Chapter 5, Problem 5.28P

(a)

Interpretation Introduction

Interpretation:

Entropy change of the gas for given steady state flow process at atmospheric pressure should be determined.

Answer to Problem 5.28P

  $\Delta S=900.86\frac{\text{J}}{\text{K}}$

Explanation of Solution

Given information:

When $800\text{kJ}$ is added to $10\text{mol}$ of ethylene initially at ${200}^{\circ }\text{C}$

  $T_0={200}^{\circ }\text{C+273}\text{.15=473}\text{.15}\text{K}$

And entropy of process at constant pressure is calculated by following formula,

  $\begin{array}{l}\Delta S=nR{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{C_P}{R}\frac{dT}{T}}\\ \text{where,}\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{C_P}{R}\frac{dT}{T}}=\left[A+\left\{B{T}_0+\left(C{T}_0{}^2+\frac{D}{{\tau }^2{T}_0{}^2}\right)\left(\frac{\tau +1}{2}\right)\right\}\left(\frac{\tau -1}{\ln \tau }\right)\right]\times \ln \tau \end{array}$

where

  $\tau =\frac{T}{T_0}$

And values of constants for ethylene in above equation are given in appendix C table C.1 and noted down below:

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ Ethylene1.42414.394-4.3920\end{array}$

For final temperature,

  $\begin{array}{l}Q=\Delta H={\displaystyle \underset{T_1}{\overset{T_2}{\int }}{C}_P{}^{ig}dT}\\ \Rightarrow Q=nR{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}dT}\end{array}$

Where

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

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

Put the values

  $\begin{array}{l}\frac{Q}{nR}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ \\ \frac{800\text{kJ}\times \frac{1000\text{J}}{1\text{kJ}}}{10\text{mol}\times \text{8}\text{.314}\frac{\text{J}}{\text{mol K}}}=1.424\times 473.15(\tau -1)+\frac{14.394\times {10}^{-3}}{2}\times {473.15}^2({\tau }^2-1)+\frac{-4.392\times {10}^{-6}}{3}\times {473.15}^3({\tau }^3-1)\\ \\ 9622\text{K}\text{=673}\text{.76}(\tau -1)+1611.199({\tau }^2-1)-155.07({\tau }^3-1)\\ \\ \Rightarrow \tau =2.905\end{array}$

Hence final temperature is:

  $\begin{array}{l}\tau =\frac{T}{T_0}\\ \\ T=2.905\times 473.15\text{K}\\ \\ T=1374.5\text{K}\end{array}$

Now, the entropy change is:

  $\begin{array}{l}{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}}=\left[A+\left\{B{T}_0+\left(C{T}_0{}^2+\frac{D}{{\tau }^2{T}_0{}^2}\right)\left(\frac{\tau +1}{2}\right)\right\}\left(\frac{\tau -1}{\ln \tau }\right)\right]\times \ln \tau \\ {\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}}\\ =\left[1.424+\left\{14.394\times {10}^{-3}\times 473.15+\left(-4.392\times {10}^{-6}\times {473.15}^2\right)\left(\frac{2.905+1}{2}\right)\right\}\left(\frac{2.905-1}{\ln 2.905}\right)\right]\times \ln 2.905\\ \\ {\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}}=10.8355\end{array}$

Therefore entropy change is

  $\begin{array}{l}\Delta S=nR{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{C_P}{R}\frac{dT}{T}}\\ \\ \Delta S=10\text{mol}\times \text{8}\text{.314}\frac{\text{J}}{\text{mol K}}\times 10.8355\\ \\ \Delta S=900.86\frac{\text{J}}{\text{K}}\end{array}$

(b)

Interpretation Introduction

Interpretation:

Entropy change of the gas for given steady state flow process at atmospheric pressure should be determined.

Answer to Problem 5.28P

  $\Delta S=2657.75\frac{\text{J}}{\text{K}}$

Explanation of Solution

Given information:

When $2500\text{kJ}$ is added to $15\text{mol}$ of 1-butene initially at ${260}^{\circ }\text{C}$

  $T_0={260}^{\circ }\text{C+273}\text{.15=533}\text{.15}\text{K}$

And entropy of process at constant pressure is calculated by following formula,

  $\begin{array}{l}\Delta S=nR{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{C_P}{R}\frac{dT}{T}}\\ \text{where,}\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{C_P}{R}\frac{dT}{T}}=\left[A+\left\{B{T}_0+\left(C{T}_0{}^2+\frac{D}{{\tau }^2{T}_0{}^2}\right)\left(\frac{\tau +1}{2}\right)\right\}\left(\frac{\tau -1}{\ln \tau }\right)\right]\times \ln \tau \end{array}$

where

  $\tau =\frac{T}{T_0}$

And values of constants for1-butene in above equation are given in appendix C table C.1 and noted down below:

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ 1-Butene1.96731.630-9.8730\end{array}$

For final temperature,

  $\begin{array}{l}Q=\Delta H={\displaystyle \underset{T_1}{\overset{T_2}{\int }}{C}_P{}^{ig}dT}\\ \Rightarrow Q=nR{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}dT}\end{array}$

Where

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

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

Put the values

  $\begin{array}{l}\frac{Q}{nR}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ \\ \frac{2500\text{kJ}\times \frac{1000\text{J}}{1\text{kJ}}}{15\text{mol}\times \text{8}\text{.314}\frac{\text{J}}{\text{mol K}}}=1.967\times 533.15(\tau -1)+\frac{31.63\times {10}^{-3}}{2}\times {533.15}^2({\tau }^2-1)+\frac{-9.873\times {10}^{-6}}{3}\times {533.15}^3({\tau }^3-1)\\ \\ \Rightarrow \tau =2.652\end{array}$

