Chapter 7.13, Problem 215RP

a)

To determine

The final temperature.

Answer to Problem 215RP

The final temperature is $315.3\text{K}$.

Explanation of Solution

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

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

Here, energy change in to the system is $\Delta E_{in}$, energy change exit from the system is $\Delta E_{out}$ and change in energy of the system is $\Delta E_{system}$.

Write the expression for the initial pressure in the cylinder using ideal gas law.

$P_1=\frac{m_{1}R{T}_1}{{\nu }_1}$ (II)

Here, initial pressure is $P_1$, initial mass is $m_1$, gas constant is R, initial temperature is $T_1$ and initial volume is ${\nu }_1$.

Write the expression for the mass of the air at final stage using ideal gas law.

$m_2=\frac{P_2{\nu }_2}{R{T}_2}$ (III)

Here, final mass is $m_2$, final pressure is $P_2$, final specific volume is ${\nu }_2$, gas constant is R, and final temperature is $T_2$.

Write the expression for the mass entering the cylinder $\left(m_i\right)$.

$m_i=m_2-m_1$ (IV)

Conclusion:

Substitute $m_i{h}_i$ for $\Delta E_{in}$, $W_{b,out}$ for $\Delta E_{out}$ and $m_2{u}_2-m_1{u}_1$ for $\Delta E_{system}$ in Equation (I).

$\begin{array}{l}{m}_i{h}_i-W_{b,out}=m_2{u}_2-m_1{u}_1\\ m_i{c}_p{T}_i-W_{b,out}=m_2{c}_v{T}_2-m_1{c}_v{T}_1\end{array}$ (V)

Here, work done during the process is $W_{b,out}$.mass flow is $m$, specific heat at constant pressure is $c_p$ , specific heat at constant volume is $c_v$, mass entering the cylinder is $m_i$, final mass is $m_2$ and initial mass is $m_2$.

From the Table A-2” Ideal-gas specific heats of various common gases” obtain the following properties for air.

$\begin{array}{l}R=0.2870\text{kJ}/\text{kg}\cdot \text{K}\\ c_p=1.005\text{kJ}/\text{kg}\cdot \text{K}\\ c_v=0.718\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Substitute $\text{1}\text{.3 kg}$ for $m_1$, $0.287\text{kJ}/\text{kg}\cdot \text{K}$ for R, $30°\text{C}$ for $T_1$, and $0.40{\text{m}}^3$ for ${\nu }_1$ in Equation (II).

$\begin{array}{c}{P}_1=\frac{\left(1.3\text{kg}\right)\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)30°\text{C}}{0.40{\text{m}}^3}\\ =\frac{\left(1.3\text{kg}\right)\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\left(30+273\right)\text{K}}{0.40{\text{m}}^3}\\ =282.6\text{kPa}\end{array}$

When system undergoes constant pressure process, $P_2=P_1$

Substitute $282.6\text{kPa}$ for $P_2$, $1.5\times 0.40{\text{m}}^3$ for ${\nu }_2$, $0.287\text{kJ}/\text{kg}\cdot \text{K}$ for $R$ in Equation (III).

$\begin{array}{c}{m}_2=\frac{\left(282.6\text{kPa}\right)\left(1.5\times 0.40{\text{m}}^3\right)}{\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)T_2}\\ =\frac{590.8}{T_2}\end{array}$ (VI)

Substitute $\frac{590.8}{T_2}$ for $m_2$ and $\text{1}\text{.3 kg}$ for $m_1$ in Equation (IV).

$\begin{array}{c}{m}_i=\frac{590.8}{T_2}-1.3\text{kg}\\ \text{=}\frac{590.8}{T_2}-1.3\end{array}$ (VII)

Substitute $\frac{590.8}{T_2}-1.3$ for $m_i$, $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, $70°\text{C}$ for $T_i$, 56.52 kJ for $W_{b,out}$, $\frac{590.8}{T_2}$ for $m_2$, $0.718\text{kJ}/\text{kg}\cdot \text{K}$ for $c_v$, 1.3 kg for $m_1$ and $30°\text{C}$ for $T_1$ in Equation (V).

$\begin{array}{l}\left\{\begin{array}{l}\left(\frac{590.8}{T_2}-1.3\right)\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\\ \times \left(70°\text{C}\right)-56.52\text{ kJ}\end{array}\right\}=\left\{\begin{array}{l}\left(\frac{590.8}{T_2}\right)\left(0.718\text{kJ}/\text{kg}\cdot \text{K}\right)T_2\\ -\left(1.3\text{kg}\right)\times \left(0.718\text{kJ}/\text{kg}\cdot \text{K}\right)\left(30°\text{C}\right)\end{array}\right\}\\ \left\{\begin{array}{l}\left(\frac{590.8}{T_2}-1.3\right)\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\\ \times \left(70+273\right)\text{K}-56.52\text{ kJ}\end{array}\right\}=\left\{\begin{array}{l}\left(\frac{590.8}{T_2}\right)\left(0.718\text{kJ}/\text{kg}\cdot \text{K}\right)T_2-\left(1.3\text{kg}\right)\\ \times \left(0.718\text{kJ}/\text{kg}\cdot \text{K}\right)\times \left(30+273\right)\text{K}\end{array}\right\}\end{array}$

$\begin{array}{l}\left(\frac{590.8}{T_2}-1.3\right)344.715-56.52=424.19-282.82\\ \frac{203657.6}{T_2}-504.65=141.3742\\ T_2=\frac{203657.6}{646.02}\\ =315.3\text{K}\end{array}$

Thus, the final temperature is $315.3\text{K}$.

b)

To determine

The amount of mass entered the cylinder.

