Chapter 13.3, Problem 57P

An insulated tank that contains 1 kg of O₂ at 15°C and 300 kPa is connected to a 2-m³ uninsulated tank that contains N₂ at 50°C and 500 kPa. The valve connecting the two tanks is opened, and the two gases form a homogeneous mixture at 25°C. Determine (a) the final pressure in the tank, (b) the heat transfer, and (c) the entropy generated during this process. Assume T₀ = 25°C.

FIGURE P13–56

To determine

The pressure of the mixture.

Answer to Problem 57P

The pressure of the mixture is $\underset{\_}{444.6\text{kPa}}$.

Explanation of Solution

Refer to Table A-2, obtain the constant-volume specific heats of the gases at room temperature.

$\begin{array}{l}{c}_{v,{\text{O}}_{\text{2}}}=0.658\text{kJ/kg}\cdot °\text{C}\\ c_{v,{\text{N}}_{\text{2}}}=0.743\text{kJ/kg}\cdot °\text{C}\end{array}$

Write the equation to calculate the volume of the oxygen tank.

$V_{1,{\text{O}}_{\text{2}}}={\left(\frac{mR{T}_1}{P_1}\right)}_{{\text{O}}_{\text{2}}}$ (I)

Here, mass of oxygen tank is $m_{{\text{O}}_{\text{2}}}$, gas constant of oxygen tank is $R_{{\text{O}}_{\text{2}}}$, initial temperature and pressure of oxygen tank are $T_{1,{\text{O}}_{\text{2}}},P_{1,{\text{O}}_{\text{2}}}$.

Calculate the mass of nitrogen gas.

$m_{{\text{N}}_{\text{2}}}={\left(\frac{P_1{V}_1}{R{T}_1}\right)}_{{\text{N}}_2}$ (II)

Here, initial temperature and pressure of nitrogen gas is $T_{1,{\text{N}}_{\text{2}}},P_{1,{\text{N}}_2}$, gas constant of nitrogen gas is $R_{{\text{N}}_2}$, and initial volume of nitrogen gas is $V_{1,{\text{N}}_2}$.

Calculate the total volume.

$V_{\text{total}}=V_{1,{\text{O}}_{\text{2}}}+V_{1,{\text{N}}_2}$ (III)

Calculate the mole numbers of ${\text{O}}_{\text{2}}\text{and}{\text{N}}_2$.

$N_{{\text{O}}_{\text{2}}}=\frac{m_{{\text{O}}_{\text{2}}}}{M_{{\text{O}}_{\text{2}}}}$ (IV)

$N_{{\text{N}}_2}=\frac{m_{{\text{N}}_2}}{M_{{\text{N}}_2}}$ (V)

Here, molar mass of ${\text{O}}_{\text{2}}\text{and}{\text{N}}_2$ are $M_{{\text{O}}_{\text{2}}}$ and $M_{{\text{N}}_2}$ respectively.

Calculate the mole number of the mixture.

$N_m=N_{{\text{O}}_{\text{2}}}+N_{{\text{N}}_2}$ (VI)

Calculate the pressure of the mixture.

$P_m={\left(\frac{N{R}_{u}T}{V}\right)}_m$ (VII)

Here, universal gas constant of the mixture is $R_{u,m}$, volume of the mixture is $V_m$, and temperature of the mixture is $T_m$.

Conclusion:

Refer to Table A-1, obtain the gas constants of ${\text{O}}_{\text{2}}\text{and}{\text{N}}_{\text{2}}$.

$\begin{array}{l}{R}_{{\text{O}}_{\text{2}}}=0.2598\text{kJ/kg}\cdot \text{K}\\ R_{{\text{N}}_2}=0.2968\text{kJ/kg}\cdot \text{K}\end{array}$

Substitute 1 kg for $m_{{\text{O}}_{\text{2}}}$, $0.2598\text{kJ/kg}\cdot \text{K}$ for $R_{{\text{O}}_{\text{2}}}$, $15°\text{C}$ for $T_{1,{\text{O}}_{\text{2}}}$, and 300 kPa for $P_{1,{\text{O}}_{\text{2}}}$ in Equation (I).

$\begin{array}{c}{V}_{1,{\text{O}}_{\text{2}}}=\left(\frac{\left(1\text{kg}\right)\left(0.2598\text{kg/kmol}\right)15°\text{C}}{300\text{kPa}}\right)\\ =\left(\frac{\left(1\text{kg}\right)\left(0.2598\text{kg/kmol}\right)\left(15+273\right)\text{K}}{300\text{kPa}}\right)\\ =0.25{\text{m}}^{\text{3}}\end{array}$

Substitute $0.2968\text{kJ/kg}\cdot \text{K}$ for $R_{{\text{N}}_2}$, $50°\text{C}$ for $T_{1,{\text{N}}_2}$, 500 kPa for $P_{1,{\text{N}}_2}$, and $2{\text{m}}^{\text{3}}$ for $V_{1,{\text{N}}_2}$ in Equation (II).

