Chapter 6, Problem 6.54P

Interpretation Introduction

Interpretation:

Determine the entropy change for propane gas caused by the given process.

Concept Introduction:

Since at first the gas is not ideal gas, so the actual path of the process is divided into two step process in which at first step the gas will convert from real gas into ideal gas at given $T_1$ and $P_1$ . The properties will change accordingly. In the second steps the ideal gas changes state from $(T_1,P_1)$ to $(T_2,P_2)$ during throttling process.

The entropy change is calculated as:

  $\Delta S={\displaystyle \underset{T_1}{\overset{T_2}{\int }}{C}_P{}^{ig}\frac{dT}{T}-R\ln \frac{P_2}{P_1}+}{S}_2{}^R-S_1{}^R$....(1)

And the temperature change for an ideal gas during throttling is zero because throttling process is isenthalpic and for an ideal gas enthalpy is the function of temperature only, so the temperature does not change during throttling process for an ideal gas.

Answer to Problem 6.54P

Entropy change is

  $\Delta S=24.65\frac{\text{J}}{\text{mol K}}$

Explanation of Solution

Given information:

It is given that propane is at $22\text{bar}$ and $423\text{K}$ and is throttled in a steady state flow process to $1\text{bar}$ .

It is also given that at final state gas is assumed to be an ideal gas.

Step 1,

A hypothetical process that transforms a real propane gas into an ideal gas at $T_1$ and $P_1$ before throttling process,

The entropy and enthalpy change for this process are:

  $H_1{}^{ig}-H_1=H_1{}^R$

and

  $S_1{}^{ig}-S_1=S_1{}^R$

Where $H_1{}^R$ and $S_1{}^R$ are calculated from

  $H_1{}^R={\left(H_1{}^R\right)}^0+\omega {\left(H_1{}^R\right)}^1$

And

  $S_1{}^R={\left(S_1{}^R\right)}^0+\omega {\left(S_1{}^R\right)}^1$

For pure species propane, the properties can be written down using Appendix B, Table B.1

$\omega =0.152$, $T_c=369.8\text{K}$, $P_c=42.48\text{bar}$, $Z_C=0.276$, $T_n=231.1\text{K}$, $V_C=200\frac{{\text{cm}}^3}{\text{mol}}$

  $\begin{array}{l}{T}_r=\frac{T_1}{T_c}\\ T_r=\frac{423\text{K}}{369.8\text{K}}\text{=1}\text{.144}\end{array}$

  $\begin{array}{l}{P}_r=\frac{P_1}{P_c}\\ P_r=\frac{22\text{bar}}{42.48\text{bar}}\text{=0}\text{.518}\end{array}$

So, at above values of $T_r$ and $P_r$, The values of ${\frac{\left(H^R\right)}{R{T}_C}}^0$ and ${\frac{\left(H^R\right)}{R{T}_C}}^1$ can be written from Appendix D

$T_r=\text{1}\text{.144}$ lies between reduced temperatures $T_r=\text{1}\text{.1}$ and $T_r=\text{1}\text{.15}$ and $P_r=\text{0}\text{.518}$ lies in between reduced pressures $P_r=\text{0}\text{.4}$ and $P_r=\text{0}\text{.6}$ .

At $T_r=\text{1}\text{.1}$ and $P_r=\text{0}\text{.4}$

${\frac{\left(H^R\right)}{R{T}_C}}^0=-0.367$, ${\frac{\left(S^R\right)}{R}}^0=-0.23$

At $T_r=\text{1}\text{.1}$ and $P_r=\text{0}\text{.6}$

${\frac{\left(H^R\right)}{R{T}_C}}^0=-0.581$, ${\frac{\left(S^R\right)}{R}}^0=-0.371$

At $T_r=\text{1}\text{.15}$ and $P_r=\text{0}\text{.4}$

${\frac{\left(H^R\right)}{R{T}_C}}^0=-0.334$, ${\frac{\left(S^R\right)}{R}}^0=-0.201$

At $T_r=\text{1}\text{.15}$ and $P_r=\text{0}\text{.6}$

${\frac{\left(H^R\right)}{R{T}_C}}^0=-0.523$, ${\frac{\left(S^R\right)}{R}}^0=-0.319$

And

At $T_r=\text{1}\text{.1}$ and $P_r=\text{0}\text{.4}$

${\frac{\left(H^R\right)}{R{T}_C}}^1=-0.251$, ${\frac{\left(S^R\right)}{R}}^1=-0.229$

At $T_r=\text{1}\text{.1}$ and $P_r=\text{0}\text{.6}$

${\frac{\left(H^R\right)}{R{T}_C}}^1=-0.381$, ${\frac{\left(S^R\right)}{R}}^0=-0.35$

At $T_r=\text{1}\text{.15}$ and $P_r=\text{0}\text{.4}$

${\frac{\left(H^R\right)}{R{T}_C}}^1=-0.199$, ${\frac{\left(S^R\right)}{R}}^0=-0.183$

