Chemical and Engineering Thermodynamics- Solutions Manual

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Preface

This manual contains more or less complete solutions for every problem in thebook. Should you find errors in any of the solutions, please bring them to my attention.

Over the years, I have tried to enrich my lectures by including historicalinformation on the significant developments in thermodynamics, and biographicalsketches of the people involved. The multivolume Dictionary of Scientific Biography,edited by Charles C. Gillispie and published by C. Scribners, New York, has beenespecially useful for obtaining biographical and, to some extent, historical information.[For example, the entry on Anders Celsius points out that he chose the zero of histemperature scale to be the boiling point of water, and 100 to be the freezing point.Also, the intense rivalry between the English and German scientific communities forcredit for developing thermodynamics is discussed in the biographies of J.R. Mayer, J. P.Joule, R. Clausius (who introduced the word entropy) and others.] Other sources ofbiographical information include various encyclopedias, Asimovs BiographicalEncyclopedia of Science and Technology by I. Asimov, published by Doubleday & Co.,(N.Y., 1972), and, to a lesser extent, Nobel Prize Winners in Physics 1901-1951, byN.H. deV. Heathcote, published by H. Schuman, N.Y.

Historical information is usually best gotten from reading the original literature.Many of the important papers have been reproduced, with some commentary, in a seriesof books entitled Benchmark Papers on Energy distributed by Halsted Press, a divisionof John Wiley and Sons, N.Y. Of particular interest are:

Volume 1, Energy: Historical Development of the Concept, by R. Bruce Lindsay.

Volume 2, Applications of Energy, 19th Century, by R. Bruce Lindsay.

Volume 5, The Second Law of Thermodynamics, by J. Kestin and

Volume 6, Irreversible Processes, also by J. Kestin.

The first volume was published in 1975, the remainder in 1976.

vi

Other useful sources of historical information are The Early Development of theConcepts of Temperature and Heat: The Rise and Decline of the Caloric Theory by D.Roller in Volume 1 of Harvard Case Histories in Experimental Science edited by J.B.Conant and published by Harvard University Press in 1957; articles in Physics Today,such as A Sketch for a History of Early Thermodynamics by E. Mendoza (February,1961, p.32), Carnots Contribution to Thermodynamics by M.J. Klein (August, 1974,p. 23); articles in Scientific American; and various books on the history of science. Ofspecial interest is the book The Second Law by P.W. Atkins published by ScientificAmerican Books, W.H. Freeman and Company (New York, 1984) which contains a veryextensive discussion of the entropy, the second law of thermodynamics, chaos andsymmetry.

I also use several simple classroom demonstrations in my thermodynamics courses.For example, we have used a simple constant-volume ideal gas thermometer, and aninstrumented vapor compression refrigeration cycle (heat pump or air conditioner) thatcan brought into the classroom. To demonstrate the pressure dependence of the meltingpoint of ice, I do a simple regelation experiment using a cylinder of ice (produced byfreezing water in a test tube), and a 0.005 inch diameter wire, both ends of which aretied to the same 500 gram weight. (The wire, when placed across the supported cylinderof ice, will cut through it in about 5 minutes, though by refreezing or regelation, the icecylinder remains intact.This experiment also provides an opportunity to discuss themovement of glaciers.) Scientific toys, such as Love Meters and drinking HappyBirds, available at novelty shops, have been used to illustrate how one can makepractical use of the temperature dependence of the vapor pressure. I also use someprofessionally prepared teaching aids, such as the three-dimensional phase diagrams forcarbon dioxide and water, that are available from laboratory equipment distributors.

Despite these diversions, the courses I teach are quite problem oriented. Myobjective has been to provide a clear exposition of the principles of thermodynamics, andthen to reinforce these fundamentals by requiring the student to consider a greatdiversity of the applications. My approach to teaching thermodynamics is, perhaps,similar to the view of John Tyndall expressed in the quotation

It is thus that I should like to teach you all things; showing you the way toprofitable exertion, but leaving the exertion to youmore anxious to bring outyour manliness in the presence of difficulty than to make your way smooth bytoning the difficulties down.