$\begin{array}{c}{m}_{{\text{N}}_{\text{2}}}=\left(\frac{500\text{kPa}\left(2{\text{m}}^{\text{3}}\right)}{\left(0.2968\text{kg/kmol}\right)\left(50°\text{C}\right)}\right)\\ =\frac{500\text{kPa}\left(2{\text{m}}^{\text{3}}\right)}{\left(0.2968\text{kg/kmol}\right)\left(50+273\right)\text{K}}\\ =10.43\text{kg}\end{array}$

Substitute $0.25{\text{m}}^{\text{3}}$ for $V_{1,{\text{O}}_{\text{2}}}$ and $2{\text{m}}^{\text{3}}$ for $V_{1,{\text{N}}_2}$ in Equation (III).

$\begin{array}{c}{V}_{\text{total}}=0.25{\text{m}}^{\text{3}}+2{\text{m}}^{\text{3}}\\ =2.25{\text{m}}^{\text{3}}\end{array}$

Refer to Table A-1, obtain the molar mass of ${\text{O}}_{\text{2}}\text{and}{\text{N}}_{\text{2}}$.

$\begin{array}{l}{M}_{{\text{O}}_{\text{2}}}=31.999\text{kg/kmol}\\ M_{{\text{N}}_2}=28.013\text{kg/kmol}\end{array}$

Substitute 1 kg for $m_{{\text{O}}_{\text{2}}}$ and $31.999\text{kg/kmol}$ for $M_{{\text{O}}_{\text{2}}}$ in Equation (IV).

$\begin{array}{c}{N}_{{\text{O}}_{\text{2}}}=\frac{1\text{kg}}{31.999\text{kg/kmol}}\\ =0.03125\text{kmol}\end{array}$

Substitute 10.43 kg for $m_{{\text{N}}_2}$ and $28.013\text{kg/kmol}$ for $M_{{\text{N}}_2}$ in Equation (V).

$\begin{array}{c}{N}_{{\text{N}}_2}=\frac{10.43\text{kg}}{28.013\text{kg/kmol}}\\ =0.3725\text{kmol}\end{array}$

Substitute $0.03125\text{kmol}$ for $N_{{\text{O}}_{\text{2}}}$ and $0.3725\text{kmol}$ for $N_{{\text{N}}_2}$ in Equation (VI).

$\begin{array}{c}{N}_m=0.03125\text{kmol}+0.3725\text{kmol}\\ =0.40375\text{kmol}\end{array}$

Substitute $0.40375\text{kmol}$ for $N_m$, $8.314\frac{\text{kPa}\cdot {\text{m}}^3}{\text{kmol}\cdot \text{K}}$ for $R_{u,m}$, 298 K for $T_m$, and $2.25{\text{m}}^{\text{3}}$ for $V_m$ in Equation (VII).

$\begin{array}{c}{P}_m=\frac{\left(0.40375\text{kmol}\right)\left(8.314\frac{\text{kPa}\cdot {\text{m}}^3}{\text{kmol}\cdot \text{K}}\right)298\text{K}}{2.25{\text{m}}^{\text{3}}}\\ =444.6\text{kPa}\end{array}$

Thus, the pressure of the mixture is $\underset{\_}{444.6\text{kPa}}$.

To determine

The heat transfer.

Answer to Problem 57P

The heat transfer is $\underset{\_}{\text{187}\text{.2}\text{kJ}}$.

Explanation of Solution

Write the equation of energy balance for a closed system.

$\begin{array}{l}{E}_{\text{in}}-E_{\text{out}}=\Delta E_{\text{system}}\\ -Q_{\text{out}}=\Delta U\\ Q_{\text{out}}=\Delta U_{{\text{O}}_{\text{2}}}+\Delta U_{{\text{N}}_{\text{2}}}\\ Q_{\text{out}}={\left[m{c}_v\left(T_1-T_m\right)\right]}_{{\text{O}}_{\text{2}}}+{\left[m{c}_v\left(T_1-T_m\right)\right]}_{{\text{N}}_2}\end{array}$ (VIII)

Here, heat output is $Q_{\text{out}}$, energy at inlet and exit of the system is $E_{\text{in}},E_{\text{out}}$, change in the energy of the system is $\Delta E_{\text{system}}$, change in the internal energy of the system is $\Delta U$, change in internal energy of ${\text{O}}_{\text{2}}$ and ${\text{N}}_2$ is $\Delta U_{{\text{O}}_{\text{2}}}$ and $\Delta U_{{\text{N}}_2}$, specific heat at constant volume of ${\text{O}}_{\text{2}}$ and ${\text{N}}_2$ are $c_{v,{\text{O}}_{\text{2}}}$ and $c_{v,{\text{N}}_2}$ respectively.

