Eliot Lira Intro to Thermodynamics in Chemical Engineering

P. 8.1) CO2 at 15MPa and 25C is throttled to 0.1Mpa. Determine the temperature andfraction vapor. (NOTE: there is a typo. Part (c ) makes no sense at 1.5MPa.)Solution: Ebal for a valve, 0=H(a) assume ideal gas,( ) KTTCpH 2980 2 ===(b) by PREOS.XLS with Ref=224.1K,0.1MPa, Liquid.H2 = H1 = 9695 J/mol, then T2sat = 184.1K. q=(9695-0)/(17066-0)= 57%Note: This is just an estimate, because CO2 is not really liquid at 0.1MPaNote2: If you solved this using 1.5MPa, you get 281K.( c ) Note: for the problem statement at 1.5MPa, there is no difference from (b).At 5.27bar, 216.6K, Hsolid=HsatVap-Hsub,Hsub=Hvap+Hfus=17829-2420+43.2*44*4.184=23362 Hsolid=17829-23362= -5533.C-Ceq ln(Psat/Pref)= -(Hsub/R)*(1/T-1/Tref) 1/T=1/216.6-ln(1/5.27)/(23362/8.314)=0.005208 T=192At 1bar, 192K=HsatVap=17322 Hsolid=17322-23362= -6040 q = (9695+6040)/(23362) = 67%FYI: This one way to make dry ice. You throttle/spray it into bag and compress itinto a block. See also CO2 chart in Perrys Handbook, Chapter 3.

P8.2) CO is liquefied from 300bar,150K to 1bar. Compute q and Sgen.Soln: Ref=82.0K,0.1MPa,liq. H(150,30)=4139. Ebal:H=0; q=(4139-0)/(6048-0)=68%Sf=0.68*73.83+0 = 50.2; S(150,30)=27.365 Sgen=50.2-27.4 = 22.8 J/mol-K.P8.3) Same CO using 90% efficient turbine.Srev=0 q=27.635/73.83=0.374; Wrev=0.374*6048-4139= -1877J/molWact=-1689Hout=4139-1689=2450 q = 2450/6048=40.5% 1-q=59.5%P8.4) Methane at 300K, 250bar is liquefied (~Linde style) with outlet pressure = 30bar.Compute fraction liquefied.Soln: Ref=111.8K,1bar; qH8+(1-q)H6=H3=10654; Tsat(30bar)=176.7KHsatVap=8289;HsatLiq=4104; H8=H(300,3MPa)=13455;q=(10654-4104)/(13455-4104)=70% 1-q=30%

Chapter 10 Practice Problem Solutions

To accompany Introductory Chemical Engineering Thermodynamics J.R. Elliott, C.T. Lira, 2001, all rights reserved. (07/06/01) 1 of 4

(P10.5) (a) Perform bubble P calculations 1-CO2, 2- ethylene. For kij = 0 Output from PRMIX.exe bp COMPONENT IS CARBON DIOXIDE ID NO. IS 909 COMPONENT IS ETHYLENE ID NO. IS 201 T(K)= 222.00 P(MPa)= .8725 ZL= .2138E-01 ZV= .8674 ID LIQUID X VAPOR Y Yi/Xi 909 .5000 .4057 .8114 201 .5000 .5943 1.189 repeating across the composition range: x1 y1 P(MPa) 0.0 0.0 1.02 0.1 0.0795 0.995 0.5 0.406 0.873 0.9 0.834 0.7012 1. 1. 0.642

End points were determined using bubble point pressure calculation using PRMIX for a single component. No azeotrope exists. Answer is 0.87MPa

(b) Output from prmix.exe: Kij MATRIX MODIFIED 909 201 201 .1100

bp COMPONENT IS CARBON DIOXIDE ID NO. IS 909 COMPONENT IS ETHYLENE ID NO. IS 201 T(K)= 222.00 P(MPa)= 1.126 ZL= .2867E-01 ZV= .8358 ID LIQUID X VAPOR Y Yi/Xi 909 .5000 .4237 .8475 201 .5000 .5763 1.153

For kij = 0.11 x1 y1 P(MPa) 0.0 0.0 1.02 0.1 0.139 1.09 0.5 0.424 1.126 0.9 0.691 0.882 1. 1. 0.642

at small x1, x1y1. Also, P maximum in mixture. Therefore, maximum P (minimum T) azeotrope will exist. Answer is 11.3 bar.



(P10.6) (a) .Pentane 7 Acetone 1051 THE DEFAULT Kij MATRIX IS 7 1051 1051 .0000

bp COMPONENT IS n-PENTANE ID NO. IS 7 COMPONENT IS ACETONE ID NO. IS 1051 T(K)= 305.05 P(MPa)= .7785E-01 ZL= .3247E-02 ZV= .9697 ID LIQUID X VAPOR Y Yi/Xi 7 .7280 .8345 1.146 1051 .2720 .1655 .6083

Answer: 0.78 bar, y1 = 0.83

(b) x=0.134 using option KI, perform bubble pressure calcs until converge on experimental pressure of 1 bar at 0.728. ki Kij = 0.11 REQUIRED NUMBER OF ITERATIONS WAS: 7 COMPONENT IS n-PENTANE ID NO. IS 7 COMPONENT IS ACETONE ID NO. IS 1051 T(K)= 305.05 P(MPa)= .9927E-01 ZL= .4199E-02 ZV= .9634 ID LIQUID X VAPOR Y Yi/Xi 7 .7280 .7167 .9845 1051 .2720 .2833 1.041

Kij = 0.117 REQUIRED NUMBER OF ITERATIONS WAS: 5 COMPONENT IS n-PENTANE ID NO. IS 7 COMPONENT IS ACETONE ID NO. IS 1051 T(K)= 305.05 P(MPa)= .1011 ZL= .4282E-02 ZV= .9629 ID LIQUID X VAPOR Y Yi/Xi 7 .7280 .7078 .9723 1051 .2720 .2922 1.074

Intermediate Answer: kij = 0.117 to fit bubble pressure, azeotrope composition not matched exactly.

bp COMPONENT IS n-PENTANE ID NO. IS 7 COMPONENT IS ACETONE ID NO. IS 1051 T(K)= 312.75 P(MPa)= .1120 ZL= .3881E-02 ZV= .9626 ID LIQUID X VAPOR Y Yi/Xi 7 .1340 .5453 4.070 1051 .8660 .4547 .5250

Answer: kij = 0.117, BP = 1.12 bar



(P10.7)

a) using shortcut K-ratio equation at 298K, predict the vapor pressure of components.

# COMPOUND Psat ( MPa) 1 CO2 6.44 2 METHANE 32.6 3 PROPANE 0.956 4 ETHANE 4.21

Note: methane is supercritical so the vapor pressure is extrapolated. yiP = xiPisat, xi = yiP/Pisat ! guess P until xi = 1. PNEW = POLD / Xi

P(MPa) X1 X2 X3 X4 Xi 3 0.14 0.028 0.628 0.143 0.939 3.19 0.149 0.029 0.667 0.152 0.997 3.2 0.15 0.0295 0667 0.153 0.9995

(b) there isnt a DP routing, use dt routine, guess P until DT = 298 REQUIRED NUMBER OF ITERATIONS WAS: 6 COMPONENT IS CARBON DIOXIDE ID NO. IS 909 COMPONENT IS METHANE ID NO. IS 1 COMPONENT IS PROPANE ID NO. IS 3 COMPONENT IS ETHANE ID NO. IS 2 T(K)= 286.67 P(MPa)= 3.000 ZL= .9407E-01 ZV= .7509 ID LIQUID X VAPOR Y Yi/Xi 909 .2207 .3000 1.360 1 .0715 .3000 4.198 3 .5197 .2000 .3848 2 .1882 .2000 1.063

REQUIRED NUMBER OF ITERATIONS WAS: 6 T(K)= 295.61 P(MPa)= 4.000 ZL= .1286 ZV= .6876 ID LIQUID X VAPOR Y Yi/Xi 909 .2410 .3000 1.245 1 .0958 .3000 3.132 3 .4637 .2000 .4313 2 .1995 .2000 1.003

REQUIRED NUMBER OF ITERATIONS WAS: 5 T(K)= 298.45 P(MPa)= 4.400 ZL= .1434 ZV= .6624 ID LIQUID X VAPOR Y Yi/Xi 909 .2486 .3000 1.207 1 .1064 .3000 2.819 3 .4421 .2000 .4524 2 .2029 .2000 .9856



P = 4.4MPA, DT = 298.5 close enough.

P(MPa) x1 x2 x3 x4 4.4 0.249 0.107 0.442 0.203

PR predicts much different P and composition. PR should be improved by using non-zero kij from binary data fits for even more accuracy.

(P10.8)

bcbZ

+=141 ,

=

d

bdb

=

=

b

TVig

bcbdb

c

RTAA

0)1ln(4)()1(4

)(.....Eqn. 7.27

( ) Zn

RTAART

ijnVTi

TVigig

iii ln

/)()(ln

,,

=

=

.Eqn. 10.16

( ) Zn

bnc

ijnVTii ln

))1ln(*4(ln

,,

=

Zn

bb

ncn

ncbijij nVTinVTi

ln)(1

14)()1ln(4,,,,

+

=

ccxn

nc

iiji

nVTi ii

=

2,,

....Eqn. 10.29

inVTi

bn

nb

ii

=

,,

... Eqn. 10.22

second term in iln becomes equal to )1(14

14 =

=

Zbb

bcb

bb

bb

cn iii

ZZbb

bcxc ji

ijij ln)1()1ln(24ln +

=


To accompany Introductory Chemical Engineering Thermodynamics J.R. Elliott, C.T. Lira, 2001, all rights reserved. (02/11/02) 16

(P14.1) An equimolar mixture of H2 and CO can be obtained by the reaction of steam with coal. Compute the equilibrium compositions at 550 C based on an equimolar feed of H2, CO, and H2O. The reaction is H

O + CO = H

+ CO

. Gf data at 550 C are given.

!"# $%

Compound In Out H2O 0.333 0.333 CO 0.333 0.333 H2 0.333 0.333+ CO2 0 Total 1 1

Ka = 2

2

( )(0.333 )*(0.333 )*(0.333 )*

PP

+

= 3.369

= 0.176; Ans. y1 = 0.157, y2 = 0.157, y3 = 0.509, y4 = 0.176

& '(')('*+*)),-*'+,',)+)(

*

)+')-.G46,)'))(0'

,)) -+)1

2"

3

32

4)56.-,),7,,)+,8(,2

8(,

3

Compound In Out N2 1 1 C2H2 1 1 HCN 0 2 Total 2 2 !"#

2 2

2 2

(2* ) *(1 ) *

PP

0

7, %$0

%$0

$



(P.14.3) Butadiene can be prepared by the gas-phase catalytic dehydrogenation of 1-Butene: C4H8 = C4H6 + H2. In order to suppress side reactions, the butene is diluted with steam before it passes into the reactor. (a) Estimate the temperature at which the reactor must be operated in order to convert 30% of

the 1-butene to 1,3-butadiene at a reactor pressure of 2 bar from a feed consisting of 12 mol of steam per mole of 1-butene.