Which appeared in The Forms of Water, published by D. Appleton (New York, 1872).

Solutions to Chemical and Engineering Thermodynamics, 3e vii

Finally, I usually conclude a course in thermodynamics with the following quotation by Albert Einstein:

A theory is more impressive the greater the simplicity of its premises is, themore different kinds of things it relates, and the more extended its area ofapplicability. Therefore, the deep impression classical thermodynamics madeupon me. It is the only physical theory of universal content which, within theframework of the applicability of its basic concepts, I am convinced will neverby overthrown.

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3.11 (a) This is a Joule-Thomson expansion =( ) = = ( ) = ( )=

$ ? $ . $

.

H T H T H T70 bar, 10133 400 1 400

32782

bar, C bar, C

kJ kg

and T = 447 C , $ .S = 6619 kJ kg K

(b) If turbine is adiabatic and reversible &Sgen = 0c h , then $ $ .S Sout in kJ kg K= = 6619 and P = 1013.bar. This suggests that a two-phase mixture is leaving the turbine

Let fraction vapor kJ kg K

kJ kg K

V

Lx

S

S=

=

=

$ .$ .

7 3594

13026

Then x x7 3594 1 13026 6619. . .( ) + ( )( ) = kJ kg K or x = 08778. . Therefore the enthalpy offluid leaving turbine is

$ . . . . .$ $

HH H

= + ( ) =( ) ( )

08788 26755 1 08778 417 46 2399 6V L satd, 1 bar satd, 1 bar

kJ

kg

Energy balance

0 = + +& $ & $M H M Hin in out out &Q0 + &W Ps

dV

dt

0

but & &M Min out=

= =&

& . . .W

Ms

in

kJ

kg3278 2 2399 6 8786

(c) Saturated vapor at 1 bar

$ . ; $ .

& . . .

%.

..

&

& . . .

S H

W

M

S

M

s

= =

= =

( ) = =

= =

7 3594 26755

32782 26755 602 7

602 7 100

8786686%

7 3594 6619 0740

kJ kg K kJ kg

kJ kg

Efficiency

kJ Kh

in Actual

gen

in

(d) 0

0

0

1 2 2 1

1 1 2

1 1 2

= + =

= + +

= + +

& & & &

& $ $ & &

& $ $&

&

M M M M

M H H W Q PdV

dt

M S SQ

TS

sc h

c h gen

Simplifications to balance equations

&Sgen = 0 (for maximum work); PdV

dt= 0 (constant volume)

& &Q

T

Q

T=

0

where T0 25= C (all heat transfer at ambient temperature)

W

Q

Water1 bar25 C

Steam70 bar447 C

$ .H sat'd liq, 25 CkJ

kg( ) = 10489 ; $ .S T sat'd liq, 25 C kJ

kg K= ( ) = 03674

&

&$ $Q

MT S S= 0 2 1c h ; = + =

&

&$ $ $ $ $ $ $ $

max

W

MH H T S S H T S H T Ss 1 2 0 2 1 1 0 1 2 0 2c h c h c h

=

= + =

&

& . . . . . .

. . .max

W

Ms 32782 29815 6619 10489 29815 03674

1304 75 4 65 1309 4 kJ kg

3.12 Take that portion of the methane initially in the tank that is also in the tank finally to be in thesystem. This system is isentropic S Sf i= .

(a) The ideal gas solution

S S T TP

P

NPV

RTN

PV

RTN

P V

RT

N N N

f i f if

i

R C

ii

if

f

f

f i

p

= = FHGIKJ =

FH

IK =

= = = =

= =

30035

701502

1964 6 mol; 1962

17684

8 314 36..

. .

.

.

K

= mol

mol

(b) Using Figure 2.4-2.

70 bar 7 MPa, T = 300 K $ . $S Si f= =505 kJ kg K

$ . ,.

.

.V m

N

i i

i

= = =

=

=

0019507

00195

3590 kg.

