input to the evaporator

An ideal refrigeration cycle uses R134a as the working fluid. The refrigerant enters the

iii) The coefficient of performance for the heat pump

compressor at state 1, shown in Fig. 10.17, at

200 kPa as a saturated vapor. It leaves the condenser as a saturated liquid at 1 MPa. Calculate:

An ideal refrigeration cycle uses R134a as the working fluid. The refrigerant leaves the evaporator at —24°C and enters the throttle at 45°C. If it produces 40 kJ/s of heating, calculate:

iv) The volume rate of flow entering the compressor 10.13

i) The work input to the compressor ii) The cooling load from the evaporator iii) The heat rejection from the condenser

An ideal refrigeration cycle, using R134a as the working fluid, operates between saturation temperatures of —20°C and 50°C. If the compressor requires 10 hp, calculate: i) The mass flow rate

iv) The coefficient of performance for the system

ii) The heat input to the evaporator

v) The Carnot coefficient of performance

ili) The coefficient of performance for a cooling system

cond (to Surroundings)

10.14

An ideal refrigeration cycle uses R134a as the working fluid. The known properties are:

State 1: 160 kPa

@]

If the mass flux is 0.8 kg/s, calculate:

P

Compressor

\

Throttle

i) The lowest temperature in the cycle < Weomp

@

ii) The horsepower input to the compressor

ili) The cooling load iv) The coefficient of performance for the

Evaporator

system

evap (from refrigerated space)

Figure 10.17

State 3: 60°C

3

Chapter 10 10.15

An ideal air-conditioning system uses R134a__—-10.19 A large refrigeration system uses ammonia as the refrigerant. The compressor has an as the refrigerant. The compressor, with inlet pressure of 20 psia and an exit pressure an isentropic efficiency of 90%, requires of 260 psia with a mass flow rate of 0.6 lbm/s. 150 kW. The ammonia enters the compressor Refer to Fig. 10.17 and determine: at 120 kPa as a saturated vapor and leaves i) The lowest cycle temperature the compressor at 800 kPa. It leaves the condenser very close to the saturated

ii)) The highest te

cyceaele tet tas

iii) The cooling load from the evaporator

10.16

determine:

i) The mass flow rate of ammonia

v) The compressor horsepower

ii) The rate of heat rejection by the con-

vi) The coefficient of performance for the system The compressor in an ideal refrigeration cycle

denser iii) The coefficient of performance for this system

40,20 The refrigeration cycle shown in Fig. 10.18

Referring to Fig. 10.4, state 1 has a quality

used R134a as the working fluid. The

of 1.0 and state 4 has a quality of 0.4. For a

following data were measured at each state:

cooling load of 8 kJ/s, calculate:

i) The necessary mass flow rate of refrigerant ii) The power input to the compressor

iii) The coefficient of performance for this system The ideal air-conditioning system in Problem 10.16 is operated as a heat pump. Determine the heating load and the coefficient of

, _ performance. 10.18

oe state. If the liquid cooling load is 600 kJ/s,

iv) The heat rejection from the condenser

using R134a has an inlet pressure of 120 kPa.

10.17

Refrigeration Cycles

An air conditioner using R134a as the refrigerant has a compressor exit temperature of 160°F and an exit pressure of 200 psia. The refrigerant enters the compressor at 20 psia and 0°F. The condenser cools the refrigerant to 120°F and 190 psia. For a cooling load of

State 1: 120 kPa, x, = 1.0 State 2: 1200 kPa, 60°C State 3: 1190 kPa, 58°C

State 4: 1170 kPa, 44°C

State St 1100 Re State 6: 130 kPa State 7: 128 kPa, x, = 0.42

State 8: 122 kPa,x,= 0.95 Assuming an adiabatic compressor, determine the compressor efficiency and the coefficient of performance of a refrigerator.

one

15 Btu/s, determine:

i) The mass flow rate ii) The isentropic efficiency of the compressor

iii) The power input to the compressor iv) The rate of heat rejection by the condenser v) The coefficient of performance for this system

Figure 10.18

ev

Part 2 Applications

@@

Cascade Refrigeration Systems

10.21

A two-stage, ideal cascade refrigeration system shown in Fig. 10.19 and Fig. 10.20 uses R134a as the working fluid in both cycles. The mass flow rate through the first stage is 8 kg/min. The known properties are:

10.22

Replace the two-stage system of Problem 10.21 with an ideal single-stage system that produces the same cooling load. Determine the minimum cycle temperature, the mass flow rate of the R134a, the compressor horsepower, and the COP. Compare with the results of Problem 10.21 and state your observations.

10.23

A two-stage, ideal cascade refrigeration system shown in Fig. 10.19 uses R134a as the working fluid in both cycles and replaces the ideal single-stage cycle of Problem 10.14. The mass flow rate through the first stage is 0.8 kg/s. The known properties are:

State 1: 120 kPa State 7: 40°C

Determine: i) The cooling load ii) The mass flow rate in system #2

iii) The net compressor power required iv) The overall coefficient of performance

State 1: 160 kPa State 7: 60°C

Qout (to surroundings)

Determine:

i) The cooling load ii) The mass flow rate in system #2

iil) The net compressor power iv) The coefficient of performance

@

v) The percentage increase in the COP and the percentage decrease in compressor power compared with that of

Heat exchanger

Problem 10.14

A Throttle V

#1

10.24

Replace the two-stage system of Problem 10.23 with an ideal single-stage cycle that uses R134a and produces the same cooling load. Determine the maximum cycle temperature, the mass flow rate of the R134a, the compressor horsepower, and the COP.

10.

An ideal two-stage cascade refrigeration system shown in Fig. 10.19 uses 0.8 kg/s of R134a as the working fluid in the first cycle and ammonia as the working fluid in the

On (from cooled space)

Figure 10.19

second cycle. Use Eq. 10.4 for the intermediate pressure. The known properties are: State 1: R134a at 160 kPa

State 7: Ammonia at 60°C Determine:

The cooling load The mass flow rate of the ammonia

The total compressor horsepower

Figure 10.20

The coefficient of performance for this system

Chapter 10 10.26

An ideal three-stage cascade system uses R134a in all three cycles with a low pressure of 100 kPa and a high pressure of 1000 kPa and with intermediate pressures of 200 kPa and 600 kPa. The mass flux in the lowpressure cycle is 0.8 kg/s. Sketch the cycle on a T-s diagram and determine:

i) The cooling load -

Refrigeration Cycles

10.28 Air from a jet engine intake is diverted to an ideal-gas air-conditioning system at 100 kPa and —20°C, as shown in Fig. 10.21. The air leaves the compressor at 600 kPa. The air enters the turbine at 25°C. Assuming constant specific heats and ideal components, calculate the low-cycle temperature and the cycle COP.

For a cabin cooling load of 3000 kJ/min,

:

li) The mass flux in each cycle iii) The total compressor power

iv) The COP of this cascade system v) Also, for a single-stage cycle between the same high and low pressures, find the compressor power and the COP for the

same cooling load.

calculate the necessary mass flow rate of air.

10.29 Air enters an ideal reverse Brayton cycle air conditioner at 14.7 psia and 35°F with a mass flux of 4 lbm/s. The compressor has a pressure ratio of 4. The temperature at the inlet of the turbine is 100°F. Assuming constant specific heats and ideal components, calculate:

i) The low-cycle temperature ii) The power requirement of the cycle

€ B Gas Refrigeration Systems

ili) The rate of heat flow from the heat exchanger

10.27

Carbon dioxide is used in the ideal-gas refrigeration system of Fig. 10.21. The gas enters the compressor at 150 kPa and — 15°C. The gas leaves the compressor at 1.2 MPa. The CO, enters the turbine at 40°C. Calculate the temperature at state 4 and the coefficient of performance for this system. Assume constant specific heats and ideal components. out

Figure 10.21

iv) The coefficient of performance for this system

TON,

pans

Srey! Uy

WR

Stat ordi

eda be

FH EP ate Ne

Ps

Se

Ne

nee

>

Rel SORES

=

n

iy ~~

=< SS

=

; ~

=
P;

(11.9)

er Part 2 Applications

Rance

Mixture ola § Va ae Ve

The Dalton law: Each component occupies the same volume and temperature as the mixture.

The Amagat law: Each component is at the same pressure and temperature as the mixture.

The partial pressure of a component is found, using the ideal-gas law (see Eq. 2.14), to be Recall: R= 8.314 kJ/kmol-K = 1545 ft-lbf/ Ibmol-°R

NR,T

(11.10)

sy;

In this equation N, is the number of moles of gas i, R,, is the universal gas constant, T is the absolute temperature of the mixture, and V is the volume of the mixture. For the complete mixture, the ideal-gas law is P=

Note:

(11.11)

With the Dalton

law, the volume, as well

ta tay

as the temperature, isthe same for all components.

(11.12)

—

The properties of mixtures of gases are typically determined on a molar basis. The internal energy, enthalpy, and entropy can be calculated based on the mole fraction of each component and the molar specific values of each property. The molar properties for several common gases are presented in the Appendix in Tables F-2 through F-5. The properties of water vapor treated as an ideal gas are presented in Appendix F-6. The equations for calculating the average properties for a gas mixture are Units: kJ/kmol and kJ/kg

u=

dy, u,

Lu,

C, = Dee.

(11.13)

h= >dyih,

h= >x;h,

Ce, On

(11.14)

=

s=

Btu/lbmol and Btu/lbm

Units: kJ/kmol and kJ/kg Btu/lbmol and Btu/lbm Units: kJ/kmol-K and kJ/kg:K Btu/lbmol-°R and Btu/lbm-°R

Di;

)'x;5,

(11.15)

where the overbar indicates a molar basis for the property. Molar specific internal energy and molar specific enthalpy have units of kJ/kmol in the SI system and Btu/Ibmol in the English system. Molar specific entropy has units of kJ/kmol-K in the SI system and Btu/lbmol-°R in the English system. Molar properties can be related to the mass-based properties by

a= Mu,

h=Mh,

5,=Ms,

C,,=M,C,,

(11.16)

Chapter 11.

Mixtures and Psychrometrics

Partial pressures

Example

A rigid tank contains 6 kg of oxygen and 4 kg of nitrogen at 800 kPa and 60°C. Calculate the partial pressure of each component and the gas constant of the

11.2 ‘Ooo

mixture; see Fig. 11.4.

an

800 kPa, 60°C

Solution

The partial pressures are related to the mole fractions by Eq. 11.12. To find

Figure 11.400

the mole fractions, the number of moles of each gas must be known. They are,

respectively, No

m,

6 kg

© My 32 kg/kmol ==

=

-

Ny

Ake

= — = =

eM.

sy

= 0.875

2108 kg/kmol

aN

No, + Nx, = 0.330 kmol

:

p y

= 0.1429

The mole fractions are then

0.1875 kmol 0S

0.1429 = = = 0.433

Yr ~ “0.330 kmol

¥2 ~ “0.330

The partial pressures, using Eq. 11.12, are

Po, = P, = y,P = 0.568 x 800 = 454 kPa Py, = P, = y,P = 0.433 X 800 = 346 kPa 2

The molar mass is

M = y,M, + y.M, = 0.568 X 32 + 0.433 X 28 = 30.3 kg/kmol The mixture’s gas constant is found to be

Hee ee ‘= —4 = > = 0.274 kJ /kg-K fee os L It is in between the gas constants of oxygen and nitrogen.

Properties of SPL

amixture

Example

11.3

STILE

A gas mixture containing 5 Ibm of nitrogen and 8 Ibm of carbon dioxide exists

at 40 psia and 80°F; see Fig. 11.5. Determine: i) The mass fraction of each component ii) The mole fraction of each component iii) The amount of heat required to raise the temperature of the mixture to 150°F at constant pressure iv) The change in the entropy of the mixture for the process of (iii) (Continued)

5 bm No

8 Ibm CO, 40 psia, 80°F

en Part 2 Applications

Example 11.3

(Continued) NELLIS

Solution m, = S\bm,

m, = 8lbm

“mm = 13 Ibm

Does ee Ee Ven

== = 0615 Oo

m, = 5 Ibm iOe ) N,=— Ms oh omdbmol = 0.179 Ibmol on

a= —2 = © = 0.182 ae Ibmol N=N, + N, = 0.179 + 0.182 = 0.361 Ibmol 0.179 = 0.50 0.182 = 0.50 = = = —““ Yer Gael ee = ge

Note: Mass fractions are

iii) The first law (for P = const):

used since Ci; |S Known on amass an if ie,were

|

known, the mole acca

=

=e

would be used.

Q = AH = mC,AT. Using

Dai

C2

P2

= 0.385 X 0.248 + 0.615 x 0.205 = 0.222 Btu/lbm-°F

2.Q = mC,AT = 13 X 0,222(150 — 80) = 202 Btu iv) From Eq. 6.19, the entropy change in an ideal gas for this constant-pressure process is, using C,, from part (iii),

Remember:

AS

The

yn

(c, ep ae fe

temperatures must be absolute.

13%

Example

11.4

Rl 7 2) C4 P, ee

2

oe

10 .0:222 x « M9 = = 0.352 Btu/°R

A mixture flowing through a turbine A gas mixture, with the mole fractions listed below, enters the turbine of Fig. 11.6 at 1200 kPa and 600 K with an entering volume flow rate of 1.2 m*/s. The mixture leaves the turbine at 300 K. Determine the power produced by the turbine.

Gas

Mole Fraction y, 7

Carbon dioxide

0.20

Oxygen

0.02

Nitrogen

0.78

Chapter 11

Mixtures and Psychrometrics

1200 kPa 600 K

300 K

Solution First, calculate the average molar mass of the mixture: M=

Sy,

= 0.2 x 44 + 0.02 X 32 + 0.78 X 28 = 31.3 kg/kmol

Next, the mass flow rate of the mixture is found, using Eq. 4.10, to be

Units: We know 2, which

the ideal-gas law gives as

n=

$7

hes8

sae

v

v = RTIP, has units of m/kg.

P

ae Be

RL SNES 9

(R,/M)T,

m

3

ent Loe

12002523133

\ke

Se 00d k (= x Pale 2/s

Now we can calculate the power output using the molar enthalpies found in Tables F-2, F-3, and F-4. Using Eq. 11.14, we have

Units: Using the ideal-gas tables in the Appendix

provides A in kJ/kmol, so mole fractions are necessary.

W = ZS y[F(600) - 7,300) kg

ea

~ —£|

02) 280 — 9431) + (0.02)(17 929 — 8736)

Ses inal

+ (0.78)(17 563 — $723) | (0.78)(

kmol

= ( 0.289 Amel (o64oi) = 2790 kW Ss

kmol

Pah aaa c-

[{ You have completed Learning

(2)

ea Part 2 Applications

11.2 Air—Vapor Mixtures

Psychrometry ____and Dry air: Air that contains no water vapor.

Atmospheric air: Air that contains dry air and water

vapor.

Note: Water vapor in air is assumed to be an

ideal gas at relatively low temperature and near atmospheric pressure.

©

Observation: Atmospheric air can be treated as a mixture of two ideal gases: dry air and water vapor.

PIP oP

C-1 and C-2. It's the

enthalpy change that is of interest, so where the O-datum is selected is not

significant.

(11.17)

where P, is the partial pressure of the dry air and P,, is the partial pressure of the vapor, both pressures being absolute. In the above example the partial pressure of the vapor is the saturation pressure of water at 20°C, which is P,, = 2.338 kPa from Table C-1. So, if P,,,, = 100 kPa, the partial pressure of the dry air in the atmosphere would be P, = 100 — 2.338 = 97.66 kPa. The enthalpy and internal energy of the water vapor depends on temperature only (being an ideal gas), and thus it is assumed that

Note: /h, is taken to be zero at O°C by the heating and air-conditioning community. That's quite different from the A value listed in Tables

This section deals with the mixture of two important gases: dry air and water vapor. Air with no water vapor is referred to as dry air. Atmospheric air contains water vapor that must be accounted for when analyzing problems involving the conditioning of air in houses and commercial buildings, and in the analysis of combustion. The air in problems considered in earlier chapters was considered to be dry air, but the small amount of water vapor in atmospheric air does not significantly influence heat transfer, enthalpy change, internal energy change, or the other properties of interest for a process. However, when considering comfort in a building or the combustion process, the amount of water vapor must be included. Combustion will be considered in Chapter 12. Water vapor exists in the atmosphere as a superheated gas. In fact, it can be modeled as an ideal gas, providing the pressure is near atmospheric pressure and the amount of superheat is not extensive. For example, Table C-1 indicates that saturated water vapor has a specific volume of 578 m°/kg at 20°C, whereas the ideal-gas law provides v = RT/P = 0.462 X 293/2.338 = 57.9 m/kg, using the gas constant for water vapor (steam) from Table B-2 and the partial pressure of the water vapor presented next to the temperature of 20°C in Table C-1. Consequently, atmospheric air can be treated as a mixture of two ideal gases: dry air and water vapor. The Dalton law is again used so that

u,(T) = u,(T)

A(T) = h,{T) iS

(11.18)

The ideal-gas assumption for the vapor in atmospheric air is acceptable from about 0°C to 50°C. | A key element in the field of heating, ventilating, and air conditioning (HVAC) is how to maintain comfortable levels of humidity in our homes and workplaces. Too much water in the atmosphere makes it uncomfortable, that is,

Remember: Water vapor in atmospheric air can be treated as an ideal gas up to about 50°C (125°F).

Psychrometry: The study of air-water vapor mixtures.

makes it difficult to lose heat by perspiration. Too little moisture irritates the skin and tissue in our respiratory systems. The remainder of this chapter is a detailed analysis of HVAC. %

11.2.1 Terminology and definitions The study of air-water vapor mixtures is called psychrometry. There are two primary quantities that specify the amount of water vapor in the air. The specific

Chapter 11

Mixtures and Psychrometrics

humidity w, or humidity ratio, is the mass of water vapor per unit mass of dy air, defined b y

vapor

Bes

(11.19)

m

This is not the definition of humidity with which we are familiar. The humidity that the weather channel presents is the “relative humidity,” defined below. The humidity ratio is more useful to engineers because it clearly defines the ratio of water mass to air mass. The humidity ratio will be required for engineering calculations. Relative humidity ¢ is the ratio of the mass m, of the water vapor actually contained in the air to the mass m, of water vapor required for the air to be completely saturated at the same temperature. It is defined to be

ss

(11.20)

mM,

For dry air that contains no water vapor, the relative humidity is zero, as is the humidity ratio. For humid air in which the water vapor is at the saturation condition, the relative humidity is 1 (or, often stated as 100%). In the following analysis, the humidity ratio is related to the relative humidity. In doing this, the water vapor, both m, and m,, will be treated as an ideal gas. Consequently, the relative humidity is related to the ratio of partial pressures by

fe Oe i,

Tae AES

jis,

Se

a

=

P Jesh dik = 622 PVR T= RP P

; mass of air./ It has units of kg vapor/kg dry air.

The term “humidity” refers to the relative humidity.

Note: The quantity m, is the maximum amount of water vapor air can hold. If the amount of water vapor exceeds this value, it

condenses out, like dew in

the morning (in Michigan) or on the outside of a glass of ice water.

(11.21)

Ne

The volume and temperature are assumed the same for both conditions: when the air contains some water vapor and when it is saturated. The humidity ratio takes the form

si

Humidity ratio: The mass of water vapor in a unit

m

(11.22) a

The gas constant for water vapor (steam) is given in

Appendix B.2. We have not used this gas constant very often up to now since the properties of water vapor have been found in Appendix C.

where R,/R, = 0.622. A combination of Eqs. 11.21 and 11.22 relates the two functions:

cv)

oP (0.622 + w)P,

Poy

0.6226P, P — $P,

(11.23)

The humidity ratio is used more often than relative humidity in calculations.

The pressure P is related to P,, and P, by Eq. 11.17

The condensation of water vapor Air at 100 kPa, 25°C, and a relative humidity of 75% is compressed isothermally until water droplets begin to appear, as in Fig. 11.7 Determine the pressure at this final state assuming no change in the amount of water in the air. (Continued)

Exampie

11.5

ce Part 2 Applications

Example 11.5

(Continued) 55DDIED DODDRIED

L ANE

DLL

Solution The states of the water are:

State 1: P, = 100 kPa, 7, = 25°C, @ = 0.75 State 2: T, = T, = 25°C, x, = 1, P= Ey = ouljekral

The initial pressure of 100 kPa is a total pressure equal to the sum of the partial

pressure P, of the air and the partial pressure P, of the water vapor. From To = 25°C Xo 5} 4-40)

Table C-1, the saturation pressure of water at 25°C is 3.17 kPa. Equation 11.21 provides the partial pressure of the water vapor at state 1:

Saturated air

Roa oor,

= 00

x oi rae

2.38 kPa

Figure 11.7 The humidity ratio follows from Eq. 11.23 to be

_ 0.6226P, Rather than write kg vapor/kg dry air, we will

write kg,,/kg,.

w)

eh

OP

0.622 x 0.75 X 3.17

100075

= 0.0152 kg,/kg,

17

where we will use kg, to be the mass of the water vapor and kg, to be the mass of dry air.

At state 2,@, = w, = 0.0152 kg,/kg, since the amount of vapor in the air remains unchanged. Using Eq. 11.22, with P, = 3.17 kPa since ¢ = 1,

nee 0.622 x P, A

@

Py

0.622 x 3.17 kPa =

0.0152

Po +P = SAF 4 1297

= 129.7kPa

1320 kPa

In an industrial drying process, wet fabric enters an air dryer with a mass flow rate of 0.5 Ibm/s and a moisture content of 80% by mass. It leaves the dryer with a moisture content of 10%. Air enters the dryer at 180°F with a relative humidity of 20% and leaves the dryer at 140°F with a relative humidity of 70%. See the schematic of Fig. 11.8. Estimate the mass flow rate of air needed to accomplish this process. Assume the vapor to be an ideal gas. Air, moisture out

Clothes Air in

Figure 11.8

Clothes

out

Chapter 11 =Mixtures and Psychrometrics a

Solution

;

Conservation of mass:

For the aie entering the dryer:

m,, = 0.5 lbm/s

wet clothes

Myi, = 0.8 X 0.5 = 0.4 lbm/s

water in fabric

“My; = 0.1 Ibm/s

Fabric:

Min = Meo

Watery

1, = Ma,cat

Ait

Mai,in = Mairour

fabric

Entering air: T, ,, = 180°F and ¢,, = 0.2. ..P, = 7515 psia. The humidity ratio is given by Eq. 11.23 to be a

ee

0.6226P, ie

OR

0.622 x 0.2 x 7.515

=

te |

:

O2E xX Po}

= 0.0708

Ibm

eee roe of water vapor Ibm, =

-

Ibm,

~ pounds of air

Fabric and air leaving the dryer:

Meout = Mein = 0.1 Ibm/s

Tricot = 140°F

and

$ = 0.7

..P, = 2.892 psia

The humidity ratio leaving is then

Ode

0.6226P, 0.622 x 0.7 X 2.892 Ibm 1 engine is runnina rich

ey

Part 2 Applications cool down as they flow out a chimney or outa

tailpipe. If they cool down too much,

and and the temperature falls below the dew point, the water dvapor hwill condense deniatenson

It =is important that the may result in damage from the acids that are contained in the conde temperature remains above the dew point when __ it’s important that the temperature remains above the dew point all the way to

exhausting the products

the exit. In high-efficiency furnaces, the exhaust temperature from the furnace is

hice cheatin damage pipes or chimneys. BERRA ROR

up a chimney butoerare condensed and sent down a drain. An example will illustrate how the dew point is calculated.

_ of combustion so thet —_indeed below the dew point. Therefore, the products of combustion are not sent

Example 12.1

Stoichiometric combustion ofnatural gas ANNES

Natural gas is 70% methane, 15% ethane, and 15% propane, by volume. Write a chemical equation for the stoichiometric combustion of natural gas in air, as demonstrated by Fig. 12.3. Fuel: Methane, ethane, propane

Air: O + 3.76N>

Combustion chamber

Products

Solution

The first objective is to determine the stoichiometric combustion equations for each constituent of natural gas. The equations are, following the procedure for propane outlined above with Eqs. 12.3: Methane: Ethane: Propane:

CH, + 2(0, +: 3.76N,) > COs + 2HOTt a7 CcH, + 3.5(0O, + 3.76N,) => 2CO, S8 3H, Or AS:“16N, C‘H, + S(O, + 3.76N,) = SCO; + 4H, Ot: 8N,

Next, multiply the equations by their percent content in the natural gas and add them together, that is,

gas = 0.15[propane] + 0.15[ethane] + 0.70[methane] The combustion equation results for 1‘mole of propane:

0.15C,H, + 0.15C,H, + 0.7CH, + 2.675(0, + 3.76N,) => 1.45CO, + 2.45H,O + 10.06N,

Chapter12

Combustion

_ Combustion with excess air Example

[EY

12.2

ETE IERIE

Propane is burned with 20% excess air. Determine i) the air-fuel ratio, ii) the volume percentage of water vapor in the products, and iil) the dew-point temperature of the products. Assume complete combustion; that is, there is no CO in the products. Solution

The stoichiometric reaction was presented in Eq. 12.4 to be C5H, + 5(O, + 3.76N,) > 3CO, + 4H,O + 18.8N, With 20% excess air, the “5” is multiplied by 1.2 to give 20% more air. The combustion equation is then

C,H, + 12 x 5(O, + 3.76N,) = ACO, + BH,O + CN, + DO, Balance the atoms:

i

Sree)

H: N: O:

8=2B (ek 3 Sik Ae OIE 62 =2A +B

AS ~B=4 oC = 22.56 “D=

22D

The combustion, expressed schematically in Fig. 12.4, is then represented by

Ch

oO,"

3.76N,) = 3CO, + 46

i) The air-fuel ratio is mass air _

pe

~ massfuel

Oe

4

Or+ 22.56N, + O,

ome ereageead Owe 29

xy

sient at

Be, of+ atl ¢ wl A, tle wats

N, =

1 X (36 + 8)

7a x4, 3

(6 X 4.76)kmol, x 29kg./kmol,, = = = = 18.82 kg /kg... 1kmol,,., X (36 + 8)kg,../kmol,,., ii) The volume percentage is a ratio of moles. It is

9 ae

=

moles water moles of products

4 0 = =~ X 100 = 13.1% 30.56 aa

Air: 6(O> + 3.76N>) Fuel: C3Hg

Combustion chamber

3CO, + 4H2O0 + Op + 22.56N5

(Continued)

Dat

i.

430,

Part 2 Applications

Example 12.2

(Continued) SAAD

iii) The partial pressure P,, of the water vapor is found using its mole fraction (see Eq. 11.12) of the products. It is The atmospheric pressure is assumed to be 100 kPa unless otherwise stated.

eneeeeeneeeseeseeeeseesececeseeeees

P=y vapor @ ;

4

Fam =~—"— 30.56 x 100 | = 13.1kP “ea

:

:

;

The dew-point temperature is interpolated in Table C-2, or is found, using the IRC Calculator (use quality = 1.0), to be

Loe SL2°6 If the temperature of the products, such as the temperature of the exhaust leaving the exhaust pipe of a vehicle, is less than T,,, the water vapor would condense, along with some nasty acids that are always present in combustion products (in negligibly small but corrosive amounts).