At $T_r=\text{1}\text{.15}$ and $P_r=\text{0}\text{.6}$

${\frac{\left(H^R\right)}{R{T}_C}}^1=-0.296$, ${\frac{\left(S^R\right)}{R}}^0=-0.275$

Applying linear interpolation of two independent variables, From linear double interpolation, if $M$ is the function of two independent variable $X$ and $Y$ then the value of quantity $M$ at two independent variables $X$ and $Y$ adjacent to the given values are represented as follows:

  $\begin{array}{l}{X}_{1}X{X}_2\\ Y_1{M}_{1,1}{M}_{1,2}\\ YM=?\\ Y_2{M}_{2,1}{M}_{2,2}\end{array}$

  $\begin{array}{l}M=\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{1,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{1,2}\right]\frac{Y_2-Y}{Y_2-Y_1}\\ +\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{2,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{2,2}\right]\frac{Y-Y_1}{Y_2-Y_1}\\ \\ {\frac{\left(H^R\right)}{R{T}_C}}^0=\left[\left(\frac{0.6-\text{0}\text{.518}}{0.6-\text{0}\text{.4}}\right)\times -0.367+\left(\frac{\text{0}\text{.518}-\text{0}\text{.4}}{0.6-\text{0}\text{.4}}\right)\times -0.581\right]\times \frac{\text{1}\text{.15}-1.144}{1.15-1.1}\\ +\left[\left(\frac{\text{1}-\text{0}\text{.815}}{\text{1}-\text{0}\text{.8}}\right)\times -0.334+\left(\frac{\text{0}\text{.815}-\text{0}\text{.4}}{\text{1}-\text{0}\text{.8}}\right)\times -0.523\right]\times \frac{1.144-1.1}{1.15-1.1}\\ \\ {\frac{\left(H^R\right)}{R{T}_C}}^0=-0.451\end{array}$

  $\begin{array}{l}M=\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{1,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{1,2}\right]\frac{Y_2-Y}{Y_2-Y_1}\\ +\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{2,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{2,2}\right]\frac{Y-Y_1}{Y_2-Y_1}\\ \\ {\frac{\left(S^R\right)}{R}}^0=\left[\left(\frac{0.6-\text{0}\text{.518}}{0.6-\text{0}\text{.4}}\right)\times -0.23+\left(\frac{\text{0}\text{.518}-\text{0}\text{.4}}{0.6-\text{0}\text{.4}}\right)\times -0.371\right]\times \frac{\text{1}\text{.15}-1.144}{1.15-1.1}\\ +\left[\left(\frac{\text{1}-\text{0}\text{.815}}{\text{1}-\text{0}\text{.8}}\right)\times -0.201+\left(\frac{\text{0}\text{.815}-\text{0}\text{.4}}{\text{1}-\text{0}\text{.8}}\right)\times -0.319\right]\times \frac{1.144-1.1}{1.15-1.1}\\ \\ {\frac{\left(S^R\right)}{R}}^0=-0.276\end{array}$

And

  $\begin{array}{l}M=\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{1,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{1,2}\right]\frac{Y_2-Y}{Y_2-Y_1}\\ +\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{2,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{2,2}\right]\frac{Y-Y_1}{Y_2-Y_1}\\ \\ {\frac{\left(H^R\right)}{R{T}_C}}^1=\left[\left(\frac{0.6-\text{0}\text{.518}}{0.6-\text{0}\text{.4}}\right)\times -0.251+\left(\frac{\text{0}\text{.518}-\text{0}\text{.4}}{0.6-\text{0}\text{.4}}\right)\times -0.381\right]\times \frac{\text{1}\text{.15}-1.144}{1.15-1.1}\\ +\left[\left(\frac{\text{1}-\text{0}\text{.815}}{\text{1}-\text{0}\text{.8}}\right)\times -0.199+\left(\frac{\text{0}\text{.815}-\text{0}\text{.4}}{\text{1}-\text{0}\text{.8}}\right)\times -0.296\right]\times \frac{1.144-1.1}{1.15-1.1}\\ \\ {\frac{\left(H^R\right)}{R{T}_C}}^1=-0.265\end{array}$

  $\begin{array}{l}M=\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{1,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{1,2}\right]\frac{Y_2-Y}{Y_2-Y_1}\\ +\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{2,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{2,2}\right]\frac{Y-Y_1}{Y_2-Y_1}\\ \\ {\frac{\left(S^R\right)}{R}}^1=\left[\left(\frac{0.6-\text{0}\text{.518}}{0.6-\text{0}\text{.4}}\right)\times -0.229+\left(\frac{\text{0}\text{.518}-\text{0}\text{.4}}{0.6-\text{0}\text{.4}}\right)\times -0.35\right]\times \frac{\text{1}\text{.15}-1.144}{1.15-1.1}\\ +\left[\left(\frac{\text{1}-\text{0}\text{.815}}{\text{1}-\text{0}\text{.8}}\right)\times -0.183+\left(\frac{\text{0}\text{.815}-\text{0}\text{.4}}{\text{1}-\text{0}\text{.8}}\right)\times -0.275\right]\times \frac{1.144-1.1}{1.15-1.1}\\ \\ {\frac{\left(S^R\right)}{R}}^1=-0.245\end{array}$