35.90 kg1282

m

kg so that

m

mkg

1000gkg

28g

mol

mol

3 3

3

At 3.5 bar = 0.35 MPa and $ .S Tf = 505 138 K. kJ kg K Also,

$ . ,.

.

.

..

V m

N

f f

f

= = =

=

=

019207

0192

3646 kg.

3646 kg1302

m

kg so that

m

mkg

1000gkg

28g

mol

mol

3 3

3

N N Nf i= = = 1302 1282 11518 mol. .

3.13 dS CdT

TR

dV

V= + eqn. (3.4-1)

S a R bT cT dT eT

dT

TR

dV

V= ( ) + + + +LNM

OQP + zz 2 3 2

so that

S T V S T V a RT

Tb T T

cT T

dT T

eT T R

V

V

2 2 1 12

12 1 2

212

23

13

22

12 2

1

2

3 2

, , ln

ln

a f a f a f c h

c h c h

= ( ) + +

+ +

Now using

PV RTV

V

T

T

P

P

S T P S T P aT

Tb T T

cT T

dT T

eT T R

P

P

= =

= + +

+

2

1

2

1

1

2

2 2 1 12

12 1 2

212

23

13

22

12 2

1

2

3 2

, , ln

ln

a f a f a f c h

c h c h

Finally, eliminating T2 using T T P V PV2 1 2 2 1 1= yields

S P V S P V aP V

PV

b

RP V PV

c

RP V PV

d

RP V PV

eRP V PV R

P

P

2 2 1 12 2

1 12 2 1 1

2 2 22

1 12

3 2 23

1 13

2

2 22

1 12 2

1

2

3

2

, , ln

ln

a f a f a f

a f a f

a f a f

d i d i

=FHG

IKJ +

+

+

3.14 System: contents of valve (steady-state, adiabatic, constant volume system)

Mass balance 0 1 2= +& &N N

Energy balance 0 1 1 2 2= + +& &N H N H &Q0 +Ws

0P dV

dt

0

=H H1 2

Entropy balance0 1 1 2 2= + + +& & &N S N S Sgen&Q

T

0

= =S S SS

N2 1&

&gen

(a) Using the Mollier Diagram for steam (Fig. 2.4-1a) or the Steam Tables

T

P

P

H

T

S1

1

2

2

2

2

600 K

35

7

30453

293

7 277

==

==

= bar

bar

J g

C

J g K$ . $ .

$ $ .H H1 2 30453= = J g . Thus $ .S1 65598 = J g K ; Texit C= 293

$ $ $ .S S S= =2 1 0717 J g K(b) For the ideal gas, H H T T1 2 1 2 600 K= = =

S S T P S T P CT

TR

P

P

RP

P

S

= =

= =

=

2 2 1 12

1

2

1

2

1

1338

0743

, , ln ln

ln .

$ .

a f a f p

J mol K

J mol K

3.15 From the Steam Tables

At 200oC,

P

V V

U U

H H

S S

H S

L V

L V

L L

L V

=

= =

= =

= =

= =

= =

15538 MPa

0001157 012736 m

85065 25953

852 45 27932

2 3309 64323

19407 41014

.$ . / $ . /

$ . / $ . /$ . / $ . /

$ . / $ . /

$ . $ .

m kg kg

kJ kg kJ kg

kJ kg kJ kg

kJ kg K kJ kg K

kJ / kg kJ / kg K

3 3

vap vap (a) Now assuming that there will be a vapor-liquid mixture in the tank at the end, the properties

of the steam and water will be

At 150oC,

P

V V

U U

H H

S S

H S

L V

L V

L V

L V

=

= =

= =

= =

= =

= =

04578 MPa

0001091 03928 m

63168 kJ 25595

632 20 kJ 27465

18418 kJ 68379

2114 3 4 9960 kJ

.$ . / $ . /

$ . / $ . /$ . / $ . /

$ . / $ . /

$ . / $ . /

m kg kg

kg kJ kg

kg kJ kg

kg K kJ kg K

kJ kg kg K

3 3

vap vap

(b) For simplicity of calculations, assume 1 m3 volume of tank.Then

Mass steam initially = 0.8 m

0.12736 m kg kg

Mass water initially = 0.2 m

0.001157 m kg

Weight fraction of steam initially =6.2814

179.14

Weight fraction of water initially =6.2814

179.14

3

3

3

3

/.

