Chapter 7.13, Problem 175RP

a)

To determine

The final temperature of the helium

Answer to Problem 175RP

The final temperature of the helium is $\underset{\_}{321.7\text{ K}}$.

Explanation of Solution

Write the relation between the pressure and temperature in the isentropic process:

$\frac{T_2}{T_1}={\left(\frac{P_2}{P_1}\right)}^{\frac{k-1}{k}}$

$T_2=T_1{\left(\frac{P_2}{P_1}\right)}^{\frac{k-1}{k}}$ (I)

Here, temperature at state 2 is $T_2$, temperature at state 1 is $T_1$, pressure at state 1 is $P_1$ , pressure at state 2 is $P_2$, ratio of specific heat is $k$ .

Conclusion:

From the Table A-2, “Ideal-gas specific heats of various common gases table”, select the specific heat ratio $\left(k\right)$ for Helium gas as $1.667$.

Substitute $20°\text{C}$ for $T_1$, $120\text{ kPa}$ for $P_2$, $95\text{ kPa}$ for $P_1$, and $1.667$ for $k$ in equation (I).

$\begin{array}{c}{T}_1=20°{\left(\frac{120\text{ kPa}}{95\text{ kPa}}\right)}^{\frac{1.667-1}{1.667}}\\ =\left(20°+273\right)\text{ K}{\left(\frac{120\text{ kPa}}{95\text{ kPa}}\right)}^{\frac{1.667-1}{1.667}}\\ =321.7\text{ K}\end{array}$

b)

To determine

The final volume of nitrogen

Answer to Problem 175RP

The final volume of nitrogen is $\underset{\_}{0.2838{\text{m}}^3}$.

Explanation of Solution

Write the expression for calculating the initial volume of helium $V_{He,1}$.

$V_{He,1}=\frac{mR{T}_1}{P_1}$ (II)

Here, mass of the helium is $m$ and gas constant is $R$.

Write the expression for calculating the initial volume of helium $V_{He,1}$.

$V_{He,2}=\frac{mR{T}_2}{P_2}$ (III)

Here, mass of the helium is $m$ and gas constant is $R$.

Write the expression for the final volume of the nitrogen $V_{N_2}{,}_2$:

$V_{N_2}{,}_2=V_{N2,1}+V_{He,1}+V_{He,2}$ (IV)

Here, volume of the nitrogen at initial state 1 is $V_{N_2,1}$.

Conclusion:

From the Table A-2, “Ideal-gas specific heats of various common gases table”, select the gas constant for the helium gas as $\text{2}\text{.0769 }\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}$.

Substitute $0.1\text{ kg}$ for $m$, $\text{2}\text{.0769 }\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}$ for $R$, $20°\text{C}$ for $T_1$, and $95\text{ kPa}$ for $P_1$ in equation (II).

$\begin{array}{c}{V}_{He,1}=\frac{0.1\text{ kg}\times \text{2}\text{.0769 }\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\times \left(20°\text{C}\right)}{95\text{kPa}}\\ =\frac{0.1\text{ kg}\times \text{2}\text{.0769 }\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\times \left(20+273\right)}{95\text{kPa}}\\ =0.6406{\text{ m}}^3\end{array}$

Substitute $0.1\text{ kg}$ for $m$, $\text{2}\text{.0769 }\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}$ for $R$, $321.7\text{K}$ for $T_2$, and $120\text{ kPa}$ for $P_2$ in equation (III)

$\begin{array}{c}{V}_{He,1}=\frac{0.1\text{ kg}\times \text{2}\text{.0769 }\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\times \left(321.7\text{ K}\right)}{120\text{kPa}}\\ =0.5568{\text{ m}}^3\end{array}$

Substitute $0.2{\text{ m}}^3$ for $V_{N2,1}$, $0.6406{\text{ m}}^3$ for $V_{He,1}$, and $0.5568{\text{m}}^{\text{3}}$ for $V_{He,2}$ in equation (IV).

