Chemical Engineering Thermodynamics: An Introduction to Thermodynamics for Undergraduate Engineering Students [1 ed.] 0471055905, 9780471055907

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Atomic Weights and Numbers Element

Symbol

Atomic Number

Atomic Weight*

Element

Mercury Molybdenum

Actinium Aluminum

Ac Al

89 13

27) 26.98154

Americium

Am

95

(243)

Antimony Argon Arsenic

Sb Ar As

sil 18 33

121.757 39.948 74.9216

Astatine

At

85

(210)

Barium Berkelium Beryllium Bismuth Boron Bromine Cadmium

Ba Bk Be Bi B Br Cd

56 97 4 83 5 35 48

13733 (247) 9.01218 208.9804 10.811 79.904 112.41

Ca Cf (Ss

20 98 6

Calcium Californium Carbon

Symbol

Atomic Number

Atomic Weight*

Hg Mo

80 42

200.59 95.94

Neodymium

Md

60

144.24

“Neon Neptunium Nickel

Ne Np Ni

10 93 28

2OAIOT. 237.0482 58.6934

Niobium

Nb

41

92.9064

Nitrogen Nobelium Osmium Oxygen Palladium Phosphorus Platinum

N No Os O Pd P Pe

7 102 75 8 46 15 78

14.0067 (259) 186.207 15.9994 106.42 30.97376 195.08

40.078 (251) 12.011

Plutonium Polonium Potassium

Pu Po K

94 84 19

Praseodymium

(244) (209) 39.0983

Cerium

Ce

58

140.12

Pr

59

140.9077

Cesium Chlorine

Cs Cl

55 7

132.9054 Promethium 35.4527 — Proctactinium

Pm Pa

61 91

(145) 231.0359

Chromium Cobalt Copper Curium Dysprosium

Cr Co Cu Cm Dy

24 2 29 96 66

51.996 58.9332 63.546 (247) 162.50

Radium Radon Rhenium Rhodium Rubidium

Ra Rn Re Rh Rb

88 86 75 45 cy

226.0254 (222) 186.207 102.9055 85.4678

Einstienium

Es

99

252)

Ruthenium

Ru

44

101.07

Erbium Europium

Er Eu

68 63

167.26 151.965

Samarium Scandium

Sm Sc

62 ZA

150.36 44.9559

Fermium

Fm

100

(257)

Selenium

Se

34

78.96

Flourine

F

18.99840

Silicon

Si

14

28.0855

9

Francium

Fr

87

(223)

Silver

Ag

47

107.868

Gadolinium Gallium Germanium Gold Hafnium Helium Holmium Hydrogen Indium Iodine Iridium Iron Krypton

Gd Ga Ge Au Hf He Ho H In I Ir Fe Kr

64 31 52 79 Ye 2 67 1 49 53 77 26 36

157-25 69.723 ZO 196.9665 178.49 4.00260 164.9303 1.00794 114.82 126.9045 192.22 55.847 83.80

Sodium Strontium Sulfur Tantalum Technetium Tellurium Terbium Thallium Thorium Thulium Tin Titanium Tungsten

Na Sr S Ta Te ie Tb Tt Th Tm Sn ‘Ee W

i 38 16 73 43 D2 65 81 90 69 50 22 74

22.98977 87.62 32.066 180.9479 (98) 127.60 158.9253 204.383 232.0381 168.9342 118.710 47.88 183.85

Atomic Weights and Numbers (continued)

Element

Symbol

Lanthanum Lawrencium Lead Lithium Lutetium

La er Pb Li Lu

Magnesium Manganese

Mg Mn

Mendelevium

Md

Atomic Number

Weight*

al 103 82 3 71 12 pb 101

138.9055 (260) 207.2 6.941 174.967 24.305 54.9380 (258)

Atomic

Symbol

Element

Uranium Vanadium Xenon Ytterbium Yttrium Zinc Zirconium

Atomic Number

U V Xe Yb ys Zn Zr

a2 23 54 70 39 30 40

Atomic Weight*

238.029 50.9415 131.29 173.04 88.9059 65.39 91.224

*Based upon carbon-12. ( ) Indicates most stable or best known isotope.

Factors for Unit Conversions

Quantity

Equivalent Values

Mass

1 kg = 1000 g = 0.001 metric ton = 2.20462 Ib,, = 35.27392 oz

Length

1m = 100 cm = 1000 mm

1 Ib,, = 160z = 5 X 10-4 ton = 453.593 g = 0.453593 kg

= 10° microns (um) = 10° angstrom (A)

= 39.37 in. = 3.2808 ft = 1.0936 yd = 0.0006214 mi 1 ft = 12m. = 1/3 yd = 0.3048 m = 30.48 cm Volume

bam = 1000L = 10° cm? = 10° mL = 35.3145 ft? = 220.83 imperial gal = 264.17 USS. gal = 1056.68 qt 1 ft = 1728 in.” = 7.4805 gal = 0.028317 m> = 28.317 L = 28317 cm?

Force

Pressure

1 N = 1 kg- m/s” = 10° dynes = 10° g- cm/s? = 0.22481 Ib,

1 Ib,

32.174 Ib, - ft/s* = 4.4482 N = 4.4482 x 10° dyn

1 atm =

101325 % 10° N/m (Pa) = 101.325 kPa = 1.01325 bar

1.01325 x 10° dyn/cm? 760 mmHg at 0°C (torr) = 33.9 ft H,O at 4°C II 29.921 in.Hg at 0°C Energy

1J = N-m = 10’ dyn-cm = 10’ergs 2.778 X 1077 kWh = 0.23901 cal 0.7376 ft - lb = 9.486 X 10°-* BTU

Power

1W = 13J/s = 0.23901 cal/s = 0.7376 ft - Ib,/s = 9.486 X 10°* BTU/s II

Gas constant (R)

1.341 x 1073hp

= 8.314 m’ - Pa/mol - K = 0.08314 L- bar/mol - K = 0.08206 liter - atm/mol - K = 82.06 cm? - atm/mol - K = 62.36 L- mmHg/mol - K = 0.7302 ft? - atm/Ib - mol -°R = 10.73 ft - psia/Ib - mol -°R = 1545 ft- Ib,/Ib - mol - °R

= 8.314 J/mol- K = 1.987 cal/mol- K = 1.987 BTU/Ib- mol - °R 0

= .001049 kWh/Ib - mol - K = 0.000780 hp - h/Ib - mol - °R EEE

Example: The factor to convert yd to in. is:

EEE

EEE

ee ———————

Nomenclature UPPER CASE Helmholtz free energy Second virial coefficient Heat capacity at constant pressure Heat capacity at constant volume Energy Voltage samemMandwp Faraday’s constant: 96,500 coulombs/equivalent G Gibbs free energy H Enthalpy I Initial; Ideal gas U Joules K Temperature, Kelvin

M,,

Molecular weight Number Pressure Vapor pressure Power Heat Gas constant Entropy Temperature Internal energy Volume a2< VZnovyvua Cn Work

LOWER CASE Specific Helmholtz free energy Activity of component i Boundary Fugacity Specific Gibbs free energy Specific enthalpy Henry’s constant of component i Mass

Thermal expansion coefficient Surface tension Symmetric activity coefficient of component i Asymmetric activity coefficient of component i Change: (out-in) or (end-beginning) Efficiency Surface area fraction Compressibility Joule-Thomson coefficient Chemical potential of component i Stoichiometric number Density Surface area Fugacity coefficient, volume fraction Extent of reaction

Moles Heat per unit mass Ratio of Cp to C, Relative molecular volume Second, specific entropy Time, throat

Specific internal energy Specific volume Velocity Mole fraction NeKX< THM Se Compressibility op Ewa MOS

factor

SUBSCRIPTS Liquid phase

atm Atmosphere Beginning Boundary Cold

Mechanical, maximum, minimum

Out Produced, isobaric Reaction Reduced Isentropic, solid phase Isothermal Turbine

Critical, crystal, compressor

Dump End Electrical

Hot, high In

Component i

Constant volume

yO8B op 47444 Vapor phase

Low pt te de for) ler Tey Pleasant jesl he!

SUPERSCRIPTS Time rate of change (as m would be

ic E

Combinatorial Excess , Ideal Pure component oO Standard state xmW Residual

time rate of change in m)

Z|

Partial molar property (M is generic extensive property: G, H, etc) Infinite dilution

AN

teiow cmt uitvoy

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nvudndtigen ee

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Tee

Then

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MUO

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ce 1, (i “« Gontatiedds

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hon,

or

iene

et

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bt Ree aay

ae! ed

mw

wn

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or

tie

eee

aii ee

od|

a my s tf els

\

sul hie atid,

OAT HTET

DENT A, Wf

~%, begyaby ey

il

0 4 H

Fabia

'

|

lath

‘(fer ibe?

:

LHe

ime |

pee

oe

Te, >

yt Dalnee g

1

LY < aR) Aa

Died +o

a

i

j

1

4

'

—

if

pahagil

-

y

?

U

wish

Wsth

;

r

’

;

:

aig i ib Maey agit

Pern “O4hian i

a)

+3; Papen)

{ee

a

0d tal(ett nhiaee Al,

on

oe

Gaia:

gan ‘sen pee i

6+

et laity

moda ligated

a,

ip

By

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ath renal

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ee

C8

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reales

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>

Chemical Engineering Thermodynamics An Introduction to Thermodynamics for Undergraduate Engineering Students

Jack Winnick Professor of Chemical Engineering Georgia Institute of Technology

John Wiley & Sons, Inc. New York Toronto

Chichester Singapore

Brisbane Weinheim

ACQUISITIONS EDITOR MARKETING MANAGER

Cliff Robichaud Jay Kirsch

PRODUCTION MANAGER _ Charlotte Hyland PRODUCTION SERVICE _Ingrao Associates DESIGNER Kevin Murphy MANUFACTURING MANAGER _ Dorothy Sinclair

ILLUSTRATION COORDINATOR _ Jaime Perea This book was set in 10/12 Times Roman by CRWaldman Graphic Communications, and printed and bound by Hamilton Printing. The cover was printed by Lehigh Press.

Recognizing the importance of preserving what has been written, it is a policy of John Wiley & Sons, Inc. to have books of enduring value published in the United States printed on acid-free paper, and we exert our best efforts to that end. The paper in this book was manufactured by a mill whose forest management programs include sustained yield harvesting of its timberlands. Sustained yield harvesting principles ensure that the number of trees cut each year does not exceed the amount of new growth. Copyright © 1997, by John Wiley & Sons, Inc.

All rights reserved. Published simultaneously in Canada. Reproduction or translation of any part of this work beyond that permitted by Sections 107 and 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Requests for permission or further information should be addressed to the Permissions Department, John Wiley & Sons, Inc.

Library of Congress Cataloging in Publication Data:

Winnick, J. (Jack), 1937Chemical engineering thermodynamics : an introduction for

undergraduate students / by Jack Winnick. p. cm. Includes bibliographical references. ISBN 0-471-05590-5 (cloth : alk. paper) 1. Thermodynamics. 2. Chemical engineering. I. Title. TP149.W526 1996 660'.2969—dc20 96-32832 (ue Printed in the United States of America

LOVORS 685 43302

Wiany

Mia) Fy eerie

orisieary loves

fr

conicees

|

al vaty ded ae tien 25

wel!

Peuctreal eatore

its syothioar

vi

daw ts

i

ca rou! as

o

an! Hhimew: (
= Sp

The negative entropy production in the latter example is evidence of its impossibility.

*The volume of the block remains nearly constant.

60

Chapter 3 The Entropy Balance

Look now at the piston—cylinder in Example 3.3. To simplify things we’ll insulate the setup to avoid heat transfer and focus only on mechanical work. For a differential accounting period, the reversible case, Fig. 3.3a, has as its energy balance:

dU =

—dW.,,

=

—P dV

(3.10)

and as its entropy balance:

dS = Sp

(3.11)

But, dU = TdS — P dV (always). So, combining this result with eq. (3.10) we see dS

must be zero (since thermodynamic temperature, 7, is always real and positive). Equation (3.11) then says Sp = 0. Now, looking at Fig. 3.3b, the energy balance says:

dU =

—dW =

—P, dV

(3.12)

and the entropy balance is:

dS = Sp

(3.13)

Again, our property relation between U and S says:

dU = TdS — P dV Then, from eq. (3.12): I aS

Pay, =

— Py dv.

(3.14)

and

T dS = (P — P,) dV If, as in this example, the gas expands, dV is positive and, for a real process, P > P,. The term T dS is then positive, making dS positive and, from eq. (3.13),

Sp > 0 It should now be evident that for a compression, dV is negative, as is P — P,. Once again, for a real process, Sp is positive. In all cases, if P = P,, the expansion (or compression) is reversible and S$, = 0.

3.1.2

Statement of the Second Law We can now state the second law of thermodynamics as follows:

Sp 2 0

(3.15)

It is zero only for the practically unobtainable reversible case and is positive for all real processes.

3.2

Heat Engines

61

The value of the second law, and in fact the entropy balance, comes not in analyzing

these obvious situations but in analyzing complex ones, where the ‘“‘reality’’ is unforeseeable.

3.2

HEAT ENGINES A common use of the second law is the analysis of ‘‘heat engines.’’ A heat engine is a device that absorbs heat at a high temperature, produces some work, usually mechanical, and discharges some heat at a low temperature. We are all familiar with these devices: the coal-burning power plant, the internal combustion engine, and so on. Laypeople are often dismayed by the fact that devices that turn heat into work cannot perform this duty with 100% efficiency, or at least something approaching it. After all, conversion of mechanical work to electricity, as in the standard electric generator, or conversion of potential energy to electricity, as in the hydroelectric example in Chapter 2, have no such limitations. Without going into details about the operation of heat engines (we’ll do that in Chapter 7) let’s analyze one using just a ‘‘black box’’ and our new tool, the second law of thermodynamics. Figure 3.5 symbolizes a heat engine operating at steady state, accepting heat, Qy, from a high-temperature source, at 7,,, (H indicates hot) delivering some work, W, and discharging some heat, Q,, at T., where the ‘‘C’’ indicates cold. An energy balance written around the “‘box’’ as the system, with an accounting period taken as the time for these quantities of energy to flow, will be:

0=Q2,+2.-

W

(3.16)

Notice no restriction on Qc. is demanded by this energy balance; if all the Q,, is converted to W, Qc = 0 and the balance is satisfied. Now look at the entropy balance, written on the same system and accounting period: + Qc/Tc + Sp

(3.17)

Qc = —(Qyu/TWTc — TcSp

(3.18)

0 =

Qy/Ty

so that,

This places a lower limit on Qc as — Q},(Ty;/T¢), for the best, ‘‘reversible’’ process. Any entropy production will act to increase OQ. (remember it’s a negative number since this is heat leaving the system) and then, from eq. (3.16), lower the work output, W.

Figure 3.5 Schematic of heat engine.

62

Chapter 3 The Entropy Balance

We commonly like to examine the thermal efficiency of these engines, expressed as:

1 = W/Qu

(3.19)

We can solve eqs. (3.16) and (3.18) for this ratio: ni

W/iQw=

ad tTe/Taes

TSp/Qu

(3.20)

In the simplest case, that of reversible operation, this reduces to just

n=

1 —Té/Tx

(3.21)

but in any real case Sp will be positive and the efficiency will be decreased from this maximum value. What does this result tell us about obtaining work from heat? Look at a typical coalburning electric power plant.

Find the maximum efficiency of a furnace operating at 500°C with cooling (air or water) available at 25°C.

SOLUTION This means the best efficiency possible is:

1 = 1 - TQ/Ty =

1 — 298/773 = 0.614

which says that even in the best case nearly 40% of the heat input must be discarded. As we'll see later, practical limitations bring this figure closer to 65%, with only 35% delivered as electrical energy.

Another useful example is the “‘heat pump,’’ Fig. 3.6. Here work is used to drive heat from a low temperature to a higher one. Common examples are the air conditioner and the refrigerator. The same first- and second-law analyses as we used for the heat engine, eqs. (3.16) and (3.17), are used for the heat pump. With an air conditioner or

Qc Tc

Figure 3.6 Schematic of heat pump.

3.3.

Examples

63

refrigerator we’re most interested in how much low-temperature heat we can remove for a certain work input. This ratio is called the coefficient of performance, or COP.

COP = 0,/(-W)

(3.22)

Solving eqs. (3.16) and (3.17) for this ratio, we get:

Qc/(—W) = To/(Tx — Te) — TeSp/(-W)Tu/Te — 11

(3.23)

and for reversible operation this reduces to: Q-/(-W)

= T-/(Ty -

Te)

(3.24)

The maximum COP is greater than one; in fact, for a home refrigerator with its freezing compartment at — 10°C and the room (to which Q,, must be discharged) at 25°C, COP

= 263/(298

— 263) = 7.5

with

Sp = 0

Once again, as with heat engines, practical limitations and irreversibilities bring this value closer to 3 or 4.

3.3

EXAMPLES We will now use the entropy balance in a variety of examples.

3.3.1

Piston—Cylinder

Look now back at the very end of Example 2.3, the piston held in place with a peg, but with the remainder of the cylinder evacuated, with the system maintained isothermal (Fig. 3.7). The energy balance for this situation revealed W = 0 and AU = 0, for this ideal gas. In this case the pressure in the system (the piston plus the gas to the left) at the end of the expansion can be found from the ideal gas law:

Pz = nRT/Vz = PgV5/Vp What is the entropy change for this gas?

Figure 3.7 Piston-cylinder combination; Pb-0.

(3.25)

64

Chapter 3 The Entropy Balance

SOLUTION Equation (3.4) tells us that

dS ="dUTa

E/T) dV

(3.26)

in general, a result that is valid even though this process is irreversible. Since dU = 0 for this isothermal process with an ideal gas we find,

dS, = (P/T) dV or, with P/T = nR/V

(3.27)

for the ideal gas,

dS, = nR dV/V

(3.28)

where the subscript T means temperature is constant. The change in entropy for our ideal gas in the process above is then:

(AS); = J dS; = nRJ dV/V = nR In (V_/V3)

(3.29)

where (AS); is the change in entropy in this constant temperature process. This is obviously a positive value. Go now to the entropy balance for the expansion, once again with the system defined as the gas + piston, and: Sp =

Sg + Sp

(3.30)

The terms for mass in and out are zero (closed system) and Q = O (from the energy balance). So,

Sp

—

AS

Sp

enon (V;/Vz)

and

Sp

II

nR In (Vg/Vp)

(3.31)

This result tells us that, even though the temperature and internal energy of this gas are unchanged, the entropy of it has increased. Further, the entropy production in this process is positive, indicating it can proceed in the direction shown. Compare this result with that obtained from a process proceeding like that shown in Fig. 2.1, where the resisting pressure, P,, is always equal to the pressure of the system, P. In this case the energy balance is: AU

=Q-W

and the work is:

W = nR In (Vz/Vzg)

(3.32)

3.3.

Examples

65

Because the ideal gas is unchanged in temperature, AU = 0, and Q = W. The entropy balance is now written: bye, aad

ae O/T, + Sp

(3.33)

or ASt=

ees

a anames. =

:

¥0'0- sa

00z

0zz

ee

og

ov

ANA bfase i

ENE

O8T

edie) — SN_} 4s.

NT:

O9T

B

> b-A-";

ee,

8,8. BY

“a

1

oO

Ort

Ea

fy =

%

Qi

OzT

A:

4

Sines bis M2

(

fo)

eeeeebece

Z0'0- -) = €0'0-

ia

Adjeyjuy (QI/N.La)

0

We H U

:

ing

|

Ae JG

ties a

;

cone:

ees

ee

STH

iN

ie

sansiy 7g

oe

ee

rol

0d-

as Z WNT ey

AA

f

—

So66.

“i

sjeajweug

e®yelL-OdH

=,

yueloblijoy

aoe

2

SA

NUE YS

NY

= 00

1200°

Kd 1100°

it; Ya f

Mfa EAN Ng

if i ae ine fHT ANNANOUNES id ul fi finan eaTASES = CL

V1

900°

AAD

guecn AN i sisNN MIE NIN =o

yin ul ii NOI NAT VAIN

iiiANYri ive iyYTS 70°

Jaen ANini MigCHAIN HANAN AN ‘taiieWANTING,

ATO LAO

ore NO

(° Ta

oe AGH a mh sy AV VaANY AANA AWN veMELT

are Se nie A

AV

A ANN

He

(

i}|

ann Ath NINA eee AGAIN ate aaa nianhele WRK RANA N UNITING NANS ST LOUIE RIANA EAT ACORN SSIS es

0.50

0.75

1.00 ene

1,50 413)

Figure 5.3 Temperature—entropy diagram for steam.

120

1575

bee

5.4

Residual Properties

121

Another set of lines appears on this diagram: those of ‘‘constant superheat.’’ These show the temperature decrease needed to bring the steam to its condensation temperature at constant pressure. This is sometimes useful in heat-exchanger design calculations.

5.4

RESIDUAL

5.4.1

PROPERTIES

Volume We have seen that deviation from ideal gas behavior is often small, at times even

negligible. We therefore define ‘‘residual properties’’ as the difference between real and ideal gases and treat only this difference by means of corresponding states. For example, for volume, which we have already handled with corresponding states, the residual property, v®, is defined:

vk =v

— uv!

(5.22)

where v is the real gas specific volume and vu! the ideal gas specific volume. But we know that:

p=

tRT/P

(5.23)

DE=; ZRT/P

(5.24)

and,

where z is the compressibility factor, defined in eq. (4.9). So,

vk = = ( nD)

5.4.2

(5.25)

Gibbs Free Energy The specific Gibbs free energy, g, is of special interest because of a mathematical transformation.* Recall from calculus that:

()-(e-Ge

as

so,

ia() = i yee ar]Bag VBE or

*We treat the specific free energy; relations with the total free energy, G, are completely analogous.

(5.27)

122

Chapter 5 Property Interrelations and we know,

g =h

—

Ts

and,

dg =v dP

es dit

(5.10)

So, substituting these into eq. (5.27):

dle

ee

ee

RT

ee

=

peli

RT

RE,

from which the terms in entropy are seen to cancel, leaving:

g ni Feta vdP hdT (£)

(5.28)

Precisely the same operations, done on the ideal gas, give:

I 8gI Usage (5) = Sr

I | led Rr

(5.29)

FhR ar eRe

(5.30)

so, for the residual Gibbs free energy, 2°:

R gR USdP (2) Pl pre

Weare already able to calculate changes in enthalpy and entropy at constant pressure, so we now focus on the changes at constant temperature, eliminating the second term in eq. (5.30). We integrate from P = 0, where g® = 0, to any pressure, P: R —= = == ROE TRIE

P

P

(ges Sig 2

= == RT

P We =

Jo :

SS 12

0

ale

‘ Cry

The last integral was obtained by use of eq. (5.25) for v®. The integration can be carried out with either the z charts (numerically) or the virial coefficient correlations, depending on the reduced pressure and reduced temperature. This need only be done once for the entire range of reduced temperature and pressure; corresponding states then allow us to use the results for any substance for which we have no exact data available. But first notice that the integrand on the RHS of eq. (5.31) becomes (0/0) at P = 0. This indefinite limit is evaluated using L’ Hospital’s rule:

az =

ap

1)

oz

|,

\aP),

= (2)

oP),

B Sad a=

1

P=0

Keli

(5.32)

5.4

Residual Properties

123

where eq. (4.11) has been used to get the final result. This lower limit in the integral in eq. (5.31), B/RT, is calculated with the correlations in Chapter 4, eqs. (4.20) and

(4.21).

5.4.3.

Enthalpy* Divide eq. (5.30) by dT at constant P:

(5.33) so that

(5.34) And we have already seen from eq. (5.31) that: P

ee Re

| 0

(Cae A) |

dPy

(5.35)

so we must now differentiate eq. (5.35) with respect to T at constant P:

(5.36) so that

(5.37) As with g®, these integrations need only be carried out once, over the entire range of

reduced temperature and pressure.

5.4.4

Entropy* The residual entropy is readily calculated from the free energy and enthalpy: sR

P

:

(5.38)

—g8/T + AR/T (2x1)

ee

SRIIPH

*B. I. Lee and M. G. Kesler, AJChE J., 21 510 (1975). *B. I. Lee and M. G. Kesler, AIChE J., 21 510 (1975).

P

gy

er|

dz \ dP

—})

paNor).P—

(5.39)

124

Chapter 5 Property Interrelations

5.5

REDUCED FORM In order to use the generalized correlations for the compressibility factor in the calculations for enthalpy and entropy it is necessary to convert eqs. (5.37) and (5.39) into reduced form. This only requires substituting Pa

Pe

r=

TI.

and

where P, is the reduced pressure (P/P.) and T, the reduced temperature (7/T,) as defined in eqs. (4.17) and (4.19a), wherever P or T appear, including the limits of integration, to obtain: P.

a PS Sher = | ob

he =

(b@e \ ae ),si

: one

5.6

Changes In Properties

137

and from the definition of entropy, dh = T ds + v dP which is divided by dP; to give:

(2) ek —

—

(=) + OP), _

(5.54)

U

°

We eliminate the term involving the entropy using our ‘‘Maxwell’’ relation, eq. (5.15):

PE i Bg aa aeesane apneBT] po pelle

5.55 (5.55)

(where B is the thermal expansion coefficient) so the total enthalpy change is:

Ah = |an - ie af

Ul te— ar} |aP,

(5.56)

and the entropy change is:

C As = |ds = I= Tp — Bu |dP

(5.57)

Notice that, for complete rigor, the terms outside the pressure integrals in the above two equations should be inside. However, for most engineering applications B and v are insensitive enough to pressure that average values can be used.

Find the change in enthalpy and entropy for the water in Example 4.7.

SOLUTION The water went from 25 to 30°C and from 1 to 32.1 bar. We use eq. (5.56) to find:

= 75 75.3 mol °K Ah =

5)

(5° K) + 18.0

(1 — 303 X 107° K7~! x 303 K)

m

x 31.1 bar x

J

10 cm? - bar

= 376.5 + 50.8 = 427.3 J/mol As = 75.3 In (=) =303

x. 10>°*
:

(C, — C,) = (2) (=) = TB’v/x P

Vv

(5.62)

5.7

Changes Of Phase

139

For the ideal gas the RHS of eq. (5.62) is simply R; but for real fluids it can be much different. Take a liquid such as water at 293 K: C.P £¢

v

2937065010" °)? * 1.0/46°x

10-°

0.27 cm? - atm/g - K = 0.0065 cal/g - K about 3 of 1% of the C, of water. Solids show even less difference: copper at 25°C shows a difference of only 7 X 107° cal/g - K. For this reason, the heat capacities of condensed phases at constant pressure or constant volume are most often considered equal.

5.7

CHANGES OF PHASE Picture a phase diagram as shown in Fig. 5.10. The isotherms drawn are to be presumed rather close together in temperature. We know that: dg =u

dP —

s aT

(5.10)

and because P, = P, and T, = T>, g, must be equal to g,. Check this in the steam tables: take any point in the saturated steam section and see whether the liquid and vapor Gibbs free energy (h — Ts) match; they will to within the precision of the tables. Now imagine that we change the temperature slightly, to T; (= T,). The change in g on the a curve must match the change in g on the B curve, in order that g; = g,. So,

dg* = dg? which means:

v® dP®

—

s* dT*

=

v® dpP —

s? aTP

(5.63)

or, (v* —

v®) dP =

Figure 5.10 Two isotherms on P-v diagram.

(s* —

s®) dT

(5.64)

140

Chapter 5 Property Interrelations making the slope of the vapor pressure curve:

ET

ed)

(el

.: (u*

=

5.65

vP)

However, because g* = g®, As = Ah/T, where ‘‘A’’ here refers to “‘a — B.’’ So eq. (5.65) becomes:*

(=) = a ODE ees NAY

(5.66)

an important exact equation called the Clapeyron equation. It holds for any phase boundary, not only vapor—liquid equilibrium (VLE); but it is widely used for vapor— liquid, either to estimate the change in vapor pressure, P°, with temperature from knowledge of the heat of vaporization or to estimate the heat of vaporization from the vapor

pressure—temperature curve. At low pressures, say, below 2 atm, two approximations can be made to the Clapeyron equation for VLE. First, the vapor is nearly an ideal gas, so:

v, ~ RT/P

(5.67)

and second, the molar volume of the liquid is much smaller than that of the vapor: Ve ~ 0 so eq. (5.66) becomes, under these conditions:

dP° > = dinP® Ie

Ah dT = =.= RT

= —

Ah R

> 7d

(1 7

(5.68)

the approximate form known as the Clausius-Clapeyron equation. It indicates the linear form for the log of the vapor pressure with the reciprocal of absolute temperature so widely used. Because of the approximations in the development of eq. (5.68), a better, although empirical form of vapor pressure is the Antoine equation:

InP?

=A -—B/(T+ ©

(5.69)

where A, B, and C are tabulated constants for each substance. A representative table is

found as Appendix III.

5.8

SUMMARY We now have the tools to calculate changes in all thermodynamic properties with changes in T and P. We shall see in the following chapters that the most important are h and s; they are key in the design of compressors, turbines, and other working machinery.

*The vapor pressure, P®, is sometimes written P,,., or PS*.

5.8

Summary

141

APPENDIX Forms for du and dh as f(T, P, V) Write an equation for u as a function of temperature and volume, so as to make use of

the heat capacity at constant volume:

u = f(T, v)

(A.1)

so that,

du =

a aT +(—)a (—) oT},

OV |

dw =c,ar+

: a (+) OU /,

(A.2)

The second derivative in eq. (A.2) is found by dividing eq. (5.16) by du at constant T:

(2) = 2(2) -» OU /

as)

OU /

and then using the Maxwell relation, eq. (5.21), in eq. (A.3), eq. (A.2) becomes:

oP du = C, dT + r(2)dv — P dv

(A.4)

Similarly, for the enthalpy:

h = f(T, P)

mele

='|—)

oh

(A.5)

eon ete Neel a

ar + (—)-aP = C, aT + |—)

(2) =

os

(3) +u=

OU

-7(2) =U)

aP

A.6

A

(A.7)

this leads to:

p dP + vaP dh = C, dT - (2

(A.8)

PROBLEMS 5.1. An adiabatic turbine operates with methane as the working fluid. The flow rate of the methane is 1000 g/s. At the inlet of the turbine the pressure is 97 bar and the temperature is 495 K. At the exit the pressure is 94 bar and the temperature is 190 K. What is the power delivered by the turbine (kW)?

142

Chapter 5 Property Interrelations

Deze Ethylene gas, initially at 65°C, 1 atm is compressed isothermally and reversibly to 225 atm. a. What is its enthalpy change? b. What fraction of the enthalpy change is due to nonideal effects? c. What is its entropy change? d. What fraction of the entropy change is due to nonideal effects? Sa: An adiabatic steam turbine produces 100,000 BTU/h electrical power. Steam enters the turbine at 300 psia, 900°F. The turbine exhaust is saturated vapor steam at 5 psia. a. What is the flow rate of steam (Ib/h)? b. What is the efficiency of this turbine compared to a reversible, adiabatic turbine with steam entering at the same pressure and temperature as in (a) and exiting at 5 psia? 5.4. Carbon dioxide enters a compressor at 273.8 K, 36.9 bar with a flow rate of 1 g - mol/s. It exits at 221.4 bar, 760.5 K. The compressor is not insulated and dissipates heat to the environment at 2000 J/s. How much power is required by the compressor? Sesh A closed system machine consisting of the following four operations is constructed and uses steam as the working fluid. Sketch on a T—s diagram the four processes: a. Saturated water at 300°F with a quality of 0.3 is heated at constant volume to 700°F. b. The fluid from (a) is expanded through an adiabatic throttling valve to 650°F. c. Heat is extracted at constant pressure. d. The fluid is brought back to its original state in an adiabatic reversible compression. 5.6. One pound-mole of carbon dioxide, initially at 5000 psia, 200°F is allowed to expand against a constant resisting force of 4000 lb, distributed across a piston of 8-in.” area. The volume is permitted to increase to such a point that after the temperature of the apparatus has been returned to its initial temperature (through immersion in a constant temperature bath), the volume has doubled. How much work is done by the piston (BTU)?

Sats A cylinder is fitted with a frictionless piston and has an initial volume of 0.02 m?. The piston— cylinder contains 0.2 kg of steam at 3100 kPa. Calculate the amount of heat transferred and the work performed when the steam is heated to 600°C at constant pressure. 5.8. Carbon dioxide at 1000 psia, 800°F is expanded continuously through an ideal, adiabatic (horizontal) nozzle to 500 psia. The inlet velocity is low. Find the exit temperature and velocity. CO, is not an ideal gas at these conditions. 5.9: How much absolute error (J/mol) would be made in calculating the enthalpy of carbon dioxide at 120°C, 150 bar if it were assumed an ideal gas? 5.10. Find Q and W for a system consisting of 1 lb of steam in each of the following processes: a. Cooled at constant volume from 50 psia, 500°F to 300°F. b. Cooled at constant pressure from SO psia, 5O0°F to 300°F. 5.11. Find the change in h and s for SO, compressed isothermally from 1 atm, 500 K, to 80 atm. 5.12. A continuous, adiabatic, reversible compression takes ethylene from 200 psia to 1500 psia. If the inlet temperature is 150°F, what is the outlet temperature (°F)? How

S13;

5.14. §:15.

5.16.

much work is

done per pound-mole of ethylene (BTU)? A mixture of saturated liquid and vapor HCFC-123 at 220°F is brought to 2 psia pressure through an adiabatic throttle. The exit temperature reads 40°F. What is the inlet quality? Use tables and charts for refrigerants in Appendix VII. What is the change in entropy (BTU/lb - °R) for the HCFC-123 in Problem 5.13? One kilogram-mole of isobutane is compressed isothermally and reversibly from 80°C, 1 bar to a pressure at which the volume is one-fourth the initial volume. How much heat is transferred? What is the final pressure? Carbon dioxide is used to recover oil from depleted wells. It is repressurized as it is recovered from the product in a steady-state compression. If it enters the compressor at 25°C, 15 bar

5.8

Summary

143

and exits at 30 bar, what is the exit temperature? Assume isentropic compression. How much work is needed per kilogram-mole of carbon dioxide? Sakis In the recompression process above, the power is available at $0.05/kWh. If the oil is worth $18/bbl, how much oil must be recovered per kilogram-mole of carbon dioxide just to pay for the electricity used by the compressor? 1 ‘‘barrel’’ (bbl) is equivalent to 42 gal. 5.18. An adiabatic turbine that operates using carbon dioxide as the working fluid has an isentropic efficiency of 70%. The inlet is at 221.3 bar, 456.3 K. The exit temperature is 365 K. How much work is produced per mole of CO,? ;

OU

Os

5.19; Develop expressions for (2) and (=) as functions of only P, v, T and heat capacity; not u, Ss, or h.

, 5.20. Develop an expression for

oh ar

: : as a function of only P, v, T and heat capacity; not u,

s, or h.

oT

5.21. Develop an expression for (Z) as a function of only P, v, T and Ca Ss

OU

os

5.22. Develop expressions for (2) and (=) in terms of the constants a and b in the van der Waals equation of state, v, and C,,. :

oh\

,

:

3.25. Develop an expression for (2) in terms of the constants a and b in the van der Waals Vv

equation of state, v, and C,,. oT

;

5.24. Develop an expression for (=) in terms of the constants a and b in the van der Waals Ss

equation of state, T, v, and C,,.

OAT

oh Using the van der Waals equation of state, find expressions for the derivatives (2) and

aP

oh aP},

£

Chapter 6

Flow of Fluids

Most chemical engineering processes operate at steady state, with fluids (and some solids) transported between equipment at different pressures and temperatures. A great deal of energy is expended when large volumes of gases must be compressed and the machinery to do so is expensive. Liquids, on the other hand, can be brought to relatively high pressures at much lower cost. A look at some examples with the fluid-flow equation developed in Chapter 3 shows why this is so.

6.1 6.1.1

GASES General We begin with the general energy balance for a steady-flow process, either compression or expansion, the period being the time for 1 mol of gas to flow through:

Ah

, §Az Aw) _ Re

28.

=.