/.

.

.

=

=

=

=

62814

17286 kg

003506

096494

The mass, energy and entropy balances on the liquid in the tank (open system) at any timeyields

dM

dtM

dM U

dtM H

dM S

dtM S

MdU

dtU

dM

dtM H H

dM

dt

MdU

dt

dM

dtH U

LL

L LL V

L LL V

LL

LL

L V VL

LL L

V L

= = =

+ = =

=

& ;$

& $$

& $

$$ & $ $

$$ $

; and

or

c hAlso, in a similar fashion, from the entropy balance be obtain

MdS

dt

dM

dtS S

dM

dtSL

L LV L

L$$ $ $= =c h vap

There are now several ways to proceed. The most correct is to use the steam tables, and to useeither the energy balance or the entropy balance and do the integrals numerically (since theinternal energy, enthalpy, entropy, and the changes on vaporization depend on temperature.This is the method we will use first. Then a simpler method will be considered.Using the energy balance, we have

dM

M

dU

H U

M M

M

U U

H UM M

U U

H U

L

L

L

V L

iL

iL

iL

iL

iL

iV

iL i

LiL i

LiL

iV

iL

=

=

= +

FHG

IKJ

+ ++

+

$

$ $ ,

$ $

$ $

$ $

$ $

or replacing the derivatives by finite differences

or finally 1 1 111

So we can start with the known initial mass of water, then using the Steam Tables and the dataat every 5oC do a finite difference calculation to obtain the results below.

i T (oC) $UiL (kJ/kg K) $Hi

V (kJ/kg K) MiL (kg)

1 200 850.65 2793.2 172.862 195 828.37 2790.0 170.883 190 806.19 2786.4 168.954 185 784.10 2782.4 167.065 180 762.09 2778.2 165.226 175 740.17 2773.6 163.427 170 718.33 2768.7 161.678 165 696.56 2763.5 159.959 160 674.87 2758.1 158.2710 155 653.24 2752.4 156.6311 150 631.68 2746.5 155.02

So the final total mass of water is 155.02 kg; using the specific volume of liquid water at150oC listed at the beginning of the problem, we have that the water occupies 0.1691 m3

leaving 0.8309 m3 for the steam. Using its specific volume, the final mass of steam is found tobe 2.12 kg. Using these results, we find that the final volume fraction of steam is 83.09%, thefinal volume fraction of water is 16.91%, and the fraction of the initial steam + water that hasbeen withdrawn is(172.86+6.28-155.02-2.12)/(172.86+6.28) = 0.1228 or 12.28%. A total of 22.00 kg of steamhas withdrawn, and 87.7% of the original mass of steam and water remain in the tank.

For comparison, using the entropy balance, we have

dM

M

dS

S S

M M

M

S S

SM M

S S

S

L

L

L

V L

iL

iL

iL

iL

iL

iiL

iL i

LiL

i

=

= = + FHG

IKJ

+ ++

+

$

$ $ ,

$ $

$

$ $

$

or replacing the derivatives by finite differences

of finally vap vap

1 11

11

So again we can start with the known initial mass of water, then using the Steam Tables andthe data at every 5oC do a finite difference calculation to obtain the results below.

i T (oC) $SiL (kJ/kg K) $Si

L (kJ/kg K) MiL (kg)

1 200 2.3309 6.4323 172.862 195 2.2835 6.4698 170.863 190 2.2359 6.5079 168.924 185 2.1879 6.5465 167.025 180 2.1396 6.5857 165.176 175 2.0909 6.6256 163.367 170 2.0419 6.6663 161.608 165 1.9925 6.7078 159.879 160 1.9427 6.7502 158.18