(b) If the initial mixture consists of 50 mol% steam and 50mol% 1-butene, how will the required temperature be affected?

Gf 600K 700K 800K 900K C4H6 195.73 211.71 227.94 244.35 C4H8 150.92 178.78 206.89 235.35 Solution: Compound In Out C4H8 1 1- C4H6 0 H2 0 H2O 12 12 Total 13 13+

P = 2 bar

( ) ( )

22

2*13 *13 * 11

130.01933

PPKa

P

Ka

+ = =

+ + =

Noting that lnKa = -Gtot/RT, we can identify the temperature by fitting a trendline to the given data. ln 0.01933 = -3.95, substitute in the equation of straight line, x = -(-14.34 3.95 )/13996. x = 0.001306 = 1/T, T = 765.22 K = 492OC

y = -13996x + 14.34

-10-9-8-7

-6-5-4-3-2-100.001 0.0012 0.0014 0.0016 0.0018

1 / T

ln K

a

Compound In Out C4H8 1 1- C4H6 0 H2 0 H2O 1 1 Total 2 2+

( ) ( )CKTSimilarly

PKa

O6.57365.846,

1118.01*2

*2

==

=

+=

We need higher T.



(P14.4) The standard Gibbs energy change for ethylene oxide at 298K for the reaction is 79.79 kJ/mole. This large negative value of GT indicates that equilibrium is far to the right at 298K but what about 550K? Heat capacity expressions are given as CP= a + b T. Solution: The heat of reaction must be looked up. Referring to Apx E.6 for ethylene and the DIPPR handbook for ethylene oxide, H298 = -52.6-52.51 = -105.1 kJ/mol. Following Eqs. 14.28 and 14.30, -105100 = J + (6.57-15.4-26.65/2) 298 + (0.1389-0.0937-0.00845/2) 2982/2 -105100 = J 22.155 *298 + 0.040975*2982/2 J =-100317 J/mol -79790/(8.314*298) = -100317/(8.314*298)

79790 100317 22.155 0.040975ln 298 298 6.168.314*298 8.314*298 8.314 2*8.314

I I = + => =

100317 22.155 0.040975ln 550 550 6.16 12.68.314*550 8.314*550 8.314 2*8.314

G = + =

o

G = -57.7 kJ/mol. If T = 550 K, increasing T will give an adverse effect on equilibrium, but the reaction is still very strongly favored and the impact will be indistinguishable. (P14.5) The water gas shift is to be carried out at a specified temperature and pressure employing a feed containing only CO and H2O. Show that the maximum equilibrium mole fraction of H2 in the product results when the feed contains CO and H2O in their stoichiometric proportions. Assume ideal gas behavior. Solution: 222 HCOOHCO +=+ Compound In Out CO z z- H2O 1 1 CO2 0 H2 0 Total 1+z 1+z

( )( )( )( )

2

2

2 2

1

1 0

0 [ (1 ) ]

Kaz

Ka z

Ka z z

=

=

= + +

2(1-Ka) +Ka (1+z) - Ka z = 0 = {-Ka(1+z) + [Ka2(1+z)2+4z(1-Ka)Ka]1/2 }/[2(1-Ka)] yH2 = /(1+z) = {-Ka(1+z) + [Ka2(1+z)2+4z(1-Ka)Ka]1/2 }/[2(1+z)(1-Ka)] yH2 = /(1+z) = {-Ka+ [Ka2+4z(1-Ka)Ka/(1+z)2]1/2 }/[2(1-Ka)] To find maximum, take derivative and set equal to zero. dy/dz = 0.5[Ka2+4z(1-Ka)Ka/(1+z)2]-1/2 [4(1-Ka)Ka/(1+z)2 8z(1-Ka)Ka/(1+z)3] = 0 1 = 2z/(1+z) z = 1. QED. (P14.6) Assuming ideal gas behavior, estimate the equilibrium composition at 400K and 1 bar of a reactive mixture containing the three isomers of pentane. Formation data are given at 400K. Solution: This is best solved by the Gibbs minimization method, adapting Example 14.10 and GibbsMin from the Rxns.xls workbook, we obtain the following.

Gf(J/mole) Gf400/RT feed ni log(ni) yi ni(Gi/RT+lnyi)nPentane 40170 12.08 0 0.111 -0.95 0.111 1.100iPentane 34310 10.32 1 0.648 -0.19 0.648 6.408neoPentane 37610 11.31 0 0.240 -0.62 0.240 2.376Tot 1 1.000 9.884

Out InC-bal 5 5Hbal 12 12



(P14.7) One method for the manufacture of synthesis gas depends on the vapor-phase catalytic reaction of methane with steam according to the equation below. The water-gas shift reaction also is important. Bases on stoichiometric feed of methane and steam, compute the eq composition at 600K, 1300K and 1, 100 bars.

4 2 23CH H O CO H+ = + rxn(1)

222 COHCOOH +=+ rxn(2) Compound In Out CH4 1 11 H2O 1 112 CO 0 12 H2 0 31+2 CO2 0 2 Total 2 2+21 600 164.68 214.01 22.97 72.3 / ......................... (1)KG kJ mole rxn = + + = 600 395.14 214.01 164.68 16.45 / ...................... (2)KG kJ mole rxn = + + =

05.27600*314.8

16450exp

708.5600*314.8/72300exp

2,600

1,600

=

=

=

=

K

K

Ka

EmoleJKa

( )( )

( )( )

3 41 2 1 2

41

1 21 1 2

21

3 21 2 1 2

21 1 2 1

( )(3 ) *(2 2 )

1 1 *(2 2 )

( )(3 ) *1 1 (2 2 )

P

KaP

P

++

=

+

+=

+

( )( ) ( )( )2

2 1 22 1 2

2 21 1 2 1 1 2

( )*( )4

1 * 14

PKa

P

++

= =

Note: high pressure tends to disfavor rxn (1). Rxn 1 is negligible at 600K, and rxn (2) requires CO to run or 1-2 will be less than zero. So both reactions are zero. At 1300 K, the situation is quite different.

1300 226.94 175.81 53.30 104.73KG = + = Ka1 = 16113

1300 396.14 226.94 175.81 6.614KG = + + = Ka2 = 0.54 Solving by method of Example 14.9, 1 = 0.972 and 2 = 0.015 at 1 bar. At 100 bar, 1 = 0.451 and 2 = 0.149 at 100 bar

PP 14.7Two simultaneous reactions:CH4 + H2O =CO + 3H2H2+CO2 = CO + H2O(Details of equations described in text)P(bars) 100T(K) 1300Ka1 16113Ka2 0.54001 0.45132 0.1491y1 0.1891y2 0.1377y3 0.1041y4 0.5178y5 0.0514nTot 2.9025Objective Functionserr1 0.0000err2 0.0000



(P14.8) Is there any danger that solid carbon will form at 550C and 1 bar by the reaction: 22 COCCO S +=

molekJH K /298 molekJG K /298 IN Out CO -110.53 -137.16 2 2-2 Cs 0 0 0 0 (gas) CO2 -393.51 -394.38 0 Tot 2-

molekJHmolekJG

TK

TK

/45.17253.110*251.393/06.12016.137*238.394

,298

,298

=+==+=

Increasing T, adverse affect on equilibrium

2108.115.298*314.8

120060expexp 298298 ERTGKa KK =

=

=

Using Shortcut Vant Hoff Eq. 14.31

( )

298

298 298

1 1 172450 1 1ln ln1.08 21 8.314 823.15 298.15

ln ln 1.08 21 44.4 4.06 exp(4.06) 57.99

HKa KaKa R T T E

Ka E Ka

= = = = = = =

( )( )

( )( )

( )( )

2

2 2 2

2* * 257.99

* 2 22 22

CO

CO

y PKa

y P

= = = =

= 0.9345 ratio of carbon solid to feed is 2

9345.0 = 0.46725. There is danger.

Note: this exemplifies a very important and undesirable side reaction in many catalytic reactions know as coking. The carbon tends to clog the catalyst pores and substantially reduce its effectiveness. Because of this problem, fluidized catalytic crackers were developed (aka. Cat crackers). The solid catalyst particles are fluidized by the upflow of gaseous reactants. As they ultimately settle at the bottom, they are removed and recirculated through an oxidation zone that burns off the coke then recycles the catalyst to the top of the bed. This is a good example of how thermodynamics impacts reactor design. (P14.9) Calculate the equilibrium percent conversion of ethylene oxide to ethylene glycol at 298K and 1 bar if the initial molar ratio of ethylene oxide to water is 3. In Out(zi) EtO 3 3- Water 1 (1-) Glycol 0 Tot 4- Kw = Psat/P = 0.0425; KEtO = 1.76; KGly = 8.6E-4; yi = zi Ki /[ Ki +L/F*(1- Ki)]

2

7824exp 23.52* * 298.15*8.314

Gly

EtO W

y PKa

y y P

= = = and yi = 1 are constraints, and L/F are unknown. Guess, = 0.99, L/F = 1/3 (all glycol in liquid, all EtO in vapor).



(P14.9) Sample solution of one reaction with vle:(Details of input equations described in text by Elliott and Lira)P(bar) T(K) Ka1

1.000 298 23.52

pSat(bar) K-ratios zFeed yi xiEtO 1.76000 1.76000 0.66676 0.99961 0.56796Water 0.04250 0.04250 0.00029 1.579E-05 0.00037Glycol 0.00086 0.00086 0.33295 0.00037 0.43167

1.0000 1.0000 1.00001- 8.7055E-04 0.99913 sum(yi-xi) 0.00000L/F 0.77111 ErrKa 5.583E-07 As it turns out, the ethylene oxide is not so volatile after all and dissolves a fair amount in the liquid. The guess about the extent of conversion being high was good though. A more clever engineering approach would be to assume complete conversion and solve the simple flash. Then back out the exact conversion assuming L/F does not change. (P14.10) Acetic acid vapor dimerizes according to 2A1 = A2. Assume that no higher-order associations occur. Supposing that a value for Ka is available, and that the monomers and dimers behave as an ideal gas, derive an expression for yA1 in terms of P and Ka. Then develop an expression for PV/n0RT in terms of yA1, where n0 is the superficial number of moles neglecting dimerization. Hint: write n0/nT in terms of yA1 where nT = n1+n2. Solution:

0 1 21 1 1

2 2(1 ) 2A A AT T

n n n y y yn n

+= = + =

221 1 12

1

1 1 412

AA A A

A

y PKaKa y y PKa yy P PKa

+ += = =

0 0 1

1Ideal gas 12

T

T A

PV PV nn RT n RT n y

= = =

Note: as Ka , PV/n0RT because the monomer is converted to dimer. Note also that PV/n0RT is what we normally refer to as the compressibility factor, Z. This is an interesting result with regard to equations of state and phase equilibria. Since Ka is simply a function of temperature [ie. exp(-G/RT)], it says that we can compute Z given a pressure and temperature. This is analogous to the pressure explicit virial equation (Section 6.4), but the form of the pressure dependence is more complex. Exploring this perspective, generalizing to density-dependent equations, and adapting to multimer-forming species and mixtures is the subject of Chapter 15. Most of the physical insight contained in Chapter 15 is contained in this simple practice problem.

