Methodical manual to laboratory works on the course «The main processes and devices of chemical technology»: methodical manual 9786010418103

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AL-FARABI KAZAKH NATIONAL UNIVERSITY

D. N. Akbayeva Zh. T. Eshova

METHODICAL MANUAL TO LABORATORY WORKS ON THE COURSE «THE MAIN PROCESSES AND DEVICES OF CHEMICAL TECHNOLOGY» Stereotypical publication

Almaty «Qazaq university» 2020

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UDC 66.0 (075.8) LBC 35 я 73 А 29 Recommended for publication by the decision of the Academic Council of the Faculty of Chemistry and Chemical Technology and Editorial and Publishing Council of the Kazakh National University named after Al-Farabi (Protocol №2 dated 12.02.2016) Reviewers: doctor of chemical sciences, professor B.S. Selenova candidate philological sciences, associated professor A.A. Muldagalieva doctor of chemical sciences, head scientific worker G.O. Nurgalieva

Akbayeva D.N. А 29 Methodical manual to laboratory works on the course «The main processes and devices of chemical technology» / D.N. Akbayeva, Zh.T. Eshova. – Ster. pub. – Almaty: Qazaq university, 2020. – 80 p. ISBN 978-601-04-1810-3 In a methodical manual the theoretical material and techniques of experimental carrying out on laboratory works «Modes of the liquid current», «Hydrodynamics of the fluidized layer bulk», «Acquaintance with the multistage piston compressor of a high pressure», «Sedimentation of solid particles by gravity in the liquid environment», «Constants definition of filtering process», «Distillation process» are provided. For fixing of theoretical material on subjects of laboratory works the tasks and test tasks are also given. The methodical manual is intended for students of chemical and technological specialties of higher educational institutions. Methodical manual is published in authorial release. В методическом пособии приведены теоретический материал и методики проведения эксперимента по лабораторным работам «Режимы течения жидкости», «Гидродинамика псевдоожиженного слоя сыпучего материала», «Ознакомление с многоступенчатым поршневым компрессором высокого давления», «Осаждение под действием силы тяжести твёрдых частиц в жидкой среде», «Определение констант процесса фильтрования», «Процесс дистилляции». Для закрепления теоретического материала по темам лабораторных работ приведены также задачи и тестовые задания. Предназначено для студентов химико-технологических специальностей высших учебных заведений. Методическое пособие издается в авторской редакции.

UDC 66.0 (075.8) LBC 35 я 73 ISBN 978-601-04-1810-3

© Akbayeva D. N., Eshova Zh. T., 2020 © Al-Farabi KazNU, 2020

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Laboratory work № 1 MODES OF THE LIQUID CURRENT Work purpose: Make the visual observation over character of a liquid current and its transition from one mode to another one. Determine the critical values of Reynolds’s number (Re). Establish the dependence between pressure loss upon friction and average velocity of the stream movement. Main definitions and theory of process The pilot studies of regularities of a liquids current allowed to find the qualitative distinctions in structure of a stream depending on traffic conditions in channels of a different form. At rather small velocities of a stream in pipes the ordered character of a current of one layers concerning others is observed. Such ordered movement without hashing between layers is called as laminar, or jet. At the laminar mode the velocity profile in a stream is parabolic and it can be found theoretically. Average velocity of liquid on section is equal to a half maximum on a stream axis: wav = wmax / 2. In practice the streams in which there is an intensive hashing of the environment caused by instant change of particles velocity meet more often. Such movement is called as turbulent, or vortex. The whirl in a stream kernel is always followed laminar at a wall, and average velocity on section isn’t equal to a half maximum. In the turbulent mode the ratio wav/wmax is function of Reynolds’s criterion. The Reynolds’s criterion is used for a quantitative assessment of viscous liquids and gases currents character Re = wavd/ν. It characterizes a ratio between forces of inertia and powers of internal friction.

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Forces of inertia are defined by the velocity of a stream and its sizes, powers of internal friction – viscosity of a stream. From this it follows that turbulent flow is peculiar to the streams possessing the developed inertia forces, and laminar – is characteristic for streams in which powers of internal friction prevail over inertia forces. The values of Reynolds’s criterion corresponding to transition from one hydrodynamic mode to another are called as critical Recr. Need of studying of the modes of a current and definition of number Recr is caused by that upon transition from one mode to other stream undergoes a qualitative and quantitative changes. As a result of it the characteristics of hydraulic resistance and a heat-mass exchange change. To all modes of a current there correspond two critical Recr values. It is experimentally established that for a current average Recr values are approximately equal in round smooth pipes 2300 and 104. At Recr ≈ 2300 there is a change of the laminar mode, and at Recr ≈ 104 there comes the developed turbulent mode. The area 2300 < Re < 104 corresponds to the transitional mode in which the stream possesses at same time by properties of laminar and turbulent flow. It should be noted that in the specific conditions depending on many reasons, the values Recr can differ from the specified ones. So, in experiences it was succeeded to tighten approach of the turbulent mode up to Recr = 105 by a curve of an entrance edge of a pipe, reduction of external indignations and other ways. Thus, the small external indignation promoted the sharp change of the mode. Therefore, the current at such values of Reynolds’s number is unstable and has to be replaced by steadier form. For coils the value Recr increases depending on the ratio of a pipe diameter d to diameter of the coil D (d/D) and it can reach 7000÷8000. Existence of various modes of a current and regularity of transition from one to another can be observed if to change a liquid consumption in a transparent horizontal pipe and on its axis to enter a thin stream of the painted liquid. When transporting liquid environments in chemical technology the one their important tasks is the definition of pressure losses in the

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pipelines and devices caused by viscous friction. The pressure losses in round pipes are calculated on Darsi-Veysbakh’s formula:

P  

l wav2 , d 2

(1.1)

where λ – friction coefficient; l – length of a pipe, m; d – diameter of a pipe, m; ρ – density of the environment, kg/m3; wav – average velocity of a stream in the section, m/s. The complex of sizes λ·l/d shows, in how many time the pressure loss upon friction differs from losses of dynamic pressure. Hydraulic resistance on friction depends on the current mode. For a laminar current of liquid in pipes of round section the following dependence is fair: λ = 64 / Re.

(1.2)

In these conditions the size of friction coefficient is inversely proportional to Reynolds’s number and doesn’t depend on a roughness of walls. At a current of liquid in pipes of not round section (square, ring, etc.) the coefficient in a formula (1.2) has other values, and the number Re is calculated on the equivalent diameter of the channel. In the transitional and turbulent modes of a current the value λ depends not only on Re number, but also on a roughness of walls. For definition of the value λ in these conditions the large number of semi-empirical and empirical formulas is offered. For Re = 104÷105 and smooth pipes Blazius’s formula was widely adopted: λ = 0,316 / Re0,25,

(1.3)

corresponding to sedate distribution of a profile of velocity. Figure 1.1 contains the dependence ΔP from the average velocity of a current. The friction coefficient λ is in the transitional mode

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from graphic dependence (figure 1.2). If to present dependence ΔP from the average velocity of a stream in logarithmic coordinates, the points in figure 1.1 in which the right angle changes, will correspond to critical values Recr, i.e. transition of one mode of a current to another.

Figure 1.1. Dependence ΔP from the average velocity of a current: I – laminar, II – transitional, III – turbulent modes

Figure 1.2. Dependence of coefficient of friction λ from Re criterion: A – smooth and rough pipes; B – smooth pipes (copper, brass, lead, glass); C – rough pipes (steel, cast iron)

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Description of laboratory setup The main knots of setup (figure 1.3) are a glass tube 7 in which the mode of a current of liquid is investigated, a pressure head tank 6, a tank with dye-stuff 4, a measured vessel 8, the gate 9 and the crane 5.

Figure 1.3. The scheme of laboratory setup for studying of the modes of a liquid current and resistance of friction

Water from a water supply system via the gate 1 is pumped in a pressure head tank 6 in which thanks to existence of the drain union 2 its constant level is supplied. From a pressure head tank a water on a glass tube 7 (d = 14×1 mm) comes to the measured vessel 8 which is in a reception vessel 10. The water consumption through a tube is regulated by trailer needle gates. For visual observation of structure of a stream in a tube 7 from a tank 4 by means of capillary tubes the diluted dye-stuff is entered. The consumption of dye-stuff is regulated by the crane 5, and the velocity of the expiration of dye-stuff will be coordinated with the velocity of the stream. Water temperature is taken by the thermometer 3. Order of work performance 1. Open the water gate 1 and fill a pressure head tank 6 to the level of the drain union 2.

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2. Open the trailer gate 9, achieving the minimum consumption of water, and by means of the crane 5 give a dye-stuff. 3. Regulating dye-stuff supply, achieve an accurate outline of the tinted stream that corresponds to the laminar mode. 4. Having increased then a consumption of water, observe destruction of the laminar mode and its transition to the turbulent. In each of the set modes measure an expense and water temperature. Carry out 12–15 measurements, and finish measurements at the developed whirl. Determine a water consumption by a volume method by a measured vessel 8 and a stop watch. Processing of measurements results and content of the report 1. Determine the average velocity of the water movement in a glass tube by the expense equation: V = wсрF ,

(1.4)

where V – a consumption of water, m3/s; F – area of section of a glass tube, m2. 2. Calculate on formulas (1.2, 1.3) or find on graphics (figure 1.2) friction coefficients λ; calculate a pressure losses ΔP on a formula (1.1). 3. Show dependence of a pressure loss on the average velocity of the movement ΔP = f (wav) in logarithmic coordinates and define transition points from one mode in another. The report on work has to include: – purpose and description of work; scheme of laboratory setup; – an example of calculation of the sizes Re, λ, ΔP (coefficients of dynamic viscosity and density of water depending on temperature are given in table 1.1); – the reporting table of the measured and calculated sizes (table 1.2); – the schedule of dependence lg ΔP = f(lg wav) and the found values of Recr.

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Specify the water temperature (t, (μ Pas).

o

C), and viscosity of water Table 1.1

Physical properties of water (on the line of saturation) t, C

o

ρ, kg/m3

с, kJ/ (kgК)

λ102, W/ (mК)

а107, m2/s

μ106, Pаs

ν106, m2/s

β104, K-1

σ104, kg/s2

Pr

0 1000 4,23 55,1 1,31 1790 1,79 0,63 756 13,7 10 1000 4,19 57,5 1,37 1310 1,31 0,70 762 9,52 20 998 4,19 59,9 1,43 1000 1,01 1,82 727 7,02 30 996 4,18 61,8 1,49 804 0,81 3,21 712 5,42 40 992 4,18 63,4 1,53 657 0,66 3,87 697 4,31 50 988 4,18 64,8 1,57 549 0,556 4,49 677 3,54 60 983 4,18 65,9 1,61 470 0,478 5,11 662 2,98 70 978 4,19 66,8 1,63 406 0,415 5,70 643 2,55 80 972 4,19 67,5 1,66 355 0,365 6,32 626 2,21 90 965 4,19 68,0 1,68 315 0,326 6,95 607 1,95 100 958 4,23 68,3 1,69 282 0,295 7,5 589 1,75 110 951 4,23 68,5 1,69 256 0,268 8,0 569 1,58 120 943 4,23 68,6 1,72 231 0,244 8,6 549 1,43 130 935 4,27 68,6 1,72 212 0,226 9,2 529 1,32 140 926 4,27 68,5 1,72 196 0,212 9,7 507 1,23 150 917 4,32 68,4 1,72 185 0,202 10,3 487 1,17 160 907 4,36 68,3 1,72 174 0,191 10,8 466 1,10 170 897 4,40 67,9 1,72 163 0,181 11,5 444 1,05 180 887 4,44 67,5 1,72 153 0,173 12,2 424 1,01 o 3 The accepted designations: t, C – temperature; ρ, kg/m – density; c, kJ/(kgК) – specific heat capacity; λ102, W/(mК) – heat conductivity coefficient; а107, m2/s – diffusion coefficient; μ106, Pаs – dynamic viscosity; ν106, m2/s – kinematic viscosity; β104, K-1 – coefficient of volume expansion of a stream; σ104, kg/s2 – superficial tension; Pr – Prandtl’s criterion.