Conclusion:

Substitute 1 kg for $m_{{\text{O}}_{\text{2}}}$, $0.658\text{kJ/kg}\cdot °\text{C}$ for $c_{v,{\text{O}}_{\text{2}}}$, $15°\text{C}$ for $T_{1,{\text{O}}_{\text{2}}}$, $25°\text{C}$ for $T_{m,{\text{O}}_{\text{2}}}$, $50°\text{C}$ for $T_{1,{\text{N}}_2}$, 10.43 kg for $m_{{\text{N}}_2}$, $0.743\text{kJ/kg}\cdot °\text{C}$ for $c_{v,{\text{N}}_2}$, and $25°\text{C}$ for $T_{m,{\text{O}}_{\text{2}}}$ in Equation (VIII).

$\begin{array}{c}{Q}_{\text{out}}=\left[\begin{array}{l}\left(\text{1}\text{kg}\right)\left(0.658\text{kJ/kg}\cdot °\text{C}\right)\left(15°\text{C}-25°\text{C}\right)+\\ \left(\text{10}\text{.43}\text{kg}\right)\left(0.743\text{kJ/kg}\cdot °\text{C}\right)\left(50°\text{C}-25°\text{C}\right)\end{array}\right]\\ =187.2\text{kJ}\end{array}$

Thus, the heat transfer is $\underset{\_}{\text{187}\text{.2}\text{kJ}}$.

To determine

The entropy generation.

Answer to Problem 57P

The entropy generation is $\underset{\_}{0.962\text{kJ/K}}$.

Explanation of Solution

Write the equation of entropy balance.

$\begin{array}{l}{S}_{\text{in}}-S_{\text{out}}+S_{\text{gen}}=\Delta S_{\text{system}}\\ -\frac{Q_{\text{out}}}{T_{b,\text{surr}}}+S_{\text{gen}}=m\left(s_2-s_1\right)\\ S_{\text{gen}}=m\left(s_2-s_1\right)+\frac{Q_{\text{out}}}{T_{\text{surr}}}\\ S_{\text{gen}}=m{\left(s_2-s_1\right)}_{{\text{O}}_{\text{2}}}+m{\left(s_2-s_1\right)}_{{\text{N}}_2}+\frac{Q_{\text{out}}}{T_{\text{surr}}}\end{array}$ (IX)

Here, entropy at inlet and exit is $S_{\text{in},}{S}_{\text{out}}$, change in the entropy of a system is $\Delta S_{\text{system}}$, surrounding temperature is $T_{\text{surr}}$, entropy per unit mass at inlet and exit are $s_1,s_2$ respectively.

Calculate the mole fraction of ${\text{O}}_{\text{2}}\text{and}{\text{N}}_{\text{2}}$.

$y_{{\text{O}}_{\text{2}}}=\frac{N_{{\text{O}}_{\text{2}}}}{N_m}$ (X)

$y_{{\text{N}}_2}=\frac{N_{{\text{N}}_2}}{N_m}$ (XI)

Calculate the value of ${\left(s_2-s_1\right)}_{{\text{O}}_{\text{2}}}$.

${\left(s_2-s_1\right)}_{{\text{O}}_{\text{2}}}={\left(c_p\ln \frac{T_2}{T_1}-R\ln \frac{y{P}_{m,2}}{P_1}\right)}_{{\text{O}}_{\text{2}}}$ (XII)

Here, the partial pressure of mixture at state 2 is $P_{m,2}$ and partial pressure of ${\text{O}}_{\text{2}}$ at state 1 is $P_{1,{\text{O}}_{\text{2}}}$.

Calculate the value of ${\left(s_2-s_1\right)}_{{\text{N}}_2}$.

${\left(s_2-s_1\right)}_{{\text{N}}_2}={\left(c_p\ln \frac{T_2}{T_1}-R\ln \frac{y{P}_{m,2}}{P_1}\right)}_{{\text{N}}_2}$ (XIII)

Here, the partial pressure of mixture at state 2 is $P_{m,2}$ and partial pressure of ${\text{N}}_2$ at state 1 is $P_{1,{\text{N}}_2}$.

Conclusion:

Substitute $0.03125\text{kmol}$ for $N_{{\text{O}}_{\text{2}}}$ and 0.40375 kmol for $N_m$ in Equation (X).

$\begin{array}{c}{y}_{{\text{O}}_{\text{2}}}=\frac{0.03125\text{kmol}}{0.40375\text{kmol}}\\ =0.077\end{array}$

Substitute $0.3725\text{kmol}$ for $N_{{\text{N}}_2}$ and 0.40375 kmol for $N_m$ in Equation (XI).

