Chapter 8.8, Problem 106RP

(a)

To determine

The final temperature in the cylinder at equilibrium condition.

Answer to Problem 106RP

The final temperature in the cylinder at equilibrium condition is $57.2°\text{C}$.

Explanation of Solution

Write the ideal gas equation to calculate the mass of the gas $\left(m\right)$.

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

Here, initial pressure of the gas is $P_1$, initial volume of the gas is $V_1$, gas constant is $R$, and the initial temperature is $T_1$.

Write the energy balance equation for the entire system considering it as a stationary closed system.

$\begin{array}{l}{E}_{\text{in}}-E_{\text{out}}=\underset{=0}{\underbrace{\Delta E_{\text{system}}}}\\ 0=\Delta U\end{array}$

$0={\left[m{c}_v\left(T_2-T_1\right)\right]}_{{\text{N}}_2}+{\left[m{c}_v\left(T_2-T_1\right)\right]}_{\text{He}}$ (II)

Here, net energy input to the system is $E_{\text{in}}$, net energy output to the system is $E_{\text{out}}$, the change in net energy is $\Delta E_{\text{system}}$, change in internal energy is $\Delta U$, constant volume specific heat is $c_v$, final temperature is $T_2$, and the initial temperature is $T_1$.

Conclusion:

Refer the Table A-1E of “Molar mass, gas constant, and critical-point properties”, obtain the gas constants of Nitrogen and Helium as

$\begin{array}{l}{R}_{{\text{N}}_2}=0.2968\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}\\ R_{\text{He}}=2.0769\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}\end{array}$

Refer the Table A-2E of “Ideal-gas specific heats of various common gases”, obtain the specific heats of Nitrogen and Helium as

$\begin{array}{l}{c}_{p,{\text{N}}_2}=1.039\text{kJ}/\text{kg}°\text{C}\\ c_{v,{\text{N}}_2}=0.743\text{kJ}/\text{kg}°\text{C}\\ c_{p,\text{He}}=5.1926\text{kJ}/\text{kg}°\text{C}\\ c_{v,\text{He}}=3.1156\text{kJ}/\text{kg}°\text{C}\end{array}$

Substitute $500\text{kPa}$ for $P_1$, $1{\text{m}}^3$ for $V_1$, $0.2968\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}$ for $R_{{\text{N}}_2}$, and $353\text{K}$ for $T_{1,{\text{N}}_2}$ in Equation (I).

$\begin{array}{c}{m}_{{\text{N}}_2}=\frac{500\text{kPa}\times 1{\text{m}}^3}{0.2968\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}\times 353\text{K}}\\ =4.772\text{kg}\end{array}$

Substitute $500\text{kPa}$ for $P_1$, $1{\text{m}}^3$ for $V_1$, $2.0769\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}$ for $R_{\text{He}}$, and $298\text{K}$ for $T_{1,\text{He}}$ in Equation (I).

$\begin{array}{c}{m}_{\text{He}}=\frac{500\text{kPa}\times 1{\text{m}}^3}{2.0769\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}\times 298\text{K}}\\ =0.8079\text{kg}\end{array}$

Substitute $4.772\text{kg}$ for $m_{{\text{N}}_2}$, $0.8079\text{kg}$ for $m_{\text{He}}$, $0.743\text{kJ}/\text{kg}°\text{C}$ for $c_{v,{\text{N}}_2}$, $3.1156\text{kJ}/\text{kg}°\text{C}$ for $c_{v,\text{He}}$, $80°\text{C}$ for $T_{1,{\text{N}}_2}$, and $25°\text{C}$ for $T_{1,\text{He}}$ in Equation (II).

$\begin{array}{l}0=\left\{\begin{array}{l}\left[4.772\text{kg}\times 0.743\text{kJ}/\text{kg}°\text{C}\left(T_2-80°\text{C}\right)\right]\\ +\left[0.8079\text{kg}\times 3.1156\text{kJ}/\text{kg}°\text{C}\left(T_2-25°\text{C}\right)\right]\end{array}\right\}\\ T_2=57.2°\text{C}\end{array}$

Thus, the final temperature in the cylinder at equilibrium condition is $57.2°\text{C}$.

