Solutions to Chemical and Engineering Thermodynamics, 3e

4

4.1 Using the Mollier diagram

=

FH IK = FH IK =( )

= =

T

P

T

PH H

510 490

1241 10 7 929 10

4 463 10 4 463

7 6

6

C

Pa

C Pa C MPa

. .

. .

c h

SS S

T

P

T

P=

FH IK FH IK =( )

= =

510 490

1069 10 9 515 10

1702 10 1702

7 6

5

C

Pa

C Pa C MPa

. .

. .

c h

=

=

=

= = ( )

H S

H S

H T

S T

S P

H P

H T

H P

S P

S T

T H

P H

P S

T S

T

P

S

a fa f

a fa f

a fa f

a fa f

a fa f

a fa f

a fa f

,

,

,

,

,

,

,

,

,

,

,

,.

0262 unitless

4.2 (a) Start from eqn. 4.4-27

H T P H T P RT Z TdP

dTP dV

VV

V

, ,( ) ( ) = ( ) + FH IK LNM

OQP=z

IG 1

PRT

V b

a T

V bV b=

( )+ 2 22

;

FH IK = + P

T

R

V b

da dt

V bV bV 2 22 so

H T P H T P

RT Z TR

V b

da dt

V bV b

RT

V b

a T

V bV bdV

RT Z a Tda

dT

dV

V bV b

V

V

V

V

, ,( ) ( )

= ( ) +

+

RSTUVW +

( )+

LNM

OQP

= ( ) + +FH IK +

=

=

zz

IG

12 2

12

2 2 2 2

2 2

From integral tables we have

dx

a x b x c b a c

a x b b a c

a x b b a c + + =

+

+ + z 2 2

2

2

1

4

2 4

2 4ln for 4 02

Solutions to Chemical and Engineering Thermodynamics, 3e

H T P H T P

RT Za Tda dT

b

V b b

V b b

V b b

V b b

RT Za Tda dT

b

V b

V b

V V

, ,

ln ln

ln

( ) ( )

= ( ) + + + +

+ + +

LNMM

OQPP

= ( ) + +

+ +

=

IG

12 2

2 2 2 2

2 2 2 2

2 2 2 2

2 2 2 2

12 2

1 2

1 2

a f

a f d id i

or finally

H T P H T P RT ZTda dT a

b

Z B

Z B, , ln( ) ( ) = ( ) +

+ +

+

LNMM

OQPP

IG 12 2

1 2

1 2

a f d id i

(b) This part is similar except that we start from eqn. (4.4-28)

S T P S T P R ZdP

dT

R

Vdv

R ZR

V b

da dT

V bV b

R

VdV

R Z RV b

V

da dT

b

V b

V b

R Z Bda dT

b

Z B

Z B

VV

V

V

V

V

V

V

V

, , ln

ln

ln ln ln

ln ln

( ) ( ) = + FH IK LNM

OQP

= +

+

LNM

OQP

= + ++ +

+

= ( ) ++ +

+

FHG

IKJ

=

=

= =

zz

IG

2 22

2 2

1 2

1 2

2 2

1 2

1 2

d id i

d id i

4.3 Start with eqn. (4.2-21): dU C dT TP

TP dV

V

= + FHGIKJ

LNMM

OQPPV

. Thus

U

TC

V

FHG

IKJ = V ;

U

TC T

P

TP

V

TP V P

FHG

IKJ = +

FHG

IKJ

LNMM

OQPPFHG

IKJV and

U

T

U

TT

P

TP

V

TP V V P

FHG

IKJ

FHG

IKJ =

FHG

IKJ

LNMM

OQPPFHG

IKJ .