Ethane is burned mass percentage Assume that CO lack of sufficient

with 85% theoretical air, as shown in Fig. 12.5. Calculate the of carbon monoxide in the products and the air-fuel ratio. is the only additional compound in the products due to the oxygen. 85% theoretical air

Combustion chamber

Products

ACO, + BHO + CN, + DCO

Solution

The stoichiometric reaction was presented in Example 12.1 so, using 85% theoretical air, the combustion equation is ¥

C,H, + 0.85 x 3.5(O, + 3.76N,) = ACO, + BH,O + CN, + DCO

Chapter 12

Combustion

_

Balance the atoms:

H: N: C

O85:0C395.

O:

= 238 3.716% 2 = 2C. 2=A+D

Wo Se Pye,Si 3uRO

5.95 =2A |

ee

Sea

The combustion equation is then

CH, + 0.85) x°3:5(O, +. 3.76N,) = 0.95CO, + 3H,O + 11.2N, + 1.05CO The mass percentage of CO is calculated to be % CO b

ee

Or

=

1.05428

0.05 X44

3 18 4 110 KOR

0s KR

= O08

The air—fuel ratio is air mass

Ou

By

~ fuelmass _

SK AIO

OS WS.

ay 13.7 BA KBidKB KBi

x30

Combustion with moist air SES

SURI NITRIDE

Determine the chemical equation for propane, burned in the combustor of Fig. 12.6, with 40% excess of air that has a specific humidity of 0.01. Assume a constant pressure of 100 kPa and also determine the dew point. 40% excess air

@ = 0.01 C3Hg

Qout

Combustion chamber

ACO, + BH,O + CO, + DN>

Solution We use the stoichiometric combustion equation with dry air presented by Eq. 12.4:

C,H, + 5(O, + 3.76N,)

=

3CO, + 4H,0 + 18.8N, (Continued)

Example

12.4

432

Part 2 Applications

Example 12.4

(Continued) Using 40% excess air, the 5 is multiplied by 1.4, and the combustion equation

with dry air becomes

C,H, + 1.4 x 5(O, + 3.76N,) = 3CO, + 4H,O + 20, + 26.32N,

When moisture is present in the air, we ignore it and balance the reaction equation with dry air. The moisture is then added

since it simply tags along, like nitrogen. It doesn't affect the other terms in the equation (except to add to the water), but it does alter

For every mole of propane burned, 2 moles of oxygen are not used and we are now using 26.32 moles of nitrogen. The specific humidity was defined in Chapter 11 as the ratio of the mass of water in the air to the mass of air. For the purposes of this analysis, we need to change this to a molar ratio. The molar mass of water is 18 kg/kmol, and the molar mass of air is 29 kg/kmol. The molar specific humidity @ is determined to be 1 f wat a o(M. 7) : ooi(2) — 9.0161 Moles of water

18

“water

mole of air

where the bar indicates a molar basis. The number of moles of water that enters the reaction with the air is now calculated to be

the heat transfer from the combustion chamber. (Heat

transfer will be considered in Section 12.3.)

eae

=

ON...

= 0.0161

x 1.4 x 5 X (4.76) = 0.536 mol of water

The complete chemical reaction for combustion with moist air is shown in Fig. 12.7 and is represented by

C,H, + 7(O, + 3.76N,) + 0.536H,O = 3CO, + 4.536H,O + 20, + 26.32N, Note that the water vapor in the air, like the nitrogen, simply tags along. Its dew point can be found by determining its partial pressure (see Eq. 11.12), which is P,= v

4.54

YH,O

P...atm = —— 35.86

x

=

100

12.

IORI AS Ea

Using Table C-2 or the IRC Calculator, we find the dew point to be

dee = 51°C

If the temperature of the products falls below 53°C, condensation will occur.

Air: 7(O + 3.76N,) + 0.536H,O

Energy Combustion chamber

44

3CO, + 4.54H,O + 20, + 26.32N> Figure 12.7

Chapter 12

Determine the AF from the products

Combustion

Example 12.5

Benzene is burned with dry air, and a volumetric analysis is performed* on the dry products (no water vapor is measured) of combustion. It is found that

the dry products contain 80.69% N,, 8.58% CO,, and 10.73% O,. The moles of

Air Benzene

each gas are shown in Fig. 12.8. Determine the composition of the air-fuel ratio and the percent excess air.

Solution Let us assume that an unknown supply of fuel provides 100 moles of dry products so that the reaction equation is

AC,H, + B(O, + 3.76N,) = 8.58CO, + CH,O + 10.730, + 80.69N,

8.58CO., + 10.7305 + 80.69N, Plus water

in which the moles of fuel, air, and water are not known. The products must contain water so that term is included, but the number of moles is unknown

Figure 12.8

- since it was a dry analysis. The atoms are balanced:

C: N: H: O:

6A 3.76B 6A 2B

= = = =

8.58 80.69 2C 1716+ C+ 2146

A = B= C= 42.92 =

143 21.46 4.29 42.91

lm

Note: A “dry analysis” of | the products means that =hal water is not included in the

analysis, but it is always

present.

Divide all terms in the reaction equation by A so the equation is in standard form:

C,H, + 15(O, + 3.76N,)

=

6CO, + 3H,O + 7.50, + 56.43N,

Observe: The oxygen balance is a check on the

The air—fuel ratio is the ratio of the mass of air to the mass of fuel using the molar mass of each. There results the following: AF =

air mass

=

1

OK AKG

BS

* fuel mass © 1 < (6 x 12:46 x 1)

=

SSS)

values obtained for A, B, and C.

kg aif K Suet

For theoretical air, the reaction equation can be shown to be

Molar mass: Me = 6X 12+6X 1

= 78 kg/kmol

C,H, + 7.5(O, + 3.76N,) = 6CO, + 3H,O + 28.2N, Thus, 15/75 = 2, yielding 200% theoretical air. The percent excess air, referring tog. 12-5, 18 % Excess air = 200 — 100 = 100%

(1) content. 3An Orsat gas analyzer, used to analyze the products, does not measure the water vapor

434

Part 2 Applications

12.3 The Enthalpy of Formation and the Enthalpy of In Section 12.2 we consideredonly the chemical reaction and did not attempt to predict the amount of heat emitted by a particular reaction. In this section, the heat emitted by a particular reaction, which is the primary

The combustion chamber.

objective in our study of combustion, will be determined. The amount of heat Q,,, given off in the combustion reaction shown in Fig. 12.9 is determined by applying the first law of thermodynamics to the process. ee This is represented by

Note: A molar basis is used since the tables present enthalpy on a molar basis.

O-

P=. Comments, The symbol "” after a compound signifies the

Soh

= 2Ne

= Jabs

(12.8)

where H 2 is the enthalpy of the reactants and He is the enthalpy of the products, on a molar basis. The changes in kinetic and potential energy across the combustion chamber (the control volume) are, as usual, assumed to be insignificant, except in certain applications such as a rocket nozzle, which will be illustrated in Example 12.8. Also, it is assumed that there is no work performed. So, to find

liquid phase, and the symbol _—_the heat emitted from the combustion chamber, H, and H, must be determined.

“g” signifies thegaseous

phase, which is assumed if

the 0" is not plesent. Fess e eee ee ee eeeeeeeeeeeeeeeeeere es

—-T) determine the enthalpies,5 a reference state is established. ; It is agreed that the 5

reference state shall be 25°C (77°F) at standard atmospheric pressure and a superscript “°” on a property will signify such a state. For instance, h°, would be the enthalpy on a molar basis of a substance at the reference state, usually at the inlet.

285 830 kJ

To illustrate the application of Eq. 12.8, consider the constant-pressure combustion of hydrogen and oxygen, represented by

1 mol Hp 25° 1 atm

1 mol

Combustion

| 420(4

chamber

2H, + O,=>2H,0(¢)

(12.9)

25°

= mol O>

1 atm

with the inlet and exit conditions both being at the reference state of 25°C and a pressure of 1 atm, hence liquid water. In order for the exit temperature to be at 25°C, the heat transfer for this combustion process must be* —285 830 kJ for each kmol of H,O(¢) formed, as illustrated in Fig. 12.10. : ; ae The reactants H, and O, are stable since they exist at standard conditions ‘ : nN : in the atmosphere in that form. So, at 25°C and 1 atm, it takes no energy

(Figure 1210.0 The combustion of hydrogen and oxygen.

a Enthalpy of

to form those two elements. Thus, the heat given off by the reaction of H, and O, is called the enthalpy of formation i as listed in Table B-7 for H 0(0).

formation: Theenergy emitted or required to form

The enthalpy of formation for H, is zero since it is a stable element. The enthalpy of H is listed as a positive 218 000 kJ/kmol because it takes energy to break apart

a compound from i stable elements.

—_H,. Observe in Table B-7 that some compounds have a negative h°, which indicates that they emit heat (an exothermic reaction), while others have a positive h°,

Exothermic reaction: A —_which indicates that they require heat to form (an endothermic reaction).

eee

Patan

eis ae

ae me ‘

Endothermic reaction:

reaction that requires heat.

The enthalpy of a particular compound or element is the sum of two enthal-

pies. The enthalpy of formation h° is the energy needed to take the constituent atoms and form them into a compound at the standard conditions of 25°C (77°F) A = and 1 atm. Elements such as nitrogen, hydrogen, oxygen, and carbon that exist 4

.

.

.

.

.

.

.

.

The minus sign is due to our sign convention: Heat transfer leaving a control volume is negative.

Chapter 12 naturally have an enthalpy of formation of zero. The other component of the enthalpy is the amount of energy needed to bring the material to a temperature different from the standard state. For example, the enthalpy of nitrogen at 127°C is hy, = hy Ee Ah, = 0 + 11 640— 8669= 2971 kJ/kmol, using values at 298 K and 400 K in Table E2. Tables F-2 through F-6 show the enthalpies of formation and enthalpy increments for common gases in SI units, and Tables F-2E through F-6E show enthalpies using English units, all on a molar basis. Table B-7 lists the enthalpies of formation for fuels and common compounds. It must be noted that the enthalpies of formation in Table B-7 are referenced to a temperature of 25°C rather than 0 K, as is done in Tables F-2 through F-6. If the enthalpy of formation of propane is desired, Table B-7 gives a value of he = —103850 kJ/kmol. This is the energy needed to combine three atoms of serien and eight atoms of hydrogen to form one mole of propane at 25°C at a pressure of one atmosphere. Enthalpy changes to other temperatures are not given for propane or other fuels in the Appendix. If the molar enthalpy of carbon dioxide at 700 K (427°C) is desired, the enthalpy of formation is added to the enthalpy increment from 298 K to 700 K from

Combustion

A compound consists of two or more elements. H, is an element since it consists of only hydrogen, as are O,

and N,.

A good approximation: Ah, = CAT = (1.042 x 28) (400 -— 298)= 2976 kJ/kmol, less than a 0.2% error. For high temperatures, the error would be unacceptable. The tables will be used exclusively.

Table F-4. This is expressed, for carbon dioxide, as

Heiecetieeih\ als 700 K == 393920

4927125

(for CO,) — 6601 = —372 996 kJ/kmol

(12.10)

All the digits may be used in the terms of an equation, but the answer should be rounded off to three or four significant digits, never five

or more. We're engineers!

The above number could be rounded off to —373 MJ/kmol when used in engineering calculations. If units of kJ/kg are desired, simply divide by the molar mass,

which for the CO, above would be —372 996 kJ/kmol/44 kg/kmol = —8477 kJ/kg. The enthalpy of formation is the energy required to form a compound. The energy emitted when a compound undergoes stoichiometric combustion at constant temperature and atmospheric pressure is the enthalpy of combustion. In Fig. 12.10 the heat transfer is —285 830 kJ/kmol for the combustion of hydrogen with oxygen at 25°C and atmospheric pressure. It is listed in Table B-8 and is referred to as the higher heating value (HHV) since water in the products at 25°C is in liquid form. If the water is in gaseous form, the /ower heating value (LHV) will result.Tomake sure that the water is a vapor, the LHV uses an exit tempera-

ture of perhaps 150°C. Because the exit temperature is not uniformly accepted to be 150°C, we most often use the HHV. The enthalpy of vaporization, which for water at 25°C is hy = 44 010 kJ/kmol (2445 kJ/kg), is also listed in Table B-7 The enthalpies of combustion (the HHVs) for various compounds and fuels are listed in Table B-8 in the Appendix. The first law, stated in Eq. 12.8, can be stated for any number of components

making up the reactants and products. In an expanded form, neglecting kinetic and potential energy changes and shaft work, the first law, for a steady-flow con-

Enthalpy of combustion: The energy emitted when a compound undergoes stoichiometric combustion

at constant temperature and atmospheric pressure.

Higher heating value (HHV): When the water in the products is in liquid form. Lower heating value (LHV): When the water in the products is in gaseous

form.

trol volume, takes the form

Oa, DN prod

R ie h—= ke), =

> Ne +

h —

h’).

(12.11)

react

where N, represents the number of moles of substance 1, with i summed over all substances in the products and the reactants. If the shaft work or kinetic energy

Note: The HHV and the LHV are listed as positive heat transfers, so they are

“negative the enthalpy of combustion.

036 | Part 2 Applications

Bomb calorimeter

Wy

a

Figure 12.11 A bomb calorimeter.

Note: This is the first law applied to a steady flow through a control volume with no shaft work.

change is not zero, as in a combustion jet turbine, it is simply included appropriately in the equation. If the control volume has no inlet or outlet, as in the bomb calorimeter of Fig. 12.11, it is a system and the first law becomes

Q=U, - U, =

N

Gah Sh = Po)

prod

Recall: u = h — Po

Note: This is the first law applied to a rigid volume, such as a bomb calorimeter.

NGG,

i

ee

react

where enthalpy # = u + Pv has been used since tables list / and not u. Because the volume of any solid or liquid is very small when compared to the volume of a gas, We retain the Pu term for a gas only. Assuming an ideal gas, Eq. 12.12 fora rigid volume takes the form

Or DN prod

Gish his

= RD)

NG

holst eRe

react

In the above three equations, an ideal gas will be assumed to find h — h° using the tables in Appendix F. For vapors, tables such as the steam tables will be used. For solids and liquids, we will use h — h° = CAT.

Example

12.6

Enthalpy of formation of CO, One mole of carbon and one mole of oxygen O, enter a combustion chamber

at 25°C and 1 atm, combustion occurs, and the CO, that is formed exits at 25°C and | atm. What is the enthalpy of formation of CO,? Refer to the steady-flow combustion chamber sketched in Fig. 12.12.

Combustion chamber r) Note:

Two moles enter, one mole leaves.

Chapter 12

Combustion MESHES

Solution

Repeated comment: The

The combustion occurs according to the reaction

numbers listed in tables are often given to five or more significant digits. Material properties are typically temperature dependent, and such properties are not known to four significant

C + 0,=CO, The steady-flow combustion chamber is flow through a control volume, so Eq. 12.11, using entries from Table F-4, is applied as follows:

CO] lI NG

Ph)

= NG

prod

digits in most, if not all,

i

problems. Data is typically collected, a best-curve fit

react

is attached to the data,

[1 s (he a Ise a hog, |prod

ie

ai)

= 1 X\( —393520-+

9304

ei x ee) —

9364)

—

0 —

0 =

and then numbers are generated to as many digits as the researcher cares to provide, as is done to generate the steam tables. So, to give answers to five or more significant digits in an engineering problem involving material properties and dimensions is simply misleading and wrong! Usually three digits, or possibly even only two, may have significance.

| =393520 kJ /kmol

where the two zeros are for the reactants C and O, since h® = 0 and (h — h°) = 0 _ at 25°C for each reactant. The heat transfer, —393 500 kJ/kmol, is the enthalpy

of formation of CO,, since that is the heat emitted when the compound is formed. The heat transfer is negative, so it is leaving the control volume. We could say that 393 500 kJ/kmol left the control volume, in which case Q out would be written as a positive quantity.

Enthalpy of combustion of liquid propane

Example

12.7

ELLE DELO

Using the enthalpies of formation from Table B-7 determine the enthalpy of combustion of liquid propane. The reactants and products are all at 25°C and 1 atm of pressure, as shown in Fig. 12.13. Qout

Combustion

Products

chamber

Note: When burned with air, it is the same as

Solution First, determine the reaction equation. It is found to be C,H,(2) + 5(O,.+ 3.76N,) = 3CO, + 4H,O(¢) + 18.8N, The water must be in liquid form since it is at 25°C and 1 atm. The combustion chamber is assumed to be a steady flow through a control volume, so Eq. 12.11 (Continued)

burning with oxygen since | the nitrogen simply tags

along. If the temperatures ~

are different, the nitrogen

does influence the heat transfer. So, the nitrogen

is simply ignored in this equation.

438

Part 2. Applications

Example 12.7 pS 2S coniewentl h® for air is zero since it is composed of oxygen and nitrogen.

(Continued) is applied as follows:

O= SNe

+ kW), -

S Nile + bP),

prod

react

[N° ]co, + The LHV would result if for

SX

H,0, —241 820 (vapor) was

[Nhe] +0

(3393520)

5

+4

VRE

es =. LNs

(285830)

—

12x

2

[tet

(—103 850)

— 2 220 000 kJ/kmol of C,H,(g)

used rather than —258 830.

The enthalpy of combustion /, of C,H,(¢) is then found to be Note: The h. for propane

h. = —Qou = — 2.220 000 kJ/kmol

gas would not include

h,, = 15060 ki/kmol, so the LHV would be —2205 000 kJ/kmol.

Example 12.8

This is the HHV since the water is in liquid form. It could also be written as —2220/44 = —50.45 MJ/kg, common units for the enthalpy of combustion.

Combustion in a jet engine Air with negligible kinetic energy (compared to the exiting kinetic energy) and liquid octane, both at 25°C and | atm, enter an insulated engine. The exhaust products exit at 900 K and 1 atm of pressure, as shown in Fig. 12.14. Estimate the velocity of the exhaust products assuming no shaft work. Assume stoichiometric combustion. Combustion chamber

See

Air

25°C

ee

ee

=

Vv

“——— =

Bite io eae

aes

Products

Fuel

900 K

Solution

Stoichiometric combustion occurs according to the reaction

CsH,9(€) + 12.5(0, + 3.76N,) =>8CO, + 9H,O + 47N, The water is obviously in vapor form since the products are at 900 K. The insulated (Q = 0) combustion chamber is assumed to involve steady flow through a control volume, so Eq. 12.8 takes the form

V2 0 = A, + id

Hp

Chapter 12

Combustion

where m VoD is the kinetic energy of the products per mole of fuel. The mass of the products per mole of fuel is found to be

m,=8

X 44+ 9 X 18 + 47 X 28 = 1830 kg/kmol fuel

Let’s find H,. It is

Pee

tH

= 8(— 393520 + 37405 — 9364) + 9(-241810 + 31828 — 9904) + 47(0 + 26890 — 8669) = —4046000 kJ/kmol fuel Next, H, is found to be a

Ht

Enthalpy of formation

AcH.(0 eon

Hy,

for liquid octane:

= 1(-249910) + 0 + 0 = —249910kJ/kmol fuel

age = —249

once

910 kJ/kmol

The first law, represented by Eq. 12.8, now allows the velocity to be found.

OTGis D Vv’ = gts

=

a

Hp)

D 1830 enor

kJ J 249910 + 4046000) —— x 1000 — ) kmol kJ

= 4149 x 10° m2/s?

eRe Units:

“.V= 2040 m/s

Combustion with excess air

kg.

=

Example

Nt

Bo are =

12.9

are

Propane gas, flowing at 0.04 kg/min, is burned with 40% excess air in the steadyflow combustion chamber of Fig. 12.15, both of which enter at standard conditions of 25°C and 1 atm of pressure. The air has a relative humidity of 70%. Determine the rate of heat transfer from the chamber if the products exit at 800 K and 1 atm.

Solution

ee

Stoichiometric combustion occurs according to the reaction C,H, + 5(O, + 3.76N,) = 3CO, + 4H,O(g) + 18.8N,

Recall: Dey a oaeeied unless otherwise stated. Moisture in the air would definitely influence the result in this example.

One

CgHe(?) 0.04 kg/min

25°C

Combustion

Products

chamber

40% excess air

(Continued)

s*

440

Part 2 Applications

Example 12.9

(Continued) leading to the equation

C,H, + 7(O; + 3.76N,) = 3CO, + 4H,0 + 20, + 26.32N, To find the mass flux of the air into the chamber, first AF is found to be

AF =

mass air

7 X 4.76 X 29k¢...

mass fuel —

1 x 44kg.4

= 21.96 kg air/kg fuel

Then, the mass flux of the air can be found knowing the mass flux of the fuel. It is Note: The mass of

Mair = Ar

water vapor in the air has been ignored since it is

X Meet

k i = 21.96 Bair x 0.04 sotuel = 0.878 kg _/min

negligibly small.

Kon

min

From the psychrometric chart in Appendix G for air at 25°C and a relative humidity of 70%, the humidity ratio is 0.014. The mass flow rate of water entering the combustion chamber can be calculated to be :

water

water

eID air ONO)

eee

air

II 0.878

< 0.014

x a = 29

002

KS anor

——— min

The number of moles of water in the air per mole of dry air is: Ne

a

) OK

N.air

M,.

= 0.014

Zo

x 18 = 0.0226

water

The combustion equation can now be modified to include the moisture content of the air.

CH, + (0;

3. 76N,) 4X:

4.76 x; 0,0226)F.O

= 3CO, + 4.75H,O + 20, + 26.32N, The first law, Eq. 12.8, applied to the combustion chamber takes the form

On Hee = {3(-393520 + 32179 — 9364) + 4.75(-241810 + 27896 — 9904) + 2(0 + 24523 = 8682) + 26:32(0 + 23714 — 8669)} —~ {103 850 + 0.753(—241810

or.

O.. =

1470000 kJ/kmol

Wis unal

+ 9904) } =

—1470000 kJ/kmol fuel

= 33 400 kJ/kg,,.3CO0O, + 4H,O The final pressure can be found using the ideal-gas law. It takes the forms

PVp= ee

PAY

PV

ji ae NRT Pea tice P,

= NN Ril

100 6 X 298 Pr OOO P

NRT

-P,, = 783 kPa

The heat transfer can be found by applying the first law, represented by Eq. 12.13. Using the ideal-gas tables in the Appendix, using T,, = 2000°C, the first law provides

Oe Sia

ART)

prod

= {3(— 393520

Nh

a he

RT),

react

+ 100804

—

9364

—

8.314

x 2000)

+4(—241810 + 82593 — 9904 — 8.314 x 2000)} — {1(— 103850 — 8.314 x 298) + 5(0 — 8.314 x 298)}

Note: Our sign

. oe

we)

out is negative.

= —1580000 kJ/kmol fuel The heat transfer from the bomb is then

Q.,= -O= le on ee

= 35920 kI/ke fuel

™ You have completed Learning Outco

(2)

Part 2

Applications

Adiabatic flame temperature: The temperature after combustion of a constant-pressure flow

through an insulated

If all of the heat developed during combustion remains in the products of combustion and if kinetic and potential energy is negligible for a constant-pressure flow through the combustion chamber, the temperature of the products is called the adiabatic flame temperature since no heat is removed from the control volume. To determine the adiabatic flame temperature, we apply the first law of thermodynamics to the control volume knowing that Q = 0. This modifies Eq. 12.8 to the simpler form

combustion chamber.

lohe= sel,

(12.14)

The total enthalpy of the reactants can be determined from information on the fuel and oxidizer used. To match the enthalpy of the products to the enthalpy of the reactants, we must use an iterative technique to find the correct temperature of the products of combustion. The adiabatic flame temperature is maximum when theoretical air is mixed with the fuel in the insulated combustor. The adiabatic flame temperature can be reduced by adding excess air to the mixture. This addition of air does not alter the first-law equation expressed by Eq. 12.14. The temperature of the products of combustion could also be lowered to a specified flame temperature by transferring heat from the combustor; an example of that calculation was given by Example 12.9. If a specified amount of heat is transferred but the temperature is not known, a trial-and-error procedure is also required. Examples will illustrate.

Example 12.11

Maximum adiabatic flame temperature Liquid octane (gasoline) and stoichiometric air enter an insulated combustion chamber at 25°C and 100 kPa, as shown in Fig. 12.17 Find the temperature after combustion in the exiting gases, neglecting kinetic and potential energy changes and assuming no shaft work. Solution

The first step is to determine the stoichiometric equation for the combustion

of octane (C,H,,) in air. It is Dry air is assumed unless

CyH,s(€) + 12.5(0, + 3.76N,) = 8CO, "+ 9H,O + 47N,

otherwise stated.

Combustion chamber

Figure 12.17

Chapter 12

Combustion

One mole of gasoline (octane) produces 8 moles of carbon dioxide and 9 moles

of water. The 47 moles of nitrogen heat up, so it affects the exiting temperature T, of the products. Let’s determine the molar enthalpy of the reactants. From Table B-7 the enthalpy of formation for liquid octane is listed as —208 450 — 41 460= —249 910 kJ/kmol. The oxygen and nitrogen are at 298 K, so h, + (h — h°) = 0 for both of those reactants. Hence, the total molar enthalpy of the reactants is

hy = Wo

= —249910kI/kmol

f CHL,

Next, the sum of the enthalpies of the three components that make up the products at the unknown final temperature 7, forms h,. The enthalpy for each component is listed as follows:

CO,

—

ah

= 93520

A, = 9304 ee

PO

8

ING ies

hg ig

Sg = ALSO Bu

th

= 994

by, = 8069

Using h, = i, the above are combined into the first law which states that the energy contained in the fuel and air at 25°C must equal the energy contained in the products at T,, expressed by Eq. 12.14. The equation simplifies to

5 650 000 = 8heo, + Wyo + 47hy, The objective is to find 7, that error procedure. Guess a value will undoubtedly be incorrect. usually be obtained. Two of our

Hint: A sketch of the interpolation may be

helpful, as in Fig. 12.18.

6200

balances the above equation. It is a trial-andfor T, and check the equation. The first guess But with two or three guesses, an answer can guesses follow:

2200 K: 5650000

2 8 x 112900 + 9 xX 92900 + 47 X 72000 = 5120000

2600 K: 5650000

2 8 X 137400+9

X 114300+47

2600 — 2200)

)

Figure 12.18

x 86600 = 6200000

The temperature is in between 2200 K and 2600 K and is interpolated to be (see Fig. 12.18)

5650 — 5120 Le oPAN Vex 5120 a

Observe: At 2200 K, the rhs is too low. At 2600 K, the rhs is too high. It must be between the two temperatures.

+ 2200 = 2400K Serna

This answer represents the maximum adiabatic flame temperature since stoichiometric air was used. Excess air can be used to decrease the exiting temperature, a desired result if the temperature cannot be as high as 2400 K because of certain materials exposed to the products of combustion. This will be illustrated in the following example.

Note: The answer would be 2396 K, but that’s four significant digits that are not supported by the three significant digits in the numbers used to

"generate the interpolation. _ We should not use more significant digits in an answer than are allowed by the numbers used in the calculations.

aaa | Part 2 Applications

Example

12.12

Adiabatic flame temperature with excess air The octane liquid of Example 12.11 undergoes complete combustion with 200% excess air in the insulated combustion chamber, using the same inlet conditions and the same assumptions. Estimate the temperature of the exhaust gases leaving the chamber of Fig. 12.19. The lower temperature gases would have less damaging effects on the metallic parts of an engine located after the combustion process.

Solution: The stoichiometric equation for the combustion of octane (C,Hg) in air is

CH, (€) + 12.5(0, + 3.76N,) = 8CO, + 9H,O + 47N, Note:

For 200% excess

air, the moles of air are tripled (see Eq. 12.5).