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Table 1.2 Reporting table

Volume V, m3

Time τ, s

Expense V, m3/s

Velocity w, m/s

Re

Observable mode of stream

Friction coefficient λ

Losses pressure ΔP, Pa

Questions for self-checking 1. Main modes of a liquid current. Nature of the streams movement. 2. How to determine the average velocity of the stream moving laminar? 3. What is the ratio between average and maximum velocity of a stream at turbulent flow? 4. What is Reynolds's criterion? What its physical sense? 5. What are the critical values of Re depending on pipeline type (a pipe, a coil)? 6. How nature of the movement of liquid changes: – when replacing water by glycerin/air; – at increase/fall of temperature of a stream; – at change of diameter of a pipe. 7. Equivalent diameter, its calculation for sections of the wrong form (a square, a rectangle, a ring). Literature 1. Reynolds А.J. Turbulent flows in engineering appendices: Translated from English. – M.: Energy, 1979. 2. Shlikhting G. Theory of an interface. – M.: Khimiya, 1974. Tasks for independent solution 1. Gas velocity 0,01 m/s, its density 1,91 kg/m3, viscosity 8,3510-6 Pаs are known, and diameter of a pipe is equal to 0,75 m. Calculate a Reynolds’s number. 2. Liquid velocity 0,77 m/s, its density 1150 kg/m3, viscosity 1,210-3 Pаs are known, and diameter of a pipe is equal to 25 mm. Calculate a Reynolds’s number. 3. Average velocity at the laminar mode of the liquid movement is 10 m/s. Calculate a maximum velocity of liquid (in m/s).

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4.

5. 6.

7. 8. 9.

10.

11. 12.

13. 14.

15.

Liquid velocity 2 m/s, its density 1200 kg/m3, viscosity 2,0510-3 Pа·s are known, and diameter of a pipe is equal to 0,88 m. Calculate a value of Reynolds’s criterion. Average velocity at the turbulent mode of the liquid movement is 15 m/s. Calculate a maximum velocity of liquid (in m/s). Diameter of a pipe 0,046 m, density of air 1,2 kg/m3 and its dynamic coefficient 0,018 mPаs are known. Calculate an air velocity (in m/s) at a transition period. Reynolds’s number is 1700. Calculate a friction coefficient. Reynolds’s number is 15 000. Calculate a friction coefficient. The diameter of a steel pipe 432,5 mm, diameter of a coil round 1 m, liquid density 1200 kg/m3, velocity of the liquid expiration on a tube 1 m/s and coefficient of friction 0,03 are known. Calculate a pressure loss upon overcoming of friction in a direct tube (in Pa). Diameter of a pipe 0,025 m, length of a tube 10 m, velocity of the liquid expiration on a tube 1,8 m/s and coefficient of friction 0,04 are known. Calculate a pressure loss on overcoming of friction in a direct tube (in m). Diameter of a coil round 1 m, internal diameter 39 mm, and loss of pressure in a direct tube 13,1 kPa are known. Calculate a pressure loss upon friction in a coil (in Pa). The loss of pressure upon friction in a coil 14,91 kPa and dimensionless correction coefficient  = 1,138 are known. Calculate a pressure loss upon friction in a direct pipe (in Pa). Solution passes on pipe space with a velocity 0,3 m/s. Calculate a high-speed pressure in tubes (in N/m2), if density of solution is 1100 kg/m3. Diameter of a pipe 0,034 m, length of a pipe 2 m, velocity of the gas expiration on a pipe 5 m/s and coefficient of friction 0,035 are known. Calculate a pressure loss on overcoming of friction in a direct pipe (in m). Gas passes on pipe space with a velocity 9 m/s. Calculate a highspeed pressure in pipes (in m).

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Laboratory work № 2 HYDRODYNAMICS OF THE FLUIDIZED LAYER BULK Work purpose: Experimentally determine the velocity of pseudo-liquefactions and a soaring of particles under the constrained conditions in a stream of air and compare them with the calculated values. Track the conditions of transition of a granular layer from a motionless state in weighed and pneumotransport mode. Main definitions and theory of process Now a number of processes of chemical technology at which there is an interaction of gas or liquid to high-disperse solid material (drying, roasting, adsorption, catalytic processes), are carried out in devices with the so-called weighed (fluidized), or boiling layer. In such devices the specified processes significantly are intensified. If through a motionless layer of the solid particles lying on a lattice pass from below up a stream of gas or liquid and thus gradually to increase its velocity, at some velocity of the environment called by critical (Wcr), all layer of solid particles passes into a weighed condition. At further increase of velocity of the environment the volume of the weighed layer increases in the device. Such extended weighed layer in which there is an intensive hashing of solid particles, in many respects reminds the boiling liquid – it «flows», takes the vessel form, through it the gas bubbles are gurgled; therefore it often is also called the boiling or fluidized layer. The single solid particle in the stream of gas or liquid directed up is affected (figure 2.1) by gravity G, carrying (Archimedean) power A and force of dynamic (high-speed) pressure of a stream P.

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For a spherical particle by diameter d (m) these forces are calculated on the following formulas:

G

d 3 6

d 3

 p g,

(2.1)

env g ,

(2.2)

P  W 2  env d 2 .

(2.3)

A

6

Here ρp and ρenv – densities of a particle and environment, kg/m3; W – velocity of a stream (environment), m/s; ψ – the dimensionless coefficient depending on the mode of a current of the environment; g – acceleration of gravity, m/s2.

Figure 2.1. Forces operating on a solid particle in the stream directed up

If G-A>P, then particle falls down; if G-A> ρl and expressions for criteria become simpler:

Ly  W f3  l2  l  s. p. g ,

(2.10)

Ar  d s3. p.  l  s. p. gl2 .

(2.11)

In engineering practice it is important to establish area of existence of a fluidized layer within Wcr÷Wsoar. For this purpose it is necessary to find the sizes Wcr and Wsoar which are calculated on Todes's formulas:

Re cr 

Ar , 1400  5,22 Ar

(2.10)

Ar , 18  0,61Ar

(2.11)

Re soar 

Wsoar  Re soar   l d eq  l  ,

(2.12)

where dr – diameter of a particle, m; μenv – dynamic coefficient of viscosity of the environment, Pа·s; deq – diameter of the device, m. Description of laboratory setup The laboratory setup for studying of hydrodynamics of a fluidized layer (figure 2.5) consists of a transparent glass column 1 (d = 20×1 mm) in which on a basic lattice the layer of granular material is filled. Behind a column the bottle 6 for catching of particles in the pneumotransport mode is established. The setup is equipped with the blower 3, rotameter 4 for definition of an air consumption and the differential manometer 2 for the measurement of pressure difference on a column. Air supply in a column is regulated by the gate 5.

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Order of work performance, processing of measurements results and content of the report 1. At the set of air expenses established by the gate 5 through the divisions 0,5 of the rotameter scale 4, measure the pressure difference on a column 1. Count an air consumption on graphics (figure 2.6). 2. Enter the obtained data in table 2.1 and construct the dependence Δp = f(Wf).

Ablation

Weighed layer

Motionless layer

Figure 2.4. Dependence of Ly criterion on Ar criterion and porosity of layer ε

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in atmosphere

Air Figure 2.5. The scheme of laboratory setup for studying of hydrodynamics fluidized layer Table 2.1 Reporting table

№ п/п

Air consumption V, m3/s

Velocity of air in a column, m/s

Observed mode

Pressure difference in a column, mm of water column/Pa

3. Determine the Wcr by the constructed plot, calculate the Lycr on a formula (2.7), determine the Ar (figure 2.5) at ε = 0,4 by graphic dependence of Ly on Ar and ε and calculate the diameter of particles on a formula (2.7), accepting ρp = 1,8 g/cm3. Values of density and viscosity of air, necessary for calculations, are given in table 2.2. 4. Calculate the ablation velocity of particles of this diameter on value Ly at ε = 1. Determine the weight of particles in a layer. 5. Set visually (3–4 times) the mode of a soaring and the beginning of ablation, measure thus a consumption of air and pressure difference in a layer.

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Table 2.2 Physical properties of dry air at atmospheric pressure с, ρ, t, μ106, λ102, ν106, Pr 3 C kg/m kJ/kgК m2/s W/mК Pаs 0 1,293 1,005 2,44 17,17 13,28 0,707 10 1,247 1,005 2,51 17,66 14,16 0,705 20 1,205 1,005 2,59 18,15 15,06 0,703 30 1,165 1,005 2,67 18,64 16,00 0,701 40 1,128 1,005 2,76 19,13 16,96 0,699 50 1,093 1,005 2,83 19,62 17,95 0,698 60 1,060 1,005 2,89 20,11 18,97 0,696 70 1,029 1,009 2,96 20,60 20,02 0,694 80 1,000 1,009 3,05 21,09 21,09 0,692 90 0,972 1,009 3,13 21,48 22,10 0,690 100 0,946 1,009 3,21 21,88 23,13 0,688 130 0,898 1,009 3,34 22,86 25,45 0,686 140 0,854 1,013 3,49 23,74 27,80 0,684 160 0,815 1,017 3,64 24,52 30,00 0,682 o 3 The accepted designations: t, C – temperature; ρ, kg/m – density; c, kJ/(kgК) – specific heat capacity; λ102, W/(mК) – heat conductivity coefficient; μ106, Pаs – dynamic viscosity; ν106, m2/s – kinematic viscosity; Pr – Prandtl's criterion. o

6. After averaging of an air consumption corresponding to the beginning of particles ablation on the equation of an expense, calculate the average velocity of a soaring Wsoar.exp and Resoar.exp.. From dependence (2.11) to calculate Rsoar.calcld. (the criterion Ar is defined in the step 3 earlier). W Re soar.exp 7. Using the ratio soar.exp  , find the Wsoar.calcldч. and Wsoar.calcld Re soar.calcld compare Wsoar..

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8. Write the obtained data in table 2.3. Figure 2.6. contains the schedule of determination of size of an expense. Table 2.3 Reporting table № п/п 1 2

3 4 5

Determined parameters

Values

Diameter of particles d, mm Critical velocity of pseudo-liquefaction Wcr, m/s: – for air; – for combustion gases Soaring velocity experimental Wsoar.exp., m/s Soaring velocity calculated Wsoar.calcld., m/s: – for air; – for combustion gases Weight of particles in a layer G, N

The report on work has to contain the purpose and an order of work performance; the scheme of laboratory setup, an example of calculation values of Wcr и Wsoar, plots and the tables of the experimental and calculated values (tables 2.1, 2.2). Questions for self-checking 1. What forces affect on a solid particle in a stream of gas or liquid? 2. What is the soaring velocity and on what it depends? 3. How are the fictitious, valid and critical velocities of gases differed from each other? 4. What is the porosity of a layer? In what limits the porosity of the weighed layer changes? 5. How the valid velocity in the motionless and weighed layers changes with increase of fictitious velocity of gas? 6. What character has the dependence of a layer porosity on a gas consumption? 7. From what the values the critical velocity and velocity of ablation depend?