10 155 1.8925 6.7935 156.5311 150 1.8418 6.8379 154.91

So the final total mass of water is 154.91 kg; using the specific volume of liquid water at150oC listed at the beginning of the problem, we have that the water occupies 0.1690 m3

leaving 0.8310 m3 for the steam. Using its specific volume, the final mass of steam is found tobe 2.12 kg. Using these results, we find that the final volume fraction of steam is 83.10%, thefinal volume fraction of water is 16.90%, and the fraction of the initial steam + water that hasbeen withdrawn is(172.86+6.28-154.91-2.12)/(172.86+6.28) = 0.1234 or 12.34%. A total of 22.11 kg of steamhas withdrawn, and 87.7% of the original mass of steam and water remain in the tank.

These results are similar to that from the energy balance. The differences are the result ofround off errors in the simple finite difference calculation scheme used here (i.e., morecomplicated predictor-corrector methods would yield more accurate results.).

A simpler method of doing the calculation, avoiding numerical integration, is to assume thatthe heat capacity and change on vaporization of liquid water are independent of temperature.Since liquid water is a condensed phase and the pressure change is small, we can make thefollowing assumptions$ $ $ $ $

$ $;

$

U H H H H

dU

dt

dH

dtC

dT

dt

dS

dt

C

T

dT

dt

L L V L

L LL

L L L L

=

and

and

vap

PP

With these substitutions and approximations, we obtain from the energy balance

MdU

dt

dM

dtH U M

dH

dt

dM

dtH

M CdT

dt

dM

dtH

C H

C

H

dT

dt M

dM

dt

C

H

M

M

LL L

V L LL L

L LL

L

L

L

L

LfL

iL

$$ $

$$

$

$

$

$ ln

= =

=

=

( ) =FHG

IKJ

c h

Now using an average value of and over the temperature range we obtain

or

vap

Pvap

Pvap

Pvap

Pvap

1

150 200

and from the entropy balance

MdS

dt

dM

dtS M

C

T

dT

dt

dM

dtS

C S

C

T S

dT

dt M

dM

dt

C

S

M

M

LL L

LL L

L

L

L

L

LfL

iL

$$ $

$

$

$ln

.

.ln

= =

=

++

FH

IK =

FHG

IKJ

vap P vap

Pvap

Pvap

Pvap

Now using an average value of and over the temperature range we obtain

or1

150 27315

200 27315

From the Steam Table data listed above, we obtain the following estimates:

CU T U T

CT

TS T S T

CS T S T

L

L

L

P

o o

o o

P

P

o o

C C

C - C

kJ

kg K

or using the ln mean value (more appropriate for the entropy calculation) based on

C C kJ

kg K

= = = = =

FHG

IKJ =

= = =++

FH

IK

= FH

IK

=

$ ( ) $ ( ) . ..

ln $ $

$( ) $( )

ln..

. .

ln..

.

200 150

200 150

852 45 632 20

504 405

200 150200 27315150 27315

2 3309 184184731542315

4 3793

2

12 1a f a f

Also, obtaining average values of the property changes on vaporization, yields

$ $ $ . . .

$ $ $ . . .

H H T H T

S S T S T

vap vap o vap o

vap vap o vap o

C CkJ

kg

C CkJ

kg K

= = + = = + =

= = + = = + =

1

2150 200

1

22114 3 19407 20275

1

2150 200

1

24 9960 41014 45487

b g b g

b g b gWith this information, we can now use either the energy of the entropy balance to solve theproblem. To compare the results, we will use both (with the linear average Cp in the energybalance and the log mean in the entropy balance. First using the energy balance

C

H

M

M

M

M

LfL

iL

fL

iL

Pvap $

ln.

..

exp . .

150 2004 405 50

20275010863

010863 089706

( ) =FHG

IKJ =

=

= ( ) =

Now using the entropy balance

ln $ ln.

.

.

.ln

.

.. ln

.

.

.

..

.