Chapter 3 Practice Problems


(P3.1) (a) the number of microstates is N2 (pg 91, typo in given answer, printings 1-3) (b) 3 particles total

3!1!*2

!3}1,2{ == THp number microstates of specific arrangement (macrostate)

probability = (# microstates of specific arrangement)/(total # of microstates)

83

23

3 ==prob

(c ) # microstates.

20!3!*3

!6

15!2!*4

!6

10!2!*3

!5

6!2!*2

!4

}3,3{

}2,4{

}2,3{

}2,2{

==

==

==

==

TH

TH

TH

TH

p

p

p

p

(d) macrostate

H T # of microstates*

0 8 1 1 7 8 2 6 28 3 5 56 4 4 70 5 3 56 6 2 28 7 1 8 8 0 1

* number of microstates = )!8(!!8

mm

total number of microstates is 28 = 256, which is the same as the sum from the table. portion of microstates (probability) for requested configurations: {5:3} = 56/256 = 0.219 = 22% {4:4} = 70/256 = 0.273 = 27% {3:5) = 22% like {5:3} probability of any one of the three most evenly distributed states = 22% + 27% + 22% = 71% (e) for 8 particle system, Stirlings approx will not apply S/k = ln(p{4:4}/p{5:3}) = ln (70/56) = 0.223



(P3.3) 15 molecules in 3 boxes, molecules are identical

!

!

1ij

i

jm

Np=

= ....Eqn. 3.4

75075!2!4!9

!151 ==p

756756)!5(!1532 ==p

31.2ln1

2=

=

pp

k

S

(P3.4) two dices. ??=

kS for going from double sixes to a four and three.

for double sixes, we have probability of 1/6 for each dice.

( ) ( )!61!*61!2

1 =p

for one four and one three probability applied for 1/6 for each one in each dice,

( ) ( ) 2*!61!*61!2

2 = p

693.02lnln1

2==

=

pp

kS

(P3.5) S = ?? Assume Nitrogen is an Ideal gas RTPV = . Eqn. 1.15

( )( ) MPaLcmmoleL

KKmoleMPacmP 108.0)1/1000(*/23300*/*314.8

3

3

1 =

=

Similarly MPaP 00723.02 =

1

2

1

2 lnlnPP

RTTCpS = Eqn. 3.23

27RCp = ..(ig)

KkgkJKmoleJRS === /07.1/88.30108.0

00723.0ln*314.8300400ln*

27

(P3.6) (a) m-balance: dnin = -dnout S-balance:



dtdnS

dtnSd outoutin

=

)( outoutinininin dnSdnSdSn =+

But physically, we know that the leaking fluid is at the same state as the fluid in the tank; therefore, the S-balance becomes:

0 so and ,)()( ===+ SdndnSdnSdnndS outinsideoutinside from the steam table .

By interpolation, implies T = 120.8OC (P3.7) (a) Steady-state flow, H = Ws

Start 1 mole basis: 221121 ,,667.0,333.0 CpxCpxCpadiabaticxx +=== , Cp for each is the same anyway.

)/(66.102*667.0)1612(333.02211 molegMWxMWxMW =++=+= R = 1.987BTU/lbmol-R.

( )

hBTUWHlbmol

BTUlb

lbmolh

lbH

lbmollbMWhlbhtonm

lbmolBTUH

RRCpdTWH

S

T

TS

/10*3.1000,305,1

5.6954*

66.10*

2000/66.10&

./2000/1&/5.6954

1001100**27

6

2

1

===

=

=

==

=

=== &

(b) ??= of the compressor. To find the efficiency of the compressor, S1 = S2 But the enthalpy and the internal energy will change which gives a change in the

Work. ??'

==

S

S

WW

At 1 bar = 0.1 MPa T S

100 7.361 120.8 7.4669

150 7.6148

State P(Mpa) TOC H(kJ/kg) S(kJ/kg*K) 1(in) 1 400 3264.5 7.4669

2 (out) 0.1 120.8 2717.86 7.4669



RT

RT

TPP

T

PP

TT

PP

RTTCp

PP

RTTCpS

CpR

RCp

1315

559*5

100

*

lnln

lnln0

2

72

2

11

22

1

2

1

2

1

2

1

2

1

2

1

2

=

=

=

=

=

==

lbmolBTUHTTCpH

/5258)5591315(95.6)(& 12

=== 76.0

69555258

==

=HH

(P3.8) Adiabatic, steady-state open system Q = 0, &(Cp/R = 7/2) ig kgkJ

kgkmolekmolekJRCpdTW /76.33728

1*/175.9457)300625(*

27625

300

==== ??=

( ) ( )molkJH

TTCpH

KTPP

TTS

CpR

/77.6794

3005.533*2314.8*7

5.533

0

12

2

1

2

1

2

=

==

=

=

=

%8.71

718.0/28*/76.337

/77.6794

=

==

=

kmolkgkgkJ

kmolkJHH

(P3.9) work required per kg of steam through this compressor? By looking at the steam table in the back of the book

P(MPa) T(OC) H(kJ/kg) S(kJ/kg-K) 0.8 200 2839.7 6.8176

4 500 3446 7.0922

kgkJHW /3.6067.28393446 === now find W = ??



S = 0 (reversible), look in the steam table (@P = 4.0MPa) to find a similar value for S = 6.8176kJ/kg-K, if this value is not available so find it by interpolation.

7.3246',7714.69386.68176.69386.6

5.32142.3331'5.3331

=

=

HH

kgkJWH /4077.28397.3246'' ===

%67,67.03.606

407===

(P3.10)@ P = 2.0 MPa & T = 600OC, H = 3690.7 kJ/kg, S = 7.7043kJ/kg-K (Steam table)

kgJkHqHH VapL /49.2493)68.2441(*98.046.1006)( =+=+= kgkJHWS /21.119749.24937.3690 ===

??= , '' W

WHH

=

= ,

S = 0 ( reversible), look for S in the satd temp. steam table and find H by interpolation, kgkJW /0.1408'=

%85,8503.00.14082.1197

===

(P3.11)

CTMPaP

dSatMPaP

O

vap

110010

',1.0

2

2

1

=

=

=

State P(MPa) T(OC) H(kJ/kg) S(kJ/kg-K) 1 0.1 99.61 2674.95 7.3589

2' 10 4062.53 7.3589 2 10 1100 4870.3 8.0288

interpolation for above table:

H(kJ/kg) S(kJ/kg-K) 3214.5 6.7714

H' = ?? S' = 6.8176 3331.2 6.9386

TOC HL(kJ/kg) Hvap(kJ/kg) steam table 20 83.91 2453.52 Interpolation 24 100.646 2444.098 steam table 24 104.83 2441.68



kgkJWH S /35.219595.26743.4870 === mass flow rate = 1 kg/s

hpWhpwatt

wattskJW

S

S

01.2944001341022.01&

2195350/35.2195

=

=

==

&

&

%2.63

63.035.219558.1387

/58.138795.267453.4062&

=

==

=

==

HH

kgkJH

(P3.13) Through the valve outin HH = MPaPin 3= MPaPout 1.0= KCT Oout 15.383110 ==

(By interpolation) Find outH from steam table.

8.26756.27766.2776

100150110150

=

outH

outH = 2695.96 kJ/kg

At 3MPa table use same value for Hin to find Sin

By interpolation 1856.62893.6

2893.62.28035.2856

96.26955.2856

=

inS

Sin = 5.976kJ/kg-K The process should be irreversible. To find Sout, interpolate using temperature at 0.1 MPa:

3610.76148.76148.7

100150110150

=

outS

Sout = 7.4118 kJ/kg-K, since Sout>Sin entropy has been generated. The entropy balance is:

genoutoutinin SmSmS &&& +=0

H'2 = 4062.53 (interpolation) H(kJ/kg) S(kJ/kg-K)

3992 7.2916 4062.53 7.3589 4114.5 7.4085



(P6.2) 15,30 == rr TP (a) Use virial equation of state.

rr TPBBZ /)(1 10 ++= ..Eqn. 6.6

Where,

2.41

6.10

172.0139.0

422.0083.0

r

r

TB

TB

=

=

. Eqns. 6.8 & 6.9

)(041.0&138998.0&077459277.0

1

0

bookBB

=

=

=

14.11530

*)138998.0*041.0077459277.0(1 =+= Z (b) ??=

MPaPPPKTTT

Cr

Cr

59.7930*653.2*6664.44*15*===

===

3

3

/254.093.311

/93.3179.20*59.79

666*314.8*14.1*

**,

cmgV

gcmMWP

TRZVRTPVZ

===

====

(P6.4)

KCFTMPaatmP

KT

OO 15.29825771.01

111

2

1

1

===

=

=

Use PREOS.XLS,

Use Solver, and set target cell on the volume and make it equal to 33.639114*2 = 67.278228,

Then by changing the cell of pressure, making sure that T2 = 298.15K

Current State Roots T (K) 298.15 Z V P (MPa) 33.839895 cm3/gmol

Current State Roots

Stable Root has a lower fugacity

T (K) 111 Z V fugacity P (MPa) 0.1 cm3/gmol MPa

answers for three 0.9670679 8924.6249 0.096803root region 0.0263855 243.49944

0.0036451 33.639114 0.093707



answers for three #NUM! #NUM! root region #NUM! #NUM!

#NUM! #NUM! & for 1 root region 0.9184568 67.278228

MPaP 84.332 =



(P7.1) TSHG

RSS

RTHH

RTGG igigig = Eqn. 7.21

( ) 11231

02/5

0

+

+

=+

= ZdbRT aTZdTZTRTHH

ig

note: )(log1 baxabax

dxe +=+

( ) ( ) 11ln12

311ln12

32/3

02/3 +

+=+

+= Zb

bRTaZ

Rab

bT

( ) ( ) ZdZTZTRSS ig ln10 + =

..Eqn. 7.23

( ) ( )( ) ( ) ( ) Zb

RTabb

bRTa

ZdbRT

ab

bdbRT

a

ln1ln1ln1ln123

ln1112

3

2/32/3

02/3

02/3

++++

+=

+

+

+=

( ) ( ) ( ) ( )( ) ( ) ZZb

bRTab

RTGG

ZbRT

abbbRT

aZbbRT

aRSS

RTHH

RTGG

ig

igigig

ln11ln1ln

ln1ln1ln1ln12311ln1

23

2/3

2/32/32/3

++=

++++

++

+

=

Or by using Eqn. (7.26)

( ) ( ) ZZdZRT

GG ig ln11

0

+=

( ) ( ) ZbbRT

abRT

GG ig 1ln1ln 2/3 ++= ( ) ( ) ZZ

dbRT

ab

bRT

GG ig ln1110

2/3 +

+=



(P7.2) rTbZ /1 =Depature function 1

0

+

=

ZdTZTRTHH

ig

Eqn. 7.24

rrr

r

rrr

ig

Tb

Tb

TbT

Tbd

TbT

RTHH

2

220

==

=

(P7.3) (a) UCvdtT

T

=21

, CvTU

V

=

dTZT

RUU ig

=

0

2

dT

ZTdTdZT

TU

V

)2( 22

2

=

dT

ZTdTdZT

RCvCv ig )2( 2

22

0

=

(b) ( ) +++= )exp(11 TabbZ( )

( ) ( ))exp()exp(2)exp(

4

2

32

2

2

Ta

Ta

Ta

Ta

TZ

Ta

Ta

dTdZ

=

=

( ) ( ) ( ) d

Ta

TaT

Ta

TaTT

aTaT

RCvCv ig })exp()exp(2)exp(2{ 4

22

32

02 ++=

First two terms cancel, integral with respect to is simple.

