Chapter 6, Problem 6.60P

Interpretation Introduction

Interpretation:

Determine the final temperature and work produced of the compressedn-butane gas, compressed isentropically in a steady state process.

Concept Introduction:

The entropy change by appropriate generalized correlations is

  $\Delta S={\displaystyle \underset{T_1}{\overset{T_2}{\int }}{C}_P{}^{ig}\frac{dT}{T}-R\ln \frac{P_2}{P_1}+}{S}_2{}^R-S_1{}^R$....(2)

Answer to Problem 6.60P

Temperature is

  $T=381.317\text{K}$

Work done is

  $W=-6332.15\frac{\text{J}}{\text{mol}}$

Explanation of Solution

Given information:

It is given that n-butane gas compressed isentropically from $1\text{bar}$ and $323.15\text{K}$ to $7.8\text{bar}$ .

Process is isentropically, so $\Delta S=0$

From equation (2),

  $\begin{array}{l}\Delta S={\displaystyle \underset{T_1}{\overset{T_2}{\int }}{C}_P{}^{ig}\frac{dT}{T}-R\ln \frac{P_2}{P_1}+}{S}_2{}^R-S_1{}^R\\ R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}-R\ln \frac{P_2}{P_1}+}{S}_2{}^R-S_1{}^R=0\\ R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}=R\ln \frac{P_2}{P_1}-}{S}_2{}^R+S_1{}^R\end{array}$

Initial state

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

$\omega =0.2$, $T_c=425.1\text{K}$, $P_c=37.96\text{bar}$, $Z_C=0.276$

  $\begin{array}{l}{T}_r=\frac{T_1}{T_c}\\ T_r=\frac{323.15\text{K}}{425.1\text{K}}\text{=0}\text{.7602}\end{array}$

  $\begin{array}{l}{P}_r=\frac{P_1}{P_c}\\ P_r=\frac{1\text{bar}}{37.96\text{bar}}\text{=0}\text{.0263}\end{array}$

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

$T_r=\text{0}\text{.76}$ lies between reduced temperatures $T_r=\text{0}\text{.75}$ and $T_r=\text{0}\text{.8}$ and $P_r=\text{0}\text{.026}$ lies in between reduced pressures $P_r=\text{0}\text{.01}$ and $P_r=\text{0}\text{.05}$ .

At $T_r=\text{0}\text{.75}$ and $P_r=\text{0}\text{.01}$

${\frac{\left(H^R\right)}{R{T}_C}}^0=-0.017$, ${\frac{\left(S^R\right)}{R}}^0=-0.015$

At $T_r=\text{0}\text{.75}$ and $P_r=\text{0}\text{.05}$

${\frac{\left(H^R\right)}{R{T}_C}}^0=-0.088$, ${\frac{\left(S^R\right)}{R}}^0=-0.078$

At $T_r=\text{0}\text{.8}$ and $P_r=\text{0}\text{.01}$

${\frac{\left(H^R\right)}{R{T}_C}}^0=-0.015$, ${\frac{\left(S^R\right)}{R}}^0=-0.013$

At $T_r=\text{0}\text{.8}$ and $P_r=\text{0}\text{.05}$

${\frac{\left(H^R\right)}{R{T}_C}}^0=-0.078$, ${\frac{\left(S^R\right)}{R}}^0=-0.064$

And

At $T_r=\text{0}\text{.75}$ and $P_r=\text{0}\text{.01}$

${\frac{\left(H^R\right)}{R{T}_C}}^1=-0.027$, ${\frac{\left(S^R\right)}{R}}^1=-0.029$

At $T_r=\text{0}\text{.75}$ and $P_r=\text{0}\text{.05}$

${\frac{\left(H^R\right)}{R{T}_C}}^1=-0.142$, ${\frac{\left(S^R\right)}{R}}^1=-0.156$

At $T_r=\text{0}\text{.8}$ and $P_r=\text{0}\text{.01}$

${\frac{\left(H^R\right)}{R{T}_C}}^1=-0.021$, ${\frac{\left(S^R\right)}{R}}^1=-0.022$

At $T_r=\text{0}\text{.8}$ and $P_r=\text{0}\text{.05}$

${\frac{\left(H^R\right)}{R{T}_C}}^1=-0.11$, ${\frac{\left(S^R\right)}{R}}^1=-0.116$

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:

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

  $\begin{array}{l}M=\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{1,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{1,2}\right]\frac{Y_2-Y}{Y_2-Y_1}\\ +\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{2,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{2,2}\right]\frac{Y-Y_1}{Y_2-Y_1}\\ \\ {\frac{\left(H^R\right)}{R{T}_C}}^0=\left[\left(\frac{0.05-\text{0}\text{.026}}{0.05-\text{0}\text{.01}}\right)\times -0.017+\left(\frac{\text{0}\text{.026}-\text{0}\text{.01}}{0.05-\text{0}\text{.01}}\right)\times -0.088\right]\times \frac{\text{0}\text{.8}-0.76}{0.8-0.75}\\ +\left[\left(\frac{0.05-\text{0}\text{.026}}{0.05-\text{0}\text{.01}}\right)\times -0.015+\left(\frac{\text{0}\text{.026}-\text{0}\text{.01}}{0.05-\text{0}\text{.01}}\right)\times -0.078\right]\times \frac{0.76-0.75}{0.8-0.75}\\ \\ {\frac{\left(H^R\right)}{R{T}_C}}^0=-0.04436\end{array}$

  $\begin{array}{l}M=\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{1,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{1,2}\right]\frac{Y_2-Y}{Y_2-Y_1}\\ +\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{2,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{2,2}\right]\frac{Y-Y_1}{Y_2-Y_1}\\ \\ {\frac{\left(S^R\right)}{R}}^0=\left[\left(\frac{0.05-\text{0}\text{.026}}{0.05-\text{0}\text{.01}}\right)\times -0.015+\left(\frac{\text{0}\text{.026}-\text{0}\text{.01}}{0.05-\text{0}\text{.01}}\right)\times -0.078\right]\times \frac{\text{0}\text{.8}-0.76}{0.8-0.75}\\ +\left[\left(\frac{0.05-\text{0}\text{.026}}{0.05-\text{0}\text{.01}}\right)\times -0.013+\left(\frac{\text{0}\text{.026}-\text{0}\text{.01}}{0.05-\text{0}\text{.01}}\right)\times -0.064\right]\times \frac{0.76-0.75}{0.8-0.75}\\ \\ {\frac{\left(S^R\right)}{R}}^0=-0.03884\end{array}$

And

  $\begin{array}{l}M=\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{1,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{1,2}\right]\frac{Y_2-Y}{Y_2-Y_1}\\ +\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{2,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{2,2}\right]\frac{Y-Y_1}{Y_2-Y_1}\\ \\ {\frac{\left(H^R\right)}{R{T}_C}}^1=\left[\left(\frac{0.05-\text{0}\text{.026}}{0.05-\text{0}\text{.01}}\right)\times -0.027+\left(\frac{\text{0}\text{.026}-\text{0}\text{.01}}{0.05-\text{0}\text{.01}}\right)\times -0.142\right]\times \frac{\text{0}\text{.8}-0.76}{0.8-0.75}\\ +\left[\left(\frac{0.05-\text{0}\text{.026}}{0.05-\text{0}\text{.01}}\right)\times -0.021+\left(\frac{\text{0}\text{.026}-\text{0}\text{.01}}{0.05-\text{0}\text{.01}}\right)\times -0.110\right]\times \frac{0.76-0.75}{0.8-0.75}\\ \\ {\frac{\left(H^R\right)}{R{T}_C}}^1=-0.06972\end{array}$

  $\begin{array}{l}M=\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{1,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{1,2}\right]\frac{Y_2-Y}{Y_2-Y_1}\\ +\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{2,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{2,2}\right]\frac{Y-Y_1}{Y_2-Y_1}\\ \\ {\frac{\left(S^R\right)}{R}}^1=\left[\left(\frac{0.05-\text{0}\text{.026}}{0.05-\text{0}\text{.01}}\right)\times -0.029+\left(\frac{\text{0}\text{.026}-\text{0}\text{.01}}{0.05-\text{0}\text{.01}}\right)\times -0.156\right]\times \frac{\text{0}\text{.8}-0.76}{0.8-0.75}\\ +\left[\left(\frac{0.05-\text{0}\text{.026}}{0.05-\text{0}\text{.01}}\right)\times -0.022+\left(\frac{\text{0}\text{.026}-\text{0}\text{.01}}{0.05-\text{0}\text{.01}}\right)\times -0.116\right]\times \frac{0.76-0.75}{0.8-0.75}\\ \\ {\frac{\left(S^R\right)}{R}}^1=-0.07576\end{array}$