Complete combustion does not have CO in the products,

From the definition of excess air, 300% theoretical air enters the combustion

chamber, so the reaction equation with the excess air becomes

C,H,, + 37.5(O, + 3.76N,)=8CO, + 9H,O + 250, + 141N, Observe the extra oxygen in the products. Let’s determine the molar enthalpy of the reactants. From Table B-7 the enthalpy of formation for liquid octane is —208 450 — 41 460 = —249 910 kJ/kmol. The oxygen and nitrogen are at 298 K,so h? + (h — h°) = 0 for both of those reactants. Hence, the total molar enthalpy of the reactants is

hy = hey = —249910 kJ/kmol fuel Next, the enthalpy of each product at the unknown final temperature is listed as follows:

COs)

hg = AG thoy = legge = aa

he

et

H,O: hyo = he + hy — hyp, = —241810 + hy. — 9904 Of hy. = Reh

N; hy =

+h, N, —h

he

bea I|

rex = /ty,

682 |

8669

The sum of the four terms forms / p Using h, =f p» the above are combined into Eq. 12.14, which states that the eneggy contained in the fuel and air at 25°C must equal the energy contained in the products at 7,. The equation is

6 673 000 = 8hgg + 9hyo + 50ho + 141hy

Chapter 12

Combustion

The objective is to find T, that balances the preceding equation. Guess a value for T, and check the equation. With two or three guesses, an answer can usually be obtained. Two guesses, one above h , and one below h,, are

1200 K: 6 673 000 2 8 x 1000 K: 6 673 000 2 8 x

x 53800 + 9 x 43400 + 50 x 38400 + 141 36800 = 7 930 000

X 42800 + 9 x 35900 + 50 x 31400 + 141 30 100 = 6 480 000

The temperature is closer to 1000 K than 1200 K and is interpolated to be

z

_ 6673 — 6480 =

7930 — 6480 1200

=

1000) + 1000 = 1030K a

—

The exit temperature has been substantially reduced, which is one reason why excess air is often used. Without excess air, the exit temperature may be too high for the materials encountered by the products.

Flame temperature with heat transfer eS

Gaseous propane reacts with 100% theoretical air both at 25°C and atmospheric pressure in the steady-flow combustion process displayed in Fig. 12.20. Determine the flame temperature if heat is transferred from the combustor at the rate of 20 MJ/kg of fuel. Neglect kinetic and potential energy changes as well as any shaft work.

Solution The stoichiometric equation for the combustion of propane (C,H,) in air is

C,H, + 5(O, + 3.76N,)=3CO, + 4H,O + 188N, One mole of gaseous propane produces 3 moles of carbon dioxide and 4 moles of water. The inert nitrogen does require some heat to reach the exiting temperature. (Continued)

Example

12.13

ane | Part 2 Applications

Example 12.13

(Continued) From Table B-7 the enthalpy of formation for gaseous propane 1is listed as —103 850 kJ/kmol. The oxygen and nitrogen are at 298 K so h; + (h —h°)=0 for both of those reactants. Hence, the total molar enthalpy of Ane reactants 1s h, =h?.,f,CH, =

—103850kJ/kmol

Next, the enthalpy of each product at the final temperature is listed as follows:

CO

Sh

HO

Hh, og= hig =

Ned

th = 03520

eo

ie

i,

fh, = 8007

The energy balance g = h, — h,, with the heat transfer Note: The heat transfer g

eas

g = —20000 x 44

= — 880000 kJ/kmol, simplifies to

is negative since it leaves

ne

the combustor.

1400000

=

sie

aloo, ae Ane 6 +

Ba

18.8 ie

The objective is to find 7, that balances the above equation. Guess a value for 7, and check the equation. Two guesses, one above h, and one below h,, are

1400 K: 1400000

2

= 3 X 65300+4

X 45700+18.8

x 43600 = 1200000

1800 K:

2

= 3 X 88800+4

x 60400+18.8

x 57700 = 1600000

1400000

The temperature is halfway between 1400 K and 1800 and is close to 1600 K. If the exit temperature is known and the heat transfer is desired, that problem does not require a trial-and-error solution.

aT Observation: Thetwo vapor trails made by those

In Chapter 9 we analyzed internal combustion engines assuming that the ©Hgine used air as the oxidizer and that air exited the engine as exhaust. Now | We know that this is only approximately correct. The nitrogen passes through

high-flying airplanes are due to the water vapor

the engine unchanged, but the oxygen is converted to carbon dioxide and water. = Knowing this and using the tools developed in this chapter, we could perform

from the engines and

a more accurate analysis of engine cycles. The error that would be found using

the atmosphere that gets

:

:

:

collected into the two ‘te more accurate analysis would be relatively small since the water and carbon extremely strong vortices dioxide make up a small percentage of the products. So, the comparisons made in created by the two wings of | Chapter 9 are valid. However, it is now obvious why water vapor is always present the aircraft. in the products of combustion and can be observed under certain conditions from the tailpipe of a motor vehicle, or the contrails behind a high-flying aircraft.

You have completed Learnin

(3)

Chapter 12

This chapter has demonstrated that the products created in the stoichiometric combustion of a hydrocarbon fuel are carbon dioxide and water. So where do the pollutants come from that exist in automobile exhaust? At very high temperatures, other reactions are taking place during the combustion process. These reactions are different from the combustion reactions previously in that the reaction is a two-way process. In these equilibrium reactions, a compound will dissociate into smaller compounds or atoms at high temperature. At the same time, these smaller compounds will re-associate back into the original compound. Both the dissociation and re-association processes take place simultaneously. The relative amounts of the various compounds will be a function of the temperature of the substances. To describe these equilibrium reactions, use will be made of the Gibbs function defined in Chapter 7: (pS 5) mosis

(12.15)

The differential of this equation is dG—

dH —

TdS — Sdf

(12.16)

From Chapter 6 we have

TdS = dH — VdP

(12.17)

Combining these two equations yields the following relationship:

dG = VdP — SdT

(12.18)

Under equilibrium conditions, both pressure and temperature are assumed constant; Eq. 12.18 then provides dG = 0 sothat

G = const

(12.19)

An equilibrium reaction is a process during which the Gibbs function is constant. It can be expressed as Reactants

Step 3 Myosin fiber

Actin fiber released Step 4

Vesa:

ge UIE Muscle fiber actions.

g l™ You have completed LearninOu

14.4 Temperature Regulation in Biological System Living organisms live in a temperature range that extends from 0°C at the low end to 42°C at the high end. Life cannot exist at temperatures below 0°C without cryopreservation techniques because water in the body will freeze. Above 42°C,

(3) and (4)

492

Part 3

Contemporary Topics

Thermoregulation: Balancing the heat to and from a body with that generated within the body. Metabolism: All the chemical reactions that produce energy in the body.

Homeothermic: Maintaining a relatively constant body temperature. Poikilothermic: Having a varying body temperature.

Endothermic organism: Organism that uses the internal generation of heat as its primary means of thermoregulation. Basal metabolic rate: Minimal rate of energy production to support life.

Piloerection: A process whereby mammals and birds reduce heat loss. Hypothermia: When the human body gets below

Sle (Sse) Hyperthermia: When the human body gets above

38°C (100°F).

proteins undergo a denaturing process that alters the structure of the protein. The subject of maintaining acceptable temperatures in living organisms is called thermoregulation, which is achieved by balancing the rate of energy production within the body and the rate of heat intake from outside the body with the rate of heat loss from the body. The term metabolism addresses all the chemical reactions that produce energy in the body. Some biological systems, like the digestive system, produce energy. Others produce energy in the form of work. Living organisms use a wide variety of processes to obtain and remove body heat. Organisms that maintain a relatively constant body temperature are called homeothermic: mammals and birds. Organisms that have body temperatures that can vary considerably are called poikilothermic: lizards and fish.

14.4.1 Endothermic organisms Organisms that use the internal generation of heat as their primary means of thermoregulation are endothermic organisms. Mammals fall into this classification. The minimal rate of energy production needed to support life in an endothermic organism is called the basal metabolic rate. This assumes that the organism is at rest and that no digestion is occurring when its temperature is measured. The basal metabolic rate accounts for the energy needed to support breathing, to circulate blood, and to maintain electrical activity to support the nervous system. The metabolic rate in an organism is affected by physical activity, eating, body growth, reproduction, and heat production. The metabolic rate is lowest when sleeping since nonessential muscles are not working. The metabolic rate during extreme exercising can be 10 to 20 times the basal metabolic rate. Heat loss in endothermic organisms occurs by convection and evaporation from the skin to the surrounding atmosphere. If your skin temperature is greater than the atmospheric temperature around you, you lose heat primarily by convection heat transfer from the surface of the skin. Endothermic organisms can gain heat by immersing themselves in warm water or cool off by immersing themselves in cold water. Mammals and birds can reduce heat loss in cold weather by a process called piloerection, where hair or feathers are lifted away from the skin to produce additional insulation near the skin. Since humans do not have much hair,

we notice this phenomenon as “goose bumps.” If your skin temperature is less than the atmospheric temperature, you lose heat by sweat evaporating from the surface of your skin. The phase change releases large amounts of heat. Bull frogs achieve this same effect by secreting mucus from their skin. If the cooling rate or heating rate of the body is insufficient to maintain a constant body temperature, the temperature of the body will change, with serious consequences. Table 14.2 lists the physical effect of either increasing or decreasing the body temperature. Hypothermia occurs when the human body temperature gets below 35°C (95°F). Hyperthermia occurs when the human body temperature reaches 38°C (100°F) or higher.

14.4.2 Ectothermic organisms Ectothermic organism: The organism is heated from sources external to the body.

An ectothermic organism is heated from sources external to the body. Lizards will absorb energy by convection from the surrounding air, by conduction from laying on warm rocks or gravel, or from direct solar radiation. Fish absorb heat energy from the surrounding water. Fish tend to be homeothermic since they live

Chapter 14 Table 14.2

Thermodynamics of Living Organisms

Physical Effects of Changing the Human Body Temperature

Body Temp

Symptoms

44°C (111°F)

Death

43°C (109°F) 41°C (106°F)

Brain damage and possibly death Fainting, dehydration, vomiting, headaches, or dizziness

38°C (100°)

Sweating and uncomfortable feelings

37°C (99°F)

Normal body temperature

36°C (97°F) 34°C (93°F)

Shivering Severe shivering, loss of movement in the extremities, and skin color changes

33°@ (91°F)

Confusion, drowsiness, slow heartbeat, and shallow breathing

31°C (88°F)

Loss of consciousness and change in heart rhythm

26°C (79°F)

Death

_in environments where the water temperature is stable. Both fish and lizards are poikilothermic. Since their body temperature will vary and they receive external heat, their metabolic rates are very low. The body temperature of an ectothermic organism will largely be a function of the environmental temperature. Cold weather will decrease the body temperature and thus reduce physical activity, whereas warm weather will increase body temperature and physical activity. Ectothermic organisms will seek to regulate body temperature by altering their physical environment: Lizards sit on warm, sunny rocks to gain energy and move into shady areas when they are too warm; fish move to warmer or colder water depending on their individual needs. Bees and other insects can raise their body temperature in anticipation of flying by exercising their flying muscles.

14.4.3 Temperature regulation in plants Some plants have the ability to raise their temperature above the ambient temperature by a process called thermogenesis. This can make them resistant to frost. Thermogenesis is the development of thermal energy as a part of a chemical reaction in the cellular respiratory process. The sacred lotus plant shown in Fig. 14.5, which is native to Southeast Asia, can remain up to 20°C above the ambient temperature. In this plant, heat is generated by breaking down starches that are stored in its roots.

khanh/Shutterstock.com tuan vu

Sacred lotus plant.

[EE

oe

Part 3 Contemporary Topics

In this chapter, the energy that can be generated from the sun through the action of photosynthesis in plants and algae was examined. Then the generation of energy from the foods eaten by both animals and humans was presented. Finally, the energy required by the muscular systems in the bodies of animals and humans was investigated, along with the regulation of temperature in biological systems. The following additional terms were introduced or emphasized in this chapter: Calorimetry: An experimental method for determining the energy of oxidation, or heating value. Chlorophyll: The pigment that absorbs sunlight. Ectothermic: Energy gained from an external source.

Ectothermic organism: An the body.

organism

heated from sources

external to

Endothermic: Energy generated internally. Endothermic organism: An organism that uses the internal generation of heat. Homeothermie: An organism that has a constant body temperature.

Metabolism: The chemical reactions that produce energy in the body. Photosynthesis: The process by which sunlight converts plants and algae into usable energy. Poikilothermic: An organism that has a varying body temperature. Thermogenesis: The production of heat in living organisms. Thermoregulation: Processes that are used to control body temperature.

Two equations were utilized in this chapter: Photon energy: Frequency of light:

€ B Energy Conversion

in Plants

14.1

Plot the energy of a photon as a function of the wavelength of light for the range of visible wavelengths.

14.2

If a plant could be engineered that would convert gamma rays (A = 0.01 nm) into chemical energy, what would be the percent

increase in energy over plants that use light with a wavelength of 660 nm? 14.3

Using Eq. 4.1, determine how many grams of sugar are produced if one kilogram of carbon dioxide is used in the photosynthesis process. How much water is needed?

Chapter 14

Thermodynamics of Living Organisms

€ B Energy Conversion in Animals 14.4 For the bomb calorimeter described in Example 14.2, determine the maximum temperature the bomb would achieve if 20 g of apple juice and 0.005 g of fuse wire were burned. 14.5 For one day, record what types of food you ate and roughly how much of each type you consumed. Calculate your energy intake for that day in joules and in food calories. 14.6 Design a menu that will achieve the following caloric intake: a)

1000 calories per day for a diet

b)

1500 calories per day for a smaller person

c) 2000 calories per day for a larger person d) 6000 calories per day for a Tour-de-France bike rider

14.7

Compare the daily caloric intake for breakfast, lunch, and dinner at a university dining hall to the typical meals you would eat at home. Which diet has a higher fat content? Carbohydrate content?

14.8

Develop a daily diet that would reduce your amount of fat intake.

14.9

A very popular diet encourages people to eat a high-protein diet. Develop a daily high-protein diet and compare it to your normal daily diet. Did you increase or decrease the fat and carbohydrate intake?

€ B The Generation of

Biological Work 14.10

Most joint actions are the combination of one or more muscles tightening and other muscles relaxing at the same time. Describe the action of three human joints that act in this manner.

14.11

Give one example of smooth muscle tissue in the human body and describe how it functions.

14.12 Perform an online search and report on what happens to cardiac muscle tissue during a heart attack.

© B Temperature Regulation in Biological Systems 14.13

If the ambient temperature is 100°F, estimate the heat transferred from your body by the evaporation of 0.01 Ibm of sweat.

14.14

List three types of fish that live in cold water and three types of fish that live in warm

water. 14.15 List three plants that have the ability to regulate their temperatures. 14.16

During the first expeditions to the Antarctic, explorers found out that short, stocky men had a much better survival rate than tall athletic men. Why is this?

14.17

List five ways to cool the human body in case of overheating. What is the best method in an industrial plant?

14.18

Find an online Basic Metabolic Rate calculator and use it to determine your BMR.

14.19

Using the BMR calculator from Problem 14.18, determine the percent drop in your BMR if you lost 10 pounds on a diet.

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THE APPENDICES

‘wn Appendix C

Steam Tables

wi} Appendix D

Properties of R134a

mm

4 Appendix E

Properties of Ammonia

Ti

Appendix F

ldeal-Gas Tables

n> Appendix G

Psychrometric Charts

wii Appendix H

Compressibility Chart

Ti

Appendix

|

Enthalpy Departure Charts

Tim

Appendix

J

Entropy Departure Charts

|

1 cm = 0.3937 in 1m = 3.281 ft 1 km = 0.6214 mi 1 in = 2.54cm 1 ft = 0.3048 m 1 mi = 1.609 km

1 Ibf = 0.4448 x 10° dyne 1 dyne = 2.248 x 107° Ibf

1 0z = 28.35g

1 kip = 1000 Ibf 1N = 0.2248 lbf

1 slug = 32.17 Ibm 1 slug = 14.59 kg

1N = 10° dyne

1 kg = 2.205 Ibm

1 1 1 1 1

Pressure

Volume

1 psi = 2.036 in Hg

1 it =7 481 gal Ws) Vit =30,02832407 1 gal (U.S.)=-231 in?

1 lbm = 0.4536 kg

mph = 1.467 ft/s mph = 0.8684 knot ft/s = 0.3048 m/s km/h = 0.2778 m/s knot = 1.688 ft/s

1 mi = 5280 ft

| 1 mi = 1760 yd

Work and Heat

1 J = 10’ ergs

1 Cal lI= 0.003968 Btu

1 hp 1 hp 1 hp 1W

1 Btu = 1055J

1 W = 10’ dyne.cm/s

1 Btu = 0.2930 W.hr

1 W = 3.412 Btu/hr

1 Btu = 778 ft-lbf

1 kW = 1.341 hp

i ttelot = 1.356 J 1 Cal = 3.088 ft-lbf

= = = =

550 ft-Ibf/s 2545 Btu/hr 0.7455 kw 1J/s

1 kWh = 3412 Btu

1 ton = 12,000 Btu/hr

1 therm = 10° Btu iequad== 10" Btu

1 ton = 3.517 kW

1 W = 0.7375 ft-lbf/s

1 1 1 1 1 1 1 1 1 1

psi = 27.7 in H,O atm = 29.92 in Hg

atm = 33.93 ft H,O atm = 101.3 kPa atm = 1.0133 bar atm = 14.7 psi in Hg = 0.4912 psi ft H,0 = 0.4331 psi psi = 6.895 kPa kPa = 0.145 psi

1 Torr

=

1mm Hg

1 gal (Brit.) = 1.2 gal (U.S.)

ea Omeetie 1 = 000s538Ie 1L = 0.2642 gal 1m? = 264.2 gal Lime =oop mrt fita=28e2.. 1 in? = 16.387 cm=

j .

|

= 5

“Cw.

Properties of the U.S. Standard Atmosphere

P, = 101.3 kPa py = 1.225 kg/m? Temperature ai ©

;

—

1,000

1.000

0.8870 0.7846 0.6920 0.6085 0.5334 0.4660 0.4057 0.3519

0.9075 0.8217 0.7423 0.6689 0.6012 0.5389 0.4817 0.4292

0.1915 0.1399

0.2546 0.1860

0.2615

0.3376

0.1022

|

0.1359

0.07466 0.05457 0.01181 0.2834 x 10°77

|

0.09930 0.07258 0.01503 0.3262 x 10°77

0.7874 x 10°?

|

0.8383 x 10

O2217-« 107 0.5448 x 10°47

0.2497 x 10°? 0.7146 x 107

9p

ol

| Appendix B_

Table B-1E_

Material Properties

Properties of the U.S, Standard Atmosphere

P, = 14.7 psia, py = 0.0763 Ibm/ft? Altitude

ft eee eee es 0 1,000 2,000 5,000 10,000 15,000 20,000 25,000 30,000 35,000 36,000 40,000 50,000 100,000 110,000 150,000 200,000 260,000

Temperature

Pressure

Density

oF

PIP,

p!po

59.0 55.4 51.9 41.2 23.4 5.54 —12.3 —30.1 —48.0 —65.8 —67.6 —67.6 —67.6 —67.6 —47.4 113.5 160.0 —28

1.00 0.965 0.930 0.832 0.688 0.564 0.460 0.371 0.297 0.235 0.224 0.185 0.114 0.0106 0.00657 0.00142 038710 0.351 x 10°47

1.00 0.975 0.945 0.865 0.743 0.633 0.536 0.451 0.376 0.311 0.299 0.247 0.153 0.0140 0.00831 0.00129 0.26710 0.477 57105:

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The Appendices

4

Appendix B_ Material Properties

| Table B-3__Critical-Point Constants

| Molar | Formula

| | Ammonia | Argon

NH, Ar

| Butane

GH,

| Benzene

| Carbon dioxide

CoHe aon

| Carbon monoxide |CO

| Carbon tetrachloride |CC, | Ethane CH, Ethylene

Helium Hydrogen

Methane |Neon | Nitrogen

| Oxygen | Propane | Propylene

| Ri34a | Sulfur dioxide Water

GH,

He wy

/CH, Ne N, O, CoH, | GH,

| CCH |so, 4,0

Mass

| K

/'

m*/

| 2897 |

|

1703 | 405.6 | 7298 | 11.28 | 3994] 151 272 486

| 1636 .| 0.0724 | 705 |0.0749

73.11 | 562 58.12]

4252

44.01 | 3042] 28.01 | 133

153.84 | 30.07] 28.05 | 400} |} 202]

ft*/

,

| 1012

492

| 714

| 7652]

380

|

5475 | 7.39 | 1070

|0.0943

240

3.50

| 0.0930

5564 | 10015] 3055] 5498] 2824]

5083]

53 333]

95] 599]

| 1604] 191.1} 3439] 20.18 | 445 | 80.1}

28.02 | 1262 | 2271 |

32.00} 1548] 2786] 44.09 | 3700] 6659] 42.08 | 365.0 | 6569]

456 488 5.12 023 130

551

507

661 708 | 742 | 332 | 188

464 | 673 273 | 395 339 | 492

5.08 | 736 426 | 617. 462 | 670

| 102.03 | 3743 | 6137] 407 | 596 | 6406] 4307] 7752] 7.898 | 1143 18.02) 6474 | 11653 | 221 | 3204

|0.2603

| 0.2547

| 0.2759 |0.148 | |01242 | 199 |0.271 /0.0578 | 0.926 |0.302 |0.0649 | 1.04 |0.304

| 0.0993 | 159 | 0.290 |0.0417 | 0.668 |0.308 |0.0899

| 1.44 | 0.291

[0.078 | 1.25 |0.1998 | 320 |0.1810 | 290

|0.308 |0.277 |0.276

|02478 | 296 | 0324 |01217 | 195 | 0,269 | 0.0568 | 090 | 0.233

and Kose 1959. 117-226 K. trom A. FE. Rew, Cres. Based Lynn, 52: data Js, on

The Appendices Table B-4

Specific Heats and Densities of Liquids and Solids CG, ki/kg-°C

op kg/m?

ed Liquids Substance

| State

C,

Water

1 atm, 25°C

Ammonia

ieatim, —20°E

1 atm, —50°C

1000 665 602 1440

iT ean, ZSAE

1210

1 atm, 15°C

879

imatine25-©

R-134a Benzene

Density p | Substance

State

C,

Density p

Glycerin

1 atm, 10°C

OME Ys

1261

Mercury

1 atm, 10°C

0.138

13 600

Propane

1 atm, 0°C

2.41

529

2.43

789

| Ethyl Alcohol | 1 atm, 25°C Solids

Substance

TecG

Ice

C,

2.033 2.10 0.699

—20 0

Aluminum

—100

Density p | Substance

920 917 2700

Lead

Copper

0

0.870

2700

100

0.941 0.448 0.233

2700 7800 10 500

Iron

20

Silver

20

12S

0

—100

0 100

C,

Density p

0.124

11 310

0.328

8940

0.381 0.393

8940 8940

Based from Thermodynamic data Wark, Kenneth McGraw-Hill, ed., rd York, New 1981. on

Table B-4E_

Specific Heats and Densities of Liquids and Solids

C, Btu/lbm-"F_

_p |bm/ft?

Liquids

| Substance Water Ammonia

| R-134a

State

CG

dis Density p | Substance

‘een, WANE

1.00

= Ale 80°F

1.08 | 272

62.4 41.3 By iho.

=(|S4F

0.294

86.0

90°F

0.348

73.4

+

Glycerin

State

za

1 atm, 50°F

Benzene Mercury

1 atm, 60°F 1 atm, 50°F iPelienesvale Propane Ethyl Alcohol | 1 atm, 77°F

C, 0.555

0.410 0.0330 0.604 | 0.681

| Density p 78.5

re

Solids

| Substance ;

| Aluminum

| Iron

i-E

0

| Ice |

|

54.7 849 33 49.2

'

C,

Density p | Substance

0.502

32

0.212

CG,

*ie

Density p |

0.030

705

= (5x0

0.0785

550

170

0

0.0893

550s

200

0.0938

550

Lead

0.471

BZ

|

SY fs.

200

0.224

170

32

0.107

490

Copper

32

McGraw-Hill, New 3rd Thermodynamic Wark, 1981 York, Kenneth ed., from data n

Appendix B-

Table B-5

Material Properties

Ideal-Gas Specific Heat of Several Common Gases as a Cubic Function of Temperature

GC: = a+ bT+cT?+dT? with Tin K and C. in kJ/kmol-K Gas

a

Air N, oO; H, co NO HO CO; NO,

28.11 28.9 25.48 29.11 28.16 29.34 32.24 22.26 22.9

Table B-5E_

b

0.001967 —0.001571 0.0152 —0.001916 0.001675 —0.0009395 0.001923 0.05981 0.05715

(a

d

0.4802 x 10° 0.8081 x 10°. =O7 155 < 100 0.4003 x 10°° 05372 10 0.9747 x 10° (OSS. 10m —3.501 x 10° eve a AO

—1.966 x 10°? —2.873 x 10°” 1312x407 —0.8704 x 10°? =2 02K 10 —4.187 x 10°? —3.595 x 10° 7.469 x 10°? Top CTO

G. Based from data Fundamentals Sonntag, E. R. and Wylen Van J. Classical of on Thermody York, New Wiley, 1976.

Ideal-Gas Specific Heat of Several Common Gases as a Cubic Function of Temperature __ ie = a+bT+cT?+dT? with Tin °R and ie in Btu/Ibm-°R b

Air N, O, H, ao) CO, N,O NO NO,

6.713 6.903 6.085 6.952 6.726 5.316 7.700 7.008 5.48

0.02609 x 10°? =0;02085-< 105 02017108 =0,02542 x 107 OBR EON O79861-x. 107 0102552 Mi0= =0,01247.< 107 0.7583 x 10°?

0.03540 x 10° 0.05957 x 10°° 0105275. .1080.02952 x 107° 0.03960 x 10~° =0.2581 x 10° OWT TSO Or 0.07185 x 10°° =01260- 10

0.08052. 410m, =O 176E< 10m 0.05372) Alm. = 0.035655. 40" —0.09100 x 10~? 0.3059 x 10°° =0:1472 x 10" Os 15 210m 0322-105

Chemical Hall, Prentice Thermodynam Kyle, G. Process and Based B. from data on Cliffs, 1984. NJ, Englewoo

The Appendices Table B-6__

Ideal-Gas Specific Heats of Several Common Gases at Different Temperatures

a Temperature, K

Temperature, K

G

G

k

C,

C,

k

2

kJ/kg:K

kJ/kg-K

kJ/kg:K

E

Air

Carbon dioxide, CO,

Carbon monoxide, CO

2

0.716 0.718 0.721 0.726 0.733 0.742 02753 0.764 0.776

0.791 0.846 0.895 0.939 0.978 1.014 1.046 1075 1,402

0.602 0.657 0.706 0.750 0.790 0.825 0.857 0.886 0.913

1314 1.288 1.268 1,252 1.239 1.229 1.220 L213 1.207

1.039 1.040 1.043 1.047 1.054 1.063 12075 1.087 1.100

0.788

1.126

0.937

1.202

0.800 0.812

1.148 1.169

0.959 0.980

1.197 1.193

0.834

1.204

1.015

0.855

1.234

1.045

Hydrogen, H,

1.400 17399 1.398 1.395 1392" 1.387 1.382 1.376 153710

les

0.816

1.364

g

1.126 1.139

0.829 0.842

1.358 1353

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Saturated Solid-Vapor

v m?/kg Sat. Solid v; x10

u kJ/kg

Sat. Vapor! 3

Vg

Sat. Solid :

Subl. Uig

TOOT 206.4 HOOT 42063 1:090m =24467 1 C90E 228353 1.090 334.2 1.089 394.4 1.089 466.7 1089 9°" 553-7 1.088 658.8

2702 2709 2 10 BYN\2 Diy 33 2714 2716 Ze Ned 2718

786.0 1129 1640 2414

2719 Di a7) 2724 2726

1.088 1.087 1.087 1.086

h kJ/kg

s kJ/kg.K

Subl. 3

ie

Subl. ,

Sig

2835-9250 pe lee Orsi 2635, 225017 alee? [eer S 2635) 72498") -— 1237/0456 2830: S24A94eiT 1253, *10'536 28367 24905) 1-268 10616 2837 2487 | -1.284 10.698 3937 62453 — 1 299) 10.7on 203/. 24/9913 15) 101865 2038. 0247/68) 1331, 10/950 111.036 1.346 92472) 2030) SNe 93/7, 2838-72464" 11.394 2457 | —1.408 2839 11.580 2450 | —1.439 2839

Based from data Hill, Keyes, Keenan, Moore, Wylen end Steam Wiley, Tables, York, New G. 1969; Van J. on and Sonntag, E. R. Classical of Fundamenta Thermodyn York, New Wiley, 1973.