(a) Ideal gas PV RT=

TP

TP

U

T

U

TV P V

FHG

IKJ =

FHG

IKJ =

FHG

IKJ0

(b) van der Waals gas

PRT

V b

a

V=

2 ;

P

T

R

V bV

FHG

IKJ = ;

FHG

IKJ

LNMM

OQPP =T

P

TP

a

VV

2

Also: dPRdT

V b

RT

V bdV

a

VdV=

( )+

2 3

2

Solutions to Chemical and Engineering Thermodynamics, 3e

4.6 d SC

TdT

V

TdP

P

= FHGIKJ

P

[eqn. (4.2-20)]

For the ideal gas d SC

TdT

R

PdPIG P=

*

. Thus, at constant temperature

d S SR

P

V

TdP

P

= FHGIKJ

LNM

OQP

IGd i

and

S T P S T P S T P S T PR

P

V

TdP

PP

P

, , , ,( ) ( ) =( ) =( ) = FHGIKJ

RSTUVW=z

IG IG0 00

However, S T P S T P, ,=( ) =( ) =0 0 0IG , since all fluids are ideal at P = 0 . AlsoPV Z T P RTr r= ,a f . Thus

V

T PRZ T P RT

Z T P

TPr r

r r

P

FHG

IKJ = +

FHG

IKJ

RSTUVW

1,

,a f a f

and

R

P

V

T

R

PZ T P

RT

P

Z

T

S T P S T P RZ

P

T

P

Z

TdP

RZ T P

P

T

P

Z

TdP

Pr r

P

PT P

T P

r r

r

r

r r Pr

T P

T P

rr r

r r

FHG

IKJ =

FHG

IKJ

( ) ( ) = + FHGIKJ

LNM

OQP

=

+FHG

IKJ

LNMM

OQPP

=

=

zz

,

, ,

,

,

,

,

,

a fk p

a f

1

1

1

0

0

IG

4.7 (a) Ideal gas

PV NRT

N

=

=( )

+( ) =

50 bar 100 m

27315 150 8314 10142 12

3

3K bar m kmol K kmol

c h. .

.

Energy balance, closed nonflow system

U Q PdV Q W= = +z .However, for ideal gas U = 0 since T is constant (isothermal). Thus

W Q PdVNRTV

dV NRTVV

NRTPP

Q

= = = = =

= +( ) FH IK= ==

z z ln ln. . ln

. .

.

2

1

1

2

3

142 1 27315 15050

300

895 9 10 895 9

895 9

kmol 8.314 J mol K

kJ MJ

MJ

Also, by Ideal Gas Law at fixed T and N

Solutions to Chemical and Engineering Thermodynamics, 3e

PV PV V VP

P1 1 2 2 2 11

2

100 m50

3001667= = = =3 3 m.

(b) Corresponding states

T P ZH H

TS Sr r

C

IGIG

initial state

final state 1.391300

73.76cal

mol K

cal

mol K

+ = =

=

150 27315

304 21391

50

73760679 094 07 0 4

4067 0765 45 2 4

.

..

.. . . .

. . . .

Number of moles of gas = = = =NPV

ZRT

142 1

094151 2

.

.. kmol ( 142 1. =

PV

RT from above)

Final volume = =V ZNRTPf f

= +( )0765 151 2 8 314 10 273 15 150

300

2. . . .=1356 m. 3

Energy balance on gas: U Q W= +

Entropy balance on gas processes in gas are reversible: SQ

TS S= + =gen gen 0 . Therefore

SQ

T= or Q T S=

S S S N S S N S S S S S S

NP

P

N N

Q T S

W U Q N U U Q N H H N P V PV Q

f i f i f f f i i i

f

i

f i f i f f i i

= = = +

= ( )RSTUVW

= RSTUVW =

= = +( ) ( ) =

= = =

c h c h c h c hm rIG IG IG IG

J K

kmol MJ

2 4 4184 8 314 04 4184

8 368 8314300

502326

27315 150 151 2 2326 14885

. . . ln . .

. . ln .

. . . .

=

+LNMMN T

H H

TCf f

C

IGd iH Hf

IGiIGc h

0Since process

is isothermal.

+OQPP T

H H

TZ RT Z RT QC

i i

Cf f i i

IGd i

= ( )( )

+( ) ( )LNM

OQP +

= + + = +=

151 2304 2 45 07 4184 8314

273 15 150 0 765 09414885

151 2 10 48365 615 7 14885 638 2 14885

850 3

3

.. . . . .

. . ..

. . . . . .