$\begin{array}{c}{V}_{N_2}{,}_2=0.2{\text{ m}}^3+0.6406{\text{ m}}^3+0.5568{\text{m}}^{\text{3}}\\ =0.2838{\text{ m}}^3\end{array}$

Thus, the final volume of nitrogen is $\underset{\_}{0.2838{\text{m}}^3}$.

c)

To determine

The heat transferred to the nitrogen.

Answer to Problem 175RP

The heat transferred to the nitrogen is $\underset{\_}{46.042\text{ kJ}}$.

Explanation of Solution

Write the expression for calculating the mass of the nitrogen $m_{N_2}$.

$m_{N2}=\frac{P_1{V}_1}{R{T}_1}$ (V)

Write the expression for calculating the temperature of the nitrogen $T_{N_2,2}$.

$T_{N_2,2}=\frac{P_2{V}_2}{mR}$ (VI)

Write the formula for the change in internal energy of the nitrogen.

$\Delta U_{N_2}=m{c}_V{}_{N_2}\left(T_2-T_1\right)$ (VII)

Here, specific heat capacity of nitrogen at constant volume is $c_V{}_{N_2}$.

Write the formula for the change in internal energy of the Helium.

$\Delta U_{He}=m{c}_V\left(T_2-T_1\right)$ (VIII)

Here, specific heat capacity of Helium at constant volume is $c_V{}_{{}_{He}}$.

Write the formula for the energy balance equation for the system:

$Q_{in}=\Delta U_{N2}+\Delta U_{He}$ (IX)

Conclusion:

From the Table A-2, “Ideal-gas specific heats of various common gases table”, select the gas constant for the helium gas  and specific heat constant at constant volume for the helium gas as $\text{2}\text{.0769 }\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}$ and $\text{3}\text{.1156}\frac{\text{kJ}}{\text{kg}\cdot \text{K}}$.

From the Table A-2, “Ideal-gas specific heats of various common gases table”, select the gas constant for the Nitrogen gas  and specific heat constant at constant volume for the helium gas as $\text{0}\text{.2968}\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}$ and $0.743\frac{\text{kJ}}{\text{kg}\cdot \text{K}}$.

Substitute $95\text{ kPa}$ for $P_1$, $0.2{\text{ m}}^3$ for $V_1$, $\text{2}\text{.0769 }\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}$ for $R$, and $20°\text{C}$ for $T_1$ in equation (V).

$\begin{array}{c}{V}_{He,1}=\frac{95\text{kPa}\times 0.2{\text{ m}}^3}{\text{2}\text{.0769 }\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\times \left(20°\text{C}\right)}\\ =\frac{95\text{kPa}\times 0.2{\text{ m}}^3}{\text{2}\text{.0769 }\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\times \left(20+273\right)\text{K}}\\ =0.2815\text{ kg}\end{array}$

Substitute $\text{120 kPa}$ for $P_2$, $0.2838{\text{m}}^{\text{3}}$ for $V_2$, 0.1 kg for $m$, and $\text{2}\text{.0769 }\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}$ for $R$ in equation (VI).

$\begin{array}{c}{T}_{N_2,2}=\frac{120\text{kPa}\times 0.2838{\text{ m}}^3}{0.2185\text{kg}\times \text{0}\text{.2968 }\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}}\\ =\frac{120\text{kPa}\times 0.2838{\text{ m}}^3}{0.2185\text{kg}\times \text{0}\text{.2968}\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}}\\ =525.1\text{ K}\end{array}$

Substitute $0.2185\text{ kg}$ for $m$, $\text{0}\text{.743}\frac{\text{kJ}}{\text{kg}\cdot \text{K}}$ for $c_V$, $521.5\text{ K}$ for $T_2$, and $20°\text{C}$ for $T_1$.

