Chapter 8, Problem 163RQ

a)

To determine

The change in entropy of helium.

Explanation of Solution

Given:

The volume of the helium gas $\left(v\right)$ is $15{\text{ft}}^3$.

The initial pressure of the helium gas $\left(P_1\right)$ is $25\text{psia}$.

The initial temperature of the helium gas $\left(T_1\right)$ is $70°\text{F}$.

The final pressure of the helium gas $\left(P_2\right)$ is $70\text{psia}$.

The initial temperature of the helium gas $\left(T_2\right)$ is $300°\text{F}$

The surrounding temperature $\left(T_{surr}\right)$ is $70°\text{F}$.

Calculation:

Refer Table A-1E, “the molar mass, gas constant and critical–point properties table”,

The gas constant of helium $\left(R\right)$ as $2.6805\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}$.

Calculate the mass of helium $\left(m\right)$ using the ideal gas equation.

  $\begin{array}{c}m=\frac{P_1{\nu }_1}{R{T}_1}\\ m=\frac{\left(25\text{psia}\right)\left(15{\text{ft}}^3\right)}{\left(2.6809\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}\right)70°\text{F}}\\ =\frac{\left(25\text{psia}\right)\left(15{\text{ft}}^3\right)}{\left(2.6809\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}\right)\left(70+460\right)\text{R}}\\ =0.264\text{lbm}\end{array}$

Refer Table A-2E, “Ideal-gas specific heats of various common gases”,

The specific heat at constant pressure $\left(c_p\right)$ of helium is $1.25\text{Btu}/\text{lbm}\cdot \text{R}$.

The specific heat at constant volume $\left(c_v\right)$ of helium is $0.753\text{Btu}/\text{lbm}\cdot \text{R}$.

Calculate the change in entropy of helium.

  $\begin{array}{c}\Delta S_{helium}=m\left(c_p\ln \frac{T_2}{T_1}-R\ln \frac{P_2}{P_1}\right)\\ \Delta S_{helium}=0.264\text{lbm}\left(\begin{array}{l}\left(1.25\text{Btu}/\text{lbm}\cdot \text{R}\right)\ln \frac{\text{300}°\text{F}}{\text{70}°\text{F}}\\ \left(-2.6805\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}\right)\ln \frac{70\text{psia}}{25\text{psia}}\end{array}\right)\\ =0.264\text{lbm}\left(\begin{array}{l}\left(1.25\text{Btu}/\text{lbm}\cdot \text{R}\right)\ln \frac{\left(300+460\right)\text{R}}{\left(70+460\right)\text{R}}\\ \left(-2.6805\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}\right)\left(\frac{1\text{Btu}}{5.40395\text{psia}\cdot {\text{ft}}^{\text{3}}}\right)\ln \frac{70\text{psia}}{25\text{psia}}\end{array}\right)\\ =0.264\left(0.4505-0.5108\right)\\ =-0.016\text{Btu}/\text{R}\end{array}$

Thus, the change in entropy of helium is $-0.016\text{Btu}/\text{R}$.

b)

To determine

The entropy change of the surrounding.

Explanation of Solution

Calculate the final volume of the helium $\left({\nu }_2\right)$ using ideal gas relation.

  $\begin{array}{l}\frac{P_1{\nu }_1}{T_1}=\frac{P_2{\nu }_2}{T_2}\\ \frac{\left(\text{25}\text{psia}\right)\left(15{\text{ft}}^3\right)}{\text{70}°\text{F}}=\frac{\left(\text{70}\text{psia}\right){\nu }_2}{\text{300}°\text{F}}\end{array}$

  $\begin{array}{c}{\nu }_2=\frac{300°\text{F}\left(25\text{psia}\right)}{70°\text{F}\left(70\text{psia}\right)}\left(15{\text{ft}}^3\right)\\ =\frac{\left(300+460\right)\text{R}\left(25\text{psia}\right)}{\left(70+460\right)\text{R}\left(70\text{psia}\right)}\left(15{\text{ft}}^3\right)\\ =7.682{\text{ft}}^3\end{array}$

Write the expression for the polytropic process.