.

kmol MJ

J MJ MJ

MJ

(c) Peng-Robinson E.O.S.Using the program PR1 with T = 27315. , P = 1 bar as the reference state, we obtainT = 150 C , P = 50 barZ = 0 9202. ; V = 06475 10 3. m mol3 ; H = 470248 . J mol ; S = 1757. J mol K .T = 150 C , P = 300 bar

Solutions to Chemical and Engineering Thermodynamics, 3e

Z = 0 7842. ; V = 0 9197 10 4. m mol3 ; H = 6009. J mol ; S = 4124. J mol .

NV

V= =

=

100 m

06475 10154 443

3

3 m mol kmol

..

Q TN S= = +( ) ( )( ) = 273 15 150 154 44 4124 1757 1546 9. . . . . MJ

W U Q H PV H PV Q

N H PV H PV Q

f i

f i

= = ( ) ( )

= ( ) ( )

=

+

L

N

MMMM

O

Q

PPPP +

=

154 44

6009 300 0 9197 10

10 4702 48

50 0 6475 10 10

10 1546 9 10

885 25

4

5

3 5

3 6.

. .

.

.

.

.

J bar m

MJ

3

{Note that N, Q and W are close to values obtained from corresponding states.}

4.8

T

P

T S

P S

T S

P T

P T

P S

S T P T

S P T PS

FHG

IKJ =

( )( )

=( )( )

( )

= ( ) ( )( ) ( )

,

,

,

,

,

,

, ,

, ,a f =

= =

S PS T

V T

C T

V T

CT

P

Pa fa f

a fP P

and

S

T

S

T

V P

V V dP

V V dP

V S P S

V T P T

V S

V T

P T

P S

S V

T V

T P

S P

S

T

T

S

C

T

T

C

C

C

= =( ) ( )( ) ( )

=( )( )

( )( )

=( )( )

( )( )

= FHGIKJ

FHG

IKJ = =

1

1

a fa fa fa f

, ,

, ,

,

,

,

,

,

,

,

,V

P

V

P

4.9 (a)

H

V

H T

V T

H T

P T

P T

V T

H

P

P

VT T T

FHG

IKJ =

( )( )

=( )( )

( )( )

= FHGIKJ

FHG

IKJ

,

,

,

,

,

,

Since

P

V T

FHG

IKJ 0 (except at the critical point)

H

V T

FHG

IKJ = 0 if

H

P T

FHG

IKJ = 0

(b)

S

V

S P

V P

S P

T P

T P

V P

S

T

T

V

C

T VV

dT

dV

C TV

V V T

C

TV

S

V

P P P

P P P

FHG

IKJ =

( )( )

=( )( )

( )( )

=FHG

IKJ

FHG

IKJ

= FHGIKJ = =

FHG

IKJ

,

,

,

,

,

,

~P P P1

11

a fa f4.10 (a) We start by using the method of Jacobians to reduce the derivatives

Solutions to Chemical and Engineering Thermodynamics, 3e

TV

T HV H

T HT P

T PT V

T VV H

H T

P T

P T V T

H V T V

H

P

P V

H T

H V

d H T

H

T

T

V

T

V

FHG

IKJ =

( )( )

=( )( )

( )( )

( )

= ( )( )

( ) ( )( ) ( )

= FHGIKJ

=

,,

,,

,,

,,

,

,

, ,

, ,

a f

a fa f

a fa f

Now from Table 4.1 we have that

H

PV T

V

TT P

FHG

IKJ =

FHG

IKJ and

HT

C V TVT

PTV P V

FHG

IKJ = +

FHG

IKJ

LNM

OQPFHG

IKJP

alternatively, since H U PV= +

H

T

U

T

PV

TC V

dP

dTV V V V

FHG

IKJ =

FHG

IKJ +

( )FHG

IKJ = +

FH IKV

Thus

T

V

P V V T V T

C V P T

V P V T P T

C V P TH

T P

V

T V

V

FHG

IKJ =

+

= +

+a f a f

a fa f a f

a fV V

Note: I have used

P

V

V

T

P

TT P V

FHG

IKJ

FHG

IKJ =

FHG

IKJ .