M

M

C

S

M

M

fL

iL

L

fL

iL

FHG

IKJ =

++

FH

IK =

FH

IK =

FH

IK

= FHIK =

Pvap

150 27315

200 27315

4 3793

45487

42315

4731509628

42315

47315

42315

47315089805

0 9628

Given the approximations, the two results are in quite good agreement. For what follows, theenergy balance result will be used. Therefore, the mass of water finally present (per m3) is

M M

V M V

MV

L L

L L o 3

3

V3

V o

3

3

final initial

occupying final C m

Therefore, the steam occupies 0.8308 m , corresponding to

final0.8308 m

C

0.8308 mmkg

kg

( ) = ( ) =

= ( ) = =

( ) = = =

0897 15506 kg

150 15506 0001091 01692

150 039282115

. .$ . . .

$.

.

b g

b g

So the fraction of liquid in the tank by mass at the end is 155.06/(155.06+2.12) = 0.9865,though the fraction by volume is 0.1692. Similarly the fraction of the tank volume that issteam is 0.8308, though steam is only 2.12/(155.06+2.12) = 0.0135 of the mass in the tank.

(c) Initially there was 6.28 + 172.86 = 179.14 kg of combined steam and water, and finally fromthe simpler calculation above there is 155.06 + 2.12 = 157.18 kg. Therefore, 87.7% of thetotal amount of steam + water initially in the tank are there finally, or 12.3% has beenwithdrawn. This corresponds to 21.96 kg being withdrawn. This is in excellent agreementwith the more rigorous finite difference calculations done above.

3.16 (a)dN

dtN N N N= = + = 0 1 2 2 1& & ; & & or

dU

dtN H N H W Q P

dV

dtW N H N H

W

NH HS S

S= = + + + = + 0 1 1 2 2 1 1 1 21

2 1& & & & & & &

&

& or = -

dS

dtN S N S

Q

TS

S

NS S= = + + = 0 1 1 1 2

0 12 1

& &&

&&

&gengen CP

J

mol K=

37151.

= - = KP298.15K

P

&

& .W

NH H C dT C TS

T

f

f

12 1 29815z = c h if the heat capacity is independent of

temperature. First consider the reversible case,

S SC

TdT R

dP

PT

T

i

f

2 11

10

0 gives = = zz P The solution is 499.14K. Then

KJ

molP&

& . .W

NCS

rev

1

49914 29815 7467= ( ) = . The actual work is 25% greater

W W C T

T

act Srev

f

f

= = =

=

125 9334 29815

549 39

. .

.

J

molK

The solultion is K

P c h

(b) Repeat the calculation with a temperature-dependent heat capacity

C T T T TP( ) = + + 22 243 5977 10 3499 10 7 464 102 5 2 9 3. . . .

Assuming reversibility Tf = 479.44K. Repeating the calculations above with the temperature-dependent heat capacity we find Wact = 9191 J, and Tf =520.92K.So there is a significant difference between the results for the constant heat capacity and variableheat capacity cases.

3.17 T T Pi f i = 300 K, = 800 K, and = 1.0 bar

C T T T TP-2 -5 2 -9 3( ) = 29.088 - 0.192 10 + 0.4 10 - 0.870 10

J

mol K

C T

TdT P

dP

P

P

W C T dT

T

T

P

P

f

revT

T

i

f

i

f

i

f

P

K

K

6

PK

K

Calculated final pressure = 3.092 10 Pa.

J

mol

( )

( ) .

=

=

=

=

=

z z

z

=

= =

300

800

1

300

80041458 10

3.18 Stage 1 is as in the previous problem.Stage 2Following the same calculation as above.

Stage 2 allowed pressure = 9.563 10 Pa

= 1.458 10 J

mol = Stage 2 work

,7

rev-4

P

W

f 2

Stage 3

Following the same calculation method

= 2.957 10 Pa = Stage 3 allowed pressure.

= 1.458 10J

mol = Stage 3 work

,-9

rev4

P

W

f 3

Question for the student: Why is the calculated work the same for each stage?