Final temperature at equilibrium condition is same even if the piston is restricted from moving.

(b)

To determine

The amount of wasted work potential for the process.

The amount of wasted work potential for the process when piston is restricted from moving.

Answer to Problem 106RP

The amount of wasted work potential for the process is $9.03\text{kJ}$.

The amount of wasted work potential for the process when piston is restricted from moving is $6.23\text{kJ}$.

Explanation of Solution

Write the expression to calculate the total number of moles in the cylinder $\left(N_{\text{total}}\right)$.

$N_{\text{total}}=N_{{\text{N}}_2}+N_{\text{He}}$

$N_{\text{total}}={\left(\frac{m}{M}\right)}_{{\text{N}}_2}+{\left(\frac{m}{M}\right)}_{\text{He}}$ (III)

Write the expression to calculate the pressure from ideal gas expression.

$P_2=\frac{N_{\text{total}}{R}_{u}T}{V_{\text{total}}}$ (IV)

Here, universal gas constant is $R_u$ and the total volume of the cylinder is $V_{\text{total}}$.

Write the entropy generation $\left(S_{\text{gen}}\right)$ equation for the process from entropy balance.

$\begin{array}{l}{S}_{\text{in}}-S_{\text{out}}+S_{\text{gen}}=\Delta S_{\text{system}}\\ {\dot{S}}_{\text{gen}}=\Delta S_{{\text{N}}_2}+\Delta S_{\text{He}}\end{array}$

${\dot{S}}_{\text{gen}}={\left[m\left(c_p\ln \frac{T_2}{T_1}-R\ln \frac{P_2}{P_1}\right)\right]}_{{\text{N}}_2}+{\left[m\left(c_p\ln \frac{T_2}{T_1}-R\ln \frac{P_2}{P_1}\right)\right]}_{\text{He}}$ (V)

Here, entropy input to the system is $S_{\text{in}}$, entropy exiting out is $S_{\text{out}}$, change in the entropy system is $\Delta S_{\text{system}}$,

Write the expression to calculate the exergy destroyed $\left({\dot{X}}_{\text{dest}}\right)$.

${\dot{X}}_{\text{dest}}=T_0{\dot{S}}_{\text{gen}}$ (VI)

Here, the surrounding’s temperature is $T_0$.

Write the formula to calculate the entropy generation when the piston is restricted to move.

${\dot{S}}_{\text{gen}}={\left[m\left(c_v\ln \frac{T_2}{T_1}\right)\right]}_{{\text{N}}_2}+{\left[m\left(c_v\ln \frac{T_2}{T_1}\right)\right]}_{\text{He}}$ (VII)

Conclusion:

Refer the Table A-1E of “Molar mass, gas constant, and critical-point properties”, obtain the molar masses of Nitrogen and Helium as

$\begin{array}{l}{M}_{{\text{N}}_2}=28\text{kg}/\text{kmol}\\ M_{\text{He}}=4\text{kg}/\text{kmol}\end{array}$

Substitute $4.772\text{kg}$ for $m_{{\text{N}}_2}$, $0.8079\text{kg}$ for $m_{\text{He}}$, $28\text{kg}/\text{kmol}$ for $M_{{\text{N}}_2}$, and $4\text{kg}/\text{kmol}$ for $M_{\text{He}}$ in Equation (III).

$\begin{array}{c}{N}_{\text{total}}=\left(\frac{4.772\text{kg}}{28\text{kg}/\text{kmol}}\right)+\left(\frac{0.8079\text{kg}}{4\text{kg}/\text{kmol}}\right)\\ =0.3724\text{kmol}\end{array}$

Substitute $0.3724\text{kmol}$ for $N_{\text{total}}$, $8.314\text{kPa}\cdot {\text{m}}^3/\text{kmol}\cdot \text{K}$ for $R_u$, $330.2\text{K}$ for $T_2$, and $2{\text{m}}^3$ for $V_{\text{total}}$ in Equation (IV).

