chapter 3 practice problems - michigan state universitylira/supp/practice/chap3prac.pdf · chapter...

Chapter 3 Practice Problems

To accompany Introductory Chemical Engineering Thermodynamics © J.R. Elliott, C.T. Lira, 2001, all rights reserved. (2/3/2009) 1 of 7

(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(!

!8mm −

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, Stirling’s approx will not apply ΔS/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

j

m

Np

∏=

= ………………………………………....Eqn. 3.4

75075!2!4!9

!151 ==p

756756)!5(!1532 ==p

31.2ln1

2 =⎥⎦

⎤⎢⎣

⎡=

Δ−

pp

k

S

(P3.4) two dice.

??=ΔkS for going from double sixes to a four and three.

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

⇒ ( ) ( )!61!*6

1!2

1 =p

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

( ) ( ) 2*!6

1!*61

!22 =⇒ p

693.02lnln1

2 ==⎟⎟⎠

⎞⎜⎜⎝

⎛=

Δpp

kS

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

⇒ ( )( ) MPa

LcmmoleLKKmoleMPacmP 108.0

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

3

3

1 =−

=

Similarly ⇒ MPaP 00723.02 =

1

2

1

2 lnlnPPR

TTCpS −=Δ ……………………………………… Eqn. 3.23

2

7RCp = …..(ig)

KkgkJKmoleJRS −=−=−=Δ /07.1/88.30

108.000723.0ln*314.8

300400ln*

27

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

S-balance:



dt

dnSdtnSd out

outin

−=)(

⇒ 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 ,)()( =Δ−=−=+ SdndnSdnSdnndS outinsideoutinside from the steam table . By interpolation, implies T = 120.8OC (P3.7) (a) Steady-state flow, ΔH = 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

===Δ⇒

=Δ⇒

===

=Δ⇒

°−===Δ ∫

(b) ??=η 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

η

At 1 bar = 0.1 MPa T S

100 7.361120.8 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

TPPT

PP

TT

PPR

TTCp

PPR

TTCpS

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

=′

⎟⎠⎞

⎜⎝⎛=′⇒

⎟⎟⎠

⎞⎜⎜⎝

⎛=′⇒

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛ ′⇒

=′

⇒

−′

==Δ

lbmolBTUHTTCpH

/5258)5591315(95.6)(& 12

=′Δ⇒−=−′=′Δ 76.0

69555258

==Δ

′Δ=⇒

HHη

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

kgkJkg

kmolekmolekJRCpdTW /76.33728

1*/175.9457)300625(*2

7625

300

==−== ∫

??=η

( ) ( )

molkJH

TTCpH

KTPP

TTS

CpR

/77.6794

3005.533*2314.8*7

5.533

0

12

2

1

2

1

2

=′Δ⇒

−⎟⎠⎞

⎜⎝⎛=−′=′Δ⇒

=′⇒

⎟⎟⎠

⎞⎜⎜⎝

⎛=

′⇒=Δ

%8.71

718.0/28*/76.337

/77.6794

=⇒

==Δ

′Δ=⇒

η

ηkmolkgkgkJ

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 =−=Δ= now find W’ = ??



ΔS = 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'' =−==Δ⇒

%67,67.03.606

407=⇒==⇒ ηη

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

L /49.2493)68.2441(*98.046.1006)( =+=Δ+= ⇒ kgkJHWS /21.119749.24937.3690 =−=Δ=

??=η , '' W

WHH

=ΔΔ

=η ,

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

⇒ %85,8503.00.14082.1197

=⇒== ηη

(P3.11)

CT

MPaP

dSatMPaP

O

vap

1100

10

',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' = 6.81763331.2 6.9386

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



kgkJWH S /35.219595.26743.4870 =−==Δ⇒ mass flow rate = 1 kg/s

hpW

hpwattwattskJW

S

S

01.2944001341022.01&

2195350/35.2195

=⇒

===⇒

%2.63

63.035.219558.1387

/58.138795.267453.4062&

=⇒

==Δ

′Δ=⇒

=−=′Δ

η

ηHH

kgkJH

(P3.12) Ethylene gas is to be continuously compressed from an initial state of 1 bar and 20° C to a final pressure of 18 bar in an adiabatic compressor. If compression is 70% efficient compared with an isentropic process, what will be the work requirement and what will be the final temperature of the ethylene? Assume the ethylene behaves as an ideal gas with CP = 44 J/mol-K (ANS. 13.4 kJ/mol, 596K) Ebal: ΔH = W.

Sbal: ΔSrev = 0 => 2 2

1 1

Rrev CpT P

T P

⎛ ⎞⎜ ⎟⎝ ⎠⎛ ⎞ ⎛ ⎞

=⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

=> T2rev = (20+273)*18^(8.314/44) = 506K.

Wrev = Cp(T2rev -T1) = 44*(506-293) = 9372J/mol => Wact = 9372/0.85 = 13.4kJ/mol

Wact = Cp(T2act -T1) = 13400 => T2

act = (13400/44)+293 = 597K

(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

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

3992 7.29164062.53 7.35894114.5 7.4085



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

chapter 3 practice problems - michigan state universitylira/supp/practice/chap3prac.pdf · chapter...

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