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Steam Tables Appendix C Ww90]N

Based on data from Keenan, Keyes, Hill, and Moore, Steam Tables, Wiley, New York, 1969.

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Table D-I

Saturated R134a—Temperature Table Specific Volume

m3/kg

Liquid

Sat. Vapor

Sat. Liquid

0;* 10°

%

Urs

Ug

hy

0.7055 Olds Oy717-2 0.7233 0.7265 0.7296 0.7328 0.7361 O37395 0.7428 0.7498 0.7569 0.7644 O72 0.7801 0.7884 0.7971 0.8062 0.8157 0.8257 0.8309 0.8362 0.8417 0.8473 0.8530 0.8590 0.8651 0.8714 0.8780 0.8847

0.3569 0.2947 0.2451 0.2052 0.1882 0.1728 0.1590 0.1464 0.1350 0.1247 0.1068 OL0919 0.0794 0.0689 0.0600 0.0525 0.0460 0.0405 0.0358 0.0317 0.0298 0.0281 0.0265 0.0250 0.0236 0.0223 0.0210 0.0199 0.0188 0.0177

—0.04 4.68 9.47 14.31 Meyers) IQA 21.68 24.17 26.67 AS ANS 34.25 39.38 44.56 49.79 55.08 60.43 65.83 TARPS) 76.80 G2.3)7/ 85.18 88.00 90.84 23,00, 96.58 99.47 102.38 105.30 108.25 Ma teZ 2

204.45 206.73 209.01 P| eeAS) 212.43 213.57 214.70 215.84 216.97 218.10 220.36 222.60 224.84 227.06 DIES) Pdf 231.46 735.05 ABS). IE: 23791 240.01 241.05 242.08 243.10 244.12 Zao. 2 246.11 247.09 248.06 249.02 249.96

0.00 4.73 9.52 (7 16.82 19.29 Tae 24.26 26.77 29.30 34.39 39.54 44.75 50.02 55.35 60.73 66.18 71.69 77.26 82.90 85.75 88.61 91.49 94.39 97.31 100.25 103.21

Pressure / Liquid kPa

51.64 63.32 77.04 93}0)5 101,99 111.60 121.92 132299 144.83 157.48 185.40 217.04 252.74 292.82 33}// (8) 387.56 442.94 504.16 571.60 645.66 685.30 20/5 770.06 815.28 862.47 911.68 962.98 1016.4 1072.0 1129.9

Enthalpy kJ/kg

Sat. Vapor

Sat.

Temp °G

Internal Energy kJ/kg Sat.

106.19 109.19 Migs

Entropy kJ/kg-K

Evap. h fg

Sat. Vapor hg

API RE 220.67 218.37 216.01 214.80 213.57 21232 211.05 209.76 208.45 205.77 203.00 200.15 197.21 194.19 191.07 187.85 184.52 181.09 177.55 175.73 173.89 172.00 170.09 168.14 166.15 164.12 162.05 159.94 157.79

222.88 225.40 227.90 230.38 ED 232.85 234.08 235.31 23653 237.74 240.15 242.54 244.90 247.23 249.53 251.80 254.03 OD. 258.36 260.45 261.48 262.50 263.50 264.48 265.45 266.40 267.38 268.24 269.14 270.01

Se

Sat. Vapor Ssg

0.0000

0.9560

0.02001

0.9506

0.0401

0.9456

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Liquid

0.0600

0.9411

0.0699

0.9390

0.0798

0.9370

0.0897

09351

0.0996 0.1094

0.9332 0.9315

0.1192

0.9298

0.1388

: 0.9267

0.1583

0.9239

Oh WH

0:92:13

0.1970

0.9190

0.2162

0.9169

0.2354.

0.9150

0.2545

0.9132

O2735

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0.2924

0.9102

Osis

0:9089

0.3208

0.9082

@.3302

0.9076

0.3396

0.9070

0.3490

0.9064

0.3584

0.9058

0.3678

0.9053

0.3772

0.9047

0.3866

0.9041

0.3960

0.9035

0.4054

0.9030

(Continued)

Wilson P. D. from “Thermodynam Basu, S. R. Stratospherical D-1 D-3 based and of Properties New Safe Table through Fluid—Refriger Working ASHRAE 134a,” Trans., equations are 2095-2118. aon 1988, 2, Pt. 94, Vol. pp.

Appendix D_

Table D-|_

Properties of R134a

(Continued) Specific Volume

Enthalpy kJ/kg

Internal Energy

m?/kg Pressure

kPa

252.6 1385.1 WZ 7S 1681.3 PVNO2 2632.4 3243.5 3974.2

Sat. Liquid v, X 10°

Sat. Vapor

0.8989 0.9142 0.9308 0.9488 1.0027 1.0766 1.1949 1.5443

0.0159 0.0142 0.0127 0.0114 0.0086 0.0064 0.0046 0.0027

Sat.

Vapor

°%

Liquid hy

DNS) 253.55 252,23 256.81 260.15 262.14 261.34 248.49

138.35 124.58 130.93 137.42 154.34 172.71 193.69 224.74

Evap. h fg

1593}-335) 148.66 143.75 138.57 124.08 106.41 82.63 34.40

Entropy kJ/kg-K Sat. Vapor

The Appendices Table D-2.

Saturated R134a—Pressure Table Specific Volume

_ Internal Energy

m?/kg

kJ/kg

Enthalpy

Entropy

kJ/kg

kJ/kg:K

Pressure

Temp.

Sat. Liquid

kPa

Sat. Vapor

5G

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v, X 10°

Sat. Vapor

Vy

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u;

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h,

hey

h,

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Si;

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S122

OSI

OOIZi

IS252

P2500

(S402

14.37

AY

1800

6229) es 09631

OOS

O49

Asyesty

AW Z2

Ws)

1000 1200

.

Ss

OAWA

Osis

A733}

Osa!

Wrsssy

2000

67.49

0.9878

0.0093

148.02

259.41

149299

127-95)

9277.94

Oro Sa

OL8984

2500

VIS

=NO5E2

OO0OS8S

10543

Aol

Iesi2

Wis

27.1/

Ose

Ostia

3000

86,22

Viale

O@OSs

Weilets

Aoraits

iiss.s0

278.01

0.6156

0.8735

QW \

D-1 Table P. D. D-3 through from Properties “Thermodynami Basu, S. R. and Wilson based of Stratospherical New Safe equations Fluid—Refriger Working 134a,” are on a ASHRAE Trans., 1988, 2, Pt. 94, Vol. 2095-2118 pp.

Appendix D_

Table D-3

Properties of R134a

Superheated R134a

m?/kg

u

h

s

v

u

h

s

kJ/kg

kJ/kg

kJ/kg:K

m3/kg

kJ/kg

kJ/kg

kJ/kg-K

0.9520 1.0062 1.0371 1.0675 1.0973 1.1267 1.1557 1.1844 1.2126 1.2405 1.2681 1.2954 1.3224

0.19170 0.19770 0.20686 0.21587 0.22473 0.23349 0.24216 0.25076 0.25930. 0.26779 0.27623 0.28464 0.29302

P = 0.06 MPa (—37.07°C) Sat. -20 -10 0 10 ‘ 20 30 40 50 60 70 80 90

0.31003 0.33536 0.34992 0.36433 0.37861 0.39279 0.40688 0.42091 0.43487 0.44879 0.46266 0.47650 0.49031

206.12 217.86 224.97 232.24 239.69 247.32 255.12 263.10 271.25 279.58 288.08 296.75 305.58

DAT 2 237.98 245.96 254.10 262.41 270.89 279.53 288.35 297.34 306.51 315.84 325.34 335.00

P = 0.10 MPa (—26.43°C)

P = 0.14 MPa (—18.80°C) Sat. —10 0 10 20 30 40 50 60 70 80 90 100

0.13945 0.14519 0.15219 0.15875 0.16520 0.17155 0.17783 0.18404 0.19020 0.19633 0.20241 0.20846 0.21449

216.52 223.03 230.55 238.21 246.01 253.96 262.06 270.32 278.74 287,32 296.06 304.95 314.01

236.04 243.40 251.86 260.43 269.13 277.97 286.96 296.09 305.37 314.80 324.39 334.14 344.04

212.18 216.77 224.01 231.41 238.96 246.67 254.54 262.58 270.79 279.16 287.70 296.40 305.27

231.35 236.54 244.70 252.99 261.43 270.02 278.76 287.66 296.72 305.94 315.32 324.87 334.57

0.9395 0.9602" 0.9918 1.0227 WOSE I 1.0829 1A 22 1a4it 1.1696 1.1977 1.2254 12528 1.2799

P = 0.18 MPa (—12.73°C) 0.9322 0.9606 0.9922 1.0230 1.0532 1.0828 T1120 1.1407 1.1690 1.1969 1.2244 1.2516 1.2785

0.10983 0.11135 0.11678 0.12207 0.12723 0.13230 0.13730 0.14222 0.14710 0.15193 0.15672 0.16148 0.16622

P = 0.20 MPa (—10.09°C)

219.94 222.02 229.67 237.44 245.33 253.36 261.53 269.85 278.31 286.93 295.71 304.63 313.72

239.71 242.06 250.69 259.41 268.23 DTA, 286.24 295.45 304.79 314.28 323.92 333.70 343.63

i) |§ 5 te a = Ve |¢ a |% 12 z 3

0.9273 0.9362 0.9684 0.9998 1.0304 1.0604 1.0898 1.1187. 1.1472 1.1753 1.2030 1.2303 1.2573

P=.0,24\MPaib=5.37°C)

= |= 2 |2 |= |@ y |2 | 12 | :

3

Sat. —10 0 10 20 30 40 50 60 70 80 90

0.09933 0.09938 0.10438 *Os10922 ~=0.11394 0.11856 0.12311 0.12758 0.13201 0.13639 0.14073 0.14504

221.43 221.50 229.23 237.05 244.99 253.06 261.26 269.61 27EA0 286.74 295.53 304.47

241.30 241.38 250.10 258.89 267.78 276.77 285.88 295.12 304.50 314.02 323.68 333.48

0.9253 0.9256 0.9582 0.9898 1.0206 1.0508 1.0804 1.1094 1.1380 1.1661 1.1939 ipod

0.08343

224.07

244.09

0.08574 0.08993 0.09399 0.09794 0.10181 0.10562 0.10937 0.1130A 0.11674 0.12037

228.31 236.26 244.30 252.45 260.72 269.12 277.67 286.35 295.18 304.15

248.89 P5184 266.85 275.95 285.16 294.47 303.91 313.49 323.19 333.04

| Z 0.9399 g 09721045 1.0034 |§ 1.0339 |& 1.0637 |z OSsOe |= 1A DT en 1.1501 : iat7eGe 4S has |e

100

0.14932

3113.57)

343.43

1.2483

0.12398

Sie.

343.03

1.2326

0.9222

The Appendices Table D-3_

537

(Continued)

v

u

h

s

v

u

h

s

m3/kg

kJ/kg

kJ/kg

kJ/kg-K

m?/kg

kJ/kg

kJ/kg

kJ/kg:K

0.9197 0.9238 0.9566 0.9883 1.0192 1.0494 1.0789 1.1079 1.1364 1.1644 1.1920 1.2193 1.2461 1.2727

0.06322

P = 0.28 MPa (—1.23°C) Sat. 0.07193 226.38 246.52 Oi, BOO7240: 2227.37 247.64 10 0.07613 235.44 256.76 20 0.07972 243.59 265.91 30 0.08320 251.83 SEND 40 0.08660 260.17 284.42 50 0.08992 268.64 293.81 60 0.09319 277.23 303.32 70 0.09641 285.96 312.95 80 0.09960 294.82 322.71 90 0.10275 303.83 332.60 100)’ |0.10587 —=°312.98 342.62 Win. 1010897" ©< 322.27 352.78 O20: ee 23371 363.08 P = 0.40 MPa (8.93°C) Sat. 0.05089 231.97 252 32 10 0.05119) 232.87 253.35 20 "0.058975 241.37 262.96 30 0.05662 249.89 272.54 40 0.05917 = 258.47 282.14 50 0.06164 267.13 291.79 60 0.06405 275.89 301.51 70 0.06641 284.75 311.32 80 0.06873 293.73 321.23 90 0.07102 302.84 331.25 100M) 10:07327 312.07 341.38 TOm ¥0:07550 321.44 351.64 120. 0.07771 330.94 362.03 130 0.07991 340.58 372.54 140 0.08208 350.35 383.18 P = 0.60 MPa (21.58°C) eee 0.03408 238.74 259.19 30 ~=©0.03581 246.41 267.89 40 0.03774 255.45 278.09 50 0.03958 264.48 288.23 298.35 273.54 0.04134 60 308.48 282.66 0.04304 70 318.67 291.86 0.04469 80 328.93 301.14 0.04631 90 339.27 310.53 0.04790 100 349.70 320.03 0.04946 110 360.24 329.64 0.05099 120 370.88 339.38 0.05251 130 381.64 349.23 0.05402 140 392.52 359.21 0.05550 150 403.51 369.32 0.05698 160

0.06576 0.06901 0.07214 0.07518 0.07815 0.08106 0.08392 0.08674 0.08953 0.09229 0.09503 0.09774

0.9145 0.9182 0.9515 0.9837 1.0148 1.0452 1.0748 1.1038 1.1322 1.1602 1.1878 1.2149 1.2417 1.2681 1.2941

0.04086 0.04188 0.04416 0.04633 0.04842 0.05043 0.05240 0.05432 0.05620 0.05805 0.05988 0.06168 0.06347 0.06524

0.9097 0.9388 0.9719 1.0037 1.0346 1.0645 1.0938 (11225 1.1505 1.1781 1.2053 1.2320 1.2584 1.2844 1.3100

0.02918 0.02979 0.03157 0.03324 0.03482 0.03634 0.03781 0.03924 0.04064 0.04201 0.04335 0.04468 0.04599 0.04729 0.04857

P = 0.32 MPa (2.48°C) «40228:43 248.66 234.61 255.65 242.87 264.95 251.19 274.28 259.61 283.67 268.14 293.15 276.79 302.72 285.56 312.41 294.46 32222 = 303.50 332.15 312.68 342.21 322.00 352.40 331.45 362.73 P = 0.50 MPa (15.74°C) 235.64 256.07 . 239.40 260.34 248.20 270.28 256.99 280.16 265.83 290.04 =.274.73 299.95 283.72 309.92 292.80 319.96 302.00 330.10 311.31 340.33 320.74 350.68 330.30 361.14 339.98 371.72 349.79 382.42 P = 0.70 MPa (26.72°C) 241.42 261.85 244.51 265.37 -. 253.83 275.93 263.08 286.35 296.69 272.31 307.01 281.57 31735 290.88 327.74 300.27 338.19 309.74 348.71 319.31 359.33 328.98 370.04 338.76 380.86 348.66 391.79 358.68 402.82 368.82

0.917 0.942 0.974 1.006 1.036 1.066 1.095 1.124 {52 1.180 1.207 1.234 1.261 0.911 0.926 0.959 0.991 (2022 1.053 1.082 sea 1.139 1.167 1.194 1.224 1.248 wae! eal 0.9080 0.9197 0.9539 0.9867 1.0182 1.0487 1.0784 1.1074 1.1358 1.1637 1910 2179 2444 1.2706 2963 (Continued )

Appendix D_

Table D-3

Properties of R134a

(Continued)

u kJ/kg

m?/kg

h

kJ/kg

s kJ/kg:K

v m?/kg

u kJ/kg

h kJ/kg

P = 0.90 MPa (35.53°C)

P = 0.80 MPa (31.33°C)

245.88 250132 260.09 AOS) WA puIf'S), 3X0) 288.87 298.46 308.11 S32 327.62 SB pov 347.51 357/-0i 367.82 378.14 388.57

266.18 AY \\JAD 282.34 293.21 303.94 314.62 329.20 335.96 346.68 357.47 368.33 BUSY 390.31 401.44 412.68 424.02

Sat.

0.02547

243.78

264.15

0.9066

OLO2Z2.55

40

0.02691

MSD MSs

273.66

0.9374

OL02525

50

0.02846

261.62

284.39

0.9711

0.02472

60

0.02992

271.04

294.98

1.0034

0.02609

70

O.OS131

280.45

305,50

1.0345

0.02738

80

0.03264

289.89

316.00

1.0647

0.02861

90

OLOB895

299/37,

826.52

1.0940

0.02980

100

O@ssi|9)

308.93

337.08

LAZZY

0.03095

110

0.03642

B35)

SLT |

1.1508

0.03207

(ZO RN OLOS7/62

S285

358.40

1.1784

0.03316

130

=6.0.03881

338.14

369.19

| 2055

0.03423

140°

0.03997

348.09

380.07

R232]

0.03529

15Q

O43

BS 1/5

391.05

1.2584

0.03633

160

0.04227

368.32

402.14

1.2843

0.03736

170

0.04340

378.61

AN 3} 3}3)

1.3098

0.03838

180

0.04452

389.02

424.63

1ESS5ul

0.03939

0.01663

25:1 08

2HO9

254.98 265.42 275.59 285.62 295.59 305.54 315.50 325.51 335.58 345.73 355.95 366.27 376.69 387.21

275.52 287.44 298.96 310.24 321.39 332.47 343.52 354.58 365.68 376.83 388.04 399.33 41070 422.16

P = 1.20 MPa (46.32°C)

P = 1.00 MPa (39.39°C) Sat.

0.02020

247.77

267-97,

0.9043

40

0.02029

248.39

268.68

0.9066

50

0.02171

258.48

280.19

0.9428

0.01712

GON

OL0230)

268.35

BS\| SS

0.9768

0.01835

70

~=—0.02423

DUS A

302.34

1.0093

0.01947

80

=©0.02538

287.82

313.20

1.0405

0.02051

90

0.02649

DSSS:

324.01

1.0707

OZ

(CQ

OOL755

307.27

334.82

1.1000

0.02244

110

0.02858

317.06

SAD FOS)

1.1286

0,02335

120

0.02959

326.93

350,57

il.USO/

0.02423

e,

130

~=©0.03058

336.88

367.46

1.1841

0.02508

140

0.03154

346.92

378.46

le Zalelel

0.02592

150

OOB2S0

357.06

389.56

23S

0.02674

160

0.03344

3073}

400.74

1.2638

0.02754

170

0.03436

377.66

412.02

1.2895

0.02834

0.03528

388.12

423.40

1.3149

0.02912

Si

kJ/kg-K

The Appendices \

Table D-3_ T,' °G

(Continued) v

u

h

Ss

m?/kg

kJ/kg

kJ/kg

kJ/kg-K

v m?/kg

P = 1.40 MPa (52.43°C)_

u kJ/kg

h kJ/kg

S kJ/kg-K

P = 1.60 MPa (57.92°C)

Sat. 0.01405 60 0.01495 FOR 001603 80 0.01701 Br 0701792 100 0.01878 110 0.01960 420° - 0.02039 150nr 0.02015 140 0.02189 150), 0.02262 160 0:02333

253.74 262.17 272.87 283.29 293155 303773 31388 324.05 B3Ap25 344.50 354.82 365.22

273.40 283.10 295.31 307-10 318.63 330.02 341.32 Bb 2259 363.86 325.15 386.49 397.89

170

0.02403

BDI |

409.36

1.2576

1805 190)

| 0.02472 0.025411

386.29 396.96

420.90 432.53

1.2834 1.3088

200

0.02608

407.73

444.24

1.3338

0.02263

0.9003 0.9297 0.9658 0.9997 1.0319 1.0628 1.0927 e218 1.1501 WAT AZ 1.2048 e235

0.01208 0.01233 0.01340 0.01435 0.01521 0.01601 0.01677 0.01750 0.01820 0.01887 0.01953 0.02017

256.00 258.48 269.89 280.78 291.39 301.84 312,20 322.53 832.87, 343.24 353.66 364.15

Zt 535 278.20 291338 303.74 31572 327.46 339.04 350:53 361.99 373.44 384.91 396.43

0.02080

374.71

407.99

1.2445

0.02142 0.02203

385.35 396.08

419.62 431.33

1.2704 1.2960

406.90

443.11

S212

0.8982 0.9069 0.9457 0.9813 1.0148 1.0467 10773 1.1069 131357 1.1638 1.1912 1.2181

Appendix D_

Table D-1E

Properties of R134a

Properties of Saturated R134a—Temperature Table Specific Volume

ft?/Ibm Pressure

Liquid

Sat. Vapor

psia

Uf

%

7.490 9.920 12.949 14.718 16.674 18.831 | AOS} 23.805 26.651 29.756 33). BY 36.809 40.788 49.738 60.125 HT2A0 97. 85.788 101.37 OGEOZ. 118.99 128.62 138.83 149.63 161.04 I73NC 185.82

0.01130 0.01143 0.01156 0.01163 0.01170 0.01178 0.01185 OLONDs 0.01200 0.01208 0.01216 0.01225 0.01233 0.01251 0.01270 0.01290 0.01311 0.01334 0.01346 0.01358 0.01371 0.01385 OHONSIS 0.01414 0.01429 0.01445 0.01520 0.01617 0.01758 0.02014 0.02329

DA WHS 4.3911 3.4173 3.0286 2.6918 2.392 2.1440 1.9208 (|.JZ51. iRoS29) 1.4009 1.2666 1.1474 0.9470 0.7871 0.6584 0.5538 0.4682 0.4312 0.3975 0.3668 0.3388 O.3131

Sat.

100 105 110 (is 120 140 160 180 200 210

243.86 314.63 400.22 OBZ OS)

0.2896 0.2680 0.2481 0.1827 0.1341 0.0964 0.0647 0.0476

Enthalpy Btu/Ibm

Internal Energy Btu/Ibm Sat. Liquid

Sat. Vapor

Sat. Liquid

Uy

hy

=0:02

87.90

2.81

89.26

5.69 7A4 8.61 10.09 11.58 13.09 14.60 1Gal3 17.67 19.22 20.78 23.94 2714

90.62 91.30 91.98 92.66 0333 94.01 94.68 95.35 96.02 96.69 97.35 98.67 99.98

310). 39

Oey

33.68 1102/54 3/02 03.78 36:72 ——1104739 40.42 105.00 42.14 105.60 43.87 106.18 45.62 106.76 A739 ee lO7. 38 49.17 107.88 510.97

108.42

58.39 110.41 66:26 5 weit 1.97

0.00 2 IBS) D1 TeMd 8.65 10.13 nikGS 13.14 14.66 16.20 17.74 19.30 20.87 24.05 DY XS 30.56 33.89 327 38.99 40.72 42.47 44.23 46.01 47.81

Evap. h fg

95.82 94.49 93). 92.38 91.64 90.89 90.12 89.33 88.53 Sie/l 86.87 86.02 85.14 83.34 81.46 79.49 7744 UDLE 74.17 F303 71.86 70.66 69.42 68.15

Entropy Btu/Ibm-°R

Sat. Vapor hg

Sat. Liquid S,

5.87 O7reZ 98,81 J) 55 100.29 101.02 ONES 102.47 OSRI9 103.90 104.61 (05,32 106.01 107.39 108.74 110.05 Miles 112.56 113.16 Dts 114.33

0.0000 0.2283 0.0067 0.2266 010133 0.2250 0.0166 0.2243 0.0199 0.2236 0:0231 0:2230 0.0264 0.2224 0.0296 0.2219 0.0329 0.2214 0.0361 0.2209 0.0393 0.2205 0.0426 0.2200 0.0458 0.2196 0.0522 0.2189 0.0585 0.2183 0.0648 0.2178 0.0711 02173 0.0774 0.2169 0.0805 0.2167 0.0836 0.2165 0.0867 0.2163 0.0898 0.2161 0.0930 0.2159 0.0961 0.2157 0.0992 0.2155 Ox'023, O7153 O50 30.2143 0.1280 0.2128 CMATLZ: BOrAlOn Oi575. - 052044) 0.1684 0.1971

114.89 115.43 15.9

49.63 5.47 59.08

66.84 65.48 3S) 5/

116.47 11695 118.65

67.20

52.58 43.78 S092 IO.

119.78 19:95) MAIZE SS: 113.45

74.83

W277

84.90

111.66

HONS 86.77

91.84

108.48

94.27

Sat. Vapor 5

S. R. and Wilson P. D. from equations of Properties “Thermodyna Basu, Fluid—Refrige Working Safe Stratospherica New D-iE Table D-3E based ASHRAE through 134a,” Trans., are 2095-2118 aon 1988, 2, Pt. 94, Vol. pp.

The Appendices Table D-2E

Properties of Saturated R134a—Pressure Table Specific Volume

_ Internal Energy

ft?/Ibm

Btu/Ibm

Enthalpy

Entropy

Btu/Ibm

Btu/Ibm-°R —————=

Pressure

Temp

Sat. Liquid

psia

SF

Vs

Sat. Vapor

Sat. Liquid

Sat. Vapor

Sat. Liquid

Vg

Uy

U,

h, 35/5 2.91 TAO 10:89" 1624 20:5) 2A NA 27 20 30.01 3253.)

8.3508 "4.3584 "2.9747 0.0118 —2AS “22601 1.5408 15.38 0.01209 29-047 10.012329"1.1692— AQ 770.01 2522770,94225 ) sOseST= © AD S9 2.0 01270 95,50 ="0.01286)0'6778 > 65.958 10:01 302)0' 59333

Sat. Vapor

hig

ne

S;

S;

-3.74 SOF 2.89 89.30 7.36 91.40 93.00 10.84 95.40 16.24 20:48 97.23 24 0? 98.71 227.10 99.96 101205 829.85" 9632.53 = 102 02

97253 94.45 92:27 90.50) 87.65 85°31 G329°* SB VAS 79.82. 1S: 256

93:79" 97.37 99.66 101289 103.96 105.88 1107-43 0872 109.83 i038 |

= 0:0090 0.0068 0.0171 0.0248 0.0364 0.0452 030523) 0.0584 0.0638 0.0686

0:23 11 0.2265 0.2242 0.2227 0.2209 0.2197 {02189 0.2183 0.2179 0.2175

Opa27e

34.62

102.89

34.84

76.84

111.68

O:0/29

Or AW2

OLS 2. “OAV

3657/5

103.68

36.99

YL

112.46

0.0768

0.2169 0.2165

VAS = OCIS WIN

Sat. Liquid

a

—53.48 0.01113 2971 O01 (AS =l47550:01164

100 120 140 160 180 200 220 240 260 280 300 350 400 450 500

Evap.

