Chapter 8, Problem 161RQ

a)

To determine

The final temperature of the helium

Explanation of Solution

Given:

The volume of the nitrogen is 0.2 m³.

The mass of the helium $\left(m\right)$ is 0.1 kg.

The initial temperature of the gases $\left(T_1\right)$ is 20°C.

The initial pressure of the gases $\left(P_1\right)$ is 95 kPa.

The final temperature of the nitrogen is 500°C.

The final pressure of the helium $\left(P_2\right)$ is 120 kPa.

Calculation:

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

The specific heat ratio $\left(k\right)$ for Helium gas is $1.667$.

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

  $\begin{array}{c}\frac{T_2}{T_1}={\left(\frac{P_2}{P_1}\right)}^{\frac{k-1}{k}}\\ T_2=T_1{\left(\frac{P_2}{P_1}\right)}^{\frac{k-1}{k}}\\ T_1=20°{\left(\frac{120\text{ kPa}}{95\text{ kPa}}\right)}^{\frac{1.667-1}{1.667}}\\ =\left(20°+273\right)\text{ K}{\left(\frac{120\text{ kPa}}{95\text{ kPa}}\right)}^{\frac{1.667-1}{1.667}}\\ =321.7\text{ K}\end{array}$

b)

To determine

The final volume of nitrogen

Explanation of Solution

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

The gas constant for the helium gas is $\text{2}\text{.0769 }\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}$.

Calculate the initial volume of helium $\left(V_{He,1}\right)$.

  $\begin{array}{c}{V}_{He,1}=\frac{mR{T}_1}{P_1}\\ V_{He,1}=\frac{0.1\text{ kg}\times \text{2}\text{.0769 }\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\times \left(20°\text{C}\right)}{95\text{kPa}}\\ =\frac{0.1\text{ kg}\times \text{2}\text{.0769 }\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\times \left(20+273\right)}{95\text{kPa}}\\ =0.6406{\text{ m}}^3\end{array}$

Calculate the final volume of helium $\left(V_{He,2}\right)$.

  $\begin{array}{c}{V}_{He,2}=\frac{mR{T}_2}{P_2}\\ V_{He,1}=\frac{0.1\text{ kg}\times \text{2}\text{.0769 }\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\times \left(321.7\text{ K}\right)}{120\text{kPa}}\\ =0.5568{\text{ m}}^3\end{array}$

Calculate the final volume of the nitrogen ${\left(V_{N_2}\right)}_2$.

  $\begin{array}{c}{\left(V_{N_2}\right)}_2={\left(V_{N2}\right)}_1+{\left(V_{He}\right)}_1+{\left(V_{He}\right)}_2\\ {\left(V_{N_2}\right)}_2=0.2{\text{ m}}^3+0.6406{\text{ m}}^3+0.5568{\text{m}}^{\text{3}}\\ =0.2838{\text{ m}}^3\end{array}$

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

c)

To determine

The heat transferred to the nitrogen.

Explanation of Solution

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

The gas constant for the helium gas ${\left(R\right)}_{He}$ is $\text{2}\text{.0769 }\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}$

The specific heat constant the helium gas ${\left(C_V\right)}_{He}$ is $\text{3}\text{.1156}\frac{\text{kJ}}{\text{kg}\cdot \text{K}}$.

The gas constant for the helium gas ${\left(R\right)}_{N_2}$ is $\text{0}\text{.2968}\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}$

The specific heat constant the helium gas ${\left(C_V\right)}_{N_2}$ is $0.743\frac{\text{kJ}}{\text{kg}\cdot \text{K}}$.

Calculate the mass of the nitrogen $\left(m_{N_2}\right)$.

  $\begin{array}{c}{m}_{N2}=\frac{P_1{V}_1}{R{T}_1}\\ m_{N_2}=\frac{95\text{kPa}\times 0.2{\text{ m}}^3}{\text{2}\text{.0769 }\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\times \left(20°\text{C}\right)}\\ =\frac{95\text{kPa}\times 0.2{\text{ m}}^3}{\text{2}\text{.0769 }\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\times \left(20+273\right)\text{K}}\\ =0.2815\text{ kg}\end{array}$

Calculate the temperature of the nitrogen $T_{N_2,2}$.

  $\begin{array}{c}{T}_{N_2,2}=\frac{P_2{V}_2}{mR}\\ T_{N_2,2}=\frac{120\text{kPa}\times 0.2838{\text{ m}}^3}{0.2185\text{kg}\times \text{0}\text{.2968 }\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}}\\ =\frac{120\text{kPa}\times 0.2838{\text{ m}}^3}{0.2185\text{kg}\times \text{0}\text{.2968}\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}}\\ =525.1\text{ K}\end{array}$

Calculate the change in internal energy of the nitrogen.

  $\begin{array}{c}\Delta U_{N_2}=m{c}_V{}_{N_2}\left(T_2-T_1\right)\\ \Delta U_{N_2}=0.2185\text{ kg}\times \text{0}\text{.743}\frac{\text{kJ}}{\text{kg}\cdot \text{K}}\left(521.5\text{ K}-20°\text{C}\right)\\ =0.2185\text{ kg}\times \text{0}\text{.743}\frac{\text{kJ}}{\text{kg×K}}\left(521.5\text{ K}-\left(20+273\right)\text{ K}\right)\\ =37.1\text{ kJ}\end{array}$

Calculate the change in internal energy of the Helium.

  $\begin{array}{c}\Delta U_{He}=m{c}_V\left(T_2-T_1\right)\\ \Delta U_{He}=0.1\text{ kg}\times \text{3}\text{.1156}\frac{\text{kJ}}{\text{kg}\cdot \text{K}}\left(321.7\text{ K}-20°\text{C}\right)\\ =0.1\text{ kg}\times \text{3}\text{.1156}\frac{\text{kJ}}{\text{kg×K}}\left(321.7\text{ K}-\left(20+273\right)\text{ K}\right)\\ =8.942\text{ kJ}\end{array}$

Calculate the heat transferred to the nitrogen using the energy balance equation for the system.

  $\begin{array}{c}{Q}_{in}=\Delta U_{N2}+\Delta U_{He}\\ Q_{in}=37.1\text{ kJ}+8.942\text{ kJ}\\ =46.042\text{ kJ}\end{array}$

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

d)

To determine

The entropy generation during the process

Explanation of Solution

Write the expression for the entropy generation $\left(S_{gen}\right)$ 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}$$\begin{array}{c}{S}_{gen}=0.2185\text{ kg}\times \left[\begin{array}{l}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}}\end{array}\right]+\left[\left[\frac{46.042\text{ kJ}}{500°\text{C}}\right]\right]\\ =0.2185\text{ kg}\times \left[\begin{array}{l}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}}\end{array}\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 $\underset{\_}{0.057\text{kJ}/\text{K}}$.