T

V

T S

V S

T S

V T

V T

V S

S T

V T

T V

S V

S

V

T

S

T

C

P

T

S

T V V

FHG

IKJ =

( )( )

=( )( )

( )( )

= ( )( )

( )( )

=FHG

IKJFHG

IKJ =

FHG

IKJ

,

,

,

,

,

,

,

,

,

,

V

(b) For the van der Waals fluid

P

T

R

V bV

FHG

IKJ = ,

P

V

RT

V b

a

VT

FHG

IKJ =

( )

+2 32

Thus

T

V

RTV V b a V RT V b

C V R V bH

FHG

IKJ =

( ) + + ( )

+

2 22n sV

after simplification we obtain

T

V

a V b RTV b

C V b V R V b VH

FHG

IKJ =

( )

( ) + ( )

2 2 2

2 2 3C

Solutions to Chemical and Engineering Thermodynamics, 3e

and

T

V

RT

C V bS

FHG

IKJ = ( )V

4.11 There are a number of ways to solve this problem. The method I use is a little unusual, but thesimplest that I know of. At the critical point all three roots of V are equal, and equal to V C .

Mathematically this can be expressed as V V C =a f3 0 which, on expansion, becomes

V V V V V VC C C3 2 2 33 3 0 + = (1)

compare this with

PRT

V b

a

V V b b V b

RT

V b

a

V bV b=

+( ) + ( )=

+ 2 22

which multiplying through by the denominators can be written as

V V bRT

Pb

bRT

P

a

PV b

RTb

P

ab

P3 2 2 3

2

32

0+ FH IK + +FH IK + + FHG

IKJ = (2)

Comparing the coefficients of V in Eqns. (1) and (2) gives TC , PC

V bRT

PVC

CC

2 3: = (3)

V bbRT

P

a

PVC

C CC: + =3

232 2 (4)

V bRT

Pb

ab

PVC

C CC

0 3 2 3: + = (5)

From Eqn. (3)

P b

RT

P V

RTZC

C

C C

CC =

= 1 3 3 or P b

RTZC

CC= 1 3 (6)

For convenience, let y ZC= 1 3 or Zy

C =13

. Then

P b

RTyC

C

=

From Eqn. (4)

Solutions to Chemical and Engineering Thermodynamics, 3e

FHG

IKJ

FHG

IKJ + =

+ + =( )

3 2 3

3 23 1

9

2

22

22

2

P b

RT

P b

RT

aP

RTZ

y yaP

RT

y

C

C

C

C

C

CC

C

C

a f

a for expanding and rearranging

aP

RTy yC

Ca f c h221

310 4 1= + + (7)

Finally from eqn. (5)

P b

RT

bP

RT

P b

RT

aP

RTZC

C

C

C

C

C

C

CC

FHG

IKJ +

FHG

IKJ

FHG

IKJFHG

IKJ =

3 2

23

a fy y y y y y3 2 2 3

1

310 4 1

1

271+ + + = ( )c h

or

64 6 12 1 03 2y y y+ + = (8)

This equation has the solution y = 0077796074.

=b RTP

C

C

0077796074. (from Eqn. (6))

aRT

PC

C

= 04572355292

.a f

(from Eqn. (7))

Also Zy

C =

=1

30307401309. .

Note that we have equated a and b to TC and PC only at the critical point. Therefore these functions

could have other values away from the critical point. However, as we have equated functions of V ,

we have assumed a and b would only be functions of T. Therefore, to be completely general wecould have

aRT

P

T

T

bRT

P

T

T

C

C C

C

C C

=FHG

IKJ

=FHG

IKJ

0457235529

0077796074

2

.

.

a f

with T

TC

FHG

IKJ 1 as

T

TC 1 and

T

TC

FHG

IKJ 1 as

T

TC 1 .

In fact, Peng and Robinson (and others) have set = 1 at all temperatures and adjusted a s afunction of temperature to give the correct vapor pressure (see chapter 5).

4.12 (also available as a Mathcad worksheet)

Solutions to Chemical and Engineering Thermodynamics, 3e

dN

dtN N N N

dUdt

N H N H QQN

H H

= + = =

= + + = =

& & & &

& & &&

1 2 2 1

1 1 2 21

2 1

0

0

Also, now using the program PR1 with T = 27315. , P = 1 bar reference state we obtain

T = 100 C T = 150 CP = 30 bar P = 20 barZ = 0 9032. Z = 0 9583.