$\begin{array}{c}{y}_{{\text{N}}_2}=\frac{0.3725\text{kmol}}{0.40375\text{kmol}}\\ =0.923\end{array}$

Refer to Table A-2, obtain the constant-pressure specific heats of the gases at room temperature.

$\begin{array}{l}{c}_{p,{\text{O}}_{\text{2}}}=0.918\text{kJ/kg}\cdot \text{K}\\ c_{p,{\text{N}}_{\text{2}}}=1.039\text{kJ/kg}\cdot \text{K}\end{array}$

Substitute 0.077 for $y_{{\text{O}}_{\text{2}}}$, $0.918\text{kJ/kg}\cdot \text{K}$ for $c_{p,{\text{O}}_{\text{2}}}$, $25°\text{C}$ for $T_2$, $15°\text{C}$ for $T_1$, $0.2598\text{kJ/kg}\cdot \text{K}$ for $R_{{\text{O}}_{\text{2}}}$, 444.6 kPa for $P_{m,2}$, and 500 kPa for $P_{1,{\text{O}}_{\text{2}}}$ in Equation (XII).

$\begin{array}{c}{\left(s_2-s_1\right)}_{{\text{O}}_{\text{2}}}=\left[\begin{array}{l}\left(0.918\text{kJ/kg}\cdot \text{K}\right)\ln \frac{25°\text{C}}{15°\text{C}}-\\ \left(0.2598\text{kJ/kg}\cdot \text{K}\right)\ln \frac{\left(0.077\right)\left(444.6\text{kPa}\right)}{500\text{kPa}}\end{array}\right]\\ =\left[\begin{array}{l}\left(0.918\text{kJ/kg}\cdot \text{K}\right)\ln \frac{\left(25+273\right)\text{K}}{\left(15+273\right)\text{K}}-\\ \left(0.2598\text{kJ/kg}\cdot \text{K}\right)\ln \frac{\left(0.077\right)\left(444.6\text{kPa}\right)}{500\text{kPa}}\end{array}\right]\\ =0.5952\text{kJ/kg}\cdot \text{K}\end{array}$

Substitute 0.923 for $y_{{\text{N}}_2}$, $1.039\text{kJ/kg}\cdot \text{K}$ for $c_{p,{\text{N}}_2}$, $25°\text{C}$ for $T_2$, $50°\text{C}$ for $T_1$, $0.2968\text{kJ/kg}\cdot \text{K}$ for $R_{{\text{N}}_2}$, 444.6 kPa for $P_{m,2}$, and 500 kPa for $P_{1,{\text{N}}_2}$ in Equation (XIII).

$\begin{array}{c}{\left(s_2-s_1\right)}_{{\text{N}}_2}=\left[\begin{array}{l}\left(1.039\text{kJ/kg}\cdot \text{K}\right)\ln \frac{25°\text{C}}{50°\text{C}}-\\ \left(0.2968\text{kJ/kg}\cdot \text{K}\right)\ln \frac{\left(0.923\right)\left(444.6\text{kPa}\right)}{500\text{kPa}}\end{array}\right]\\ =\left[\begin{array}{l}\left(0.918\text{kJ/kg}\cdot \text{K}\right)\ln \frac{\left(25+273\right)\text{K}}{\left(50+273\right)\text{K}}-\\ \left(0.2968\text{kJ/kg}\cdot \text{K}\right)\ln \frac{\left(0.923\right)\left(444.6\text{kPa}\right)}{500\text{kPa}}\end{array}\right]\\ =-0.0251\text{kJ/kg}\cdot \text{K}\end{array}$

Substitute $-0.0251\text{kJ/kg}\cdot \text{K}$ for ${\left(s_2-s_1\right)}_{{\text{N}}_2}$, $0.5952\text{kJ/kg}\cdot \text{K}$ for ${\left(s_2-s_1\right)}_{{\text{O}}_{\text{2}}}$, 1 kg for $m_{{\text{O}}_{\text{2}}}$, 10.43 kg for $m_{{\text{N}}_2}$, 187.2 kJ for $Q_{\text{out}}$, and $25°\text{C}$ for $T_{\text{surr}}$ in Equation (IX).

$\begin{array}{c}{S}_{\text{gen}}=1\text{kg}\left(0.5952\text{kJ/kg}\cdot \text{K}\right)+10.43\text{kg}\left(-0.0251\text{kJ/kg}\cdot \text{K}\right)+\frac{187.2\text{kJ}}{25°\text{C}}\\ =1\text{kg}\left(0.5952\text{kJ/kg}\cdot \text{K}\right)+10.43\text{kg}\left(-0.0251\text{kJ/kg}\cdot \text{K}\right)+\frac{187.2\text{kJ}}{\left(25+273\right)\text{K}}\\ =0.962\text{kJ/K}\end{array}$

Thus, the entropy generation is $\underset{\_}{0.962\text{kJ/K}}$.