Sat. Vapor

90:54

0.01360

0.3941

40.61

105.06

40.91

72.94

113.82

0.0839

100.56

0.01386

0.3358

44.07

106.25

44 43

LOPS?

114.95

0.0902

0.2161

09°56

10014112.

0.29116

47.23

107.28

47.65

68.26

tilssh

OOS:

OL7i'57/

72 ZAO!O1 433

502569

50.16

108.18

50.64

66.10

116.74

0.1009

0.2154

125.23

OWI46s

Ores

52.90

108.98

53.44

64.01

117.44

0.1057

0.2151

132.277

40,01489'"90.2056

D548)

= 10916S—)

56.09

9561.96—=)

4113.05

0.1101

0.2147

1ISS.79

OOS:

Wyle;

Dyes

110.30

58.61

59) SIE

118.56

0.1142

0.2144

1447

1001 S41

"0:1695e

"960.28 ~ 110/84,

61.02

57.97.

118.99

0.1181

0.2140

(5O:7O

OO

Oxis0)

O23

US

63.34

56.00

WS), 335

OAS

OAIss

156.17

0.01596

0.1424

64.71

WAZ

65.59

54.03

| Sie

OnI2Z54

Or ils2

1GS.72

OOM)

Wi iKexe

69.88

112.45

TAOS)

49.03

120.00

0.1338

0.2118

22S"

OWl/ss

OOsas

74.81

W277

76.11

43.80

119.91

ONAN

~~ OAIO#

190.12

0.01863

0.0800

USISS

112.60

81.18

38.08

119.26

On493 5 0:2079

199.38

0.02002

0.0657

84.54

1UAAS

86.39

31.44

WAS:

0.1570

xs:

0.2047

from D-1E Table D-3E through based Wilson P. D. S. R. and Properties “Thermodynami Basu, of Stratospherical New equations Safe Working Fluid—Refriger are on 134a,” a ASHRAE Trans., Pt. 94, Vol. 1988, 2, 2095-2118 pp.

sea

§=6Appendix D_

Table D-3E__ T,

Properties of R134a

Superheated R134a Vapor v

u

h

Ss

v

u

h

s

ft?/Ibm

Btu/Ibm

Btu/Ibm

Btu/Ibm-°R

ft?/Ibm

Btu/Ibm

Btu/Ibm

Btu/Ibm-°R

P= 10 psia (—29.71°F)

P= 5 ose: (142579)

Sat.

4.3581

89.30

WES!

0.2265

2.9747

91.40

99.66

0.2242

—20 0 20 40 60 80 100 120 140 160 180 200

4.4718 4.7026 4.9297 5,1 538 5.31 e 5.5959) 5.8145 6.0318 6.2482 6.4638 6.6786 6.8929

90.89 94.24 9767 ON 104.80 108.50 1122 116.18 120.16 124.23 128.38 132.63

SI) 7/ 102.94 106.79 ORs. 114.74 118.85 [23105 127.34 131.72 139, NS 140.74 145.39

0.2307 0.2391 0.2472 O25535 0.2632 0.2709 0.2786 0.2861 O2E35 0.3009 0.3081 ORlSZ

3.0893 3.2468 3.4012 SID 555) 3.7034 3.8520 5998 4.1456 4.2911 4.4359 4.5801

93.84 97.28 100.89 104.54 108.28 i240 116.01 120.00 124.09 128.26 is2252

102.42 106.34 110.33 114.40 118.56 12279 274 (eS 136.00 140.57 145.23

0.2303 0.2386 0.2468 0.2548 0.2626 0.2703 02778. 0.2854 0.2927 0.3000 0.3072

Sat. 0 20

2.2661 2.2816 2.4046

93.00 93.43 96.98

lO TSS 101.88 105.88

0.2227 0.2238 0.2323

1.5408

95.40

103.96

0.2209

1.5611

96.26

104.92

0.2229

40 60 80 100 120 140 160 180 200 220

2.5244 2.6416 DH XSS, 2.8705 2a 829 3.0942 3.2047 3.3144 3.4236 3,323

100.59 104.28 108.05 111.90 115.83 ORS 123),8)5 128.13 132.40 136.76

109.94 114.06 125 (22,352 126.87 (oe 135.81 140.40 145.07 149.83

0.2406 0.2487 0.2566 0.2644 0.2720 0.2795 0.2869 O29 27 0.3014 0.3085

1.6465 (R293 1.8098 1.8887 1.9662 2.0426

99.98 103.75 107.59 111.49 115.47 WEDS:

(I@Q12 W335 IWA6S IZESS |Zo.3¢ 130.87

0.2315 0.2398 0.2478 0.2558 0.2635 OZ 7aal

PAA 2|\SPE. A ASI | 2.3407

123.66 127.88 S257, (36:55

135.42 140.05 144.76 149.54

0.2786 0.2859 0.2932 0.3003

P = 20 psia (—2.48°F)

P = 30 psia (15.38°F)

P = 40 psia (29.04°F)

P = 50 psia (40.27°F)

Sat.

1.1692

105.88 108.26

WAIST 0.2245

98.71

107.43

0.2189

1.2065 (2728:

Yi) 23 Shee) (3,20

0.9422

40 60

2102

O233)|

0.9974

102.62

ess

0.2276

80 100 120

B57 eS OV 1.4575

WOW 111.08 i Set

117.00 121.42 125.90

0.2414 0.2494 O5/8

1.0508

106.62

116.34

0.2361

O22 1520

110.65 114.74

120.85 125339

0.2443 O.AS23

140— 160 180

{|SN@S 1.5746 1.6319

WIZ (233,358) 127.62

130.43 135.03 1599/0

0.2650 O25 0.2799

1.2007 1.2484 122953

118.88 123.08 127.36

129,98 134.64 139.34

0.2601 0.2677 0.2752

200 220

1.6887 1.7449

TSit94 136.34

144.44 149.25

OST Z 0.2944

1.3415

Ista

144.12

0.2825

1.3873

VO, 12

148.96

0.2897

240

1.8006

140.81

154.14

OOS

1.4326

140.61

VS337

OPASKSe)

260 280

1.8561 (SZ

145.36 149.98

SS), (Ke 164.13

0.3085 6.3154

1.4775 a2

145.18 149.82

158.85 163.90

0.3039 0.3108

of Properties “Thermodyna Basu, S. ASHRAE 134a,” Fluid—Refrig Working Safe Stratospheric New R. and Wilson P. D. from Vol. Trans., Pt. D-3E equations 94, 1988, 2, 2095-2118. through D-1E Table based app. on are

The Appendices Table D-3E

(Continued) v ft/lbm

u Btu/Ibm

h Btu/Ibm

Ss Btu/Ibm-°R

v ft?/Ibm

P = 60 psia (49.89°F)

u Btu/Ibm

h Btu/Ibm

s Btu/Ibm-°R

P = 70 psia (58.35°F)

Sat.

0.7887

99.96

108.72

0.2183

0.6778

101.05

109.83

0.2179

60

0.8135

102.03

111.06

0.2229

0.6814

101.40

1)NO2s:

0.2186

80

0.8604

106.11

115.66

0.2316

0.7239

105.58

114.96

0.2276

100

0.9051

WO

120526

0.2399

0.7640

109.76

120

0.9482

114.35

124.88

0.2480

0.8023

113.96

119.66 124.36

0.2361 0.2444

140

6.9900

118.54

W2O 5s

0.2559

0.8393

118.20

160

1.0308

122.79

(B4e25

0.2636

0.8752

122.49

WACHOV, 133.82

0.2524 0.2601

1

180

1.0707

NZI AG)

138.98

OAT WZ

0.9103

126.83

138.62

0.2678

200

1.1100

131.47

143.79

0.2786

220

1.1488

135.91

148.66

0.2859

0.9446 0.9784

1 B4)233 135.69

143.46 148.36

0.2752 0.2825

240

1.1871

140.42

153.60

0.2930

1.0118

140.22

(5Ss55

0.2897

260

e225

145.00

158.60

0.3001

1.0448

144.82

158.35

0.2968

280

12627,

149.65

163.67

0.3070

1.0774

149.48

163.44

0.3038

300

1.3001

154.38

168.81

Omiss

1.1098

154.22

168.60

OSOH

P = 80 psia (65.93°F)

P = 90 psia (72.83°F)

Sat.

0.5938

102.02

110.81

OAS

0.5278

102.89

iMees

OPW

80

0.6211

105.03

114.23

0.2239

0.5408

104.46

iS:47,

0.2205

100

0.6579

109.30

119.04

0). 2327

0.5751

108.82

iiS239

0.2295

120

0.6927

SHES

S282.

0.2411

0.6073

S345)

(23.27

0.2380

140

0.7261

MESS

128.60

0.2492

0.6380

117.50

2S.

0.2463

160

0.7584

122.18

eer

0.2570

0.6675

LAWNS

132.98

0.2542

0.2620

180

0.7898

126.55

NSS-25

0.2647

0.6961

126.28

WS

200

0.8205

130.98

143.13

O2722.

0.7239

NOs:

142.79

220

0.8506

(85:47)

148.06

0.2796

0). /5u2

135.25)

147.76

O27

240

0.8803

140.02

5305

0.2868

0.7779

139.82

(S277

0.2843

260

0.9095

144.63

158.10

0.2940

0.8043

144.45

157.84

0.2914

280

0.9384

149.32

1O37 1

0.3010

0.8303

149.15

162.97

0.2984

300

0.9671

154.06

168.38

0.3079

0.8561

153.91

168.16

0.3054

320

0.9955

158.88

173.62

OsT4y7

0.8816

158278

7 32u

0.3122

P = 100 psia (79.17°F)

P=

Sat.

0.4747

103.68

112.46

0.2169

80

0.4761

103.87

112.68

0.2173

100

0.5086

108.32

UULYS

0.2265

120 140

160

180

0.5388 0.5674

0.5947

0.6210

(ZW UWS

(ASS

125.99

122.70 127.63

132255

137.49

OM2352 0.2436

OAS

7

0.2595

ASST)

0.2696

120 psia (90.54°F)

0.3941

105.06

Sey

0.2165

0.4080

107.26

NG.

0.2210

111.84

SZ.

0.2301

WO. 37/

126.61

0.2387

120.89

131.66

0.2470

0.5082

125.42

129.97

136.70 ARNE TES)

0.2550 0.2628

OsS5 0.4610

0.4852

200

0.6466

130.48

142.45

0.2671

0.5305

220

0.6716

135.02

(AYES

0.2746

05520

134.56

146.82

0.2704

OPS 73

139.20

ales

0.2778

0.2891

0.5937

143.89

0.2962

0.6140

148.63

le 720y 62.26

0.2850 0.2921

0.3031

0.6339

153.43

67.51

0.2991

0.6537

158.29

Wiel

0.3060

240 260

0.6960 0.7201

139.61 144.26

152.49 5/259 7A:

280

0.7438

148.98

GZ

300

0.7672

Seis

167.95

320

0.7904

158.59

IAS

0.2819

0.3099

(Continued )

Appendix D_

Table D-3E

Properties of R134a

(Continued)

ie

v

u

h

s

v

u

h

s

oF

ft?/Ibm

Btu/Ibm

Btu/Ibm

Btu/Ibm-°R

ft?/Ibm

Btu/Ibm

Btu/Ibm_

Btu/Ibm-°R

P = 140 psia (100.6°F)

P = 160 psia (109.6°F)

Sat.

0.3358

106.25

114.95

0.2161

0.2916

107.28

115.91

02157

120

0.3610

110.90

120.25

0.2254

0.3044

109.88

118.89

0.2209 0.2303

140

0.3846

Sexe

125.54

0.2344

0.3269

AAS

124.41

160

0.4066

120

130.74

0.2429

0.3474

119.49

129.78

0.2391

180

0.4274

124.82

135.89

OZ5iIT

0.3666

124.20

135.06

0.2475 0.2555

200

0.4474

129.44

141.03

0.2590

0.3849

128.90

140.29

220

0.4666

134.09

146.18

0.2667

0.4023

13326

145252

0.2633

240

0.4852

138.77

Si s4

0.2742

0.4192

138.34

150.75

0.2709

260

0.5034

143.50

156.54

0.2815

0.4356

(4S ai

156.00

0.2783

280

ORS 7a

148.28

161.78

0.2887

0.4516

147.92

161.29

0.2856

300

0.5387

Sse

167.06

0.2957

0.4672

152.78

166.61

0.2927

320

Oi559

157.99

172.39

0.3026

0.4826

157.69

171.98

0.2996

340

0.5730

162.93

Ween

0.3094

0.4978

162.65

(WELSS,

0.3065

360

0.5898

167.93

183.21

0.3162

0.5128

167.67

182.85

GO: 31382

P = 180 psia (117.7°F)

P = 200 psia (125.3°F)

Sat.

0.2569

108.18

116.74

0.2154

120

0.2595

108.77

117.41

0.2166

0.2288

108-98

117.44

02151

140

0.2814

WSs}

1286924

160

0.3011

118.74

128.77

0.2264

0.2446

112.87

121.92

0.2226

O23855

0.2636

117.94

127.70

180

0.3191

23°56

023271

134.19

0.2441

0.2809

122.88

13328

200

0.3361

0.2410

128.34

139.53

0.2524

0.2970

12976

138875

220

0.2494

Om523

Issel

144.84

0.2603

0.3121

132.60

144.15

0.2575

240

0.3678

137.90

USO

0.2680

0.3266

137.44

149.53

0.2653

260

0.3828

ea

155.46

0.2755

0.3405

142.30

154.90

0.2728

280

0.3974

A725

160.79

0.2828

0.3540

147.18

160.28

0.2802

300

0.4116

152.44

166.15

0.2899

0.3671

152.10

165.69

0.2874

320

0.4256

7

ales

0.2969

0.3799

157-07

iis

0.2945

340

0.4393

162.36

177.00

0.3038

0.3926

162.07

176.60

0.3014

360

0.4529

167.40

182.49

0.3106

0.4050

levels

1S2e12

0.3082

Sat.

0.1424

Mae:

119.62

OP132

0.0965

qs

160

0.1462

112.95

ZeOW

O2155

le

sks}

Ss

P = 300 psia (156.2°F)

P = 400 psia (179.9°F) 7/9/

119.91

0.2102 0.2102

180

On6s3

118.93

128.00

0.2265

0.0965

112.79

119.93

200

Ona

124.47

134.34

0.2363

0.1143

120.14

128.60

02735

220

0.1905

129.79

140.36

0.2453

OTS

126.35

135.79

0.2343

240

0.2021

134.99

146.21

0.2537

0.1386

(SPL

142.38

0.2438

260

0.2130

140.12

151.95

0.2618

0.1484

137.65

148.64

0.2527

280

0.2234

145.23

572638

0.2696

0.1575

143.06

15472.

0.2610

300

O23 33)

510),S33!

163.28

OP

0.1660

148.39

160.67

0.2689

320

0.2428

155.44

168.92

0.2845

0.1740

153.69

166.57

0.2766

340

O57

OO

57

174.56

0.2916

0.1816

158.97

172.42

0.2840

360

0.2611

165.74

180.23

0.2986

0.1890

164.26

178.26

0.2912

380

0.2699

170.94

185.92

OS055

0.1962

169.57

184.09

0.2983

400

0.2786

176.18

191.64

0.3122

0.2032

174.90

189.94

0.3051

Table E-I

Saturated Ammonia

Specific volume

Enthalpy

m?/kg

kJ/kg

Ut

Vg

0.001424 S15) 57.69

Entropy

kJ/kg.K

h,

hig

Stg

Sg

2.6254

= AVAL}

1416.7

e724:

—011942

6.3502

6.1561

0.001434

2.1140

= 26.0

1405.8

le7S2

Om 156

6.1902

6,0746

0.001439

1.9032

= 73

1400.3

138235

—0.0768

6.1120

64.42

0.001444

1.7170

15 p8)

(Bay

1385.8

—0.0382

6.0349

6.0352 5.9967

Witeita

0.001449

ipo oZal

0.0

1389.0

1389.0

0.0000

5) 589)

5.9589

79.80

0.001454

1.4085

8.9

13333

|3Q2Z2

0.0380

5.8840

5:20)

88.54

0.001460

2st

17.8

1377.6

S954.

0.0757

5.8101

5.8858

98.05

0.001465

5 SY,

26.8

1371.8

1398.5

ON s7

SyisVZ

5.8504

h,

S¢

108.37

0.001470

1.0562

S15)

1365.9

1401.6

0.1504

5.6652

5:8156

SESS

0.001476

0.9635

447

1360.0

1404.6

0.1873

5.5942

Asi be)

131.64

0.001481

0.8805

Si

1354.0

1407.6

0.2240

5.5241

5.7481

144.70

0.001487

0.8059

62.6

1347.9

1410.5

0.2605

5.4548

‘I (SS

158.78

0.001492

0.7388

VAS

1341.8

1413.4

0.2967

5.3864

5.6831

IWS93

0.001498

0.6783

80.7

ISS o16

1416.2

OB B27

5.3188

505 I> 5.6205

190.22

0.001504

0.6237

89.7

1329.3

1419.0

0.3684

Sy AeyA0)

207.71

0.001510

0.5743

98.8

1322.9

© 14217

0.4040

5.1860

5d, DIO

226.45

0.001515

0.5296

107.8

1316.5

1424.4

0.4393

5), AO

5.5600

246.51

0.001521

0.4889

116.9

1310.0

1427.0

0.4744

5.0561

5.5305

267.95

0.001528

0.4520

126.0

OSS

142935

0.5093

49922

5.5015 5.4730

290.85

0.001534

0.4185

18592

1296.8

1432.0

0.5440

4.9290

Bi 5Z5

0.001540

0.3878

144.3

290%)

1434.4

0.5785

4.8664

5.4449

SAIeZ5

0.001546

Or5599

5535

1283.3

1436.8

0.6128

4.8045

5.4173

368.90

0.001553

0.3343

162.7

1276.4

143971

0.6469

4.7432

5.3901

398.27

0.001559

0.3109

VALS

1269.4

1441.3

0.6808

4.6825

533653

429.44

0.001566

0.2895

181.1

1262.4

1443.5

0.7145

462.49

0.001573

0.2698

190.4

25522.

1445.6

0.7481

4.6223 4.5627

IAS 5.3108

497.49

0.001580

OAS

199.6

1248.0

1447.6

0.7815

4.5037

5.2852

5S Ano

0.001587

O235i

208.9

1240.6

1449.6

0.8148

4.4451

DYE

573.64

0.001594

0.2198

218.3

2S eZ

Syl

0.8479

4.3871

Zs oU

(225m,

1453.3

0.8808

Al SVBS)S

5.2104 5.1861 5.1621

614.95

0.001601

0.2056

DRUMS BED

1218.1

1455.1

0.9136

4.2725

0:T805

246.4

1210.4

1456.8

0.9463

4.2159

0.1693

255.9

1202.6

1458.5

0.9788

A

NSIS

5elson

0.001631

0.1590

265.4

1)WSyaty

1460.0

1.0112

4.1039

Salis

0.001639

0.1494

274.9

1186.7

(AGES

1.0434

4.0486

5.0920

658.52

0.001608

0.1926

704.44

0.001616

USATS,

0.001623

803.66 857.12

(Continued)

of Tables No. Circular Standards Ammonia Properties Thermodynamic Based Bureau National from 142, data on

Appendix E

Table E-|_

T°C

Properties of Ammonia

(Continued)

P, kPa

913.27 | 24 97219) 265 1033.97.) BS NOLNS71) || 30 =1166.49 | B27 123741a Beam S008 21559,0sale 861) 1.469.925) 40 1554.33} 42 1642.35] 44 1734.09 | 46 1829.65] 258 1929: 13)|" pO 42032.62"

Specific volume

Enthalpy

Entropy

m3/kg

kJ/kg

kJ/kg.K

VF 0.001647 01001655 0.001663 OWOSA 0.001680 01001689 OLOOE9S 0.001707 0.001716 0.001726 0.001735 0.001745 0.001756 0.001766 0001777

hy ON B22 0.1245 O73 0.1106 0.1044 0.0986 0.0931 0.0880 0.0833 0.0788 0.0746 0.0707 0.0669 0.0635

Neg

h,

Sr

Sty

Ss,

3,98)3// 3.9392 3.8850

5.0467 5.0244

294.0 303.6

1170.3 1162.0

1464.3 1465.6

mROVOS 1.1075 1.1394

Sule 3279 332.6 342.3 352.1

1153.6 1145.0 1136.4 22.68" TSE

1466.8 1467.9 1469.0 146929 IZORS

ie FA 1.2028 1.2343 1.2656 1.2969

BS) 1 SVT 3.7246 3.6718 3.6192

5.0023 4.9805 4.9589 4.9374 4.9161

361.9 SYAlE 7 Bono 39). 401.5 411.5

ICO 11005 HO .2 10817 1072.0 WG22

WATS 4727 Wa 147382 1473.5 WavS7

1.3281 1.359) 1.3901 1.4209 1.4518 1.4826

3.5669 3.5148 3.4630 3.4112 3:3 9D BS 079

4.8950 4.8740 4.8530 4.8322 eo 4.7905

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The Appendices

TABLE E-1E

Saturated Ammonia

Specific Volume, ft?/Ibm Sat. Liquid

Us | | | | | | | 20) | =15 |

Diop 6.54 POF 8.95 10.41 12205 13.90 15.98 18.30 20.88

=11C

23,74)

Enthalpy, Btu/Ibm

Sat. Vapor

Vfy

10:0228 944,707 |O10229-=-38.375 |010230 =33'057 |G023.1F FF28 597 |0°02322> 24.837 721.657 |O1023389 |0.0235 18.947 |0.0236 16.636 |0.0237 14.656 | 0.02381 12.946

Vy

Sat. Liquid

Entropy, Btu/Ibm-°R

Sat. Vapor

at h,

hy

BAYT 34 a= 21,2 30136 |= 15.9 35.08 106 28.62 = 5.3 24.86 0 5,3 21.68 18.97 1037 16.66 | 16.0 14.68 | 21.4 12.97 | 26]

610.8 607.5 604.3 600.9 597.6 594.2 590.7 587.2 583.6 580.0

23916" 59120] 09377 995,06 597.6 5995 601.4 603.2 605.0 606.7

Sat. Liquid

Sat. Vapor

h, nSse

Sty

So

|=O1051 ee 125286 44769 |e OOSSOummeleS Oe lead |= Ol0250m 47 5ou 1.44907, | O002/tee 4495 = 14868 0.000 1.4242 1.4242 0101269" =1.3994 ~ 14120 813751 = “HA00T OL0250" "1.38386 0:0374" ~~17,3512 1.3774 1.3277 0:0497 1.3664 1.3044 0.0618

576.4

608.5

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IOS.

2.4939 2.520| 1148 2.2857 2.312| 1205 2.0985 2.125] 1262 1.9283 1.955| 132.0 1.7741 1.801| 1378 1.6339 1.661| 1435 1.5067 1.534| 149.4 1419/1552 1.3915 1.313 | 161.1 1.2853 oON io 1.217 1.1891 173.0 1.128] 1.0999 110186 9) 1.0471 179.0 0.973 | 185.1 0.9444

518.1

627.3

513.4 5086 503.7 4987 4936 488.5 483.2 477.8 472.3.