(6.1a)

There is usually no significant difference in elevation in this machinery, and unless the fluid is near sonic velocity, no significant kinetic energy change. Further, this type of machinery is usually near adiabatic, so,

Ah=@-W

(6.1b)

and, for the general (nonideal gas) case, Ah (outlet — inlet) will be function of both T

and P. The general problem is one in which the inlet pressure and temperature are known, as is the outlet pressure. This leaves two unknowns: outlet temperature and work. We have, however, the entropy balance at our disposal:

As = —= + Sp = 0

(6.2a)

T,

which adds another unknown, the entropy production, Sp. The best we can do at this point is to find the reversible work, with S, = 0, the minimum work for adiabatic compression or maximum work from an adiabatic expansion:

144

6.1

ee

Sp = 0

Gases

145

(6.2b)

T,

The performance of real machinery is then compared with this benchmark. Solution of eq. (6.2b) involves finding the outlet temperature that satisfies the isentropic condition; this means interpolating in a table of thermodynamic properties, or finding the root, 75, from our corresponding-states correlations: 270 c P As = —SR+ cp aT — Rin + sf)= fl) = 0

(6.2c)

1

where C> is the ideal gas heat capacity. The energy balance, eq. (6.1b), then is used to find the work; this, in the absence of tabulated data, will also require using residual enthalpies, as described in Chapter 5: Z

Ah = —hR + | cars hk where the superscript R indicates residual.

6.1.2

Ideal Gases For an ideal gas, the residual properties are zero, and the solution of eq. (6.2b) for the outlet temperature is simpler. With constant heat capacities, we can solve for T, explicitly. Adiabatic, Reversible, Compression /Expansion

Calculate the power needed to compress 1 mol/s of a diatomic ideal gas isentropically (adiabatically and reversibly), with a constant* C, = $R, from atmospheric pressure and temperature to 10 atm.

SOLUTION The energy balance written over this compressor is:

Ah = —W

(6.1b)

The enthalpy change from inlet to outlet is:

an = |an = [oar + [ras

*Using the more correct C, = f(T) does not change the method of attack; it only involves more algebra.

146

Chapter 6 Flow of Fluids But, for our isentropic case, As = 0, so eq. (6.1b) becomes: out

an

= | vudP = —W in

and for an ideal gas, To

[ Goal,

Ah

TI

where we find 7, from:

As = 0 = C2 ln (To/T,) —R In (Po/P) = 0 T(Po/P)*/ = 575.3 K To and

W

=

—4R(575.3

—

298) =

—8070J

in | s, so:

P = 8.07 kW

As an exercise, show that the steady-state, isentropic work of compression for 1 mol

of an ideal gas (with constant heat capacity) can also be represented by:

= |

-Po( 5) ey”) We

(6.3)

sire) (Bah where r is the ratio of Cp/Cu. As a help, proceed as we did in the development of eq. (2.30) for the batch, isentropic work of compression for an ideal gas. Isothermal, Reversible Compression

Next, consider the isothermal, reversible compression of an ideal gas. Here,

Ah=q-W=0 and,

AS

4)

Dey

6.1

Gases

147

so,

W = T As =

—RT In (P,/P;)

(6.4)

which for our example gives a work of — 5705 J, considerably less than the adiabatic, reversible case. Staged Compression As we will see in more detail in Chapter 8, it is difficult to design a compressor or turbine to be isothermal; there is no simple way to achieve the needed heat transfer. However, the compression can, and usually is, done in adiabatic stages with heat transfer inbetween. Let us say the above compression is done in two adiabatic stages, the first to 5 atm and the second to 10, with heat transfer back to 298 K between stages. The work in the first stage is:

W

_ =R - 298(1.4) ;

0.4

(5.0°78° —

1.0) =

—5062. J

1.0) =

—1899. J

and in the second, starting again at 298 K, —R - 298(1.4)

es

0.4

(2.0°-78° —

for a total of —6961 J, greater than that of isothermal compression but considerably less than that required by a single adiabatic stage. The process is illustrated in Fig. 6.1.

P

Figure 6.1 Work for adiabatic, staged compression

148

Chapter 6 Flow of Fluids

The work of isothermal compression from P, to P, is the integral under the curve a— d—f, whereas that of single-stage adiabatic compression is that under a—b—c. The intermediate work of two-stage adiabatic compression with intercooling is under a—b—d-e.

6.2

LIQUIDS The work of compression for liquids is far less than that for gases; this is so because the molar volumes are so much smaller. We will calculate the work necessary to bring liquid water at room temperature from 1 atm to 10 atm. For 1 mol of water, nearly incompressible, eq. (6.1a) becomes:

W = —Ah = —v AP =

—18cm?/mol

X 9 atm =

—16 W

or 0.2% of that for the ideal gas.

6.3

STEAM Steam is the working fluid in an enormous number of industrial processes; many for the production and use of mechanical and electrical energy. Working engineers will encounter design and ‘‘tech service’’ problems dealing with steam at many points in their careers. For this reason it is essential that we become familiar with the form and use of the steam tables, Appendix VI. They are not only useful in themselves; they also are a universal model for tables of thermodynamic data for other substances.

What is the minimum power (hp) to adiabatically compress 200 lb/min of saturated (vapor) steam from an inlet condition of 25 psia to an exit of 95 psia? If the compressor efficiency is 70%, what is the power requirement and what is the state of the outlet steam (7, h, s)?

SOLUTION Take, as the system, the compressor. As accounting period, take 1 min. First, for the

minimum power, the energy balance is:

min

where:

m, = 200 Ib h, = 1160.6 BTU/lb ~ from steam tables and the entropy balance is: n

0 = m(s; — So)

+0 +0

no Q, no Sp

6.3

Steam

149

so, So = S; = 1.7141 BTU/Ib - °R At 95 psia, this value of entropy is located by interpolation to be at 498.5°F. The enthalpy at this point, also by interpolation, is 1280.0 BTU/Ib. The energy balance is then solved to give:

Ah = 1280. —

1160. = 120 BTU/lb

At a mass flow rate of 200 lb/min, AH = m Ah/At = 200 X (120.) = 24,000 BTU/min

=

—P

a minimum power requirement, — P,,;,, of 566 hp. The actual power will be more, because of the 70% efficiency: P, = 566/0.70 = 809 hp and the exit conditions will have to satisfy the energy balance for the real compressor:

0 = 200(4g —

1160.) + 34,280.

so that,

ho = 1331. BTU/Ib which, at 95 psia, corresponds to an outlet temperature of 600°F with an outlet specific

entropy of 1.7645 BTU/Ib-

R.

Notice the real compressor, although still adiabatic, shows a positive entropy production (write the entropy balance for the real case), which, of course, it must if we

have done everything correctly.

A rigid tank 3 m? in total volume initially contains 1200 kg of saturated liquid water; the remainder of the volume is filled with saturated vapor steam. The initial temperature is 200°C; at that time 800 kg of liquid water at 60°C are added to the tank, along with enough heat to keep the temperature at 200°C. a. What is the mass of saturated vapor steam at the end of the process? b. How much heat was added?

SOLUTION At first glance this appears to be a difficult problem; there seems no way to get directly to the answers. It is with such problems as these that the balance techniques show their value.

150

Chapter 6 Flow of Fluids

Choose, as the system, the tank and contents at any time. As accounting period, take the time it takes to add the water and bring to equilibrium. This is not a steady-state problem; we return to the general energy balance: U,

=

U,.+

A,

+

2

What terms have been canceled? First, there is no work done on or by the system—its volume is constant. Next, no mass leaves the tank. We neglect any potential or kinetic energy of the incoming water; likewise, the tank and contents do not change position or velocity during the accounting period. Now to the solution: First, arrange everything we do know.

P, = Py = 1554.9 kPa 800 kg my he = 2501 ki/ke

(Neglect the effect of pressure on the liquid enthalpy; we now know it to be very slight. Thus, we use h of the saturated water at 60°C.)

He =H

the O00

a0kt

Vz = 3m? = m,-u, + m,,-v, = [1200.(1.156) + m,,(127.2)] X 10° m3 Notice the specific properties of the saturated water and steam are the same at beginning and end; the T and P are the same.

We can solve for the initial vapor mass: m,, =

12.68 kg

and then the initial internal energy: Us

SS

Mp

£ Uu; ae

m,,

s Uu,

a

1.054

x

10° kJ

Now to calculate the internal energy at the end: Up

=

=m,"

II

u;

+

mM, > u;

2012.68 kg = m, + m,,

SMe Dy tt

0,

Solve these last two for the masses of liquid and vapor at the end:

m),, = 2007.3 kg m,. = 5.342 kg Uz = 2007.3 X 850.6 + 5.342 x 2593.2 = 1.721 x 10°KJ

6.4

Corresponding States

151

The energy balance now gives us Q:

Q = 1.721 X 10° — 1.054 x 10° — 2.009 x 10° = 4.662 x 10° kJ

Sticking to the rigor of the energy balance and mass balance has led us directly to the desired answers. Attempting the solution with ‘‘shortcut’’ methods almost always leads to incorrect answers.

6.4

CORRESPONDING STATES With working fluids other than steam, for which tables or graphical data are unavailable, ““corresponding states’’ is used to arrive at the answer. Typically, the outlet temperature will be unknown, so the entropy balance cannot be solved explicitly. In the high-density region, a trial and error procedure is followed to arrive at the temperature that satisfies

the balance. Once this is found, the exit enthalpy is calculated directly, once again following the five-step procedure outlined in Chapter 5.

In a polypropylene manufacturing plant, excess propylene from a reactor is expanded through a turbine to produce electric power. The propylene exists the reactor at 95 bar, 450°C and must exit the turbine at no lower than 5 bar. If the turbine is 80% efficient and operates adiabatically, what is the maximum power output at a propylene flow of

1 kg - mol/s? SOLUTION Take as the system the turbine; as accounting period, | s. First, find the power of a 100% efficient turbine:

Energy Balance: Entropy Balance:

Ah = As, = 0

1 Winax

Look up critical properties of propylene and calculate reduced temperature and pressure:

T, = 365K

P, = 46.2 bar

w = 0.148

T,, =

P,, = 2.06

high density

P..

probably low density

so, 1.6

and,

T..

=?

= 0.11

152

Chapter 6 Flow of Fluids

sk = R[—-0.4 + 0.148(—0.15)]

Ash. Sy =) = Rela

a 95

=

—0.422R

(Figs. 5.4 and 5.5)

= 2.94R :

T.Be Ast = n{16m In (2) + 22.706 < 10° °(, = 723) mao

Xx 10mg

5

(Ts — nx

0.675 x 365°

sh =

ROX ee

0.722 x =)

oF 0.14s(

TS

(Eq. 5.50)

Now add these up to find the total As = 0 as in Chapter 5:

As == ish

Asher Ash

sk = 0

an implicit equation in T, that solves to give T, = 561 K. At this temperature the reduced temperature is 1.536, well into the ‘‘low-density’’ region, verifying our choice of technique for the reduced entropy calculation at the outlet. The enthalpy change can now be calculated, with this outlet temperature, using the scheme outlined in Section 5.6.2. Here Figs. 5.4 and 5.5 are used for the reduced enthalpy at the inlet (high density) and eq. (5.49) for the reduced enthalpy at the outlet (low density).

—n® + An, + hn = 323R — 2162R — 19R = —1858R a) ax |= 15,449 kW cb= II

So the actual power is 0.80 X 15,449 = 12,360 kW

The actual enthalpy change is calculated from the energy balance for the real turbine:

where Ah = by solving:

—1858R

X 0.8 =

—12,360 kJ. The actual outlet temperature is found

Ah = —hR + Anh + AR = —12,360 kJ for the outlet temperature, T,, another solution of an implicit equation in T.

6.5

THROTTLES A throttle is simply an adiabatic valve; the energy balance has been analyzed in Chapter 2 to show that it is isenthalpic. The enthalpy of the entering fluid equals that of the exiting fluid. The pressure at the exit must be lower than that at the entrance; otherwise there could be no flow. This type of isenthalpic process is often referred to as a Joule-

6.5

Throttles

153

Thomson expansion, and the derivative:

By =

oT (z)

(6.5)

is called the Joule-Thomson coefficient. We saw in Chapter 2 that an ideal gas will show no temperature change through the throttle; now look at a real fluid. The Joule-Thomson coefficient can be related to more fundamental thermodynamic properties. We start with:

eea 1; (*) is proportional to (24) Vv

(6.49)

and here the velocity increases as the area increases, along with a corresponding decrease in the pressure. Outlet pressures between that which satisfies eq. (6.48) and that which satisfies eq. (6.49) cause a shock wave, that is, an abrupt pressure increase with kinetic energy converted to thermal energy, in the diverging section. The corresponding entropy production causes nonisentropic conditions to exist, and eq. (6.30) is no longer valid. This,

164

Chapter 6 Flow of Fluids

in simple terms, is what happens when the air flowing over an airfoil (e.g., a wing) at supersonic velocity suddenly rejoins the stationary air behind. A shock wave is created perpendicular to the airfoil, one that follows the moving body at all times during supersonic flight—not, as is sometimes thought, only when the air plane ‘‘breaks the sound barrier.”’

6.9

SUMMARY We have seen how to calculate the work needed to compress fluids, both gases and liquids, as well as the work obtained when these fluids are expanded through a power turbine. Not all expansion leads to work; the throttle is an example of ‘‘lost work.”’ There is no analogue to the throttle for compression; one must put in at least the minimum work required by the second law. The principles in this chapter are only the rudiments needed for design of machinery to perform these tasks. In a following chapter we will look in more detail at the types of equipment available.

APPENDIX

Adiabatic Compressibility The adiabatic compressibility, k,, is directly and simply related to the isothermal compressibility, k, through the heat capacities. The relationship is developed here. We begin with the derivative of the volume with pressure at constant entropy, which, with use of the chain rule, can be broken down as:

(2) - aT) (av

me

oP) \aP} \ar/.

Ad)

Each derivative on the RHS of eq. (A.1) is then split into its entropy derivatives using the same mathematics as used in the development of eq. (4.44):

() aP}.

_ (3) (as

O5jJp\ OP)

So

and

(2) _ _ (80) (as Ofte

mNaseon

far

The temperature derivatives of entropy are related to the heat capacities as: Os

|)

=

C

and

Os

C

—)}=—

and Maxwell’s equations, (5.15) and (5.21), allow for the replacement of the other entropy derivatives with P—v—T derivatives:

s)--@) = @-e oP),

aT}.

a

ar}, \aP},

aes

Problems

165

Substituting eqs. (A.4) and (A.5) into eqs. (A.2) and (A.3), and these into eq. (A.1), we get:

(2 _ _1 (av) (ar oP}.

PNOT)

NOP),

ne

a

which is equivalent to:

av) _ 1 (a OB mamTeNOby

ae

Ag

erly) pets, f.0) ology v\aP)rlu\aP).|or®

)

Finally, therefore:

ye

PROBLEMS 6.1 A high-pressure polymerization process requires a continuous supply of sulfur dioxide at 40. MPa. The sulfur dioxide is available at 2.6 MPa and 158°C. If a compression could be carried out isothermally and reversibly, what rate of heat transfer would be necessary for a flow rate of 1 kg - mol/s of SO,? What would be the minimum power for this isothermal compressor? 6.2. Ethane is being compressed adiabatically and reversibly from 1.72 MPa, 32°C to 7.36 MPa. Downstream, the ethane is cooled back to 32°C in a heat exchanger. What is the horsepower of the compressor for a flow of 10° SCFH [standard cubic feet/hour (15°C, 0.1 MPa)]? 6.3. Steam at 5.51 MPa, 410°C enters an adiabatic turbine from which it discharges at 0.207 MPa.

6.4.

6.5.

6.6.

6.7.

A sample of the discharge steam is conducted through an adiabatic throttling valve that exhausts to the atmosphere (1 bar); the discharge steam temperature reads 104.4°C. a. If the steam rate is 50 kg/h, what is the horsepower output of the turbine? b. What is the rate of entropy production through the turbine per kilogram steam? Through the valve? c. Calculate the efficiency of this turbine as compared with a reversible turbine operating to the same outlet pressure. A mixture of saturated steam and water at 4100 kPa is flowing in a high-pressure line. When a small portion of this mixture is expanded through an adiabatic throttle down to 101.3 kPa, the temperature of the steam exiting is found to be 125°C. What is the quality of the steam in the high-pressure line? A well-designed nozzle operates with steam entering at 528 K, 600 kPa. The steam flow rate is 1.0 kg/s, entering with a negligible velocity. If the exit pressure is 450 kPa, what is the exit velocity? Assume isentropic operation. What is the exit temperature? A compressor operates adiabatically, compressing propylene from an inlet condition of 298 K, 11 bar to 19 bar. Find the power needed to handle a flow of 2000 g- mol/s if the compressor efficiency is 75%. What is the outlet temperature? An insulated pipe has a partly opened valve in it. An ideal gas flow through it at steady state. The diameter of the pipe is 0.5 ft and is uniform throughout its length. The inlet velocity is

166

Chapter 6 Flow of Fluids

10 ft/s and the temperature is 500°R. A pressure drop and velocity increase results from a partly open valve in the pipe. The exit velocity is 30 ft/s.

GP =

193" (ft

1b/1b R)

C, =

138 (ft - Ib,/Ib°R)

What is the exit temperature from the pipe? 6.8. A rigid tank 20 m? in volume contains 2000 kg of a mixture of saturated steam and water at 240°C. A steam line admits 100 kg of saturated vapor steam at 3348 kPa. How much heat must be removed to have the tank end up at its original temperature and pressure? What are the masses of liquid and vapor at both the beginning and the end of the process? 6.9. Steam enters an adiabatic turbine at 500 psia, 700°F and exits at 67. psia. A small portion of the exhaust is passed through an adiabatic flow calorimeter at atmospheric pressure. The calorimeter indicates the expanded exhaust steam is at 250°F. Find: a. The quality of the exhaust steam b. The power output of the turbine with a steam flow rate of 10,000 Ib/h c. The efficiency compared to an ideal turbine with the same exhaust pressure 6.10. Superheated steam at 300°C, 1.0 MPa fills a perfectly rigid steel vessel 1 m? in volume. How much heat must be transferred to the surroundings to bring the steam to 200°C? What is the final pressure? Neglect the heat capacity of the vessel? Compare with the result found if steam were an ideal gas. 6.11. An insulated tank holds 100.0 kg of liquid water at 20°C just at its vapor pressure. At time zero, 8.0 kg of steam at 10.0 bar is admitted from a line, raising the temperature of the contents to 65°C. If the heat losses, the heat capacity of the tank, and the small amount of vapor above the liquid at the beginning, are neglected and the tank is considered perfectly rigid, what was the quality of steam in the line? 6.12. A steel SCUBA tank weighing 22 lb, 0.6 ft? in volume, is initially evacuated. If it is filled to 2000 psia from a line with air (ideal gas with C, of 3R) at 2100 psia, 70°F, what is the final temperature of the air that fills the tank at the end of the filling process? Ignore heat transfer to the tank. Repeat the calculation, considering the tank to reach thermal equilibrium with the air. The heat capacity of the steel can be taken as 0.12 BTU/lb - °R. Repeat both calculations for the tank filled to 3000 psia from a line at 3100 psia, 70°F. 6.13. Tertiary recovery of oil involves injection of a gas into a depleted well. The gas dissolves in the remaining oil, causing swelling and a lowering of the oil’s viscosity; both these effects aid in bringing the oil to the surface for collection. Carbon dioxide is commonly used for this purpose. In one installation, CO, is taken from a pipeline at 10 bar, 300 K, and compressed to 30 bar. If the compression takes place isentropically, what is the exit temperature, and work per kilogram-mole of gas? 6.14. A chemical reactor requires 10 kg - mol/h propane at 40 bar; it is available at 30°C, 1 bar. a. Find the power needed by an isentropic, single-stage compressor. i b. Compare with that required by a three-stage isentropic compressor with intercooling to

35°C between stages The:ratios? P,/P33="P,/ P= Ps)Py 6.15. A nozzle is designed to provide a high-velocity stream of CO, by taking a high-pressure stream and allowing it to exit at a lower pressure. If the upstream conditions are 350 K, 150 bar and the downstream pressure is 15 bar, what is the exit velocity? The inlet velocity is 5 m/s and the ratio of upstream to downstream nozzle areas is 10. Assume isentropic behavior. Find the result for the ideal gas first; then compare with that obtained for the real gas using corresponding states correlations.

Problems

167

6.16. A similar nozzle to that in Problem 6.15 is adiabatic but not isentropic. Here the outlet temperature is 200 K. What is the outlet velocity? 6.17. For the nozzle in Problem 6.16, find the entropy production for both the ideal gas and the real gas.

6.18. Calculate the speed of sound in water at 25°C, 1 bar. 6.19. Calculate the speed of sound in steam at 25°C, 1.5 kPa. 6.20. Isobutane is to be expanded from 40 bar, 600 K down to 20 bar. How much power can be obtained from an isentropic turbine? The flow rate of the isobutane is 100 kg - mol/h. Compare the outlet temperature with that reached using an adiabatic throttle. What is the entropy production in the latter process? 6.21. An isentropic nozzle operates using steam at 550°F, 100 psia; the exit pressure is 70 psia. What is the velocity at the exit? What is the temperature? Neglect inlet velocity and any potential energy change. 6.22. The same isentropic nozzle as in Problem 6.20 uses air at 550°F, 100 psia. What will be the exit temperature and velocity with an exit pressure of 70 psia? Neglect inlet velocity and potential energy effects. 6.23. A steady flow 100 kg - mol/h of n-butane is to be expanded from 40 bar, 600 K down to 5 bar. How much power could be obtained from an isentropic turbine (kW)? What is the outlet temperature (K)? 6.24. A pump can bring water to 500 kPa with 0.6 kW of power, with the inlet at 1 bar. If the horizontal pipe is 1 in. in diameter, what nozzle size is needed to achieve a minimum velocity

at the outlet of 50 ft/s? 6.25. For the pump in Problem 6.24, with the nozzle attached, how much more power is needed to achieve a spray of droplets whose average diameter is 1 pm (1 pm =

107° m)?

6.26. A ‘‘snow machine’’ operates by spraying cold water through a nozzle into air whose temperature is below O0°C. If enough time for heat transfer is allowed before the drops hit the ground, they form ice particles. How much power is needed for a pump that brings 10 gal/min water from a well through a nozzle that disperses the water into 10-jm drops, 20 ft above the well surface? The drops must have a velocity of 3 ft/s on leaving the nozzle. Neglect frictional losses in the pipe and nozzle. Two perfectly insulated, rigid tanks are connected by an insulated valve, as shown below. 6.27.

Tank A initially contains 1.3 lb of steam at 40 psia, 800°F. Tank B, with a volume of 5.0 ft?,

contains a mixture of saturated liquid water and steam at 90% quality, 90 psia. The valve is opened and the entire system comes to equilibrium. If the process is entirely adiabatic, what is the final pressure (psia)? The valve and connecting lines are of negligible volume.

Chapter 7

Power Production

The availability of electricity at the flick of a switch is taken for granted in virtually all of the United States, and throughout much of the world. The electric power industry is enormous, serving the general public as well as industry and commerce. The first commercial electricity-generating stations, built over 100 years ago, used fuel-burning steam boilers as the source of energy. Despite a truly phenomenal century of scientific and engineering advances, this scheme is still the most practical means for electricity production on a large scale. The only real alteration has been the replacement of fossil fuels (coal, oil, natural gas) with nuclear-fission fuels at some locations. When nuclear

plants were first envisioned there were claims that power would be “‘too cheap to meter.’ Although the nuclear-fission reactor has proved to be a dependable source of heat for steam power plants, costs unrecognized at the time have pushed the net cost of nuclear-generated power above that for fossil plants. There are, and have been for many years, alternatives to the standard steam power plant. These include hydroelectric, wind, and solar collection. All of these are highly desirable, environmentally benign, apparently inexpensive means of producing electricity. Each, however, has its drawbacks. Hydroelectric and wind plants are limited to

topographically suitable locations; contrary to popular belief, it is relatively expensive to ‘‘ship’’ electricity through high-voltage transmission lines over long distances. Solar collectors appear a viable alternative; the sun illuminates the earth daily with far more energy than we need. Unfortunately, the diffuse nature of this input makes collection, conversion, and storage an expensive task. But there have been striking advances in this technology in the past decade that indicate its time may be soon upon us. Fusion energy, that used by the sun and by hydrogen bombs, is an apparently limitless source of ‘“‘clean’’ energy. In the years immediately after the successful testing of the first fusion weapon, there were predictions by those involved in its development that commercial fusion power was “‘about ten years away.’’ This prediction has been repeated every 10 years or so; we seem no closer to this goal than ever. Two other technologies now appear to have a very real chance at making significant inroads into the picture. Fuel cells utilize fossil fuels in a manner that bypasses the production of heat and the subsequent restrictions of the second law; they are currently being commercially implemented in limited applications. The ‘‘combined-cycle’’ plant uses fossil fuel as the energy source; but through a more complex scheme of burning and energy conversion, it is very much more efficient than the simple Rankine cycle.

168

7.1

The Rankine Cycle

169

This means not only that more electricity is produced per pound of fuel, but also that fewer emissions are generated: sulfur dioxide, nitrogen oxide, and even carbon dioxide. We will look at both of these alternatives later in this chapter.

7.1

THE RANKINE CYCLE Any process that converts heat to other, more useful forms of energy is subject to the limitations of the second law. The Rankine cycle is that used by virtually all fossil and nuclear installations. Heat is transmitted to a process fluid, usually water, at a high temperature (and pressure) inside a furnace; work is extracted from the fluid by a power turbine; the fluid is condensed; and finally the fluid is recompressed to the high pressure.

7.1.1

Ideal Rankine Cycle In Fig. 7.1 we have a representation of the temperature-entropy diagram for water. The actual numerical values of the pressure and temperature have been left off for simplicity; the complete 7—s diagram for water should be consulted for these. Let’s first envision the best possible way to get work from the high-temperature heat at 7). Starting at point 0, heat is added to the saturated, high-pressure, high-temperature liquid sufficient to bring it to point 1, the saturated vapor at the same T and P. From here, there follows an isentropic expansion through an ideal turbine to point 2. This is a mixture of saturated steam and water at T¢. It is further condensed in the next step, to point 3, and then compressed mechanically back to 0. Let’s analyze each step for the most efficient energy conversion possible, that is, with zero entropy production everywhere. Each piece of process equipment will serve as our system in each step and 1 s as accounting period. The numerical subscripts in the following equations refer to the points on the diagram. In each step, the accounting period is taken as is.

A. Boiler Energy Entropy

Of O=

Hoa, + On $5) — S, + On/Ta

where the subscript ‘‘H’’ refers to the “‘hot’’ temperature (boiler).

Figure 7.1 Ideal Rankine cycle.

170

Chapter 7 Power Production B. Turbine Energy Entropy

0 0=S,

— S,

where W, is the work output of the turbine. C. Condenser

Energy Entropy

eae =

Dia daly Sy =

Qc

Sangh QW TaGy

Set

icGs, —a55)

(7.7)

making the efficiency:

N=

1 + [To(s3 — s2)1/[Tx(s) — 59)]

(7.8)

7.1.

Oy

condenser}

The Rankine Cycle

171

Oc

Figure 7.2 Rankine cycle power plant schematic.

Notice in Fig. 7.1 that (s; — s3) = —(s, — So), because of the two isentropic processes, so eq. (7.8) becomes:

n=

1- TQ/Ty

(7.9)

just as we saw in Chapter 3 eq. (3.21), for the generic ideal heat engine. This is the absolute best efficiency obtainable (Carnot efficiency) with high-temperature heat available at Tj; and a low-temperature sink available at T.. For example, if the boiler were operating at 370°C and cooling water were available at 25°C, the maximum efficiency would be:

nN =

1 — 298/643 = 53.6%

with the other 46.4% of the energy supplied to the boiler discarded as waste heat.

7.1.2.

Real Rankine Cycle A real Rankine cycle must operate within the confines of real working fluids. First, the compression step will be highly inefficient if it is to handle the mixture of liquid and vapor shown in Figure 7.1. Instead, the mixture should be completely condensed to exploit the low work needed to pump a liquid. This, of course, means that more heat will be discarded in the condenser, as shown in Fig. 7.3. Next, the pumping process cannot go vertically all the way to the boiling temperature, for this would bring the pressure to an exceedingly high level (see the 7—s diagram for water). Instead, the saturated water at the condenser temperature is pumped just to the pressure inside the heat-exchanger tubes that make up the boiler. This subcooled (at a temperature below saturation at this pressure) liquid is heated in the boiler and subsequently converted to vapor, all at constant pressure. This vapor is still at a temperature

172

Chapter 7 Power Production

_ condense Ye

HN

_

YNZ

Figure 7.3 T-s diagram for Rankine cycle.

below that available in the hottest part of the furnace, so more heat is added, super-

heating (bringing it to a temperature above saturation at the existing pressure) the vapor to the maximum temperature allowed by the materials of construction. The superheater is another tubular heat exchanger, which, like the boiler, is exposed to the hot com-

bustion gases inside the furnace. All three steps after the water exits the pump thus follow an isobar on the diagram. The superheated steam has the maximum enthalpy possible entering the turbine. A cut-away view of a commercial steam turbine is shown in Fig. 7.4. The high-pressure steam enters through the set of valves at the left of center, above the turbine. It is

Figure 7.4 Commercial steam turbine showing high-and low-pressure stages and valve sets. (Courtesy Dresser-Rand.)

7.1

The Rankine Cycle

173

directed through the first four sets of turbine blades and then through the second valve set, just right of the first. Here the medium-pressure steam is divided into two parts: one part is bled to other use, such as heating or process use. The other part, which may well be the majority or even all of the steam, is directed to the low-pressure turbine stages, to the right of the reentry port, underneath the second valve set. The exhaust from the last set of turbine blades, at the far right, is at the condenser pressure: the vapor pressure of steam at the condenser temperature. As we know from analyses of turbines in the preceding chapters;

for the most common, adiabatic case. This means we want the exit enthalpy to be as low as possible. What determines this? The entropy balance:

in the best case (Sp = 0). So that the minimum exit enthalpy is fixed by the entropy at the exit pressure. At the same value of entropy, the enthalpy decreases with decreasing pressure, so we want to have as low a pressure as possible at the turbine exit. There is only one way to decrease the exit pressure; by condensing the steam at low temperature, the turbine exhaust can be brought well below atmospheric pressure. For example, with cooling water at 70°F, the steam might be condensed at 80°F (allowing 10°F heattransfer driving force); this would bring the turbine exhaust pressure to 0.5 psia. Look at the difference in work output this accomplishes, compared with exhausting to the atmosphere:

Steam at 800°F, 1000 psia is produced in the boiler of a power plant. Compare the work delivered per pound of steam with an ideal, adiabatic turbine for the case of a condenser operating at 70°F with the case of direct discharge to the atmosphere.

SOLUTION System: turbine Accounting Period:

time for 1 lb steam (1/m)

A. Condenser at 70°F, condenses the steam at 80°F (0.5 psia).

0 = h, — ho — Wr = (1389.6 — ho — W,) BTU/Ib where W, is the work delivered by the turbine O

So

=

5;

—

So

1.5677 BTU/Ib -°R = x1, + xy5, = x,(0.0932) + (1 — x,)2.0359

174

Chapter 7 Power Production

where x, and x, refer to mass fraction of saturated liquid and vapor respectively.

x, = 0.2410 ho = 0.2410 X 48.037 + 0.7590 X 1096.4 = 843.7 BTU/Ib W, = 545.9 BTU/Ib B. Exhaust to atmosphere (14.7 psia).

O'= hy — he — Wy =91389.6

= he — W,) BRU/Ib

0 = 5; — So

So. =I S67

7-BIU/lbie R*=

x5) pa soe—

1 (O3121) oe x. (E7568)

x, = 0.1309 ho = 0.1309 X 180.17 + 0.8691

x 1150.5 =

1023.5 BTU/Ib

W, = 366.1 BTU/Ib

There is a 49% increase in delivered work if the steam is condensed! Of course, this condensate must be repressurized, but this work is very slight, as we saw in Chapter 6:

System: pump Accounting Period:

time for 1 lb steam

Wp =

—v AP =

=

—0.016 ft?/Ib x 999.5 psia

—16.0 psia - ft/lb =

—2.94 BTU/Ib

The enthalpy of the liquid leaving the pump and entering the boiler is thus at

ho = 48.037 + 2.94 = 50.98 BTU/Ib To find the overall thermal efficiency we need only to determine the heat added in the boiler and superheater. System: boiler Accounting Period: On

time for | lb steam

ht =

Qy = M1 =

oer

1389.6

(Wy +

Of

—

50.98

W,)/Qy

=

=

1338.6 BTU/Ib

(545.9

—

2.94)/1338.6

=

40.5%

which is much lower than the Carnot efficiency, Nnax, between the two extreme temperatures 800°F and 80°F:

Tees =

= Tofi =

40)

160m

e507.

7.1

The Rankine Cycle

175

s

Figure 7.5 Comparison of Rankine and Carnot cycles.

The difference is substantial, even though we have treated the realistic cycle as having ideal individual steps. What causes this loss in efficiency? Look at Fig. 7.5. The Carnot efficiency is calculated for the process enclosed by the bold lines, as given by eq. (7.6); the more realistic process is confined to that in Fig. 7.3, reproduced here. The ‘‘heat in’’ occurs over a temperature range, with only the last bit of superheat at the maximum temperature available. This lowers the net work from that given in eq. (7.7), for which all heat in

is at the maximum temperature. Is there anything that can be done to alter the ‘‘real’’ process to approach more closely the Carnot? There are, in fact, several modifications that have been made over

the years. The most influential is “‘reheat,’’ shown in Fig. 7.6. Here the steam is expanded through one, smaller turbine to some intermediate pressure between the boiler pressure and the condenser pressure. It is then recycled back to the furnace at this lower pressure, as shown in Fig. 7.7, where it is heated to the maximum temperature, in this case, 800°F. Refer to the labels to identify the streams in this somewhat more complex scheme.

RY

Figure 7.6 Rankine cycle with reheat.

176

Chapter 7 Power Production

Boiler

Condenser

Qc

Qn

Figure 7.7 Rankine cycle with reheat schematic.

A. The first turbine now exhausts to an intermediate pressure near saturation; here we have chosen 200 psia. The entropy balance around this turbine still shows isentropic conditions:

Sy = 5, =

1.5677 BTU/Ib - °R

But, at this new pressure the exit enthalpy is found in the steam tables by searching for this entropy at 200 psia. It is found to be at a temperature between 410 and 420°F; interpolation gives the enthalpy:

hy =

1217.0 BTU/Ib

The work of the first turbine is now:

Wy, = 1389.6 —

1217.0 =

172.6 BTU/Ib

B. On reheat of this steam to 800°F, 200 psia, the enthalpy rises to:

h; =

1425.5 BTU/lb

and

S; =

1.7663 BTU/Ib - °R

and the heat added is:

Qo, = hy — hy = 1425.5 — 1217.0 = 208.5 BTU/Ib C. Expansion of this steam through the second turbine to the condenser pressure is isentropic: S4 = 53 =

1.7663 BTU/Ib - °R

7.1.

The Rankine Cycle

177

and we solve for the quality of this saturated mixture: 1.7663 = x(0.0932) + (1 — x,)(2.0359) x; = 0.1388* and then the exit enthalpy:

h, = 0.1388 X 48.037 + 0.8612 eurnes

Figure 7.8 Typical fossil-fuel fired furnace.

Modern Rankine-cycle plants run at even higher temperature and pressure than those used in the previous example, offering ideal efficiencies greater than the 38% calculated. But, as we know, isentropic turbines and pumps are unachievable; however, actual ‘‘coal-pile to bus-bar’’ efficiencies do reach 35%.

7.2

BRAYTON CYCLE The Brayton cycle is that used by gas turbines. In it, air is compressed using some of the power from the turbine, as shown in Fig. 7.9a. Ambient air is compressed using some of the power delivered by the turbine. Fuel is injected and the mixture ignited, producing high-temperature, high-pressure combustion gases. These pass through the turbine and exhaust to the atmosphere. As in all heat engines, the efficiency rises with the temperature at which heat is added. The maximum temperature in the Brayton cycle is T3; it is limited by the materials of construction in the turbine.

7.2.

The Brayton Cycle

179

Air

Figure 7.9a Simple Brayton cycle.

This cycle is analyzed just as we have done for other cycles. Before we analyze the individual parts, let’s look at the overall cycle. The thermal efficiency is defined as the net work divided by the heat input: dy WatrhiWe Oin

where for adiabatic, but not necessarily reversible machines, the steady-state energy balances reduce to;

The cycle is modeled as if air only is in a closed loop, receiving heat from the combustion of the fuel and rejecting heat at the exhaust from the power turbine. This is shown schematically in Fig. 7.9b. Using this ‘‘air-standard’’ engine, with the air an ideal gas with constant heat capacity, these enthalpy changes are: ti

ng —

CAT;

ieedg)

and h,

—

h,

=

Wret

Figure 7.9b Air-standard Brayton cycle.

180

Chapter 7 Power Production and the heat input in the combustor leads to an enthalpy gain of: OF

Ci(T3 —

>)

Substituting these into the equation for the efficiency above,

raeones Nace Weg Leek in

0;

et (T;

=

eta

(7.10a)

T,)

where W,, is the net work, W. the work required by the compressor and W; the work delivered by the turbine, a result that is valid for any combination of compressor and turbine, ideal or not, as long as they are adiabatic.