  $P_2{\nu }_2^n=P_1{\nu }_1^n$        (IV)

Rewrite the above Equation to obtain the exponent n.

  $\begin{array}{l}\left(\frac{P_2}{P_1}\right)={\left(\frac{{\nu }_1}{{\nu }_2}\right)}^n\\ n=\frac{\ln \left(\frac{P_2}{P_1}\right)}{\ln \left(\frac{{\nu }_1}{{\nu }_2}\right)}\\ n=\frac{\ln \left(\frac{70\text{psia}}{25\text{psia}}\right)}{\ln \left(\frac{15{\text{ft}}^3}{7.682{\text{ft}}^3}\right)}\\ n=1.539\end{array}$

Calculate the boundary work for the polytropic process $\left(W_{b,in}\right)$.

  $\begin{array}{c}{W}_{b,in}=-\frac{mR\left(T_2-T_1\right)}{1-n}\\ W_{b,in}=-\frac{\left\{\begin{array}{l}\left(0.264\text{lbm}\right)\left(2.6805\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}\right)\\ \left(300°\text{F}-70°\text{F}\right)\end{array}\right\}}{1-1.539}\\ =-\frac{\left\{\begin{array}{l}\left(0.264\text{lbm}\right)\left(2.6805\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}\right)\\ \left(\frac{1\text{Btu}}{5.40395\text{psia}\cdot {\text{ft}}^{\text{3}}}\right)\left[\left(300+460\right)\text{R}-\left(70+460\right)\text{R}\right]\end{array}\right\}}{1-1.539}\\ =55.9\text{Btu}\end{array}$

Write the expression for the energy balance equation of the system.

  $E_{in}-E_{out}=\Delta E_{system}$        (I)

Here, net energy transfer inside the system is $E_{in}$, net energy transfer from outside to system is $E_{out}$, and change of internal energy in the system is $\Delta E_{system}$.

Substitute $W_{b,in}$ for $E_{in}$, $Q_{out}$ for $E_{out}$ and $m\left(u_2-u_1\right)$ for $\Delta E_{system}$ in Equation (I) to obtain the heat transfer output is $\left(Q_{out}\right)$.

  $\begin{array}{l}{W}_{b,in}-Q_{out}=m\left(u_2-u_1\right)\\ -Q_{out}=m\left(u_2-u_1\right)-W_{b,in}\\ Q_{out}=W_{b,in}-m{c}_v\left(T_2-T_1\right)\end{array}$

  $\begin{array}{c}{Q}_{out}=55.9\text{Btu}-\left(0.264\text{lbm}\right)\left(0.753\text{Btu}/\text{lbm}\cdot \text{R}\right)\left(300°\text{F}-70°\text{F}\right)\\ =55.9\text{Btu}-\left(0.264\text{lbm}\right)\left(0.753\text{Btu}/\text{lbm}\cdot \text{R}\right)\left[\left(300+460\right)\text{R}-\left(70+460\right)\text{R}\right]\\ =10.2\text{Btu}\end{array}$

Calculate the entropy change of the surrounding.

  $\begin{array}{c}\Delta S_{surr}=\frac{Q_{out}}{T_{surr}}\\ \Delta S_{surr}=\frac{10.2\text{Btu}}{70°\text{C}}\\ =\frac{10.2\text{Btu}}{\left(70+460\right)\text{R}}\\ =0.019\text{Btu}/\text{R}\end{array}$

Thus, the entropy change of the surrounding is $0.019\text{Btu}/\text{R}$.

c)

To determine

Whether this process is reversible, irreversible, or impossible.

Explanation of Solution

Calculate the total entropy change during the process.

  $\begin{array}{c}\Delta S_{total}=\Delta S_{system}+\Delta S_{surr}\\ \Delta S_{total}=-0.016\text{Btu}/\text{R}+0.019\text{Btu}/\text{R}\\ =0.003\text{Btu}/\text{R}>0\end{array}$

Thus, the total entropy change during the process is $0.003\text{Btu}/\text{R}$.

The value of the total change in entropy $\left(\Delta S_{total}\right)$ is positive so the system is irreversible.