Chapter 7, Problem 7.34P

Interpretation Introduction

Interpretation:

To find final temperature, T

To find work required, W

Concept Introduction:

Propylene:

T_c= 365.6K

P_C= 46.65bar

$\omega =0.140$

$T_0=303.15$ K

$P_0=11.5$ bar

$P=18$ bar

For isenthalpic process, $\Delta S=0$

For the heat/capacity of Propylene

$A=1.637$

$B=22.706\times {10}^{-3}{K}^{-1}$

$C=-6.915\times {10}^{-6}.K^{-2}$

Use generalized second-virial correlation:

The entropy change is given by

$\Delta S={\displaystyle {\int }_{T_1}^{T_2}{C}_P^{ig}}\frac{dT}{T}-R.\ln (\frac{P_2}{P_1})+S_2^R-S_1^R$ Combined with

  ${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}\frac{dT}{T}=A.\ln \tau +[B{T}_0+(C{T}_0^2+\frac{D}{{\tau }^2{T}_0^2})(\frac{\tau +1}{\tau })](\tau -1)$ ; D = 0;

We know that

  $\Delta S=R.\left[A.\ln (\tau )+\left[B.T_{{}_0}+C.T_0^2.\left(\frac{\tau +1}{2}\right)\right].(\tau -1)-\ln \frac{P}{P_0}\mathrm{...}\right]+SRB\left(\frac{\tau -T_0}{T_C},\omega P_r\right)-SRB(T_{r0},\omega P_{r0})$

$\begin{array}{l}\Delta H=R.\left[A.T_0.(\tau -1)+\frac{B}{2}.T_0{}^2.({\tau }^2-1)+\frac{C}{3}{T}_0{}^3.({\tau }^3-1)\mathrm{...}+T_c.\left(HRB\left(\frac{\tau .T_0}{T_c},\omega P_{r,}\right)-HRB(T_{r0},\omega P_{r0},)\right)\right]\\ \end{array}$

$\Delta H=R.\left[A.T_0(\tau -1)+\frac{B}{2}{T}_0^2({\tau }^2-1)+\frac{C}{3}{T}_0^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\right]=R\times ICPH(T_0,T,A,B,C,D)$

Where

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=A.T_0(\tau -1)+\frac{B}{2}{T}_0^2({\tau }^2-1)+\frac{C}{3}{T}_0^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })=ICPH(T,T_0,A,B,C,D)$

$\frac{S^R}{R}=-P_r\left[\frac{0.675}{T_r{}^{2.6}}+\omega \left(\frac{0.722}{T_r{}^{5.2}}\right)\right]=SRB\left(T_{r,}\omega P_r\right)$

$\frac{H^R}{R{T}_C}=P_r\left[0.083-\frac{1.097}{T_r{}^{1.6}}+\omega \left(0.139-\frac{0.894}{T_r{}^{4.2}}\right)\right]=HRB(T_r\omega P_r)$

Where,

$\Delta S$ = 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

$\eta$ = 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

$\Delta H$ = Actual Change in Enthalpy

$\Delta H_{ig}$ = Ideal Gas Change in Enthalpy

W = Power required of the compressor

T_c= Critical Temperature = 365.6 K

P_c = Critical Pressure = 46.65 bar

  $\omega$ = 0.140

m = 1000 mol/sec

Answer to Problem 7.34P

T = 327.2 K

W = 1205 kW

Explanation of Solution

$T_{r0}=\frac{T_0}{T_c}$

$T_{r0}=0.8292$

  $P_r=\frac{P}{P_c}$

$P_{r0}=0.2465$

$P_{r0}=\frac{P_0}{P_c}$

$P_r=0.386$

Use generalized second-virial correlation:

The entropy change is given by

$\Delta S={\displaystyle {\int }_{T_1}^{T_2}{C}_P^{ig}}\frac{dT}{T}-R.\ln (\frac{P_2}{P_1})+S_2^R-S_1^R$ Combined with