7.2.1

Ideal Machinery The reversible, adiabatic compressor and turbine are isentropic. With this knowledge,

which has been determined from the entropy balance in Chapter 3, we can find the intermediate temperatures. We take the compressor first, with an accounting period as the time for 1 mol of air to pass through the system. 1. Compressor Energy Balance:

0 =k

= ho Oa

The subscript ‘‘s’’ on the work term refers to the isentropic nature of the compressor. The air is an ideal gas with constant heat capacity; this means the enthalpy change from inlet to outlet is just C, AT:

One Co

Dueck) ee WVcx

so the work is:

We = 6

= 1)

2. Combustor The combustor is modeled as heat transfer to the air equivalent to the energy value of the fuel, sufficient to bring the air to the maximum temperature 7}. Energy Balance:

0 =5C

Gost,

O..,

sO, One

CAT.

~

7;)

7.2

The Brayton Cycle

181

3. Turbine

The turbine is also adiabatic and reversible and so is isentropic. The enthalpy loss of the air is equal to the work done.

Energy Balance:

Ss

4. Whole Process The efficiency of the Brayton cycle is the net work done divided by the heat in, as shown in eq. (7.10a). It remains for us to determine T,, Q,,, and T,. The isentropic compressor with an ideal gas with constant heat capacity has been analyzed using the entropy balance in Chapter 3:

T, _ (B\7 (ica

Py

as well as the turbine:

But, in the simplest case,

P, = P,and P, = P; That is, the exhaust is to the same pressure as the intake (atmospheric) and there is insignificant pressure drop through the combustor and the connecting lines. This allows us to find the temperature ratios:

a cereP, TP: r=

tel

eeT

which we can solve for T; and substitute into eq. (7.10a):

which is:

[i

3

II

—

|

(oIae

poh

oe

(7.11a)

182

Chapter 7 Power Production

So the efficiency of the ideal Brayton cycle depends only on the pressure ratio in the compressor (and turbine).

Find the efficiency of an ideal Brayton cycle with a compressor that develops a pressure ratio of 5 and a maximum temperature of 1100 K. Also find T, and T,. The inlet is at 298 K, 1 bar.

SOLUTION Equation (7.11a) gives us the efficiency directly:

i

=

1

—_—

5Ne

—

=

B 0.369

T, and T, are found from the pressure ratio:

®

By ‘

i

0.286

Al)

= 472 K

() D 0.286

(2)

= 694 K

P;

and we can check the efficiency calculated above, using instead the temperatures just found in the general equation (7.1 1a): Bs (7; re: T,) "

(T; rs T,)]

=

0.369

(T; rary T))

in agreement with that found above from the pressure ratio. The heat input in the combustor is:

On = CTs

7.2.2

1yP=! C100)

=" 4712) = G23

Real Machinery The efficiency drops sharply when the characteristics of real machinery are used. A nonideal compressor requires more work than the ideal, and the nonideal turbine is less capable of supplying it. The net effect is to decrease the net work more than just that of each unit individually.

EXAMPLE 7.3 In the previous example, what will the overall efficiency be if the compressor efficiency is 85% and the turbine efficiency 90%?

7.2

The Brayton Cycle

183

SOLUTION We must go through the entire cycle, just as in Example 7.2, but accounting for the nonideal machines. 1. Compressor Energy Balance:

where we find the work required by the compressor, Wc, from that of the isentropic case; Wo =

Wee 0.85

(Tent) ROS

(298 — 472) te. eS 5

so that the temperature rise through the real compressor is 205 K and T, = T, + 205

= 503K

2. Combustor The maximum temperature, T, is the same as in the isentropic case, but because

T, is higher the heat input is smaller. Energy Balance:

Omaha

lames near erik LOO, Gap Qos

ap

3. Turbine The energy balance is the same as that for the isentropic turbine, except here the work output is 90% of the ideal work. Energy Balance:

Wr

aa

0.90W,,.

=

0.90

.

CiAT3

=

T,),

where we evaluate the work of the isentropic turbine from its outlet temperature: Pp

T,4 =

VpT,(—

0.286

= 694K

5}

=

|=

(1100 —

694)- C, = 406.-C,

and then the work of the real turbine is 90% of this,

Wy = 0.90W;, = 365-C,

184

Chapter 7 Power Production Therefore, the loss in enthalpy through the adiabatic turbine:

hy —~ hg = Wr = CT, ISIS ODMEN

— Ty)

C,, making the temperature at the outlet,

Tye) Fe

65=

1100) =. 3654S RK

Now that we have all the temperatures we can find the thermal efficiency from eq. (7.10d):

Title 7 = (Fp St (aE)

COO ae 13S 29 8a en (11005503)

503)

= 0.268

Notice the fraction of the turbine work required by the compressor in this real cycle is 205/365 = 56%, compared with 174/406 = 42.8% in the ideal cycle.

7.2.3

Regenerative Cycle The previous examples show that a great deal of energy T,. Some of this energy can be recovered by means of illustrated in Fig. 7.10. The compressed air is preheated in and then mixed with fuel before going to the combustion

is lost in the exhaust gas, at a gas—gas heat exchanger as a regenerative heat exchanger turbine. The power from this

expansion, less that used for air compression, is delivered as shaft work.

The temperature—entropy diagram, based on 1 mol of air going through the system and with isentropic compressor and turbine, is shown in Fig. 7.11. We can make an estimate of the efficiency of this device, assuming ideal gas behavior with constant heat capacity. We must also make some estimate of the efficiency of the regenerative heat exchanger, eg; that is, what fraction of the maximum possible enthalpy gain for the gas leaving the compressor, through the regenerator, is actually achieved:

Air

Exhaust

Regenerative heat exchanger

Figure 7.10 Brayton cycle with regeneration.

7.2

The Brayton Cycle

185

Figure 7.11 T-s diagram for ideal Brayton cycle with regeneration.

Here, h; is the maximum exit enthalpy achievable with regeneration. With our assumption of constant heat capacity, this becomes; La Fel Nreg i

T; —

T,

where the ‘‘ideal’’ outlet temperature, T;, is equal to the return gas temperature from the turbine, 7;. For the ‘‘ideal regenerator,”’ this fraction is 1.0.

Ideal Machinery

We can find a simple expression for the cycle efficiency with the ideal regenerator. Starting with the definition Wet

Bhan tigi

=

Wr

as Wo

O,,

and the assumption of constant heat capacity

BG

eee

areas €,(T, —

22)

Ua — 75) +

Ts)

i, —

— D)) 13)

and recalling from Example 7.2, for the ideal regenerator,

The expression for the coefficiency simplifies to: ep aaa

vit

ite),

——— eo

4 = 13)

oy

IES,

186

Chapter 7 Power Production

and we can eliminate the intermediate temperatures with the ideal machinery (isentropic compressor and turbine), using: faa

Pa \y 7 (2)

T=

pall

Pa\ Paves (pe r(2) = r(2)

and

ial

1

Substituting for the temperatures, T, and T3 in the efficiency expression: r-1

Ne DD

Lies 7(2)

Ges ae 1a Lat

r

f ee

N=

27—1)

a

=

i

Tigh Ps \ex

ao

(7.11b)

which is valid only for ideal machinery and ideal regenerator. This ideal regenerator would have zero AT between passing streams, clearly an unachievable limit, requiring infinite heat-transfer area. Real regenerators have maximum efficiencies, N,.g, of around 80%. Equation (7.11b) thus gives a value for the maximum possible efficiency, one that is never achieved in practice.

Real Machinery

The 7-s diagram for a Brayton cycle with regenerator and nonideal compressor and turbine is shown as Fig. 7.12. The efficiency becomes:

n

Wy ,.

Webbe icgeiee psn

7 one

where the subscript ‘‘n’’ refers to net work out.

Laon

ECn=

DE)

(7.10b)

7.2

The Brayton Cycle

187

Figure 7.12 Brayton cycle with regenerator and nonideal machinery.

A. Estimate the 7.3 and the temperature B. What would ences?

efficiency of a cycle using the same nonideal machinery as in Example same maximum temperature, but with a regenerator requiring 100 K differences between heat-transfer streams. the cycle efficiency be with a regenerator requiring only 50 K differ-

SOLUTION A. The large temperature difference, 100 K, between the inlet and exhaust streams, as shown below:

Regenerator with 100 K temperature differences.

is needed because gas—metal heat transfer is relatively poor. Because we know T, = 503 K, 7, must be 603 K; likewise T; is 735 K, so T; must be 635 K. This makes the regenerator efficiency:

_eee 635 = 503 157 _

piee Ss,745 1508

at 57%, a rather low efficiency for a regenerator.

188

Chapter 7 Power Production The efficiency of the cycle,

is now:

Bas AC

LF

Pee ee

bie

JOS

CUM

CCTs

ee tees

(298 — 503) Sey

(1100 — 635)

so that we have recovered a significant amount of the loss incurred because of the less than ideal machinery. B. With the more efficient regenerator, requiring only 50 K between passing streams.

685—

Des)

5503

2 EC 9935029503

yg oh

and the cycle efficiency becomes,

= C1003 == 9735) (298) =U — 9503) ~ 0.386 (1100 — 685) a significant gain over the less efficient regenerator in part A, and even more over the simple cycle of Example 7.3.

Higher efficiencies are achieved in the Brayton cycle with more efficient regenerative heat exchanger and gas reheat, but the requirement of more heat-exchange area comes with a significant cost penalty; the materials for high-temperature duty tend to be expensive. Further, it is evident from the shape of the 7—s diagram that efficiencies comparable with that of a Carnot cycle operating between the same two extreme temperatures will not be possible. Still, with higher turbine temperatures (up to 2500°F is anticipated), efficiencies of 40% are possible.

7.3

OTHER CYCLES There are other heat-to-work cycles, besides the Rankine and Brayton. The most common of these are the Ericsson and Stirling. Neither is as often found in actual application, but they are of interest because their ultimate efficiencies are the same as that of the Carnot cycle.

7.3.1

Ericsson The ideal Ericsson cycle, in principle, operates using isothermal compressor and turbine. Imagine a closed loop of, say, air, which enters the turbine in state 1, shown in

Fig. 7.13, at a high pressure, P,,. Heat is transferred to the air in the turbine from a high-temperature source (Tj;). The exhaust from the turbine, state 2, is at the same

7.3,

Other Cycles

189

regenerator

Wret

Figure 7.13 Ericsson cycle. temperature as state 1, but at lower pressure, P,. In the ideal regenerator, 7, also

equals T;. The temperature—entropy and pressure-volume graphs for this ideal cycle are shown in Fig. 7.14. There are two isothermal steps and two constant pressure steps, as shown.

Figure 7.14 Ideal Ericsson cycle.

190

Chapter 7 Power Production

As in the Carnot cycle, the heat transfer is isothermal at both high and low temperatures. However, there is no isentropic compression or expansion; the cycle is completed with the isobaric heat exchange. We can analyze this cycle for the efficiency in our usual way, taking one step at a time. First, considering the turbine, the energy balance taken over a time period equivalent to the flow of 1 mol of air:

(i= he — fo

Ova,

which, using the state numbers in the diagrams, is: O=

ht, — ht, + Oy —

Wy

with constant heat capacity for the air:

OR Cie. and, because T; =

TO

Ve

T>, the first term is zero, leaving:

(7.12)

Oras Now the entropy balance is:

H

where the entropy production, Sp, is zero in the ideal turbine. For our ideal gas, the entropy change through the turbine is substituted:

if P 0 = Gin — Rint + T; foe els and remembering there are only two temperatures and two pressures:

ed

bytes Ere

Ls

Nie dre Hein

ery, aed tay IE,

we get for the heat (and work) in the turbine: P.

Of

Wee

eRT an ae

(7.13)

i

The same procedure is used to find the work needed by the isothermal compressor. The energy balance is: 0 =

bh, =

fa se Oi

ae

7.3

Other Cycles

191

which results in, with the constant heat capacity, Oz=

CAT;

=

T,) me Or

>

We

and, since T; = T,, Oe

owe

(7.14)

The entropy balance for the compressor is:

On= 783"

S47

gu Sp

qT

leading to:

0=C,mn—-

RI eee 4

P,

he

and, with T; = T,, P.

P.

Py

lay

Q, = Wo = RT, In— = —RT, In

(7.15)

The net work is Wy + We: Pu Wie = Wr + Wo = R (Ty — T,) In pe

(7.16)

1

and we can find the efficiency as:

SC

n=

On

Roy =

i

syd, ln Py

eae a i-— = T,

T,

(7.17)

ee the same as for a Carnot cycle operating between the same two temperatures.

7.3.2

Stirling Cycle The Stirling cycle, Fig. 7.15, is similar to the Ericsson, except that the regenerative heat exchange is accomplished at constant volume instead of constant pressure. The analysis is similar to that for the Ericsson cycle; the efficiency is identical, given by eq. (7.17). Both these cycles have the advantage of using external combustion, so that the nature of the fuel is not important; the burning occurs continuously. But both suffer from the need for complex equipment to accomplish the regenerative heat exchange, so that neither has found broad application.

192

Chapter 7 Power Production

Figure 7.15 Ideal Stirling cycle.

Operating in reverse, the Stirling cycle is used for refrigeration. We will discuss this in Chapter 10.

7.4

FUEL CELLS AND BATTERIES The inherent second law limitations of the Rankine and Brayton cycles have led to the quest for ways to bypass the conversion to heat and obtain electric energy directly from chemical energy. The most successful have been fuel cells and batteries. The principle of operation in both these devices is the splitting of a chemical reaction into two parts. Instead of allowing the reactants to come into direct contact with each other, as in the

combustion of fuel with air, they are separated by a material that transfers ions but not electrons. The electrons involved in the reaction are then transferred through an external circuit, where electrical work is extracted from them. In fuel cells, the reactants are

supplied from external sources; in batteries, they are contained within the device. We will treat them separately.

7.4

7.4.1

Fuel Cells and Batteries

193

Fuel Cells In the standard burning of a fuel to obtain heat—let’s say hydrogen with oxygen to water:

2H, + O, = 2H,0

(7.18)

What happens at a molecular level is the transfer of two electrons from the hydrogen into a molecular bond with the oxygen. The energy of the electrons in this molecular bond is lower than in the hydrogen by 243. kJ/mol. In the transfer, this energy is given off as heat. A fuel cell splits the transfer so that the electrons are removed from the hydrogen at a site remote from the oxygen. The electrons, stripped from the hydrogen at the higher energy level, equivalent to about 1.2 V more negative than in the water molecule, can perform electrical work on their path to the oxygen. The reaction is completed through ionic transport via a wetted mat. The process may be more clear in Fig. 7.16. In this scheme the ions are hydroxide, OH, generally from sodium or potassium hydroxide dissolved in water. This solution, which conducts the reacting ions between the electrodes, is called the electrolyte. The mat can be as simple as a few sheets of filter paper supported between two electrodes, electronic conductors that allow intimate contact between gas and liquid; nickel screen is often used. It is at the three-phase interface of screen—liquid—gas that the electrochemical reactions occur. When the two half-cell reactions, those occurring on the top and the bottom electrode, are added the overall reaction (7.18) is obtained.

The electrical work is: W. = JE At

(7.19)

when E£ is the voltage. The current, J, is given by:

I = n¥ nh (Coulombs/mol)

(mol/s)

(7.20)

where n is the moles of electrons in the reaction per mole of reactant, n is the molar

flow rate of reactant, and ¥ is Faraday’s constant: 96,500 C/mol of electrons. For example, if in the above fuel cell we have 1 g - mol/s of hydrogen reacting, the current will be:

I = (4/2) X 96500 X 1 =

O> emp

H> eee

0 + 2H + 4e7-~ ———>

2H + 40H)

>

Figure 7.16 Aqueous alkaline fuel cell.

193,000 C/s =

4OH-

4H0 + Fe

193,000 Amperes (amps, A)

194

Chapter 7 Power Production

The theoretical voltage is found from thermodynamics; because this reaction occurs at constant temperature and pressure, the equilibrium voltage is determined from the molar standard Gibbs free energy of reaction, Agg, which is the stoichiometric sum of the molar standard Gibbs free energies of formation, g$, of the 7 products and reactants, all

at 298 K:

(7.21)

Age = dv8%

where v is the stoichiometric coefficient. We take the reaction (7.18), noting that v of the product, water, is +2, whereas those of the reactants are

— 2 (hydrogen) and — 1

(oxygen). The molar standard free energy of water is found from the Handbook of Chemistry and Physics* to be — 228.6 kJ/mol; those of the elements are zero. The calculation is, for the reaction as written in eq. (7.18),

Age = 2 X< (~228:6,k)/mol),—

2.

(0-— Ii x. 0 =

457.2 k)

and, recalling that, at constant temperature and pressure, the negative of the change in Gibbs free energy is equal to the maximum work, eq. (5.4b), W max = we can combine voltage**:

S/N)

= ji S< Jzmax >< ING

this result with eq. (7.20) to arrive at the theoretical maximum

Emax = —A8p/(nF)

(7.22)*

which, in this case, with n = 4, is 1.184 V4 We use the symbol, Agr, instead of Agp in eq. (7.22) for the more general case where we are not necessarily at standard conditions; the difference between these conditions is taken up in Chapter 15. The maximum power of this fuel cell is then:

P =

1.184 V X 193,000

A = 228.5 kW

The energy balance for a fuel cell operating at steady state, as shown in Fig. 7.17, is written over an accounting period equal to the time for 1 mol of fuel to flow through the system:

*Some Standard Free Energies of Formation are listed in Appendix IV. See Handbook of Chemistry and

Physics, op. cit., for more complete listing. **We could have written the reaction: H, + 30, = H,0O, and the free energy of reaction would be one half that calculated here. But then the number of electrons transferred would also be halved, so that the equilibrium voltage is unchanged.

‘The lower case ‘‘g”’ signifies the molar property.

*We have taken the product water to be a gas; if it is a liquid, the voltage becomes 1.23 V.

7.4

fuel

exhaust

att

depleted air

Fuel Cells and Batteries

195

Figure 7.17 Fuel cell schematic.

Where Qp is the heat transferred to the surrounding, which is at Tp and the entropy balance, over the same accounting period:

oO

ll

ASE

=) soit ee gS

(7.24)

D

Combining egs. (7.23) and (7.24), the heat loss to the surroundings is:

Qp = Tp As — TpSp

(7.25a)

Opin = Tp As

(7.25b)

which, at a minimum,”* is;

For the present example,

MP

=

g —Ag°

h [. 228.6 + Ah° 6 — 1h

241.8t

giving the maximum work (Sp = 0), with An =

W, max =

—Ah

+ Tp As =

Lada AEE

298

mol - K

1.0 moles in the accounting period:

—Ag = E,,,y/ At

= Eat

n At = E,,,nk

(7.26a)

where E is voltage, and n = 1. The actual work is:

W =

—Ah + Tp As — TpSp = EnF

(7.26b)

The difference in these works is just the lost work resulting from entropy production:

W, max

— W = TySp = (E — Emaxn¥

(7.27)

*We will see in Chapter 15 that As on oxidation for most fuels is negative, so that the heat flow is also negative (out of the system).

196

Chapter 7 Power Production showing up as an addition to the heat loss:

Op =ily AS

(Eee ee

(7.28)

The difference between actual and maximum cell voltage, (E — E,,,,), 1s always negative; it is because of irreversibilities at the electrodes and resistive losses throughout. The actual voltage achieved is usually about 0.8 V at a flux of 0.2 A/ cm?. Our example fuel cell would require 143 m? of cell area arranged in stacks of repeating elements,

each about | m7”. This type of fuel cell is used in manned spacecraft; similar types have been employed in urban generating stations, using ‘‘reformed’’ natural gas to supply the hydrogen:

Clie

0

On=_ CO,

4H,

(7.29)

and employing an acid-based electrolyte in place of the hydroxide. When the power output of this fuel cell is compared to that obtained in a Rankinecycle plant, it is found to be considerably more efficient at producing electricity from fuel, even including the irreversibilities. The capital costs, however, are still somewhat

higher than that necessary to make widespread application economically feasible. Newer versions, still under development, operate at higher temperatures and show promise of even higher efficiencies. 7.4.2

Batteries In a battery, a substance containing electrons with relatively high energy (such as the fuel in a fuel cell) comprises the anode electrode; here electrons are removed

and

allowed to flow through a work-extracting device (e.g., a light bulb). At the cathode the electrons are accepted by a material that can accept the lower energy electrons. The generic reactions are shown in Fig. 7.18. The material comprising the anode, ‘‘red’’ (reduced), gives up an electron, and in so doing is ‘“‘oxidized’’ to material ‘‘ox.’’ At the cathode, electrons are accepted, reducing the ‘‘ox’’’ material to ‘‘red’.’’ The circuit

is completed by ions; cations, +, moving from right to left, and anions, —, moving left to right. An example will help clarify the process.

Ox'+e — > red’

Figure 7.18 A simple battery.

red

—

ox +e—

7.4

Fuel Cells and Batteries

197

In the Daniell cell, historically the first working battery, the anode is made of zinc (Zn) and the cathode of copper (Cu). The electrolyte is an aqueous solution of ZnSO, and CuSO,. Nothing happens until the circuit is completed externally, either through a load, or simply short-circuiting the two metals. In the latter case, no work is obtained and the process is called corrosion. In either case the reactions occurring are:

Zn — Zn*+

+ 2e7

(7.30)

Cu2*+ + 2e7 > Cu

(7.31)

at the anode, and:

at the cathode. The maximum voltage available is found from the Gibbs free energy for the entire process:

Zn + CuSO, > Cu + ZnSO,

(7.32)

or, alternatively, from a table of standard reduction potentials,* as shown in Table 7.1. These are the so-called half-cell reactions which, when combined with a standard, or “‘reference,’’ half-cell oxidation reaction, gives a full reaction from which a Gibbs free

energy can be calculated. The most common reference half-reaction is the reduction of protons to hydrogen. For example, our reactions in the Daniell cell above, eqs. (7.30) and (7.31), when combined with the reference are, first at the anode:

Zn + SO%7 — Zn** + SOZ + 2e7 2H* + SO%” + 2e° > H, + SOZ7

Zn + H,SO, = ZnSO, + H, The first two equations are added to obtain the third, with the sulfate at the left in the first cancelling with that on the right in the second. From this, Agr for the third can be found and the potential calculated from eq. (7.22), yielding E = —0.7628 V for the reduction of Zn** [the reverse of eq. (7.30)], as shown in Table 7.1. The negative sign means that zinc ions will not spontaneously be reduced in acid solution; Agr is positive. At the cathode,

Cie SO? 4 22 = Cu + SO2Poe sSO7me— 1SO, + 2eCuSO, + H, = Cu + H,SO, the same process yields

E=

+0.34 V for the reduction of Cu?* vs. hydrogen.

*Sometimes called ‘‘electromotive force.’’ This is only a small sample of the half-cell reduction potentials available; see, for example, the Handbook of Chemistry and Physics.

198

Chapter 7 Power Production Table 7.1

Standard Reduction Potentials

a

Potential, V

Reaction

Ag” +e= AgCl + e

Ag = Ag + Cl™

Ag,O + H,O + 2e- = 2Ag + 2(0H)>

+ 0.7996 + 0.345

+0.344

Av sixes =sAu Bry + ems =" Bae

+ 1.68 + 1.065

Cdittewer a= Cd Cd(OH), + 26> = Cd + 20H=

— 0.4026 — 0.81

Ce?+ + 3e° = Ce

= 21335

ie = Com

Cele

+ 1.4430

CL,.+ 2e- =2ci-

+ 1.3583

Cut +e" = Cu Cu tet Demon

+0.522 +0.34

Cu(OH), + 2ee2=- Cin Fe?+ + 2e7 = Fe

Fe?+ +e”

+0.770 + OH™

= Fe(OH),

ogee 22s

2H 3HOY

— 0.224 — 0.409

= Fe?*

+ e

Fe(OH),

20OR-

—().56

0.0000 ~ 0.8277

On

Hg*>+ e~ = Hg

+ 0.7961

lit

— 3.045

by “ees

+0.535

aoe

er=ii

Ms(OH), + 2e~ = Mg + 20H~ 2MnO, + H,O + 2e7 = Mn,O, + 20HNi(OH), + 2e” = Ni + 2OH™ NiOOH + H,O + e- = Ni(OH),

Ont 2H

Zen

+ OH™

1,0;

=9.67 +0.20

— 0.66 + 0.49

+ 0.682

O, + 4H+ + 4e- = 2H,0 PbO, + 4H* + 2e- = Pb?+ + 2H,0 PbO} + 2H*!4HjSO/h2e* SPbSO; 4 DEO PbSO, + 2H 44 9¢> = Phe HSO)

+ 1.229 +1.46 + 1.685 ~ 0.3505

Ti

— 0.7628

43 en

Zn(OH),

+ Ze"

-="Zn

20H

= DAS

The total cell voltage is found by adding the two half-cells, the oxidation of zinc and the reduction of copper, to get:

EF. = +0.76 + 0.34 = +1.10V The same result is obtained from the free energy of reaction of eq. (7.32). The theoretical maximum energy obtainable from the battery is calculated from the maximum cell voltage and total number of electrons available for transfer from anode to cathode:

7.4

W, max SpE

a

et Ati= Ee,

Fuel Cells and Batteries

AC

199

(7.33)

where AC is the total charge transferrable, in Coulombs (A - s);

AC = neAn

. Fy is the force of gas on blades.

described in Chapter 7; in automobile engine application, however, there is no heat recovery. The advantage of smooth operation is negated by the low efficiency and complexity in converting the high speed of the turbine to the variable-speed drive wheels. The action of the gas on the turbine blades is illustrated in Fig. 9.13(a). As the gas expands through the turbine, from left to right, the momentum change drives the turbine blades upward, as shown. A simplified engine design is shown as Fig. 9.13(b). Wankel Engine

The Wankel engine, Fig. 9.14, is the only rotary engine to succeed in automobile applications. The shape of the combustion chamber is complex, allowing the planetary

238

Chapter 9 Motive Power

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d(— (m dP+{(—) (my + (Sy >

>

dn;

(11.60)

NA;

or, using eq. (11.59):

M a: dM = ( |—) |— oe dp + Gi aT + 2M,hs dn,

(11.61)

which should be compared to eq. (11.13) for a pure component. The difference is seen

*Sometimes referred to as ‘‘partial molal.’’

11.3.

Partial Molar Properties

303

to be that we now have a sum of the partial molar properties, rather than the pure molar property as the third term.

11.3.2

The Gibbs-Duhem Equation We will now show two important results: (1) the Gibbs-Duhem equation, which is a

criterion for thermodynamic consistency to be used later in discussion of phase equilibria, and (2) that eq. (11.61) integrates to:

M = >Mn,

(11.62)

Take, for example, the internal energy, U:

dU = TdS — PdV + >, dn,

(11.36)

Imagine we have a large container of a complex mixture such as gasoline, on the left in Fig. 11.2, along with a much smaller container, on the right, holding a portion of the exact same mixture, all at constant temperature and pressure. Now, at time zero, we

toss the contents of the small container in with the large. If the internal energy of the contents of the large container initially is U, and that of the small AU, after they are mixed:

U + AU = kU where k is a number slightly larger than 1, say 1.01.

AU = kU —-U=(k

—

1)U

(11.63a)

and AU is different from U only because of the difference in the extensive properties of the two subsystems, that is, their size.

AU = TAS — PAV + d\n, An,

U+AU=kU

AU

Figure 11.2 Mixing of portions of identical mixtures.

(11.63b)

304

Chapter 11 Phase Equilibrium—Fundamentals The important fact here is that the intensive variables, T, P, and w, are unchanged in

the mixing process. This is only true because the two subsystems have identical chemical compositions and are at the same T and P. If we now substitute eq. (11.63a) into eq. (11.63b):

(eS

RUrSrT

and the factor (k —

eae Sy

See eri

Sp,

k —

1)n;

(11.63c)

1) can be divided out, leaving:

U = TS — PV + Spin,

(11.63d)

From the definition of the enthalpy, H, we get,

H=U

+ PV=TS

+ > pn;

(11.63e)

and for the Helmholtz free energy,

A=U

— TS = —PV + S\pyn,

(11.63f)

and, most important, the Gibbs free energy:

G=H — TS = dpjn,

(11.64a)

which, on a molar basis (dividing through by the total number of moles, 7) is:

f=

ye Gy,

(11.64b)

where y, is the mole fraction of component i, n,/n, and G, is the partial molar Gibbs free energy of component i. The Gibbs free energy in the form:

G = DGn,

(11.65)

dG = >'G; dn, + Din, dG, = Diu, dn, + Din; dy,

(11.66)

is differentiated to:

which, on comparison with eq. (11.43), shows that

—SdT + VdP — din, dp, = 0

(11.67a)

this, at constant temperature and pressure becomes

Dn; du; = 0

(11.67b)

an important result [in either form, eq. (11.67a) or eq. (11.67b)], known as the Gibbs-

11.4

Ideal Mixtures

305

Duhem equation. Now for the second result, we use as an example the total volume, V, and take its total differential, dV:

av av aVv ay = (2) aT + (=) dP + >(2) P,n

oP T.n

dn,

an; T.P.nj4;

At constant temperature and pressure, just as we showed for the internal energy above, an incremental increase in the size of the system results in a change in total volume, AV: AVitp = Dy, An;

where V, is the partial molar volume of component i. For a (k —

1) increase:

(k — 1)V = DVKk — In, from which the factor (k —

1) can be divided out, leaving:

V => Vi

(11.68)

or, on a molar basis (dividing through by total moles, 7),

v = DV,

(11.69)

The analogous results are found for H, A, and G, as given in the general case by eq.

(11.62); for the last, however, the chemical potential is also equal to the partial molar

property.

Ma Gea

dG (=) on;L

(11.70)

TP

so it is arrived at either in this manner or as shown in eq.(11.64a) above. This is not the case for any of the other partial molar properties (U;, H,, etc.).

11.4

IDEAL MIXTURES There are two classes of extensive properties: those that are functions of the entropy (S, A, G), and those that are not (V, U, H). For the latter class, if the mixture is ideal,

the partial molar properties are equal to the pure molar properties; the mixture volume, for example, is just the mole fraction—weighted sum of the pure molar volumes, eq. Clio): This is not the case with the free energies, G or A, or the entropy, S, even for ideal

mixtures. The Gibbs free energy is:

G=

DP: C= Dn( + Yen]

;

306

Chapter 11 Phase Equilibrium—Fundamentals which, in terms of the fugacities, is:

G = dwin, + RTDn, In f,

(11.71)

i

and for an ideal mixture, represented by G’, the fugacities are equal to the partial pressures, so:

G' = Dmin, + RTD/n,; In p;

(11.72)

Now imagine a process wherein a number of ideal gases all at temperature, 7, and initial partial pressures, p,,, are mixed. The total free energy before mixing is the sum of the component free energies:

G = > win, + RTD nin p,

(11.73)

Thus the change on mixing, AG™, of these ideal gases, eq. (11.73) subtracted from CUE Z) As:

AG™ = RT>‘n, In (2)

(11.74)

Pio

In the simplest case, imagine a mixing process where each pure gas is at the same initial pressure, p;,; the final pressure will then be that same pressure,

Pep.

(11.75)

Dp; = y;P

(11.76)

But the final partial pressures will be

where y; is n;/n. So eq. (11.74) becomes

AG™ = RTD, In (y,)

(11.77)

and is always negative. It should be clear that this must be so. Remember, a system will always spontaneously find a minimum in free energy; gases will tend to mix, not unmix. The entropy change on mixing is found from the definition of G:

G=H

— TS

sO,

AGM = AHM

—

T ASM

(11.78)

11.4

Ideal Mixtures

307

and, for an ideal gas mixture there is no change in enthalpy (or internal energy or volume) on mixing, for the enthalpy of an ideal gas, or mixture, is dependent only on temperature. This means the ideal entropy change on mixing at constant T and P is:

AG™ SS oe ce

—RD'n; In y,

(11.79)

and is always positive. For nonconstant T and P, we go back to eqs. (11.74) and (11.78).

An entropy balance on this process, including all gases as the system, is: Sz =

Sp + $, —

$0

+

O/T,

+ Sp

and since the only difference in the system between beginning and end is the fact that the gases go from unmixed to mixed.

and there is always a positive entropy production, as expected. We can now see why, even for an ideal mixture, the entropy-dependent partial molar properties (S, A, G) are different from the pure molar properties (s, a, g): it is the effect of the ideal entropy of mixing.*

ee Find the ideal change in G and S when 3 mol of Argon (1), initially at 1 atm, 298 K, are allowed to mix with 2 mol of Xenon (2) at the same initial conditions.

SOLUTION The total moles are 3 + 2 = 5, so that,

AG’

3 2 = ra In 5 He ein 2|= 2478[—3.365]

=

—8339. J

and

AS = +28. J/K = Sp For the ideal gas mixture,

ie

RT inp = ww; +.RT InP + RT In y,

(11.80)

where 1; is the chemical potential of component i at 1 atm and temperature, 7, in concert with eq. (11.47).

*For a statistical interpretation of the ideal entropy of mixing, see K. Denbigh, The Principles of Chemical Equilibrium, 3rd ed., London: Cambridge University Press, 1971, pp. 48-56.

308

11.5 11.5.1

Chapter 11 Phase Equilibrium—Fundamentals

REAL GAS MIXTURES Fundamentals In real gas mixtures, where chemical effects must be accounted for, the partial molar properties must be used, just as we have used the molar properties to find changes in pure component fugacities. Expanded forms of the differential properties, dG, dH, dU,

and dA lead to new Maxwell’s equations. For example, we can use eq. (11.14) to get:

EDaG

a BN

gas

aT

OP ys

act

(11.81)

and then,

O2G\

apaT/,

tay

(OV.

\aTh,

(11.82)

so,

(11.83a) and likewise,

(11.83b)

(11.83c)

(11.83d) if we do these same operations not holding n constant; so,

(Hr) ss

=

a

-S

ae

oT Pn

=.

on eS

bes

(11.84a)

(11.84b)

on TP

as

(11.84c)

on TP

and a similar partial differentiation with respect to P and n leads to:

(2 _ (¥ oP Tn

On }yp

The RHS of eq. (11.84d) for the pure substance is the molar volume:

(11.84d)

11.5

OV (2)

Real Gas Mixtures

=U

on TP

309

(11.85)

and for the component in a mixture it is the partial molar volume: OV () On; TP

= J;

=

(11.86)

which, for a component in an ideal mixture, becomes equal to the pure molar volume, as in eq. (11.85).

11.5.2

Temperature and Pressure Dependence: General In calculating phase equilibria, it is usually the case that fugacities (or chemical potentials) for the components in question are not available at the temperature and pressure in question. Thermodynamics has, as one of its greatest strengths, the structure permitting reasonable extrapolation from conditions for which data are available. Here we develop the exact relations to be used in this type of extrapolation. Phases in equilibrium must be at the same temperature and pressure. So the fundamental equilibrium equation, eqs. (11.31), can be written: a

el LE TT

B

(11.87)

and any change in 7, P, or x that results in a new equilibrium state must affect both

sides of eq. (11.87) equally: ee

YAN Sai OX

The

imo

|ae if

(11.88)

From the definition of fugacity, eq. (11.44), Bes=ap

se shRT In fF

(11.89)

noting that the fugacity must be expressed in atm (or bar) in order to have the proper units. And, by definition,

i, =a)

+ FP, x)

So, when we take the total differential of both sides of eq. (11.87),

op?

dy,

(11.90)

310

Chapter 11 Phase Equilibrium—Fundamentals

where mole fraction is expressed as x in the alpha phase and as y in the beta phase. The pure component chemical potential, 5, is the same for both phases; its derivatives with T therefore cancel. From the Maxwell equations, eqs. (11.82) through (11.85),

Ee oT

ek We NEE P,n

eee

(11.91)

On; T.P,n;

en

V,

(11.92)

One and also,

Lb; =

= dG =G; (*) Ni/T.P.n;

(11.93)

Recall the definition of the enthalpy, H:

H=U+

ec

PV

so that, on differentiating with respect to n,, 0H

0U

ov

WM: )TPn,

WMT Pn,

Oni) t Pn,

this, from the definition of partial molar properties, is the same as:

H, =U. + Py,

(11.95)

This same relationship between pure molar and partial molar properties is universal; so, for example, for the Gibbs free energy,

Go

See

(11.96)

and substituting eq. (11.91) above for the partial molar entropy,

bi;.= H.; + T|— ( oT is

: (11.97)

This is the same as:

——|

=

:

(11.98)

11.5

Real Gas Mixtures

311

[Carry out the partial differentiation in eq. (11.98) to see that you, indeed, get back to eq. (11.97).]