Now, from equation,

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

Or

  $\begin{array}{l}\frac{H_1{}^R}{R{T}_C}=\frac{{\left(H_1{}^R\right)}^0}{R{T}_C}+\omega \frac{{\left(H_1{}^R\right)}^1}{R{T}_C}\\ \frac{H_1{}^R}{R{T}_C}=-0.04436+0.2\times -0.06972\\ \frac{H_1{}^R}{R{T}_C}=-0.058304\\ H_1{}^R=-0.058304\times 8.314\frac{\text{J}}{\text{mol K}}\times 425.1\text{K}\\ H_1{}^R=-206.063\frac{\text{J}}{\text{mol}}\end{array}$

And

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

Or

  $\begin{array}{l}\frac{S_1{}^R}{R}=\frac{{\left(S_1{}^R\right)}^0}{R}+\omega \frac{{\left(S_1{}^R\right)}^1}{R}\\ \frac{S_1{}^R}{R}=-0.03884+0.2\times -0.07576\\ \frac{S_1{}^R}{R}=-0.053992\\ S_1{}^R=-0.053992\times 8.314\frac{\text{J}}{\text{mol K}}\\ S_1{}^R=-0.4489\frac{\text{J}}{\text{mol}\text{K}}\end{array}$

Considering at final state gas is an ideal gas.

Hence, $H_2{}^R=0$ and $S_2{}^R=0$

Now,

  $\begin{array}{l}{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}=\ln \frac{P_2}{P_1}-}\frac{S_2{}^R}{R}+\frac{S_1{}^R}{R}\\ {\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}=\ln \frac{7.8}{1}-}0+\frac{\left(-0.4489\frac{\text{J}}{\text{mol}\text{K}}\right)}{8.314\frac{\text{J}}{\text{mol K}}}\\ {\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}=}2\end{array}$

Hence,

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

  $\tau =\frac{T}{T_0}$

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

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ n-bu\tan e1.93536.915-11.4020\end{array}$

  $\tau =\frac{T}{T_0}$

  $\begin{array}{l}{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}}=\left[A+\left\{B{T}_0+\left(C{T}_0{}^2+\frac{D}{{\tau }^2{T}_0{}^2}\right)\left(\frac{\tau +1}{2}\right)\right\}\left(\frac{\tau -1}{\ln \tau }\right)\right]\times \ln \tau \\ {\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}}=\\ \left[1.935+\left\{36.915\times {10}^{-3}\times 323.15+\left(-11.402\times {10}^{-6}\times {323.15}^2+\frac{0}{{\tau }^2\times {323.15}^2}\right)\left(\frac{\tau +1}{2}\right)\right\}\left(\frac{\tau -1}{\ln \tau }\right)\right]\times \ln \tau \\ \\ \\ 2=\left[1.935+\left\{36.915\times {10}^{-3}\times 323.15+\left(-11.402\times {10}^{-6}\times {323.15}^2\right)\left(\frac{\tau +1}{2}\right)\right\}\left(\frac{\tau -1}{\ln \tau }\right)\right]\times \ln \tau \\ \\ \tau =1.18\end{array}$

  $\begin{array}{l}\tau =\frac{T}{T_0}\\ 1.18=\frac{T}{323.15\text{K}}\\ T=381.317\text{K}\end{array}$

Now, for work produced

  $W=\Delta H$

And enthalpy change by generalized correlations is given by,

  $\Delta H={\displaystyle \underset{T_1}{\overset{T_2}{\int }}{C}_P{}^{ig}dT+}{H}_2{}^R-H_1{}^R$

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

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

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\\ =1.935\times 323.15\times (1.18-1)+\frac{36.915\times {10}^{-3}}{2}\times {323.15}^2\times ({1.18}^2-1)+\frac{-11.402\times {10}^{-6}}{3}\times {323.15}^3({1.18}^3-1)\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=-786.41\text{K}\end{array}$

Hence,

  $\begin{array}{l}W=\Delta H=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}dT+}{H}_2{}^R-H_1{}^R\\ W=8.314\frac{\text{J}}{\text{mol K}}\times -786.41\text{K+}0-\left(-206.063\frac{\text{J}}{\text{mol}}\right)\\ W=-6332.15\frac{\text{J}}{\text{mol}}\end{array}$

Conclusion

Temperature is

  $T=381.317\text{K}$

Work done is

  $W=-6332.15\frac{\text{J}}{\text{mol}}$