6282 629.1 6299 630.7 631.4 632.0 632.6 633.0 633.4 633.7 633.9 634.0 634.0

aN D o> ~S

460.9 455.0 448.9

| | | | | | | | | | | | |

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0.2322

0.9972

0.2430 0.2537 0.2643 0.2749 0.2854 0.2958 0.3062 0.3166 0.3269 0.3372 0.3474 0.3576 0.3679

0.9786 0.9603 0.9422 0.9242 0.9064 0.8888 0.8713 0.8539 0.8366 0.8194 0.8023 0.7851 0.7679

i Spy 1 S157, 1 3062

1 .2790 ] .2704 1 2535 1 2453

1 DANG 1 2140 1 .2065 1 991 1 AGS 1 .1846 1 AWS | .1705 1 OSS 1 .1566 1 .1497 1 1427

No. Circular Standards Tables Ammonia of Properties Thermodynami Bureau National 142, from data on

1 238 5aS

Appendix E_

Table E-2E

Properties of Ammonia

Superheated Ammonia Temperature, °F

P, psia

(iF)

0

20

40

v 28.58 29.90 31.20 10 h 618.9 629.1 639.3 (—41.34) 5 1.477 1.499 1.520 7, 18,92=419'82 2070 15 h 617.2 627.8 638.2 e222) as AD 450 Aya wm 614.099 14.78 A545. 20 h 615.5 626.4 637.0 (S16:64)" 5 1.391 91414-1436 Oe \ieOme 75: i280. 25 h 613.8 625.0 635.8 (—7.96) 5 1.362 1.386 1.408 WAD25 99731 A020 30 h 611.9 623.5 634.6 els 91337 1262 1285. v 8.287 8695 35 h 622.0 633.4 (5.89) 5 1341) 1865" v 7.203 7.568 40 h 620.4 632.1 G66) nya es 1.323 1.347 v 6.363 6694 AS h 618.8 630.8 WO:37)) mars 1807) 1331 v 5.988 50 h 629.5 (21.67) 5 1317) v 4.933 60 h 626.8 | 0.21) S 1299

60

80

100

120

32.49 33.78 35.07 36.35 649.5 659.7 670.0 680.3 1.540 1.559 1.578 1.596 215872244 23:31 2447 648.5 658.9 669.2 679.6 1,491eRI5119 1529 4548 16:)129°16.78~ 17.43 48:08 647.5 658.0 668.5 678.9 1.456 1.476 4495 1518 125849913 37. 138.90" Wa4e 646.5 657.1 667.7 678.2 1.429 1.449 1.468 1.486 1052511. TOS ASS We09 645.5 656.2 666.9 677.5 1 A0emn 426 i446 aiiG4 9.093 9.484 9869 10.25 6444 655.3 666.1 6768 1:386021.407> 11427 11445" 7.922 8.268 8.609 8.945 643.4 654.4 665.3 676.1 1.369 1.390 1.410 1.429 7.014 7.326 7.632 7.934 642.3 653.5 664.6 6755 15354921.375 "91395 Sie 6.280 6.564 6843 7.117 641.2 652.6 663.7 674.7 S40" 82361 esS2 aA! 5.184 5428 5.665 5.897 639.0 650.7 662.1 673.3 13315ea1.337 5 Ve5steys)

140

160

180

200

37.62 38.90 690.6 701.1 1.614 1.631 25:03) 25.88 690.0 700.5 e666 01583 18:73 49.37 689.4 700.0 1.531 91.549 1495.015.47 688.8 699.4 1.504 1.522 126438 412.87 688.2 698.8 18492 11500 10.63 11.00 687.6 698.3 12464) Ste431 9278 9.609 686.9 697.7 1.447 1.465 8.232 8.528 686.3 697.2 19438 81450 7.387 7.655 685.7 6966 1A20, Sede 6126 6.352 684.4 695.5 ikon 14s

40.17

41.45 P222 1.664 27259 TP lee 1.616 20.66 WPANJ 1.582 16.50 720.8 t>>> S373 720.3 e555 {ey5 RRS) 1.515 10.27 719.4 1.499 QS IMSS, 1.485 8.185 TANS 1.472 6.789 7 WD 1.449

711.6 1.647 26.74 FAVA SA je529 20.02 710.6 11,559) 299 710.1 1-539 13.30 709.6 Lat 7/ 1S 709.1 1.498 OI938 708.5 1.482 8.822 708.0 1.468 UZ 707.5 1.455 6.576 706.5 1.432

2ieS 732;0 1.598 17.02 Z31E6 SY) 14.16 731 1.550 WAN VEO. I25Sil 10.59 1303 EDS 9.406 1298 13504 8.448 729.4 1.488 7.019 728.6 1.466

No. Circular Standards of Bureau National Based from data Properties Thermodynami Tables 142, Ammonia on

The Appendices Table E-2E

(Continued) Temperature, °F

P, psia

a

°F)

60

80

100

120

140

160

180

#200

240

#=280

#«320

~&360

0 4401 4615 4.822 5.025 5.224 5.420 h 636.6 648.7 660.4 671.8 683.1 694.3 See eI4e 07 |1.350 1.350) Mes77 = 1395. U5 3.012) “4605 =4:190. 4371) 4548" 4.722 MgnOs4322040)/ 2656.) 6/04 68fio693.2 S27 e298 9520) 1 340m es6Or e728 0 3.353> 3.529 3.698 3.862—4.021 4178 h 631.8 644.7 657.0 668.9 680.5 692.0 Ps 71.257 1.261~ 1304-19325 144 1.363 O 298583449 3304 31454531600! (3.748 h_2629'3 642. 6—655.2 667.3 679.2 690.8 s 1.244 1.266 1.289 1.310 1.331 1.349 v ZGGrrZ2 288" 2-404 21515: 2.622 h 633.8 647.8 661:1 673.7 686.0 s 1294 1240 1-263" 412284 A305 v e720" 1.818 91-910) 11.999 h 639.9 654.4 668.0 681.0 s 1:199> 40225 1248 1.269" v 1aa3 S525 1601 h 647.3 662.0 675.8

5.615 5.807 6.187 6.563 705.5 7166 738.9 761.4 1A1s 21490. TA4Gsen1 404 “4.893 51063 65 3988 S178 7044 715.6) (20a 7607 1,396, 914146 177s 4.332 4.484 4.785 5.081 703.4 714.7 737.3 760.0 1391 1-400' 4432 464 3-888 4.071 64.094 8562 702.3 713.7 736.5 759.4 1.368 1.385 1.419 1.451 21727 2.88028 .030 3227 698.0 709.9 733.3 756.7 19324-4242) 11376 12409" 2.0840: 167" 2.328" 2.484) 693.6 705.9 730.1 753.9 1.289. 1.308 1344-1377 1.675.- 1.745— 1.881) 2.012 689.1 701.9 7268 751.1

3 420 780.0 1.440 637 777.7 1.408 2:1740) 775.3

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1.587 699.8

1.714 725.1

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2.069 798.4

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1.305

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Table F-I

Properties of Air 5°

h kJ/kg

280 290 300 310 320 340 360 380 400 420 440 460 480 500 520 540 560 580 600 620 640 660 680 700 720 740

P.

u kJ/kg

v,

kJ/kg-K

199.97 0.3363 142.56 1707 1.29559 219:97 0.4690 156.82 1346 1.39105 240.02 0.63555—171.13 1084 -1.47824 260.09 0.8405 185.45 887.8 1.55848 280.13 1GSS9) 01999752" 738.0%" 1563279 290.16 123, 20695 676.4) *166802 300.19 133608, 21407 Bee212 1-70208 310.24 ISeAGUS-27 2505 5723" “1.73498 320.29 tielor. 223;43 82528,.6) ”1276690 340.42 2 1AD 9 242°32 CE 4544" 1782790 360.58 21026 Ge 257-245 6393-4" 1.62543 380.77 3.176 ~~ 271.69 * °343.4— 1.94001 400.98 3.806' |~286.16 © 301.6 1.99494 421.26 4.522 300.69 266.6 2.04142 441.61 Dosen Sloss05 82500 §2.083870 462.02 6745 mee 52919) OY 2 Trae 2 AG 482.49 7.268 344.70 189.5 2.17760 503.02 8.411 359,497 W706" 2.24952 523.63 9.684 374.36 154.1 2.25997 544.35 11.19 389.34 139.7 2.29906 Soon | 2.66 404.42 127.0 2.33685 586.04 14.38 41955 al 150/237 348 607.02 16.28 434.78 105.8 2.40902 628;07 18.36 450.09 96.92 2.44356 649.22 20.65 465.05 88.99 2.47716 670.47— 23.13 481.01.) 81.89" 2.50985 691,82 7 25.85 A96'62, 975.50" 2.54175 N32 2 “28.80 S128 3 EOIN6. 2.57277 734.82 732.02 528.146453 2.60319 756.44 35.50 544.02 - 59.82 2.63280

5° TK 780 820 860 900 940 980 1020 1060 1100 1140 1180 1220 1260 1300 1340 1380 1420 1460 1500 1540 1580 1620 1660 1700 1800 1900 2000 2100 2200

h kJ/kg

P,

u kJ/kg

v,

kJ/kg-K

800.03 43.35 576.12 51.64 843.98 52.49 608.59 4484 888.2 /=— 63209), 641/40) 39512 932,932 7 5290) 4.557e54-3) OTL.92= 780S 108,08 8022

2.69013 2.74504 2.79783 2.84856 2.89748

WO2325,

105.2

741.98

26.73

1068;39" 1114.86 1 toW:077 1207.57 1254734

=(28:40 143.9 1671e6 19351) 222295

776410 81062 845133 26830:35" 77915557

(SOS

Asay

951.09

13.747

3.19834

1348.55 1395.97" 1443.60 1491.44

290.8 “3309, 375.3 424.2

986.90 (1022.82 1058.94 1095.26

12.435 (i275 10.247 9.337

3.23638 327345 3.30959 3.34474

928-72 21.14 BS 896= 169465 215,241

2.94468

2.99034 3.03449 =3107732 senses 3115916

ISS IAAT S47 SON S/S

520s

790

iaesves)

Savile)

Wo)

sway

1685:972 1684.51" 1733-177 1782.00"

7001208 67284 7500) 834.1

9120549) 242-435 2127965 1316:96

(ai aio2 86.569" 756.046" 5.574

3 44516 9347712 3.50829 3.53879

NSO

SHS(G

Sisal)

|S i\aly/

3}'sY5ysK5)//

4761 3.944 3.295

3.5979 3.6684 ©3173541 Shi SU

1880. 1 2003.3" 2127.4

10252 213924] “13 10F= 1487.2 1655" 115826

22571

ZO06SE

SOV SH7,

IIS)

2377.4

BSS)

NG SyS'

BI5\8) | -3}1310\0)5)

2503.2"

-3138

(18/724

2.012"

“39191 J. Based from data Keenan H. Tables, Gas Kaye, J. and Wiley, York, New 1945. on

AppendixF

Table F-2__

Ideal-Gas Tables

Molar Properties of Nitrogen, N, h, = 0 kJ/kmol

h kj/kmol

u kJ/kmol

s° kJ/kmol-K

TK

0 6391 6975 7558 8141 8669 8723 9306 9888 10 471 il OSs 11 640 (|2 225 12 811 13 399 13 988 14 581 (S172 15 766 16 363 16 962 17 563 18 166 18 772 19 380

0 4562 4979 5396 5813 6190 6229 6645 7061 7478 7895 8314 8733 9153 9574 9997 10 423 10 848 Wil 2ay 11 707 12 138) 12 574 1S Ot 13 450

0)

1000

iOr991 20 604 21 220 21 839 22 460 23 085 23714 24 342 24 974 25 610 26 248 26 890 Tf S37) 28 178 28 826 29 476

182.639

1020

185.180

1040

WS7514

1060

189.673

1080

191.502

1100

191.682

1120

P3592

1140

195.328

1160

196.995

1180

198.572

1200

200.071

1240

201.499

1260

202.863

1280

204.170

1300

205.424

1320

206.630

1340

2OVAT IZ

1360

208.914

1380

209.999

1400

211.049

1440

212.066

1480

2ils2055

1520

214.018

1560

13 892 14 337 14 784 15 234 15 686 16 141 1or599 17 061

214.954

1600

215.866

1700

216.756

1800

220.907

2300

17 17 18 18 19

221.684

2400

222.447

2500

223.194

2600

PPE,

2700

224.647

2800

19 883

J2INS), BSS)

2900

20 362 20 844 JE| 32S)

226.047

3000

226./28

3100

UD

3200

524 990 459 931 407

217.624

1900

218.472

2000

219.301 DOO

2100 S

SRE

2200

h kJ/kmol

u kJ/kmol

30 129 30 784 31 442 32 101 32 762 33 426 34 092 34 760 35 430 36 104 36 777 38 129 38 807 39 488 40 170 40 853 41.539 42 227 42915 43 605 44 988 46 377 47 771 49 168 50 571 54 099 57 651 61 220 64 810 68 417 72 040 75 676 79 320 82 981 86 650 90 328 94014 97 705 101 407 105 115 108 830

21815 22 304 22 795 23 288 23 782 24 280 24 780 25 282 25 786 26 291 26 799 27 819 28 331 28 845 29 361 29 878 30 398 30 919 31 441 31 964 33014 34 071 35 133 36 197 37 268 39 965 42 685 45 423 48 181 50 957 53 749 56 553 59 366 62 195 65 033 67 880 70 734 73 593 76 464 79 341 82 224

s° kJ/kmol-K

228.057 228.706 229.344 229.973 VisSU) are We Be72. 251799 2372. 338)\ 232:973 233.549 234.115 239223 235.766 236.302 236.831 2390353 237.867 238.376 238.878 A338) Sls; 240.350 241.301 242.228 243.137 244.028 246.166 248.195 250.128 251.969 253.726 255.412 257.02 258.580 260.073 AS) 53||72 262.902 264.241 265.538 266.793 268.007 269.186

=

NSRDS-NBS1971 Tables, Thermochemi JANAF from Based data on

The Appendices Table F-3

Molar Properties of Oxygen, O,

h; = 0 ki/kmol TK

h kJ/kmol

@)

u kJ/kmol

6)

s° kJ/kmol-K

0

6)

|

TK

h kJ/kmol

ukJ/kmol

— s° kJ/kmol-K

1020

32088

23607

244.164

1040

32789

24142

244 844

220

6404

4575

240

6984

4989

198.696

1060

33490

24677

245.513

260

7566

5405

201.027

1080

34194

JES) |\i

246.171

196.171

280

8150

5822

203.191

1100

34899

25753

246.818

298

8682

6203

205.033

1120

35606

26294

247.454

300

8736

6242

205.213

1140

36314

26836

248.081

320

9325

6664

207-2.

1160

37023

27379

248.698

340

9916

7090

208.904

1180

37734

27923

249.307

249.906

360

10511

7518

210.604

1200

38447

28469

380

11109

7949

P2222.

1220

39162

29018

250.497

400

11711

8384

INS SSS

1240

39877

29568

PS AOH/S,

420

12314

8822

215.241

1260

40594

30118

251653

440

12923

9264

216.656

1280

41312

30670

252.219

460

(B5s5

9710

218.016

1300

42033

31224

252.190

480

14151

10160

219.326

1320

42753

31778

RES 225, 253.868

500

14770

10614

220.589

1340

43475

32334

520

15395

11071

221.812

1360

44198

32891

254.404

540

16022

11533

D222 QOi7,

1380

44923

33449

254.932 255.454

560

16654

11998

224.146

1400

45648

34008

580

17290

12467

225.262

1440

47102

35129

256:4/5

600

17929

12940

226.346

1480

48561

36256

257.474

620

18572

13417

227.400

1520

50024

37387

258.450

640

19219

13898

228.429

1540

50756

37952

258.928

660

19870

14383

229.430

1560

51490

38520

259.402 260.333

680

20524

14871

230.405

1600

52961

39658

700

21184

15364

23}||SS

1700

56652

A2BA/,

262.571

720

21845

15859

232.291

1800

60371

45405

264.701

740

22510

16357

233.201

1900

64116

48319

266.722

760

DENS

16859

234.091

2000

67881

5258

268.655 270.504

780

23850

17364

234.960

2100

71668

54208

800

24523

17872

Days) 19)\\0

2200

75484

57192

PUY IL ATES

820

25199

18382

236.644

2300

79316

60193

273.981

840

UX] T/

18893

237.462

2400

83174

63219

275.625

860

26559

19408

238.264

2500

87057

66271

PLILTE AAS

880

27242

19925

239.051

2600

90956

69339

278./38

900

27928

20445

239.823

2/700

94881

72433

280.219

920

28616

20967

240.580

2800

98826

75546

281.654

940

29306

21491

BA SES:

2900

102793

78682

283.048

960

29999

22007;

242.052

3000

106780

81837

284.399

980

30692

22544

242.768

3100

110784

85009

285.713

1000

31389

23075

243.471

3200

114809

88203

286.989

Poni JANAF NSRDS-NBS-37, Tables, Thermochemical 1971 daté

Appendix F_Ideal-Gas Tables

Table F-4_

Molar Properties of Carbon Dioxide, CO,

h, = —393 520 kJ/kmol =

hkJ/kmol

ukJ/kmol

— s° kJ/kmol-K

3335337)

0 202.966 205.920 208.717 211.376 213.685 213.915 216.351 218.694 220.948 223.122 225.225 227.258 229.230 231.144 233.004 234.814 236.575 238.292 239.962 241.602 243.199 244.758 246.282 247.773 249.233 250.663 252.065 253.439 254.787 256.110 257.408 258.682 259.934 261.164 262.371 263.559 264.728 265.877 267.007 268.119

34455

BODINE

0 4772 5285 5817 6369 6885 6939 7526 8131 8752 BB9Z. 10046 10714 iS95 12091 12800 IBS2i 14253 14996 (S751 16515 Zon 18076 18869 19672 20484 e21s805 22134 PDS. 23817 24669 BSI) 26394 ALAS! 28125 29031 ZOD. 30818 SANS 32625

TK

h kJ/kmol

1020 1040 1060 1080 1100 1120 1140 1160 1180 1200 1220 1240 1260 1280 1300 1320 1340 1360 1380 1400 1440 1480 1520 1560 1600 1700 1800 1900 2000 2100 2200 2300 2400 2500 2600 2700 2800 2900 3000 3100 3200

43859 44953 46051 47153 48258 49369 50484 51602 52724 53848 54977 56108 57244 58381 59522 60666 61813 62963 64116 65271 67586 69911 72246 74590 76944 82856 88806 94793 100804 106864 112939 119035 125152 131290 137449 143620 149808 156009 162226 168456 174695

u kJ/kmol

35378 36306 37238 38174 39112 40057 41006 41957 42913 43871 44834 45799 46768 47739 48713 49691 50672 51656 52643 53631 55614 57606 59609 61620 63741 68721 73840 78996 84185 89404 94648 99912 ~ 105197 110504 115832 121172 126528 131898 137283 142681 148089

s° kJ/kmol-K

270,293 271.354 272.400 273.430 274.445 275.444 276.430 277.403 278.361 AY 3IOJ 280.238 281.158 282.066 282.962 283.847 284.722 285.586 286.439 287.283 288.106 289.743 2S) ||BSS 292.888 294.411 295.901 299.482 302.884 306.122 309.210 312.160 314.988 Sil 7 99)5 3201302 322.308 B2O 222 327.549 329.800 333) | SS 334.084 336.126 SSou09

1971 NSRDS-NBS-3 Tables, Thermochemi JANAF from Based data on

The Appendices Table F-5

Molar Properties of Carbon Monoxide, CO

h; = —110 530 kJ/kmol TK 0

h kJ/kmol

u kJ/kmol

220-

0 6391

0 4562

240

6975

4979

260

7558

280

8140

300

8723

320 340

cy ee

TK

h kJ/kmol

u kJ/kmol

s° kJ/kg-K 235.728

0)

1040

31 688

23 041

188.683

1060

32 335)//

23 544

236.364

1080

5396

IG 227 GBR 54

B5029 33 702

236.992 237.609

3x8) 2

193e/ 43

1120

34 377

24 049 24 557 25 065

6229

(977233

1140

35 054

ADS

238.817

9306

6645

199.603

1160

9889

7062

ZONRSTAl

1180

3D) 1333 36 406

26 088 26 602

239.407 239.989

360

10 473

7480

203.040

1200

380

11 058

7899

204.622

1220

5/095 37 780

PAH A A\'3 27 637

240.663 241.128

400

11 644

8319

206.125

1240

420

W2Z232

8740

207.549

1260

38 466 39 154

28 426 28 678

241.686 242.236

1100

440

12 821

9163

208.929

1280

460

13 412

9587

210.243

1300

480

14 005

10014

Dain OA

500

14 600

10 443

PZ ING)

520

15 197/

10 874

540

oy 727)

560

I 3S)

BRS I1\\ If

39 844

29 201

242.780

40 534

WS) WS

243.316

1320

41 226

30 251

243.844

1340

41919

30 778

244.366

213.890

1360

42 613

31 306

244.880

1] 307

215.020

1380

43 309

31 836

245.388

11 743

Zoi:

1400

44 007 45 408

32 367 33 434

245.889 246.876

—

580

17 003

Iasi

PE TEATLS

1440

600

aon

12 622

218.204

1480

46 813

34 508

247.839

620

18 221

13 066

AG AOS

1520

48 222

35 584

248.778

640

18 833

(3512

220.179

1560

49 635

36 665

249.695

660

19 449

13 962

DIN V2

1600

5053

By 70

AN), SZ

680

20 068

14414

dejaje (Qs

1700

54 609

40 474

Ne S|

700

20 690

14 870

DU

1800

58 191

43 225

254.797

720

J! SiS

15 328

223.833

1900

61 794

45 997

256.743

'S\3\3

740

21 943

15 789

224.692

2000

65 408

48 780

258.600

760

Dd ifs:

16 255

22.5

2100

69 044

51 584

260.370

780

23 208

6 VZ3

2IAS\.3)5)/)/

2200

72 688

54 396

262.065

5

800

23 844

17 193

227.162

2300

76 345

Sf 222

263.692

2

820

24 483

17 665

iY

2400

80 015

60 060

265.253

2

840

25 124

18 140

228.724

2500

83 692

62 906

266.755

2

860

25 768

18 617

229.482

2600

87 383

65 766

268.202

: 3

YSy2

880

26 415

19 099

230.227

2700

Sl O77

68 628

269.596

900

27 066

19 583

230.957

2800

94 784

71504

270.943

|2

920

Deyh GANS!

20 070

231.674

2900

98 495

74 383

272.249

:

3000

1OZ2TO

UI AS)

273.508

:

3100

1105 QS

80 164

274.730

S =

940

28 375

2OSS9

AZ

960

29038

2 O51

233 OZ

3UE)

980

DNS

21545

B33) NDE

3150

107 802

81 612

PAIS). AS

1000

30 355

22 041

234.421

3200

109 667

83 061

275.914

| 1020

31 020

22 540

Zao, OU9

|3 ;

Appendix F_Ideal-Gas Tables

Table F-6

Molar properties of hydrogen, H,

|

h, = 0 kJ/kmol

ie ah K

0 260 270 280 290 298 300 320 340 360 380 400 420 4A0 A60 480 500 520 560 600 640 680 720 760 800 840 880 920 960 1000 1040 1080 120 1160 1200 1240 1280 1320 1360 1400

hkJ/kmol

—u kJ/kmol

5° kJ/kmol-K

0 7370 7657 7945 8233 8468 8522 9100 9680 10 262 10 843 11 426 12 010 12 594 13179 13 764 14 350 14.935 16 107 17 280 18 453 19 630 20 807 21 988 23171 24 359 25 551 26 747 27 948 29 154 30 364 31 580 32 802 34 028 35 262 36 502 37 749 39 002 40 263 41 530

0 5209 5412 5617 5822 5989 6027 6440 6853 7268 7684 8100 8518 8936 9355 9773 10 193 10611 11 451 12 291 13 133 13 976 14 821 15 669 16 520 17 375 18 235 19 098 19 966 20 839 2ALTAT 22 601 23 490 24 384 25 284 26 192 27 106 28 027 28 955 29 889

0 126.636 127.719 128.765 129.775 130.574 130.754 132.621 134.378 136.039 137.612 139.106 140.529 141.888 143.187 144.432 145.628 146.775 148.945 150.968 152.863 154.645 156.328 157.923 159.440 160.891 162.277 163.607 164.884 166.114 167.300 168.449 169.560 170.636 171.682 172.698 173.687 174.652 175.593 176.510

5° kJ/ TK

1440 1480 1520 1560 1600 1640 1680 1720 1760 1800 1840 1880 1920 1960 2000 2050 2100 2150 2200 2250 2300 2350 2400 2450 2500 2550 2600 2650 2700 2750 2800 2850 2900 2950 3000 3050 3100 3150 3200 (Ee

‘

hkjJ/kmol

wukJ/kmol

kmol-K

42 808 44 091 45 384 46 683 47 990 49 303 50 622 51 947 53 279 54 618 55 962 57 311 58 668 60 031 61 400 63 119 64 847 66 584 68 328 70 080 71 839 73 608 75 383 77 168 78 960 80 755 82 558 84 368 86 186 88 008 89 838 91671 93512 95 358 97 211 99 065 100 926 102 793 104667 106 545

30 835 31 786 32 746 33713 34 687 35 668 36 654 37 646 38 645 39 652 40 663 41 680 42 705 43 735 44771 46 074 47 386 48 708 50 037 51 373 52 716 54 069 55 429 56 798 58 175 59554 60 941 62 335 63 737 65 144 66 558 67 976 69 401 70 831 72 268 73 707 75 152 76 604 78 061 79 523

177.410 178.291 179.153 179.995 180.820 181.632 182.428 183.208 183.973 184.724 185.463 186.190 186.904 187.607 188.297 189.148 189.979 190.796 191.598 192.385 193.159 193.921 194.669 195.403 196.125 196.837 197.539 198.229 198.907 199.575 200.234 200.885 201.527 202.157 202.778 203.391 203.995 204.592 205.181 205.765

The Appendices Table F-7_

Molar Properties of Water, HO

h; = —241 810 ki/kmol TK

ee

h kJ/kmol

¢)

u kJ/kmol

¢)

0)

s* kJ/

5° kJ/

kmol-K

TK

h kJ/kmol

u kJ/kmol

6)

1020

36 709

28 228

DES74 15

178.576

1040

Sif sya

28 895

234.223 235.020

kmol-K

220

7295

240

7961

5965

181.471

1060

38 380

29 567

260

8627

6466

184.139

1080

39 223

30 243

235.806

280

9296

6968

186.616

1100

40 071

30 925

236.584

298

9904

7425

188.720

11/20

40 923

Sl Gili

23/52.

300

9966

7472

188.928

1140

41 780

32 301

238.110

320

10 639

7978

191.098

1160

42 642

32 997

238.859

340

ii) 314

8487

193.144

1180

43 509

33 698

239.600

360

11 992

8998

195.081

1200

44 380

34 403

240.333

380

12 672

O53

196.920

1220

45 256

35) (IZ

241.057

400

13 BS

10 030

198.673

1240

46 137

35 827

241.773

420

14 043

10551

200.350

1260

47 022

36 546

242.482

440

14 734

iwOW5

201.955

1280

47 912

3 LO

DAS NSS

460

15 428

11 603

203.497

1300

48 807

38 000

243.877

49 707

33/3

244.564

5466

480

16 126

12 135

204.982

1320

500'

16 828

12 671

206.413

1340

50 612

39 470

245.243

520

17 534

13 24

207.799

1360

Syl S32

40 213

245.915

540

18 245

IS WSS)

209.139

1400

53). 335}

AM) TA

247.241

560

18 959

14 303

210.440

1440

55 198

43 226

248.543

580

19 678

14 856

AA. OZ

1480

57 062

44 756

249.820

600

20 402

15 413

212.920

1520

58 942

46 304

251.074

620

ZA 130)

15975

DA AD

1560

60 838

47 868

252.305

640

21 862

16 541

ZA SEZ85

1600

62 748

49 445

253), 535

660

22 600

1H WZ

216.419

1700

67 589

5Sr455

256.450

680

23347.

17 688

ANS

1800

VDSS}

SVR 47,

259.262

700

24 088

18 268

218.610

1900

HA SH

61 720

261.969

720

24 840

18 854

219.668

2000

82 593

65 965

264.571 267.081

740

25 597

19 444

220.707

2100

37) WSS)

70 275

760

26 358

20 039

221.720

2200

92 940

74 649

269.500

780

DH NAS

20 639

BED ANY

2300

98 199

79 076

271.839

800

2/7 896

21 245

223.693

2400

103 508

S15} S)sjs}

274.098

820

28 672

PAP E35)

224.651

2500

108 868

88 082

276.286 278.407

840

29 454

22 470

DDO?