The analog of eq. (11.98) for the pure component is:

=A len = 2

aT

(11.99)

Substituting eqs. (11.91) through (11.98) into eq. (11.90) we get:

— H*

Tp? OP

Ve

T dE

R ain fe

iby dx;

— HP

ve

R a In f?

7 ee7

~

dy, dy,

(11.100) :

This result can be applied to a common situation: a pure gas in equilibrium with the same gas in solution in a solvent (e.g., CO, in water). For the pure gas phase (y =

1.0), the RHS of eq. (11.100) is:

RHS

Sen

als 7

Deta7 dP

: (11.101)

and if the gas is presumed ideal,

v; = RT/P Therefore, —he

RHS

=

—

7 ar s+ K din P

(11.102)

and eq. (11.100) becomes

— H°

T?

ve

ENE SS eat | OS

ip

R dln fe

ere 1

nT i

—hP

“dT +RdinP_

Dy

Ay(11.10 ae?

This states, in effect, that the term involving the solubility of i in phase a (term 3) is dependent on temperature through the difference in enthalpy of component i in the dissolved phase versus the gas phase (terms 1 and 4); and it is dependent on pressure through the partial molar volume in the dissolved phase versus the ideal gas (terms 2 and 5). This result will be seen to be useful in determining the effects of temperature and pressure on phase equilibria in real mixtures, which we will treat in Chapters 12 and 13.

312

11.6 11.6.1

Chapter 11 Phase Equilibrium—Fundamentals

IDEAL LIQUID MIXTURES Fundamentals

The simplest liquid mixture is one in which the fugacity of each component is equal to its mole fraction—weighted pure fugacity:*

fi = £3;

(11.104)

with f, the fugacity of pure liquid i at the T and P of the solution. At low pressures,

(ee use

(11.105)

In either case, the chemical potential in an ideal solution is:

pe =

+ RT in x,

(11.106)

If, in addition, the vapor is an ideal gas mixture,

(=p

aay

ee

(11.107)

This equation is known as Raoult’s law when applied to all components in the mixture.

Find the vapor composition above a liquid mixture of 50 mol% benzene (1)—hexane (2) at 25°C if it is assumed that both the liquid and the vapor form ideal solutions. The vapor pressures of benzene and hexane at 25°C are 93.9 and 149.2 mmHg, respectively.

SOLUTION The vapor mole fractions are determined by rearranging eq. (11.107) for each component, 4 aes

0.5 xX 93.9 P

2s 0.5 x 149.2 Voy = P

and, the total pressure is the sum of the partial pressures:

P = y,P + y.P = x,P, 4xP;-=11215 mmHg — 0459 ann and we calculate the vapor mole fractions as:

y, = 0.386

y> = 0.614

*This is the ‘“‘Lewis and Randall fugacity rule,’’ named for G.N. Lewis and M. Randall, who introduced it. See Chapters 12 and 13.

11.6

11.6.2

Ideal Liquid Mixtures

313

Phase Diagrams The entire phase diagram for a binary, ideal solution is calculated as in Example 11.3, using a number of liquid mole fractions. The result will look like Fig. 11.3. We can complete the benzene—hexane diagram at 25°C in this manner. The point, P;, will be at 93.9 mmHg and P3 will be at 149.2 mmHg. The equilibrium y for a given x is shown by any horizontal line. If the pressure is slowly lowered on a mixture of overall composition z in the figure, in this case a subcooled liquid, it will first show a bubble of vapor when the pressure reaches the upper curve, the ‘‘bubble point.’’ This bubble will have the composition y shown by the point ‘‘bp’’ on the diagram. If the pressure is lowered further, at any point the equilibrium compositions are given by the intersections with the phase boundaries of a straight horizontal line at that pressure. Finally, when the pressure reaches the lower boundary, we are at the ‘‘dew point.’’ The vapor now has composition z and there is left only a drop of liquid, with composition at ‘‘dp.”’ An analogous situation is arrived at for a system at constant pressure. Here the partial pressures must always sum to this total pressure:

Pe= pyr py =v

Pk yP = xP) + %P,

(11.108)

Of course, the temperature must vary across the range of mole fractions. As an example,

take the system: n-pentane + n-heptane, which is nearly ideal. At a total pressure 101.32 kPa, we can draw the vapor—liquid equilibrium (VLE) phase diagram once know the vapor pressures of the pure components. These are available in the form Antoine equations with the constants tabulated in many sources. A partial list can found as Appendix III.

Pressure

Figure 11.3 Typical vapor-liquid equilibrium (VLE) diagram (isothermal).

of we of be

314

Chapter 11 Phase Equilibrium—Fundamentals

NW

;

B

aier G,

B In PS = A, —- Tia 6,

(11.109a)

(11.109b)

For n-pentane (1) and n-heptane (2), these are:

A, B, C, A, By Cy

= = = = = =

13.8183 2477.07 233.21 13.8587 2911.32 216.64

with the temperature, 7’ in °C. First, we need to set the temperature limits; for an ideal system, all solution temperatures at constant pressure will lie between the pure component boiling points at this pressure. We solve eqs. (11.109) for each of these:

T;

36.04°C

T;

98.42°C

Now we set the temperature at any arbitrary value between 36 and 98°C and solve eq. (11.108) for x, (with x, = 1 — x,). Then we solve eq. (11.110) for y:

ye= LLP.

(11.110)

In this way we get the following values at 40, 70, and 80°C: y ipa

40 70 80

©

xy

yy

0.860 0.250 0.142

0.983 0.700 OSig

A graph similar to Fig. 11.4 is obtained in this way. In this figure, a liquid of overall composition z is heated until it reaches the lower phase boundary; at this point a bubble of vapor first appears (bp). Continued heating vaporizes more and more of the mixture; at any point in the two-phase region the equilibrium compositions lie on the horizontal. Finally, at the top boundary, there is only a drop of liquid left; this is the dew point (dp). The same compositions will be found if a vapor is cooled, originating above the two-phase region. The amounts in the two phases are found from the mass balance (level rule).

11.6

Ideal Liquid Mixtures

315

Temperature

0.0 x,y

Figure 11.4 Isobaric VLE phase diagram (ideal).

Find the compositions and fractions of the total moles in the equilibrium phases for overall composition: z, = 0.45 (n-pentane) with n-heptane (2), at P = 101.32 kPa, when the temperature is 70°C.

SOLUTION At this temperature, we have already found the phase compositions:

x, = 0.250

y, = 0.700

A mass balance (in moles) over both phases is:

EV

i= £0

(11.111)

where L and V are the moles of liquid and vapor, respectively, at equilibrium, and, on each component,

xL + yV = z, 1.0

(11.112)

Solving eqs. (11.111) and (11.112) for L and V, we get, with the values of x, y, and z above,

L = 0.556

V = 0.444

both in mole fractions. a

ttt

316

11.6.3

Chapter 11 Phase Equilibrium—Fundamentals

Multicomponent Flash Calculations A multicomponent liquid mixture at high pressure can be brought to a low pressure by simply flowing it through a valve. In this manner an isenthalpic expansion will cause a separation into a liquid and a vapor phase. These will be in equilibrium, so their compositions will be different. This separation can be useful in obtaining streams highly different in volatility, for example, in petroleum refining. It is known

as a ‘‘flash’’

because of the instantaneous partial vaporization that occurs just downstream of the valve. When the components are chemically similar, they behave according to Raoult’s law, eq. (11.107).

i

ey ee

(11.113)

Ke,

where K; is known as the equilibrium constant. In some systems, such as mixtures of hydrocarbons, this constant will be known for each component as a function of tem-

perature and total pressure, even though it will not be precisely equal to the ratio of vapor pressure to total pressure. A calculation of the equilibrium phases at a known temperature and pressure then becomes a matter of solving a set of equations for, say, all the liquid mole fractions and the fraction that is vapor, V. Each z; is:

A

1s oie VeVi

an

laa Vi) toy,

V.

(11.114)

eka)

(11.115)

which can be rewritten

V) EE KV

4, = (Le

IE ae

or, in terms of x:

=! Gi x, = id —W) — Vv)++ KY] KV]

(11.116)

and all x’s must sum to 1.0, so:

Di OF amen rare

1.0

(11.117)

If all the z’s and K’s are known, we guess at V until eq. (11.117) is satisfied. Alternatively, a hand calculator can be used to find the root; many have this feature prepro-

grammed. Eq. (11.116) is then used to find each x, and eq. (11.113) for each y. A similar solution exists for finding bubble points and dew points in these mixtures. In Fig. 11.3, at the bubble point pressure the liquid has the same composition as the overall mixture:

4, = &i

(11.118)

11.6

Ideal Liquid Mixtures

317

and, if eq. (11.107) is valid, each y; is equal to x,P;/P; so*

P°

Dy; = 1.0 = Saag

(11.119)

where P is now the bubble-point pressure, P,,,,:

Pyur = xP;

(11.120)

At the dew point, y, = z; and YP

Jes So,

Dx Se= 10 Se = > ee PRTpS

(11.121)

L

where P is now the dew point pressure, P4.,,:

(11.122)

11.6.4

Free Energy of Mixing An ideal liquid solution is defined as one in which the component chemical potentials are given by eq. (11.106):

ri

aes a «Ya

Maa

(11.106)

The total Gibbs free energy of the ideal solution is then:

G = din, = Dina; + RTDIn; In x,

(11.123)

where the first term on the RHS is the contribution of the pure liquids:

DG, = Dw;

(11.124)

the second is then the change in free energy on mixing:

AG™ = RT)7, Inx,

*Otherwise, K; is substituted for P;/P.

(11.125)

318

Chapter 11 Phase Equilibrium—Fundamentals and for all ideal solutions, there is no enthalpy change on mixing,

AA

=")

AG ©

= —R>n, In x;

SO,

AS™ = ——— je

(11.126)

This is in complete analogy with the same relation for ideal gas mixtures, eq. (11.79).

11.6.5

Ideal Gas Solubility Gas solubility in a liquid solvent is determined by equating vapor and liquid fugacities of the species. If an ideal gas forms an ideal solution, these fugacities are: {pak

= aP;

(11.127)

Notice that the solubility in this ideal situation is independent of the properties of the solvent. This, of course, will rarely yield accurate values since there will always be nonideal effects in the liquids; we will see how to deal with those later. As an example of the correspondence between actual and ideal solubility, Table 11.1 shows the actual mole fraction of CO, in some varied solvents at 20°C, 1.0 atm.*

The vapor pressure of CO, at 20°C is 56.3 atm; its ideal solubility is therefore:

1.0/56.3 = 0.0178 mol CO,/mol solvent Table 11.1

CO, Solubility at 20°C, 1 atm

Solvent

Mole fraction CO,

Carbon tetrachloride

0.0100

Pyridine

0.0129

Toluene Chloroform Amy] acetate

0.0107 0.0123 0.0270

The ideal solubility of 0.0178 is seen to be good for at least order of magnitude estimates; in the absence of any data it is quite useful. When the solution is at a temperature above the critical temperature of the solute gas, a hypothetical vapor pressure can be estimated from the slope of the vapor pressure curve through the critical point. This type of calculation is useful only as a very approximate estimate of the gas solubility; more accurate methods, using a modicum of actual data, are described in the next chapter.

*From K. Denbigh, Principles of Chemical Equilibrium, London: Cambridge University Press, 1971.

11.6

11.6.6

Ideal Liquid Mixtures

319

Colligative Properties Properties of dilute solutions that depend only on the number of solute species present and not their chemical nature are called ‘‘colligative’’ properties, from the Latin colligatus, meaning ‘‘collected together.’’ A relatively simple treatment is used here for ideal solutions; in the next chapter we look briefly at the effect of nonideality. Ideal Solid Solubility Picture a crystalline solid in equilibrium with the same dissolved component in solution; when there is solid present at equilibrium the solution must be saturated. At this point, the chemical potential of the pure solid is equal to that of the component in solution, and a new term, activity, is introduced:*

a, eRe = 4 =

he

so that,

Pi, = Bi, = Bb, + RT nts =p, + RTIna, = wp, + RTInx,

(11.128)

1

where \1,, refers to the pure crystalline solid, .,, to the solute in solution, and 1}, to the solute as a pure liquid. In keeping with the ideal solution, the activity of the solute, a,, is equal to its mole fraction, x,. Because this is the mole fraction at saturation at this particular T and P, it is called the solubility. Rearranging eq. (11.128):

Inx, =

vil RT.

RT

(11.129)

The temperature dependence of the solubility is far more important than the pressure dependence; as we have seen, the pressure dependence on condensed phases is very slight. To determine the effect of temperature on solubility, we first differentiate eq. (11.129) at constant P: (? In 1) ‘ i A(w1,/T) 2s Au /T) oT ke trol: RK. of. Pp

(11.130)

The two terms on the RHS are found from eq. (11.99) to be the negative of the respective enthalpies divided by the temperature squared, so (11.130) becomes:

Galas

aaa)

iesereiftacuat, ft

RT?

(11.131)

*The effect of pressure on the condensed phase is negligible, so that the pure component fugacity at any pressure is considered the same as at the standard state, 1 bar.

320

Chapter 11 Phase Equilibrium—Fundamentals

The numerator on the RHS is simply the enthalpy of melting of the pure solute, Ah,,,, so,

Me 8 Sia)

Ah of eed al aR

(11.132a)

The ideal solubility is then independent of the properties of the solvent, and it always increases with temperature, as the enthalpy of melting is always positive. Departure from this simple equation is due to nonideal effects, that is, interactions between solute and solvent molecules. The ideal solubility at temperature T can be estimated if the latent heat of melting is assumed constant in the range from the melting point of the pure solute to T, by integration of eq. (11.132a):

Inxy; =

PTE

Ree

Ie

=| 1

Tes

Ree

A

Uy

(11.132b)

m

This assumption, in addition to the assumption of ideal solution behavior, severely limits the accuracy of this prediction to temperatures near the melting point of the solute and in solvents of similar chemical nature to the solute. An example of relatively good prediction is that of naphthalene, with a melting point of 80°C, in organic solvents at 20°C.* The latent heat of melting is 18.58 kJ/mol; the solubility at 20°C, from eq. (115132b),is:

Xoge¢ = 0.273 and is independent of solvent. The actual solubility of naphthalene in some nonpolar solvents are listed below: Mole fraction

Solvent Hexane Chlorobenzene Carbon tetrachloride Benzene Toluene

naphthalene at saturation, 20°C 0.090 0.256 0.205 0.241 0.224

Showing that the ideal solubility provides a reasonable estimate when solute and solvent are chemically similar. Freezing-Point Depression

Salt (NaCl) is commonly sprinkled on roads and sidewalks covered with ice. The desired effect is the lowering of the melting point such that at temperatures not far below 0°C,

*J. H. Hildebrand and R. L. Scott, The Solubility of Nonelectrolytes, 3rd ed., New York: Reinhold, 1950.

11.6

Ideal Liquid Mixtures

321

a mixture of salt solution and ice is formed, a ‘‘slush’’ that is more amenable to vehic-

ular and foot traffic than solid ice. The effect, qualitatively, of the dissolved salt (the solute) is to lower the activity of the water (the solvent) in the liquid phase primarily by lowering its mole fraction, x, from unity to some slightly lower value. This lower liquid activity is, at equilibrium, equal to the pure solid (ice) activity at a temperature lower than the normal freezing point. The following analysis quantifies the effect. Picture the solvent (2) at its freezing point. The chemical potential of the solvent (containing the dissolved solute) is equal to that of the pure solid solvent at any temperature at which they are in equilibrium:

Vo, = bo + RT In x,

(11.133)

Solving for x, the mole fraction of solvent in the liquid phase,

cP

M2.

=

Mo

me

RTA

11.134 ae

AERT:

and then differentiating with respect to T at constant P:

al

CIN

oT

1CEO AM2,/T)

1 d(p5/T)Dn _ ha, — Io, |Aly,2

EGO

Rena

Rel

RT2

— RT?

(11.135)

we get an equation for the solvent that is identical in form to eq. (11.132) for the solute. Here, however, when eq. (11.135) is inverted,

Mc

oT

deadly

(11.136)

we have the dependence of solvent freezing point on solute content of the liquid. Over small ranges in temperature the Ah of the solvent is nearly constant and eq. (11.136) can be integrated between its normal melting point, T,,, where x, = 0, and any arbitrary point, T: In (1— 1)

i

mize

Ah,

din(i

— x) =

1

In (—

R

ae

aT

Jam T? Ahy,,|

1

Ah,

ae

Rae

1

— — Fs = ee) = —| F EB

If x, is small ( 0915)"

= 0.732

and

f = oP = 0.732 X 140. = 102.5 bar From the steam tables at this temperature and 14,000 kPa:

w=

g =h

5S ty ll C=

—

Ts = 2675.7 ts)

— 1964.8 kJ/kg

—

= 3155.4

613. X 5.4348 —

(at 1 bar)

=

615.. X 8.3520

—655.83 kJ/kg

336

Chapter 12 Nonideal Gas Mixtures V2

tell

1.0

0.8

6° 0.7

0.6

0.5

0.4

0.3

oe PALIN |

LAAN Aon Pst NRG SA WN SATHEN LAMY

\ N

1.4

vale KE He aaaia ies}

SS | MEE

1.2

as Weal

** 1.0

Figure 12.2 Correlation for 6°; high-pressure range.

Using eq. (11.52):

Ap = f =

1309. kJ/kg = (8.314 X 613./18.) In f 101.8 bar

As with the low-density calculation, we obtain excellent agreement between the actual and estimated values.

12s

1.10

i 1.50-1.80 1.40-2.00 3.00 1.20 4:00

1.05 } 5 oe

1.10

1.05 1.0

Figure 12.4 Correlation for '; high-pressure range.

Pure Components

337

338

12.2

Chapter 12 Nonideal Gas Mixtures

REAL GAS MIXTURES The fugacity of a component in a real mixture is defined by:

dp; = RT

d\n f;

(12.16)

just as for a pure, real gas, eq. (11.44). But now p,; and f; depend on the other constituents of the mixture as well as T and P. The fugacity approaches the partial pressure as the total pressure approaches zero (and the gas becomes ideal): fi===>

||

(12.17)

Da so that:

b, = ce,

(12.18)

Pi

12.2.1

Lewis and Randall Rule

In the simplest case for a real gas mixture, the fugacity coefficient is found for a component in a mixture in the same way that the pure-component ¢ has been determined, at the same temperature and total pressure. This method of determining the component fugacities includes the effect of T and P but neglects the effect of the other components on component /; it is called the Lewis and Randall rule, for the two people who first applied it.*

Estimate the fugacity of 20 mol% ethylene (1) in a mixture with nitrogen (2) at 50°C, 50 atm using the Lewis and Randall fugacity rule.

SOLUTION We find the pure component fugacity of ethylene at 50°C, 50 atm from

li

a

The method of calculating @ depends on the density region. At 50°C, 50 atm, the reduced temperature and pressure for ethylene are:

*G. N. Lewis and M. Randall. For full description, see G. N. Lewis, M. Randall, K. S. Pitzer, and L. Brewer, Thermodynamics, New York: McGraw-Hill, 1961.

12.2

Real Gas Mixtures

339

which is just into the low-density region, allowing us to use the virial equation for the fugacity coefficient. From these we get:

B® = —0.259

B® = 0.040

B, = —0.256

B = BRT./P, = —119 cc/mol

Using eq. (12.7):

ob, = exp(BP/RT)

= exp{(—119.

x 50.)/(82.06 X 323.)} = 0.799

and the fugacity in the mixture is:

= 0.799 X 0.20 X 50. = 7.99 atm

f; = y,P

where y, is the vapor-phase mol fraction of component 1, ethylene. This is in fair agreement with the experimentally determined value of 8.76 atm.

To achieve better agreement with nonideal gas mixtures, more exact methods, incorporating the effects of the nonideal interactions between unlike molecules, will have to be employed.

12.2.2

Exact Expression for Fugacity The calculation of the fugacity of a gas in a mixture of real gases, where intermolecular forces are significant, requires the development of a new framework of thermodynamic equations. We begin with the formal relation for the internal energy, U, of the mixture: 44 au ce I (=) dV er\eV

en

(12.19)

where the total internal energy, U, is integrated at constant temperature from the ideal gas state, U' (where

P =

0, V =

©), to the real gas state at

P =

P, V =

V. The

derivative at constant temperature and mole numbers is evaluated using the defining equation for entropy, eq. (3.4), and the Maxwell relation, eq. (5.21):

0) a OV)

a eae OV) r

ap (=) ap era

(12.20)

Equation (12.19) becomes: V

C= | E (2) - P|dV + Sind SS

oT

Vin

(12.21)

i

where the last term is the ideal gas internal energy, dependent only on T. The entropy of the mixture, in the same format, is:

340

Chapter 12 Nonideal Gas Mixtures

1-95-12) @J¢ V

I

2L\V rn — R>in, In YP) + Dons)

OV rn

(12.22)

The first term on the RHS of eq. (12.22) is the residual entropy, calculated at constant temperature. The last two terms are the ideal gas contributions: the entropy of mixing and the pure component values. Using Maxwell’s equations for the derivatives and substituting nRT/V for the ideal gas pressure,

av ae

NOT on

OV)

ae \ole) oe

man,

we alrive at:

cee I“I

(aP

nR

*)

co

oT

— —|

Wir

dV — Rin, In(y,P) + dns!

V

i

(12.23)

i

We can now express the Helmholtz free energy:

A=U—TS=

“l nRT a

ie

dV

V

—- RTDn, In (=) —}+

(ul — Dn

Ts! i) a |12.24 )

Here the term (n,RT/V) has been substituted for its equivalent (for the ideal gas), (y,P). We can now use the equation for the chemical potential in terms of A, eq. (11.41):

aA Nos (4)

(11.41)

Carrying out the differentiation at constant temperature and volume under the integral as well as in the second term,

Vv s

=]

RT

3

(oP

AV PSERT

/>-(—

[|V

v

||

|e

eae

| Eee on;Ul fn (=) v

oa oP Slot |

Teele)

with:

Ol

sont, Sa) |een ae

on; oo

M; = [eytia

ls ON u

9

WRT)

dV.T

\ien

RT |n

n,RT

is nRT

|

+ RT + (uj(ul — Ts! 5;)

eae

12.27 (12.27)

12.2

Real Gas Mixtures

341

which is an exact equation, independent of any approximations for the gas properties. Recall that the ideal gas chemical potential is:

gi = pw; = ul + Pul — Tsl = ul + RT — Ts!

(12.28)

and Bw; —

*

fi

»; = RT In fF

(12.29)

with

wi; = RT In fj Now, f% is defined as 1 atm, so that we can use eq. (12.27) to obtain, for the fugacity of a component in a mixture,

RT In f

Ffne! o| V

——

—"

;

a Ni)L py a

—_

aye

—

IRE

n

a n,;RT

——

(12.30) a

and then, subtracting RT In (y,P) = RT |n (n,P/n) from both sides,

ap PV RT In(4 (2+)is = rTInd, Dry = [| |RT ie - \(=an, dv —RTin(—} Lea ENGI) U

(12. gees

The last term in parentheses is seen to be the mixture compressibility factor, z, so the final general form for the fugacity coefficient is: Vv

RT In $; = | | = (=) be

GY e—eRI-1ni(Z)

(12.32)

i/ TVn;

1

At this point we must use an equation of state to evaluate the derivative inside the integral. As with z, h, and s, we must use two different techniques, depending on the density region. Virial Equation Here the virial equation, eq. (4.13), truncated at the second term, is presumed valid:

B z=l1l+c— U

(12.33)

Therefore,

:

RT

BRT

v

v2

nRT _ Bn?RT

V

Vv?

(

)

342

Chapter 12 Nonideal Gas Mixtures where this value of B is for the mixture and is a mole fraction—weighted average of the individual B’s*:

= DD yyBy

(12.35)

tj

For example, in a binary mixture,

(12.36)

B = yiBy, + 2y,y2Bi2 + Y3Bo

The symbol B,, should be read ‘‘Boje-one” and not ““Bejeyen,’ because it arises from the nonideal behavior caused by collisions between molecules of species 1. It is calculated exactly as we calculated pure-component B’s; the same is true for B,, and, in multicomponent mixtures, for all other like coefficients. However, a new situation has arisen: that of B,. In order to use eqs. (4.20), (4.21), and (4.22), a value for T, must be chosen. Without going into the statistical mechanics involved, the best simple choice for this parameter is:

= (TT)

(12.37)+

The other necessary mixture parameters are estimated in the following manner from the pure-component values:

oO. =

Oia

@;

(12.38)

2

Lo

Loy OV oo ap

(12.39) 3

Veet; Uo, =

5

(12.40)

bel Show re = aeee

(12.41)

ty

Before we can calculate a component fugacity, we need to evaluate the integral in eq. (12.32). To do this, we first need to express the partial derivative, (@P/dn TY. ny? in terms of the virial parameters by differentiating eq. (12.34): oP

fa

RT

tess!V

0

> ae A eee an; b 7 nnB;

RT

eee al

(12.42) Vn;

*This form arises as a result of the origin of B. It is a measure of the effect of binary collisions; the number of these collisions will be proportional to the number of each of the participant molecules. So, for 1-1 collisions, it will be proportional to n7/n, etc. +When available, an empirical factor, k,,,ij? is used as: T,. = y

(TT) (l — iat

|

kj)

12.2

Real Gas Mixtures

343

Say, for example, we need to find $,:

RTS SeRT

oP

+ n3By,

+ 2n,B,,

+ n,By

ony TV typ ae Vv AP yp [n,Byy

+--+]

(12.43)

>dy,B;

2.45)

or, in general,

aP on,l

T,V,n;

RI

ORT

V

V Jj

When eq. (12.44) is put into eq. (12.32), the integration can proceed:

fe (#) Jaw oo

V

on;

4 —

2nRT>\y,Bi;

TV.n,j

dVime2RE v2 =

al

U

co

J

so, finally,

In b; = ome = DyBy — In @ WS)

(12.46)

él

where v and z refer to the mixture properties.

Estimate the fugacities of ethylene and nitrogen in the mixture described in Example 12.3, using eq. (12.46).

SOLUTION We find the necessary Appendix I.

critical properties

a og

for ethylene

(1) and nitrogen

282.4 K

a

12672) K

49.7 atm

ee

33.4 atm

w, = 0.085

®, = 0.040

Ue, = 129. cc

Ue, = 89.5 cc

A da 0.276

i, = 0.290

At T = 323 K, P = 50. atm, T. = 1.14 a) II 1.006

T. = 2.56 P, = 1.50

both in the low-density region. We then use eqs. (12.37)—(12.41) to get:

ae ] N

=

|=

0.0625 188.8 K

Ley = 0.283 re = 40.8 atm

Use

LOCC

(2) in

344

Chapter 12 Nonideal Gas Mixtures and calculate the individual virial coefficients from eqs. (4.20)—(4.22):

B,, =

[—0.259

+ 0.085(0.040)]

x 82.06 x 282.4/49.7

=

—119. cc

By, =

[—0.011

+ 0.040(0.136)]

< 82.06 x 126.2/33.4 =

—1.66 cc

By =

[—0.096

+ 0.0625(0.121)]

x 82.06 x 188.8/40.8

=

—33.6 cc

We then find the mixture virial coefficient, B, from eq. (12.35):

B = 0.2%-119.)

+ 2 X 0.2 X 0.8(—33.6)

+ 0.87(— 1.66) =

—16.6 cc

and z of the mixture from eq. (4.14),

z =

1 + BP/RT = 0.969

The mixture volume, v, is,

U2= 97

Pas NOE) ICG

Now we have all the parameters needed for use of eq. (12.46):

In; = ee (By + y2By) — Inz » = p57 1020-119.) + 0.8(—33.6)] — In 0.969 = —0.166 so that:

b, = 0.847 and

Ff, = diye = O847tx 0.28% 50: = 8.47 atm which is within about 3% of the true value, 8.76 atm. This is about as good as we can expect with these simple rules, eqs. (12.37)-(12.41), and is better than the result in Example 12.3 (7.99 atm), where we just used the Lewis and Randall rule. Even though these equations may not seem ‘“‘simple,’’ they are referred to this way because they require no information other than pure-component parameters, which are readily available. More exact methods are available, but they require correction factors for each specific mixture; these are widely used in industry, but we will not detail them here.

To finish our example, we calculate , in the same manner as o;:

n b>

Te Pe!

(y,B12

SF Y7B>>)

—

Jia i

0.00015

but notice the slight difference in the coefficients used, as designated by eq. (12.46).

12.3.

Real Gas Solubility

345

The fugacity of the nitrogen is much closer to ideal than that of the ethylene:

b»

1.00015

fr

1.00015 x 0.8 x 50. = 40.0 atm

Equations of State In the high-density region, there is more than one way to quantify real gas mixture behavior. However, in this introductory text, we will just use the equation of state (EOS) method, as it has gained the most use in industrial practice. At that, we will limit the

treatment to only one of several equations of state (EOS) methods, as it has gained the most use in industrial practice. At that, we will limit the treatment to only one of several acceptable equations: the Soave-Redlich-Kwong (SRK) already used for pure gases. Furthermore, because VLE calculation entails solving the EOS for both equilibrium phases simultaneously, we will postpone this topic until Chapter 14.

12.3. 12.3.1

REAL GAS SOLUBILITY Henry’s Law In the discussion of ideal liquid mixtures, it was seen that the fugacity of each component was linear in mole fraction, with the proportionality constant equal to the purecomponent fugacity, f;,at the temperature of the solution (Raoult’s law): ig = 1%

(12.47)

At low pressures, f; can be taken to be the vapor pressure, P°, and eq. (11.107), a

commonly used form of Raoult’s law, is recovered. In no case is Raoult’s law followed over the entire range of mole fractions; however, in all cases it has been found empirically that, in the very dilute region, 1 pee eee

(12.48)

This linear dependence is followed out to some x,, although in highly nonideal solutions, this may be very small, perhaps 10~‘ or less. In any case, the linearity at low concentrations is known as Henry’s law, and k; is known as Henry’s constant for the particular system and temperature. Henry’s law is especially valuable for sparingly soluble gases and solids; here the linearity may extend as far as is needed in practice. The plot in Fig. 12.5 shows how the partial pressure of oxygen above water is linear with the mole fraction of oxygen in the water. At these low pressures and ambient temperature,

Por To: and xo, is the solubility in mole fraction, which can be directly converted to other units,

such as mg/L. The intercept on the right ordinate in Fig. 12.6 is Henry’s constant; the

346

Chapter 12 Nonideal Gas Mixtures 1000

800

600

400

Partial of kPa pressure oxygen,

200

Ole

"6.7 x 10°

1.3 x 10-4

2.0 x 10-4

Mole fraction oxygen in water

Figure 12.5 Solubility of oxygen in water at 23°C.

5.0 x 10°®

Partial of kPa pressure oxygen,

Mole fraction oxygen in water

Figure 12.6 Henry’s constant for oxygen in water at 23°C.

line is an extrapolation of the line in Fig. 12.5. This limit can never actually be reached, or even approached, since oxygen can never be liquefied above its critical temperature,

119°C Henry’s constant, k, is a strong function of temperature. For gases, the usual trend is to rise with increasing T (decreasing solubility); for solids, the usual trend is the opposite. These trends are seen to follow Le Chatelier’s principle, that a system moves in a direction so as to relieve a stress placed upon it: Dissolution of a gas will normally be accompanied by heat release, in the simplest case, that corresponding to condensation. Raising the temperature thus places a stress antagonistic to dissolution, lowering

12.3 Table 12.1

Real Gas Solubility

347

Henry’s Constants, k, in kPa, for Gases in Water at 23°C

Gas

ke Om?

Argon Oxygen Nitric oxide

Gas

3.61 4,29 2.81

oolOne

Nitrogen Hydrogen sulfide Carbon dioxide

8.52 SP 0.156

the solubility. In the case of solids, dissolution is similar to melting, so heat must be provided; thus the increase in solubility with temperature. Some Henry’s constants, taken from the International Critical Tables, Vol. Il,* are

listed in Table 12.1. All these are for gases dissolved in water at 23°C; however, these tables are an excellent source for many other systems and temperatures. The Gibbs-Duhem equation, eq. (11.67), has as one of its uses the determination of the fugacity of one component in a binary solution, such as a gas dissolved in a liquid, if the fugacity of the other is known. If we take eq. (11.67b), substitute RT In f; for each of the chemical potentials, then divide both sides by RT and n, the total moles,

we arrive at:

x, dinf,; + x,dInf, = 0 Now substitute our expression for Henry’s law for component dissolved gas:

(12.49) 1, which will be the

x, dln (%k,) + x» dln f, = 0 which is rearranged to: _X1 Ukx)) =

din f, = SSG (k,x;) =

Xq

2D

kyxy

—dx, xX

Note in the above expression that k, is constant and so can be brought out of the differential and canceled with the k, in the denominator. Now, since: i

eeone | dx,

—

—

dx,

we get:

dit

f= = ="d

*International Critical Tables, McGraw-Hill, New York, 1926.

In x,

348

Chapter 12 Nonideal Gas Mixtures

Fugacity

Mole fraction 1

Figure 12.7 Limiting fugacities in a binary solution.

which can be integrated from the pure 2 state to some arbitrary mixture composition, io9

Q

>

=F

—sN

|

| Solo? OIN N JT cea

Q

=y

& iS)

= Ry

fy = $35

(12.50)

a result that becomes Raoult’s law for species 2 at low pressures. Our conclusion from this exercise is that, when one species is in its Henry’s law range of concentration, the other must be in its Raoult’s law region. This behavior is shown in Fig. 12.7. In multicomponent mixtures, it can be shown in this same way, that if all but one species are in their Henry’s law range, the remaining species must be in its Raoult’s law region. For example, if a number of gases are dissolved in a solvent, the solvent fugacity can often be estimated in this way.

12.3.2

Temperature and Pressure Dependence In many instances solubility data will not be available at the temperature and pressure of a specific process. However, one or more data points for the system in question at other conditions may be used to estimate the solubility. The thermodynamics involved in the extrapolation is exact; however, the partial molar properties involved are rarely

12.3.

Real Gas Solubility

349

known with precision. What can be determined is the form of the temperature and pressure and dependence and, at least, a semiquantitative estimate for that dependence. Extrapolation from one datum is unlikely to be accurate; however, two or more data

points used with the thermodynamic framework developed below allow a reliable estimate. If the dissolved gas is in the Henry’s law region, where superscript « refers to the liquid and 8 to the gas phase. fi = kx;

(12.51)

In ff = Ink, + Inx;

0) dlnf* Ox;

fs

dlnx,; Ox;

(12.52)

Xx;

which we can use in place of the third term on the LHS of eq. (11.103), developed in the last chapter as the general result for temperature and pressure dependence:

—H* 72 NE

Ve

R

—hP p rare ar oh + RdinP

7 (0)Oak = OL ax, mee

(12.53) :

U

and, rearranging eq. (12.53),

h® — He

|

P

ye

GHP or = aP-=

din (5)= dlink;

(12.54)

i

because P = kx for a pure, ideal gas* in the Henry’s law range. From eq. (12.54) we can obtain the effect of both temperature and pressure on the Henry’s constant and, hence, solubility.

Temperature

The temperature effect (isobaric) is found by setting dP = 0 in eq. (12.54):

dink)

_

B_ He bien

Olas

(12.55a)

RT

or, equivalently,

dink) a(1/T) )p

— _|AP-S

jae AP) _ Moin

R

*That is, it is ideal in the vapor phase, but not in the liquid.

is

(12.55b)

350

Chapter 12 Nonideal Gas Mixtures The difference in enthalpies on the RHS of eq. (12.55b) is the heat release per mole of dissolution of i. Barring large chemical effects in solution, it is similar to the heat of

condensation and is positive; this means k will increase with temperature, and solubility

will fall. Even though this heat will rarely be known, it is not generally a strong function of temperature (or pressure), so In k will be expected to be nearly linear with 1/7. The measured temperature effect on the Henry’s constant for two gases, oxygen and nitrogen, in water, is shown in Figs. 12.8 and 12.9. Linearity is seen only over part of

In (atm) k

1/Tx 103 (1/K) Figure 12.8 Henry’s constant for oxygen in water at 1 atm.

TG

11.4

In (atm) k

1/Tx 103 (1/K) Figure 12.9 Henry’s constant for nitrogen in water at 1 atm.

12.3.