Now, from equation,

  $H_1{}^R={\left(H_1{}^R\right)}^0+\omega {\left(H_1{}^R\right)}^1$

Or

  $\begin{array}{l}\frac{H_1{}^R}{R{T}_C}=\frac{{\left(H_1{}^R\right)}^0}{R{T}_C}+\omega \frac{{\left(H_1{}^R\right)}^1}{R{T}_C}\\ \frac{H_1{}^R}{R{T}_C}=-0.451+0.152\times -0.265\\ \frac{H_1{}^R}{R{T}_C}=-0.4913\\ H_1{}^R=-0.4913\times 8.314\frac{\text{J}}{\text{mol K}}\times 369.8\text{K}\\ H_1{}^R=-1510.51\frac{\text{J}}{\text{mol}}\end{array}$

And

  $S_1{}^R={\left(S_1{}^R\right)}^0+\omega {\left(S_1{}^R\right)}^1$

Or

  $\begin{array}{l}\frac{S_1{}^R}{R}=\frac{{\left(S_1{}^R\right)}^0}{R}+\omega \frac{{\left(S_1{}^R\right)}^1}{R}\\ \frac{S_1{}^R}{R}=-0.276+0.152\times -0.245\\ \frac{S_1{}^R}{R}=-0.6344\\ S_1{}^R=-0.313\times 8.314\frac{\text{J}}{\text{mol K}}\\ S_1{}^R=-2.6\frac{\text{J}}{\text{mol}\text{K}}\end{array}$

For Step 2

Enthalpy will not change during throttling of propylene, so $\Delta H=0$

Enthalpy of propylene which is now in ideal state is same in final state and hence

  $H_1{}^R=-1510.51\frac{\text{J}}{\text{mol}}$

  $\begin{array}{l}\Delta H={\displaystyle \underset{T_1}{\overset{T_2}{\int }}{C}_P{}^{ig}dT-}{H}_1{}^R\\ 0={\displaystyle \underset{T_1}{\overset{T_2}{\int }}{C}_P{}^{ig}dT-}{H}_1{}^R\\ H_1{}^R=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}dT}\\ \frac{H_1{}^R}{R}={\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}dT}\end{array}$

  ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$

Where $\tau =\frac{T}{T_0}$

Values of above constants forpropane in above equation are given in appendix C table C.1 and noted down below:

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ propane1.21328.785-8.8240\end{array}$

  $\begin{array}{l}\frac{-1510.51\frac{\text{J}}{\text{mol}}}{8.314\frac{\text{J}}{\text{mol K}}}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ -181.68=1.213\times 423\times (\tau -1)+\frac{28.785\times {10}^{-3}}{2}\times {423}^2\times ({\tau }^2-1)+\frac{-8.824\times {10}^{-6}}{3}\times {423}^3({\tau }^3-1)\\ \tau =0.963\end{array}$

And

Hence

  $\begin{array}{l}0.963=\frac{T}{423\text{K}}\\ T=407.35\text{K}\end{array}$

Now for entropy change

  $\Delta S={\displaystyle \underset{T_1}{\overset{T_2}{\int }}{C}_P{}^{ig}\frac{dT}{T}-R\ln \frac{P_2}{P_1}}-S_1{}^R$

  ${\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}}=\left[A+\left\{B{T}_0+\left(C{T}_0{}^2+\frac{D}{{\tau }^2{T}_0{}^2}\right)\left(\frac{\tau +1}{2}\right)\right\}\left(\frac{\tau -1}{\ln \tau }\right)\right]\times \ln \tau$

  $\tau =0.963$

Values of above constants for propane in above equation are given in appendix C table C.1 and noted down below:

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ propane1.21328.785-8.8240\end{array}$

  $\begin{array}{l}{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}}=\left[A+\left\{B{T}_0+\left(C{T}_0{}^2+\frac{D}{{\tau }^2{T}_0{}^2}\right)\left(\frac{\tau +1}{2}\right)\right\}\left(\frac{\tau -1}{\ln \tau }\right)\right]\times \ln \tau \\ {\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}}=\\ \left[1.213+\left\{28.785\times {10}^{-3}\times 423+\left(-8.824\times {10}^{-6}\times {423}^2+\frac{0\times {10}^5}{{0.963}^2\times {423}^2}\right)\left(\frac{0.963+1}{2}\right)\right\}\left(\frac{0.963-1}{\ln 0.963}\right)\right]\\ \times \ln 0.963\\ \\ {\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}}=-0.439\end{array}$

Hence,

  $\begin{array}{l}\Delta S=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}-R\ln \frac{P_2}{P_1}-}{S}_1{}^R\\ \Delta S=8.314\frac{\text{J}}{\text{mol K}}\times -0.439-8.314\frac{\text{J}}{\text{mol K}}\times \ln \frac{1}{22}-\left(-2.6\frac{\text{J}}{\text{mol}\text{K}}\right)\\ \\ \Delta S=24.65\frac{\text{J}}{\text{mol K}}\end{array}$