2600

I

AaeZ73,

92 656

860

30 240

23 090

2265517,

2700

119 717

97 269

280.462

880

3] O32

DSS I'S

227.426

2800

125 198

101 917

282.453

900

31 828

24 345

DEE

2900

SORA

106 605

284.390

920

32 629

24 980

229.202

|

3000

136 264

Walt) S21

286.273

940

33 436

D'S (D4

230.070

3100

141 846

116 072

288.102

960

34 247

26 265

230.924

3150

144 648

118 458

288.9

980

35 061

26 913

DSi TASH

3200

aye Alsy7/

120 851

289.884

1000

35 882

27 568

2378 50)/,

3250

150 250

12 3 250

290.7

[

v

NSRDS-NBS-37 Tables, Thermochemica JANAF from 1971 data

Appendix F_ Ideal-Gas Tables

Table F-1E

Properties of Air

h, = 0 Btu/lbmol h Btu/Ibm

Pe

|

v,

uBtu/lbm

s° Btu/Ibm-°R irae

440 480 520 Sa7 540 560 580 600 620 640 660 680 700 720 740 760 780 800 820 840 860 880 900 920 940 960 980 1000 1020 1040 1060 1080 1100 TO 1160 1200 1240 1280 1320 1360 1400 1440

105.11 114.69 124.27 128.10 129.06 133.86 138.66 143.47 148.28 153:09 57-92 162773 167.56 172739 177523 182.08 186.94 191.81 196.69 201.56 206.46 2135 216.26 2271s 226.11 231.06 236.02 240.98 245.97 250:95 290.90 260.97 205.99 271203 281.14 291.30 301.52 Slie/9 S27 ciu 332.48 342.90 353237,

0.4858 0.6776 0.9182 1.2147 133593 1,3860 1.5742 1.7800 2.005 2.249 2.514 2.801 Fala (G 3.446 3.806 4.193 4.607 5.0511 5.526 6.033 6.573 7.149 7.761 8.411 O02 9.834 10.610 11.430 127298 135215 14.182 15.203 16.278 17.413 18.604 PalWals) 24.01 ZENS 30555 34.31 38.41 42.88 47.75

68.11 74.93 o1a77 88.62 Olas 92.04 95.47 98.90 102.34 105.78 109721 12767 116.12 119.58 123.04 126.51 WASLS) 133.47 136,97 140.47 143.98 147.50 T5102 154.57 158.12 161.68 165.26 168.83 172.43 176.04 179.66 1S3e29 186.93 190°5s 194.25 201.63 209,05 216:53 244.05 231,03 23929 246.93 254.66

,

305.0 240.6 123:65 158.58 146.34 144.32 131278 120.70 110.88 VO2 42 94.30 87.27 80.96 TS-2) 70.07 65.38 61.10 57.20 53.63 50.35 47.34 44.57 42.01 39.64 37.44 35.41 38,52 St 80512 20.09 Taipei 25.82 24.58 23,40 722350 20.293 18.514 16,932 15.5116 14253 Sats 12.095 PLATZ

0.52890 055172 057253 Oeei7s 0.59945 0.60078 0.60950 0.61793 0.62607 0.63395 0.64159 0.64902 0.65621 0.66321 0.67002 0.67665 0.68312 0.68942 0.69558 0.70160 0.70747 OF 1328 0.71886 0.72438 0.72979 0.73509 0.74030 0.74540 0.75042 0.75536 0.76019 0.76496 0.76964 0.77426 0.77880 0.78767 0.79628 0.80466 0.81280 0.82075 0.82848 0.83604 0.84341

1945. York, New Wiley, Tables, Gas Kaye, J. and Keenan H. from Based data on

The Appendices Table F-1E

T°R 1480 1520 1560 1600 1640 1680 1720 1760 1800 1900 2000 2100 2200 2300 2400 2500 2600 2700 2800 2900 3000 3100 3200 3300 3400 3500 3600 3700 3800 3900 4000 4100 4200 4300 4400 4500

(Continued)

h,; = 0 Btu/lbmol

h Btu/Ibm | 363.89 374.47 385.08 395.74 406.45 417.20 428.00 438.83 449.71 477.09 504.71 532.55 560.59 588.82 617.22 645.78 674.49 703.35 732.33 761.45 790.68 820.03 849.48 879.02 908.66 938.40 968.21 998.11 1028.1 1058.1 1088.3 1118.5 1148.7 1179.0 1209.4 1239.9

P. 53.04 58.78 65.00 Tile 78.99 86.82 95.24 104.30 114.03 141.51 174.00 212.1 256.6 308.1 367.6 435.7 513.5 601.9 702.0 814.8 941.4 1083 1242 1418 1613 1829 2068 2330 2618 2934 3280 3656 4067 4513 4997 5521

uBtu/lbm 262.44 270.26 278.13 286.06 294.03 302.04 310.09 318.18 326.32 346.85 367.61 388.60 409.78 431.16 452.70 474.40 496.26 518.26 540.40 562.66 585.04 607.53 630.12 652.81 675.60 698.48 721.44 744.48 767.60 790.80 814.06 837.40 860.81 884.28 907.81 931.39

|

v, 10,336 9.578 8.890 8.263 7.691 7.168 6.690 6.251 5.847 4.974 4.258 3.667 3.176 2.765 2.419 2.125 1.876 1.662 1.478 1.318 1.180 1.060 0.955 0.8621 0.7807 0.7087 0.6449 0.5882 0.5376 0.4923 0.4518 0.4154 0.3826 0.3529 0.3262 0.3019

| 5° Btu/lbm’R 0.85062 0.85767 0.86456 0.87130 0.87791 0.88439 0.89074 0.89697 0.90308 0.91788 0.93205 0.94564 0.95919 0.97123 0.98331 0.99497 1,00623 1.01712 1.02767 1.03788 1.04779 1.05741 1.06676 1.07585 1.08470 1,09332 1.10172 1.10991 1.11791 1.12571 1.13334 1.14079 1.14809 1.15522 1.16221 1.16905

Appendix F_ Ideal-Gas Tables

Table F-2E

Molar Properties of Nitrogen, N, h; = 0 Btu/lbmol h Btu/

300 320 340 400 440 480 520 a3) 540 560 580 600 620 640 660 680 700 720 740 760 780 800 820 840 860 880 900 920 940 960 980 1000 1020 1040 1060 1080

2082.0 2221.0 2360.0 DUDA 3055.1 3933.1 Soles SUZS)S 3/50),8 3889.5 4028.7 4167.9 4307.1 4446.4 4585.8 4725.3 4864.9 5004.5 5144.3 5284.1 5424.2 5564.4 5704.7 5845.3 5985.9 6126.9 6268.1 6409.6 5512 6693.1 6835.4 BIT Tao 7120.7 7263.8 7407.2 WS

u Btu/ 1486.2 UXISD 1684.8 1982.6 2181.3 LSYS)S, 2578.6 2663.1 2678.0 2777.4 2876.9 2976.4 S059 3117'5.5 SLU De 3374.9 3474.8 3574.7 3674.7 3774.9 XSL 3319/5), J/ 4076.3 4177.1 4278.1 4379.4 4480.8 4582.6 4684.5 4786.7 4889.3 4992.0 5095.1 5 I'SS,5 D302, 2 5406.2

h Btu/

U Btu/

s° Btu/

Ibmol-°R

T°R

Ibmol

Ibmol

Ibmol-°R

41.695 42.143 42.564 43.694 44.357 44.962 45°519 45.743 45.781 46.034 46.278 46.514 46.742 46.964 47.178 47.386 47.588 47.785 47.977 48.164 48.345 48.522 48.696 48.865 49.031 49.193 49.352 49.507 49.659 49.808 49.955 50.099 50.241 50.380 50.516 50.651

1100 1120 1160 1200 1240 1280 1320 1360 1400 1440 1480 1520 1560 1600 1640 1680 1720 1760 1800 1900 2000 2200 2400 2600 2800 3000 3100 3200 3300 3400 3600 3700 3800 3900 5300 5380

7695.0 USSS).3) 8129.0 8420.0 8712.6 9006.4 9301.8

551105 5615.2 5825.4 6037.0

50.783 50°912 5d 67

6250.1 6464.5 6680.4

5ilk653 51.887 52.114

9598.6 9896.9 10196.6 10497.8 10800.4 11104.3 11409.7

6897.8 7116.7 HIST VDDSd. TINS 8006.4 S25 773

52335) 52255) D263 52-969 Soma 55.869 mele

11716.4 12024.3 |Z3337/ 12644.3 1295603 13741.6

8459.6 8688.1 8918.0 9149.2 Exsi| 7/ 9968.4

Seo 582956 54.118 54.297 54.472 54.896

14534.4 16139.8 WHUSTS: 19415.8 21081.1

10562.6 OS 13001.8 14252.5 15520.6

255.05 56.068 Dowel, 57.436 58.053

22761-5 23606.8

16803.9 17450.6

58.632 58.910

24455.0 25306.0 26 59).7/ 278744 PASH D3) | ZS) IS 30462.8

18100.2 iS7S257/ 19407.7 LOP2ZS.S '21387.4 2205 1.6 ZAUNVAS)

SANS 59.442 3S)(SS)// 60.186 60.422 60.562 60.877

42728.3 43436.0

B2Z057 S27 525!

63.563 63.695

s° Btu/

|

51.413

York, New Wiley, 1945. Tables, Gas Kaye, J. and Keenan H. from Based data on

The Appendices Table F-3E_

Molar Properties of Oxygen, O,

h,; = 0 Btu/lbmol re

h Btu/

u Btu/

s° Btu/

Ibmol

Ibmol

Ibmol-°R

Ler

s° Btu/ h Btu/Ibmol

u Btu/Ibmol

Ibmol-°R

300

2073.5

1477.8

44.927

1280

9254.6

(| 2.7

55.386

B20)

DENPAS

WTA

45.375

1320

9571.6

6950.2

55.630

340

DE5yley/

1676.5

45.797

1360

9890.2

7189.4

55.867

400

2769.1

1974.8

46.927

1400

10210.4

7430.1

56.099

420

2908.3

2074.3

47.267

1440

10532.0

7672.4

56.326

440

3047.5

PANTS

47.591

1480

10855.1

7916.0

56.547

480

33265

D333}

48.198

1520

11179.6

8161.1

56.763

520

3606.1

DEV

48.757

1560

11505.4

8407.4

56.975 SV aoe

Saif!

B/25el

2658.7

48.982

1600

eS S2e5

8655.1

540

3746.2

2673.8

49.021

1640

12160.9

8904.1

DROS

560

3886.6

2774.5

49.276

1680

12490.4

9154.1

57.582

580

4027.3

JAS] 5,5)

49.522

1720

12821.1

9405.4

SI HHT

600

4168.3

2976.8

49.762

1760

IST53C

9657.9

57.968

620

4309.7

3078.4

49.993

1800

13485.8

640

4451.4

3180.4

50.218

1900

AZ|

9911.2

XS. (|IS

10549.0

58,607

660

4593.5

3282.9

50.437

2000

15164.0

11192.3

59.039

680

AFS6) 72

3385.8

50.650

2200

16862.6

124937 13813.1

59.848 60.594

700

4879.3

3489.2

50.858

2400

18579.2

720

5022.9

3593.1

51.059

2600

20311.4

15148.1

61.287

740

5) (7/0)

3697.4

5). 257/

2800

22057.8

16497.4

61.934

760

5311.4

3802.2

51.450

3000

DEON TA

17860.1

62.540

780

5456.4

3907.5

51.638

3100

24702.5

18546.3

62.831

800

5602.0

AO

5){|82

3200

25590.5

|}Q2EI5).7/

63), 11S

820

5748.1

ANS \S).7/

52.002

3300

26481.6

19928.2

63.386

840

5894.8

4226.6

SPAS)

3400

27375.9

20623.9

63.654

860

6041.9

4334.1

2

a

3600

29173.9

22024.8

64.168

880

6189.6

"4442.0

Syn Syn

3700

500/725

22729.8

64.415

900

6337.9

4550.6

52.688

3800

30984.1

DSASTES

64.657

$

920

6486.7

4659.7

52. {SoZ

3900

31893.6

24148.7

64.893

=

940

6636.1

4769.4

X30) 2

4100

BIAS

25579.5

65.350

3

960

6786.0

4879.5

53), ||FAO

4200

34639.9

26299.2

O)59)// 1

:

980

6936.4

4990.3

55526

4300

35561.1

27021.9

65.788


, 0.000308 m3 0.0064 m° b) 818 kPa

Ch apter 3 3.1 3.2 3.3 3.4 3.5 3.6

C D B B °C C

2.42 2.42 2.43 2.44

b) 0.008 m? d) 0.7849 m3 b) 652°C 424kPa

wh 3.8 3.9 3.10

a}

A B A

2.48 3294 kJ/kg, 3290 kJ/kg, 3290 kJ/kg

3.11 C

2.49 b) 9100 Btu

3.12 D

2.50 b) 0.073 m*/kg, 900 kJ/kg 2.51 b) 0.646 m*/kg, 11 500 kJ, 12 800 kJ 2.52 a) 4174 kJ

3.13 D 3.14 A 3.15 C

2.53 b) 1070 Btu/lbm, 1160 Btu/Ibm

3.16 B

2.54

3.18

4580 kJ

300 ft-lbf

2.56 Saturated mixture, 0.288% 2.58 ii) 24 380 kJ

3.20 8002 kg, 392.5 kJ 3.22 1977 J

2.60 225 kJ/kg

3.24 b) —30,100 ft-lbf

2.62 0.313 Ibm, 1.12 ft

3.26 —31.9kJ

2.64 28.8 kg

3.28 3125 kJ

2.66 Helium

3.30 —0.0125 J

2.68 ii) 12.5 kg/m’, 1230 N iv) 1337 kPa, 1960 N

3.32 —0.1125 kJ

2.70 0.0157 lbm/ft?

3.34 739 J

22

3.36 320 J, 76.4 rpm

Daya

LOl pear

2.74 b) 662 kPa

3.38 135 kW/m?

2.74 d) 664 kPa

3.40 941 Btu/s

2.76 ii) 0.921 m*/kg iv) 0.923 m*/kg

3.42 50.9 Btu/hr-ft”, 55.0 Btu/hr-ft, 8.06%

2.78

b) 448.1 kJ/kg

2.78 d) 722 kJ/kg 2.79 b) 370 kJ/kg, 513 kJ/kg

3.46

32°C, no work

3.48 25 MW 3.50 Q,, = —200kJ, W,., = 400 kJ,

2.80 b) 3500 Btu, 4850 Btu

Q,, = —600 kJ, Q,, = 1000 kJ

2.80 d) 3530 Btu, 4860 Btu 2.82 b) 2.20 kJ/kg-K

3.52 b) 0.23 kg melts so 0°C 3.54 b) 130°F

2.83 b) 0.510 Btu/lbm-°R 2.84 14,000 Btu, 14,000 Btu

3.56 153°C 3.58 61.3 ft", ft?. 170°F, 170°F 95.1 95.1 psia psi 3.58 61.3

2.86 5820 kJ

3.60

a) 120 kPa, 0.00877 kg, 104.80°C, 1100 kPa, 188 J

Answers to Selected Problems 3.61 b) 4.9 cm 3.62 96.9 MJ

4.14

3.63 b) —17 600, —4180 kJ 3.64 b) 1193 K, 154 kJ, 537 kJ 3.66 b) —6480 kJ, —1030 kJ

4.16

3.68 42.9°C 3.70 b) -2450 kJ 3.72 3.46 kJ, 22.7 kJ 3.74 669°C 3.76 a) 0.731, 0, 1130 kJ/kg 3.78 a) 630°C 3.80

200.) k J

4.15 4.17 4.18 4.19 4.20 4.21 Pp TAA Far > 4.22 0.00141 m°/s, 1.41 kg/s, 178 m/s

0.0856 Ibm/s, 1.25 ft?/s 4.26 b) 1.172 m/s 4.24

4.28 b) 2560 kPa gage

3.82 400°C, —54.5 kJ 3.83 Diao ks

4.29

3.84 5.54 ft?/Ibm, —36.8 Btu/lbm 3.85 b) 365 kJ/kg, 1 kJ/kg

4.32

3.86 3.87

C787 iI b) 3750 kJ, —5310 kJ

3.88 b) 200 kPa, 0.339 m*/kg, 0.483 m°/kg, 200 kJ, 3.89

576 kJ, 57 kJ b) 132 Btu, 0.8 Btu

3.90 b) 1148 kJ, 0, —294 kJ, 854 kJ 5.91 b) 0, 148 kJ, —113 kJ, 35 kJ 3.92 b) 73. kJ, 128 kJ, —208 kJ, 47 kJ 3.93 b) 178,200 ft-lbf, 229 Btu

Chapter 4

4.42

b) 2090 kW 447 kw 10,200 hp b) 201 kW 0.0833 kW b) 0.0957 kW b) 23.7kW 110 hp —139kJ/s

4.44

1.37 hp

4.30 4.34 4.36 4.37 4.38 4.40

15.3%, 99.6°C DOs 4.48 b) 18% 4.50 —29°C, 0.946 m°/s 4.46

4.47

4.52 273 kg/s, 51.6 m/s

4.3

b) 0.359 382°F 4.55 b) 293°F

4.4

4.56

8.85 kg/s

4.5

4.58

4.6

4.60

0.098 30.9°C

4.7

4.62

1008 m/s, 5.83 cm

4.8

4.64

4.9

4.66

0.963 0.578 kg/s, 2.33 m/s, 18.4 m*/s, 28.8 kJ/s

4.10

4.67 b) 512 m/s

4.11

445°C, 61.7 kg 4.69 b) —33.6 kJ 4.70 b) 127°C, 3.48 kg

4.1 4.2

4.12 4.13 SRO YS VUPPAORPTAR

4.53 4.54

4.68

579

Answers to Selected Problems

4.72

Dy 129° CS

4.73

b) 203.5 kPa

4.74

BD) 2s.Spsia, 156° F

4.75

4.78

b) 7940 kW, 53.4 hp, 26.8% a) 7980 hp, 67 hp, 21.6% b) 70.8 kJ/s, 2.36 b) 229 Btuls, 6.47

4.79

b) 651 kW, 50.7%, 1075 kJ/s, 37.7%

4.80

b) 1243 hp, 41.5%, 1434 Btu/s, 38%

4.76 4.77

Chapter5 5.1

5.42 339 hp 5.44 Impossible 5.45 b) 125 kJ/s, 75 kW, 460°C 5.46 0.333 5.48 171°C 5.50 b) 572 hp, 657 kJ/s 5.52 3.96 hp 5.54 174 hp

Chapter6

b) 4440 Btu/s

6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9 6.10 6.11 6.12 6.13 6.14 6.15 ee ee eee a ee ee

957 kJ/s, 68.9 kg/hr

6.16

Zero

b) 176,000 Btu/hr, 64.8%

6.18 6.20 6.22 6.22 6.23 6.24 6.24 6.25 6.26 6.27 6.28 6.29 6.30 6.31

—0.200 kJ/kg-K 0.0711 kJ/kg-K, —0.211 m*/kg b) 1.56 kJ/kg-K d) 19.5 kJ/kg-K c) 15.3 m° b) 357°C, —173 kJ/kg d) 485°C, —112 kJ/kg b) 285°F, 36,500 ft-Ibf/lbm b) 379°C, 3.98 kJ/K b) 136°C, 0.557 kI/kg b) 220 kJ, 220 kJ, 0.623 kJ/K b) 1838 kPa, 400°C b) 400°C, 1.18 kJ/K b) 249°C, 1.81 kI/K

o)

5.2 Do 5.4

rPRTEOMASP AOS 678 W: only (1)

b) 9.89 kJ/s, 67% BAI ONO e108 key

b) 92.8 hp b) 14.1% 20.6 X 10° MJ/day b) 22 700 hp 6.70 hp b) 16.7 kJ/s, 2.24 33.5 kJ/s, 26.8 kJ/s 108,000 Btu/hr, 5.05 b) 19.1 hp 741% 68.9% 60%, 236°C 209 hp

Answers to Selected Problems 6.32 b) 350°C, —473 kJ 6.33 b) 219°F, 117,000 ft-Ibf/lbm 6.34 431°C, 1.16 kJ/K

yall

6.36

7.2

—0.1066 kJ/K, 0.2396 kJ/K

Chapter 7

6.38 9.77°C, 0.170 kJ/K 6.39 b) 28.7°C, 0.102 kJ/K

13

6.40 1.475, 0.6981 kJ/kg:K 6.41 b) —488 kJ/kg, —1.19

15

6.42 b) 18.36 Btu/°R 6.43 b) 0.346 kJ/K 6.44 1.49 kJ/K 6.46 b) 487°C, —446 kJ/kg 6.47 b) 4510 kJ 6.48 15.55 MJ, 20.4 kJ/K 6.50 por”ONS 9d © 6.52 b) 1.49 kJ/kg-K 6.52 d) 350°C, —167 kJ/kg 6.52 f) 369°C, 4.04 kJ/K 6.54 —30°C, 0.057 kJ/kg-K 6.56 975 MJ/s, 147 kJ/K-s 6.58 3480 hp

7.4

kJ/kg

7.6

rPoORPaADA

gf |

7.24

A 0.0260 m*/kg, 0.00184 m*/kg —0.201 Btu/Ibm-°R 40.1°C a) 397°C 760°F 1406 kJ/kg, 2.453 kJ/kg-K, 0.25%, 0% b) 52°F

7.26

C ee ae Rin

7.8 7.14 7.18 7.19 7.20 7.22

eT:

CATES AL

Gain ub

ean

1

S1

et o(+ = 0;

vga

Se ie enw

*2 1

~) 05

ae

6.59 b) 15 MW, 8 kJ/K:s 6.60 b) 84 500 hp 6.60 d) 88 100 hp

—11.4 kJ/kg, -1.0 kJ/kg-K, —0.976 kI/kg-K 3.86 X 10° *K ',2.33 GPa, 0.109 kJ/kg-K

6.61 b) 42.3 hp 6.62 331 hp, 0 kJ/K-s 6.64 1.08 Btu/°R-s

0.26 x 10-*K ',230 MPa, 3.3 X 10 °kJ/kg-K Die2ee 0.041 °C/kPa, —11°C

6.66 0.188 Btu/°R-s 6.67 b) 3.5 kJ/K 6.68

10.58 MW

6.70 b) 774% 6.70 d) 775% 6.71 b) 22,800 hp, 80.8% 6.72 567°C, 98.9% 6.74 b) 503 kJ/kg 6.76 79.4% 6.78

113 hp

6.80 208 m/s, 64°C 6.82

194 kJ/kg

6.83 b) 246 kJ/kg 6.84

131 hp

0.0075 °C/kPa, 2.39 kJ/kg-K, 2.38 kJ/kg-K

b) —141 kJ/kg, —173 kJ/kg b) 0, —48.3 kJ/kg b) 0.170 Btu/Ibm-°R, 0.170 Btu/Ibm-°R b) 0.173 kI/kg-K, 0.173 kI/kg-K, i) 500 kJ/kg, —0.364 kJ/kg:K, iii) 532 kJ/k —0.318 kJ/kg-K, i) 62.5 kJ/kg, —0.433 kJ/kg:K iii) 62.5 kJ/kg, —0.522 kI/kg-K i) 3380 kW ii) 3380 kW i) 1600 kPa ii) 1600 kPa gg

—66 Btu/Ibm, —0.291 10,900 ft-lbf/lbm

Btu/Ibm-°R,

582

Answers to Selected Problems

3.01 kg/s, 74.1 MJ/s, 30 MW, 185°C, 44.2 MJ/s, 212 hp, 40.5%

Chapter8

8.36

8.1

8.38 42.5% 8.40 272 MJ/s, 2.36 kg/s, 11.1 MW, 40.8%

8.2 8.4

8.42 2.93 kg/s, 29.1 MW, 59.9 MJ/s, 94.1%, 30.8 MJ/s, 383 hp, 48.6%

8.5

8.44

6.97 lbm/s and 5.83 Ibm/s, 49,000 hp, 72,000 Btu/s, 163°F, 37500 Btu/s, 617 hp, 479%

8.46

1227 Btu/lbm, 297 Btu/Ibm, 390 Btu/Ibm, 540 Btu/Ibm, 56%

8.48

141.2 MJ/s, 39 MW, 45.6 MJ/s, 28.3 MJ/s, 80%

8.3

8.6 8.7 8.8 8.9 8.10

Chapter9

8.11 Gl53> Pp BS’ & Op © 8.12 ere 8.14 b) 8440 kW, 16.8 MJ/s, 42.6 hp, 25.2 MJ/s, 33.5%

8.14 d) 9560 kW, 16.5 MJ/s, 42.8 hp, 26 MJ/s, 36.8%

9.1 9.2 9.3

8.15

b) 9150 kW, 177 Mi/s, 42.5 hp, 26.8 MJ/s, 34.1%

9.4

8.16

b) 10 MW, 16.4 MJ/s, 107 hp, 26.4 MJ/s, 379%

9.6

8.18

b) 10,300 23,600 b) 10,700 24,800 b) 11,600 23,400

9.7

8.19 8.20

hp, 16,400 Btu/s, 44.7 hp, Btu/s, 30.7% hp, 16,600 Btu/s, 44.9 hp, Btu/s, 33.2% hp, 15,200 Btu/s, 90.2 hp, Btu/s, 35.2%

8.22 26.6%, 27.3% 28.0%, 28.6% 8.24

72.6 MJ/s, 15.7 MJ/s, 38.5 KW, 49.8 MJ/s, 130°C, 214 hp, 43.6%, 298 kg/s

95

9.8 9.9

oO CeCe & CP 9.10 SO

195 kJ/kg, 541 kJ/kg, 36.0% 9.14 i 75.4in° 9.15 b) 56.5%, 982.4 kJ/kg, 555 kJ/kg 9.16 b) 20.3, 329.3 Btu/Ibm, 231 Btu/Ibm 9.12

9.18

a) 2001°C and 4660 kPa, 51.2%, 614 kJ/kg, 876 kPa

9.19

b) 553 kJ/kg, 46.1%, 789 kPa

9.20

b) 216°C 74,000 Btu/s, 16,500 Btu/s, 32000 Btu/s, 53,600 Btu/s, 339°F, 250 hp, 40.8%, 766 lbm/s

3783°F, 1188 psia, 60.2% ,234,000 ft-lbf/lbm, 137 psia :

9.21

b) 3670°F, 925 psia, 56.5%, 219,700 ft-lbf/Ibm, 132 psia

b) 74,100 Btu/s, 19,000 Btu/s, 37000 Btu/s, 56,100 Btu/s, 449°F, 250 hp, 39.7%, 801 Ibm/s

9.22

b) 179,200 ft-lbf/Ibm, 46.0%, 113 psia

9.24 3015395

b) 72.6 MJ/s, 18.3 MJ/s, 38.9 kW, 52 MJ/s, 182°C, 214 hp, 42.8%, 311 kg/s b) 74.9 MJ/s, 15.7 MJ/s, 42.9 kW, 47.6 MJ/s, 98.5%, 214 hp, 47.4%, 285 kg/s

68.7 MJ/s, 7360 kJ/s, 84.8 kW, 36.4 MJ/s, 58.2°C, 268 hp, 42.9% b) 58.4 MJ/s, 3.5 kg/s, 24.4 kW, 210 hp, 875%, 41.8%

b) 58,200 Btu/s, 9.32 Ibm/s, 31,000 hp, 245 hp, 88.8%, 376%

9.26

b) 1521°C and 6627 kPa, 65.8%, 527 kJ/kg, 644 kPa b) 436 kJ/kg, 54.5%, 540 kPa

2639°F, 1114 psia, 66.0%, 154,000 ft-lbf/lbm, 82.6 psia b) 66.0%, 2640 kJ/kg, 1740 kJ/kg

Answers to Selected Problems

9.34 9,36 9.38 9.39

54.3%, 589 kJ/kg, 1080 kJ/kg, 491 kJ/k 11.3, 24.7676 ; b) 72.2%, 5720 kJ/s, 520 kW, 64.1% b) 28.7%

9.40 63.4%

9.42 b) 20.8%, 5.13% 9.44 b) 71.8%, 3700 kW, 1040 kW, 28.2%, 3500 kJ/s 9.45 b) 56.7% 9.46 68.4%, 31.6% 9.48 b) 776% 9.50 12 MW, 40%, 11.6 MJ/s, 3560 kW, 30.8%, 51.4% 11-6 MJ/s, BTS

28.3%, 41.4% 9.54 b) 10,700 hp, 4.52 Ibm/s, 3130h re oo a i

kW,

Chapter 10 Hd 0.3 B 0p dra 10.5 D 0.6 D

C D B B A D B D

5) 0.5, 0.125, 0.375, 0.374, 0.164, 0.462, ) 0.398 kJ/kg-K

11.14 d) 0.3,0.4,0.3, 0.227,0.531, 0.242, 0.394 kI/kg-K 11.15 b) 0.3, 0.4, 0.3, 0.227,0.531, 0.242, 73.2 ft-Ibf/lbm-°R 11.15

d) 0.5, 0.125, 0.375, 0.541, 0.077, 0.382,

11.16

b) 0.266 kJ/kg-K, 5.77 kg

89.3 ft-Ibf/lbm-°R

11.18 11.20 11.22 11.24 11.26

kJ/kg, 169 kJ/kg,

10.12 10.14 10.16 10.18 10.20 10.22

11.6 kW, 28.4 kJ/s, 3.45, 0.0412 m°/s —15.6°C, 52 hp, 81.6 kJ/s, 2.1 0.062 kg/s, 2.79 kW, 2.86 0.296 Ibm/s, 84%, 10.5 hp, 22.4 Btu/s, 2.03 91%,2.13 11.1 kg/min, —22.3°C, 11.1 hp, 2.87