Real Gas Solubility

351

the temperature range; for both gases the implied [from eq. (12.55b)] heat of dissolution from the slopes in the linear part of these graphs is about 2000 cal/mol. This is considerably higher than the heats of condensation of these gases at their normal boiling points (where data are already available); these heats are 1629 cal/mol for oxygen and 1336 cal/mol for nitrogen. So, if we had just used the heat of condensation to predict the temperature effect on the Henry’s constant, we would have been significantly in error. However, if we had two data points (k vs. T) for the Henry’s constant, we could interpolate or extrapolate to another temperature, using eq. (12.55b), with some confidence. You will recognize the form of eq. (12.55) as identical to the Clausius-Clapeyron equation for vapor pressure vs. temperature. This approximate linear temperature dependence (In k vs. 1/T) is seen for many ‘‘equilibrium’’ constants, both physical, and as we will see in Chapter 15, chemical equilibrium.

Pressure

The isothermal pressure effect on the Henry’s constant is, from eq. (12.54),

gimme yo Ve

OP

cer

RT

oe

The magnitude of this effect can be demonstrated with a simple example. Take nitrogen, with a molecular weight of 28, dissolved in water. If we assume the dissolved state partial molar volume is equal to the liquid molar volume, an assumption good only for order-of-magnitude calculations,

—_

Ve~v, = M“~~ 60 — p mol

The effect on the Henry’s constant is: Alnk

AP

=

60

cc

.

mol - K

Se mol 82.06cc:atm

Mv

1

Xe 298K

=

Ones

0):

aan

atm

which gives an effect resulting from a 10 atm pressure increase about the same as that of a rise of 1°C in temperature. Note, however, that this does not mean the solubility is insensitive to pressure; only the Henry’s constant is. The solubility goes up nearly linearly with pressure, as indicated by eq. (12.48). The measured pressure effect is shown in Fig. 12.10. The effect is approximately linear at this temperature, but the implied partial molar volume (from the slope) is about 240 cc/mol, four times the pure molar volume

of liquid nitrogen. So, as with the

temperature effect, we can take two pressure data and find a third, but extrapolation using the pure molar volume for the partial molar volume is not accurate.

352

Chapter 12 Nonideal Gas Mixtures eo2

ass(o>)

(atm) In k

Pressure, atm

Figure 12.10 Henry’s constant for nitrogen in water at 25°C.

12.4

SUMMARY Nonideal gas mixtures require the application of a different form for the fugacity coefficient, even at low density. But application is not difficult and reasonable accuracy is possible even without additional empirical parameters.

PROBLEMS 12.1. At 25°C, under a pressure of 1 atm of pure oxygen, 35 cc oxygen gas dissolves in 1 mole of a certain oil (as measured at 25°C, 1 atm). What is the Henry’s constant (atm)? 12.2. The following are data for the solubility of hydrogen in water at 23°C:

Find the Henry’s constant.

Py, mmHg

Xy, X 10°

1100 2000 3000 4000 6000 8000

1.861 3.382 5.067 6.729 9.841 12.56

Problems

353

12.3. The following are data for the solubility of sulfur dioxide in water at 25°C:

Pso,, mmHg

Xs0, X 10°

5.0 10.0 50.0 100. 200. 300.

0.3263 0.5709 21329 4.213 7.448 10.20

Find the Henry’s constant.

12.4. The following are data for the solubility of carbon monoxide in water at 19°C:

Poo, mmHg

TeGrx10:

900 1000 2000 3000 4000 5000 6000 7000 8000

2.417 2.685 5,304 8.000 10.63 bys 15.80 18.28 20.67

Find the Henry’s constant.

12.5. Calculate the vapor-phase fugacity coefficient ,, for carbon dioxide (1), in an equimolar gas mixture with propane at 300 K, 17 atm.

12.6. (a) Calculate the fugacity of component (1) in a vapor mixture with (2) at 353 K; the mole fraction y, = 0.30 and the total pressure is 300 mmHg. The virial coefficients here are, in

cm?/mol:

B, = -2625

B, = —634

By, = —3668

(b) Compare with the Lewis and Randall fugacity rule. 12.7. At 60°C, 1 atm total pressure, pure CO, is in equilibrium with an aqueous solution of 0.000294 mol fraction CO,. Neglecting the effect of pressure on the Henry’s constant, estimate the mole fraction of CO, in an aqueous solution at 100 atm of pure CO, at 60°C. Nonideality can be neglected in the liquid phase, but not in the vapor phase. 12.8. Calculate the molar volume of the mixture: y, = 0.3 methane, y, = 0.3 propane, y,; = 0.4 pentane at 100°C, 1 atm. The virial coefficients are;

By 20,

By, Ds

By; = 9.9%

Boy — 241.

By, = Sek

B33 =r Ls

354

Chapter 12 Nonideal Gas Mixtures

12.9. A gas mixture exiting from a sour-gas treatment process contains 25 mol% CO, and 75 mol% H,S. For design of a separation process to separate these components it is necessary to determine the fugacity of each in the mixture. a. Calculate the fugacity of the H,S if the mixture is at 350°C, 100 atm using the appropriate correlation with k,, = 0.08. [See footnote to eq. (12-37).] b. Compare with that calculated using the Lewis and Randall fugacity rule. c. Compare with that calculated assuming a mixture of ideal gases. 12.10. Calculate the fugacity of benzene at 575°C, 129 atm. 12.11. A vapor mixture of 20 mol% H,S, 80% isopentane is in contact with its equilibrium liquid at 552.5 K, 47 bar.

a. Estimate the fugacity of the isopentane in the vapor using the virial equation with ky» = 0.11. [See footnote to eq. (12-37).]

b. Compare with that calculated from the Lewis and Randall fugacity rule. c. Compare with that found from the assumption of an ideal gas mixture. d. Estimate the fugacity of the isopentane in the liquid phase. 12.12. The Henry’s constant for propane in a nonvolatile oil (MW = 550) at 134°C is reported as 2.15 xX 10% atm. What is the solubility of propane (g/100 g oil) under 20 bar of pure propane at 134°C? Neglect departure from Henry’s law behavior in the liquid, but take into account nonideal gas behavior. 12.13. Find the fugacity of steam at 700°F, 3000 psia from: a. Fugacity coefficients from reduced property charts, Figs. 12.1 through 12.4 b. Fugacity coefficients from the virial equations c. Steam tables

Chapter 13

Real Liquid Mixtures The nonideality of the liquid phase is generally more important in phase equilibrium calculations than is the nonideality of the vapor phase. This is especially true at low pressures. Before we treat mixtures, we must first make any necessary corrections for nonideal behavior to the pure components.

13.1

FUGACITY OF A PURE LIQUID When a pure liquid exists in equilibrium with its pure vapor, its fugacity is equal to that of the vapor. Notice in Example 12.2, the fugacity of pure steam at 340°C, 140 bar, was 102 bar; yet the vapor pressure of steam at 340°C is 146 bar. Even at its own vapor pressure, P°, the fugacity is not equal to the vapor pressure, because the purevapor fugacity is not equal to the pressure (unless 6 = 1). So, in order to satisfy the basic equation of equilibrium,

rede hs (13.1)

6: P° = fp ~ P°

The liquid fugacity is only slightly dependent on pressure; this dependence is obtained from eq. (11.44) as:

dinf\ | (ans =v,

(13.2)

where the liquid molar volume, v,, is nearly independent of pressure; so eq. (13.2) is generally integrated to:

In f = In fp +

Vie —

a

PP’)

(13.3)

When the pressure above the liquid is greater than the vapor pressure, because of an inert atmosphere, for example, the liquid is subcooled, and its fugacity is increased by a factor known as the Poynting correction, the antilog of eq. (13.3): f=

fp-e

ufP — P°)

*

(13.4)

355

356

Chapter 13 Real Liquid Mixtures

Calculate the fugacity of pure liquid refrigerant R-12 at its own vapor pressure at 40°C. SOLUTION

From Table 10.1 the saturation vapor pressure of R-12 at 40°C is 9.48 atm. We find the fugacity of the saturated liquid from the fugacity of the saturated vapor: BP

fa =

(12.8)

Tiss =e

where the subscripts ‘‘sl’’ and ‘‘sv’’ refer to the saturated liquid and vapor, respectively, if we are in the low-density region. We test this with the reduced properties, found from the critical properties of R-12:

T,Cc = 384.7 K

Pc

39.6 atm

T,di = 0.813

Ppr

0.239

w = 0.158

which is close to the dividing line on Fig. 4.5. From eqs. (4.20, (4.21), and (4.22),

B® =

—0.505

B® =

—0.271

B, =

—0.548

so that:

B =

—436.9 cc/mol

and, from eq. (12.8):

fa = P+ = 9.48 exp[(—436.9 x 9.48)/(82.05 x 313.)] 9.48 x 0.851 = 8.07 atm Because the density is so close to the dividing line in Fig. 4.5, we could reasonably use the fugacity-coefficient charts, Figs. 12.1 through 12.4. From these we obtain:

6° = 0.875 b! = 0.885 b = 9

($1)? = 0.858

and

f =

9.48 X 0.858

= 8.13 atm

under 1% different from that calculated from the virial equation, (12.8).

13.2

Fugacities in Liquid Mixtures

357

Calculate the fugacity of liquid R-12 at 40°C under a pressure of 100 atm nitrogen. The density of liquid R-12 at these conditions averages 1.25 g/cc.

SOLUTION Use eq. (13.4), with the fugacity at the vapor pressure at 40°C as calculated in Example 13.1:

v, = My/p = 96.7 cc/mol v,(P—P*)

f =

13.2 13.2.1

are op

lige

[96.7(100. — 9.48)]

=

8 OF

= e

82.05 = 75.35 mmHg

P = p, + py =

122.8 mmHg

y, = p/P = 0.386

y, = 0.614

which are the same as found in Example 11.3. Thus the vapor-phase nonideality does not affect the calculated vapor-phase composition within our precision limits.

13.2.2.

Real Liquid Mixtures at Low Pressures: The Activity Coefficient Definition In describing real gases, we have seen the practicality of quantifying the departure of the real gas from that of the ideal gas at the same temperature and pressure; these departures were called the residual properties. Because the departure from ideality was, in many Cases, a correction of only percentage order, the ideal gas property became a solid base from which to calculate merely the perturbation from nonideal effects: errors in the calculation of this perturbation have a small effect on the total property calculation. A somewhat different approach to nonideality is taken for liquids from that taken for gases. Here we add a correction factor to the properties of the ideal solution. This correction factor is called the activity coefficient, -y, which must, in general, be deter-

13.2

Fugacities in Liquid Mixtures

359

mined from experimental measurements. The chemical potential of the ideal liquid mixture is given by eq. (11.106); it is modified by correcting the activity, a, by:

so in a real solution;

Bw; = p; + RT In y;x;

(13.5)

If component i is a liquid at the mixture T and P, the usual situation, 1; is the chemical potential of pure i at T and P. Therefore,

Vimrek.

aSe

xe

10

Now picture an ideal vapor («) in equilibrium with a nonideal liquid (8). As before,

pe = pP Expanding this equation with our relations for the two phases,

(w;)* + RT In (y,x,)) = (w)*® + RT In p;

(13.6)

and, at constant temperature and pressure, the pure-phase chemical potentials are constant:

(w)° — (u)* = K ¥ f@)

(13.7)

where K is dependent only on T and P. So eq. (13.6) can now be solved for the partial pressure as: Mi =

(eX/FT yx; ar Poy k

(13.8)

The exponential in the second term must be the vapor pressure, since, at x = 1.0, the partial pressure must equal the vapor pressure, and the exponential is not a function of x. Temperature and Pressure Dependence

Just as we have seen for gas properties, it will often be necessary to estimate liquid properties at a certain temperature and pressure from experimental data at other conditions. We need a thermodynamic framework to do this interpolation (or extrapolation). Equation (13.5) is divided by T: eB Fiei bee i

In y; vy. — Rinx. R In ex.

13.9 (13.9)

360

Chapter 13 Real Liquid Mixtures and differentiated with T at constant pressure and mole fraction: Li

(5) OR

Bj a

(i)

H

-

pute NG aa week cy"

oT

oT

ji

51 n

Yi

(13.10)

oT

where H, is the partial molar enthalpy of component i. Because, y; > 1.0 as x; > 1.0, atx = 1.0 the first term on the far RHS must relate to the pure component; that is, the pure molar enthalpy:

IL;

fo) AEE Ss

(%) oT

=o

h. 2

13.11 ( )

Replacing this term in eq. (13.10) and solving for the derivative of the activity coefficient,

oT

Rig

(13.12)

The enthalpy difference in eq. (13.12) is that between pure liquid i and i in solution at mole fraction x. This equation is very similar to eq. (12.55) for the temperature dependence of the Henry’s constant. In fact, all ‘‘equilibrium constants’’ have this same form of temperature dependence. Look back at eq. (5.68), the temperature dependence of the vapor pressure; the vapor pressure can be considered an equilibrium constant between pure liquid and vapor phases. In a similar manner, the pressure dependence of the activity coefficient is found to be:

oP

RT

pee!

which is seen to be zero for a component whose partial and pure molar volumes are identical.

13.3.

EXCESS PROPERTIES The extensive properties (U, H, V, S, A, G) of a real solution are related to those of an

ideal solution through terms called excess functions.* These are defined as the difference between the real solution property and the ideal solution at the same T, P, and x.

*Analogous to the residual properties for gases.

13.3.

13.3.1

Excess Properties

361

Excess Free Energy The Gibbs-Duhem equation at constant T and P, eq. (11.67b), can be written for a binary as:

)

ed)

Oxy TP

Ox TP

xX (22: & oo #2) =40

(13.14)

yw; = pw, + RT In (y;x;)

(13.15)

with

Substituting eq. (13.15) into eq. (13.14),

al el eee Ox,

TP

al tyes) Ox

TP

= 0

(13.16)*

The definition, above, of any excess property as the property of the real mixture minus that of the ideal mixture at the same temperature, pressure, and mole fraction, is equivalent to the change in the property on mixing the pure components minus the ideal change on mixing. For example,

G=

AG’ — AG™

(13.17)

where the superscript E indicates the excess property. Using the first definition for the excess Gibbs free energy, G®,

GE = Gu — Gy

(13.18)

where G,, and G,, are the free energies of the mixture and ideal mixture, respectively, and the total Gibbs free energy of the real mixture:

Gy = me, = une; + RTDIn; In x, + RT Dn; In y;

(13.19)

G = Gy — Guy = RTDn,; In y;

(13.20)

so that GF is:

*This is an approximation. For a two-phase, two-component system, for example, there are F = C — P + 2 = 2 degrees of freedom; one cannot hold T and P constant and still have mole fraction vary. The exact expression is, from eq. (11.67a),

xd He

ve dP RT

hE dT RT?

but at near-atmospheric conditions, the two terms on the right are usually negligible.

362

Chapter 13 Real Liquid Mixtures

the first two terms on the RHS of eq. (13.19) being the pure-component free energies, eq. (11.124), and the ideal free energy on mixing, eq. (11.125). These two terms are thus the total free energy of the ideal mixture, Gy,. The total differential of eq. (13.20) is:

dG® = RT>\n, dn y, + RTD[Iny{dn))]

(13.21)

but,

Sin, din y; = 0 from eq. (13.16), the Gibbs-Duhem Equation; this reduces eq. (13.21) to:

dG® = RT>.[In y{dn,)]

(13.22)

So, dividing eq. (13.22) by dn, at constant T, P, and all other n;:

(=)

= G = pF = RT Iny,

BE

i

L

—.

(13.23)

/T,P,nj

which defines the partial molar excess Gibbs free energy, G¥. From the discussion at the start of the chapter, we see that the fugacity of a component in a real liquid mixture can be expressed as:

fa = Wid ~ Wi

(13.24)

where the activity coefficient is evaluated at the system T and P and f* is the fugacity of component / in its “‘standard state’’; generally this is the pure-component vapor pressure at the system temperature and total pressure. This definition means that at x; = 1.0, y; = 1.0.*

13.3.2

Excess Enthalpy Other excess properties have forms analogous to the free energy; for example, the excess enthalpy on mixing is:

bE, =~ AWM —Ah™ = AM = Sin, — Dahe= Doe —h)

(3.25)

where the ideal enthalpy change on mixing, Ah’, is always zero.

*We are neglecting the very small effect of pressure on the pure component vapor pressure. That is, the total pressure for any composition will not be equal to the vapor pressure of either component; but this is a negligible effect at low pressures.

13.3.

Excess Properties

363

The difference between the excess properties that are entropy dependent, such as the free energies, and those that are not—the enthalpy, internal energy and volume—shows up in the change on mixing for the ideal mixture. For the non—entropy dependent properties, the ideal changes on mixing are zero. This difference is shown graphically in: Fig 13.1). Although the enthalpy of mixing is shown here as positive (endothermic), meaning this particular solution will need to absorb heat from the environment to stay at constant temperature during the mixing process, the enthalpy of mixing can be of either sign. But the ideal entropy of mixing is always positive, leading to a negative ideal free energy of mixing.

Ag = Ah — TAs as shown in the upper graph. The excess free energy on mixing can be of either sign; that shown in the graph is positive.

Figure 13.1 Excess free energy and heat of mixing.

364

13.3.3

Chapter 13 Real Liquid Mixtures

Temperature Dependence The form for the temperature dependence of the excess properties is found in the same manner as for the activity coefficients. The molar form of the excess Gibbs energy is, from eq. (13.20),

oa

RTD x; In y;

(13.26)

which, on dividing by T, eae =

Rye In y;

gives a form we can use to get the temperature dependence:

0(g"/T) ot

=

aT

R

2

pt

dIny; aaa SH)

aT

Now we can use eq. (13.12): d Iny; =

Rae

h, —

Ai;

IORIC

(13.12)

:

to reach:

0

E

AEED oT

h.

a

=H.

De

RD, eel RT

Ah,i —

ie

Saves it al

Te Te

(13.27)

where the excess enthalpy on mixing, defined as the difference between real and ideal enthalpies on mixing.

HE = by — dy = DH, — Doh

(13.28)

allows conversion of eq. (13.26) to:

6) ap

ge

;

(13.29a)

or, equivalently,

a(g®/T)

cei = OLE)

HF

(13.29b)

13.3.

Excess Properties

365

The same result can be obtained using the development in Chapter 11. The change in free energy on mixing, using eq. (11.93), is: Ag” = DxG,

=

D8:

(13.30)

which differentiates at constant T to:

Ee aT

tgs i MOU), oF

yy,2 eu) rr

(13.31)

Now, from eq. (11.98), for both pure component i,

au/T) _ Ag/T) oT

A

oT

Be

and component i in a mixture,

Au/T) _ G/T) _ _H, oT

oT

ies

converting eq. (13.31) to:

aAgM/T) _ =H, ~h, op > 2a) ~2a 1 =

me

(13.32) —fpE

-2| SA

a

Sea =

T?

bringing us to the same equation for the temperature dependence of the total change in free energy on mixing,

aAg/T) _ a(1/T)

yp

(13.33)

as for the excess free energy, g®, eq. (13.29b). The reason these are identical is that the ideal free energy change on mixing divided by T is independent of temperature:

Re T = R>

xin x, © f(T)

The enthalpy of mixing may itself be a function of temperature. That is, if the heat capacity of the mixture is different from the mole fraction—weighted average of the components, there will be an excess heat capacity, Gg,

ane

a

=

7

Cc. =

Ca

—

Dies

(13.34)

which must be determined from experimental data. These data are available for only a limited number of systems.

366

Chapter 13 Real Liquid Mixtures

13.3.4

Pressure dependence The pressure dependence of the properties of liquid mixtures is generally very slight, the formal relations are directly found from eq. (11.92) E

LE:

(13.35)

and eG

13.4

eee ei ae ait!

(13.36)*

ACTIVITY-COEFFICIENT EQUATIONS The liquid phase is much more complex than the vapor; liquid molecules constantly encounter 12 or more neighboring molecules, some like and some unlike. Liquid state theory is nowhere near ready to treat any but the very simplest of pure liquids, let alone mixtures. For this reason activity coefficients are estimated from the empirical correlations. Some of these are quote sophisticated, with interaction terms based on common groups found in the molecules. Although they are numerically complex, modern computing allows rapid calculation.

13.4.1

Restrictions on Form Before considering any of the correlations, however, we show how the Gibbs-Duhem equation, eq. (13.16), can be used to place some restrictions on the form of any thermodynamically consistent equations. Take a general expression for the activity coefficients as a power series in mole fractions: In y,

al

— x,)

+ bd

In Y2

ax, + bog +e

— x,)% + cd

— Pa)

(13.37a)

(13.37b)

We now differentiate the above expressions with respect to x, and substitute in eq. (13.16):

HOF Oe

2B gh 2B

ae SOr iy HASCKG a 3ex, eee

en

x

ey)

and then replace x, with (1 — x). The coefficients of all powers of x, must sum to zero independently; this is the only way the equation will always sum to zero, as prescribed by the Gibbs-Duhem relation. We find, for the present case,

*The exact form depends on whether the pure component fugacity, f°, is defined at the system pressureor reference pressure. In the first case, eq. (13.36) is used; in the latter case the numerator becomes simply V,.

13.4

Activity-Coefficient Equations xa

367

=())

(13.38a)

xl:—a — 2b — 3c — a’ + 2b' = 0

(13.38b)

x4:2b + 6c — 2b’ + 3c’ = 0

(13.38c)

xi:-3c — 3c’ = 0

(13.38d)

The last of these, eq. (13.38d), results in

eee

(13.39)

and using this in eq. (13.38c),

2b (3e%=92b' =) 0 or b' = 3c + 2b)/2 = b + B/2)c

(13.40)

This, when used with eq. (13.38b), leads to:

a=a'=0

(13.41)

These relations form important linkages between the coefficients; although we have only considered terms to the third power of mole fraction here, similar results follow

for higher order equations. The last result is especially important; it shows that no activity coefficient equation can have terms in the first power of x.

13.4.2.

The Wohl Expansion Although most equations for activity coefficients are basically empirical in nature, they do have some theoretical framework. One of the earliest treatments to link activity coefficients to intermolecular interactions is that of Wohl,* who theorized that the

excess Gibbs free energy depends on the number and energy of the unlike interactions in solution. Larger molecules will have contact with more molecules in solution than will smaller ones; this is accounted for by using the relative molecular volumes," 7;.

For a binary system, his expansion appears: 13)

oa Se RT(x,r;

2a;rb,6,

fe XoPp)

+ 3a) :rb7b> a 3a; yb 165 (13.42)

2H 4411120}.

*K. Wohl, Trans. AIChE, 42, 215 (1946). ‘Estimated by the pure molar volume.

a8 4213

st 641 122705 tee

368

Chapter 13 Real Liquid Mixtures

where a are interaction-energy parameters. The volume fractions, , not to be confused with fugacity coefficients, are defined by: : Xie

2 %

13.43

Dr i

The high density of liquid solutions, compared with gases at low pressure, requires consideration of ternary and higher order interactions. For each possible combination of collision, a different effective interaction energy parameter, a, is defined. That is, for two molecules of type 1 and one of type 2, the parameter a,, is used, and so on. The probabilities for each type of interaction are dependent on the size and number of each component molecule; the coefficients: 2, 3, 4, and 6 arise from the random probability

of each arrangement. The physical significance of these parameters a is not exact; what this expansion does is (1) allow for the generation of reasonable empirical forms for g® (and hence In y) and (2) permit extension to multicomponent systems. Three- and Four-Suffix Margules’ Equations A simplification of eq. (13.42) can be made for a binary mixture of molecules of similar size: r; ~ 1p. In this case ); = x; and eq. (13.42) is differentiated according to eq. (13.23),* with terms to the fourth order in composition, to obtain the four-suffix

Margules equations: Linen

In y, &=

Aeros Dy ’ A+

Verte Gia,

(13.44a)

3 ' 5B + 2C’ ’ |x;2 —

' 8 ’ 14 [B+ ne x7 + C’x}

(13.44b)

They are called four-suffix equations even though only three coefficients, A, B, and C, appear, because the coefficients of the first-order terms must always be zero as specified by eq. (13.41).

The coefficients are related to the energy parameters, a, by:

A’ = r(2ay. — 3ay29 + 60442 — 644422 + 1244412) B=

(64499

=

64112

=

841009

=

2444112

ate 244119)

(13.45a) (13.45b)

although in practice these energy parameters are not independently evaluated; the coefficients in eq. (13.44) are found empirically. The number and precision of the VLE data requisite for the four-parameter form are seldom available. For this reason, the Margules equations are usually terminated at the

*Note that the otal free energy, G*, must be used (ng®), and not the molar free energy, 9".

13.4

Activity-Coefficient Equations

369

third order by setting C’ = 0:

In'y, = A’xs + Bix

In y>

lI

(13.46a)

3 ( 7 32) — Bx

(13.46b)

or, in the equivalent but more symmetric form,

In y, = (2B — A)x3 + 2A — B)x

(13.47a)

In y, = (2A — B)x? + 2(B — A)?

(13.47b)

where: A’ =2B—A and

B' = 2(A — B) Take two binary systems with nearly equal molar volumes.* One is: Cyclohexane (1)—toluene (2)

v, = 1166 cc/mol

P = 760. mm

Hg

vy, = 113.7 cc/mol Margules’ constants:

A = 0.1587

B = 0.3600

These are displayed as Fig. 13.2 and 13.3. The other system is: Methylcyclopentane (1)-toluene (2)

v, = 120. cc/mol

P = 760. mm Hg

v, = 115. cc/mol Margules’ constants:

A = 0.1598

B = 0.3747

displayed as Figs. 13.4 and 13.5. The fits to the activity coefficients are within the experimental accuracy, as is the reproduction of the VLE. Use of the three-suffix Margules form, however, is not limited to systems with equal pure molar volumes. It can be used as a thermodynamically consistent, although em-

*H. S. Myers, Ind. Eng. Chem., 48, 1104 (1956).

—

e@

Calculated

Exp. data

(1.01325 bar)

coefficient, Activity y

0.00

0.20

0.40

0.60

0.80

1.00

Mole fraction cyclohexane (liquid)

Figure 13.2 Fit of the three-suffix Margules equation to the system: cyclohexane-toluene.

375.0

370.0

365.0

K Temperature,

360.0

355.0

Mole fraction cyclohexane

Figure 13.3 Experimental and calculated VLE for the system: cyclohexanetoluene at 1.012 bar.

370

—— Calculated e@

Exp. data

(1.01325 bar)

coefficient, Activity y

0.00

0.20

0.40

0.60

0.80

1.00

Mole fraction methylcyclopentane (liquid)

Figure 13.4 Fit of the three-suffix Margules equation to the system: methylcyclopentane-toluene.

390.0

380.0

——

Calculated

e

Exp. data

(1.01325 bar)

370.0

Temperature, K 360.0 f

350.0

Mole fraction methylcyclopentane

Figure 13.5 Experimental and calculated VLE for the system: methylcyclopentane-toluene at 1.013 bar.

371

372

Chapter 13 Real Liquid Mixtures

pirical, form for fitting more complex systems. For example, three systems that are fit quite well by this form at 50°C are*:

a) Chloroform—methanol b) Acetone—chloroform c) Acetone—methanol

A

B

0.720 — (0.829 0.720

1.805 — 0.691 0.518

The data, along with the fits for these systems, are shown as Fig. 13.6, a, b, and c. The

molar volumes of these components are not equal, as would be required from the theory; the theory just provides a consistent framework for an empirical equation. This means that the energy parameters, at least for systems with unequal molar volumes, can have no significance. Notice that the form of equations (13.47) simplifies at either end of the mole fraction range. At x, = 1.0, x, = 0.0 and In y, = 0.0. At the other end, x, = 1.0, x, = 0.0 and In y, = 0.0. This must be the case because, for the pure component, the activity coefficient must be unity. The limiting behavior for the component approaching zero mole fraction, that is, for In y, as x, approaches zero, is found by setting x. = 1.0 in eq. (13.47a):

A

eB Atay

A

Se

and similarly for In y, as x, approaches zero: In y2 ——5? 2A

B+

2B

- A=

B

XQ

These are called the activity coefficients at infinite dilution; you can see on Fig. 13.6 that the curves, when extrapolated to the end points, yield the values in the preceding table. Notice they also agree with the activity coefficients extrapolated to the infinitedilution end for the systems graphed in Fig. 13.6. The coefficients also conform to eqs. (13.39)-(13.41); if we take: b= (@2B — A)

and

¢=2(A — B)

c

=

—c

=

2(B = A)

then:

b’' = (c + 2b)/2 = [6(A — B) + 2(2B — A)]/2 = 2A — B which recovers the three-suffix Margules equations, eqs. (13.47a) and (13.47b).

Two-Suffix Margules’ Equations

The simplest nonzero form for the excess Gibbs free energy of a binary mixture that still satisfies the Gibbs-Duhem equation is:

GP = Axx, *W. H. Severns, A. Sesonske, R. H. Perry and R. L. Pigford, AIChEJ, 1, 401 (1955).

(13.48)

13.4

Activity-Coefficient Equations

373

Beets Sie See am

Peabo ed ceeof No a He ete Pra 0.5

Sains

Xacetone (b)

(a)

ae

.LULUL LULL

Ea ae Ne

Xacetone (c)

Figure 13.6 Fit of three systems with unequal molar volumes to the three-suffix Margules equation. 50°C. (a) Chloroform-methanol, (b) Acetone-chloroform, (c) Acetone-methanol.

T =

It is generated from equations (13.26), (13.44a), and (13.44b) with B’ and C’ = 0.* It is readily apparent that g® goes to zero at both ends of the mole fraction range, a primary requisite from eq. (13.20), since at one end x is zero and at the other In y is zero for each component. *In this form,

A = A’.

374

Chapter 13 Real Liquid Mixtures

A/RT

| A/RT

Iny

O.

0.5 x

1

Figure 13.7 The two-suffix Margules equation.

We find the activity coefficients from eq. (13.23): |a

te GF

=

ering,

=

aG*

(S)

on,

(A

eA

Sa) ey

BE

on,

T,P,n2

a

on,

and since,n = n, + ny,

d[n,/n]

RT |In y, = An, —— On,

ei

ore

= An, ————_ _ n

io

= A

n> n

lI A =

(13.49

2

)

and similarly:

RT In y5 = Axe

_

(13.50)

As a simple exercise, test this result with the Gibbs-Duhem equation, eq. (13.16), to see whether it conforms. The form of the activity coefficients is symmetric, as shown

in Fig. 13.7. This simple form adequately represents some nearly ideal liquid mixtures; two examples are argon—oxygen (A = 148 J/mol at 83.8 K)* and benzene-cyclohexane (A = 1268 J/mol at 303 K).* The van Laar Equation

If, in the Wohl expansion, all a parameters of higher order than the binary a,, are assumed negligible, that is, if the binary interactions can account for all the nonideality,

*R. A. H. Poole, G. Saville, T. M. Herrington, B. D. C. Shields, L. A. K. Staveley, Trans. Far. Soc. 58, 1692 (1962).

'G. Scatchard, S. E. Wood, J. M. Mochel, J. Phys. Chem. 43, 119 (1939).

13.4

Activity-Coefficient Equations

375

eq. (13.42) becomes: E § =|

20 9X|XoI11p eae ae

RE

MAT

13.51

tl)

(

)

When this is differentiated according to eq. (13.23), the van Laar equations for the activity coefficients are obtained:

A

In y,; =

5

(13.52a)

je at

Bx,

B

ln. =

(13.52b)

These van Laar activity coefficient equations form another two-constant set that is thermodynamically consistent; that is, it satisfies the Gibbs-Duhem equation. From the form of the equations, it can be seen that In y, has a limit equal to A at x, = 0, and In 2 reaches B at x, = OQ. For the pure components (x = 1) the activity coefficients become equal to unity. Two systems with similar chemical properties, well fitted by the van Laar form, are the system*: Butylbenzene (1)-nitrobenzene (2)

v, =

156.

v, =

T = 100°C

102.2

van Laar constants

A = 0.6182

B = 0.6773

displayed as Figs. 13.8 and 13.9, and the system:** p-cymene’ (1)- aniline (2)

Vv, =

156.6

P =

100.00 mmHg

vz = 91.15 van Laar constants

A = 1.0898

B = 0.8706

displayed as Figs. 13.10 and 13.11.

*J. M. Wise, M. Balish, F. Gerewith, V. Fried, Fluid Phase Equil., 4, 115 (1980).

**S_R. M. Ellis, M. Razavipour, Chem. Eng. Sci., 11, 99 (1959). ‘p-cymene is also known as 1-isopropyl-4-methyl benzene.

376

Chapter 13 Real Liquid Mixtures 2.00

——

Calculated

@ y, (Exp. data) © Yo (Exp. data)

1.60

1.40

coefficient, Activity y

1.20

1.00

0.00

0.20

0.40

|

0.60

0.80

1.00

Mole fraction butylbenzene (liquid)

Figure 13.8 Fit of the van Laar equation to the system: butylbenzenenitrobenzene at 373.15 K.

However, just as with the Margules form, systems more complex than suggested by the development may be fit by the van Laar form. For example, the aromatic—aliphatic system: benzene (1) + isooctane (2) at 45°C is well fitted by these equations with A = 0.419, B = 0.745; the activity coefficients are nonsymmetric, as seen in

Higvl3.125

13.5 13.5.1

DETERMINATION

OF ACTIVITY COEFFICIENTS

Direct Calculation from Data Activity coefficients are most often obtained from VLE data. At pressures. near atmospheric or below, the nonideality corrections to the vapor fugacity and the pure-liquid vapor pressure are often negligible. Under these conditions, f,v = yP

=

f;,Uy =

yxiP;

oO

(13.53)

and the activity coefficients can be directly calculated from experimental measurements

*S. Weismann and S. E. Wood, J. Chem. Phys., 32, 1153 (1960).

13.5

— @

Determination of Activity Coefficients

377

Calculated Exp. data

3/3.15K

bar Pressure,

0.40

0.60

Mole fraction butylbenzene

Figure 13.9 Experimental and calculated VLE for the system: butylbenzene-nitrobenzene at 373.15 K.

of T, P, x, and y:

melo ee

(13.54)

x;P ;

The following data were taken for the system ethanol—benzene at 750 mmHg total pressure*: xy

J

T

PS

V1

Y2

0

0

0.04 0.28 0.43 0.61 0.80 0.89 0.94 1.00

0.15 0399 0.450 0.503 0.610 0.724 0.815 1.00

79.7

804

(EPs 68.3 67.8 68.3 70.1 72.4 74.4 78.1

671 507 497 507 545 598 650 750

(5.70) 4.20 ZA 1.58 T2272 1.05 1.02 1.00 1.00

1.00 1.025 1.209 1.421 1.845 2.661 3.181 3.666 (4.06)

*D. Tyrer, J. Chem. Soc. (London), 101, 1104 (1912).

—

— @

Calculated y, (Exp. data)

2

Yo (Exp. data)

0.1333 bar

>

coefficient, Activity y

0.00

0.20

0.40

0.60

0.80

1.00

Mole fraction p-cymene (liquid)

Figure 13.10 Fit of the van Laar equation to the system: p cymene-aniline at 0.1333 bar. 394.0

_— Calculated

© Exp. data)

-

01833 ba

Temperature, K

Mole fraction p-cymene

Figure 13.11 Experimental and calculated VLE for the system: p cymene-aniline at 0.1333 bar.

378

13.5

Determination of Activity Coefficients

379

1.0

= 0.419

Figure 13.12 Fit of the van Laar equation to the system: benzeneisooctane at 45°C.

The last two columns, the activity coefficients, are calculated from the data in the first

five columns using eq. (13.54). The values in parentheses are extrapolated from the rest of the curve.

13.5.2

Use of Correlations Infinite Dilution Whereas the best values for the parameters in any of the activity-coefficient equations are found from a fit of all the data to the particular form, a reasonable estimate can be made from the infinite-dilution values. For example, in much the same way as the threesuffix Margules equation, eq. (13.47), the van Laar form, eq. (13.52), reduces to:

Ae M1

x10

: (0)

=A

a Bet | ]Nn

Y2

x20

2

B.

= B

(0)

A: at which are, for the ethanol-benzene system at 750 mmHg: II In yy =

B

In yz

In 5.70 II 1.74

In 4.06

1.40

where the superscript © (infinity) means ‘‘at infinite dilution’’ for that component. The

380

Chapter 13 Real Liquid Mixtures fit to the data obtained using constants obtained in this manner are shown in the following table: x}

yi

z

(Yve

(Ye

0 0.04 0.28 0.43 0.61 0.80 0.89 0.94 1.00

0 0.150 0.399 0.450 0.503 0.610 0.724 0.815 1.00

tat ee 68.3 67.8 68.3 70.1 72.4 74.4 78.1

(5.70) 4.20 201 1.58 1.22 1.05 1.02 1.00 1.00

5.70 4.82 2.20 1.59 1.22 1.05 1.01 1.00 1.00

Ware

1.00 1.025 1.209 1.421 1.845 2.661 3.181 3.666 (4.06)

(Yae

1.00 1.01 1.16 139 1.84 2.64 3.18 5.5 4.06

It can be seen that the agreement between the experimental activity coefficients, (y;)., and those calculated, (y;),, is satisfactory for most purposes. Azeotropes The ethanol (1)—benzene (2) system exhibits a common nonideal phenomenon: an azeotrope. This is a maximum or minimum in the boiling curve combined with equal compositions of the vapor and liquid. At 1 atm, this system has an azeotrope at 44.8 mol% ethanol (x = y) boiling at 68.24°C, as shown in Fig. 13.13. At mole fractions below this value, ethanol is the more volatile component, with y values greater than x; at higher mole fractions, the reverse is true. The Gibbs-Duhem equation in terms of activity coefficients is:

d In

01

(22) OX,

+ (2%) /pp

Ox,

=0

(13.55)

/pp

If we look at a binary system at low pressure, where the vapor acts ideally, all 6 = 1. Therefore,

Vi

aei

(13.54) .