3.19 The mass, energy and entropy balances aredM

dtM M M M

dU

dtM H M H Q W M H H W

W M H H

dS

dtM S M S

Q

TS M S S S

S M S S

= + = =

= = + + + + =

= +

= = + + + = + =

=

& & , & &

& $ & $ & & ; & $ $ & ;

& & $ $

& $ & $&

& & $ $ &

& & $ $

1 2 2 1

1 1 2 2 1 1 2

1 2 1

1 1 2 2 1 1 2

1 2 1

0

0 0

0 0

s s

s

gen gen

gen

c hc h

c hc h

300 05 =C, 5 bar MPa. $ .H1 3064 2= kJ kg$ .S1 7 4599= kJ kg K

100 01 =C, 1 bar MPa. $ .H2 26762= kJ kg$ .S2 7 3614= kJ kg K

&

& . .W

Ms kJ kg1

2676 2 3064 2 388 = = satisfied the energy balance.

&

&$ $ . . .

S

MS Sgen kJ kg K

12 1 7 3614 7 4599 00985= = = can not be. Therefore the process is impossible.

3.20 Steam 20 bar 2= MPa and 300 C $ .H = 30235 kJ kg$ .S = 67664 kJ kg (from Steam Tables)$ .U = 2772 6 kJ kg

Final pressure = 1 bar. For reference saturation conditions areP T= =01 99 63. . MPa, $ .U L = 417 36 $ .HL = 417 46 $ .SL = 13026$ .U V = 25061 $ .HV = 26755 $ .SV = 7 3594

(a) Adiabatic expansion valve & &W Q= =0 and 0

M.B.:dM

dtM M= + =& &1 2 0 ; & &M M2 1= ;

E.B.:dU

dtM H M H= + =& $ & $1 1 2 2 0 ; $ $H H2 1=

From Steam Tables H2 30235= . T = 250 C $ .H = 29743 kJ kg $ .S = 80333 kJ kg K

P = 01. MPa T = 300 C $ .H = 3074 3 kJ/kg $ .S = 82158 kJ/kg KBy interpolation T H= 275 C gives = 3023.5 kJ / kg$ all vapor

$ .

& $ & $ &

&

&$ $ .

S

dS

dtMS M S S

S

MS S

=

= + + =

= =

81245

0

1359

1 2 2

2 1

kJ kg K

= 8.1254 6.7664 kJ kg K

gen

gen

gen

(b) Well designed, adiabatic turbine

E.B.: & $ & $ &M H M H W1 1 2 2 0+ + = ; & $ $W H H= 2 1c hS.B.: & $ & $M S M S1 1 2 2 0+ = ; $ $S S2 1= ; $ .S2 67664= kJ kg K Two-phase mixture. Solve for fraction of liquid using entropy balance.

x x

x

( ) + ( ) == ( )7 3594 1 13026 6 7664

0 902

. . .

. not good for turbine!$ . . . . .&

& . . .

&

& .

H

W

M

W

M

2 0902 26755 0098 417 46 2454 2

2454 2 30235 569 3

569 3

= + =

= ( ) =

=

kJ kg

kJ kg

kJ kg

(c) Isorthermal turbine superheated vaporT

P

= =

OQP

300

01

C

MPa final state

.

$ .

$ .

H

S

=

=

3074 3

82158

kJ kg

kJ kg K

E.B.: & $ & $ & &M H M H Q W1 1 2 2 0+ + + =s

S.B.: & $ & $&

M S M SQ

T1 1 2 2+ + +Sgen

0= 0

&& $ & $ & $ $

&

&$ $ . . .

.&

&

&

&$ $ . . . .

Q

TM S M S M S S

Q

MT S S

W

M

Q

MH H

= =

= = +( ) ( )

=

= + = + ( ) =

1 1 2 2 1 2 1

2 1

1 2

300 27315 82158 67664

8307

8307 30235 3074 3 779 9

c h

c h

c h

kJ kg K

kJ kg

kJ kgs

get more work out than in adiabatic case, but have to put in heat.

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Chemical and Engineering Thermodynamics- Solutions Manual