( ))exp(22 TaTaRCvCv ig =



(P7.4) Helmholtz Energy, A = ??Depature function,

RSS

RTUU

RTAA igigig = .Eqn.7.20

dTZT

RTUU ig

=0

.Eqn. 7.22

RTbNd

RTbNT AA 5.95.9

02

=

= ( ) ZdZd

TZT

RSS ig ln1

00

+

=

Eqn. 7.23

( ) ZRT

bNbRT

bN

ZdRT

bNdbbd

RTbNT

AA

AA

ln5.9

1ln45.9

ln5.91

45.9

0002

+++=

++

=

( ) ZbRT

bNRT

AA Aig

ln1ln45.9 =

or by using Eqn. (7.25)

( ) ZbRT

bNRT

AA

ZdRT

bNdbb

RTAA

ZdZRT

AA

Aig

Aig

ig

ln1ln45.9

ln5.9

14

ln1

00

0

=

=

=

(P7.5) Compute H,S,U,V of 1,3 butadiene from 25bar,400K to 125bar,550K.For State 2: T (K) 550 Z V H-Hig U-Uig S-SigP (MPa) 12.5 cm3/gmol J/mol J/mol J/molK & for 1 root region 0.67898224 248.3825681 -8054.423 -6586.505 -11.10052

State 1 has three real roots. Take the more stable root (lower fugacity value).For State 1 T (K) 400 Z V fugacity H-Hig U-Uig S-SigP (MPa) 2.5 cm3/gmol MPa J/mol J/mol J/molKanswers for three 0.668475 889.2328 1.871218 -3460.25 -2357.73 -6.24206root region 0.180657 240.3178 0.103098 137.145 2.008636 -14993.7 -12011 -35.6649



( ) ( ) ( )( ) ( ) ( )igigigig

igigigig

SSSSSSSSSHHHHHHHHH

11122212

11122212

+==+==

Find ??12 = igig HH

( ) ( ) ( ) ( )4142313221221232

12

432

)(2

1

2

1

TTDTTCTTBTTA

dTDTCTBTACpdTHHT

T

T

T

igig

+++=

+++==

A B C D-1.687 3.42E-01 -2.34E-04 6.34E-08

moleJHH igig /81.1717312 =Similarly for = 2

1 1

212 ln

T

T

igig

PPRdT

TCpSS

( ) ( ) ( )

+++

=

1

231

32

21

2212

1

212 ln32

lnPPRTTDTTCTTB

TTASS igig

KmoleJSS igig = /87.2212KmoleJS

moleJH=

=/01.18&

/63.12579

( )1221

)( TTRHdTTCvUT

T

== , RCpCv =( )

( ) ( ) ( )moleJU

UUUUUUUUUU

moleJUUigigigig

igig

/94.1169773.2357505.658671.15926

&

/71.15926400550*314.881.17173

11122212

12

=+=

+====

molcmVVVV

/8.640

2.8894.2483

12

===

(P7.6) Ethane tank leaks to turbine: molesnmVMPaPKT 28301,10,425 311 ==== ;Solution: Ebal: H=W; Sbal: S=0.(a) Compute ToutInitial. Sout=Sin=78.2J/Mol-K



(P7.7) Compute W for 80% eff turbine on Ethylene from 350C,50bar to 2bar. Compare Tf forthis process to Tf of 100%eff turbineSolution: Ebal: H=W; Sbal: 12 SS = =9.1954; Ref=298.15K,1bar, id gas.T (K) 623.15Z V H U SP (MPa) 5 cm3/gmol J/mol J/mol J/molK

& for 1 root region 0.986361 1022.041 18750.12 13639.92 9.1954For Rev: Use Solver at 0.2MPa and S2=9.1954J/mol-K then we can find HandH

T (K) 404.71Z V H U SP (MPa) 0.2 cm3/gmol J/mol J/mol J/molK

& for 1 root region 0.99499 16739.56 5234.912 1887 9.1954

molJH

HHHHH

WW

HH

moleJH

s

s

/95.793712.18750912.5234

12.187508.0

8.08.0

/21.1351512.18750912.5234

2

2

12

12

=

====

===

molJHHHH

/166.1081212.1875095.793712

===

T (K) 452.012672 fugacity H U SP (MPa) 0.2 MPa J/mol J/mol J/molK

& for 1 root region 0.199324 7937.95 4192.62 15.50748631KT 4522 =

(P7.8) Rankine on methanol, see figure 4.3 page 143. Ref=336.7,0.1Liq.State T(K) P(Mpa) H S4sat vap 337.4 0.1027 37853 112.165 sat liq 337.4 0.1027 73 0.233 610 4.087 51458 112.16

2647.0)7351458(

)3785351458( ===

HQW (Note: neglecting pump work.)

(P7.9) Use the energy equation to get (U-Uig)/RT.a. For SW fluid for g=10-5x. x=r/;

( )5.1

1

4332

5.1

1

32

0 45

310

244)510(

24

2

=== xx

TkNdxxx

TkNdrrrg

RTuNN

RTUU

B

A

B

AAAig

NA3 = 1; /kBT = 1 (U-Uig)/RT = -5.7.b. For Sutherland potential with g = 1+2/x2.

( )

=

+== 1

533

6

2

12

32

0 52

424421

24

2xx

TkNdx

xx

xTkNdrrrg

RTuNN

RTUU

B

A

B

AAAig NA3

= 1; /kBT = 1 (U-Uig)/RT = -3.

Chapter 9 Practice Problems (p9.01) The stream from a gas well consists of 90 mol% methane, 5% ethane, 3% propane and 2%

n-butane. This stream is flashed isothermally at 233 K and 70 bar. Use the shortcut K-ratio method to estimate the L/F fraction and liquid vapor composisitons. (ANS. L/F = 0.181)

Solution By short-cut vapor pressure eqn.

yx

KT

Pi

ii

ii

r i

=

+ FHGIKJ

LNM

OQP

LNMM

OQPP

10 73

1 1 1^?,

,

a f

y x Ki i i= Though not required, the table below also shows bubble T and dew T calculations at 70 bar. For the bubble calculations, in each column, the temperature at the top of the column is used to

calculate the K-ratio. Then yi = xiKi. The temperature is adjusted until the sum of ys is unity. This is an iterature calculation.

To accompany Introductory Chemical Engineering Thermodynamics 1 of 8 J.R. Elliott, C.T. Lira, 2004, all rights reserved, (3/17/04)

For the dew T calculations, in each column, the temperature at the top of the column is used to calculate the K-ratio. Then xi = yi/Ki. The temperature is adjusted until the sum of xs is unity. This is an iterative calculation.

For the isothermal flash calculation, the T = 233K and the pressure is 70 bar, so the K-ratio is

fixed, as tabulated in the column under 233. Equation 9.63 is programmed in the cells below the value of L/F=0.181 below the Flash title. Each row holds the value of the term Di = zi(1-Ki)/[Ki + (L/F)(1-Ki)] from equation 9.63. These values are summed at the bottom of the column. The criteria for the isothermal flash is that L/F is adjusted until the sum goes to zero, as is shown at L/F = 0.181. Once the value of L/F is found, the xi values and yi values in the last columns are generated separately using equations 9.57 and 9.58 respectively.

pMPa= 7.000 BUBT DEWT FLASH

z Tc Pc w 210.9 y 300.5 x 233 0.181 x yC1 0.9 190.6 4.6 0.011 1.108 0.997 4.7949 0.188 1.767 -0.424 0.553 0.977C2 0.05 305.4 4.88 0.099 0.049 0.002 0.6332 0.079 0.111 0.1632 0.184 0.020C3 0.03 369.8 4.25 0.152 0.006 0.000 0.1456 0.206 0.016 0.152 0.155 0.002nC4 0.02 425.2 3.8 0.193 8E-04 0.000 0.0379 0.527 0.003 0.1088 0.109 0.000

1.0000 0.9997 5E-07 1.000 1.000


(p9.2) An equimolar mixture of n-butane and n-hexane at pressure is isothermally flashed at 373K. The liquid-to-feed ratio is 0.35. Use the shortcut K-ratio method to estimate the pressure and liquid and vapor compositions. (ANS. P=0.533MPa; xC6=0.78). Solution By short-cut vapor pressure eqn.

yx

KT

Pi

ii

ii

r i

=

+ FHGIKJ

LNM

OQP

LNMM

OQPP

10 73

1 1 1^?,

,

a f

y x Ki i i= For the isothermal flash calculation, the T = 373K. Equation 9.63 is programmed in the cells

below the value of L/F=0.35 below the Flash title. Each row holds the value of the term Di = zi(1-Ki)/[Ki + (L/F)(1-Ki)] from equation 9.63. The value of Ki requires P which is to the left of the table. These values are summed at the bottom of the column. The criteria for the isothermal flash is that P is adjusted until the sum goes to zero, as is shown at P = 0.533. Once the value of P is found, the xi values and yi values in the last columns are generated separately using equations 9.57 and 9.58 respectively.


Though not required, the table below also shows the bubble T and dew T calculations for P = 0.533 MPa.

For the bubble calculations, in each column, the temperature at the top of the column is used to calculate the K-ratio. Then yi = xiKi. The temperature is adjusted until the sum of ys is unity. This is an iterature calculation.