Then, eq. (13.55) becomes:

YP + x) dIn (225) P = 0 x, dn (2) xP;

(13.56)

which we can expand to:

x, dln (y,P) — x, ad In x, +x; d In (y,P) — x, dina But the second and fourth terms cancel:

=0

(13.57)

13.5

Determination of Activity Coefficients

381

°C Temperature,

O

0.2

0.4

0.6

0.8

1.0

x,y

Figure 13.13 Experimental and calculated VLE for the system: ethanol-benzene at 1.0 atm. Fit is with the van Laar equation.

dx sh 35ON

—dx reed

(13.58)

leaving:

x, d|In (Q,P) + x, dln QP) = 0

(13.59)

(@@, + x) dinP + dy, + 2 dy, = 0

(13.60)

This can be expanded to:

v1

and, since (x, + x,) = 1, and dy, =

2

—dy,, eq. (13.60) becomes:

ain Pg, = dy, = 0 y1

(13.61)

2

which, when solved for d In P is: Binip cco

2

(13.62)

382

Chapter 13 Real Liquid Mixtures Finally, rearranging eq. (13.62):

GES gin a) dy, iy y,d

—

(13.63)

y)

we see that measurements of dP/dy, P, and either y or x (at constant 7) can be used to obtain the other phase composition (x or y), with eq. (13.63). This is an alternative to

measuring both x and y, in addition to P and T. This technique can be useful when it is easier to measure y by, say, vapor density measurements, than to measure x. Equation (13.63) also tells us that when x, = y, (azeotrope), dp/dy = 0; we must be at a maximum or minimum in the P—x-y curve. The inverse is also true: when at an extremum in the P—x—y curve we must be at an azeotrope. The mixture at the azeotropic composition acts thermodynamically just like a third component, even though there is no physical or chemical evidence of its existence. The phase rule predicts there will be only two degrees of freedom with two components. In fact, there are three degrees of freedom; specifying T and P within the phase envelope can lead to two separate sets of x—y equilibria, on either side of the azeotrope. This is consistent with C = 3. Furthermore, the logarithm of the azeotropic pressure varies nearly linearly with 1/7, just like a pure component vapor pressure as given by the Clausius-Clapeyron equation. There are then two VLE phase diagrams connected at the azeotropic composition: one between ethanol and the azeotrope, another between the azeotrope and benzene.

Azeotropes are found with many systems and can be either maximum or minimum boiling. Examples of each are shown in Fig. 13.14.* One of the advantages of the van Laar form of activity coefficient equation is that it reproduces azeotropic phase diagrams reasonably well. In fact, the azeotropic point can be used as a way to quickly evaluate the van Laar constants. We can rearrange eqs. (13:52)ito: In

?)

A = In af isee] x, In y, B=In ve

1 in

(13.64a) 2

ms‘Ie ] xX,

Xy In 2

(13.64b)

where the activity coefficients are determined from eq. (13.54) with each x = y, so that:

vee

P

i

only at an azeotrope.

At 68.3°C, the temperature of the azeotrope, the vapor pressures of ethanol and benzene are 506 and 517 mmHg, respectively. In this way we get:

*M. A. Rosanoff and C. W. Easley, Jour. Amer. Chem. Soc., 31, 953 (1909).

13.5

Gee

Rue

en Geeumdrae

Determination of Activity Coefficients

1.0

(in

O 2a

383

ae OiOmanO. at) EKO

x2 mole fraction CS»

Xg mole fraction chloroform

(a)

(b)

Figure 13.14 Examples of maximum and minimum-boiling azeotropes: (a) carbon disulfideacetone at 1.0 atm, (b) chloroform-acetone at 1.0 atm.

x, = 0.448

y¥, = 760/506

=

1.502

In y, = 0.407

x, = 0.552

Y. = 760/517

=

1.470

In y, = 0.385

Therefore, from eqs. (13.64a) and (13.64b):

A =

1.909

B =

1.329

These constants are somewhat different from those obtained from the extrapolation to infinite dilution in Section 13.5.2.1, and they will be different from those found by fitting all the data. However, no analytic form will fit experimental data perfectly; any of these sets is far better than use of Raoult’s law, which can never predict an azeotrope. Furthermore, there is some compensation of error between sets of A and B, so that the

activity coefficients calculated from the two different sets are actually close to the experimental values. For example, take the point at 0.8 mole fraction ethanol. Equations (13.52) can be solved using each of the two sets of A and B. First, with the infinite dilution values:

1. A = 1.74, B = 1.40, y, = 1.050, y, = 2.639 Then with the azeotrope values:

2. A = 1.909, B = 1.329, y, = 1.043, y. = 2.623 Both sets of constants yield activity coefficients acceptably close to the true values of 1.05 and 2.661, respectively.

384

Chapter 13 Real Liquid Mixtures Best Fit of all Data

Often it will be necessary to ‘‘fit’’ a number of data points to a specific equation that is nonlinear in form. The example we use here is the fit of experimental activity coefficient data to the three-suffix Margules equation, eq. (13.47). Let us say we have 27 sets of x, y, P data and we wish to find the constants, A and B, in the three-suffix

Margules equation as well as see how well it fits the data. If we can neglect the vapor-phase nonideality, a good assumption near atmospheric pressure, the experimental gammas (referred to in the program to follow as GIEX and G2EX) are found from each datum as: 1,

=

Py,

Pix, ie

ji

Py,

eee

The calculated gammas (G1CAL and G2CAL) are found from eqs. (13.47) using constant values of A and B with each experimental value of x.

In fitting data to equations by least-squares analysis, the fundamental criteria are those that minimize the difference between the experimental and calculated independent variable at all values of the dependent variable. In simpler terms, it means finding those values of the parameters in the fitting equation, such as A and B in eq. (13.47), that

minimize the error in the fitting equation. In mathematical terms, this is written:

o|Sem y;,, — In 1 mA

| Sem Nee 0B

| 3 = 0

: (13.65a)

II =>

(13.65b)

| A

where the subscript “‘c’’ means calculated, and ‘‘e’’ means experimental. These rela-

tions ask: In changing the values of A and B independently, when does the sum of difference between calculated and experimental in y at all i data come to a minimum? With linear fitting equations, the above two relations give rise to explicit solutions for the “‘best’ values of A and B.* However, when the fitting equations are nonlinear, such

as our eqs. (13.47), explicit solutions are not arrived at. Instead, complex relations implicit in A and B, result. These make analytical solution for the ‘‘best’’ parameters impossible; instead, it is necessary to use numerical means. The IMSL® program titled UNLSF modifies A and B until the sum of the squares of the differences between these experimental and calculated activity coefficients, for

*See any text on linear least-squares analysis.

13.5

Determination of Activity Coefficients

385

both components, is at a minimum. In much the same way as NEQNF works, a subroutine we write (called FUNC) generates the calculated activity coefficients for each

datum with each set of A and B supplied by UNLSF. In the program outlined below, many of the variables are internal in UNLSF and must be dimensioned, but they are never used by us. There are M sets of data input and N parameters to be found. The program calls these parameters X(1) through X(N). In our example, N = 2; X(1) is A and X(2) is B. Here is the strategy involved: I. Set up integer and real variables. Specify number of parameters, N, and number of data points, M. M must be larger than N. Put variables in common blocks. Identify external subprograms. II. Read in data, initial guesses of parameters A and B.

Ii. Call UNLSF. This program will, internally, call FUNC to test various values of A and B for goodness of fit. IV. The subroutine FUNC, written by us, must generate the difference between the calculated and experimental activity coefficients at each datum. Here, whatever equation we choose can be used: van Laar’s, Margules’, etc.

An example of a program using UNLSF is given below. Here the three-suffix Margules equation (two parameters) is used to regress 27 data points for the mixture ethanol (1)-chloroform (2) at one temperature.* In addition, a FORTRAN program titled, ‘“GETAS.FOR”’ is included in the disc enclosed with this text. This program will determine from experimental data the best

coefficients in three activity coefficient equations: the Margules, van Laar and Wilson.

Find the best parameters in the three-suffix Margules equation from the following set of 27 P—x-y data at 35°C:*

Ethanol (1)—Chloroform (2), 35°C xy

J1

0.0 0.0384 0.0400 0.0414 0.0440 0.0685 O15) 01577 0.1735 0.2254 0.3217

0.0 0.0586 0.0597 0.0615 0.0637 0.0839 O27 0.1248 0.1302 0.1446 0.1673

P, mmHg

295.11 303.91 303.69 304.17 304.87 306.05 306.25 305.12 305.39 303.05 296.93

*G. Scatchard and C. L. Raymond, J. Am. Chem. Soc., 60, 1278 (1938).

386

Chapter 13 Real Liquid Mixtures Ethanol (1)—Chloroform (2), 35°C (continued)

is

yy

P, mmHg

0.3815 0.5154 0.5173 0.5616 0.6078 0.6155 0.6773 0.6986 O77 127 0.7639 0.8270 0.8891 0.9406 0.9458 0.9703 0.9759 0.9938 1.0

0.1819 0.2188 0.2203 0.2354 0.2588 0.2630 0.2991 0.3130 0.3253 0.3793 0.4696 0.6115 0.7657 0.7846 0.8790 0.9009 0.9746 1.0

DOVWOS 274.46 274.04 267.65 255.28 29399 236.50 229.24 225.06 205.68 177.60 148.26 125.82 123.45 113.61 111.31 104.87 102.78

The program, using these data and the routine UNLSF, is reproduced below. The results are printed out afterwards. program

main

integer ldfjac, m, n parameter (ldfjac=27,m=27,n=2) integer

iparam(6)

real fjac(ldfjac,n), fscale(m), x(n), xguess(n), xscale(n)

common

/pres/

common,

/gammay/maigisa(27

Gommom

oily

datanr

fvec(m),

rparam(7),

pcall(27),pcal2 (27) \rig2s (27) .qlic(2i))c2Zce(27))

xl (QW),

wi(2Q7),

jorcess (27)

ie canker. 2ieales0Y/

external

func,

unlsf

epeni (unit =, mamen= Yexpsdats, type=‘old’ ) read(1,*), (l(a )y, ey ine epress Galen p=) 7) print*,’ Type in guesses for read*, xguess(1), xguess(2)

a

and

b

:

'

13.5

call

unlsf(func,m,n,xguess,xscale,

fscale,iparam, rparam,x,fvec,fjac,1ldfjac)

Calle

£une (man, sc.t)

point*,; / Doimt*, ”

area, - xi) (aj) 5 oaarma 41°) 2) iparan(s) aterations=. 9"? iparamis) of functwons eval=

# of

Preinte*),

Peane, Gow 10

Determination of Activity Coefficients

0)

ky =sele

WiElee

am

(6, %), Salk)

glix( kK)

egle(k))-

g2x(k) , g2c(k)

stop end

subroutine func(m,n,x,£) integer m,n common /gamma/glx(27),g2x(27),gl¢e(27) ,g2c (27) common /pres/ pcall(27),pcal2 (27) common, /blkil/tx 0, In yj > 0

(13.82)

so, using eq. (13.78),

lniy; = In y, — in-yy = In y,

A

(13.83)

and combining this with eq. (13.80), we get a simple, thermodynamically valid expression for the asymmetric activity coefficient:

In yi = AGZ.— 1)

(13.84a)

398

Chapter 13 Real Liquid Mixtures In practice, the best value for A is found from:

tt _ py

(13.85)

x}

where the fugacity coefficient, ¢, in the vapor may be appreciable when extending into higher pressures. After taking the logarithms of both sides of eq. (13.85),

In (224) = Ink, + AGZ — 1)

(13.86)

Xx)

data in the form of liquid and vapor mole fractions, as well as the fugacity coefficient in the vapor (from correlations) are regressed to yield A. The data shown in Fig. 13.19, for carbon dioxide in water at 50°C, when treated in this way, yield values of:

k, = 2894. bar

A=

— 7.932

The straight line is of slope k,; with no activity-coefficient correction, there is considerable error at the higher pressures. However, with the parameters above the fit is within experimental error. To find the form of the activity coefficient of the solvent when the solute is described by eq. (13.84a), we use the Gibbs-Duhem equation in the form:

CO» bar fugacity,

0)

2

4

6

8

Mole fraction CO» in water x 103 Figure 13.19 Solubility of CO, in water.

10

12

13.8

Immiscibility

399

We note that, from eq. (13.83),

Iny, =Inyi +A so that:

alin,

= d tncy,

and the Gibbs-Duhem equation becomes:

xy

diny; nV

+ 2X aN n Y2

dx,

dx,

and, differentiating eq. (13.84a) with x,,

din yi

SS

dx,

SSN

2

Thus, the Gibbs-Duhem equation solves for the solvent term:

dln y>

re

= 2Ax,x,/x, = 2Ax,

which we can integrate from x, = 0 to x: x1

ny

- 0 = 24| AN peAA:

0

where, at the lower limit,

x, = 0

x, = 1.0

yy.= 1.0

In y, = 0

Therefore the form for the solvent,

In y, = Ax?

(13.84b)

is consistent with eq. (13.84a) for the solute.

13.8

IMMISCIBILITY As we have seen with ideal mixtures, the free energy is minimized by the mixing process. The ideal free energy of mixing for a binary system is given by eq. (1 £425), rewritten here on a molar basis:

Ne

hon IX, 4 Xx, I xX,

(13.87)

400

Chapter 13 Real Liquid Mixtures

Figure 13.20 Gibbs free energy on mixing for the ideal mixture [eq.(13-87)].

This equation is plotted as Fig. 13.20. Note that any mixtures formed from pure components, or from other mixtures of the components, will always lie between the components (or prior mixtures), and so, from the shape of the curve, will always have

negative changes in free energy on mixing.

13.8.1

Partial Miscibility Real mixtures often have unlike (1-2) interactions that are more energetic, in the repulsive sense, than like interactions (1-1, 2—2). The energy required to maintain these unlike pairings in solution can be so great that the entropy increase on mixing [which, eq. (11.126), is the entire cause of the ideal free energy change] is overcome. Large size or shape differences between components, such as a polymer with a small solvent molecule, can also produce large negative entropy changes on mixing because of the difficulty such mixtures have in spatial accommodation. In either case, if complete mixing were to produce a system with a free energy more positive than one with two separate liquid phases of different compositions, this phase separation will occur. An example of such a free energy diagram is shown as Fig. 13.21. Here a mixture of composition c will split into phases of compositions a and b, whose total Ag@/RT lies at c’, which is lower than the single phase, at c’. The two equilibrium phases, whose compositions lie on the tangent to the curve, have a total free energy lower than any other possible combination of phases. The sizes of the phases are set by the lever rule between a—c and c-b. Thermodynamic description of phase separation is aided by use of the activity, a;, defined in Chapter 11, where:

and, since:

AgM = Ag’M + g®

(13.89)

13.8

Immiscibility

401

Mole fraction

Figure 13.21 Gibbs free energy on mixing with liquid—liquid phase separation.

and the equation for the excess Gibbs energy on a molar basis is:

g® = RTDx,In y,

(13.90)

using eq. (13.87) for Ag’, eq. (13.89) is:

Ngee

GO ye RIN

(13.91)

In the two-phase region, the activity of each component in each phase will remain the same, independent of the size of the phases; the activity of each component will also be equal in the two phases:

fi =f" (TE

(13.92a) OR

a, =a;

ok

(13.92b)

Thus, the partial pressure of each component will be fixed in the two-phase region, because:

a; = yx, = 7

(13.93)

where we have used the partial pressure as the vapor fugacity; this is the ideal vapor mixture assumption, good at low pressures.

402

Chapter 13 Real Liquid Mixtures

One liquid phase

Two liquid phases

Figure 13.22 Liquid—liquid phase diagram for a system with an upper critical solution temperature (UCST).

Figure 13.22 illustrates the liquid—liquid phase diagram for a partially miscible binary system. This is the most common type of T7—x phase diagram, one that shows a UCST, an upper critical solution temperature. That is, as the temperature rises, the region of partial miscibility shrinks until it disappears altogether at the UCST. At a temperature below the UCST, say at T on the diagram, the two phases marked x’ and x” are in equilibrium. Any mixture made up between these two compositions will split into these two phases. The activities of the components, as we have seen, will be constant between x’ and

x"; only the sizes of the phases will change as we move in between these limits. Figure 13.23 demonstrates graphically how the activity of each component varies across the

Activity

Mole fraction

Figure 13.23 Activities in a binary liquid system with limited miscibility.

13.8

Immiscibility

403

Activity

+

Figure 13.24 Activity prediction for partially-miscible solutions.

entire mole fraction range. Thus there exists a region of constant partial pressure of each component across this range of mole fraction. Not all empirical equations for activity coefficients are able to predict liquid—liquid phase separation. The primary criterion for partial miscibility is a maximum in the free energy of mixing curve, Fig. 13.21. For this to occur the predicted individual activities must also show maxima, as shown in Fig. 13.24. As noted earlier, the actual activity

and free energy curves are straight and horizontal in the region of immiscibility, so predicted activities, such as shown in Fig. 13.24, have a hypothetical region between points b and c on the diagram; the actual behavior must lie along the dashed line, as shown in Fig. 13.24. At a pressure low enough that a vapor phase is in equilibrium with the liquids, behavior like that shown in Fig. 13.25 is seen. At temperatures below the dashed line in Fig. 13.25, there is no vapor phase. The liquid phases in the central portion of the mole fraction range are composed of varying amounts of I and II; the amounts in each phase are determined by the lever rule. To the left and right of the two-phase region in this temperature range, the liquids are miscible. The minimum temperature for a vapor phase is that corresponding to the dashed line; the only vapor composition is M, and it is in equilibrium with liquid phases of compositions a and B. At higher temperatures, several possibilities exist. Starting at the left ordinate and moving to the right, the liquids are miscible until we reach the left vapor—liquid region. In this area, there is vapor—liquid I equilibrium along a horizontal line connecting the phase boundaries. From the right ordinate, moving left, the liquids are miscible until the phase boundary is reached; from there on the vapor—liquid II phases are in equilib-

rium.

13.8.2

Total Immiscibility Mixtures of extremely unlike materials, such as n-heptane—water, form two distinct phases, each more than 99.9% pure. Although total immiscibility is theoretically im-

404

Chapter 13 Real Liquid Mixtures

Vapor

Liquid I+ vapor

_/

Liquid 1 of

Temperature

Liquid +0.

XA Liquid—liquid vapor-phase diagram for partially-miscible liquids.

Figure 13.25 Liquid—liquid—vapor equilibrium (LLVE) diagram for partially miscible systems.

possible, because there must be a sufficient number of molecules of each component to achieve equal activity with the nearly pure phase, for our purposes this assumption is valid when the phases are more than 99% pure. The phase diagram of temperature vs. mole fraction appears as in Fig. 13.26. This type diagram is reached when points a and B on Fig. 13.25 move horizontally toward the ordinates, as the components become more and more mutually insoluble. At some point the liquid phase boundaries essentially merge with the pure-component ordinates, as in Fig. 13.26. The minimum boiling temperature at the total pressure for which this diagram is drawn, at point M, separates the region of two “‘pure’’ liquid phases from regions where one or the other liquid is in equilibrium with vapor of varying composition. The curved lines meet the ordinates at the boiling temperatures of the pure components at the system total pressure. Any horizontal line drawn in the region above M, but below the ends of the two-phase regions, will indicate the vapor composition in equilibrium with either pure liquid 2, on the left ordinate, or pure 1, on the right. If, in addition, the vapor in equilibrium is nearly ideal,

Vie = WP

YoX2P2 = yoP

-

(13.94a)

(13.94b)

and the two partial pressures add up to the total pressure. When we consider that both phases are pure, both x’s and both y’s are unity, so,

P = Putar,

(13.95)

13.8

Immiscibility

405

*11Y1

Figure 13.26 Liquid—liquid—vapor equilibrium (LLVE) diagram for totally miscible systems.

and, along the curve jig — M,

ne

Ie

Yi cag

(13.96)

where 7’; is the temperature at which P; = P, whereas along the curve T,; — M,

YP = P;

Bate ts

(13.97)

This behavior is useful when a thermally unstable organic needs to be distilled. From eq. (13.95) we see that 1 atm pressure boiling can be attained well below the normal boiling point of the organic, say component 1, by adding steam. Take, as an example, indene (C)Hg), which normally boils (at 1 atm) at 181.6°C. We may need to distill this indene out of a mixture with other, much higher boiling hydrocarbon contaminants. To find at what temperature steam distillation will vaporize indene, we need to find the temperature at which eq. (13.95) adds up to 1 atm. The vapor pressure of indene in this temperature range is well fitted by:

In P°(mmHg)

=

—5405./7T(K) + 18.53

By trial and error, using the steam tables, we find that at 97.5°C, where indene has a

vapor pressure of 52 mmHg, water has a vapor pressure of about 707.5 mmHg, so that steam added to liquid indene at 1 atm will cause boiling at about 97.5°C. The resulting

406

Chapter 13 Real Liquid Mixtures 181.6

1.0 XxIndene

Figure 13.27 Liquid—liquid—vapor equilibrium (LLVE) diagram for water-indene at | atm.

mixture of water and liquid indene will produce a vapor with 52/760 = 6.8% indene (plus 93.2% steam), which when condensed, is easily separated from the water condensed with it. This vapor composition will remain constant until the organic liquid phase is completely distilled away. Point M on Figure 13.23 will lie only 6.8% off the water ordinate, as shown in Fig. 13.27.

13.8.3.

Quantitative Analysis The onset of liquid—liquid phase separation is marked by the appearance of an inflection point in the free energy of mixing curve; a totally miscible mixture will exhibit a curve something like Fig. 13.20, whereas that of a partially miscible mixture will look like Fig. 13.21. The exact representation of this behavior is complex; we will see some examples in the next chapter. Here we will just describe the simplest type of liquid— liquid equilibrium: a binary system showing an upper critical solution temperature (UCST). Precisely at the UCST in Fig. 13.22, the curve of free energy on mixing vs. mole fraction will show a flat spot right at the composition where phase separation appears, as shown in Fig. 13.28. As the temperature is lowered, the curve will rise further, and two equilibrium liquid phases will appear. This is mathematically equivalent* to

?AgM Ae

= 0

:

(13.98)

and at temperatures below the UCST, where two phases of unequal composition are in equilibrium, such as in Fig. 13.13,

*The partial differentiation means that temperature and pressure are being held constant. In differentiating with respect to x, in a binary system, it is impossible to hold x, constant; these are not shown as partials.

13.8

2A

( g

M

Immiscibility

ra|+ a 4]= rn) xy

X4

arly

(13.108)

13.9

Advanced Correlations—The Unifac Method

409

The largest value of the product x x, occurs at x = 0.5; this will produce the smallest value of the parameter A that will predict phase splitting: A = 2RT

(13.109)

Larger values of A correspond to mixtures with high repulsive forces and positive deviations from Raoult’s law: partial pressures in excess of those in an ideal solution. The activity coefficients generated from large positive values of A (y >> 1) reflect these repulsive forces; when they become strong enough, the mixture ‘‘prefers’’ to exist as two liquid phases. That is, the total free energy of such a two—liquid phase system is lower than that of one liquid phase.

13.9

ADVANCED

CORRELATIONS—THE

UNIFAC METHOD

Newer correlations for activity coefficients utilize the contributions of interactions between functional groups, such as —CH,, =O, —OH, etc., rather than those between entire molecules, as with the Wilson equation. In this way, activity coefficients can be

predicted for mixtures for which no actual data exist, by ‘‘assembling’’ the pure components from the individual groups and assessing the contributions of their interactions. The most successful of these is the UNIFAC* (universal functional activity coefficient) method, one that is quite intricate but simple to apply to computer use. We have seen that, at low pressures, the overwhelming

nonideal contribution to

vapor—liquid equilibria comes from the liquid phase. Calculation of the liquid-phase nonideality using activity coefficients from one of the equations shown so far has been relatively simple; but the determination of the parameters in these equations depends on having data for, at least, all possible binaries (in a multicomponent mixture).

Determining characteristic parameters for the hundred or so individual components of most industrial interest requires an extensive, but achievable, database. However,

when considering all possible binaries from these components, the database becomes practically unachievable. If we look at the compounds of interest, we see that they are composed of many of the same groups of constituent atoms. For example, all the hydrocarbons have CH;—, CH=, CH=, or =C= groups. Other groupings might include —OH, O=, —Cl, etc. For 70 years or so, it was recognized that a more reasonable way to correlate liquidphase activity coefficients would be to base them on the way these groups, rather than the molecules, interacted. In this way the number of empirical parameters would be drastically reduced. Even so, the task was too formidable until fast and easy computation became available. This is because huge amounts of data had to be regressed to find the ‘‘best’’ form for the equation as well as the best parameters to use for each group. Although several other attempts have been made, UNIFAC has caught on as the accepted method worldwide. A large database has been developed for the molecular and group parameters needed to use this method, and whereas the equations appear

*A detailed description of the basis and use of this method can be found in A. Fredenslund, J. Gmehling, P.

Rasmussen, Vapor-Liquid Equilibrium Using UNIFAC, New York: Elsevier, 1977.

410

Chapter 13 Real Liquid Mixtures

formidable, computation is direct (no iteration) and simple even with just a hand calculator. We will here go through a very simplified description of the method along with an example of its use. More detailed treatments can be found elsewhere.* Activity coefficients are derived from the excess free energy, as we have seen:

and, on a molar basis, the excess free energy is made up of energetic as well as entropic contributions:

This is important to the development of a theoretical framework because the entropic part is not describable by merely considering the interactions of groups. The way the groups are bound in the molecules determines their freedom to move in solution; a CH,— group on ethane is more mobile than one on the end of a decane molecule. This difference in freedom of movement is seen macroscopically in the nonideality of mixtures that are chemically very similar (same groups) but are very different in the size and shapes of the component molecules. In these solutions the enthalpy of mixing can be near zero while the excess Gibbs energy is highly positive. This means that the excess entropy on mixing is negative, a deviation from random mixing caused by size and shape difference of the molecules in solution. The UNIFAC method begins with the splitting of the excess Gibbs energy into a “‘combinatorial’’ part, superscript C, and a ‘‘residual’’ part, superscript R:

ge = gi + gk

(13.110)

The combinatorial part is based on molecular parameters that are developed from the individual groups but do not have any interaction terms. That is, they depend only on the size and shape of the molecule as constructed from the groups. The residual part reflects the difference in interaction energies between the different groups. The combinatorial part derives from a measure of the relative surface areas and volumes of the constituent molecules, as arrived at from independent spectroscopic data: Cc

ae

ob;

go = Da; ln i=1

Xj

z

us

SD gyn 2 i=1

0;

;

-

(13.111)

where the volume fraction, ¢, and surface area fractions, 0, are defined, respectively, by:

*A. Fredenslund, J. Gmehling, and P. Rasmussen, op. cit.

13.9

Advanced Correlations—The Unifac Method

b, =z

411

(13.112)

2)

and

ae

(13.113)

2%

In these equations, the molecular volume, Vp» is defined by the sum of its constituent groups as: N

y= Dr

R

(13.114)

where v4, is the number of k groups in molecule j, and R, is the volume of group k. The molecular surface area, q;, is found by summing the individual group areas in the molecule:

(13.115)

qe = Dvh-

where Q, is the group surface area. Values of R and Q for some selected groups and subgroups are reproduced in Table 13.1. For the complete list see the book by Fredenslund et al., already noted. The residual part of the free energy in eq. (13.110) is represented by the sum of all the group interactions among all the molecular species (i = 1 — M) in solution: M

on rae

0 Dd ax

i=1

M

>» 9; é vs) In (

j=1

(13.116)

where the group energy interaction parameters, w,,, are defined by:

We A large number of these mined; a small sample is only on the main groups Now the form for the ferentiation to:

Cane

a; * a;

(13.117)

binary interaction constants, a;,, have been empirically detergiven in Table 13.2. Notice that the group interactions depend involved; the subgroups do not have individual a;;. activity coefficients is obtained from the free energy by dif-

In y, = In yf + In y®

(13.118)

where the first two parts follow the definitions in eqs. (13.111) and (13.116). First, the configurational is:

412

Chapter 13 Real Liquid Mixtures Table 13.1

Group Surface Areas and Volumes Numbers of Bonds

Group

| Subgroup

CHx CH, CH

C=C

CH=CH CH=C CH,=C

COOH | CH,CH,OH CHOHCH, CHOHCH, CH,CH,OH CHCH,OH_

oon |e

of Groups

ee

0.9011 0.6744

Butane

2CH;,2CH,

0.4469 0.2195

2-Methylpropane 2,2-Dimethylpropane

3CH;, 1CH 4CH;, 1C

1.3454

1-Hexene

| 1.1167 0.8886 1.1173

| 1.8788 | 1.878 | 1.6513 | 2.1055 | 1.6513

1CH;, 3CH,, 1CH,—=CH

2-Hexene 2-Methy]-2-butene 2-Methy]-1-butene

2CH,, 2CH,, 1CH=CH 3CH;, 1CH=C 2CH;, 1CH,, 1CH,=C

1-Propanol 2-Butanol 3-Octanol Ethanol 2-Methyl-1-Propanol

1CH;, 1CH,CH,OH 1CH;, 1CH,, 1CHOHCH, 2CH;, 4CH,, ICHOHCH, 1CH,CH,OH 2CH;, 1CHCH,OH

CH,CO |CH,CO CH,CO

1.6724 1.4457

2-Butanone 3-Pentanone

1CH;, 1CH5, 1CH,;CO 2CH;, 1CH,, 1CH,CO

CH,OH | CH,0H

1.4311

Methanol

1CH,0H

Water

1H,O

H,O

H,O

0.92

COOH

| COOH HCOOH

1.3013 Po2s

Acetic acid Formic acid

1CH;, 1COOH 1HCOOH

CELA!

|CERCE CHCI, Cals

2.2564 2.0606 1.8016

Dichloromethane 1,1-Dichloroethane

1CH,Cl, 1CH;, 1CHCI, 2CHeueC

CH,Cl CHE! CCl

1.4654 £238 OF791

e€Ci

2,2-Dichloropropane

1CH;, 2CH,, 1CH,Cl 2CH,, 1CHCI 3CH;, 1CCl

1-Chlorobutane 2-Chloropropane 2-Chloro2-methylpropane

é

0;

Inyy = In

+ ate Pos +1,+ i

i

o; >
; 1220.1

The figure produced from these data is shown in Fig. 14.6. The program listing is given in Appendix B, following this chapter.

14.2.6

Adiabatic Flash When a liquid mixture is flashed, that is, reduced in pressure by flowing through a throttling valve, the exit temperature is not generally known. In this case the energy balance must be used in conjunction with the mass balance and equilibrium criteria. As we have seen, the energy balance for steady-state flow through a throttle is: Hy =5Ho

For 1 mol of entering liquid, with mole fraction, z,, of each component, this is:

hy = (hi + 2hb) = ho = WLOyht + xh) + Vinh + yohb)lo

4-15)

440

Chapter 14 Phase Equilibrium—Nonideal

- —

@

Calculated.

Exp. (0.508 bar)

KTemperature,

Mole fraction 2-propanol

Figure 14.6 Experimental (points) and calculated (curve) VLE for the

system: 2propanol-water at 0.508 bar. Wilson parameters from literature.

with the terms:

hi = hi + CLT — T°)

h? = hi + An? b= Oe V where T°, the reference temperature, is conveniently set equal to the inlet temperature, T;. This defines the inlet enthalpy as zero, if the pressure effect on the liquid enthalpy can be neglected, generally a good assumption below 20 bars. Equation (14.15) is applied, with eqs. (14.14) and (14.13) for each component, in

addition to the mass balance:

Dx — 1.0 =0 giving us four equations (for the two-component system) and four unknowns: V (or L), X;, X, and T. Equation (14.13) requires an activity coefficient representation that is temperature dependent. Although the UNIFAC (see Section 13.9) equations for the activity coefficients give the best predictive representation of temperature-dependent equilibria, and the Wilson equation could be used, a somewhat similar form is also applicable. This is the non-

14.2

Lower Pressures

441

random two liquid (NRTL) equation,* which for a binary mixture gives the following for the activity coefficients:

In y,

In

ea

=

ve

2

INE21

‘eo

x

=75 a

eee

ral

Tale 12/\12

a

2

IN Ee

14.1

(x. + x,Ajo)

Ay

a ao

Tsu

pee 3tee)

(x, + X,N5,)"

(

se

14.1

oy

with the definitions:

Ke

ef %iTi)

where the parameters A;, and a,; must be supplied from a fit of data. These are available for a large number of mixtures in the compilation by Gmehling.* The four equations are best solved using NEQNF or equivalent routine.

A mixture of 0.05 mol% methyl formate (1) in methanol (2) is produced in a pressurized reactor at 110°C, 20 bar. To perform a preliminary separation of the products, this liquid is throttled down to 1 bar through an adiabatic valve. Calculate the equilibrium compositions, phase distribution (L and V), and temperature.

SOLUTION The literature shows the following NRTL parameters for the mixture;

Ay = 633.9246 cal/g- mol

A,, = 313.3162 cal/g- mol

aj. = a, = 0.2960

and for the pure components;

Ch, = 3.30 cal/g - mol - K Ch, = 2.08 cal/g - mol - K Ah® = 6708. cal/g - mol

Ah} = 8430. cal/g - mol

*H. Renon and J. M. Prausnitz, AIChE J., 70, 1785 (1968). The NRTL equation is very similar to Wilson’s

but has a third empirical parameter, a, the nonrandomness factor, which allows for the fitting of highly nonideal mixtures.

tJ. Gmehling, U. Onken, and J. R. Rarey-Nies, Vapor—Liquid Equilibrium Data Collection, Frankfort/Main: DECHEMA, 1988.

442

Chapter 14 Phase Equilibrium—Nonideal

We can now write a program FLSH with four equations and four unknowns, V, x), x2, and temperature, and then let NEQNF solve them. The four functions to be minimized are shown in bold:

program main ‘‘*flsh’’ calcs.

flash

of

MeF-MeOH

parameter (n=4) real errel, ingl j;lng?/1 j/k1

kero,

g270l2;, get

integer k, nout real fnorm, x(n), xguess(n), f£f(n) external func, neqnf Ceinnoyal/Clejeay/ il pw, Cll, G2, sail pRB, WAL, WA Ep BL AS GOmmoni/ Oui sch le ect Ap echo cna errel=0.0001 itmax=100 xguess(1)=0.2 xguess(2)=0.2 xguess(3)=0.8 xguess(4)= 350. call negnf(func,errel,n,itmax,xguess,x, fnorm) print*, ‘Flash of MeF-MeOH’

10

jopeabake jonsaliaess, jorcaliave, jonealiMent, joreshiate*, continue

baile Allyn Seeee fo) SUS pea Siler ee “Wee, Ay, “Migs? al |See Sea AV ie? Stil, stD, ses, seed!