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8.

Why with increase in fictitious velocity of air the hydraulic resistance of a motionless layer grows, and the weighed layer remains constant? 9. How the size of particles influences on porosity of a layer? 10. How to calculate the hydraulic resistance of the weighed layer, knowing the weight of a motionless layer? 11. Call the main advantages of the weighed (fluidized) layer for organizing of technological processes.

L/h

Divisions of a scale of the rotameter

Figure 2.6. Schedule for determination of an expense size Literature 1. Davidson J., Harrison of D. Pseudo-liquefaction. Tranlated from English. – M.: Chemistry, 1973. 2. Pseudo-liquefaction / Under the editorship of V.G. Aynstein, A.P. Baskakov. – M.: Chemistry, 1991. Tasks for independent solution 1. The bulk density of silica gel bulk = 650 kg/m3, density of particles  = 1100 kg/m3 are known. Calculate a porosity of a motionless layer.

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2. 3. 4.

5. 6. 7. 8. 9. 10.

11.

12. 13.

The height of a motionless layer 0,2 m, porosity of a motionless layer 0,4 are known. Density of particles is 1200 kg/m3. Calculate a hydraulic resistance of the weighed layer of particles (in Pa). The height of a motionless layer 0,2 m, porosity of a motionless layer 0,4 are known. Porosity of the weighed layer is 0,47. Calculate a height of the weighed layer (in m). The area of cross section of a particles layer 0,5 m2, height of a layer of solid particles 0,6 m and free volume of a particles layer 0,4 are known. Calculate a volume occupied by a layer of solid particles (in m3). The volume occupied by a layer of solid particles 0,37 m3, density of particles 1400 kg/m3 are known. Calculate a weight of this layer of solid particles. The working velocity of air 0,36 m/s and porosity of the weighed layer 0,43 are known. Calculate a valid velocity of air in the free section of a layer (in m/s). Calculate the velocity of an ascending stream of air in the air separator, necessary for office small (d < 1 mm) apatite particles from larger. Air temperature is 20 °C. Density of apatite is 3230 kg/m3. How the settler productivity will change if to increase the temperature of water suspensions from 15 to 50 °C? In both cases of Re < 0,2. Calculate density of the water suspension containing 10 % (mass.) of solid phase. The relative density of a solid phase is equal 3. Determine the precipitation velocity in water at 25 °C of oblong particles of coal ( = 1400 kg/m3) and lamellar particles of slate ( = 2200 kg/m3), having equivalent diameter 2 mm. Determine the velocity of the air necessary to start formation of the weighed layer of particles granulated alumosilicagel under following conditions: air temperature is 100 °C; density alumosilicagel (seeming) equal to 968 kg/m3; diameter of particles is 1,2 mm. What hydraulic resistance, if height of a motionless layer of 400 mm will be? In the conditions of the previous task define the porosity and height of the weighed layer if the velocity of air exceeds critical by 1,7 times. Determine the greatest diameter of the granulated particles of coal, starting to pass to a suspension in air at a velocity it in device of 0,2 m/s. Temperature is 180 °C. Define it is also volume concentration of particles if the velocity of air increases to 0,4 m/s. Coal density (seeming) equal to 660 kg/m3.

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Laboratory work № 3 ACQUAINTANCE WITH THE MULTISTAGE PISTON COMPRESSOR OF A HIGH PRESSURE General information Compressors are the devices intended for compression of gases in the machines, in which extent of compression – the ratio between final pressure of gas to initial one more than three

p2  3, p1

(3.1)

where р1 – the initial pressure of gas (absorption pressure); р2 – the final pressure (forcing pressure). In contrast of vacuum pumps in which р2  1 ат, and р1 < 1 ат, in compressors forcing pressure always is more than the atmospheric. In a course of thermodynamics theoretically different processes of compression of gas from initial pressure р1 (a point A) up to the final pressure р2 are considered (figure 3.1): 1) if all heat which is allocating during compression of gas, continuously is completely taken away (at reversible process), for example, by the water cooling walls of the compressor, process of compression happens at a constant temperature Т1 – isothermal – the line AB; 2) if during gas compression heat isn't taken away and not brought – the compressor ideally heat-insulated, friction and heat exchange in it are absent, process of compression happens at invariable entropy S1 – adiabatically – the line AC;

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3) if being allocated heat is taken away not completely, and only partially, the process of compression will be polytropic at which the indicator of a polytrope т is less than an indicator of an adiabatic curve k – the line AD. If in the compressor there is additional allocation of heat as a result of friction and arrival of heat exceeds its branch, there will be a polytropic process of compression, at which т > k – the line AE.

Figure 3.1. Chart T-S

Apparently from the chart (figure 3.2), the final temperature of gas after its compression up to the pressure р2 for various processes of compression will be different. The work spent at these processes for compression of gas will be various also. As at isothermal process AB the work expense the smallest, such process of compression would be also the most economic, however practically it is impossible to carry out reversible isothermal process of gas compression in actual practice. The valid working process in the cooled compressor goes usually on some polytrope. Final temperature of gas Т2 (in K) after its polytropic compression up to the pressure р2 can be defined from the following equation

T2  p2    T1  p1 

m1 m

,

(3.2)

– 26 –

where р1 and Т1 – the initial pressure and gas temperature; m – polytrope indicator. In figure 3.2 it is presented theoretical р-V chart (indicative) of the one-stage piston compressor.

A

B

Figure 3.2. Chart p-V

On this chart the volume Vo is the geometrical (working) volume described by the piston at its movement from extreme left position A before the extreme right position B. The line ab represents process of gas absorption and the compressor cylinder with a constant pressure р1, the line bc – gas compression at a piston stroke from right to left, the line cd – pushing out of the compressed gas from the cylinder with a pressure р2, the line da – an expansion at a piston stroke to the right compressed gas which has remained in harmful (dead) space, from pressure р2 up to the pressure р1, i.e. before opening in a point and the soaking-up valve. The volume V – the gas volume which has arrived with a pressure р1 in the cylinder of the compressor during a

– 27 –

absorption course. The ratio V/Vо = о is called volume efficiency coefficient (e.c.) of the compressor. Apparently from the chart (figure 3.2), with a increasing of final pressure from р2 to р'2, i.e. with increase of degree compression from р2/р1 to р'2/р1 volume the value of e.c. of the compressor decreases:

V' V  , V0 V0

(3.3)

In figure 3.3 the scheme of the one-stage piston compressor of double action (the cylinder works with two parties) with water cooling shirt is given. In order to avoid the considerable fall volume e.c. i.e. the compressor productivity, and inadmissible temperature increase in it, the extent of gas compression in the compressor cylinder shouldn't exceed some limit. In one-stage piston compressors with cooling water shirts the ratio р2/р1 usually makes 3-5.

Water

Figure 3.3. Scheme of the one-stage piston compressor of double action: 1 – piston; 2 – rod; 3 – the soaking-up valve; 4 – delivery valve; 5 – cooling shirt; 6 – oil-seal.

– 28 –

If it is required to compress the gas to more high pressure, the multistage compressors with intermediate cooling of gas between steps in portable water refrigerators are applied. Thus extent of compression in each step doesn't exceed the limit stated above. In figure 3.4 the scheme of the two-level piston compressor with intermediate cooling of gas between steps is submitted. In parallel located cylinders of simple action (work with one party), but can be and double.

Figure 3.4. Scheme of the two-level piston compressor: 1 – cylinder of Ist step; 2 – cylinder of IInd step; 3 – intermediate refrigerator; 4 – final refrigerator after the IInd step; 5 – receiver (collector-oil separator); 6 – cranked shaft; 7 – filter

In multistage piston compressors other arrangement of cylinders – horizontal, consecutive, V-shaped is applied also. Accepting that in all n steps of the multistage piston compressor the extent of compression i is identical, and disregarding pressure losses between steps in valves and intermediate refrigerators, we will receive the following approximate dependence

– 29 –

in 

p final pinit

,

(3.4)

where рinit – absorption pressure; рfinal – forcing pressure. Work purpose: acquaintance with operation of the piston compressor as the solution of theoretical tasks on the teacher's recommendation. Solution of tasks on this theme. Tasks for independent solution 1. Determine by an analytical way and by the chart T - S the temperature air after its adiabatic compression from initial pressure (absolute) 1 kgf/cm2 up to the final pressure of 3,5 kgfs/cm2. Reference temperature is 0 °C. Define also expense of work for compression of 1 kg of air. 2. Determine the capacity consumed by carbon dioxide piston compressor with a productivity of 5,6 m3/h (under absorption conditions). Compressor compresses carbon dioxide from 20 to 70 kgfs/cm2 (pressure absolute). Initial temperature is – 15 °C. Coefficient of efficiency of the compressor to accept equal 0,65. Solve this task by analytical way, and using the chart T - S for carbon. 3. Define a volume coefficient of efficiency of compressor of the previous task, if the harmful space makes 6 % from the volume described by the piston, and the polytrope indicator of expansion m = 1,2. 4. Determine a productivity and the spent capacity for the one-stage piston compressor according to the following data: diameter of the piston is 250 mm, a piston stroke is 275 mm, volume of harmful space is 5,4 % from volume, described by the piston, the of rotation frequency is 300 rpm. The compressor squeezes atmospheric air to рabs = 4 kgfs/cm2. A polytrope indicator of expansion for 10 % it is less than adiabatic curve indicator. Reference temperature of air is 25 °C. The general coefficient of efficiency of the compressor is 0,72. 5. How does the productivity and compressor power consumption change in the previous task if the blower pressurization to рexec = = 0,4 kgfs/cm2 will be given? Final pressure (absolute) is 4 kgfs/cm2. 6. At what forcing pressure the volume coefficient of efficiency of the one-stage the piston compressor compressing ethylene, will fall to 0,2? Pressure of absorption is 1 kgf/cm2. Gas expansion from harm-

– 30 –

ful space consider adiabatic. Volume of harmful space makes 7 % from the volume described by the piston. 7. Proceeding from a condition that compressor lubricant oil allows without noticeable deterioration of greasing temperature in the cylinder not above than 160 °C, determine limit value of forcing pressure in the one-stage piston compressor: a) for air, b) for ethane. Pressure of absorption is 1 kgfs/cm2. Initial temperature is 25 °C. Consider process of compression adiabatic. 8. Define a demanded number of steps of the piston compressor, which has to compress nitrogen from 1 to 100 kgfs/cm2 (pressure absolute), if the allowed temperature at the end of compression shouldn't exceed 140 °C. Consider process of compression the adiabatic. Reference temperature of nitrogen is 20 °C. 9. Define theoretical expense of work for hydrogen compression from 1,5 to 17 kgf/cm2 (pressure absolute) at one-stage and two-level compression. Reference temperature of hydrogen is 20 °C. 10. The compressor at test forced atmospheric air in a cylinder the volume of 42,4 dm3. In 10,5 min the pressure in a cylinder raised from 0 to 52 kgfs/cm2 (pressure superfluous), and air temperature in a cylinder rose from 17 to 37 °C. Determine compressor productivity in m3/h (under normal conditions). 11. Determine the power consumption and a consumption of water on refrigerators of the piston compressor which squeezes 625 m3/h (under normal conditions) ethylene from pressure (absolute) 9,81×104 to 176,6×104 Pa. Coefficient of efficiency of the compressor 0,75. Cooling water heats up in refrigerators on 13 °C. Initial temperature of gas of 20 °C. Tests for independent work 1. The machines intended for movement and compression of gases are called: A) Pumps. B) Compressors. C) Centrifuges. D) Piston pumps. E) Centrifugal pumps. 2. Productivity of the compressor is measured in: A) m3/s.

