Chapter 7.13, Problem 177RP

A piston–cylinder device initially contains 15 ft³ of helium gas at 25 psia and 70°F. Helium is now compressed in a polytropic process (PVⁿ = constant) to 70 psia and 300°F. Determine (a) the entropy change of helium, (b) the entropy change of the surroundings, and (c) whether this process is reversible, irreversible, or impossible. Assume the surroundings are at 70°F.

To determine

The change in entropy of helium.

Answer to Problem 177RP

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

Explanation of Solution

Write the expression for the ideal gas equation to calculate the mass of helium.

$m=\frac{P_1{\nu }_1}{R{T}_1}$ (I)

Here, mass of helium is m, initial pressure is $P_1$, initial volume is ${\nu }_1$, initial temperature is $T_1$ and gas constant is R.

Write the expression to calculate the change in entropy of helium.

$\Delta S_{helium}=m\left(c_p\ln \frac{T_2}{T_1}-R\ln \frac{P_2}{P_1}\right)$ (II)

Here, mass of helium is m, specific heat at constant pressure is $c_p$, initial temperature is $T_1$, final temperature is $T_2$, gas constant is R, initial pressure is $P_1$, and final pressure is $P_2$.

Conclusion:

From Table A-1E, “the molar mass, gas constant and critical–point properties table”, select the gas constant of helium as $2.6805\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}$.

Substitute $\text{25}\text{psia}$ for $P_1$, $15{\text{ft}}^3$ for ${\nu }_1$, $2.6809\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}$ for R and $70°\text{F}$ for $T_1$ in Equation (I).

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

From Table A-2E, “Ideal-gas specific heats of various common gases”, select the specific heat at constant pressure $\left(c_p\right)$ and specific heat at constant volume $\left(c_v\right)$ of helium as $1.25\text{Btu}/\text{lbm}\cdot \text{R}$ and $0.753\text{Btu}/\text{lbm}\cdot \text{R}$ respectively.

Substitute $\text{0}\text{.264}\text{lbm}$ for m, $1.25\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_p$, $\text{300}°\text{F}$ for $T_2$, and $\text{70}°\text{F}$ for $T_1$, $2.6805\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}$ for R, $\text{25}\text{psia}$ for $P_1$, and $\text{70}\text{psia}$ for $P_2$ in Equation (II).

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

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

To determine

The entropy change of the surrounding.

Answer to Problem 177RP

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

Explanation of Solution

Write the expression to calculate the ideal gas equation for initial and final condition.

$\frac{P_1{\nu }_1}{T_1}=\frac{P_2{\nu }_2}{T_2}$ (III)

Here, final volume is ${\nu }_2$.

Write the expression for the polytropic process.

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

Rewrite the Equation (IV) to calculate exponent n.

$\begin{array}{l}\left(\frac{P_2}{P_1}\right)={\left(\frac{{\nu }_1}{{\nu }_2}\right)}^n\\ n=\frac{\ln \left(\frac{P_2}{P_1}\right)}{\ln \left(\frac{{\nu }_1}{{\nu }_2}\right)}\end{array}$ (V)

Write the expression to calculate the boundary work for the polytropic process $W_{b,in}$.

$W_{b,in}=-\frac{mR\left(T_2-T_1\right)}{1-n}$ (VI)

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

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

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 $\Delta E_{system}$.

Write the expression to calculate the entropy change of the surrounding.

$\Delta S_{surr}=\frac{Q_{out}}{T_{surr}}$ (VIII)

Here, surroundin temperature is $T_{surr}$ and heat transfer output is $Q_{out}$.

Conclusion:

Substitute $\text{300}°\text{F}$ for $T_2$, and $\text{70}°\text{F}$ for $T_1$, $\text{25}\text{psia}$ for $P_1$, and $\text{70}\text{psia}$ for $P_2$ and $15{\text{ft}}^3$ for ${\nu }_1$ in Equation (III).

$\frac{\left(\text{25}\text{psia}\right)\left(15{\text{ft}}^3\right)}{\text{70}°\text{F}}=\frac{\left(\text{70}\text{psia}\right){\nu }_2}{\text{300}°\text{F}}$

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

Substitute $\text{25}\text{psia}$ for $P_1$, and $\text{70}\text{psia}$ for $P_2$ , $15{\text{ft}}^3$ for ${\nu }_1$ and $7.682{\text{ft}}^3$ for ${\nu }_2$ in Equation (V).

$\begin{array}{c}n=\frac{\ln \left(\frac{70\text{psia}}{25\text{psia}}\right)}{\ln \left(\frac{15{\text{ft}}^3}{7.682{\text{ft}}^3}\right)}\\ =1.539\end{array}$

Substitute $\text{0}\text{.264}\text{lbm}$ for $m$ , $2.6805\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}$ for R, $\text{300}°\text{F}$ for $T_2$, and $\text{70}°\text{F}$ for $T_1$, and $1.539$ for n in Equation (VI).

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

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 (VII).

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

Here, heat transfer output is $Q_{out}$ , specific heat at constant volume is $c_v$, initial internal energy is $u_1$ and final internal energy is $u_2$.

Substitute $\text{55}\text{.9}\text{Btu}$ for $W_{b,in}$, $\text{0}\text{.264}\text{lbm}$ for $m$ , $0.753\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_v$, $\text{300}°\text{F}$ for $T_2$, and $\text{70}°\text{F}$ for $T_1$ in Equation (IX).

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

Here, the entropy change of the surrounding is $\Delta S_{surr}$ and surrounding temperature is $T_{surr}$.

Substitute $\text{10}\text{.2}\text{Btu}$ for $Q_{out}$ and $\text{70}°\text{F}$ for $T_{surr}$ in Equation (VIII).

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

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

To determine

Whether this process is reversible, irreversible, or impossible.

Answer to Problem 177RP

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

The system is irreversible.

Explanation of Solution

Write the expression to calculate the total entropy change during the process.

$\Delta S_{total}=\Delta S_{system}+\Delta S_{surr}$ (X)

Here, the total entropy change in the system is $\Delta S_{total}$.

Conclusion:

Substitute $-0.016\text{Btu}/\text{R}$ for $\Delta S_{helium}$ and $0.019\text{Btu}/\text{R}$ for $\Delta S_{surr}$ in Equation (X).

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

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

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

Thus, the system is irreversible.