Sil,

Sal

stop

end

subroutine func(x,f,n) heraell , inv, dlsave A, il, Vell eA), jill, GO ,GulLea- oil x(n) -e (nk, neo), mco2mnit smo Cleiiinery/CeNcey/ i Wy Gib GA, Sll SB Wl, WA ye, lk pid common/out/xf1,xf2,xf3,xf4 EDQHIES . =i), QE 1eeyelh neal

Methyl

inlet

formate

mole

=

1,

fractions

methyl

alcohol

=

2

14.2

Lower Pressures

N=) OS I= AN (OR abaal NRTL

parameters

alZ=633). 9246 EVANS Sh13}, Shale alp=0.2960 enthalpies dh1=6708. dh2=8430.

of

vaporization

and

heat

capacities

COlL=s) 3) 0 OZ —Aj0eS

Callenmactivil ty Gocktr1erents el ASel A esi Cala iy eyie gl2=exp (-alp*t12) g21=exp(-alp*t21) Tao Alp: rtSORE (GOP hoe (oR (bielba: Dire PAM)) Vetta (t12*g12/ (x2+x1*g12) **2) ) Abc

— Slee

a (Gtedlen Malin) (Ex? tae A call) \y)) De

GEZais GZ desler Dil) *2 )i)

gl=exp (l1ng1) g2=exp (lng2) Antoine equations

for

methyl

plo=exp

(10.718-3257./t)

p20=exp

(12.7445-4301.17/t)

VLE

formate

and

constants

l= \oiloy/is kK2=02*p20/p WAl=Sleilesoril WDM OD stream enthalpies lod k—=eqoyil-2 (ie S110) dz il=CoAs (ie= =O) hiv=hil1+dh1 h2v—=h2i1+dh2 minimization equations £ (1) =x1-21/ ((1.-v) +k1*v) £(2) =x2-22/((1.-v) +k2*v) £ (3) =x1+x2-1.0 £ (4) =1* (x1*h11+x2*h21)+(y1*hlvt+y2*h2v) *v return

end

The results are: S$ run

flsh

Flash

of

VL

MeF-MeOH

Se OOR=02eP=

In. 010

methanol

443

444

Chapter 14 Phase Equilibrium—Nonideal T= 332.2 *7= 2.6LB=02 yi= 10.204 V= 0.135 L= 0.865 X,= 0.974 yo= 0.796 3.1851232E-07 -6.1988831E-06 0.0000000E+00 FORTRAN

-0.1166992

STOP

They show a drop in temperature from feed to exit of some 51°C to 332 K. The vapor phase is 20.4% methyl formate, considerably enriched from the feed (5%). The last line shows the values of f(1) through f(4) at the end of the routine; the small values indicate that convergence has been achieved.

14.2.7

Multicomponent Mixtures The utility of the procedures discussed in the previous sections is seen more clearly when calculations for multicomponent mixtures are needed. The exact same approach is used for each of the above examples; now, however, (n + 1) > 3. Both the Wilson

and the NRTL forms for the activity coefficients are readily extended to multicomponent mixtures; here we use the Wilson.

EXAMPLE 14.7 Find the dew point pressure and liquid composition for a vapor mixture of acetone (1)—methyl acetate (2)—methanol (3) at 323 K. The vapor composition is:

y, = 0.3 yo = 0.3 yz = 04 and the Wilson parameters at 323 K are*:

A,, Ay, Aj, Azo A,3 Az,

= = = = = =

0.5781 1.3654 0.6370 0.4871 0.6917 0.7681

SOLUTION The activity coefficient of component 1 is found from eq. (13.72):

*Tbid.

14.2

=

=== 4 XN) Kk KeNyo + Xghyg.) (XpAgy + x + Nau

4:

Lower Pressures

445

3/31 iegs, 31 be ApNgo 2th3o +

3

with equations for the other two activity coefficients found similarly. We again use NEQNF with the 4 = (n + 1) simultaneous equations: ee

iP!Masog Pi

P YoP> 1? TELS Nice

kovitw Agel

OO

The solution is relatively insensitive to the initial guesses; in this case we use all x = y and P = 0.6 bar to reach the answers:

x, = 0.270 X, = 0.227 x3; = 0.503 P = 0.83 bar

These are within experimental error of the data.*

14.2.8

Fit of Data To Find Parameters In Chapter 13, we fitted experimental VLE data to a specific activity coefficient equation to find the ‘“‘best’’ parameters in that equation (Example 13.4). In that example, we assumed all fugacity coefficients in the vapor phase were unity. Although this assumption is generally good at pressures near or below | bar, more accurate fits require use of calculated fugacity coefficients. At low pressures these fugacity coefficients are calculated from eq. (12.46); these are functions of vapor composition and pressure, as well as temperature. This lengthens the solution somewhat; each calculation of a vaporphase fugacity requires a fugacity coefficient. The overall strategy, however, is identical to that used in Example 13.4: I.

Set up integer and real variables. Specify number of parameters, NV, and number of data points, M. M must be larger than N. Put variables in common blocks. Identify external subprograms.

*W. H. Severns, A. Sesonske, R. H. Perry, R. L. Pigford, AIChE J., 1, 401 (1955).

446

Chapter 14 Phase Equilibrium—Nonideal II. Read in data, initial guesses of parameters A and B. III. Call UNLSF. In this case we have used the double-precision form, DUNLSF, to allow for the greater precision permitted by the use of the calculated fugacity coefficients. This program will, internally, call FUNC to test various values of

IV

A and B for goodness of fit. The subroutine FUNC, written by us, must generate the difference between calculated and experimental activity coefficient at each datum. Here, whatever equation we choose can be used: van Laar’s, Margules’, etc. Within FUNC, the

fugacity coefficients are calculated from eq. (12.46), using the current values of y,, P, and T. In Example 14.8, the Wilson equation is used*; the program is called GETAS.FOR and can be accessed in the disk accompanying the text. Use of the subroutine DUNLSF requires accessing the IMSL library. The following data is typical input for program GETAS.FOR; it is for 2-propanol—water at 0.508 bar. The complete listing is in Appendix B, following this chapter.

Find the best Wilson parameters for the system: 2-propanol—water at 0.508 bar. Plot the resulting T—x—y diagram.

SOLUTION The following input file is used: Data

file:

com.dat

5087.3, 4/267,,0.6065, 220.00, 0.240, 0.025 64703 (221.270 3447 Dy yO en SS, Oreo 823/829, 20102337252 0350 3) OG IES IL, LUO) SOS p ASS su Ae Data

file:

wexp.dat

0.508 68229, 0. 0610; 0. coo? 68.04,0.0614,0.4732 66.2550 209 94,0 1s Uo 66258, OU Oiio 5 OL amon 66).027, 0,529,007 o208 65.22, Un2 07 OD eou 64.66,0.4010,0.5737 65 457024629 70590 64.15,0.4994,0.6010 64.22,0.5159,0.6069

*The Wilson, Margules or van Laar equation can be selected.

14.2

Lower Pressures

447

64.08,0.5492,0.6172 63 29970 7L490. 7005 64.80,0.8856,0.8448

The output from GETAS.FOR follows; the system is 2-propanol—water at 0.508 bar. The

values

for

A

and

B

are:

962.646 1183 .543 The residuals are:

4.283E-02 8.219E-02 4.647E-02 Ze 29 OR Tye SISBeOS i, 2658-02) J. 8 58h —02

6.139E-02 9.3618-02 3.210E-02 OS 1). 862H=—02 92...6 108-02

These compare well with the Wilson parameters obtained from the literature: 964.3974 and 1220.1455. Figure 14.7 shows the slight difference between the curves generated with the two sets of parameters.

—

\

Wilson parameters, from literature

964.3974 1220.1455

©

Experimental values

_—_

Wilson parameters, from GETAS.FOR 962.6461

1183.5439

K Temperature,

Mole fraction 2-propanol

Figure 14.7 Experimental (points) and calculated (curve) VLE for the system: 2propanol-water at 0.508 bar. Wilson parameters from literature and from fitting data.

448

14.3

Chapter 14 Phase Equilibrium—Nonideal

HIGH PRESSURE At high pressures, say above 20 bar or so, one or more components in a mixture may be above their critical temperatures. The simplest mixtures will have critical temperatures and pressures—that is temperature—pressure points at which vapor and liquid compositions coincide—that lie on a line between the critical points of the pure components. As you can imagine, the situation becomes very complex for all but these most simple systems. For a more comprehensive treatment see the book by Prausnitz et al.*

14.3.1

Qualitative Description Figure 14.8 shows the P—T-—x,y behavior for a simple binary. The critical point of component | lies on the back plane, and that of component 2 on the front plane. The dotted line is the locus of mixture criticals. The ellipses are x—y diagrams similar to those we have seen at low pressure; here they extend beyond the critical point of complement 1. On Fig. 14.9 we see slices at three constant temperatures: the first at a temperature, T,, below the critical of either pure component; the second is above T,,; and the third is above both T,, and T,,. In this case the locus of mixture critical points actually reaches temperatures, such as T;, above the critical of either pure component. The dashed lines are the lines connecting vapor (the lower curve) and liquid (upper) compositions at any pressure at the particular temperature. As with pure component, the vapor and liquid phases become more and more alike as we approach the critical point. The mixture critical points, labeled CP, are at the highest pressure at each temperature, where the liquid and vapor phases become identical. On a P-T diagram (Fig. 14.10) we see liquid—vapor envelopes for three different

\ isa area Pressure

a \

To,

—_—_—_c_cc— >

Temperature

Figure 14.8 P-T-x-y diagram for simple binary mixture.

*J. M. Prausnitz, R. N. Lichtenthaler, and E. G. de Azevedo, Molecular Thermodynamics of Fluid-Phase Equilibria, 2nd ed., Englewood Cliffs, NJ: Prentice-Hall, 1986.

14.3

Pressure

Mole fraction A

Figure 14.9 Constant temperature VLE points of substances A and B.

near and above critical

Figure 14.10 Multicomposition phase diagram: locus of mixture critical points: Z123 Map

mixtures of various compositions phase equilibrium points

High Pressure

449

450

Chapter 14 Phase Equilibrium—Nonideal

Constant-quality lines

Figure 14.11 Phase diagram for a mixture at constant composition.

constant-composition mixtures, z,5 3.At the point of intersection, m,, vapor y, is in equilibrium with liquid x,; at m,, y> is in equilibrium with x3. One of these constant-composition envelopes is expanded in Fig. 14.11. In this particular case the mixture critical point lies to the left (lower temperature) of the maximum pressure; however, it may lie on either side. Notice the lines (dashed lines) of constant

quality (% vapor). Inside the phase envelope there exist separate liquid and vapor compositions in equilibrium with each other. An unusual event occurs when the mixture critical point lies below the maximum temperature on the phase envelope, as shown in Fig. 14.12. A mixture at point A is all vapor; yet, reducing the pressure at constant temperature, as shown by the arrow, brings

Figure 14.12 Retrograde condensation: M is the point of maximum liquid content; A and B are both completely vapor.

14.3

High Pressure

451

Figure 14.13 Azeotropes in the critical region: (a) minimum pressure azeotrope; (b) maximum pressure azeotrope.

about partial condensation until point M is reached. This corresponds with the minimum-quality line broken through with a vertical path. A separation of some of the higher boiling components in a complex mixture such as this liquid can be obtained in this way. An actual situation in which this occurs is natural gas production. The gas mixture emerging from the well at high pressure often exhibits the type of behavior shown in Fig. 14.12. On reduction of the pressure the correct amount, some liquids can be collected; this is important because these liquids are more valuable than the lower-

boiling components. Some of the remaining vapor can then be recompressed and returned to the well, and the process can be repeated. Separation in this manner is more economical than distillation or similar processes. The discussion above encompasses only the simplest type of high-pressure VLE. There can be a minimum- or maximum-boiling azeotrope that extends into the critical region, leading to a critical curve that extends beyond either pure component, as shown in Fig. 14.13. When liquid—liquid immiscibility is present at high pressure, the situation becomes more complex. We will discuss these more complex systems at the end of this chapter.

14.3.2

Mathematical Description At high pressures it is difficult, if not impossible, to describe the liquid phase with eq. (14.2), using activity coefficients with a pure liquid phase as the reference or “‘standard state.’’ Although there are other methods for quantitatively treating phase equilibrium in this region,* it is now common to use an equation of state for both phases, with eq. (11.53) as the equilibrium criterion:

*Ibid.

452

Chapter 14 Phase Equilibrium—Nonideal fi.

(11.53)

oiy,P = ohh

(14.17)

or, in terms of fugacity coefficients,

Solving for the fugacity coefficients from eq. (12.32): RT In () = RT 1n ob; = | E =

iP

(=)

dV — RT In(z)_

(42.32)

an; T,V.nj

=

where V is the total volume of the mixture, with an equation of state that is satisfactory for the system in question. In Chapter 12, we showed how the truncated virial equation, when used with eq. (12.32), gave a relatively simple result for the fugacity coefficient, eq. (12.46). The virial equation, however, is not descriptive of the liquid phase, so, as shown in Chapter 4, a cubic equation of state, at least, must be used. There are many of these, but we will limit our discussion to the Soave-Redlich-Kwong equation, eq. (4.33):

LgRT om

gee)

v—b

viv + b)

Sie)

When this is differentiated and integrated, as indicated in eq. (12.32), eq. (14.18) results:

In by = Ae = b= ne

By ~ $[20% 88) in (1+2) (14.18) a

with the following definitions in the mixture:

P De A = 0.42748 ab al

(14.19a)

B = 0.08664 oe

(14.19b)

incorporating the following mixing rules:

a= Dona,

(14.20)

ay = Vaja,(1 — k,)

(14.21)

with

where ki is an empirical correction factor, and

b = Dixb;

(14.22)

14.3.

High Pressure

453

The parameter a in eq. (14.19) is calculated from eqs. (4.36) and (4.37). The only parameter not available from pure-component information is the correction factor, kj. This must be evaluated empirically for each binary. However, in many cases reasonably good phase equilibrium predictions can be made with k, = 0.

Pure Component

For a pure component, a slightly different procedure, shown in Appendix A at the end of this chapter, leads to the fugacity of the component in either phase:

nb = @-D-ne@~ BH - 40 (2*2) B

Zz

(A.9)

with

ee

aP

(RT)?

piesRT The fugacities are equal in the two phases, so the fugacity coefficients must be also, since:

ii

i

and the pressure is, of course, the same for each equilibrium phase, leaving: b;, =

o,

Equation (A.9) must be solved for the pressure at temperature T that produces this equality. The phases will have different values for v, and hence z, the compressibility factor, at this J and P, so the Soave-Redlich-Kwong equation of state (SRK EOS) is:

RT B=

a(T) Ip

v(v + b)

This must be solved for these two real volume roots as part of the process. A simple program using NEQNF is listed below. Just as we did in Chapter 11, using equality of free energies in the two phases, we use equality of fugacity, an equivalent criterion here. Notice that, in both cases, we had to integrate v dP at constant T using an equation of state. Here, this integration has already been done (Appendix A). program

G

calc

VAPSRK

vapor

parameter external

pressure (n=3)

fen,

negnf

from

SRK

454

Chapter 14 Phase Equilibrium—Nonideal iateyelll,

icigvonam,

>

xgquess\(n)y As) 7xiG), tizya2

external fcn,neqnf common/data/aingt,aing2?, xf, aki,ak2,112,12tgl7 a2

eeimurerm//1o)//lo), loll, ,lo2 ,loll ©y, lo SHO, 4 IDLO; 127 O, W Gommony Calle fteci tcl noc, DEZ OMis oma nze le Zea WZ 7 WAL pSV common/phi/phil,phi2,phlo,ph2o0,val,pc12,tcl12, JoLPAe),, loll Dal Ww ,(exe xguess

(1)

0.

xia(A)

=

160

xguess(3)

=

350.

val

=

SW

Oil

errel

=

0.0001

itmax

=

100

do7999

3=15 LOL

Wal

=

V2

ele

SislssO)2 al

val

call

negnf(fcn,errel,n,itmax,xguess,x, fnorm)

Ch

2x3) 27 3:05

=

Vrasieen (soe) “Se((al\ rn Sy, Hostal 2 poe (Bh) jorcaliness , SY print*,* 2-propanol-water at 0.508 bar’ print*,” The answers areé:’,; yl, y2, x, v,he

(15.23)

U

where the individual hes are the enthalpies of formation; they are found in the same

references as the free energies of formation. Combining eqs. (15.21), (11.99), and (15.23), we see the relation between free energy and enthalpy of reaction:

rc RT

Ane = ERE RT?

kee,

which can be integrated as well, assuming a constant Ahp;

K.

In = K,

Aha}

1

1

——8]— - — R E al

: we

In fact, the enthalpy of reaction, Ahg, is never constant, but in many cases it is a fair assumption; we find the more exact expression later. For our example reaction, we can add up the enthalpies of formation according to eq. (15.23) to get:

Aha = +83.000 kcal/mol Now let’s say we want a rough estimate of the free energy of reaction at 1000 K. Applying eq. (15.25),

Meslidoo . 01544 R000)

83,000]

= R(298)

R

‘1 298

1 1000

Agrliooo = — 22,554 cal/mol Notice that there has been a large negative change in the free energy, so much so that the free energy of reaction is now negative, and the equilibrium constant is, at this high temperature, 22,554

Kio09 =

€1:987*1000

=

8.5 X

10+

indicating a preponderance of products over reactants. In fact, any time there is a positive enthalpy of reaction, such as in our example, the reaction equilibrium will shift toward the products as temperature rises. This is another example of Le Chatelier’s principle: the positive or endothermic enthalpy of reaction means that as the temperature is raised, the system will act to relieve this ‘‘stress’’ by absorbing heat. It can do this

514

Chapter 15 Chemical Reaction Equilibria

only by driving the reaction from left to right. The opposite is true also: the equilibrium for an exothermic reaction will move to the left with increasing temperature. Exact Form

Equation (15.25) was obtained by the approximate integration of eq. (15.24), assuming the enthalpy of reaction was constant. This will never be exactly true, because each of the individual enthalpies is a function of temperature: a aT

‘

= C,Pi

(15.28)

with each heat capacity represented by a polynomial in temperature such as:

C6Pj =arebr

pcr

at

(15.29)

Referring back to equation (15.23), the enthalpy of reaction at any temperature T is then*: a

Ah; = Ah; + >) v,

|_C,, dF

(15.30)

To

and the exact form for the dependence of the free energy of reaction on temperature is: iie Ag? 1 (7i ART) i.a{—=-) (=) = -=aioe

(15.31)

When the integrations indicated in eq. (15.30) are carried out, each becomes, of

To

b;

C;

1

C,, aT = n|ocraoe) ah ene ale) + Ze slots

where the coefficients a;,ize b;,Ue meth c;,

1

a(t“3 *)|

d; are found in Appendix II. These results are substituted

coae

into eq. (15.30) and then into (15.31) to give: i

Ag?

J (2) = To,

Nw

‘i

ecre

a

boas

(am, [ S+ Tel

+ R

D ve; (7

+ | TD,

aleI TdT

T

eD na —R

Pret

Taek

D

Ts

dT

OT

Dvds,|| =a)

> vb;

va, + T2R—

ar

— +R

>») VC;

+ TI—

:

1

- =

d;

(15.32)

E dT

=

*To simplify the appearance of the equations, the subscript R (reaction) is omitted from both Ag and Ah.

|

15.2

Equilibrium

515

Because none of the terms in the parentheses are f(T), eq. (15.32) is readily integrated, and we get:

Agr

Agr,

RD

ORT,

SS

( 1

1)

ee

|

rae hee i

=

Sy

it

gues

2

Ve i

T — T3)

T

Db;

(7

fs

— 7,

rm

3B v,d;

1

2

(4

0) (15.33)

- ——-|s5-s

0)

1

*)

where the factor Fy is a constant [~ f(T)], defined as:

Ahy, R

fy =

i

rea Be >) v,a, + TZ

Zz

] i te » vid; 0

(15.34)

This calculation has been carried out for a large number of species; a representative

table of both Aga and Ah, for a number of reactions as a function of temperature is attached as Appendix V.

Find the exact equilibrium constant at 1000 K for the reaction in Example 15.1.

SOLUTION In Example identified:

15.1, the following

v Hydrogen

CO Water Ethane

stoichiometric

and heat capacity constants were

a

BEX 19?

3.249

0.422

0

2 ne 4310 —2) 3.470 all ited Wed

0.557 1.450 19275

0 0 OL

5

ve ox10%

VaxK10-* 0.083

= 0.031 0.121 0

so that

Dy Wedel 4026

De

> vb; = —18.901 x 1077

5.5610" >

d=

0:595-< 10°

and with J, = 298, T = 1000, with Aha = + 83.000 kcal/mol at 298, the exact value for the free energy of reaction at 1000 K is:

Agiooo = — 28,300. cal/mol and

Kio00 =

1.533

X 10°

516

Chapter 15 Chemical Reaction Equilibria so that a 20% error in the free energy of reaction (28,300 — 22,554) caused by the use of the constant enthalpy of reaction (at 298 K) brought about a huge error in the equi-

librium constant: 8.5 X 104 vs. 1.533 x 10°!

15.2.3.

Equilibrium Conversion The simplest method for finding the mole fractions of reactants and products present at equilibrium is as follows: 1. Write the reaction with correct stoichiometric coefficients. 2. Underneath each component fill in the starting, reacting, and equilibrium number of moles. 3. Fill in the expression for the equilibrium constant with the equilibrium terms. Remember that the total number of moles is needed to convert moles of each species to mole fraction.

The moles of any species reacting, n,, will be related to the moles reacting of the species taken as the “‘key’’ component v,, by stoichiometry: Vv:

he = te

(1535)

Vy

the moles of each species at equilibrium is just the moles at start minus the moles reacted:

Choosing the key component can be important in solving the resulting equation. It is generally useful to pick that component that is expected to have the lowest mole fraction; this will be evident from the value of the equilibrium constant. Large values of K indicate a preponderance of products and vice versa. An example is clearly called for. We will use the same reaction we have been working with: the steam reforming of ethane.

Calculate the equilibrium composition when a mixture of 2 mol of steam and 1 mol of ethane is fed to a reactor at 1000 K, 1 atm. As we have seen, the equilibrium constant here is 1.533 x 10°. Our key component can be either ethane of steam; here we have chosen ethane. SOLUTION We write:

C,H; St Rx Eg.

1 1 x

+

2H,O0O

2 ee (eee) Ds

=

2CO 0 =2(beSax) 2s)

=

5H, 0 = 5 (less yy Bil S589)

15.2

Equilibrium

517

The terms in the Rx. row come from eq. (15.35); those in Eq. from eq. (15.36). The total moles, ny = 7 — 4x. (Add all the terms in the bottom row.) The equilibrium constant is now expressed* as: 2

5

Veo

Vin 4 &

4

2

Ve

with the subscripts E for ethane,

5

4

No * Ny, « P

K = ———__

= ——*

wi

Ng:

(15.37)

Nw: Np

W for water, and T for total moles. We can substitute

the terms from the Eq. row above:

ehh

ai

is

Letras) ices

(aged

x + (2x)*-(7

oled

—

=

4x)

1.533 x 10°

(15.38)

Because of the large size of K, we can assume x H, The hydrogen can now be eliminated using reaction (2):

which leaves reaction (3) as:

15.3

Multiple Reactions

519

This is simplified to:

3C,H, = CjH, + 4CH, so that only one equation is needed to describe the equilibrium among these three components. C = N — Ry, in this case is: 2 = 3 — 1; two components are the minimum number to form any equilibrium mix of the three.

Look at a more complex system, one consisting of hydrogen, ethane, water, carbon monoxide, and methanol.

SOLUTION 1. Formation equations:

(1) 2C + 3H, = CH, (2) H, + 40, = H,O (3) C+ 40, = CO (4) C + 2H, + 40, = CH,OH 2. Eliminate carbon using reaction (1): Ge

(3) C,H,

3C,H,

a

3H,

— 3H, + 30, = CO

then solve reaction (2) for O,, and eliminate in reactions (3) and (4): 30) =O

1,

(4) 3C,H, + 4H, + H,O — H, = CH,OH These last two can be rewritten:

C,H, + 2H,O = 2CO + 5H, C,H, + 2H,O = 2CH,OH + H, so that RR = 2 and

C = 5 —

2 = 3.

The second of these last two equations can be subtracted from the first to give another equally valid pair:

(a) CO + 2H, = CH,0H (b) C,H, + 2H,O = 2CH,0OH + H,

520

Chapter 15 Chemical Reaction Equilibria We can show that the conditions for equilibrium among these components can be described using only the changes in mole numbers of two key components. The way we have written reactions (a) and (b), these key components are CO and ethane. First, for the change in hydrogen:

From (a):

dny, = 2 dnco

(15.40a)

From (b):

dny, =

—dng

(15.40b)

Net:

dny, = 2 dncg — dng

(15.40c)

Next, for the change in moles of water,* which appears only in reaction (b): From (b):

dnw =

2 dng

(15.41)

and finally for the methanol (subscript Me): From (a):

dny. =

—dnco

(15.42a)

From (b):

dny. =

—2 dng

(15.42b)

Net:

diye =

—dnco — 2 dng

(15.42c)

We can now write, for a change in Gibbs free energy for the system at constant T and P:

dGyp = My dng, + Me dng + pw dnw + co dNco + Mme Any.

(15.43)

and substitute for the changes in moles of hydrogen, water, and methanol using eqs. (15.40) through (15.43) above:

dGrp = By(2 dncg — dng) + bg dng + py(2 cng)

(15.44)

+ co ANco + Bme(— Aco — 2 dng) Now, because reactions (a) and (b) are independent, we can take the partial derivative

of eq. (15.44) with respect to both dngg and dnz, and both must be equal to zero at equilibrium:

(0G

ie

ee

eee

(15.45a)

ONco TP

dG

S| Ong

T.P.nco

= —Hun + Be + 2hw — 2Bme = 0-

— (15.45b)

and these are exactly the equations we would get if we merely set

> vib; = 0 l

for each of reactions (a) and (b) at the start. *Here, subscript W refers to water, Me to methanol and E to ethane.

(15.14)

15.3

15.3.2

Multiple Reactions

521

Equilibrium Compositions The equilibrium compositions when multiple reactions are occurring can be found in the same way as with a single reaction. The best way to illustrate the procedure is with an example. In Chapter 7 we discussed the idea of coal gasification. Here coal is treated with air (or oxygen) and steam to produce a mixture of hydrogen, carbon monoxide, water vapor, carbon dioxide, nitrogen, and traces of hydrogen sulfide, ammonia, and tars. We will consider only the main components in this analysis. In gasifiers, the coal is fed on a moving bed and the steam—air mixture is blown

through. Alternatively, the coal is crushed or pulverized, and blown through the combustor along with the steam—air mix. Higher temperatures are achieved in the latter

processes.

We will look at the gasification process in a simplified manner; the coal will be assumed to be pure carbon, with an activity of unity at all times.* For a start, let’s assume the feed gas is 1 mol of steam + 2.38 mol of air (0.5 mol oxygen, 1.88 mol nitrogen). We now use the rules set out for independent reactions.

SOLUTION 1. The CO, water vapor, and carbon dioxide are formed from their elements:

(a) 2C + O, = 2CO (b) 2H, + O, = H,O (c) C + O, = CO, 2. Carbon is not eliminated, since it is present in the reacting system; but, with its

activity equal to 1.0, the following independent equilibria arrive: 2

K, = 22. P ~10" a

Yoo os

K, =

—

~ 10"

YorYHa * P

Ke

1014 Yor

The values of the K’s were determined from free energies of reaction, as we have done before. It is clear from the large values of the K’s that no free oxygen is present in the equilibrium gas.' Whereas it cannot be completely absent (we will calculate its exact equilibrium concentration later), it is so low as to be unimportant in the stoichiometry. *A pure liquid or solid has an activity of unity, by definition. The gas from a gasifier is highly reducing in nature. This is because insufficient oxygen is provided for

complete combustion. This is a quite different case from a Rankine-type combustor, where the goal is heat production. Here extra oxygen is provided, beyond stoichiometric, so that all fuel value is obtained in the burning.

522

Chapter 15 Chemical Reaction Equilibria We thus eliminate oxygen from reactions (b) and (c) by solving (a) for O2. This produces

(d) HO + C = H, + CO (e) C + CO, = 2CO and we see that

C = N — Ry = 4 — 2 = 2. We will need to find the equilibrium

mole fractions, at any T and P, from the relations:

- P fees SOaae 2

(15.46a)

Yw

pee Yeo: eee P

(15.46b)

YCo2

Although these reactions are independent, CO appears in both. The simplest method of attack is to assume one reaction goes to equilibrium, then use those compositions in the equilibrium relation for the second. We choose, arbitrarily, CO, as the key component; although we have no reason to think that it will always be at low levels, it is as good a choice as any. As we did with a single reaction, we write, first for reaction (e):

(C)ipG

ton

St. — Rx. — Eq. —

CO;

=

0.5 (0.5 — x) x

2CO 0 (Oe) =) 2

Then we use these terms for any reactants in the following reactions; in this case only CO is needed in reaction (d). Here we arbitrarily choose water vapor (w) as key: (d)

"HO"

St. 1.0 og, I) == Eq. w

3G =

"H

—— ——

0 wy = Il Layee

-

CO ith Wwe = I aery

The total moles at equilibrium, nz, is the sum of all gaseous species, including nitrogen:

Ny = Ny, + Nco, + Nco + Nw + Mp II

1.88

+ x +

Q —

2x —w)

+ w+

(1 —

w) = 4.88

— x = w

Our expressions for the equilibrium constants become:

Ky

DYN ieee P90) |0dbg oon VDaodat w(4.88

eho

Saas

w)

(15.47)

(2 220 = Wye OP :

x(4.88 — x — w)

where we have substituted n,/n; for each y,. It now becomes a matter of solving two simultaneous nonlinear algebraic equations for any values of T and P in our reactor. At

15.3

high pressures it will be pressures, but for now we The easiest manner of NEQNF, as we have done

Multiple Reactions

523

necessary to go back and use fugacities instead of partial will be satisfied with the solutions for ideal gases. solution is the use of a library computer routine such as before. The program is shown schematically below. MAIN

set

‘‘n’’, number of parameters (Gn this case, n=2)

set

error tolerance, maximum number of interations, initial guesses for X(1)-X(n)

fheresh

CALL

Ri

Mat] x,

(2)

NEQNF

w) ]

NEQNF CALL

print

FUNC

answers

FUNC

define constants set £(1) - £(n) here, £(1), £(2)

are

eqs.

(15-47)

Some results follow: T,K_

P, bar

Ky

K,

Yco2

1100 1100 1100

il 20 50

Wi fle? 1.2

eA OLORS TiE4 ew Ol093 11.4 40.116

Yco

0.385 0.222 0.163

Yw

0.007 0.068 0.105

Yu

YNn2

0.202 0.170 0.145

0.393 0.447 0.471

1500

1

584.6

1514.

0.0001

0.410

0.00014

0.205

0.385

1500 1500

20 50

584.6 584.6

1514. 1514.

0.002 0.005

0.405 0.398

0.003 0.007

0.203 0.201

0.387 0.399

We have assumed the free oxygen in the equilibrium gas is zero. Now that we have the mole fractions of the main components we can calculate the exact value for the oxygen from any of the equilibria in which it is involved. For example, reaction (a), using the free energy data in Appendix V, the standard free energy of reaction at 1100 K is:

Ag, = 2x(—209,110)

— 2x(0) —

1x(0) =

—418,220 J

524

Chapter 15 Chemical Reaction Equilibria giving, at 1100 K and 1 bar,

ae

MEET

—Ag, __

II

Tp

418,220

RATA

TTOO 2

>

\|

7.25 X 1019 = —*Co— A, * Yoo

from which we can calculate the oxygen mole fraction,

0.3857

Yo. — 795 x 109 ===) (455

lnc

To show that any equation involving oxygen could have been used, we can check the result using reaction (b):

Ag, = 2x(—187,000) — 2x(0) — 1x(0)

= —374,000 J

In K, = 40.895 ve

K, = 5.76 x 107 = —-“— YH

° You

0.0072 Sa seta Sea Th iL Yon ~ 0.2022 X 5.76 X 10"

eh (4

where the small difference between the results using reaction (a) or reaction (b) is just

that resulting from roundoff of the mole fractions. If we change the oxygen rate, say by using oxygen instead of air,* some effect is seen. With an oxygen rate of 0.75 mol to 0.1 mol nitrogen and 1.0 mol of steam, the — equilibrium relations become:

(e)

Cree

St. — RX, ——~ Eq. —

CO,

=

0.75 *(0.) 5" tx) x

2CO 0 —2(Oi/5o— ox) Jigs) acs PA8

and

(dd)

H,O-+"C

St 1.0 Rx. l-w Eq w

— — a

=""H, 0 Sail A en

+

CO UES) ae otc | et) eee Vea,

*The advantages of pure oxygen, higher flame temperatures, and higher fuel value are somewhat offset by the cost of separating the nitrogen from the inlet air.

15.4

Nonideal Gases

525

The total moles are now:

Ny = Nn, + Nco, +t Nco + Nw + Ny, II O01

+x+

(25 —-2x-w)+wt+(il-w) =36—-—x-—w

At the same conditions as Example 15.6, using air, the effect of using oxygen is seen on both the composition of the gas and the total moles. T,K

15.4

P, bar

Ka

K,

Yco2

Yco

Jw

Yu

YN

1100

1

ile 2

11.4

0.036

0.642

0.016

0.276

0.029

1100

20

ibe

11.4

0.224

0.357

0.148

0.233

0.038

1100

50

ci

LLA

0.285

OW Sa een 25

0.196

0.042

1500

1

584.6

1514.

0.0003

0.694

0.0003

0.278

0.028

1500 1500

20 50

584.6 584.6

1514. 1514.

0.006 0.015

0.684 0.670

0.006 0.015

0.275 0.271"

0.028 0.029

NONIDEAL GASES Some reactions of industrial importance are carried out at high pressures. In this case it is necessary to account for possible nonideality in the gas phase. Our equation for equilibrium is still:

K = [la

(15.48)

but instead of replacing the activities with the partial pressures, we must use:

=

l

Sew

(15.49)

with the fugacity coefficients determined as in Chapter 12. The most exact method of solution will require use of an equation like (12.46), or, if the reaction is well into the high-density region, use of an equation of state. Many times such exact solutions are not needed; then the Lewis and Randall approximation can be used.

As an example of an important high-pressure reaction we will analyze the ammonia synthesis process. There are several variations, but all use an iron catalyst to form ammonia from the elements:

N, + 3H, = 2NH;

(15.50)

SOLUTION The reaction is exothermic, so increased temperature drives the equilibrium to the left. But at low temperatures the rate of reaction is too slow to be useful; a reactor would

526

Chapter 15 Chemical Reaction Equilibria

have to be immense to produce significant ammonia at room temperature. Although pressure does not affect the equilibrium constant, it has a great effect on the equilibrium concentrations when, as here, there is a change in the number of moles on reaction, v. K=

I] yridyi ay Aig

(15.51)

Equation (15.51) becomes: ie

Onn (yno Ono)

nus)

2

5:

=2

(15.52)

Vn PH)

The reaction is commonly carried out near 800 K and at pressures from 200 to 900 bar. We will analyze for the equilibrium mole fractions at 800 K, 200 bar with a feed consisting of the stoichiometric mix of 1 mol nitrogen, 3 mol hydrogen. Free energy data at high temperatures is available in the JANAF Tables.* In these tables, all elements have zero free energy and enthalpy of formation at all temperatures." At 800 K,

Pni,(800 K) = +38.662 kJ/mol and with the zero values for the nitrogen and hydrogen, the standard free energy of reaction is:

Agoo = 2Bxws — Pry — 3¥x, = 238.662) — 0 — 0 = 97.324 kT which gives an equilibrium constant of: =175324

K =

¢8314-800

= 893

x

1Q~°

However, to solve eq. (15.52) for the equilibrium mole fractions, we first need to evaluate the fugacity coefficients of all three species.

H, N, NH,

T,, K

P., bar

@

v,, cc/mol

bes

fie

P.

Bo 1262 405.6

13.0 33.9 112.8

= (0222 0.04 O25

65. 89.5 TS)

0.305 0.290 0.242

24.1 6.34 OF

15.4 Dee) edpi

The reduced temperatures (T, = T/T,) for the nitrogen and hydrogen are so high (24.1 and 6.3, respectively) that these gases will be nearly ideal, even at this pressure, as can be seen from the fugacity coefficient plots. The reduced temperature for ammonia,

*Joint Army Navy Air Force tables of thermodynamic data. Published by the American Chemical Society as part of the Journal of Physical and Chemical Reference Data. ‘This is exactly as we have defined these properties. However, in some data sources this is not always the case. Care needs to be taken in retrieving data to make certain of the reference states.

15:4

Nonideal Gases

527

however, is only 1.97 and its reduced pressure is 1.77. With the other components nearly ideal, we can use the Lewis and Randall rule for the fugacity of the ammonia. As the ammonia is just in the low-density region, we calculate the pure component fugacity coefficient from,

b

=

ekT

with, from Chapter 4,

Bo= BRIPs =

(4.22)

B° ¥ + wB! =

(0.083 —

0.422 0.172 2) + oo.» a ae)

The fugacity coefficient is calculated to be 0.975, so even for the ammonia there is not a great effect of nonideality. Now to calculate the equilibrium concentrations, we choose the key component to be ammonia, because of the small value of K:

N, St. Rx.

1.0 0:5

Eq. 1 ene

+

3H, 3.0 IL pe

Otaxg

63 =

dix

=

2NH, 0 HX

ié

OOo oy) ede = 40x.