$\begin{array}{c}{P}_2=\frac{0.3724\text{kmol}\times 8.314\text{kPa}\cdot {\text{m}}^3/\text{kmol}\cdot \text{K}\times 330.2\text{K}}{2{\text{m}}^3}\\ =511.1\text{kPa}\end{array}$

Substitute $4.772\text{kg}$ for $m_{{\text{N}}_2}$, $0.8079\text{kg}$ for $m_{\text{He}}$, $5.1926\text{kJ}/\text{kg}°\text{C}$ for $c_{p,\text{He}}$, $1.039\text{kJ}/\text{kg}°\text{C}$ for $c_{p,{\text{N}}_2}$, $0.2968\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}$ for $R_{{\text{N}}_2}$, $2.0769\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}$ for $R_{\text{He}}$, $330.2\text{K}$ for $T_2$, $353\text{K}$ for $T_{1,{\text{N}}_2}$, $298\text{K}$ for $T_{1,\text{He}}$, $511.1\text{kPa}$ for $P_2$, and $500\text{kPa}$ for $P_1$ in Equation (V).

$\begin{array}{c}{\dot{S}}_{\text{gen}}=\left\{\begin{array}{l}\left[4.772\text{kg}\left(\begin{array}{l}1.039\text{kJ}/\text{kg}°\text{C}\times \ln \left(\frac{330.2\text{K}}{353\text{K}}\right)\\ -0.2968\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}\times \ln \frac{511.1\text{kPa}}{500\text{kPa}}\end{array}\right)\right]\\ +\left[0.8079\text{kg}\left(\begin{array}{l}5.1926\text{kJ}/\text{kg}°\text{C}\times \ln \left(\frac{330.2\text{K}}{353\text{K}}\right)\\ -2.0769\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}\times \ln \frac{511.1\text{kPa}}{500\text{kPa}}\end{array}\right)\right]\end{array}\right\}\\ =0.0303\text{kJ}/\text{K}\end{array}$

Substitute $298\text{K}$ for $T_0$ and $0.0303\text{kJ}/\text{K}$ for ${\dot{S}}_{\text{gen}}$ in Equation (VI).

$\begin{array}{c}{\dot{X}}_{\text{dest}}=\left(298\text{K}\right)\times 0.0303\text{kJ}/\text{K}\\ =9.03\text{kJ}\end{array}$

Thus, the amount of wasted work potential for the process is $9.03\text{kJ}$.

Substitute $4.772\text{kg}$ for $m_{{\text{N}}_2}$, $0.8079\text{kg}$ for $m_{\text{He}}$, $330.2\text{K}$ for $T_2$, $353\text{K}$ for $T_{1,{\text{N}}_2}$, $298\text{K}$ for $T_{1,\text{He}}$, $0.743\text{kJ}/\text{kg}°\text{C}$ for $c_{v,{\text{N}}_2}$, and $3.1156\text{kJ}/\text{kg}°\text{C}$ for $c_{v,\text{He}}$ in Equation (VII).

$\begin{array}{c}{\dot{S}}_{\text{gen}}=\left\{\begin{array}{l}\left[4.772\text{kg}\left(0.743\text{kJ}/\text{kg}°\text{C}\times \ln \frac{330.2\text{K}}{353\text{K}}\right)\right]\\ +\left[0.8079\text{kg}\left(3.1156\text{kJ}/\text{kg}°\text{C}\times \ln \frac{330.2\text{K}}{353\text{K}}\right)\right]\end{array}\right\}\\ =0.02089\text{kJ}/\text{K}\end{array}$

Substitute $298\text{K}$ for $T_0$ and $0.02089\text{kJ}/\text{K}$ for ${\dot{S}}_{\text{gen}}$ in Equation (VI).

$\begin{array}{c}{\dot{X}}_{\text{dest}}=\left(298\text{K}\right)\times 0.02089\text{kJ}/\text{K}\\ =6.23\text{kJ}\end{array}$

Thus, the amount of wasted work potential for the process when piston is restricted from moving is $6.23\text{kJ}$.