– 31 –

B) m2/s. C) m/s2. D) kg/s. E) kg/m3. 3. The theoretical work spent by the multistage compressor at gas compression is measured in: A) J/kg. B) J/s. C) kW. D) kJ. E) J. 4. Extent of compression at which the volume coefficient of the compressor becomes equal to zero is called: A) Harmful space. B) Compression limit. C) Polytrope indicator of gas expansion. D) Productivity. E) Indicator power. 5. The ratio between power of the isothermal machine and the actual power of this machine working with gas cooling is called: A) Thermodynamic efficiency. B) Isothermal efficiency. C) Adiabatic efficiency. D) Isoentropic efficiency. E) Full isothermal efficiency. 6. The work of isothermal and mechanical efficiencies is called: A) Thermodynamic efficiency. B) Isothermal efficiency. C) Adiabatic efficiency. D) Isoentropic efficiency. E) Full isothermal efficiency. 7. Productivity of the compressor is calculated on a formula: A) QρgH/1000η. B) fsn. C) QγH/102η.

– 32 –

D) fsnλ. E) ρgQH. 8. Values of residual pressure of system in the conditions of high vacuum are in the following limits: A) 10-3  10-8 mm of mercury. B) 1,0  560 mm of mercury. C) 1,0  10-3 mm of mercury. D) 560  760 mm of mercury. E) 1,0  760 mm of mercury. 9. Values of residual pressure of system in the conditions of average vacuum are in the following limits: A) 10-3  10-8 mm of mercury. B) 10-8  mm of mercury. C) 1,0  10-3 mm of mercury. D) 560  760 mm of mercury. E) 1,0  760 mm of mercury. 10. Values of residual pressure of system in the conditions of low vacuum are in the following limits: A) 10-3  10-8 mm of mercury. B) 10-8  mm of mercury. C) 1,0  10-3 mm of mercury. D) 10-3  mm of mercury. E) 1,0  760 mm of mercury. Questions for self-checking 1. What kind of devices call compressors? 2. How the vacuum pumps differ from compressors? 3. What is extent of compression? 4. In what limits extent of compression in one step of the piston compressor changes? 5. How the various thermodynamic processes of compression differ? 6. In what case compression is made at the smallest expense of work? 7. How to determine the gas temperature at the end of process of polytropic compression?

– 33 –

8. 9. 10. 11. 12. 13.

How is cooling of the gas compressed in the compressor carried out? Give the image of the processes happening in the compressor, using the T-S and p-V charts. What is volume coefficient of efficiency of the compressor and how it depends on extent of compression? Why is the multistage compression applied? What ratio between extent of compression and number of steps in the multistage compressor exists? What is purpose of an oil-seal?

Literature 1. The manual to a practical training in laboratory of processes and devices of chemical technology / Under the editorship of P.G. Romankov. – L.: Chemistry, 1975. – P. 52-56. 2. Kasatkin A.G. Main processes and devices chemical technologies. – M.: Chemistry, 1973. – P. 152-175. 3. Pavlov K.F. etc. Examples and tasks of a course of processes and devices of chemical technology. – L.: Chemistry, 1987. – P. 70-71, 89-90.

– 34 –

Laboratory work № 4 SEDIMENTATION OF SOLID PARTICLES BY GRAVITY IN THE LIQUID ENVIRONMENT General resistance law of the environment Hydromechanical sedimentation process of solid particles in the liquid or gas environment is widespread in equipment and is used for separation of various two-phase or multiphase systems. Physical characteristics of solid particles movement in the environment of liquid or gas depend not only on rheological properties of systems, but also on many other factors connected, mainly, with conditions of a flow (figure 4.1), i.e. with the interface theory, and resistance law. a

b

c

Figure 4.1. The scheme of a bodies flow of a different form in various modes: a – the cylinder; b – plates with keen edges (Re = wbρ/μ = 10, where b – the geometrical size of a particle); c – plates with keen edges (Re = wbρ/μ = 250)

– 35 –

In case of quantitative determination of hydraulic resistance of the solid particles moving in a flow of liquid or gas, first of all it is necessary to establish the connection between loss of a kinetic energy and a movement mode. Generally force of resistance which is showing in a moving flow, is expressed by the equation Fr  C

w02 2

f ,

(4.1)

where Fr  resistance force; f  square of a projection of particles at the plane, perpendicular to the direction of their movement;   environment density; w0  velocity of the environment far from a particle; C  coefficient of the proportionality making sense of general coefficient of resistance. The equation (4.1) can be provided in the following type:

Fr Cw02  , 2 f

(4.2)

The received equation (4.2) connects the pressure difference overcoming by a moving particle (the ratio between resistance force and the unit area of a particle section), and the part of a kinetic energy spent for resistance of movement. As the complete force of resistance can be provided by the amount of frontal resistance and resistance to the friction,

Fr  F f .r .  F fr ,

(4.3)

that and the general coefficient of resistance  can be expressed by the equation

   f .r .   fr ,

(4.4)

– 36 –

where  f .r .  the coefficient of the front resistance, which is often called as a form coefficient;  fr  friction coefficient. In case of a laminar current the particle is smoothly flowed round by a flow, and energy is spent thus generally only for friction overcoming. In case of turbulent flow inertia forces start prevailing, and with increase in velocity the increasing role is played by the front resistance depending on a form of a streamline surface. With achievement of some value of Reynolds's number by resistance of friction it is possible to neglect and the main expenses of a flow energy will be made on overcoming of front resistance. According to the equation for calculation of resistance coefficient in case of various hydrodynamic modes can be received by handling of the experimental data in the form of the generalized dependences between criteria of hydrodynamic similarity. Thus, it is possible to write down the generalized dependence:

  f (Re).

(4.5)

The type of this function for a flow of spherical particles on the basis of a large number of experimental data is provided by diameter in figure 4.2 (graphical dependence call the Rayleigh's curve). From the schedule it is visible that there are three various modes of movement, to each of which there corresponds a certain nature of dependence  from Re : laminar mode (region of the Stokes law) approximately in case of Re < 2:

 

24 Re

transitional mode in case of Re = 2 – 500:

(4.6)

– 37 –

 

18,5 Re 0, 6

(4.7)

automodel mode (reg gion of the Newtoon's resistance sq quare law) in 5 case of 210 > Re > 500:

  0,44  const

(4.8)

Figu ure 4.2. Dependencce of resistance coeefficient of the envirronment from moode of a spherical particles flow

To dependence 24/Re there coorresponds a diirect site in logarrithmic coordinates (figure 4.2). It should be noted that t in case of a sp pherical particles flow (or sedimentation) on chartt   f (Re) transsition from a lam minar mode to thee turbulent isn’t expressed so distin nctly, as in case of o movement of floows in pipes. 5 In case of the dev veloped turbulencce in the field of 210  > Re > 500 0 can be neglectted resistance of friction, and fro ont resistance beco omes prevailing. In I case of self-sim milarity approach constants c can conssider coefficient of resistance (equaation 4.8). The considered resistance r law off the environmen nt belongs to free movement of sp pherical solid parrticles. Resistancee of the real

– 38 –

partiicles, which form m differs from a spphere (figure 4.3)), in addition depeends on a form facctor (sphericity cooefficient).

Figu ure 4.3. Resistance coefficient c of the ennvironment for a floow of particles various form: fo 1 – sphere; 2 – cylinder; 3 – disk

The factor  forr particles of the irrregular form is as a the relation of a surface of the sph here f sph having the same amountt, as well as a p to a partiicle surface f : real particle,



f sph f



4,878 8 V 2/3 f

(4.9)

The sphere surfaace with the amouunt equal to amount of a particle, is determined by the eequivalent diam meter equal 2/3 3 m where f sph   6V p /  . d eq  6V p /  . From





Thus, for the sollid particles differring in a form from a sphere, the value v of resistancee coefficient moree also depends no ot only on criterio on of Re, but also on a form factor  :

  f (Ree, ).

(4.10)

Values of a form factor for some soolids are given in table 4.1.

– 39 –

Table 4.1 Values of a form coefficient for some solids Form of particles

Ψ

sphere

1,000

cube

0,806

Cylinder ( h  3r )

where h  height;

Form of particles Prism ( а  а  2а ) cylinder ( h  10 r )

cylinder

0,860

( h  20 r )

Ψ

0,767 0,691 0,580

r  radius; а  length.

Resistance coefficient for non-spherical particles can be determined by the dependence similar to the equation (4.6) for spheres:

  А / Re,

(4.11)

where A  f ( ). For laminar sedimentation of non-spherical particles the empirical equation can be used: A

24 0,843 lg( / 0,065)

(4.12)

For spherical particles (  1) the equation (4.12) gives a value A = 24, and dependence (4.12) turns into the equation (4.6). For turbulent sedimentation the resistance coefficient doesn’t depend on Reynolds’s number, but only on a form factor:

  5,31  4,88 . For spherical particles (  1) we will receive   0,43.

(4.13)

– 40 –

The general resistance law of the environment doesn’t depend by nature forces causing movement of solid particles in this environment. Sedimentation velocity In case of sedimentation of fine dispersive solid particles in gas or the liquid, being observed in case of gas purification, separation of suspension, the main characteristic of process is sedimentation velocity. Precipitation represents process of separation in case of which fluidized in liquid or gas solid or liquid particles separate from a continuous phase by gravity, inertia forces (including centrifugal) or electrostatic forces. The precipitation which is occurring by gravity, is called as upholding. Generally upholding is applied to preliminary, rough separation of heterogeneous systems. The particle falling by gravity, will increase the velocity until resistance force of the environment won’t counterbalance the gravity. Then the particle will continue movement regularly, with a fixed velocity. This fixed velocity call the velocity of free sedimentation Ws . Thus, in case of a particle fall the three stages of its movement take place: 1) initial moment of fall; 2) movement with increase in velocity; 3) uniform movement (with a fixed speed). Velocity increase from W  0 to W  W fin  Ws happens during very short period (for example, the particle of a dust with a diameter of 10 microns and density ρsolid = 2700 kg/m3 reaches the fixed velocity of sedimentation through 0,006 s), therefore the third stage of movement of a body is of interest to technical calculations only. Under the influence of gravity it is possible to apply a general formula which is fair for laminar, transitional and turbulent modes of particles sedimentation with various sphericity for determination of sedimentation velocity:

Ws 

4  sol   gd p ,  3

(4.14)

– 41 –

wherre d p  diameterr of a being bessieged particle, m; m

sol and

  density of a solidd particle and envvironment, respecttively, kg/m3;   resistance coefficient. The equation (4.1 14) is fair for lam minar, transitional and turbulent mod des of particles sedimentation witth various sphericcity, and the resisstance coefficient can be determinedd from figure 4.4..