Be sure to follow the construction of the table through from start to finish. We begin with x as the moles of ammonia at equilibrium. This means — x has had to react (because we started with zero). From eq. (15.35), the moles reacted of the hydrogen and nitrogen

are calculated as shown. Finally, the moles of the reactants at equilibrium are found by subtracting Rx. moles from St. moles. We now substitute these terms into eq. (15.52), with dy, = dy, = 1.0:

x7(0.975)?(4._ — x)? Kk = ——$—————— = 8.93 x 10-° (1 — 0.5x)(3 — 1.5x)?P? ? Using our value of P = 200. bar, this solves using a root-finding routine, such as NEQNF, to x = 0.503. The equilibrium mole fractions are then:

Yn, = 0.144 Vy, = 0.642

yuo = 0.214

In practice, the unused reactants are separated from the ammonia and recycled. Higher reactor pressures are seen, by the form of the equilibrium constant, to give higher ammonia concentrations.

528

15.55

Chapter 15 Chemical Reaction Equilibria

SOLID-GAS EQUILIBRIA There are important reactions that involve solids and gases. Key among these are the roasting of ores to produce mineral products. Many solids are immiscible with other solids; this results in multiple solid phases in contact with each other. At high temperatures these phases will be in equilibrium, although at ambient temperatures this may not be the case. If the solid phases in equilibrium are essentially pure, their free energies are functions only of temperature, and not of the amounts of other components in the mixture. Pressure has only a very small effect on the free energy of a solid phase and can be neglected.

As an example of how this simplifies the equilibrium calculations, we take the roasting of magnesium carbonate, a component of dolomite, to obtain magnesium oxide, and

calculate the tendency for decomposition at two temperatures:

MgCo, av tm MgO

+ CO,

(15.53)

SOLUTION The criterion of equilibrium is given by

(15.14)

a

ah which in this case is:

+ Beco. —

Lugo

=

Bmgco;

(15.54)

9

and the equilibrium constant is found from the standard state chemical potentials: — Rinks

=

»

Vibhi =

Lomig0 a

Meco -

Lmgcos

(15.55)

The free energy (chemical potential) of any gaseous component is related to its fugacity by:*

WS

Aes MR

Toln &

(11.89)

but for our solids,

Mi = Bi

(15.56)

That is, the chemical potential is that of the pure phase at the temperature of the system, as discussed in the first paragraph. Substituting equations (11.89) and (15.56) into eq. (15.54),

*The standard state is the pure component at the temperature of the system and 1 atm pressure.

15.5

Lvizo ~ Hcos + RT In foo, —

Solid—Gas Equilibria

Uiniecos

529

(15.57)

and then eq. (15.55) becomes:

—RT In K =

—RT In foo,

(15.58)

or Ko =

Pcor

at pressures low enough to assume ideal behavior. At any temperature we can find the equilibrium vapor pressure of carbon dioxide over the two solids. We can calculate it at, say, 800 K. Here, from the JANAF tables,

Wco, = — 395.586 kJ

Hugo = —4.6352 kJ

Mmeco; = — 889.702 kJ

so that, from eq. (15.55),

RT In K = —889.702 + 463.452 + 395.586 = —30.664 kJ making, In K =

—4610

and

K = 0.01 bar which, assuming an ideal gas, is equal to the carbon dioxide pressure at equilibrium:

Peo, = 9.01 bar In any situation with the two solid phases present at 800 K, and a pressure of CO, above this value, the MgO will slowly be converted to MgCO,. At any lower CO, pressure, the opposite will occur. This temperature will not be high enough to provide a rapid conversion of the carbonate to the oxide. At 1000 K, however, the free energies are:

co, = —395.886kI

Ridin

= tjgo = —445.763 KJ

ttgco, = — 835.612 KI

6030

So,

In K = 0.726

which yields a K of 2.067 bar. This is a substantial carbon dioxide pressure, one that will decompose the carbonate at an acceptable rate, providing there is a stream of relatively CO,-free air blown over the solids. EE ee E 0

530

Chapter 15 Chemical Reaction Equilibria Another way to look at this is that in a closed vessel containing the two solids, this is

the carbon dioxide pressure that will exist at equilibrium at any given temperature.

15.6

ADIABATIC REACTION TEMPERATURE When the equilibrium reaction temperature is unknown, as when the reaction is carried out in an adiabatic reactor, the solution for the equilibrium conversions and temperature

must be carried out simultaneously. The equilibrium conversions in an exothermic reaction, such as the burning of a fuel in air, lie more to the left (reactants) as the

temperature increases. On the other hand, more conversion releases more heat, raising the temperature. The true equilibrium is thus dependent on a match between the two competing forces. The calculation procedure using NEQNF solves two equations: one resulting from the energy balance, the other from the equilibrium relation. For an adiabatic reactor, these arrive from:

0=

AAG

OW

(15.59)

and,

K=|] yridyi » P”

(15.19)

with the equilibrium constant, K, found from,

—RT In K = > vik, = Age

(15.21)

The energy balance, eq. (15.59) is written in terms of the individual components in the reactions, as well as any inerts:

Dy lila

aoe

(15.60)

with each component enthalpy a function of temperature, T2 (hi) 7,

=

(oss

ata

C,

aT

(15.61)

1

The temperature dependence of K is found from the temperature dependence of the free energy, just as in Section 15.2.2:

Agr, Rie

Agr, ORT

du =e

1

ail

Teer:

Lane, © Vibd op a

(2 v,a)) In

_

2

eae

(15.33)

15.6

Adiabatic Reaction Temperature

531

with F, a factor dependent only on 7,, defined as:

Sh.

°

Any F

R

4

Le

V,a;) Se Tr woes +

73

Scat V;C;) a ;

A (S vd)

(15.34)

These equations require only the enthalpies of formation of the participants in the reaction, in order to calculate Ah;, by

A

he

(15.62)

and the coefficients in the heat capacity equation,

GC) = a,+ bF+

of + e

(15.29)

for each component, i.

15.6.1

Solution for Temperature and Composition The procedure to solving for the equilibrium temperature and composition is best illustrated with an example.

A mixture of 0.5 mol/h CO + 0.5 mol/h O, at 298 K, 1 atm is fed to an adiabatic reactor, operating at 1 atm. Find the exit composition and temperature, assuming equilibrium is reached.

SOLUTION First, we write the energy balance for a system consisting of the contents of the reactor, over a time period of | h: 0) Hine ile

0

which we can write in terms of the outlet composition and temperature: > nh = » Noho

(1)

Ncohcot + Nox(hot + Mxalhxy1 = Mcohco + Noto, + Ncortco: + Mn2/n2)o where each of the individual enthalpies is:

(h)z, = (hi)r, +

.

C,, aT

532

Chapter 15 Chemical Reaction Equilibria To avoid confusion, we use states ‘‘1’’ and ‘‘2’’ in place of “‘I’’ and ‘‘O,”’ so (1)

becomes, for this case with no nitrogen or other inert in the inlet, T2

Ts

0.5heq + 0.5h6, = neo|Meo+ is Gus ar|+ ro We+ ik Ce ar|

ae

T2

+ reo

Heo+

4 Ge ar| 1

Then we can write for the equilibrium, selecting CO as the key component: CO St. RO Eq. Np ="KEE

+

50,

0.5 0.5 — x a

=

0.5 3(0.5 = 8) (0.25 + 0.5x)

O252450

5x)

(058

Co, 0 =) Se) (0.5 — x) =

0S

05x

in the equilibrium expression, 0.5

SS YcoYor

ee NcoNo;

Yeo2

Ink

=

—

NcotT

Agr z RT,

which is strongly dependent on the equilibrium temperature, T,, as given above, where

T, is the inlet value, 298 K. We can now write for K in terms of the equilibrium mole fractions:

No, = 0.25 +.0.5nco Nco>

—

Ny =

0.5

ee

0.75

(3)

Neco

(4)

+ 0.5nN¢9

(5)

with the equilibrium constant for the reaction, Yco

Kaas = YcoYor

NcoT

0.5

ree Neolo,

:

(6)

This brings us to two equations, (1a) and (6), with two unknowns: x, the mole numbers

of CO at equilibrium in the exit, and T,. We have available the free energies of formation of all three components at 298 K, as well as the standard enthalpies of formation and heat capacity constants. The procedure to solving these two equations is outlined in the following program.

15.6

Adiabatic Reaction Temperature

533

MAIN

SOE

LY Vigne

iaiiiiloysnay

One

Gelshasinvereeues:

(in this case, n=2) error tolerance, maximum number of iterations, initial guesses for X(1)-X(n) [here ACL) = noo, &(2)

set

CALL

NEQNF

NEQNF CALL

print

FUNC

answers

FUNC

define constants Skoie ae ((il)y = G21) heneyn

vi ())i

os (2)

are

‘eqs.

(la);

(6)

The program to solve for the equilibrium composition and temperature for any inlet composition, including those with an inert, N,, is given below. Double precision must

be used in the cases where one or more of the components is at very low levels; this will be the case when a large excess over stoichiometric of a reactant is present. Adiabatic

c Ee

Flame

Temperature

Calculation

program main implicit double precision acttamuilame tor co + O27 general co+02+co2 parameter (n=2) real*8

(a-h,o-z) = CO?

errel integer k, nout Realla GuEnonmn ex (ll)ysaxoquess (Mm), |£(2) 7xé (2) external func, dnegqnf common/out/co,o02,co2,eqk,xf,xco0,xo20,

xco20,en2 errel=1.0d-10 itmax=300 xguess(1)=0.2

534

Chapter 15 Chemical Reaction Equilibria xguess (2) =2000.000

call dnegnf (func,errel,n,itmx,xguess,x, fnorm) print*, ‘nco0=, no20=,nco20=,N2=',xco0,xo20, xco20,en2

DEInt*

pocOn

la)

Drink?) -ets= fo. print? ,.) “CO,) 02,4

ORSiMe, Y iiMeisiS

eer den

o (en)

LL) ye ca) ect re rae z) COZ; CO; O2,CO7, cok

“ , iaVerain

stop

end

SU rOUlEdMen tulae) (ue ta) implicit double precision (a-h,o-z) igteys ullets; >*e((iai)) 16 (2). wie, 1aVCO), ANSTO , salle, 19K” , IAKELOXS!,, Nae ((Z) nco0,nco20 real*8 no20 acres ei), Jol) ,e(A), Gla) pao) pine (2) 7S) , hs (4) ,z(4) olenimmnsigl/ ihe // CC), OF, COA, Eile, ik 7 3100), OA), COA), Cia meoU=—OR5 MmoOZO='50 neoZ0=0eL en2=10.0 xCov=neod xo20=n020 NCOAVSaexoZO

OMS ASV}

SL

YS

2) (A) 36.4 /OS 2\ (7) =7 «230

@\ (3) SIL) , ees i ((4))) =6 , SILVA lol) =O _ OOLLOGE io (2) =0 , O0LOOS Ip (3) =0 -WO2OVES 1o((44) =O. OOUILTS

c(1)=0.0 c(2)=0. c(3)=0. c(4)=0. d(1)=-0.062e05 a(2)=-0.452e05 a(3)=-2.299e05

15.6

Adiabatic Reaction Temperature

d(4)=0.0795e05

ho ho ho ho

(1) (2) (3) (4)

=-26417. =0.0 =-94056. =0.0

gl1l=-32783.. g21=0.0 g31=-94263. nco=x (1) nco2=nco20+nco0-x (1)

no2=n020-0.5*nco0+0.5*nco

dgl=g31-0.5*g21-g11

hsum=0.0

Goma

Oma =e 4

aoa) tan) be — PL) rb Ly (roe TOM) Oe te Sah) HAGE = W2cr) ata) (le/T2—-2./T1) Sia =z) 2c)

hsum

=hsum

2—-T1 #2) 72,

ths (i)

10

continue 7 (1) —i— les v(2)=-0.5 wi S)j=Hi sumg=0.0 Sa 00 sb=0.0 sc=00 sd=0.0 Comes Ome sa=sat+v(i)*a(i)*dlog(abs(T2/T1) ) Ssb=sb+vi(a))*b(a)/2-* (82-TL) SeG=CC tw (a1) wre (4) HG), be (Ber) lan he) Sc=—scev Gu) -ci(anei2). 4 (et ex 2 e/ PeKD:)

30

continue sumg=-sa-sb-sc+sd

535

536

Chapter 15 Chemical Reaction Equilibria cL =@. sa=0. Sb=0F SeG=0F Oo Sa=0F ©SS Clo 60 wSiL,3 sa=satv(i)*a(i)*T1 Sloy=fslorrya((al)) “lon(al ))7B, & (Aa)

Bio

60

eyory (GL) ei(al)) /O,

sd=sd+v (1) ?a(i)/2 continue

(Re

A Baanil

8 (e/a

22 ))

alee)

hT1=v (3) *ho(3)+v(2) *ho (2) +v(1) *ho(1) fl=-hT1/r+sa+sb+sc-sd

nt=nco2+nco+no2+1n2 Ge Pace fie /wilkstieils (AL, (WIL ll. 1k )) arisiuhaite; Cen

sCleAie“4wy

k=exp (-dg2/r/T2) £(1)=(neo0*ho(1) + no20*ho(2)+nco20*ho(3)+en2*ho (4) -hsum) ak = dlog(k) ae (6 eis O5@)) ieSO ial MEO = (iNeOA alors (aie) OL, Delays (saioX)) oe O), 5/20) (1s) )) ic (2) = (ineoOrinece))

1000.0

CO=—nIco o2=no02 COZ —nNeozZ eqk=k return

end

Although this program appears somewhat complex, it is relatively straightforward in structure. The main program calls NEQNF (in this version we use DNEQNF, the double precision form). NEQNF takes the initial guesses for the moles of CO (x[1]) and temperature (x[2]) at equilibrium and calls FUNC to evaluate the functions to be minimized, f(1) and f(2). The first of these is the energy balance, in which the enthalpy out, Ho, depends on the moles of each component at equilibrium as well as the temperature out, T,. The second is the difference between the current value of neo (x[1]) and the ngog

calculated from the equilibrium constant (nco,). A sample output and results for various inlet compositions are shown below. neo0=7noZ0=) ec

nco20—

NZ =

O10) 010)

T= Ab (OS) l= 13} 784.65 0.2000E+00) 10%000E+V0RIW S27

0.5000

TE

5.0000

bn

7

B=14

IL, OOW

15.6 CO,

O02,

COA, S

5.499 Enonrm=

B565,

537

Oe Pawo

107564307683583.2 5 .294E-23

FORTRAN

STOP

No

Nco

No,

No,

0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 5

Dex 10 leal0e 0.0554 0.2248 0.3071 IO 10-9 AD Ne 5.06060 0% —2.7ex 107

O05. W052 0.25 O25 0.10 02> 025 1,0 5:0"

05 54

.e

xguess(2)=0.0

errel=0.0001 itmax=100 call negqnf(fcn,errel,n,itmax,xguess,x, fnorm) jonengiotw, © Sei 4% se (Aly

printsy

ax (2),

Peace

15.7 DLimGsny

«CO2=47, yd

Mele

Electrochemical Equilibrium

CO=4 py25-)

Om"

573%, 2 1g

543 AES

=',y5

prints,"

Total

moles=",

t

stop

end

subroutine fcn(x,f,n) eS se (ial) 12D) sae (D) common/data/xf,y1l,y2,y3,y4,y5,t H2=3 .0* (1.-x(1) )+x(2) COZ sa) W= Ops Se (GL) —s< (2) CO=1.-x(1)-x(2) te 67.5 =e A5gi(i8) £(1)=311.9*x (1) *W*t**2-CO*H2**3 £ (2) =0.988*CO*W-CO2*H2 wel) fe Vvalk=COZr/ite WASCOe y3=W/t WARDre meen

end S$ run

reform

(Gl) ee On Oo OA S(A)= Wn 38zZ COZ= 5. SSIS07 CO= 9.494E-02 2 = 0m520 M= 1.311E-04 (Methane) Total moles = 6.498

ERAO=

0326

The value for each species shown is its mole fraction at equilibrium. Notice the CO is at 9.49%, too high for use in the PAFC; the shift reaction is then carried out at a lower temperature, 227°C, where the equilibrium lies much farther to the right.

3. Shift at 500 K

CO + H,O = CO, + H, Age = 0 — 394.94 + 155.41 + 219.05 = —20.48 kJ K,; =e

oh Ase J

=

137.9 =

n CO2!H>

(15.73)

Npo"co

St.

Rx. Eq.

CO

H,O

Co,

H,

0.6167 0.6167 — x x

2.118 06167 — x 1,5013.+ x

0.3821 — (0.6167. x) 0.9988 — x

3.3790 — (0.6167 — x) 3:9957 — %

544

Chapter 15 Chemical Reaction Equilibria

The moles of each species at the start are the mole fractions of each times the total moles (6.498). Eq. (15.73) can be solved analytically; it is simply a quadratic in x:

» (0.9988) = 13.9957

K.

DG

x)

=

i171

Yco

aS

0.0029

x(1.5013 + x) =

0.0186

==

Nco

Nco,

=

0.9802

Yeo.

=

0.1509

No

=

1.5199

YH20

—

0.2339

Ny, =

3.9771

ya=

0.6120

There is still too much CO at 2900 ppm (0.29%); the gas is now sent to a methanation reactor where, over the proper catalyst, it is reduced to acceptable levels. The reaction is the reverse of the reforming reaction; its extent is limited by the CO:

CO + 3H, = CH, + H,O

(+CO,)

Age = —219.05 — 32.74 + 155.41 + 0 = —96.38 kJ K, = 1.17 x 10!°

co 0.0186 0.0186 —x x

St. Rx. Eq.

3H, 3.9771 30.0186 —x) 3.9213 + 3x

CH, 0.00085 —(0.0186-—x) 0.0195 — x

H,O 1.5199 —-(0.0186 — x) 1.5385 — x

CO, 0.9802 0 0.9802

The carbon dioxide, which does not react, contributes to the total moles: Ny

Ho

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571

Appendix V

Standard State Free Energies and Enthalpies as Functions of Temperature

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verell £90671 £96071

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Appendix VI Steam Tables

91 LI 8I 61 0z

IZ GG €@ v7 S@

97

ST'E0€ST'P0E

STLOE

8Z O€ I€ be pels “LLYOE LZ STO0€ 67 SI'ZOE 6L89'F Ao STSOE S190 TH96'b cors’s cf Se S180

PIEST STSTE OTEST STLIE 8I€ST

ele ST OEST STI STCCE Ete ST

PCE ST STS 9tEST LZEST 87E ST

67E ST OEEST IE’ST CLE SI Eee CT

ee SI SeeCL 9¢E ST LEE ST ST8ee

OEEST OveST IveST Cre ST eve ST

pre Sl

4 LY 8Y 6V os

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9¢ LS 8S 6S 09

19 (6) £9 $9 69

99 L9 89 69 OL

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"sqy

67

V7

IT

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67 y80'LI Ll clo ILL'81 09961

91

CI 16L el cE vl COI PI 08 SI ves

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6L71'0 evel OIvT0 O8r1'0 esst0

9660°0 Lvol'0 IOIT‘O 6STTO 81710

8900 6080'0 £6800 8680°0 Sv60'0

9850°0 6190°0 £$90°0 0690°0 87L0'0

anssalg

une

LEC YI

LT IT = 087 OW = P87 6S B87BL = COT LO

SST87 96ST S9'E9T LIT HB TCLTED

ET SE = 8ET HPS THT TL = 16'9b7 = 6D'TSZ

PrEIZ LIZCO= OBIT 6677 OETLI

COT ES 961 IL 68007 LODSOZ 607997

TILT =—ED 18°SLt = 6LT66 V8l LI SER8I

SLOSS vst 76 606ST LTE9IT SPLIT

3s bry

SIEET

98C9C

TO0t9? SIZ9T SETIT TS7TIT 6909T

CII9T TEI9T 6OvI9T Q9OI9T V8I9T

L@9C SPO9T T9097 0°8097 L609

66ST L'S6S7 SLOST CO6ST OT097

Tsgst 98S7 6 988ST yOOSe TCOST

TOLST LLSZ 6 LO6LSt SI8St fesse

CLOST 0°69S7 SOLStT O'CLST VPLSC

yes dea

EVET GH =: IVET HW = O8BEET WOEET ~=—ss OVEET

T9SET =LESET = CISET B88vET €9bET

89ET = 6S9ET SENET = TI9€Z 98SET 8

SO8EC T8L¢é~ LSLET TELET OLE~ 8

WHET 9 = T06e@ LL8E~ €S8E7 TBET—@ =

HOKE) =—S = Vcore L66E~ PvLoec OS6ET

VOTH = Tvlve LIIhe €60"V~ 69077

deaq

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EITEH IDLIT I OL LOST 0E7 ST

COL CS OLYIEI 88007 90°S07 =: 60~ST

CIILI O8SLT = 866LT 8 SOT BBI HE

3s

GELS ©

V88IT = SS8IT =978IT = D6LIT LOLIT =

70%% 8 66617 OLOIT Tv6I% TIEIS

TLIC €VITT vile SB807Z LOT

=SIET 8 LBC BST7T = 67~CC TO¢7Z_—

LSPeC = 6CHTT TOvez TLEtZ CvEt

66S@% TLS@@_ THrStC VIS@Z }=60r Y8bTT

Tylec CILe? 897=F 9S9TT 8 L797

deaq

yeursjuy ‘A8ioug ‘n

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3s bry

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VCHT 8 £9CV7 QLtVT 68crvC T0erve

de,

3y/Py

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OL960

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SEP 09580 $8980 6088°0 ££68°0

S08L°0 TE6L0 85080 8180 OIE80

LYLL9O

T1169 L SE8s9L 19989 8878' = L =LIO8'9 =

ESOL O£FZOL 87669 L9969 8869L

SS6I'L €99T'L €LelL S80I'L 86L0L

LIvVLL

8918 9108 P98L'L PILL S9SLL

8r68L 06L8L €£98L LLYSL TES

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€090°8 cepO'8 79C0'8 £600°8 9766L

I8r1°8 cOET8 97IT8 0s608 9LLO'8

S6E78 60778 $7078 Cr8l8 19918

8rees PSIE8 C9678 CLLT8 €8S78

dea

(1-39) 3s

Pp9IBL SesLL 60SLL BILL = 1989°L

deagq

Vv9ILO0 6treL £67L'0 § HAETEL = C7HL' LC OV8T = OSSL0 evScTL LLOLO 8h7TL =

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Sc8s'0 L86S°0 07190 7790 £€8£9°0

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3s bry

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8001 600°T 600'T 600'T O10'T

900°T L00'T L00°T LO00'T 800'T

3s bry

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vICSEL L9°THOL 8PL9vO 70'89¢9 107997

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16S81 € 9'069L1 96891 cse09l TSLCSI

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ESelOT €POL69° 6PLO97 v0'69b9 ZOT987

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86rIT C 8°6L601 O'88r0I TZOOT ¢ L8°8LS6

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Appendices

579

18e ST

I

60rC

Sv8TL

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O€LEL SLOG

STEEL

TTISz

89897 €889¢ SSECC

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ev0'l Sv0'l 9r0'T Lv0'l 8r0'T vr0'l SP0'l

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Ley 6e

L'6607 9°9607 9°€60¢ ¢' L807 ce90r IlcOv

=—LLI

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9°060¢

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8S°18E I9°C87E LS‘O9T€ O9'T9TE LLYVvOE 08'SP0E £67 C6 VEOSO 8L°LT87 8C87C8

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Sy cole Lhe 90°6te EVE OT 98°SSe S6r7~0'L6r7 6067CCOVV €6PC9'V6v~8°

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CCLV SPLve O'9LVC

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580 Appendix VI Steam Tables

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674

Appendix VI Steam Tables Compressed Liquid, English Units

T, deg F

u, BTU/Ib

h, BTU/Ib

480 490 500 510 520 530 540 544.58

460.80 472.33 484.02 495.87 507,93 520.20 382.12 538.55

464.49 476.06 487.79 499.69 SET 524.12 536.69 542.55

fodeg F

u, BTU/Ib

h, BTU/Ib

s, BTU/(Ib-°R) 0.66309 0.67534 0.68763 0.69997 0.71238 0.72489 0.73754 0.74338

Wl) 195 17.64 PARSE 37.10 46.85 56.60 66.36 1613 85.90 95.67 105.45 115225 125.04 134.85 144.65 154.46 164.29 174.12 183.97 193.82 203.69 213.58 223.47 233.39 243.31 2ST 263.24 21323 283.24 293.28 303.35

0.019917 0.020132 0.020360 0.020603 0.020862 0.021142 0.021444 0.021591

s, BTU/(b-°R)

Pressure = 5,000 psia 12.84 — 0.00378 22.54 0.01581 32.25 0.03506 41.99 0.05398 51.74 0.07257 61.51 0.09084 71.29 0.10880 81.08 0.12645 90.88 0.14380 100.69 0.16087 110.50 0.17766 120.32 0.19418 130.16 0.21044 140.00 0.22645 149.85 0.24221 159.70 0.25775 169.57 0.27305 179.45 0.28815 189.34 0.30303 199.25 0.31771 209.16 0.33219 219.10 0.34649 229.05 0.36061 239.01 0.37455 249.00 0.38833 259.00 0.40195 269.03 0.41542 279.08 0.42873 289.15 0.44191 29925 0.45495 309.38 0.46786 319.54 0.48064 eee

30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210 220 230 240 250 260 270 280 290 300 310 320 330 340

v, ft? /Ib

v, ft?/Ib

cl

_

0.015756 0.015761 0.015773 0.015790 0.015811 0.015836 0.015865 0.015897 0.015932 0.015970 0.016011 0.016055 0.016102 0.016151 0.016203 0.016258 0.016315 0.016374 0.016436 0.016501 0.016568 0.016638 0.016710 0.016784 0.016861 0.016941 0.017024 0.017109 0.017198 0.017289 0.017383 0.017481

Appendices

675

Compressed Liquid, English Units T, deg F

350 360 370 380 390 400 410 420 430 440 450 460 470 480 490 500 510 520 530 540 550 560 570 580 590 600 610 620 630 640 650 660 670 680 690 700

u, BTU/Ib

313.45 323.58 338:13 343.93 354.16 364.43 374.74 385.09 sk bpel!) 405.95 416.46 427.02 437.65 448.34 459.10 469.94 480.86 491.88 502:99 514.20 525.54 536.99 548.58 560.32 SIZ23 584.32 596.62 609.16 621.97 635.07 648.51 662.37 677.35 691.76 706.02 F223

h, BTU/Ib

s, BTU/(1b-°R)

Pressure = 5,000 psia 329.73 339.96 350.21 360.51 370.85 381.23 391.65 402.12 412.65 423.23 433.87 444.57 455.34 466.18 477.10 488.10 499.19 510.39 521.68 533.09 544.63 556.30 568.12 580.10 92:27 604.63 617.23 630.08 643.23 656.70 670.55 684.85 700.31 715.26 730.13 746.03

TT

0.49330 0.50585 0.51829 0.53063 0.54287 0.55501 0.56707 0.57904 0.59094 0.60276 0.61452 0.62622 0.63787 0.64947 0.66103 0.67255 0.68405 0.69553 0.70701 0.71848 0.72996 0.74146 0.75300 0.76458 0.77622 0.78795 0.79978 0.81174 0.82386 0.83617 0.84871 0.86153 0.87536 0.88854 0.90153 0.91529

v, ft? /Ib 0.017581 0.017685 0.017793 0.017904 0.018019 0.018139 0.018262 0.018390 0.018523 0.018660 0.018803 0.018952 0.019107 0.019267 0.019435 0.019610 0.019793 0.019985 0.020185 0.020396 0.020618 0.020852 0.021099 0.021361 0.021640 0.021937 0.022255 0.022596 0.022965 0.023365 0.023801 0.024278 0.024801 0.025379 0.026033 0.026783 __V—K—@geee

Appendix VII

Properties of Selected Refrigerants HFC-134A Saturation Properties Entropy

Temp. | Pressure FE

—40 =e) = 3h) 25

psia

| Volume ft?/Ib Liquid Vr

Density lb/ft?

| Vapor | Liquid Vg 1/Vs

Enthalpy Btu/Ib

Vapor

Liquid

1/Ve

he

Btu/(Ib)(R)

Latent Ah,

Vapor hz

Liquid Se

Sg

ee 97.9 98.7 99.4 100.2

0.0000 0.0035 0.0070 0.0105 0.0139

0.2316 0.2306 0.2297 0.2289 0.2281

Vapor

7.417 8.565 9.851 11.287 12.885

0.0113 0.0114 0.0115 0.0115 0.0116

5.7904 5.0582 4.4366 3.9032 3.4471

88.31 87.81 87.31 86.81 86.30

0.1727 0.1977 0.2254 0.2562 0.2901

4.5 6.0

97.2 96.4 Oey7 94.9 94.2

14.659 16.620 18.784 21.163 DS

0.0117 0.0117 0.0118 0.0119 0.0119

3.0525 Tfiils) 2.4155 2.1580 1.9327

85.79 85.28 84.76 84.23 83.70

0.3276 0.3688 0.4140 0.4634 0.5174

Ve 9.0 10.6 2st ee7/

93.4 92.6 91.8 91.0 90.2

100.9 101.7 102.4 103.1 103.9

0.0174 0.0208 0.0241 0.0275 0.0308

0.2274 0.2267 0.2261 0.2255 0.2250

26.625 29739 B30129 36.810 40.800

0.0120 0.0121 0.0122 0.0123 0.0124

1.7355 1.5620 1.4090 1.2737 1.1538

83.17 82.63 82.08 81.52 80.96

0.5762 0.6402 0.7097 0.7851 0.8667

IS 16.8 18.4 20.0 21.6

89.4 88.5 Siev 86.8 85.9

104.6 105.3 106.0 106.7 107.4

0.0342 0.0375 0.0408 0.0440 0.0473

0.2244 0.2240 0.2235 0.2231 0.2227

45.115 49.771 54.787 60.180 65.963

0.0124 0.0125 0.0126 0.0127 0.0128

1.0472 0.9523 0.8675 0.7916 0.7234

80.40 79°82 79.24 78.64 78.04

0.9549 1.0501 1.1527 1.2633 1.3823

MVP! 24.8 26.4 28.0 29:7

85.0 84.1 83.1 82.1 81.2

108.1 108.8 109.5 110.2 110.9

0.0505 0.0538 0.0570 0.0602 0.0634

022234} 0.2220 0.2217 0.2214 0.2211

72.167 78.803 85.890 93.447 101.494

0.0129 0.0130 0.0131 0.0132 0.0134

0.6622 0.6069 0.5570 0.5119 0.4709

77.43 76.81 76.18 75.54 74.89

1.5102 1.6477 1.7952 1.9536 2.1234

31.4 33.0 34.7 36.4 38.1

80.2 79.1 78.1 77.0 TS9

11S WED 112.8 113.4 114.0

0.0666 0.0698 0.0729 0.0761 0.0792

0.2208 0.2206 0.2203 0.2201 0.2199

110.050 119.138

0.0135 0.0136

0.4337 0.3999

74.22 73.54

2.3056 2.5009

69:9 41.6

74.8 73.6

114.6 MNS)

0.0824 0.0855

0.2196 0.2194

0.0 IES) 3.0

Temp.

°F

— = = = =

40 35) 30 75) 74H)

Appendices

677

HFC-134A Saturation Properties _ | Pressure | Volume ft*/Ib psia

Liquid | Vapor

Vp

Vg

Entropy

Density lb/ft? | Liquid | Vapor

1/V,

1/v,

105

128.782 138.996 149.804

2.7102 2.9347 3.1754

110 115 120 125 130

161.227 173.298 186.023 199.443 213.572

3.4337 3.7110 4.0089 4.3294 4.6745

135 140 145 150 155

228.438 244.068 260.489 277.721 295.820

5.0468 5.4491 5.8849 6.3584 6.8743

160 165 170 175 180

314.800 334.696 355.547 3112393 400.280

7.4390 8.0600 8.7470 9.5129 10.3750

190 200

449.384 503.361 563.037

12.4962 IS silos% 21.1071

Enthalpy Btu/Ib

Btu/(Ib)(CR)

Liquid | Latent | Vapor

he

Ah,

h,

HFC-134a Superheated Vapor Pressure = 5.00 psia

Pressure = 10.00 psia

0.01116 8.37521

| —3.8 | —0.0093 | Sat Liq | 0.01146 05.2 0.2343 | Sat Vap | 4.37445

3.1 | 0.0074 98.8 | 0.2297

= 20 0 20 40

8.66551 9.10747 9.55110 9.98004 10.41667

4.48430 4.71254 4.93583 5.15996

| | | |

100.5 104.2 108.0 111.8

| | | |

0.2336 | —20 0 0.2419 20 0.2499 40 0.2578

60 80 100 120 140

10.84599 11.27396 11.70960 Pi43592 12.56281

5.37924 5.59910 5.81734 6.03500 6.25000

| | | |

115.8 119.8 123.9 128.0 7)7132.3'

| | | | |

0.2655 0.2731 0.2805 0.2879 0.2951

60 80 100 120 140

160 180 200

12.98701 13.40483 13.83126 —

6.46412 | 6.68003 | 6.89180 | TAOTS Ze

136.7 | 0.3022 141.1 | 0.3093 145.6 | 0.3162 0.3231 50:2

160 180 200 220

678

Appendix VII Properties of Selected Refrigerants

Pressure= 20.00 psia

0.01167 2.98686

a Sat Liq | 0.01184 11.4 | 0.0259 0.2273 | Sat Vap | 2.27635 | 102.8 | 0.2258

20 40 60 80

3.09885 3.25415 3.40716 3.55745 3.70645

229095 2.41196 2 S29] 2 2.64550 2.76014

100 120 140 160 180 200

3.85356 4.00160 4.14594 4.29185 4.43656 4.58085

2.87274 2.98418 3.09502 320513 3.31455 3.42349

100 120 140 160 180 200

220

4.72367

S52 92

220

rea

40 60 80

= 30.00pau

Sat Liq a 16.9 _ Sat Vap | 1.54895 | 105.4 | 0.2239 20 40 60 80 100

1.90621 2.00361 2.09864 2.19202 2.28363

1.56863 1.65235 1.73340 1.81291 1.89107

120 140 160 180 200

2.37417 2.46427 2.55503 2.64201 2.73000 2.81690

1.96773 2.04332 2.11864 2.19346 2.26706 2.34082

Appendices

679

“eae 0. 0.2228

60 80 100 120

1.40095 1.47254 1.54202 1.61005 1.67701

1.21242 1.27665 1.33887 1.39958 1.45900

60 80 100 120

140 160 180 200 220

1.74307 1.80832 1.87266 1.93686 2.00040

1.51745 1.57505 1.63212 1.68890 1.74520

140 160 180 200 220

1.80050

so

TE

= 50.00 ean fa

“— 24.9 1.94805 | 108.9

100 120 140 160 180 200 220

1.00180 1.05396 1.10436 1.15354 1.20178 1.24891 1.29550 1.34174 138135 1.43266

: :

— 0:2220'

Sat Liq |-Sat Yap?

eT

= 60.00re |

se }0:79390

28.0 an "110.2 |.0.2214

0.81793 0.86356 0.90728 0.94967 0.99098

100 120 140

1.03135 1.07101 1.11012 1.14903

160 180 200 220

680

Appendix VII Properties of Selected Refrigerants

0.01304 0.59787 80 100 120 140 160

0.62445 0.66020 0.69430 0.72711 0.75896

0.510 1093 0.54037 0.56838 O:59527

180 200 220 240

0.79020 0.82068 0.85063 0.88020

0.62135 0.64675 0.67155 0.69604

Sat Liq an 01465 Sat Vap | 0.23026 120 140 160 180 200 220 240

0.33259 0.35495 0.37573 0.39538 0.41418 0.43232 0.44996

0.24565 0.26428 0.28129 0.29712 0.31217 0.32658

0.46718

0.34046

125.2 125.2 140 160 180 200 220 240

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Appendix VII Properties of Selected Refrigerants

HCFC-123 Superheated Vapor—Constant Pressure Tables v = Volume in ft?/Ib_ h = Enthalpy in Btu/Ib_ s = Entropy in Btu/(Ib)(R) Pressure raat = 1. psia

Ek

= 5.00Pree

Vv : 0.0101 — 22.38 | 30.4546

40 60 80

_ 5 | 0.0081 83.8 | 0.1917

Sat Liq a | Sat Vap | 6.7836

14.5 a OL. | 0.1882

32.0682 33.5014 34.9271 36.3475 37.7641 39.1754

Pressureae = 00 psia

0.0110 2.4293

=a 6 98.8

Appendices HCFC-123 Superheated Vapor—Constant Pressure Tables v = Volume in ft?/Ib h = ey in Btu/Ib s = Entropy in po (Ib)(°R)

683

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