  ${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}\frac{dT}{T}=A.\ln \tau +[B{T}_0+(C{T}_0^2+\frac{D}{{\tau }^2{T}_0^2})(\frac{\tau +1}{\tau })](\tau -1)$ ; C = 0;

Let us assume $\tau =1.1$ 1

Given

  $\Delta S=R.\left[A.\ln (\tau )+\left[B.T_{{}_0}+C.T_0^2.\left(\frac{\tau +1}{2}\right)\right].(\tau -1)-\ln \frac{P}{P_0}\mathrm{...}\right]+SRB\left(\frac{\tau -T_0}{T_C},\omega P_r\right)-SRB(T_{r0},\omega P_{r0})$

Where

$\frac{S^R}{R}=-P_r\left[\frac{0.675}{T_r{}^{2.6}}+\omega \left(\frac{0.722}{T_6{}^{5.2}}\right)\right]=SRB(T_r,\omega P_r)$

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

  $R.\left[A.\ln (\tau )+\left[B.T_{{}_0}+C.T_0^2.\left(\frac{\tau +1}{2}\right)\right].(\tau -1)-\ln \frac{P}{P_0}\mathrm{...}\right]=SRB(T_{r0},\omega P_{r0},)-SRB\left(\frac{\tau -T_0}{T_C},\omega P_r\right)$

Solving we get, $\tau$ = 1.069

$T=\tau \times T_0$ T = 324.128 K

$T_r=\frac{T}{T_c}$

$T_r=0.887$

$\Delta H_{ig}=R\times ICPH(T_0,T,1.637,22.706\times {10}^{-3},-6.915\times {10}^{-6},0.0)$

Where

$\Delta H=R.\left[A.T_0(\tau -1)+\frac{B}{2}{T}_0^2({\tau }^2-1)+\frac{C}{3}{T}_0^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\right]=R\times ICPH(T_0,T,A,B,C,D)$

Where

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=A.T_0(\tau -1)+\frac{B}{2}{T}_0^2({\tau }^2-1)+\frac{C}{3}{T}_0^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })=ICPH(T,T_0,A,B,C,D)$

$\Delta H_{ig}=1.409\frac{kJ}{mol}$

  $\Delta H\text{'}=\Delta H_{ig}+R.T_c.(HRB(T_r,\omega {\mathrm{P}}_r)-HRB(T_{r0},\omega P_{r0}))$

$\Delta H\text{'}=964.1\frac{J}{mol}$

The actual enthalpy change from $\eta =\frac{{(\Delta H)}_S}{\Delta H}$ ;

$\eta =0.80$

$\Delta H=\frac{\Delta H\text{'}}{\eta }$

$\Delta H=1205.2\frac{J}{mol}$

Work required, W = m $\Delta H$ = 1205 kW

The actual final temperature is now found from

$\Delta H={\displaystyle {\int }_{T_1}^{T_2}{C}_p^{ig}}dT+H_2^R-H_1^R$

$$

Combined with ${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=A.T_0(\tau -1)+\frac{B}{2}{T}_0^2({\tau }^2-1)+\frac{C}{3}{T}_0^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$

Now Guess $\tau =1.1$ 1

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

$\Delta H=R.\left[A.T_0.(\tau -1)+\frac{B}{2}.T_0{}^2.({\tau }^2-1)+\frac{C}{3}{T}_0{}^3.({\tau }^3-1)\mathrm{...}+T_c.\left(HRB\left(\frac{\tau .T_0}{T_c},\omega P_{r,}\right)-HRB(T_{r0},\omega P_{r0},)\right)\right]$

Where

$\frac{H^R}{R{T}_c}=P_r\left[0.083-\frac{1.097}{T_r^{1.6}}+\omega \left(0.139-\frac{0.894}{T_r{}^{4.2}}\right)\right]=HRB\left(T_R,P_R,\omega \right)$

By trial and error we get $\tau =1.079$

$T:=\tau .T_0$ We get

T= 327.2 K

Conclusion

T = 327.2 K

W = 1205 kW