Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN: 9781259696527
Author: J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher: McGraw-Hill Education
Question
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Chapter 7, Problem 7.34P
Interpretation Introduction

Interpretation:

To find final temperature, T

To find work required, W

Concept Introduction:

Propylene:

Tc= 365.6K

PC= 46.65bar

ω=0.140

T0=303.15 K

P0=11.5 bar

P=18 bar

For isenthalpic process, ΔS=0

For the heat/capacity of Propylene

A=1.637

B=22.706×103K1

C=6.915×106.K2

Use generalized second-virial correlation:

The entropy change is given by

ΔS=T1T2CPigdTTR.ln(P2P1)+S2RS1R Combined with

  T0TCPRdTT=A.lnτ+[BT0+(CT02+Dτ2T02)(τ+1τ)](τ1) ; D = 0;

We know that

   ΔS=R.[A.ln(τ)+[B.T0+C.T02.(τ+12)].(τ1)lnPP0...]+SRB(τT0TC,ωPr)SRB(Tr0,ωPr0)

ΔH=R.[A.T0.(τ1)+B2.T02.(τ21)+C3T03.(τ31)...+Tc.(HRB( τ. T 0 T c ,ωP r,)HRB(Tr0,ωPr0,))]

ΔH=R.[A.T0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)]=R×ICPH(T0,T,A,B,C,D)

Where

T0TCPRdT=A.T0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)=ICPH(T,T0,A,B,C,D)

SRR=Pr[0.675Tr2.6+ω(0.722Tr5.2)]=SRB(Tr,ωPr)

HRRTC=Pr[0.0831.097Tr1.6+ω(0.1390.894Tr4.2)]=HRB(TrωPr)

Where,

ΔS = Change in Entropy

Cp = Molar heat capacity

R = Universal Gas Constant

T0= Initial given temperature = 303.15K

P0 = Initial given Pressure = 11.5 bar

P2 = Final Given Pressure = 18 bar

η = Given efficiency = 0.80

A, B, C, D = Constants for heat capacity of air

A = 1.637

B = 22.706 x 10-3 K-1

C = -6.915 x 10-6 K-2

D = 0

ΔH = Actual Change in Enthalpy

ΔHig = Ideal Gas Change in Enthalpy

W = Power required of the compressor

Tc= Critical Temperature = 365.6 K

Pc = Critical Pressure = 46.65 bar

  ω = 0.140

m = 1000 mol/sec

Expert Solution & Answer
Check Mark

Answer to Problem 7.34P

T = 327.2 K

W = 1205 kW

Explanation of Solution

Tr0=T0Tc

Tr0=0.8292

  Pr=PPc

Pr0=0.2465

Pr0=P0Pc

Pr=0.386

Use generalized second-virial correlation:

The entropy change is given by

ΔS=T1T2CPigdTTR.ln(P2P1)+S2RS1R Combined with

  T0TCPRdTT=A.lnτ+[BT0+(CT02+Dτ2T02)(τ+1τ)](τ1) ; C = 0;

Let us assume τ=1.1 1

Given

   ΔS=R.[A.ln(τ)+[B.T0+C.T02.(τ+12)].(τ1)lnPP0...]+SRB(τT0TC,ωPr)SRB(Tr0,ωPr0)

Where

SRR=Pr[0.675Tr2.6+ω(0.722T65.2)]=SRB(Tr,ωPr)

Substitute all the values to satisfy the following equation by trial and error,

   R.[A.ln(τ)+[B.T0+C.T02.(τ+12)].(τ1)lnPP0...]=SRB(Tr0,ωPr0,)SRB(τT0TC,ωPr)

Solving we get, τ = 1.069

T=τ×T0 T = 324.128 K

Tr=TTc

Tr=0.887

ΔHig=R×ICPH(T0,T,1.637,22.706×103,6.915×106,0.0)

Where

ΔH=R.[A.T0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)]=R×ICPH(T0,T,A,B,C,D)

Where

T0TCPRdT=A.T0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)=ICPH(T,T0,A,B,C,D)

ΔHig=1.409kJmol

  ΔH'=ΔHig+R.Tc.(HRB(Tr,ωPr)HRB(Tr0,ωPr0))

ΔH'=964.1Jmol

The actual enthalpy change from η=(ΔH)SΔH ;

η=0.80

ΔH=ΔH'η

ΔH=1205.2Jmol

Work required, W = m ΔH = 1205 kW

The actual final temperature is now found from

ΔH=T1T2CpigdT+H2RH1R

Combined with T0TCPRdT=A.T0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)

Now Guess τ=1.1 1

Substitute all the values to satisfy the following equation by trial and error

ΔH=R.[A.T0.(τ1)+B2.T02.(τ21)+C3T03.(τ31)...+Tc.(HRB(τ.T0Tc,ωPr,)HRB(Tr0,ωPr0,))]

Where

HRRTc=Pr[0.0831.097Tr1.6+ω(0.1390.894Tr4.2)]=HRB(TR,PR,ω)

By trial and error we get τ=1.079

T:=τ.T0 We get

T= 327.2 K

Conclusion

T = 327.2 K

W = 1205 kW

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