$\begin{array}{c}\Delta U_{N_2}=0.2185\text{ kg}\times \text{0}\text{.743}\frac{\text{kJ}}{\text{kg}\cdot \text{K}}\left(521.5\text{ K}-20°\text{C}\right)\\ =0.2185\text{ kg}\times \text{0}\text{.743}\frac{\text{kJ}}{\text{kg×K}}\left(521.5\text{ K}-\left(20+273\right)\text{ K}\right)\\ =37.1\text{ kJ}\end{array}$

Substitute $0.2185\text{ kg}$ for $m$, $\text{3}\text{.1156}\frac{\text{kJ}}{\text{kg}\cdot \text{K}}$ for $c_V$, $321.7\text{ K}$ for $T_2$, and $20°\text{C}$ for $T_1$ in equation (VIII)

$\begin{array}{c}\Delta U_{He}=0.1\text{ kg}\times \text{3}\text{.1156}\frac{\text{kJ}}{\text{kg}\cdot \text{K}}\left(321.7\text{ K}-20°\text{C}\right)\\ =0.1\text{ kg}\times \text{3}\text{.1156}\frac{\text{kJ}}{\text{kg×K}}\left(321.7\text{ K}-\left(20+273\right)\text{ K}\right)\\ =8.942\text{ kJ}\end{array}$

Substitute $37.1\text{ kJ}$ for $\Delta U_{He}$ and $8.942\text{ kJ}$ for $\Delta U_{N_2}$ in equation (IX)

$\begin{array}{c}{Q}_{in}=37.1\text{ kJ}+8.942\text{ kJ}\\ =46.042\text{ kJ}\end{array}$

Thus, the heat transferred to the nitrogen is $\underset{\_}{46.042\text{ kJ}}$.

d)

To determine

The entropy generation during the process

Answer to Problem 175RP

The entropy generation during the process is $\underset{\_}{0.057\text{kJ}/\text{K}}$.

Explanation of Solution

Write the expression for the entropy generation for the isentropic process:

$S_{gen}=\Delta S_{N_2}+\Delta S_{surr}$

$S_{gen}=m_{N_2}\left[c_P{}_{N_2}\ln \left(\frac{T_2}{T_1}\right)-R\ln \left(\frac{P_2}{P_1}\right)\right]+\frac{-Q_{in}}{T_R}$ (XI)

Here, specific heat capacity of nitrogen at constant pressure is $C_{P_{N2}}$ and reservoir temperature is $T_R$.

Conclusion:

Substitute 0.2185 kg for $m_{N_2}$, $1.039\frac{\text{kJ}}{\text{kg×K}}$ for $c_{P_{N_2}}$,521.5 K for $T_2$, $20°\text{C}$ for $T_1$, $0.296\text{kJ}/\text{kg}\cdot \text{K}$ for $R$, $120\text{ kPa}$ for $P_2$, 95 kPa for $P_1$, $46.042\text{ kJ}$ for $Q_{in}$, and $500°\text{C}$  for $T_R$ in equation (XI).

$\begin{array}{c}{S}_{gen}=0.2185\text{ kg}\times \left[1.039\frac{\text{kJ}}{\text{kg×K}}\times \ln \left(\frac{521.5K}{20°\text{C}}\right)-0.296\text{kJ}/\text{kg}\cdot \text{K}\frac{120\text{ kPa}}{\text{95 kPa}}\right]+\left[\left[\frac{46.042\text{ kJ}}{500°\text{C}}\right]\right]\\ =0.2185\text{ kg}\times \left[1.039\frac{\text{kJ}}{\text{kg×K}}\times \ln \left(\frac{521.5K}{\left(20+273\text{ K}\right)}\right)-0.296\text{kJ}/\text{kg}\cdot \text{K}\frac{120\text{ kPa}}{\text{95 kPa}}\right]+\left[\frac{46.042\text{ kJ}}{\left(500\text{+273}\right)\text{ K}}\right]\\ =0.057\text{kJ}/\text{K}\end{array}$

Thus, the entropy generation during the process is $\underset{\_}{0.057\text{kJ}/\text{K}}$.