Hence, final temperature is

  $\begin{array}{l}\tau =\frac{T}{T_0}\\ \\ T=2.652\times 533.15\text{K}\\ \\ T=1413.9\text{K}\end{array}$

Now, the entropy change is:

  $\begin{array}{l}{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}}=\left[A+\left\{B{T}_0+\left(C{T}_0{}^2+\frac{D}{{\tau }^2{T}_0{}^2}\right)\left(\frac{\tau +1}{2}\right)\right\}\left(\frac{\tau -1}{\ln \tau }\right)\right]\times \ln \tau \\ {\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}}\\ =\left[1.967+\left\{31.63\times {10}^{-3}\times 533.15+\left(-9.873\times {10}^{-6}\times {533.15}^2\right)\left(\frac{2.652+1}{2}\right)\right\}\left(\frac{2.652-1}{\ln 2.652}\right)\right]\times \ln 2.652\\ \\ {\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}}=21.3114\end{array}$

Therefore entropy change is

  $\begin{array}{l}\Delta S=nR{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{C_P}{R}\frac{dT}{T}}\\ \\ \Delta S=15\text{mol}\times \text{8}\text{.314}\frac{\text{J}}{\text{mol K}}\times 21.3114\\ \\ \Delta S=2657.75\frac{\text{J}}{\text{K}}\end{array}$

(c)

Interpretation Introduction

Interpretation:

Entropy change of the gas for given steady state flow process at atmospheric pressure should be determined.

Answer to Problem 5.28P

  $\Delta S=1238441.845\frac{\text{J}}{\text{K}}=1.24\times {10}^6\frac{\text{J}}{\text{K}}$

Explanation of Solution

Given information:

When ${10}^6\left(\text{btu}\right)$ is added to $40\text{lb-mol}$ of ethylene initially at ${500}^{\circ }\text{F}$

  $\begin{array}{l}{T}_0=\left({500}^{\circ }\text{F-32}\right)\left(\frac{5}{9}\right)\text{+273}\text{.15=533}\text{.15}\text{K}\\ \text{and}\\ \\ Q={10}^6\left(\text{btu}\right)=1.05\times {10}^6\text{kJ}\\ n=40\text{lb-mol=18}\text{.14}\text{kmol}\end{array}$

And entropy of process at constant pressure is calculated by following formula,

  $\begin{array}{l}\Delta S=nR{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{C_P}{R}\frac{dT}{T}}\\ \text{where,}\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{C_P}{R}\frac{dT}{T}}=\left[A+\left\{B{T}_0+\left(C{T}_0{}^2+\frac{D}{{\tau }^2{T}_0{}^2}\right)\left(\frac{\tau +1}{2}\right)\right\}\left(\frac{\tau -1}{\ln \tau }\right)\right]\times \ln \tau \end{array}$

where

  $\tau =\frac{T}{T_0}$

And value of constants for ethylene in above equation are given in appendix C table C.1 and noted down below:

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ Ethylene1.42414.394-4.3920\end{array}$

For final temperature,

  $\begin{array}{l}Q=\Delta H={\displaystyle \underset{T_1}{\overset{T_2}{\int }}{C}_P{}^{ig}dT}\\ \Rightarrow Q=nR{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}dT}\end{array}$

Where

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

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

Put the values

  $\begin{array}{l}\frac{Q}{nR}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ \\ \frac{1.05\times {10}^6\text{kJ}}{18.14\text{kmol}\times \text{8}\text{.314}\frac{\text{J}}{\text{mol K}}}=\\ 1.424\times 533.15(\tau -1)+\frac{14.394\times {10}^{-3}}{2}\times {533.15}^2({\tau }^2-1)+\frac{-4.392\times {10}^{-6}}{3}\times {533.15}^3({\tau }^3-1)\\ \\ \Rightarrow \tau =2.25\end{array}$

Hence, final temperature is

  $\begin{array}{l}\tau =\frac{T}{T_0}\\ \\ T=2.25\times 533.15\text{K}\\ \\ T=1199.587\text{K}\end{array}$

Now, the entropy change is:

  $\begin{array}{l}{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}}=\left[A+\left\{B{T}_0+\left(C{T}_0{}^2+\frac{D}{{\tau }^2{T}_0{}^2}\right)\left(\frac{\tau +1}{2}\right)\right\}\left(\frac{\tau -1}{\ln \tau }\right)\right]\times \ln \tau \\ {\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}}\\ =\left[1.424+\left\{14.394\times {10}^{-3}\times 533.15+\left(-4.392\times {10}^{-6}\times {533.15}^2\right)\left(\frac{2.25+1}{2}\right)\right\}\left(\frac{2.25-1}{\ln 2.25}\right)\right]\times \ln 2.25\\ \\ {\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}}=8.21161\end{array}$

Therefore entropy change is

  $\begin{array}{l}\Delta S=nR{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{C_P}{R}\frac{dT}{T}}\\ \\ \Delta S=18.14\text{kmol}\times \text{8}\text{.314}\frac{\text{J}}{\text{mol K}}\times 8.21161\times \frac{1000\text{mol}}{1\text{kmol}}\\ \\ \Delta S=1238441.845\frac{\text{J}}{\text{K}}=1.24\times {10}^6\frac{\text{J}}{\text{K}}\end{array}$