Figure 4.4. Depen ndence of resistancce coefficient of the environment on Reynolds's R number and a a form factor off the particles which h are besieging by gravityy

Velocity of precipitation of a spheerical single particcle in case of a lam minar mode is deteermined by a form mula:

Ws 

d p2  sol   g 18

.

(4.15)

Dependence (4.15) carries the nam me of a Stokes fo ormula and is 4 fair for f area 10  Ree  2 . To determine a mode m of precipitattion and, thereforee, to choose a form mula for calculatio on of sedimentatiion velocity, it is necessary to

– 42 –

know the size of Reynolds's criterion which is connected with Ws . Respect there to equations (4.14) and (4.15) are applicable for calculation of sedimentation velocity by the method of consecutive approximations, i.e. at the first step of calculation it is necessary to be set, for example, by a laminar mode of sedimentation, and then, having determined precipitation velocity, to check, whether Reynolds’s criterion in the area corresponding to the accepted condition lies. In case of discrepancy of results it is necessary to pass the second step of calculation before obtaining of satisfactory converence of data. However such labor-consuming procedure of calculation of sedimentation velocity can be avoided, having changed the equation (4.14) by Lyashchenko's method. Lyashchenko's method is based on substitution in the equation (4.14) of expression Ws  Re  /(d p  ) and squaring of both parts of the received equation, from where the following expression turns out: 3 2 4 d p  g  sol    Re  3 2  2

4.16)

The right member of equation (4.16) represents Archimedes criterion. Thus:

 Re 2 

4 Ar 3

(4.17)

Knowing the size of Archimedes criterion (for sedimentation of a given size particles), according to the Lyashchenko's chart (figure 4.5) it is possible to calculate easily value of Re from which the required velocity of sedimentation is determined. As the size of resistance coefficient  depends on a sedimentation mode, it is possible to establish boundary values of Archimedes criterion, corresponding to transition of one area of sedimentation to another one.

– 43 –

Figure 4.5. Depeendence Ly = f (Ar)): 1,6 – spherical paarticles; 2 – rounnded particles; 3 – aangular particles; 4 – obllong particles; 5 – laamellar particles

In the field of a laaminar mode of sedimentation (at Re R  2, i.e. in the conditions whicch are characterrized by the Stokes's S law)   24 / Re and the eqquations (4.15) annd (4.17) will becoome:

R  Re

Ar or Re  0,056 Ar 18

(4.18)

– 44 –

In the region of the Newton's law (in the conditions of automodelity of Re criterion)   0,44 and the equation (4.17) it is possible to provide so: (4.19) Re  1,74  Ar 0,5 In transitional area the upper limiting value of Archimedes criterion corresponds to Re = 500 and is calculated on the equation:

Re  0,152 Ar0,715

(4.20)

In case of a transitional mode the values of Archimedes criterion will be in the range 36  Ar  8,3  10 4. The interpolation semi-empirical dependence connecting criteria of Ar and Re for approximate calculations of sedimentation velocity of a single particle, looks like:

Re 

Ar 18  0,61Ar 0,5

(4.21)

Approximate calculation of sedimentation process of nonspherical particles can be performed as follows. A) If it is necessary to find the sedimentation velocity of nonspherical particle, at first determine the equivalent diameter of a particle of the set volume:

d э  3 6V /   1,243 M /  sol ,

(4.22)

where V  particle amount, m3; M  mass of a particle, kg; sol  density of a particle, kg/m3. Then the size of Archimedes criterion can be calculated on formula:

Ar 

d eq3 (  sol   ) g

2

(4.23)

– 45 –

By means of figure 4.5 and proceeding from the size Ar, find value of Lyashchenko's criterion on a curve for a spherical particle then calculate the sedimentation velocity of a particle of nonspherical form:

Ws   /  3 Ly (  sol  env ) g / env , 2

(4.24)

where  – the coefficient considering a form of a particle,  – dynamic coefficient of the environment viscosity, Pаs. /

The values of coefficient data are given in table 4.2.

 / determined by the experimental Table 4.2

Values of coefficient

/

for particles of a various form

Form of a particle Archimedes criterion

Spherical

Rounded

Angular

Oblong

Lamellar

15 000 20 000 40 000 100 000 200 000

1 1 1 1 1

0,805 0,800 0,790 0,755 0,753

0,680 0,678 0,672 0,650 0,647

0,610 0,595 0,590 0,564 0,562

0,450 0,441 0,433 0,420 0,408

400 000

1

0,740

0,635

0,560

0,392

B) If it is necessary to determine the size of a particle of the irregular form by a preset value of sedimentation velocity, at first calculate the size of Lyashchenko's criterion:

Ly 

2 Ws3  env  env (  sol   env ) g

(4.25)

– 46 –

Further by means of figure 4.5 on a curve for a spherical particle find the size of Archimedes criterion then calculate the equivalent diameter:

d eq   // 3

(  sol

2 Ar env ,   env )  env g

(4.26)

 // – the coefficient considering of a particle form. // The values of coefficient  , found on the experimental data,

where

are given in table 4.3. Table 4.3 Values of coefficient

 //

for particles of a various form

Form of a particle Lyashchenko's criterion

Spherical

Rounded

Angular

Oblong

Lamel lar

13 130 260 580 2600 5000

1 1 1 1 1 1

– 1,21 1,34 1,44 1,61 1,76

– 1,495 1,640 1,700 1,960 –

– 1,865 2,030 1,180 2,500 –

2,09 2,92 3,34 3,68 – –

Work purpose: a) determine the sedimentation velocity of a single particle depending on diameter, forms and from physical properties of a particle and liquid (environment); b) using the experimental value of sedimentation velocity, calculate the equivalent diameter of a particle depending on a form and physical properties of a particle and liquid (environment). Compare the calculated and experimental data.

– 47 –

Work plan: 1. Acquaintance with work. 2. Preparation of liquids. 3. Setup preparation. 4. Carrying out the sedimentation process. 5. Discussion of obtained results. 6. Registration of experimental results. Necessary devices and equipment for work: 1. Glass cylinders. 2. Stop watch. Necessary reactants: 1. Particles of a spherical and irregular form. 2. Solutions of liquids. Technique of work carrying out

The laboratory setup consists of three glass cylinders dц ≥ 50 mm filled with various liquids. Physical properties of liquids are known. In the upper and lower part of each cylinder there are tags for determination of a way of passing by a particle in the course of sedimentation. Measuring by a stop watch time of passing of a particle within the planned distance, it is possible to determine the sedimentation velocity of a particle. To prevent initial acceleration of a particle in liquid the upper tag, it is necessary to put below from the top level of the cylinder on 30÷40 mm. Particles of a spherical and irregular form slowly put on a liquid surface in the cylinder. The time of passing of a particle between two tags measure by a stop watch. The measurement repeat several times. For calculation of Re, Ar, Ly criteria find the sedimentation velocity and average diameter of a particle. It is necessary to avoid in the course of particles sedimentation a sticking to walls of the cylinder or air bubbles to particles. Otherwise do not to consider result of such measurement and experiment needs to be repeated. Results of experiment bring in the reporting table 4.4.

– 48 –

Table 4.4 Reporting table

Particle

τ, s

М, kg

ρ, kg/m3

Wprec. (exp) m/s

Re

Ar

Ly

Wprec. calcd., m/s

dэ (on mass.), m

dэ calcd., m

Treatment of the experimental data and report creation. After each measurement calculate the sedimentation velocity on which calculate the Lyashchenko's criterion. Using the value of Lyashchenko's criterion from figure 4.5 find Archimedes criterion and calculate the equivalent diameter of a particle. The value of a form coefficient by Lyashchenko's criterion take from table 4.2. On the known mass of a particle calculate the equivalent diameter of a particle, then calculate Archimedes criterion. According to Archimedes criterion from figure 4.5 find values of Lyashchenko's criterion and calculate sedimentation velocity. The value of a form coefficient by Archimedes criterion take from table 4.3. The calculated results compare to experimental data, draw conclusions, bring in the reporting table and results of work hand over to the tutor. Tests for independent work 1. Separation of liquid non-uniform systems belongs to: A) Mass-exchange process. B) Heat-exchange process. C) Chemical process. D) Physical process. E) Hydromechanical process. 2. For purification of gases from dust are applied: A) Cyclones. B) Centrifuges. C) Filter presses.

– 49 –

D) Pumps. E) Compressors. 3. Resistance of a filtering partition is calculated on a formula: A) r0  P / hdep .

B)

r0  P /  .

C) R f . p .   P / hdep w . D)

R f . p.  P / w .

E)

r0  P / hdep w .

4. Specific resistance of a deposit is calculated on a formula: A) r0  P / hdep .

R f . p .   P / w . C) R f . p .   P / hdep w . B)

D)

r0  P / hdep w .

E)

r0  P / hdep w .

5. The factor of separation is calculated on a formula: A) FK P . 2

B) Gw / gr . C) 2n / 60r . 2

D) w / gr . E) G / gr . 6. In the course of filtering the resistance of a deposit is determined by a formula: A) R  r .

B)

R  Rdep  R f . p. .

C)

R  uq .

D) R  PK . E) R  ru .

– 50 –

7. For filtering process the specific resistance of a deposit has dimension: А) Ns/m3. B) Ns/m2. C) Ns/m. D) Ns/m4. E) Nh/m3. 8. At centrifugation the factor of separation is determined by the equation:

A) Ф 

w2 r . g

wR . g C . C) P  F B) Ф 

mw2 . r Mn2 E) C  . R

D)

C

9. At centrifugation the centrifugal force is determined by the equation:

A) Ф 

w2 r . g

wR . g C . C) P  F B) Ф 

mw2 D) C  . r

– 51 –

Mn2 E) C  . R 10. In cyclones the value of centrifugal force is determined by the equation:

w2 r A) Ф  . g wR . B) Ф  g C C) P  . F mw2 D) C  . r Mn 2 . E) C  R 11. The material balance of separation process according to total amount of substances is determined by the equation: A) Gmix = Gmoth.l. + Gdep. B) Gmix xmix = Gmoth.l. xmoth.l. + Gdep xdep. C) G = Gmix xdep – xmix/xdep – xmoth.l. D) G = Gmix xmix – xmoth.l./xdep – xmoth.l. E) G = Gmoth.l. xmoth.l. – xmix/xdep – xmoth.l. 12. The material balance of separation process according to a disperse phase is determined by the equation: A) Gmix = Gmoth.l. + Gdep. B) Gmix xmix = Gmoth.l. xmoth.l. + Gdep xdep. C) G = Gmix xdep – xmix/xdep – xmoth.l. D) G = Gmix xmix – xmoth.l./xdep – xmoth.l. E) G = Gmoth.l. xmoth.l. – xmix/xdep – xmoth.l. Questions for self-checking 1. What is equation generally expressed the resistance force which is showing in a moving flow?

– 52 –

2. 3.

Of what resistances the complete force of resistance is sum up? How express the generalized dependence for determination of resistance force? 4. How express the generalized dependence for determination of resistance force of the irregular form particles? 5. What is the precipitation? What forces affect on being besieged particle? 6. What parameters are included into the equations of Reynolds's criterion of rather being besieged particle? 7. What modes are known for sedimentation process? 8. How the sedimentation velocity of the irregular form particle is determined? 9. How it is possible to determine the particle diameter if sedimentation velocity is known? 10. How the form of a being besieged particle influences on dependence Ly = f (Ar)? 11. What chart is used for determination of approximate values of sedimentation velocity and equivalent diameter in any mode of sedimentation? Literature 1. Romankov P.G., Kurochkin M.I. Hydromechanical processes of chemical technology. – L.: Chemistry, 1982. – P. 118-132. 2. Directory of the chemist. Volume 5. – M.: Chemistry, 1966. – 972 p. 3. Kasatkin A.G. Basic processes and devices of chemical technology. – M.: Chemistry, 1973. – P. 95-101.