For the dew T calculations, in each column, the temperature at the top of the column is used to

calculate the K-ratio. Then xi = yi/Ki. The temperature is adjusted until the sum of xs is unity. This is an iterative calculation.

pMPa= 0.533 BUBT DEWT FLASHz Tc Pc w 349.2 y 383.2 x 373 0.35 x y

nC4 0.5 425.2 3.8 0.193 1.766 0.883 3.5299 0.142 2.906 -0.426 0.223 0.649nC5 0 469.7 3.37 0.249 0.624 0.000 1.3902 0.000 1.11 0 0.000 0.000nC6 0.5 507.4 3.01 0.305 0.236 0.118 0.5826 0.858 0.452 0.4257 0.777 0.351nC7 0 540.3 2.74 0.349 0.097 0.000 0.2631 0.000 0.199 0 0.000 0.000

1.0010 0.9999 1E-06 1.000 1.000


(p9.3) A mixture of 25 mol% n-pentane, 45% n-hexane, and 30% n-heptane is flashed isothermally at 365.9K and 2 bar. Use the shortcut K-ratio method to estimate the L/F fraction and liquid and vapor compositions (ANS. L/F = 0.56) By short-cut vapor pressure eqn.

yx

KT

Pi

ii

ii

r i

=

+ FHGIKJ

LNM

OQP

LNMM

OQPP

10 73

1 1 1^?,

,

a f

y x Ki i i= Though not required, the table below also shows bubble T and dew T calculations at 2 bar. For the bubble calculations, in each column, the temperature at the top of the column is used to

calculate the K-ratio. Then yi = xiKi. The temperature is adjusted until the sum of ys is unity. This is an iterature calculation.




For the isothermal flash calculation, the T = 365.9 K and the pressure is 2 bar, so the K-ratio is fixed, as tabulated in the column under 365.9. Equation 9.63 is programmed in the cells below the value of L/F=0.56 below the Flash title. Each row holds the value of the term Di = zi(1-Ki)/[Ki + (L/F)(1-Ki)] from equation 9.63. These values are summed at the bottom of the column. The criteria for the isothermal flash is that L/F is adjusted until the sum goes to zero, as is shown at L/F = 0.56. Once the value of L/F is found, the xi values and yi values in the last columns are generated separately using equations 9.57 and 9.58 respectively.

pMPa= 0.2 BUBT DEWT FLASH

z Tc Pc w 358.6 y 312.48 x 365.9 0.56 x ynC4 0 425.2 38 0.193 57.73 0.000 18.804 0.000 67.18 0 0.000 0.000nC5 0.25 469.7 3.37 0.249 2.107 0.527 0.5756 0.434 2.51 -0.227 0.150 0.377nC6 0.45 507.4 3.01 0.305 0.821 0.369 0.1898 2.370 1.001 -2E-04 0.450 0.450nC7 0.3 540.3 2.74 0.349 0.348 0.104 0.0694 4.324 0.432 0.227 0.400 0.173

1 7.129 3E-08 1.000 1.000


(p9.04) A mixture containing 15 mol% ethane, 35% propane, and 50% n-butane is isothermally flashed at 9 bar and T. the liquid-to-feed ratio is 0.35. Use the shortcut K-ratio method to estimate the pressure and liquid and vapor compositions. By short-cut vapor pressure eqn.

yx

KT

Pi

ii

ii

r i

=

+ FHGIKJ

LNM

OQP

LNMM

OQPP

10 73

1 1 1^?,

,

a f

y x Ki i i= For the isothermal flash calculation, the P=9 bar. Equation 9.63 is programmed in the cells

below the value of L/F=0.35 below the Flash title. Each row holds the value of the term Di = zi(1-Ki)/[Ki + (L/F)(1-Ki)] from equation 9.63. The value of Ki requires T which is to the left under the Flash title. These values of Di are summed at the bottom of the column. The criteria for the isothermal flash is that T is adjusted until the sum goes to zero, as is shown at T = 319.4K. Once the value of T is found, the xi values and yi values in the last columns are generated separately using equations 9.57 and 9.58 respectively.

Though not required, the table below also shows the bubble T and dew T calculations for P = 0.9

MPa.


For the bubble calculations, in each column, the temperature at the top of the column is used to calculate the K-ratio. Then yi = xiKi. The temperature is adjusted until the sum of ys is unity. This is an iterature calculation.



pMPa= 0.900 BUBT DEWT FLASHz Tc Pc w 290 y 326.9 x 319.4 0.35 x y

C1 0 190.6 4.6 0.011 32.92 0.000 49.257 0.000 45.74 0 0.000 0.000C2 0.15 305.4 4.88 0.099 3.963 0.594 7.9952 0.019 7.027 -0.184 0.031 0.214C3 0.35 369.8 4.25 0.152 0.86 0.301 2.0955 0.167 1.779 -0.181 0.232 0.413nC4 0.5 425.2 3.8 0.193 0.213 0.106 0.614 0.814 0.505 0.3648 0.737 0.372

1.0016 1.0002 6E-08 1.000 1.000

4200:225 EQUILIBRIUM THERMODYNAMICS SPRING 90 TEST 1 SSN________ 1. Short Answer a) Estimate the change in entropy when one mole of nitrogen is compressed by a piston in a cylinder from 300K and 23 liters/gmol to 400K and 460 liters/gmol.(Cp=7 cal/gmol) b) Draw a sketch of the square well potential and indicate the position(s) where the force between two atoms is zero. c) Write the simplest possible form of the energy balance for application to the following process and system: water drips slowly out of a hole in the bottom of an enclosed tank; system: the tank and its contents. d) Write the simplest possible form of the energy balance for application to the following process and system: a rubber balloon being inflated; system: the balloon and its contents. 2. Estimate the density (g/cc) of liquid butane at 300 K and 38 psia. Assuming that a disposable lighter costs $1.00 and contains 2 g of butane, what is the value of the butane? ($/gal) 3. Saturated steam at 660F is adiabatically throttled to atmospheric pressure. Estimate the final condition of the steam. 4. A mixture of 1CO:2H2 is adiabatically continuously compressed from 5 atm and 100F to 100 atm and 1100F. Estimate the work of compressing 1 ton of the gas.(Cp=7Btu/lbmol-R) 5. Determine the efficiency of the compressor in problem 4. 6. An insulated cylinder is fitted with a freely floating piston and contains 1 lbm of steam at 120 psia and 90% quality. The space above the piston, initially 1 ft3, contains air at 300 K to maintain the pressure on the steam. Additional air is forced into the upper chamber, forcing the piston down and increasing the steam pressure until the steam has 100% quality. The final steam pressure is 428 psia and the work done on the steam is 91 Btu, but the air above the steam has not had time to exchange heat with the piston, cylinder or surroundings. The air supply line is at 700 psia and 300 K. What is the final temperature of the air in the upper chamber? 1a)7.4cal/mol-K (c)HdM=d(MU) (d)HdM+W=d(MU) 2).6 g/cc 3) 0.96 4) 1.3E6 5) 76% 6)360K 4200:225 EQUILIBRIUM THERMODYNAMICS SPRING 91 TEST 1 SSN________ 1. Short Answer a) 15 molecules are distributed as 9:4:2 between boxes A:B:C respectively. The partitions between the boxes are removed and the molecules distribute themselves evenly between the boxes. Compute S. b) Explain in words how the pressure of a fluid against the walls of its container is related to the velocity of the molecules. c) What is it about molecules that requires us to add heat to convert liquids to gases? d) A rigid cylinder of gaseous hydrogen is heated from 300K and 1 bar to 400K. How much heat is added to the gas? 2. Steam is produced at 30 bar and some unknown temperature. A small amount of steam is bled off and goes through an adiabatic throttling valve to 1 bar. The temperature of the steam exiting the throttling valve is 110C. What is the value of the specific entropy of the steam before entering the throttle? 3) An adiabatic compressor is used to continuously compress nitrogen (Cp/R=7/2) from 2 bar and 300K to 15 bar. The compressed air is found to have an outlet temperature of 722K. How much work is required? 4) What is the efficiency of the compressor in the previous problem? 5) As part of a supercritical extraction of coal, an initially evacuated cylinder is fed with steam from a line available at 20 MPa and 400C. What is the temperature in the cylinder immediately after filling? Answers: 1.a)2.31k 1.b)change of direction due to wall collision gives change in velocity per unit time providing the acceleration in the force per unit area that is pressure. 1.c)potential energy causes them to stick together when close to each other. 1.d)2079 J/mol. 2)5.9736kJ/kg-K 3)436J/g 4)53%5)454C 4200:225 EQUILIBRIUM THERMODYNAMICS SPRING 92 TEST 1 SSN________ 1. Short Answer a) How does heat get from the flame of a gas oven into a pizza inside the oven? b) Nitrogen (Cp/R=7/2) is adiabatically and reversibly compressed from 300K and 5 bar to 25 bar. Compute the temperature coming out of the compressor. c) Estimate the S of liquid freon-12 at -40F.(Hint: use chart) d) In a large refrigeration plant it is necessary to compress a fluid which we will assume to be an ideal gas with constant heat capacity, from a low pressure P1 to a much higher pressure P2. If the compression is to be done in two

2stages, first compressing the gas from P1 to P*, then cooling the gas at constant pressure down to the compressor inlet temperature T1, and then compressing the gas to P2, what should the value of the intermediate pressure be to accomplish the compression with minimum work? (Hint: don't derive the whole formula. If you know the answer, just write it down.) 2. An adiabatic compressor is used to continuously compress low pressure steam from 0.8 MPa and 200C to 4.0 MPa and 500C in a steady state process. What is the work required per kg of steam through this compressor? 3) Compute the efficiency of the compressor in the previous problem. 4) An ordinary vapor compression cycle is to operate a refrigerator on F-12 between -40F and 120F (coil temperatures). Compute the coefficient of performance and the heat removed from the refrigerator per day if the power used by the refrigerator is 9000 J per day. 5) Airplanes are launched from aircraft carriers by means of a steam catapult. The catapult is a well-insulated cylinder that contains steam and is fitted with a frictionless piston. The piston is connected to the airplane by a cable. As the steam expands, the movement of the piston causes movement of the plane. A catapult design calls for 270 kg of steam at 15 MPa and 450C to be expanded to 0.4 MPa. How much work can this catapult generate during a single stroke? Compare this to the energy required to accelerate a 30,000 kg aircraft from rest to 350 km per hour. Answers:1.a)air collisions b)475K c).0024 d)P1P2 2)606kJ/kg 3)67% 4)1.55,-14,000J/day 5)165224 kJ vs. 141782 4200:225 EQUILIBRIUM THERMODYNAMICS SPRING 93 TEST 1 SSN________ 1. Short Answer a) 20 molecules are contained in a piston+cylinder at low pressure. The piston moves such that the volume is expanded by a factor of 4 with no work produced of any kind. Compute S/k. b) A tank of N2 (Cp=7R/2) at 300K and 25 bars leaks adiabatically until the pressure drops to 5 bar. What is the final temperature?. c) A Carnot cycle is to operate with coil temperatures from -160F to -280F. Compute the coefficient of performance. d) As part of the air standard Otto cycle, air (Cp=7R/2) in a cylinder at 400K is compressed adiabatically and reversibly with a volumetric compression ratio of 8:1. Develop an expression relating the work required for a given compression ratio in terms of the temperatures T1 and T2 (ie. the temperatures before and after).(Hint: No numbers are necessary in your "expression".) e) A tank of air (Cp=7R/2) drives an adiabatic, reversible turbine exhausting to the atmosphere. Derive an overall entropy balance between the specific entropy of air exiting the turbine vs. the specific entropy in the tank. 2. An adiabatic turbine is supplied with steam at 2.0 MPa and 600C and it exhausts at 98% quality and 24C. Compute the work output per kg of steam.(15) 3. Compute the efficiency of the turbine in problem 2.(20) 4. An ordinary vapor compression cycle is to be operated on methane to cool a chamber to -260F. Heat will be rejected to liquid ethylene at -165F. The temperatures in the coils are -160F and -280F. Compute the coefficient of performance. (Hint: Use Chart )(20) 5. A well-insulated cylinder, fitted with a frictionless piston, initially contained 9 kg of liquid water and .4 kg of water vapor at a pressure of 1.4 MPa. 2 kg of steam at 1.6 MPa was admitted to the cylinder while the pressure was held constant by allowing the piston to expand. a) Write the energy balance for this process. (10) b) If the final volume of the contents of the cylinder was six times the initial volume, determine the temperature of the superheated steam that was admitted to the cylinder.(10). 1a)23.2(b)189(c)1.5(d)CvT(e) S =0(2)-1194J(3)85%(4).86(5) MU=W+HM,557C 4200:225 EQUILIBRIUM THERMODYNAMICS SPRING 94 TEST 1 SSN________ 1. Short Answer a) Rolling two dice (six sided cubes with numbers between 1 and 6 on each side) is like putting two particles in six boxes. Compute S/k for going from double sixes to a four and three.0.693 b) Air (Cp/R=3.5) is isothermally compressed in a piston+cylinder from 100C and 1 bar to 100 bar. Estimate the work requirement in J/mole.14000

