Chapter 7, Problem 7.33P

Interpretation Introduction

Interpretation:

To find final temperature, T

To find work required, W

To find entropy change of ammonia gas $\Delta S$

Concept Introduction:

Ammonia:

T_c= 405.7.K

P_C= 112.8.bar

$\omega =0.253$

$T_0=294.15$ K

$P_0=200$ kPa

$P=1000$ kPa

For isenthalpic process, $\Delta S=0$

For the heat/capacity of ammonia

$A=3.578$

$B=3.020\times {10}^{-3}{K}^{-1}$

$D=-{\mathrm{0.186.10}}^5.K^2$

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

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

We know that

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

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

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

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

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

Where,

$\Delta S$ = Change in Entropy

Cp = Molar heat capacity

R = Universal Gas Constant

T₀= Initial given temperature = 21^oC = 294.15K

P₀ = Initial given Pressure = 200 kPa

P₂ = Final Given Pressure = 1000 kPa

$\eta$ = Given efficiency = 0.82

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

A = 3.578

B = 3.020 x 10-3 K-1

C = 0

D = -0.186 x 10⁵K²$\Delta H$ = Actual Change in Enthalpy

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

W = Power required of the compressor

T_c= Critical Temperature = 405.7 K

P_c = Critical Pressure = 112.8 bar = 112800 kPa

  $\omega$ = 0.253

Assume m = 1000 mol/sec

Answer to Problem 7.33P

T = 447.47K

W = 5673.2 kW

$\Delta S=2.347\frac{J}{mol.K}$

Explanation of Solution

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

$T_{r0}=0.725$

  $P_r=\frac{P}{P_c}$

$P_{r0}=0.01773$

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

$P_r=0.0886$

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 $\tau =1.4$

Given

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

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,

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

Solving we get, $\tau$ = 1.437

$T=\tau \times T_0$ T = 422.818 K

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

$T_r=1.042$

$\Delta H_{ig}=R\times ICPH(T_0,T,3.578,{\mathrm{3.020.10}}^{-3},0.0,-{\mathrm{0.186.10}}^5)$

Where

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

$\Delta H_{ig}=4.826\frac{kJ}{mol}$

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

$\Delta H\text{'}=4652\frac{J}{mol}$

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

$\eta =0.82$

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

$\Delta H=5673.2\frac{J}{mol}$

Work required, W = m $\Delta H$ = 5673.2 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 $\tau =1.4$

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

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

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 $\tau =1.521$

$T:=\tau .T_0$ We get

T= 447.47 K

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

Substituting the values in the above equation

$\Delta S=2.347\frac{J}{mol.K}$

Conclusion

T = 447.47K

W = 5673.2 kW

$\Delta S=2.347\frac{J}{mol.K}$