– 53 –

Laboratory work № 5 CONSTANTS DEFINITION OF FILTERING PROCESS General information

Filtering is the separation process of heterogeneous systems by means of porous partitions which detain one phases of these systems and pass others ones. Separation of the suspension consisting of liquid and solid particles weighed in it, make by means of the filter (figure 5.1) which in the elementary type is the vessel divided into two parts by a filtering partition. Suspension place in one part of this vessel so that it adjoined to a filtering partition. In the divided parts of a vessel create a difference of pressure under the influence of which liquid passes through a time of a filtering partition, and solid particles are late on its surface. Thus suspension is divided into a net filtrate and a damp deposit. This process of suspension separation call as filtering with formation of a deposit. Filtering is hydrodynamic process and is performed by means of a difference of pressure ΔР, a filtering partition created on both parties P  P1  P2 . Difference of pressure on both parties of a filtering partition can be created by means of compressors, vacuum pumps and liquid pumps, and also at the expense of the hydrostatic pressure of the most divided suspension. The difference of pressure is in limits of 0,5-0,9 kgf/cm2 when using vacuum pumps, and when using piston compressors it constitutes to 5 kgf/cm2.

– 54 –

When filtering work as hydrodynamic factors (surface of particles, sphericity, porosity), and physical and chemical ones. The resinous and colloidal impurities in suspension letter the pores that affects influence of the electrokinetic potential arising on the section border of solid and liquid phases in the presence of ions.

Figure 5.1. General scheme of filtering: 1 – filter case, 2 – a filtering partition; I – suspension, II – filtrate, III – deposit

With increase in the size of particles influence of hydrodynamic factors amplifies, and in process of reduction of their size the influence of physical and chemical factors increases. From the conditions of filtering influencing its current, the greatest value have a difference of pressure on both parties of a filtering partition and suspension temperature. For engineering calculations of the filtering equipment it is necessary to know the so-called constants of filtering characterizing hydraulic resistance of a deposit and the filtering partition. The velocity of suspensions filtering essentially depends on physical properties and a solid particles fineness of aggregate. On degree of a solid particles fineness of aggregate the suspensions divide on: a) rough (when the sizes of particles more than 100 microns); b) thin (when the sizes of particles from 100 to 0,5 microns); c) dregs (when liquid with a size of solid particles from 0,5 of micron);

– 55 –

d) colloidal solutions (when the size of solid particles is from 100 microns and less). In practice all types of suspensions, mostly meet the different sizes of particles, i.e. polydisperse systems. When filtering suspension arrives on a porous filtering partition through which the liquid phase passes, and the weighed particles remain on a filter surface in the form of a deposit. Good operation of the filter in many respects depends on properties of a filtering partition. Filtering partitions make of various cotton fabrics (belting, coarse calico, muslin, diagonal), woolen fabrics (cloth, baize, felt), fabrics from synthetic fibers (polyvinylchloride, perchlorovinyl, polyamide, vinyon, saran, orlon, lavsan) and fabrics from fibers of a mineral origin (asbestine and glass). Recently all start applying porous metal, ceramic and ceramic-metal filtering partitions more widely. The choice of this or that filtering partition is caused by: 1) porosity (the sizes of a pores shall be such that particles of a deposit were late on a partition), 2) chemical resistance to action of the filtered environment, 3) sufficient mechanical durability, 4) heat resistance at a filtering temperature. Filtering velocity

Usually in view of the small size of a pores in a layer of a deposit and a filtering partition, and also small movement velocity of a liquid phase in a pores it is possible to consider that filtering proceeds in laminar area. As shows experiment, under such circumstances filtering velocity at each this moment it is directly proportional to a difference of pressure, but it is inversely proportional viscosity of liquid of a phase and the general hydraulic resistance of a layer of a deposit and a filtering partition. As generally in the course of filtering of value of a difference of pressure and hydraulic resistance of a layer of a deposit eventually change, the variable velocity of filtering (m/s) express in a differential form:

W 

dV Sd

(5.1)

– 56 –

According to stated the main differential equation of filtering looks like:

dV P  Sd  ( R d  R f . p. )

(5.2)

where V  amount of a filtrate, m3; S  surface of filtering, m2;   duration of filtering, sec; P  difference of pressure, Pa;  – viscosity of a liquid phase of suspension, Pas; R d  resistance of a deposit layer, m-1; R f . p.  resistance of a filtering partition, m-1. The equation (5.2) is a special case more the general law according to which the process velocity is directly proportional to a driving force and is inversely proportional to resistance. In this case the difference of pressure represents a driving force, and general resistance consists of deposit resistance ( R d ) and a filtering partition ( R f . p. ). Both specified resistances are difficult functions of many variables. So, the value Rd of subjects is more, than it is less porosity of a deposit more specific surface of solid particles constituting it; the value Rd is influenced also by the value and a form of particles. From the equations (5.1) and (5.2) follows that Rd and Rf.p. are expressed in m-1. The value Rf.p. in the course of filtering can be considered approximately as a constant, neglecting its some possible increase owing to penetration into a time of a partition of new solid particles. The value Rd with increase of thickness of a layer of a deposit changes from zero at the beginning of filtering to the maximum value at the end of process. For integration of the equation (5.2) it is necessary to establish dependence between resistance of a deposit layer and amount received filtrate. Considering proportionality of amounts of a deposit and a filtrate, designate the ratio between a deposit amount and filtrate amount through x0 . Then the amount of a deposit will be equal

– 57 –

to x0V . At the same time the amount of a deposit can be expressed by work h d S (where h d  height of a layer of a deposit in m). Hence,

x0V hd S

(5.3)

From where thickness of a uniform layer of a deposit on a filtering partition will constitute

hd  x0

V S

(5.4)

Resistance of a layer of a deposit can be expressed equality:

R d  r0 hd  r0 x0

V , S

(5.5)

where r0  specific volume resistance of a layer of a deposit, m-2. From equality (5.5) follows that the value r0 characterizes a resistance rendered to a flow of a liquid phase by a uniform layer of a deposit by a thickness in 1 m. Having added Rd value from equality (5.5) in the equation (5.2), we receive

dV W  Sd 

P V     r0 x0  R f . p .  S  

(5.6)

Having accepted that resistance of a filtering partition can be neglected ( R f . p.  0) , taking into account equality (5.4) from the equation (5.6), we will find

– 58 –

r0 

P hdW

(5.7)

In case of  = 1 Pa·s, hd = 1 m and W = 1 m/s the value r0 = P. Thus, the specific resistance of a deposit numerically is equal to the difference of pressure necessary in order that a liquid phase with viscosity (viscosity of water in case of 20 °C is equal 10-3 Pa·s) 1 Pa·s it was filtered with a velocity of 1 m/s, through a deposit in 1 m thick of layer. It is obvious that this hypothetical difference of pressure which in practice isn’t used, shall be very great. For strongly squeezed precipitation r0 value reaches 1012 m-2 and more. Having accepted V  0 that corresponds to the beginning of filtering when on a filtering partition the deposit layer wasn’t formed yet, from an equation (5.6) we will receive:

R f . p. 

P W

(5.8)

In case of  = 1 Pas and W = 1 m/s the value Rf.p. = Р. It means that the resistance of a filtering partition numerically is equal to a difference of pressure, necessary in order that a liquid phase with viscosity of 1 Pa·s passed through a filtering partition with a velocity of 1 m/s. For a number of filtering partitions the value Rf.p. has an order of 1010 m-1. Filtering equation in case of constant difference of pressure

In case of P = const and invariable temperature for the filter of this design and the chosen filtering partition all sizes entering into the equation (5.6), behind an exception V and  , are permanent. Let’s integrate this equation ranging from 0 to V and from 0 to  : V

 

   r0 x0 0

or



V   R f . p. dV   PSd S  0

(5.9)

– 59 –

r0 x0

V2  R f . p .V  PS  2S

Having divided both parts of the last equation on will finally obtain

R f . p. S

PS 2 V 2 V 2  r0 x0 r0 x0 2

(5.10)

r0 x0 / 2S , we

(5.11)

The equation (5.11) shows dependence of a filtering duration on filtrate amount; solving it concerning V, it is possible to express filtrate amount through a duration of filtration. This equation is applicable to an incompressible and squeezed precipitation as in case of P = const the values r0 and x0 also are permanent. From the equation (5.6) follows that on condition P = const in process of increase in amount of a filtrate and therefore, and filtering durations the velocity of filtering decreases. Determination of constants in the filtering equations

Under constants in the equation of filtering (5.11) understand an ratio of a deposit amount to filtrate amount х0, a specific volume resistance of a deposit r0 and resistance of a filtering partition Rf.p.. For a precipitation meeting in chemical productions and consisting, as a rule, of particles less than 100 microns in size, these sizes find experimentally. Let’s consider one of methods of determination of the specified sizes by practical consideration from the filtering equation in case of a fixed difference of pressure which is characterized by the big accuracy of obtained results. For this purpose we will lead the mentioned equation to a type:

– 60 –

 V

 AV  B

wherre

A

B

(5.12)

r0 x0

(5.13)

2PS P 2

R f . p. PS S

(5.14)

.

In case of constan nt temperature annd pressure differeence all sizes enterring into the righ ht parts of equalitiies (5.13) and (5.14), are constantt. Therefore the values A and B allso are constant, and a the equation (5.12) is the equaation of the straight line inclined too a horizontal axis at an angle which h tangent is equall A , and the cut piece p on ordinate axis will be B . For creation off the specified strraight line in coorrdinates  / V  V put a number off points based on measured in expeerience and corressponding to one aanother of values  / V and V (figu ure 5.2). Then by the schedule dettermine the valuees A and B , then from equalities (5.13) ( and (5.14) calculate ro and Rf.p.. Find the o direct measureement of amountss of a deposit value x0 as a result of

(Vо ) and filtrate: x0  Vо / V .

s/m3

v, m3

– 61 –

Figure 5.2. Dependence  / V  V for determination of specific resistance of a deposit and resistance of a filtering partition

Work purpose: carrying out process of suspension filtering in case of various values of a pressure difference and a ratio of a liquid and solid phase and determination of process constants. Work plan: 1. Acquaintance with work. 2. Suspension preparation. 3. Setup preparation. 4. Carrying out process of filtering. 5. Discussion of results. 6. Registration of experimental results. Necessary devices and equipment for work: 1. Vacuum pump. 2. Byukhner's funnel. 3. Bunsen's flask. 4. Filter paper. 5. Chemical flask. 6. Glass tubes. 7. Clips. Necessary reactants: 1. Boron-water suspension. 2. Suspension of clays. Technique of work carrying out

In certain ratios (for example, 1:2; 1:3; 1:4; 1:5; 1:8) of a solid and liquid phase prepare boron-water suspension. Prepare setup for carrying out the filtering process according to in figure 5.3. In case of the closed crane 3, turn on the vacuum pump and by means of the air crane 5, determine the diluted amount of given air. On a bottom of

– 62 –

Byuk khner's funnel pllace by the size oof filter paper. Suspension S in case of continuous hashing pour in a funnel 1 and a funnel contiusly fill with it until u filtering process will end. When W filtering nuou sincee the moment wh hen in Bunsen flaask the first dropp of a filtrate will arrive, include a stop watch and m measure filtering tiime. When in nnel a certain am mount of a filtratee will be gained (for ( example, a fun 20 ml), m write down th he indication of a stop watch in thhe laboratory journ nal.