V = 09340 10 3. m mol3 V = 01686 10 2. m mol3

H = 3609 72. J mol H = 679606 . J molS = 1584. J mol K S = 4 68 . J mol

&

& . . .Q

N= =679606 3609 72 318634 J mol

4.13 (also available as a Mathcad worksheet)

Since process is adiabatic and reversible S = 0 or S Si f= , i.e.,

S S T310 K, 14 bar 345 bar( ) = =( )?, . Using the program PR1 with the T = 27315. K and P = 1

bar reference state we obtain T = 310 K , P = 14 bar , Z = 0 9733. , V = 01792 10 2. m mol3 ,H = 109083. J mol and S = 1575. J mol K .

By trial and error (knowing P and S , guessing T) we obtain T = 34191. K , P = 345 bar ,

Z = 0 9717. , V = 08007 10 4. m mol3 , H = 188609. J mol , S = 1575. J mol K =Tf 34191. K .

System = contents of compressor

M.B.: dN

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

E.B.: dU

dtN H N H= = + +0 1 1 2 2& &

&Q 0adiabatic

+ &W PsdV

dt

0

volume ofcompressor

constant

& & &W N H N HS = +1 1 2 2 or &

& . . .W

NH HS = = =2 1 188609 109083 7952 6 J mol

4.14 (a) Pa

VV b RT

PV

RT

V

V b

a

RTV+

FHG

IKJ ( ) = = 2

i) lim limPV

V

PV

RT

V

V b

a

RTV

=

FHG

IKJ =0 1

ii) B VPV

RTV

V

V b

a

RTVPV

V= FH IK =

RSTUVW

lim lim0

1 1

= ( )

( )RST

UVW = ( ) RST

UVW = lim limV VVV V b

V b

a

RTV

bV

V b

a

RTb

a

RT

Q.

N1.

N2.

M.B.

E.B.

Solutions to Chemical and Engineering Thermodynamics, 3e

iii) C VPVRT

BV

VbV b V b

V bb V

V bb

PV

V V= FHG

IKJ =

( )

RSTUVW = =

lim lim lim0

22

21

=C b2

(b) At the Boyle temperature: limP

VPV

RTB

FH IK = =0 1 0 0

0 = ba

RTB, T

a

RB b= but a

V RTc c=9

8, b

V c=3

(Eqns. 4.6-3a)

TV RT

RVT TB

c c

cc c= = =

9 8

3

27

83375.

4.15 (a) From table: TB ~ 320 K , i.e., B T BBa f = ( ) =320 K 0The inversion temperature is the temperature at which

T

P CV T

V

T

TP T

RTP

BRP

dBdT

V TVT

RTP

BRTP

TdBdT

B TdBdT

H P

P P

P

FHG

IKJ = =

FHG

IKJ

LNM

OQP

FHG

IKJ = +

LNM

OQP = +

FHGIKJ = + =

01

P

Thus, T inv is the temperature at which B TdB

dt = 0 .

Plot up B vs. T, obtain dB dT either graphically, or numerically from the tabular data. I find

T inv ~ 600 K . Also, dB dT decreases with increasing temperature (i.e.,

dB dT ~ .456 cm mol K3 at 87.5 K and 0.027 cm mol K3 at 650 K. Presumably it is negativeat even higher temperature!)

(b) Generally

= FHGIKJ =

FHG

IKJ

RSTUVW =

RSTUVW

TP C

V TVT C

B TdBdTH P

1 1

P P

Using the data in the table it is easy to show that for T T< inv , B TdB

dT < >0 0 , while for

T T> inv , B TdB

dT > 0 and less than [Except at a phase transitionsee Chap. 5 andProblem 5.1however, has no meaning in the two-phase region], if dT dP Ha f is to be zero,then d H dP Ta f must equal zero. That is, an inversion point occurs when isotherms are parallelto lines of constant H (vertical line). This occurs at low pressures (ideal gas region) and at high

Solutions to Chemical and Engineering Thermodynamics, 3e

Pa

VV b RT

V V

T T

T

PRT

V b

a

V

ii

i

i f

f i

f

f

f

f f

+FHGG

IKJJ =

= =

=

=

= = =

=

=

d id i

a f c h

d i c h

2

5 4

5

2 4 5

6 678 10 1336 10

02283

2 2725 6678 1062 73

29315 62 73 230 42 42 73

8314 230 42

1336 10 4 269 10

02283

1336

. ; . .