10.24

69°C, 1.30 ke/s, 85 hp, 2.1

WW 26uS6 kis hOsice sand 123 kg/s,o5hp,33, 83 hp, 2.5 10.28

113° 114 115 116 11.7 11.8 11.9 11.10

11.14

b) 8660 kW, 30.1%,

eaibe nese pias 10.10 eee

Chapter 1 11.1 B 112 C

11.11 B

51.8%

9,52

[R¥:¥:

—94°C, 1.48, 0.676 kg/s

0.9664, 0.0139, 0.0147, 0.0015, 43.48 kg/kmol 573 kPa, 16.2 kPa, 9.6 kPa. 1.2 kPa 0.4, 0.6, 0.478, 0.522, 40.3 ft-Ibflbm-°R, 209 Btu 8.27 m’, 484 kJ 0.222, 0.333, 0.444, 128 Ibm, 168 Ibm, 352 Ibm, 42.9 ft-Ibf/lbm-°R 11.28 122 kPa, 3930 kJ 11.30 2530 kJ, 5.6 kJ/K 11.32 661 kJ, 0.352 kJ/K ene

11.35 11.35 11.36 11.38 11.39 11.40

b)152kW d) 143 kW 676 kPa, 482 m/s a) 72.9%, 98.3 kPa, 0.0108 kg,/kg, b) 14.44 psia, 0.0177 Ibm, /Ibm, 5) 96.31 kPa, 0.0238 kg, /kg, |

11.41

b) 14.15 psia, 0.00112 Ibm,/lbm,

«11-42. 2) 94.9 kPa, 41.4% 11.43 b) 13.86 kPa, 52.9% 11.44

b) 12.0°C

Answers to Selected Problems

11.45

b) 62.4°F

12.10

11.46 45% 11.48

b) 13.2°C, 0.0096 kg,,/kg,,

11.50 0.00852 kg,/kg,, 48.4%, 43.6 kJ/kg

12.11

d) C,H, + 75(O, + 3.76N,) > 3CO, + 4H,O + 2.50, + 28.2N, b) C,H, + 6.75(O, + 3.76N,) = 4.5CO, + 3H,O + 1.5CO + 25.38N,

11.52 Hf) SUE

12.14

d) C,H, + 4.5(O, + 3.76N,) = 2CO, + 4H,O + CO + 16.92N, by CH, + C,H, + 14.5(O, + 3.76N,) > 9CO, + 11H,O + 54.52N, d) CH, + C;Hg + C,H, + 14.5(O, + 3.76N,) = 10CO, + 9H,O + 54.52N, 12.73 Ibm,,,/Ibm,,.,, 11.15%, 49°C

11.54 b) 28.4°C, 44%, 24.8°C, 92 kJ/kg,

12.15

b) 2.95 kg,,,../min

11.56

0.019 kg. /kg,, 23 kg,,/hr

12.15

d) 1.83kg,,,./min

11.58

b) 50.4 kg/hr

12.16

3.077 K8.oxyeen/KBjuep» 66-7%,72°C 45.1kg,. ke, 4.43%, 32°C 31.0 kg, /kg;,9 180%

11.51

b) 56%, 0.0147 kg,,/kg,, 101.7 kPa

11.51

d) 69%, 93.4 kPa

12.11 12.12

11.51 f) 48%, 0.0079 kg.,/kg,, 42 kJ/kg, 11.52

b) 71%, 0.0114 Ibm,/lbm,

11.52

d) 89%, 14.3 psia

12.12

11.60 59F

12.18

11.61

12.20

b) 0.0086, 0.986, 0.014

11.62

93% 11.64 10% 11.66 b) 205 kg,,/hr 11.67 b) 13.3 kJ/s, 15%

12.22

CH,

11.68

87 kJ/s, 0.017 kg,,/s

12.25

11.69

b) 25.4°C, 50%, 160 kg/min

12.26

11.70

63°F, 74%, 230 lbm/min

12.26

11.72

82%, 4.1 lbm/min

12.27

11.73

b) 130 kg_,/s, 2.2 kg,,/s

12.27

b) CH, b) 2.31 kg,,/K2p¢5 40.5°C b) —2 220 000 kJ/kmol d) —8 100 000 kJ/kmol b) —1 245 000 kJ d) —4596 000 kJ b) 1520 kJ/s d) 1500 kJ/s

12.23 12.24 12.25

Chapter 12

12.28

b) —768,000 Btu/Ibmol

12.28

12.1

12.30

d) —2,844,000 Btu/lbmol 1140 m/s b) 4280 kJ d) 11020 kJ b) 900 kPa

A

12.2

12.32

12.3

12.32

12.4

12.33

12.5

12.34

12.6

12.36

12.7

12.37

12.42

b) 3800°F b) 1580°F b) 1560°F 1530°C 1130°C

12.43

b) 1850°C, 746 kPa, 1880 m°

12.44

b) 940°C 1020°C

12.8

SPpennas

12.38

12.9

b) C,H, + 75(O, + 3.76N,) = 6CO, + 3H,O + 28.2N, d) C,H, + 5(O, + 3.76N,) = 3CO, + 4H,O + 18.8N, b) C,H, + 11.25(O, + 3.76N,) => 6CO, + 3H,O + 3.750, + 42.3N,

12.39

12.9

12.10

—7090 Btu, 34.3 psia, 10,000 ft* b) 1500°C :

12.40

12.46

Absolute pressure, 17 Absolute temperature, 22 ZEtTO, 23,56

Absorption refrigeration, 377 Acceleration, 10

of gravity, 11 Additive pressures, Dalton’s law of, 391

Additive volumes, Amagat’s law of, 391 Adiabatic efficiency, 225 of compressors, 226 of turbines, 225 Adiabatic flame temperature, 442 with excess air, 444 Adiabatic process, 95, 106

quasi-equilibrium, 107 Adiabatic saturation process, 400 Adiabatic saturation temperature, 400

Back-work ratio, 343 Barometer, 17 19 Basal metabolic rate, 492 Base units, 10

Beattie-Bridgeman equation, 67 Bi-metallic strip, 25 Biodiesel, 464 mixtures, 466 Biofuel, 462

Black body, 96 Boiler, 158,279 Boiling, 40 Bomb calorimeter, 488 Bore, 320 Bottom dead center, 320

Bourdon gage, 21 Brayton cycle, 163, 341

Advection, 95

back-work ratio, 343

Air, 391 atmospheric, 499

pressure ratio, 343

conditioner, 129, 367 excess, 427

properties of, 501, 549, 556 theoretical, 427

Air-conditioning processes, 405

efficiency, 161, 343 with reheating, and regeneration, 349 with regeneration, 346 Brayton-Rankine cycle, 352 Btu (British thermal unit), 56 Bulk modulus, 261

adding heat, 405, 407 adding moisture, 405 evaporative cooling, 408 latent heat cooling, 406 mixing air streams, 406, 409

removing moisture, 405 Air-fuel ratio, 427

Air refrigeration, 378 Air-standard analysis, 317, 396

Algae fuel, 467 Air-water vapor mixtures, 287 Amagat’s model, 391 Ammonia, tables, 542 Atmosphere, table of, 499, 500

Atmospheric air, 396 Atmospheric pressure, 18 Availability, 230

Calorimetry, 488 Carbon dioxide, properties of, 501, 552, 560

Carbon monoxide, properties of, 501,553, 561 Carnot cycle, 112, 189 efficiency, 190 P-v diagram, 189 reversed, 191

Carnot heat pump, 191 Carnot refrigerator, 191 Cascade refrigeration, 374 Celsius temperature scale, 22 Central difference, 72 Chemical compound, 426 Chemical energy, 434 Chemical reactions, 426 Chlorophyll, 485

Index

Clapeyron-Clausius equation, 252 Clapeyron equation, 256 Clausius inequality, 205 Clausius statement of the second law, 183

Clearance volume, 320 Closed feedwater heater, 294 Coal, 425 Coefficient: Joule-Thomson, 263 of performance, 186, 368 heat pumps, 187,372 refrigerators, 186, 369 heat pumps, 372 refrigerators, 372 of thermal expansion, 24 Cogeneration cycle, 299 Cold-air standard, 317 Combined reheat-regenerative cycle, 296 Combined gas-vapor cycle, 352 Combustion, 425 complete, 426 enthalpy of, 434 in a bomb, 441 incomplete, 426 theoretical, 426 with moist air, 431 Combustor, 341 Comfort zone, 405 Complete combustion, 426 Compressed liquid, 42 Compressed-liquid region, 42 Compressibility chart, 556 Compressibility factor, 66 Compression-ignition engines, 320, 330 Compression ratio, 320 effect on thermal efficiency, 324 Compressor, 139, 367 air, 140 efficiency, 226 first-law analysis, 139 intercooling, 348 Condensate, 294 Condensation,40, 53, 401 Condenser, 280, 367 Conduction, 96 Conservation: of energy, 28 control volumes, 138 steady-flow, 138 systems, 99 unsteady-flow, 155

of mass, control volumes, 133

steady-flow, 131 Constant pressure process, 102 Constant pressure specific heat, 69 of gas mixtures, 394 table, 503 - 506 Constant temperature process, 105 Constant volume process, 101 Constant volume specific heat, 69 of gas mixtures, 392 Continuity equation, 133 Continuum, 8 Control surface, 7, 129 Control volume, 7 128 second law, 221

Convection, 95 forced, 95 Convection heat transfer coefficient, 95 Conversion of units, 345

Cooling: with dehumidification, 405 evaporative, 406

Cooling tower, 278, 410 COP (see Coefficient of performance) Counter-flow heat exchanger, 148 Critical point, 52 table of properties, 502 Cutoff ratio, 332 Cycles efficiency, 158, 185 opens 17 power, 317 performance, 161 refrigeration, 160, 367 Cycles, 158 Brayton, 163, 341 Brayton-Rankine, 352 Carnot, 112, 189, 191 diesel, 330 Ericsson, 337 Otto, 323 Rankine, 158, 282 Stirling, 337 Cyclic formula for derivatives, 260 Dalton’s model, 391 Dead state, 230 Dehumidification, 405

Dehydration, 463

Index Density, 4

Energy equation, 138 Engine:

variable, 134

Dependent property, 40 Dew-point temperature, 401 Diesel cycle, 112, 330

Carnot, 112 external combustion, 317

two-stroke, 332 Differential, 13 exact, 13 inexact, 13 Differential form of first law, 252 Diffuser, 151

knock, 331 internal combustion, 317

Dimensional homogeneity, 9 Dimensions, 9 derived, 9

heat, 182

compression-ignition, 330 diesel, 330 reversible, 112

spark-ignition, 320 turbojet, 350 thrust of, 350

English units, 11

primary, 9

Enthalpy, 57

units of, 9

of air-water vapor mixture, 396, 401

Discharge of a tank, 155 Displacement volume, 320

of combustion, 435, 508 of formation, 434, 507

Dissociation, 427 Distillation, 463 Double interpolation, 48, 63

of gas mixture, 392 of liquids, 58

Dry air, 396 Dry-bulb temperature, 401 Dry steam plant, 472 Dual cycle, 335 Ectothermic organism, 492 Efficiency, 158, 185, 282 adiabatic, 225 Carnot, 190

compressor, 226 cycle, 158 plant, 282 second-law, 231

molar, 392

of products, 434 of reactants, 434 reference state, 434 of solids, 58

of vaporization, 60, 507, 508 Enthalpy change: of gas mixture, 392 general differential expression, 259 of ideal gas, 70 five methods, 70

4

thermodynamic, 282 turbine, 225

Electric potential, 92 Electrical work, 92 Emissivity, 96

Endothermic organism, 492 reaction, 434, 486

Energy, 27 chemical, 27

conservation of, 28 equation, 28, 99, 138 internal, 27

kinetic, 27 latent,60

potential, 27 Energy conversion in plants, 485 Energy efficiency ratio (EER), 187

of incompressible substance, 68 of reacting mixture, 434 of nonideal gas, 259, 266 Enthalpy departure, 266 chart, 568 Enthalpy-pressure diagram, 54 Entropy, 207 increase-in-, principle of, 209 production, 209

units, 209 Entropy change, 210 general expression, 210 of gas mixture, 392 of ideal gas, 211 of irreversible process, 208 of liquid, 213 of nonideal gas, 254, 267 of reservoir, 214 of solid, 213 of surroundings, 214

587

588

Index

Entropy change (continued)

systems, 99, 101

of universe, 209

unsteady-flow, 155

with variable specific heat, 218

Fixed volume, 102

of water, 216

Flame temperature, 442

Entropy departure, 267

Flash steam plant, 473

chart, 570

Flow rate, 133

Entropy production, 209 Equality of temperature, 22 Equations of state: Beattie-Bridgeman, 67

Flow-work, 89 Fluid, 39 working, 39 Fourier’s law of heat transfer, 95

ideal gas, 64 van der Waals, 67

Friction, 181 Fuel, 425

virial, 67 Equilibrium:

alage, 467 cell, 470

constant, 448, 509

Fuel cell, 470

composition of products, 449

types, 471

reaction, 447 of a system, 15

Fuel-air ratio, 427 Fusion, 40

thermodynamic, 15

heat of, 60

Equilibrium ratio, 427

Equivalence ratio, 427 Ericsson cycle, 112 Evacuated tank, 155

:

Gage pressure, 19 Gas, 8 Gas constant, 65

Evaporation, 40

of gas mixture, 393

Evaporative cooling, 366, 408

table of, 501

Evaporator, 367

universal, 65

Exact differential, 13

Gas mixtures, 389

Excess air, 427 Exergy, 230 Exhaust stroke, 322

properties of, 392 Gas power cycles, 317 Gas refrigeration cycle, 377

Exhaust valve, 320

Gas turbine, 341

Exothermic reaction, 434 Extensive property, 13 External combustion, 317 External irreversibility, 205 Fahrenheit temperature scale, 22

Gas turbine cycle (see Brayton cycle) Gas-vapor mixtures, 396 Generalized compressibility chart, 566 Geothermal energy, 473 Gibbs function, 252, 447 Gravimetric analysis, 390

Feedwater heater, 293

Gravity, 11

closed, 294 open, 293 Fermentation, 463 Finite difference, 72 First law of thermodynamics,

86, 99

proportionality constant, 11 specific, 14 Heat, 95 of fusion, 60

latent, 60

control volumes, 137

sign convention, 96

for a cycle, 99

specific, 96 (See also Specific heat)

differential form, 252

of sublimation, 60

liquid flow, 138

reacting systems, 435 steady-flow, 137

transfer, 95

of vaporization, 60 Heat capacity (see Specific heat), 69

Heat engines, 185 efficiency, 118 Heat exchanger, 435

entropy change, 211 equation of state, 65

cross-flow, 148

internal energy change, 70 isentropic processes, 211

parallel-flow, 148

law, 64

shell-and-tube, 148

properties of, 501 specific heats, 501

Heating value, 435 Heat pump, 161, 187,371 coefficient of performance, 187, 372 Heat ratio, specific, 71 Heat Heat Heat Heat

reservoir, 180 sink, 180 source, 180 transfer, 95

conduction, 95 convection, 95

during combustion, 436 radiation, 96 vs work, 96 isothermal, 105, 189

quasi-equilibrium,189 units, 97

Heating, 406 with humidification, 407

Heating value of a fuel, 435 table, 508 Helmholtz function, 252

High-efficiency furnace, 428 Higher heating value, 435

Homogeneous, 38 Horsepower, 143 Humidification, 405, 407

Humidity, 397 relative, 397

specific, 397 Humidity ratio, 397 HVAC, 396 Hydrocarbon fuel, 425 Hydroelectric power, 477 dams, 477

pumped storage, 477 Hydrogen, properties of, 501, 554, 562 Hygrometer, 399 Hyperthermia, 492 Hypothermia, 492 Ice cubes, 215

Ideal gas, 64 constant, 65

enthalpy change, 70

tables, 501

Ideal gas law, 64 Ideal gas mixture, 389 Incomplete combustion, 426

Incompressible substance, 73 enthalpy change, 74 entropy change, 213 internal energy change, 74 specific heat, 74, 503

Increase-in-entropy principle, 209 Independent property, 40 Inequality of Clausius, 205 Inexact differential, 13

Intensive property, 13 Intercooling, 348 Internal combustion engine, 317 compression-ignition engines, 330 spark-ignition engines, 322 Internal energy, 27, 56

of liquids, 58 of solids, 548 Internal energy change: of gas mixture, 392 of ice, 74 general expression, 258 of ideal gas, 70 of nonideal gases, 258 Interpolation, 47 double, 48 IRC Property Calculator, 49 example, 51 Irreversibility, 232 causes of, 231 steady-flows, 233 unsteady flows, 232 Isentropic efficiency, 225 of compressor, 226 of turbine, 225 Isentropic gas flow, 211 Isentropic process, 208 of ideal gases, 211 Isentropic relations of ideal gases, 211 Isobaric process, 102

ca

Index

Isolated system, 7 Isometric process, 102 Isothermal process, 105

Molar mass, 389 table of, 501, 502

Jet engine, 438

Mole, 65, 389 number of, 389 Molecules, number of, 8 Mole fraction, 389 2 Multistage refrigeration systems, 375 Muscle, 490 cardiac, 490 skeletal, 490 smooth, 490

Joule, 88 Joule-Thomson coefficient, 263

Kelvin-Planck statement of the second law, 182 Kelvin temperature scale, 22 Kinetic energy, 27

Latent heat, 60 cooling, 406 of fusion, 60 of sublimation, 60 of vaporization, 60 Law, 6 Leaf cross section, 485 Lean mixture, 427 Light frequency, 486 Liquid, 8 compressed, 42 flow, 133, 138 subcooled, 42 Liquid-vapor mixture, 42 Liquid saturation curve, 52 Losses, 302 Lower heating value, 435 Macroscopic forms of energy, 27 Makeup water, 411 Manometer, 20 Mass, 9 conservation of, 132 molar, 64 table of, 501 Mass flow rate (flux,) 131 Mass fraction, 389 Maximum flame temperature, 442 Maxwell relations, 253 Mean effective pressure, 320 Mercury, 19 Metabolism, 492 Mixing chambers, 145, 181 Mixing of air streams, 406 Mixture, 389 lean, 427 rich, 427 Molar analysis, 390

Molar specific heats, 382 Molar specific volume, 67

Muscle fibers, 491 Myosin, 491 Natural gas, 428 Newton’s law of cooling, 95 Newton’s second law, 9

NIST, 24 Nitrogen, properties of, 501, 550, 558 Nonequilibrium process, 16 Nonequilibrium work, 92 Nozzle, 151 subsonic, 151

supersonic, 151

Ocean thermal energy, 475 Open feedwater, 295 Open cycle, 317 Open refrigeration system, 377 Osmotic power, 478 Otto cycle, 112, 322 analysis of, 323 efficiency, 325

two-stroke, 327 Overall heat transfer coefficient, 314

Overall plant efficiency, 282 Oxygen, properties of, 501,551,559 Paddle wheel work, 93 Parallel-flow heat exchanger, 148 Partial derivative, 253

Partial pressure, 391 Pascal, 18 Path function, 13 PdV work, 88

Percent excess air, 427 Percent theoretical air, 427

Perfect gas, 64 (See also Ideal gas) Performance parameter, 185, 368

Index Phase, 39 diagram, 52 Phase-change processes, 40, 256

Propane, 437 Properties of ideal gases, 501, 549 Property of a system, 12 dependent, 40

property diagrams for, 52 Phase equilibrium, 24 Phlogiston, 6 Photon energy, 486 Photosynthesis, 485

intensive, 13 Psychrometric charts, 401, 564

Photovoltaic cells, 467 Piloerection, 492

Psychrometrics, 401 Psychrometry, 396

Piston-cylinder, 87

Psychrometer, 401

Planck’s constant, 486

extensive, 13

independent, 40

Poikilothermic, 492

sling, 401 Pump, 139 efficiency, 228

Polytropic process, 110 Potential energy, 27

Pump work, 139, 281 Pure substance, 38

Power, 92

P-v diagram, 52 P-v-T diagram, 53

Plant efficiency, 282

Power cycle, 315 Power plant losses, 302 Preheater, 293 Pressure, 17 absolute, 17

atmospheric, 18 critical, 52,502 gage, 17 mean effective, 320 measurement, 21

partial, 396 ratio, 335, 343 reduced, 67 relative, 219 units, 18 vacuum, 18

vapor, 396 Pressure ratio, 335, 343

Primary dimensions, 40 Principle of entropy increase, 209 Problem solving, 98 Process, 15 adiabatic, 106

irreversible, 205 isentropic, 208 isobaric, 102 isometric, 102

isothermal, 105

polytropic, 110 quasi-equilibrium, 16 reversible, 205 Production of entropy, 209 Products of combustion, 426

Quality, 42 of a two-phase mixture, 42 Quasi-equilibrium process, 16 adiabatic, 107

R134a, 61 tables, 530

Radiation heat transfer, 96 Rankine cycle, 158, 282

cogeneration cycle, 299 combined cycle, 296 efficiency, 158, 282 regenerative, 293 reheat, 290

supercritical, 287 Rankine temperature scale, 22 Ratio of specific heats, 71 table, 501 Reactants, 426 enthalpy of, 434 Reaction, stoichiometric, 426 Real gases, 66 Reciprocating engine, 320 Reduced pressure, 67 Reduced temperature, 67 Refrigerant 134a, properties of, 530 Refrigerants, 61, 373 Refrigeration, 367 absorption, 377 multistage, 374 ton of, 376

592

Index

Refrigeration cycle, 160, 183, 367 actual, 370

ammonia-absorption, 377 gas, 377 vapor, 367 Refrigerator, 161, 183 Carnot, 191 coefficient of performance, 186, 369, 372

Regeneration: Brayton cycle, 346 Ericsson cycle, 337 Rankine, 293

Stirling, 337 Regenerative cycle, 293 Regenerator, 338, 346 Reheat Brayton cycle, 349 Reheat Rankine cycle, 290 Reheater, 349 Reheat-regenerative cycle, 349

Relative humidity, 397 Relative pressure, 219

Relative specific volume, 219 Reservoir, thermal energy, 180 Resistivity, 95 Reversed Carnot cycle, 191 Reversible engine, 189 Reversible process, 181, 205

Reversible work, 231 R-factor, 95

Rich mixture, 427

systems, 208 control volume, 143

steady-flow, 144 Second-law efficiency, 231 Seebeck coefficient, 472

Semiconductor, 467 Sensible heat, 405, 407 Shear force, 92 Shaft work, 92 SI units, 9

prefixes, 10 Sign convention: heat, 96 work, 89

Significant digits (numbers), 9, 437 Simple substance, 38, 211 Sling psychrometer, 401

Solar energy, 467 active heating, 468 passive heating, 469 Solid, 8 properties of, 503 Spark-ignition engines, 320, 322 Specific enthalpy, 57 Specific enthalpy of air-water vapor mixture, 396, 401 Specific gravity, 14 Specific heat, 69 at constant pressure, 69 at constant volume, 69

Running lean, 427

as cubic function of temperature, 504 of a gas mixture, 285

Running rich, 427

generalized relations, 258

Saturated: liquid, 42 line; 52 mixture, 42

of ideal gas, 69 of incompressible substance, 73 of nonideal gas, 260 of superheated steam, 72 table of properties, 503-506

Rotating shaft, 92

vapor, 42

vapor line, 52 Saturated liquid-vapor, 510 Saturated solid-vapor, 529 Saturation pressure, 40 Saturation temperature, 40 Seasonal EER (SEER), 188 Second law of thermodynamics, 182 Clausius statement, 183 control volume, 221 Kelvin-Planck statement, 182

steady-flow, 222

at various temperatures, 505

Specific heat ratio, 71 Specific heat relations, 258 Specific humidity, 397 Specific internal energy, 56 Specific properties, 14, 56 Specific volume, 14 critical, 502

pseudo-reduced, 67 relative, 219

Specific work, 88 Speed of light, 486

Index Spring constant, 92 Spring work, 92 State, 12

State postulate, 16,58 Statistical thermodynamics, 8 Steady-flow devices, 139

scale, 22 wet-bulb, 401

Temperature scales, 22 Theoretical air, 427

Theoretical combustion process, 426

conservation of energy, 138

Thermal conductivity, 95 Thermal efficiency, 158, 185 Thermal equilibrium, 15

conservation of mass, 131

Thermal pollution, 280

second law analysis, 222

Thermistor, 27 Thermocouple, 26

Steady-flow, 133

Steam, 41, 47

Steam generator (see Boiler), 279 Steam power plant, 282 Steam table, 44,510 Steffan- Boltznann law, 96 Steffan-Boltznann constant, 96

Stirling cycle, 112, 337 Stoichiometric: air, 309 coefficient, 447 reaction, 426 Stroke, 320

Subcooled liquid, 42 Sublimation, 53 Sublimation, heat of, 67

Supercritical state, 52 Supercritical vapor power cycle, 287 Superheat region, 48 Superheated vapor, 42, 47 Surroundings, entropy change of, 214 Sustainability, 7 279, 462

Swamp cooler, 388 System, 7, 86 isolated, 7 Tds relations, 211

Temperature, 22 absolute, 22, 23 adiabatic flame, 442 adiabatic-saturation, 400 Celsius, 22 critical, 502

dew-point, 401 dry-bulb, 401 kelvin, 23 measurement, 24 Rankine, 23 reduced, 67 relative scale, 212

Thermodynamic equilibrium, 15 Thermodynamics, 5 first law of, 86 future of, 6

history of, 6 second law of, 182 zeroth law of, 22

Thermoelectric generator, 472 Thermogenesis, 493 Thermometer, 24

bi-metalic strip, 25 Bourdon gage, 25 thermister, 27

thermocouple, 26 Thermoregulation, 492 Throttling device, 143, 376, 373

calorimeter, 143 Throw, 329

Ton of refrigeration, 375 Top dead center, 320 Torque, 92 Torricelli barometer, 19 Transducer, 21 Transient flow, 155 Triple point, 54 T-s diagram, 217 Turbine, 139, 280 efficiency, 225 work, 139 Turbojet engine, 350 Turboprop engine, 351 T-v diagram, 52 Two-stage refrigeration, 374 Uniform flow, 132 Units, 9 of heat, 96 of work, 88 table of conversions, 498

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Index

Units, conversion of, 498 Universal gas constant, 65, 389

Voltage, 93

Universe, entropy change of, 209 Unrestrained expansion, 182 Unsteady-flow, energy, 155 discharging a tank, 155 filling a tank, 155 US. standard atmosphere, 499

pseudo-reduced, 67 specific, 14 Volume expansivity, 261 Volumetric analysis, 390 Volumetric flow rate, 133

Utilization factor, 300

Wankel engine, 328

Vacuum, 18 Valves, 143, 263

Water vapor, ideal gas properties of, SUI, 505 Wet-bulb temperature, 401

van der Waals equation, 67

Work, 88

Volume, 14

constants, 509 Vapor, 39, 41 saturated, 42

battery, 88

superheated, 47 Vapor refrigeration cycle, 160, 367 multistage, 347 Vapor power cycles, 158, 282 Vapor pressure, 396

nonequilibrium, 83 paddle wheel, 93 pressure, 88 quasiequilibrium, 89 rate (power), 137 reversible, 231

Vapor saturation line, 52 Vaporization, 40, 53

shaft, 92

enthalpy or latent heat of, 60 Variable specific heats, 218 Virial coefficients, 67

Virial equation of state, 67

electrical, 92 flow, 88

sign convention, 89 spring, 92 Working fluid, 39 Zeroth law of thermodynamics, 22 Z-factor, 66

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