Figure 5.3. 1 – Byukhner's funnel;; 2 – Bunsen's flassk; 3 – clip; 4 – safety bottle; 5 – air crane; 6 – vacuum pump; 7 – differential manometer (vacuuum gage)

In the laboratory journal also writte down also vacuuum gage indicattions. When the laast drop of a filtraate will be filteredd switch off a stop watch and writee down the stop w watch indication in the magav pump. Meeasure the internaal diameter of zine.. Switch off the vacuum a fun nnel and deposit thickness by meeans of a line annd data write down n in the laboratorry magazine. Resuults of experimennt bring in the reporting table 5.4. Treatment of the experimental d data and report creation. c Acding to the obtaineed data, find the rratio between the filtering time cord and amount of a filtraate. Build the deppendence chart off the obtained  /V , ount. For ordinatee axes postpone values v data from filtrate amo and for abscissa axis of value of a filttrate amount. Usiing the found

– 63 –

points, carry out a straight line. The tangent of an inclination angle of a straight line (the relation of an opposite leg to an adjacent leg) is equal to a constant A . The piece of a straight line, cut on ordinate axis, gives a constant B . Table 5.4 Reporting table Ratio solid and liquid phases

Time, τ, s

Volume, V, m3

τ/ V, s/m3

Difference of pressure, ΔР, Pa

А, s/m6

В, s/m3

rо, 1/m2

Rf.p., 1/m

Using the found values A and B , calculate resistance of a deposit and filtering partition and compare with theoretical data. Questions for self-checking 1. What is the filtering process? 2. What size is a driving force of filtering process? 3. What materials are used as a filtering partition? 4. From what values the general resistance of filtering process sum up? 5. How to obtain the equation determining velocity of filtering process? 6. What is the physical meaning of specific volume resistance of a deposit? 7. What is the physical meaning of resistance of a filtering partition? 8. What values have sizes of specific resistance of a deposit and filtering partition? 9. How is the filtering equation expressed in case of a fixed difference of pressure? 10. Under what conditions the velocity of filtering decreases? 11. What values are included into constants in the filtering equations? 12. How to determine constants in the filtering equations? 13. What setup is used for carrying out process of filtering? 14. How carry out a filtering process in laboratory conditions? 15. What values can be determined when carrying out experiment?

– 64 –

16. What values are included into the reporting table? Literature 1. Kasatkin A.G. Basic processes and devices of chemical technology. – M.: Chemistry, 1973. – P. 191-197. 2. Guide to a practical training in laboratory of processes and devices of chemical technology / Under the editorship of Romankov P.G – L.: Chemistry, 1975. – P. 98-106. Tasks for independent solution 1. The dynamic coefficient of a suspension liquid phase 2810-3 Pаs, filtering velocity 0,0410-3 m3/(m2s) and difference of pressure 3103 Pa are known. Calculate the resistance of a filtering partition (in m-1). 2. The dynamic coefficient of a suspension liquid phase 2510-3 Pаs, filtering velocity 0,0510-3 m3/(m2s), difference of pressure 3,5103 Pа and a deposit height 0,01 m are known. Calculate a deposit resistance (in m2). 3. The dynamic coefficient of a filtrate 2010-3 Pаs, mass of dry solid substance 0,1 kg/m3 and specific resistance of a deposit 7,6109 m/kg of a dry deposit are known. Difference of pressure on the filter equal to 3103 Pa. Calculate a filtering constant (in m2/s). 4. The specific resistance of the filtering partition 2,7108 m/m2, specific resistance of a deposit 7,6109 m/kg of a dry deposit and mass of dry solid substance to 1 m3 of a filtrate 0,1 kg/m3 are known. Calculate a filtering constant (in m3/m2). 5. The mass of a deposit and the liquid which are in a drum of the centrifuge 400 kg, the rotation frequency of the centrifuge 1200 rpm, diameter of a drum 800 mm are known. Calculate a centrifugal force developed at centrifugation (in N). 6. After filtering 5 kg of suspension a damp deposit 1,5 kg is formed. Determine the mass of a filtrate (in kg). 7. After filtering 15 kg of suspension a damp deposit 5,0 kg is formed. Determine the mass of a filtrate (in kg).

– 65 –

Laboratory work № 6 DISTILLATION PROCESS General information

In chemical technology, and also petrochemical, food, pharmaceutical industries the broad variety of the liquid and gas mixes are used, which are subjected for separation into rather pure components or fractions of various structure. One of the most widespread methods of separation of the liquid uniform mixes consisting of two or a large number of flying components, the distillation is. Distillation represents the process including partial evaporation of the separating mix and the subsequent condensation of being formed vapors, carried out once or repeatedly. As a result of condensation receive the liquid which structure differs from structure of an initial mix. Separation by distillation is based on a various volatility of components of a mix at the same temperature. Therefore at distillation all components of a mix pass to a vaporous condition in the quantities proportional to their fugacity (volatility). In the elementary case the initial mix is binary, i.e. consists only of two components. Steam obtained at its distillation contains rather large number easily flying, or a low-boiling component (LBC), than an initial mix. Therefore, in the course of distillation the liquid phase is impoverished, and the steam phase is enriched with LBC. Not evaporated liquid has structure richer with a hardly flying or high-boiling component (HBC). This liquid is called as the rest, and the liquid received as a result of condensation of vapors, distillate or rectificate. Extent of enrichment of steam phase LBC other things being equal depends on a type of distillation. There are two essentially different types of distillation: 1) simple distillation and 2) rectification.

– 66 –

Simple distillation represents process of single partial evaporation of a liquid mix and condensation of being formed vapors. Simple distillation is applicable only for separation of the mixes which volatilities of components are significantly various, i.e. the relation of volatilities (a relative volatility) components is considerable. Usually it is used only for preliminary rough separation of liquid mixes, and also for cleaning of difficult mixes from undesirable impurity, pitches, etc. Significantly deeper separation of liquid mixes into components is reached by rectification. Rectification is a process of separation of homogeneous mixes of flying liquids by bilateral mass- and heat-exchange between the non-equilibrium liquid and steam phases having various temperature and moving relatively each other. Characteristics of two-phase systems liquid-steam

If the system consists of two components (C = 2) and between them there is no chemical interaction, in the presence of liquid and steam phases number of phases Ph = 2. According to the rule of phases, the number of freedom degrees (NFG) of such system makes: NFD = C + 2 – Ph = 2 + 2 – 2 = 2

(6.1)

Therefore from three independent parameters, which are completely defining a condition of system (temperatures t, pressure р and concentration c), it is possible to choose randomly any two ones. Thus value of the third parameter which can’t be any anymore will be defined. In this regard for the physical and chemical characteristic of binary systems liquid-steam it is convenient to use so-called phase charts. If to designate through x structure of a liquid phase, and through at у – structure of a steam phase, that, accepting P = const, it is possible to construct the dependence chart of boiling temperature and liquid condensation from composition of liquid and steam phases (figure 6.1). For separation an initial mix of structure Х0 heat with a constant pressure to boiling temperature Т0 (a figurative point a0) thus the first

– 67 –

vial of steam (a figurative point b0) of composition У0. Steam in comparison with initial liquid is richer with an easily flying component B, and solution is enriched with a component A and its temperature of boiling at a supply of warmth increases to Т1 (a figurative point a1). In the course of distillation the composition of liquid solution changes from Х0 to Х1, and composition of steam – from Y0 to

Y1 . If to condense all received steam, condensate (the first fraction) /

will have composition Y1 , intermediate between Y0 and Y1 .

Figure 6.1. Fractional distillation of unlimited mixing-up liquids which aren’t forming an azeotropic mix at P = const

When boiling the remained liquid (a figurative point a1) of composition Х1 the first vial of vapor is formed (a figurative point b1) of composition Y1 which has been also enriched with an easily flying component B. At change of composition of liquid solution from Х1 to Х2 the composition of steam changes from Y1 to Y2 . At condensation of this steam condensate (the second fraction) of composition

Y2/ . At further heating of the remained liquid it is similarly possible to receive the third, the fourth, etc. fractions.

– 68 –

When heating an initial mix of composition Х0 the last drop of liquid (a figurative point a3) of composition Х3 will disappear at temperature Т3. Equilibrium steam with it (the figurative point b3) has composition Х0. If to subject each of the received fractions to similar distillation, the set of the new fractions enriched with easily flying B. Similar on structure of fraction will turn out it is possible to unite and subject to further fractionation until condensate will be not represented by almost pure component B, and overtaken liquid – a pure component A respectively. Material balance of simple distillation process

Distillation subdivide on simple and equilibrium. Simple distillation call process of partial evaporation of a boiling liquid mix with continuous branch and condensation of the formed vapors. The obtained condensate is called as the distillate, not evaporated liquid – the vat rest. For equilibrium distillation evaporation of liquid part and long contact of the formed vapors with not evaporated liquid before achievement of phase balance is characteristic. By consideration of distillation processes recognize that steam formation occurs very slowly and that between liquid and steam balance is established. At simple periodic distillation the maintenance of low-boiling components in liquid and steam phases continuously falls. Let’s designate through G  amount of liquid cubed, through x  its structure. The maintenance of a low-boiling component is equal Gx in liquid. At evaporation of small amount of the liquid dG having structure x , concentration of liquid will decrease by size dx and the rest of liquid will cubed be expressed by size:

G  dG ,

(6.2)

x  dx.

(6.3)

and its composition will be

– 69 –

The maintenance of a low-boiling component in the rest of liquid will make:

G  dGx  dx.

(6.4)

The amount of distillate equally to amount of the evaporated liquid dG , and its composition y is equilibrium with x . The material balance of an easily flying component can be presented for a considered period in the following look:

Gx  G  dG x  dx   ydG or

Gx  Gx  xdG  Gdx  dGdx  ydG.

(6.5) (6.6)

Neglecting by product dGdx as the infinitesimal size of the second order we will receive:

Gdx  dG y  x 

(6.7)

dG dx  . G yx

(6.8)

or

Let’s designate amount of liquid cubed after an distillation through G1 and its composition through x1 , then the equation (6.8) it is possible to integrate ranging from G1 and x1 to G and x . Let’s obtain: G

x

dG dx G G  x y  x 1 1

(6.9)

or x

2,3 lg

G dx  . G1 x1 y  x

(6.10)

– 70 –

The integral of the right member of equation (6.10) is defined graphically. For this purpose on abscissa axis put values x , and on ordinate axis the corresponding values

1 and find the area liyx

mited to a curve, an axis x and the verticals which have been carried out through abscissa x1 and x . This area is equal to required integral (figure 6.2). Calculation can be conducted both in weight, and in molar sizes.