3c) Air (Cp/R=3.5) is adiabatically and reversibly compressed in a piston+cylinder from 100C and 1 bar to 100 bar. Estimate the work requirement in J/mole.21000 d) Suppose a particular charge of gunpowder resulted in a high pressure gas at 100 bar and 3000K in a 10 cm dia X 20 cm length chamber. This chamber is connected to 480cm barrel of the same diameter angled at 45. Assuming that the mass of the cannonball is 1 kg and the expansion of the gas is adiabatic and reversible, write the energy balance for the travel of the cannonball from its initial position to the end of the gun barrel. (You do not need to substitute any numbers.) Wdt=d[M(U+v2/2+gZ)] e) A series of three adiabatic, reversible compressors with interstage cooling to the initial inlet temperature is to compress air (assume ideal gas) from 300K and 1 bar to 100 bar. Estimate the pressure after the first compression stage assuming the series has been designed to minimize the overall work requirement. 4.6bar 2. An adiabatic compressor has been designed to continuously compress 1 kg/s of saturated vapor steam from 1 bar to 100 bar and 1100 C. Estimate the power requirement of this compressor in horsepower. (15)3000 3. Determine the efficiency of the compressor described above. (20)60 4. A cold storage room is to be maintained at 10F and the available cooling water is 70F. Assume that the cold-room coils and the condenser are of sufficient size that a 10F approach can be realized in each. The refrigerant capacity is to be 126,500 kJ/hr. Freon-22 (!!!) will be used for the vapor compression cycles. Calculate the COP for the following cases: a) Carnot cycle (5)5.75 b) Ordinary vapor compression cycle for which compressor is 100% efficient.(10)4.3 c) Ordinary vapor compression cycle for which compressor is 80% efficient.(5)3.4 (Note: This is Freon-22. Minimal credit will be awarded for Freon-12. 5. A 1 m3 tank is to be filled using N2 at 300K and 20 MPa. Instead of throttling the N2 into the tank, a reversible turbine is put in line to get some work out of the pressure drop. If the pressure in the tank is initially zero and the final pressure is 20 MPa, what will be the final temperature in the tank? How much work will be accomplished over the course of the entire process? (Hint: consider the entropy balance carefully.)(20)300K,20E6J 4200:225 EQUILIBRIUM THERMODYNAMICS SPRING 96 TEST 1 SSN________ 1. Short Answer (20) a) Two moles of oxygen are mixed with an equal volume of nitrogen at 300K and 1 bar. Estimate (assuming ideal gas) the change in entropy (J/mole-K) for the oxygen. b) Saturated liquid water at 25C is steadily pumped to a pressure of 70 bars. Estimate the work requirement in J/g. c) An initially evacuated cylinder of hydrogen is to be filled from a supply line available at 400K and 50 bars. Estimate the final temperature (K) of the hydrogen in the cylinder after filling. d) Write the most compact but applicable energy balance for the following: A surge tank is mounted at the side of an air supply line to damp out fluctuations in the air pressure. As the pressure in the supply line begins to drop, air leaks out of the surge tank into the supply line. System: the surge tank and its contents. 2) Freon-22 is adiabatically compressed in a steady state process from saturated vapor at 240K to 15 bars and 400K. Compute the specific work required (kJ/kg).(20) 3) Estimate the efficiency of the compressor from the above problem.(20) 4) A heat engine is to operate on steam. Cooling water is available such that the condenser operates at a temperature of 50C. The manufacturer of the boiler specifies that the temperature of the steam exiting the boiler may not exceed 400C. The manufacturer of the turbine requires that the quality of steam exiting the turbine must be at least 89.5%. a) Estimate the maximum thermal efficiency of a Carnot cycle operating between these upper and lower temperatures.(5) b) Estimate the maximum thermal efficiency of a single stage Rankine cycle operating between these upper and lower temperatures.(20) 5) Suppose the fluid in the fix-a-flat can was Freon-22 (MW=85.5) at 50wt% liquid and 100kg/m3. Let the can be 500 cm3 and the tire be 40,000 cm3 at 300K and 1 bar. a) write the mass balance for the overall process (2). b) write the energy balance for the overall process (2). c) write the entropy balance for the can of Freon-22 (2). d) Approximating that the tire was originally filled with Freon-22, that the contents of the tire are at such a low

pressure that they can be treated as an ideal gas, and that the change in the temperature of the tire is negligible,

4solve for the tire pressure (bars) when the pressure in the can has dropped to 4 bars. Is it enough to inflate the tire? (9)

Answers: (1)a. 11.5J/mole-K, b.7J/g, c.560K, d.Houtdn=d(nU) (2) 100J/g (3)64 (4)52%,30% (5) a.mCf + mTf = mCi + mTi b. mCf UCf+ mTf UTf = mCi UCi + mTi UTi c.S=0 d.1.25bars 4200:225 EQUILIBRIUM THERMODYNAMICS SPRING 97 TEST 1 SSN________ 1. Short Answer (15) a) The specific volume of steam at 4 MPa and 1200C in m3/kg is ______ m3/kg b) A 2000 kg automobile traveling at 25 m/s strikes a plunger in 10,000 cm3 of water, bringing the auto to a stop.

What is the maximum temperature rise, in C, of the water? c) Find the work, in kJ/kg needed to compress air isentropically from 20C and 100 kPa to 6 MPa (in a

piston+cylinder). 2. Consider the case of 8 particles distributed between two boxes. What is the fraction of microstates with 4

particles in the first box and 4 particles in the second box? (10) 3. As part of a refrigeration cycle, Freon 134a is adiabatically compressed from the saturated vapor at -60C (note the negative sign on temperature) to 1017kPa and 100C. How much work is required in kJ/kg?(15) 4. Estimate the efficiency of the compressor in problem 3.(20) 5. A Rankine cycle is to be operated on steam entering the turbine at 500C and 5MPa and expanding isentropically to 10kPa. a) Estimate the maximum work output in kJ/kg.(10) b) Estimate the maximum thermodynamic efficiency for this cycle. (10) 6. We have an isothermal 100% efficient, continuous compressor to raise steam from 0.5 MPa to 5 MPa. The process of interest involves a fluidized bed reactor that requires a volumetric flow rate of 0.1 m3/sec to maintain the fluidization. The stoichiometry for the reaction of interest requires that 1.73 kg/sec of steam be fed to the reactor. a) Write the appropriate reduced energy and entropy balances for the compressor.(6) b) Determine the temperature at which the compressor should operate. (4) c) Determine the work requirement in horsepower of the compressor. (10) Answers: (1)0.17,15C,466,(2).27(3)121(4)75%(5)1224,38%(6)S=Q/T,400C,1616hp 4200:225 EQUILIBRIUM THERMODYNAMICS SPRING 98 TEST 1 SSN________ 1. Short Answer (35)

Sketch the force between two molecules vs. dimensionless distance, r/, according to the Lennard-Jones potential. Is the value of r/ when the force is equal to zero greater, equal, or less than unity?(5) Estimate the change in entropy (J/mole-K) when 0.5 moles of helium are mixed with 0.5 moles of hydrogen at 300K.(5) Estimate the "lost work" of the process in part c above.(5) Estimate the work required (J/mole) to adiabatically and reversibly compress argon from 1 bar and 300 K to 20 bars in a steady state process. (10) Ten particles distributed between two boxes go from 9 in box A to 6 in box A. Compute the change in entropy (dimensionless S/k will suffice).(10)

2. Steam expands through an adiabatic turbine from 200 bars and 700C to 1 bar saturated vapor. Compute the work output of the turbine in kJ/kg. (15)

3. Compute the efficiency of the turbine in problem 3.(15) 4. Freon 134a is used in an OVC heat pump providing cooling to a building during the summer. The air inside the

building is to be maintained at 25C with a 5C approach temperature (cf. Index for definition of approach temperature) and the heat is to be rejected to the outside air at 35C with a 25C approach. The compressor is 80% efficient. Compute the coefficient of performance for this cycle and compare it to the value for a Carnot cycle.(20)

55. It is desired to determine the volume of an initially evacuated tank by filling it from an 80 liter cylinder of air at

300 bars and 300K. The final pressure of both tanks is 5 bars. Estimate the volume in liters.(15) Answers: 1(a) greater (b) 5.763J/mol-K (c) 1729J/mol (d) 14432 J/mol, 2. 3.04*kB 3. 1133kJ/kg 4. 83% 5. 4.24vs.7.3 6. 4720L 4200:225 EQUILIBRIUM THERMODYNAMICS SPRING 99 TEST 1 SSN________ 1. Short Answer (25) a) Derive the expression for the force between two molecules characterized by the Yukawa potential (given

below)(5)

62. A Rankine cycle operates on steam exiting the boiler at 7 MPa and 550C and dropping to 60C and 98%

quality. (a) Compute the efficiency of the turbine.(10) (b) Estimate the pump work (kJ/kg).(5) (c) Compute the thermal efficiency of the Rankine cycle.(5) (d) Compute the thermal efficiency of a Carnot cycle operating between 550 and 60C.(5)

3. 200 moles per hour of natural gas is to be adiabatically and reversibly compressed from 300K and 1 bar to 100

bars in a continuous two-stage process with inter-cooling to 300K. Natural gas may be approximated by pure ideal gas methane. (a) What pressure do you recommend between stages? (5) (b) Estimate the final temperature exiting the second stage. (5) (c) Estimate the work requirement (kJ/mole) (6) (d) Estimate the average power requirement for the compressor (hp). (4)

4. Freon 134a is to be adiabatically and reversibly compressed from saturated vapor at -25C to 3MPa.

(a) Compute the work requirement (kJ/kg) using the chart. (5) (b) Compute the heat removed in a condenser(QH) that drops the outlet of the above compressor to saturated liquid. (5) (c) Compute the Coefficient of Performance (QL/W) for an OVC cycle based on this compressor and condenser. (5)

5. Steam originally exists in a piston +cylinder at 0.4 MPa and 350C. The piston is forced down adiabatically and

reversibly till the volume is 38% of the original volume. (a) Write the energy and entropy balances for this process (6). (b) Estimate the final pressure and temperature. (6) (c) Compute the work for this process (kJ/kg) (8).