.

. ..

. . . .

. .

. .

.

.

m mol m mol

Pa m mol

J mol K m mol K

K C

3 3

6 2

3

10

8286 10 8286 bar

4 2

6

= =

c hPf . . Pa

(d) Here we will use the program PR1. Using the 273.15 K and 1 bar reference state we find that atthe initial conditions Z = 11005. , V i = 05365 10 4. m mol3 , Hi = 328185. J mol and

Si = 5912. J mol K . Therefore

U H P Vi i i i= = = 328185 500 05365 10 10 5964354 5. . . J mol

Now since U Uf i= = 596435. J mol and

V Vf i= = 2 1073 10 4. m mol3 .We must, by trial-and-error, find the temperature and pressure of the state having theseproperties. I find the following as the solution Tf = 230 9. K ; P ~ 99 bar (for which

V = 01075. m mol3 and U = 5968 4. J mol ). To summarize, we have the following answersfor the different parts of the problem:

P f T f

Ideal gas 250 bar 293.15 KCorresponding states 101.1 bar 237 Kvan der Waals 82.86 bar 230.42 KPeng-Robinson 99 bar 230.9 K

Once again, the ideal gas solution is seriously in error.

4.20 Mass balance (system = both tanks): N N Ni f f1 1 2= +

energy balance (system = both tanks): N U N U N Ui i f f f f1 1 1 1 2 2= +

entropy balance (system = portion of initial contents of tank 1, also in there finally): S Si f1 1=

Also, P P Pf f f1 2= = ; NV

Vi

i11

1

= ; N VV

ff11

1

= and N VV

ff2

2

2

=

(a) Ideal gas solution: obtain P

T

P

T

P

T

i

i

f

f

f

f1

1

1

1

2

2

= + from mass balance and

P P P P Pi f f f f1 1 272 250 bar 25 10= + = = = . Pa from energy balance

Solutions to Chemical and Engineering Thermodynamics, 3e

T TP

PTf i

f

i

R Cf

1 11

11

8 314 35 565

20 273 151

2249 3

239

=FHG

IKJ = +( )

FH

IK =

=

P

K

C from entropy balance

. .

.

. .

and

1 2 1355 9 82 7

2 1 12

T T TT

f i ff= = = . . K C

Also

N

N

P V

RT

RT

P V

f

i

f

f

i

i1

1

1 1

1

1

1 1

7

7

25 10

249 3

293 15

500 100588= =

=

.

.

.

..

and

N

N

N

N

f

i

f

i2

1

1

1

1 0412= FHG

IKJ = .

(b) Corresponding States Solution:

Initial conditions Tr = =29315

190 71538

.

.. ; Pr =

=5 104 64 10

10777

6.. ; =Z 122. ;

H H

TC

IG

J mol K

= 180 . ; S SIG J mol K = 96 . .

Mass balance:

P

Z TP

Z T Z T

i

i if

f f f f1

1 1 1 1 2 2

751 1 50 10

122 293 151398 10= +

RSTUVW

=

=

.

. .. (1)

Entropy balance:

S S S S S S S Sf if f i i

1 1 1 1 1 1 1 10 = = + IG IG, IG, IGd i d i d i

or

S S CT

RPf f f

1 11

729315 50 109 6 +

= IG Pd i ln . ln . . (2)

Energy balance:

N U N U N Ui i f f f f1 1 1 1 2 2= + but N N N N U U N U Ui f f f f i f f i1 1 2 1 1 1 2 2 2 0= + + =d i d i

or

Solutions to Chemical and Engineering Thermodynamics, 3e

P V

Z RTH P V H P V

P V

Z RTH P V H P V

Z TH H H H H H Z RT Z RT

Z TH H H H H H

f

f ff f f i i i

f

f ff f f i i i

f ff f f i i i f f i i

f ff f f i i i

1 1

1 11 1 1 1 1 1

2 2

2 22 2 2 1 1 1

1 11 1 1 1 1 1 1 1 1 1

2 22 2 2 1 1 1

0

1

1

+ =

+ +

+ +

d i d in s d i d in s

d i d i d in s

d i d i

, IG , IG , IG , IG

, IG , IG , IG , IGd in s + =Z RT Z RTf f i i2 2 1 1 0

Substituting in the known values gives

135565 29315 8314 6 406 5

135565 29315 8 314 6 406 5 0

1 11 1 1 1 1

2 22 2 2 2 2

Z TH H T Z T

Z TH H T Z T

f ff f f f f

f ff f f f f

+ +

+ + + =

, IG

, IG

d i c hn s

d i c hn s

. . . , .