1 - х to calculation yx of simple distillation

Figure 6.2. Dependence

The amount of obtained distillate G  G1  Gd and its composition xd can be defined from the equation of material balance of an easily flying component:

Gx  G1 x1  G  G1 xd , from where

xd 

Gx  G1 x1 . G  G1

(6.11)

(6.12)

– 71 –

Work purpose: carry c out the distiillation process off a binary mix on laaboratory setup for f simple distillaation, make a matterial balance of prrocess and define structure distillatee. Refractomeetry

Refractometry is based on refractioon coefficient deteermination of studiied substance. Wh hen passing a ray of light from density from one envirronment on envirronment with otheer density its direcction changes – it experiences the refraction. r Changge of the directionn of a falling beam m is connected wiith change of lighht velocity on lim mit of the section of two environmeents with a variouus density. The ratio betweeen a sinus of the falling angle of a beam and a sinuss of the angle of its refraction is caalled as coefficiennt or an indicatorr of refraction (fig gure 6.3):

n

sin i1 , sin i2

(6.13)

wherre i1  hade, i2  refraction cornerr.

Figure 6.3. Beam refraction onn border of two tran nsparent environm ments

The refractometers are used for deefinition of a refraaction indicator. There T are differen nt types of refracttometers. Amongg them the re-

– 72 –

fracttometers of Abbee's type, created m more than hundreed years ago, still remain the most widespread. Feaatures of Abbe's type t refractoi the device of a measuring prism, existence of meteers are consisted in an additional, a so callled lighting prism m, usage for meaasurements of day or o artificial light and a in a scale desiign. The Abbe's refracctometer scale is ggraduated directlyy in values of an in ndicator of refracttion nD . Need of any calculations therefore t disappeears, and all proceedure of measurem ments borrows sevveral minutes. In modern m models thee scale is projectedd in sight of a pippe and is visible at a the same time with w the boundaryy line of full internnal reflection (figu ure 6.4).

Figure 6.4. Field of visionn of an eyepiece of the t IRF-454 refraactometer

The IRF-454 refrractometer (figuree 6.5) can be an exxample of the mostt perfect design off devices of this kkind. Technique of wo ork with liquids. The refractometeer is installed in frront of the windo ow or by 40-60 w watt electric lamp so that light fell on o an entrance sid de of a lighting prrism. Some modeels have autonom mous lighting, and with them it is poossible to work in the darkened room m. Work with liqu uids, as a rule, dem mands a thermostaating to within ±0 0,1–0,200, for wh hat through shirts of prisms pass water w of constantt temperature from m the circulating tthermostat. The handle 4 (fig gure 6.5) open thee prismatic block and on a horizon ntal side of a meaasuring prism put 1-2 drops of studdied liquid by

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mean ns of a pipette, wiithout concerningg a tip of its polishhed surface of glasss. Close the block k and, rotating thhe flywheel 6, dirrect a pipe at lightt and shade bordeer. By turning of a mirror 9, achievve the best illumiination of a scale and establish an eyepiece on distiinct visibility of a sighting cross and a scale. Then, rrotating the flywhheel of jack 3, ng of the boundarry line achieve. The T flywheel destrruction of colorin preciisely establish a cross c hairs on boorder light and daark water and makee the report on sccale nD. The pricee of the smallest division of a scalee of nD makes 0,00 01.

Fig gure 6.5. IRF-454 reefractometer: 1 – eeyepiece; 2 – cap off the justified adap ptation; 3 – flywheeel jack; 4 – handle of the camera of a lighting l prism; 5 – thermometer; 6 – flywheel scale tu turn; 7 – jack scale;; 8 – gate of a window of o a lighting prism; 9 – illumination miirror; 10 – gate mirro or of a measuring prrism; 11 – justified d key; 12 – justified pplate

Upon termination n of reports open the prismatic bloock and wipe work king surfaces of cameras c and prism ms a pure soft ragg or filter paper. The polished sid de of a measuringg prism should bee wiped very careffully, without preessing it not to daamage polishing. Then prisms wash h out alcohol or air, entering som me drops it into the t block the samee as measured liq quid. After prism w washing again wiipe and leave the block b for some tim me open for dryingg.

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Technique of work carrying out

1. Prepare solution of isopropyl alcohol in water with concentration 30-40%. 2. Using the refractometer, measure the optical density of the prepared solution and according to the calibration plot ( nD i-PrOH – weight. % i-PrOH), define exact composition of solution. 3. Assemble the device according to figure 6.6.

Figure 6.6. Laboratory setup for simple distillation: 1 – Wyurts's flask; 2 – thermometer; 3 – refrigerator; 4 – flask receiver

1. Measure the weight of the prepared solution and pour in Wyurts's flask for distillation (previously it is necessary to put some slices of a boiler in a flask). 2. Start up water in the refrigerator, then heat a flask and watch the thermometer indication. 3. From the moment of emergence of the first drop of distillate it is necessary to write down every 5 minutes thermometer indications in the laboratory diary before achievement of 20-25 minutes. 4. After that disconnect heating and cool a flask to room temperature. 5. Determine the weight of the received distillate and by the refractometer measure optical density and determine percentage of distillate by the chart.

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6. The rest of initial solution from Wyurts's flask pour previously the weighed flask and determine the rest weight. 7. By means of the refractometer, measure the optical density of residual solution and according to the calibration chart ( nD i-PrOH – weight. % i-PrOH), define the rest compositon. 8. Order a workplace. Laboratory setup. Laboratory setup for simple distillation consists of a round-bottomed flask 1 for heating of initial liquid, the thermometer 2 for measurement of temperature, the direct refrigerator 3 for vapor condensation, a conic flask 4 as the receiver for collecting distillate. Processing of skilled data and drawing up report. Using the reference data for system isopropyl alcohol-water (initial data to take

from a tutor) build the chart of dependence 1/у* – х = f (x) ( y  equilibrium concentration). From the chart by experimental data determine the area of integration and a method of a trapeze, find value of integral. Using a method of graphic integration, count a molar ratio of quantity of an initial mix to quantity of the vat rest: 2,3 lg(F / W )  S  M x  M y (where S  the area determined by a *

method of a trapeze; M x  scale on an abscissa; M y  scale on ordinate). Count mass percent of alcohol in distillate and compare the received value to the percentage of distillate determined experimentally by optical density. The received results bring in the reporting table, draw conclusions and work hand over to the tutor. Table 6.1 Reporting table

F, kg

tf, ºС

Xf, kg/ kmol

τ, min

W, kg

tw, 0 С

Xw, kg/ kmol

D, kg

Xd.exp, kg/ kmol

Xd.calc, kg/ kmol

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Questions for self-checking 1. What is the distillation? What types of distillation exist? 2. What is the physical meaning and physical value of simple distillation? 3. In what cases it is favorable to carry out distillation process in vacuum? 4. Why at simple distillation full separation of components from each other isn’t reached? 5. How to explain change of boiling temperature of liquid cubed in the course of distillation? 6. How from the equation of material balance for simple distillation the composition of distillate is defined? 7. On what the refractometric method is based? 8. How to define a refraction indicator of liquid? 9. What parameters are measured in experiment process? Literature 1. Kasatkin A.G. Main processes and devices of chemical technology. – M., Chemistry, 1973. – P. 471-512. 2. Pavlov K.F., Romankov P.G., Noskov A.A. Examples and tasks of a course of processes and devices of chemical technology. – L.: Chemistry, 1987. – P. 195-198; 319-347. 3. Physical chemistry. In 2 books of Book 1. Substance structure, Thermodynamics / Under the editorship of K.S. Krasnov. – M.: Vyssh. Sch., 1995. – P. 453-455. 4. Ioffe B.V. Refraсtometric methods of chemistry. 3rd prod. the processed. – L.: Chemistry, 1983. – P. 8-9; 152-157 . 5. Kogan V.B., Friedman V.M. The directory on balance between liquid and steam in dinarny and multicomponent systems. – L.: Goskhimizdat, 1957. – 499 p. 6. Dobroserdov L.L. Phase balance liquid-steam in system isopropyl alcohol-water-chloride calcium // J. Appl. Chem. – T. 32, № 11. – 1959. – P. 2582-2584. Tasks for independent solution 1. After distillation of ethyl alcohol water solution with mass 200 kg in the vat rest contains 5 % (mass.) of alcohol. Define a quantity of the vat rest (in kg). 2. In a simple distillation cube the distillation of 1000 kg of the mix containing 68 % (mass.) of ethyl alcohol. After distillation the vat

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3. 4.

5.

rest contains 23 kg of ethyl alcohol. Calculate a mass of ethyl alcohol in distillate (in kg). After distillation of 800 kg of mix of ethyl alcohol and water the distillate with mass 590 kg is formed. Define a mass fraction of ethyl alcohol (in %) in distillate. It is known a molar part of an easily flying component in distillate 0,975, in initial liquid of a rectifying column 0,675 and in steam, equilibrium with liquid of food 0,315. Define the minimum number of a phlegm in a rectifying column of continuous action. The polluted turpentine is distilled in current of saturated water vapor under atmospheric pressure. The amount of the water vapor leaving together with turpentine steam made 2330 kg. The leaving steam with the distilled turpentine is cooled from 110,7 to 96 °C. Calculate quantity, the warmth given off (in kJ) if the specific heat of water vapor made 1,97·103 J/(kgК)).

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CONTENT Laboratory work № 1. MODES OF THE LIQUID CURRENT......................................................................................3 Main definitions and theory of process ...........................................3 Description of laboratory setup .......................................................7 Order of work performance ............................................................7 Processing of measurements results and content of the report.....................................................................................8 Tasks for independent solution .......................................................10 Laboratory work № 2. HYDRODYNAMICS OF THE FLUIDIZED LAYER BULK...........................................12 Main definitions and theory of process ...........................................12 Description of laboratory setup .......................................................17 Order of work performance, processing of measurements results and content of the report .................................................................18 Tasks for independent solution .......................................................22 Laboratory work № 3. ACQUAINTANCE WITH THE MULTISTAGE PISTON COMPRESSOR OF A HIGH PRESSURE ......................................................................24 General information ........................................................................24 Tasks for independent solution .......................................................29 Tests for independent work.............................................................30 Laboratory work № 4. SEDIMENTATION OF SOLID PARTICLES BY GRAVITY IN THE LIQUID ENVIRONMENT ...........................................................................34 General law of the environment resistance .....................................34 Sedimentation velocity ...................................................................40 Technique of work carrying out......................................................47 Tests for independent work.............................................................48

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Laboratory work № 5. CONSTANTS DEFINITION OF FILTERING PROCESS .................................................................53 General information ........................................................................53 Filtering velocity .............................................................................55 Filtering equation in case of constant difference of pressure ..........58 Determination of constants in the filtering equations .....................59 Technique of work carrying out......................................................61 Tasks for independent solution .......................................................64 Laboratory work № 6. DISTILLATION PROCESS ......................65 General information ........................................................................65 Characteristics of two-phase systems liquid-steam.........................66 Material balance of simple distillation process ...............................68 Refractometry .................................................................................71 Technique of work carrying out......................................................74 Tasks for independent solution .......................................................76

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Educational issue Dina Nauryzbaevna Akbayeva Zhaniya Turluhanovna Eshova

METHODICAL MANUAL TO LABORATORY WORKS ON THE COURSE «THE MAIN PROCESSES AND DEVICES OF CHEMICAL TECHNOLOGY» Stereotypical publication Computer page makeup and cover designer: N. Bazarbaeva IB No. 9284 Signed for publishing 31.03.2020. Format 60x84 1/16. Offset paper. Digital printing. Volume 5,0 printer’s sheet. 100 copies. Order No.878. Publishing house «Qazaq University» Al-Farabi Kazakh National University KazNU, 71 Al-Farabi, 050040, Almaty Printed in the printing office of the «Qazaq University» publishing house