6. A rigid insulated cylinder is initially divided into two compartments by a frictionless piston that does not

conduct heat. Initially, the piston separates two ideal gases (Cv/R=2 for both gases). The entire system is initially at 300K. One gas is at 200 bars and occupies 10% of the total fixed volume and the other is at 20 bars. The piston is attached to a rod such that work is adiabatically and reversibly removed as the two sides of the piston equilibrate. (a) What is the relationship between volume and pressure on each side of the piston? (5) (b) What is the ratio of the final volumes and what are the final temperatures for each gas at equilibrium? (15) (c) If the piston were able to conduct heat, but the process was conducted irreversibly, such that no work was removed, what would be the final states (P,V,T) of both sides of the piston at equilibrium? (10)

4200:225 EQUILIBRIUM THERMODYNAMICS SPRING 87 TEST 2 SSN________ 1) True or false.

____The compressibility factor Z is always less than or equal to unity.

____The critical properties Tc and Pc are constants for a given compound.

____A closed system is one of constant volume.

____A steady-state flow process is one for which the velocities of all streams may be assumed negligible.

____Gravitational potential-energy terms may be ignored in the steady-state energy equation if all streams entering and leaving the control volume are at the same elevation.

____In an adiabatic flow process, the entropy of the fluid must increase as the result of any irreversibilities within the system.

____The temperature of a gas undergoing a continuous throttling process may either increase or decrease across the throttling device, depending on conditions.

____When an ideal gas is compressed adiabatically in a flow process and is then cooled to the initial temperature, the heat removed in the cooler is equal to the work done by the compressor. (Assume potential and kinetic energy effects are negligible.)

____In the limit as P 0, the ratio f/P for a gas goes to infinity, where f is the fugacity. ____The residual Gibbs function is related to f/P by (G-Gid)/(nRT)=ln(f/p).

2) Reduce (H/S) to a form involving p, V, T, Cp, Cv, and their derivatives. 3) A house has an effective heat loss of 100,000 Btu/hr. During the heating season of 160 days the

average inside temperature should be 70F while that outside is 45 F. Freon-12 is the working fluid and an ordinary vapor-compression cycle is used. A 10 F approach on each side may be assumed. Electricity costs $0.14/KW-hr = $0.00004/Btu. a) What is the cost in $/hr if the compressor is 100% efficient? b) What is the cost if the compressor is 80% efficient?

4) Determine the horsepower required to continuously compress reversibly and adiabatically 1 lbm/min of ethylene oxide from 70F and 1 atm to 250 psia. Cp = 12.8 cal/gmol-K; Tc = 469 K; Pc = 70.1 atm; = 0.200.

5) Propylene vapor is processed from 325 K and 21.35 atm to 225 K and 1 atm. Compute the change in entropy. Suppose saturated propylene vapor at 325 K is expanded reversibly and adiabatically to 1 atm. What is the final quality? Tc = 364.8 K; Pc = 45.5 atm; = 0.142; Cp = 14.6cal/mol-K; ln(pvap) = -2238 + 9.953 where pvap is in atm and T is in K. 1.FTFFTTTTFT 2.T(1+V/Cv(dp/dT)v ) 3.a.0.44.b.0.55 4.1.96 5.19.8cal/mol-K,90%

4200:225 EQUILIBRIUM THERMODYNAMICS SPRING 88 TEST 2 SSN________ 1. Short Answer

a) In an ordinary vapor-compression cycle, what is the entropy after the throttle relative to the entropy before the throttle (higher, lower or equal) and why?

b) Estimate the vapor pressure of sulfur dioxide at room temperature (298K). Tc=430.8 K; p =7.78 MPa; =0.251

c) What is the change in entropy for an ideal gas (Cp=7 cal/mol-K) when it is raised from a temperature of 300K and pressure of .1 MPa to 320K and 5MPa?

d) After reacting H2+O2, H2O vapor at 1500 K and 11 MPa is expanded reversibly and adiabatically through a nozzle to 0.1 MPa. Estimate the outlet temperature of this exhaust stream.

e) A two-stage compressor operates adiabatically during each stage with intercooling to 60F between stages. We would like to compress helium (CP = 5cal/mol-K) from 1 atm to 16 atm. What pressure between stages minimizes the reversible work for the compression of this gas?

f) Develop an expression for (G-Gid)/RT as a function of b for a gas that can be described by: Z = 1 + 2b/(1-2b)

2. Evaluate (H/P)v in terms of only P, V, T, CP, and CV and their derivatives. Describe a physical situation to which this quantity relates.

3. Estimate the enthalpy of 1,3, butadiene at 125 atm and 530F relative to its saturated liquid at 60F. T = 425.4 K; P = 4.33 MPa; = 0.193; CP =23 cal/mol-K;

4. It is desired to design a Rankine cycle using a 70% efficient turbine and expanding the steam to saturated vapor at 224F. The boiler operates at 500 psia. Compute the thermodynamic efficiency for conversion of heat into work.(hint: neglect the pump work). Compare this to the thermodynamic efficiency of the Carnot cycle operating between the same temperatures.

5. By holding the can upside down at room temperature (298K), liquid freon-22 (MW=85.5) can be dispensed from a can through 1 ft of stainless tubing (100 g, CP =.12 cal/g-K). (a) Estimate the heat of vaporization (in cal/g) of freon-22 at its normal boiling temperature (-42C). (b) Estimate the number of moles of freon-22 that must be wasted before the first drop of liquid comes out of the pipe if the pipe is insulated. Velocity of the vapor freon coming out of the tube may be neglected. Tc = 369.8; P = 4.97 MPa; = 0.221

Answers: 1)a.higher, throttling irreversible b.0.389MPa c.-7.32cal/mol-K d.585K e.4atm f.-ln(1-2b)+Z-1-lnZ (2)Cv(T/P)v+V (3)9443cal/mol (4)21%vs46% 4200:225 EQUILIBRIUM THERMODYNAMICS SPRING 89 TEST 2 SSN________ 1. Estimate the entropy of 1 gmol of propane at 150F and 200 psia. The entropy is to be taken as zero at 1 atm ideal gas and 0F.Tc =369.8 K; Pc =4.249 MPa; =0.152 2. Estimate the vapor pressure of isobutane at 300 K. Tc =408K; Pc =3.65MPa; =0.177 3. Argon (CP =5 cal/mol-K) is compressed reversibly and adiabatically in a continuous single stage process from 0.15 MPa and 300 K to 0.90 Mpa. Estimate the work done by the compressor. 4. Suppose the argon from the preceding problem was compressed from 0.15 MPa to 0.90 MPa in a two-stage process with intercooling back to 300 K. What would be the optimum interstage pressure and the work done in that case? 5. Express in terms of in terms of only P, V, T, CP, CV and their derivatives. Your answer may include absolute values of S if it is not associated with a derivative. (G/H)P 6. The compressibility factor for a certain fluid is well-represented by: Z = 1 + ab/(1+b)2 Develop an expression for the Helmholtz energy departure function. 7. Saturated vapor propane at -40C is to be compressed to 55 atm. Estimate the work required if the compressor is adiabatic but only 50% efficient. Tc =369.8 K; Pc =4.249 MPa; =0.152 8. An ordinary vapor-compression cycle is to be designed for superconductor application using N2 as refrigerant. The expansion will be to atmospheric pressure. A heat sink is available at 105 K. A 5 K approach should be sufficient. Roughly 100 Btu/hr must be removed. Compute the coefficient of performance (COP) and compare it to the Carnot COP. Also, estimate the power requirement (hp) of the compressor assuming it is adiabatic and reversible. 1.-.7246cal/mol-K 2.0.3746MPa 3.-1572cal/mol 4.-1293cal/mol 5. -S/P 6. ab/(1+b) 7.-3563cal/mol 8.1.33,0.30hp

4200:225 EQUILIBRIUM THERMODYNAMICS SPRING 90 TEST 2 SSN________ 1.a. List in the space below all the differences between a Rankine cycle and a Carnot cycle. Order the list from most important to least important. 1.b. An inventor proposes to have developed a small power plant that operates at 70% efficiency. It operates between temperature extremes of 1000F and 100F. Develop your own analysis of the maximum possible efficiency. (Choose the power cycle that gives the highest possible value for the efficiency.) 1.c. Estimate the vapor pressure of CO2 at 14C. (Tc=304.2K; Pc=7.381Mpa; =.228). 2. Estimate the quality of Freon-12 after the throttle in an ordinary vapor-compression cycle operating between -20F and 114.3F. 3. A simple Rankine cycle operates between superheated steam at 1000xF and 400 psia entering the turbine and 2 psia entering the pump. What is the maximum possible efficiency for this Rankine cycle? 4. Express in terms of in terms of only P, V, T, CP, CV and their derivatives. Your answer may include absolute values of S if it is not associated with a derivative. (S/P)G 5. Suppose an ideal gas was continuously compressed adiabatically and reversibly from 45 psia and 70F to 1500 psia in two stages with intercooling to 70F between stages. What would be the optimal pressure between stages (for minimum work) and the temperature coming out of the second stage before it is cooled. (CP =13 cal/gmol-K). 6. Estimate the change in enthalpy (Btu/lbmole) when ethane at 70F and 400 psia is compressed to 1500 psia and 120F (Tc =305.4K; Pc =4.880MPa; =.099; CP =13 Btu/lbmol-R). 7. Suppose ethane was compressed adiabatically in a 70% efficient continuous compressor. The downstream pressure is specified to be 1500 psia at a temperature not to exceed 350F. What is the highest that the upstream temperature could be if the upstream pressure is 200 psia? (Hint: neglect the departure function for the upstream thermodynamics.) Answers: 1b) 62% 1c) 5MPa 2) 42% 3) 29% 4) CpV/TS-(dV/dT)p 5) 260psia 6) -2300Btu/mol 7) 269K 4200:225 EQUILIBRIUM THERMODYNAMICS SPRING 91 TEST 2 SSN________ 1.a. The conditions of a fluid encountered in some process calculations are at a high reduced pressure and a low reduced temperature. Would this fluid have density and enthalpy similar to a gas, a vapor, a supercritical fluid or a liquid? Why? 1.b. Why does the compressibility factor increase sharply at high density? 1.c. Estimate the value of the compressibility factor, Z, for neon at Pr=30 and Tr=15. 1.d. Estimate the density of neon at Pr=30 and Tr=15.(1.2,0.26g/cc) 2. Freon-12 is used in a heat pump operating on an ordinary vapor compression cycle with a 100% efficient compressor. The average outdoor temperature is 90F and the desired indoor temperature is 70F. The design is such that a 40F approach temperature is used on each side. Compute the Q /W for this process and compare it to the value you would expect for a Carnot cycle operating at the same conditions. What would be the power requirement of the compressor motor (in hp) to provide 12,000 Btu/hr of heating capacity?(1 hp=0.7074 Btu/s) 3. Propane is compressed from 1 bar and 0xC to 64 bar and 100C. Compute the change in molar entropy, S. 4. As part of a liquefaction process, ammonia is throttled to 80% quality at atmospheric pressure. If the upstream pressure is 100 bar, what must be the upstream temperature? (Assume Cp=8.8 cal/mol-K). 5.a. For certain fluids, the equation of state is given by Z = 1 - b/Tr Develop an expression for the enthalpy departure function for fluids of this type. ( -2b/Tr) 5.b. Evaluate ( dS/dT)G in terms of only p, v, T, C , C and their derivatives. Your answer may include absolute values of S if is not associated with a derivative.