. . . , . (3)

Eqns. (1-3) now must be solved. One possible procedure isi) Guess P f

ii) Use Eqn. (2) to find T f1iii) Use Eqn. (1) to find T f2iv) Use Eqn. (3), together with T f1 and T

f2 to see if guessed P

f is correct. If not, go back to

step i.After many iterations, I found the following solution Pf = 9787. bar ; T f1 2216 K= . ;

T f2 259 4= . K ; N Nf i

1 1 0 645= . ; N Nf i

2 1 0 355= . .(c) Peng-Robinson equation of state

Here we use the equations

N N Ni f f1 1 2= + (4)

N U N U N Ui i f f f f1 1 1 1 2 2= + with U H PV= (5)

S Si f1 = (6)

P P Pf f f1 2= =

and N V Vi i1 1 1= ; N V Vf f

1 1 1= ; N V V V Vf f f

2 2 2 1 2= = since V V1 2= (value of V1 cancels outof problem, so any convenient value may be used). Procedure I used to solve problem was as

follows. From PR1 we know V Ni i1 1c h and Si1 given initial conditions. Then1. Guess value of T f1 , find P P

f f1 = that satisfies S S

f i1 1=

2. Use T f1 , Pf and V f1 to get N

f1 ; then N N N

f i f2 1 1= so V

f2 is known.

3. From P f and V f2 find (trial-and-error with PR1) Tf

2

4. See if eqn. (5) energy balance is satisfied; if not go back to step 1. After a number of

iterations I find Pf =103 6 bar. ; T f1 222 3= . K ; Tf

2 2555= . K ; N Nf i

1 1 0 619= . ;

N Nf i2 1 0381= . .

Solutions to Chemical and Engineering Thermodynamics, 3e

Summaryideal gas(part a)

Correspondingstates (part b)

P-R E.O.S.(part c)

P f 250 bar 97.87 103.6

T f1 249.3 K 221.6 K 222.3 K

T f2 355.9 K 259.4 K 255.5 K

N Nf i1 10.588 0.645 0.619

N Nf i2 1 0.412 0.355 0.381

Clearly, the ideal gas assumption is seriously in error!

4.21 System = contents of compressor. This is a steady-state, open constant volume system.

mass balance: dN

dtN N= = +0 1 2& &

energy balance: dU

dtN H N H Q W Ps= = + + + 0 1 1 2 2& & & &

dV

dt

0

= + +0 1 1 2& & &N H H Q Wsa fentropy balance:

dS

dtN S N S

Q

T= = + + +0 1 1 2 2& &

&S gen

0

.= +&

&N S S

Q

T1 1 2a f

Thus,

& &Q TN S S= 1 1 2a f&

&Q

NQ T S S

12 1= = a f

and

& &

&W Q

NQ H HS

+= + =

12 1W

(a) Corresponding states solution

Q T S S T S S S S S S

T S S S S RP

P

= = +

= RSTUVW

2 1 2 2 2 1 1 1

2 2 1 12

1

a f d i d i d in s

d i d i

IG IG IG IG

IG IG ln

Now Tr = =37315

405 6092

.

.. ; Pr , .

~ . 1 =1

112 80 009 ; Pr , .

. 2 = =50

112 80 443 . Thus

Q RT T S S S SPT

PT

r

r

r

r

= + RST

UVW= + ( ) =

==

==

ln

. . ln . . .

, ,..

..

50

1

8314 37315 50 37315 523 0 14,088 1

2 2 0 4440 92

1 1 0 0090 92

2 1

IG IG

J mol

d i d i

and

W + = = +

=

Q H H H H H H H H

T

2 1 2 2 20

1 1 1( ) ( ) ( )IG IG

sinceconstant

IG IG