Answers: 1a.liquid 1b) close packed 1c)1.2 1d).26g/cc 2).95hp 3)-8.7cal/mol-K 4)358K 5a) 5b)Cp/T-S(dv/dT)p/V 4200:225 Eq. Thermo. SPRING 92 TEST 2 SSN_______ 1.a. Sketch the radial distribution function vs. radial distance for a low density hard sphere fluid. Describe in words why it looks like that. 1.b. The attractive contribution to the compressibility factor is:

ZkT

rdu rdr

g r dratt = 6

4 2( )

Show how this expression can be rearranged into two dimensionless groups, one group which includes the effects of well-depth () and size () (e.g. of the Sutherland potential) and another group which is a universal constant. 1.c. Estimate the vapor pressure of ethane at 244 K. 1.d. Estimate the heat of vaporization of ethane at 244 K. 2. Ethane is continuously compressed from 280 K and 1 bar to 310 K and 75 bar. Compute the change in enthalpy per gmol of ethane. (Cp/R = 5). 3. Ethane is expanded through an adiabatic, reversible turbine from 75 bar and 310 K to 1 bar. Estimate the temperature of the stream exiting the turbine and the work per gmol of ethane. (Hint: Is the exiting ethane vapor, liquid, or a little of each?)(Cp/R = 5) 4. Evaluate (dG/dV)T in terms of p, v, T, Cp, Cv, and their derivatives. Your answer may include absolute values of S if it is not associated with a derivative. 5. Develop an expression for the departure function based on the equation of state given below such that its value may be computed given T,P, and . [G(T,p)-Gig(T,p)]/nRT where Z = 1 + b/(1-b)- a/RT3/2(1+b) (This is the Redlich-Kwong (1958)eqn. -ln(1-b)-a/bRT3/2ln(1+b)+Z-1-lnZ) 6. Our space program requires a portable engine to generate electricity for a space station. It is proposed to use sodium as the working fluid in a customized form of a Rankine cycle. The high temperature stream is not superheated before running through the turbine. Instead, the saturated vapor is run directly through the (100% efficient, adiabatic) turbine. The rest of the Rankine cycle is the usual. That is, the outlet of the turbine is cooled to saturated liquid which is pumped (neglect the pump work) back into the boiler. The cycle is to operate between TL =1156 K (this is the boiling temperature of sodium) and TH =1444 K.

a) Estimate the quality coming out of the turbine. b) Compute the work output per unit of heat input to the cycle, and compare it to the value for a Carnot cycle operating between the same TH and TL .

(Tc =2300 K; Pc =195 bar; =0 ; CP/R = 2.5) Answers: 1.c)11.2 bar d)2777 cal/mol 2)-1852cal/mol 3)-745cal/gmol 4)V(dp/dV)T 5)-ln(1-b)-a/bRT3/2ln(1+b)+Z-1-lnZ 6.a)90% b).184

4200:225 Eq. Thermo. SPRING 93 TEST 2 SSN_______ 1a. Compute the specific volume (cc/mol) of saturated liquid isopentane at 1 bar. Tc =460.4; Pc =33.84 bar; =.227 1b. Estimate the work (J/mol) of adiabatically and reversibly compressing saturated liquid isopentane from 0.1 MPa to 20 MPa in a continuous process. 1c. Estimate the vapor pressure (bar) of isopentane at 400K. 1d. 1 liter of air (Cp/R=3.5) at 273K is to be compressed adiabatically and reversibly in a piston+cylinder to 0.1 liter. Estimate the final temperature. (5) 2. Ammonia is to be isothermally compressed in a specially designed flow turbine from 1 bar and 100C to 50 bar. If the compression is done reversibly, compute the heat flow needed per mole of ammonia. Tc =405.6K; Pc =112.8 bar; =0.250; CP/R=4.6 (20) 3. Express in terms of in terms of only P, V, T, CP, CV and their derivatives. Your answer may include absolute values of S if it is not associated with a derivative. (U/T)A.(10) 4. Suppose u(r) is given by the square-well potential and g(r)=10-5(r/) for r>. Evaluate the internal energy departure function where NA3=1 and /kT=1. (10) -5.7 5. A tank containing carbon dioxide (CP/R=4.5) at 350 K and 50 bar is vented until the temperature in the tank falls to 280K. Assuming no heat transfers between the gas and tank find the pressure in the tank at the end of the venting process. Tc =304.2K; Pc =73.76 bar; =.225 (20) 6. In our discussion of departure functions we derived the following expression for evaluating the internal energy departure function given any equation of state. ( )U U

RTT

ZT

dig

o

=

a) Derive the analogous expression for (Cv-Cvig)/R b) Derive an expression for (Cv-Cvig)/R in terms of a, b, , T for the EOS: Z b

ba T= +

+ 1

11

[exp( / ) ]

Answers: 1a) 112 (b) 2220 J/mol (c)12.5bar(d)686K (2)-13900J/mol(3)Cv-S/p[T(dp/dT)-p]

(4) -5.7 (5)20.8bar(6.a) (

2 2

2

2TZT

TZ

Td

o

(b) a2T-2 exp(a/T) )

4200:225 Eq. Thermo. Spring 94 TEST 2 NAME_______ 1a. Compute the density (g/cc) of liquid MethylTertiaryButylEther (MTBE) (MW=88) at 1 bar and 300K. Tc =496.4; Pc =33.7 bar; =.269(5) 1b. Estimate the vapor pressure (bar) of MTBE at 400K.(5) 1c. When the radial distribution function is equal to unity, how does the coordination number change with respect to the range of the neighborhood around the central atom?. Give an equation with no integral signs in it. (5) 2. Express in terms of in terms of only P, V, T, CP, CV and their derivatives. Your answer may include absolute values of S if it is not associated with a derivative. (A/V)P.(10) 3. Ammonia is continuously compressed from 1 bar and 100C to 50 bar and 150C. Compute the enthalpy change per mole of ammonia. Tc =405.6K; Pc =112.8 bar; =0.250; CP /R=4.6 (20) 4. Suppose u(r) is given by the 1.5 square-well potential and g(r)=exp(5/r) for r>. Evaluate the internal energy departure function where 3=1 and /kT=1. (15)

5. Vapor Freon 152a is to be adiabatically and reversibly compressed from 1 bar and 248 K to 10 bars in a continuous process. Compute the work required in Joules per mole (20) (Tc =386.7; Pc =45bar; =0.256; CP =68 J/mol-K) 6. Suppose an equation of state of the form: Z = 1 + 10 Yb*exp(2 Yb ) where Y = exp(a/Tr)-1 and a,b are constants Develop an expression for the Helmholtz energy departure function.(20) Answers: 1a0.76g/cc (b) 6.5 bars (c) 4/3 R3 2. 69J/mol 40 J/mol 3. -S(T/V)P - P 4. 5092 400 6.5[exp(2Yb)-1] 4200:225 Eq. Thermo. Spring 96 TEST 2 SSN _______ 1.a. Estimate the vapor pressure of propane at 325 K. 1.b. Estimate the saturated liquid density (g/cc) of propane at 325 K. 1.c. What is it about molecules that causes the compressibility factor to be less than unity most of the time? 1.d. Estimate the heat of vaporization (J/mole) of propane at 325 K. 2. Express in terms of P, V, T, Cp, Cv, and their derivatives. Your answer may include absolute values of S if it is not associated with a derivative. (S/T)H.(10) 3. Estimate the change in entropy (J/mole-K) for raising propane from a saturated liquid at 230K to a saturated vapor at 298K.(20) (CP /R = 8.85) 4. Suppose the radial distribution function at intermediate densities can be reasonably represented by: g ~ (1+2(/r)2) at all temperatures. Derive an expression for the attractive contribution to the compressibility factor for fluids that can be accurately represented by the Sutherland potential.(15) 5. Suppose we wanted to design a fix-a-flat system based on propane. Let the can be 500 cc and the tire be 40,000 cc. Assume the tire remains isothermal and at low enough pressure for the ideal gas approximation to be applicable. The can is ~ filled with 250 g of saturated liquid propane at 298K. If the pressure in the can drops to 0.85 MPa, what is the pressure in the tire and the amount of propane remaining in the can? Assuming 20 psig is enough to drive the car for a while, is the pressure in the tire sufficient? Could you fill another tire?(20) 6. Even in the days of van der Waals, the second virial coefficient for square-well fluids was known to be: B2/b = 4 + 9.5 [exp(NA/RT)-1]. Noting that ex ~ 1 + x + x2/2, this observation suggests the following equation of state:

Zbb

NRT

bA= + 14

19 5

.

Derive an expression for the Helmholtz energy departure function for this equation of state. (15) 1)18bar,.44,molecular attraction,12300(2)ugly(3)79(4)33/kT(5)-4ln(1-b)-9.5b/kT 4200:225 Eq. Thermo. Spring 97 TEST 2 SSN _______ 1.a. Estimate the fugacity of saturated liquid n-butane at 390 K. 1.b. Which would you expect to be higher: the density of butane at 426K and 38 bars or the density of butane at 450K and 60 bars? Why? 1.c. Why does the radial distribution function go to zero for r < ? 1.d. Estimate the heat of vaporization (J/mole) of n-butane at 390 K. 2. n-butane is isothermally compressed from 425K and 1 bar to 100 bars. Estimate the change in enthalpy (J/mole). 3. Estimate the work output (J/mole) for n-butane when it is continuously, adi

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Eliot Lira Intro to Thermodynamics in Chemical Engineering