Answer to Problem 215RP

The amount of mass entered the cylinder is $\text{0}\text{.5742 kg}$.

Explanation of Solution

Substitute 315.3 K for $T_2$ in Equation (VII).

$\begin{array}{c}{m}_2=\frac{590.8}{315.3\text{ K}}\\ =1.874\text{kg}\end{array}$

Substitute 1.874 kg for $m_2$ and 1.3 kg for $m_1$ in Equation (V).

$\begin{array}{c}{m}_i=1.874\text{ kg}-1.3\text{kg}\\ \text{=}\text{0}\text{.5742 kg}\end{array}$

Thus, the amount of mass entered the cylinder is $\text{0}\text{.5742 kg}$.

c)

To determine

The work done during the process.

Answer to Problem 215RP

The work done during the process is $56.52\text{kJ}$.

Explanation of Solution

Write the expression for the work done during the process $\left(W_{b,out}\right)$.

$W_{b,out}=P_1\left({\nu }_2-{\nu }_1\right)$ (VIII)

Here, initial volume is ${\nu }_1$ , final volume is ${\nu }_2$ and initial pressure is $\left(P_1\right)$.

Conclusion:

Substitute $282.6\text{kPa}$ for $P_1$, $\left(150\%\times 0.40{\text{m}}^3\right)$ for ${\nu }_2$, and $0.40{\text{m}}^3$ for ${\nu }_1$ in Equation (VIII).

$\begin{array}{c}{W}_{b,out}=282.6\text{kPa}\left(1.5\times 0.40{\text{m}}^3-0.40{\text{m}}^3\right)\\ =56.52\text{kJ}\end{array}$

Thus, the work done during the process is $56.52\text{kJ}$.

d)

To determine

The entropy generated for the process.

Answer to Problem 215RP

The entropy generated for the process is $0.09714\text{kJ}/\text{K}$.

Explanation of Solution

Write the expression for the entropy generated during the process.

$\begin{array}{c}{S}_{gen}=m_2{s}_2-m_1{s}_1-m_i{s}_i\\ =m_2{s}_2-m_1{s}_1-\left(m_2-m_1\right)s_i\\ =m_2\left(s_2-s_i\right)-m_1\left(s_1-s_i\right)\\ =m_2\left(c_p\ln \frac{T_2}{T_i}-R\ln \frac{P_2}{P_i}\right)-m_1\left(c_p\ln \frac{T_1}{T_i}-R\ln \frac{P_1}{P_i}\right)\end{array}$ (IX)

Here, entropy generation is $S_{gen}$, initial entropy is $s_1$,entropy entering the cylinder is $s_i$ , final entropy is $s_2$ , pressure entering the cylinder is $P_i$ , and final pressure is $P_2$ , temperature entering the cylinder is $T_i$ , final temperature is $T_2$ , gas constant is $R$ , specific heat at constant pressure is $c_p$, mass entering the cylinder is $m_i$, final mass is $m_2$ and initial mass is $m_2$.

Conclusion:

Substitute 1.874 kg for $m_2$, $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, 315.3 K for $T_2$, 70 C for $T_i$, $0.287\text{kJ}/\text{kg}\cdot \text{K}$ for R, $282.6\text{kPa}$ for $P_2$, $\text{500 kPa}$ for $P_i$, 1.3 kg for $m_1$, 30 C for $T_1$, and $282.6\text{kPa}$ for $P_1$ in Equation (IX).

$\begin{array}{c}{S}_{gen}=\left\{\begin{array}{l}\left(1.874\text{ kg}\right)\left(\begin{array}{l}\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \frac{315.3\text{K}}{70°\text{C}}\\ -0.287\text{kJ}/\text{kg}\cdot \text{K}\ln \frac{282.6\text{kPa}}{500\text{kPa}}\end{array}\right)\\ -\left(1.3\text{kg}\right)\left(\begin{array}{l}\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \frac{30°\text{C}}{70°\text{C}}\\ -0.287\text{kJ}/\text{kg}\cdot \text{K}\ln \frac{282.6\text{kPa}}{500\text{kPa}}\end{array}\right)\end{array}\right\}\\ =\left\{\begin{array}{l}\left(1.874\text{ kg}\right)\left(\begin{array}{l}\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \frac{315.3\text{K}}{\left(70+273\right)\text{K}}\\ -0.287\text{kJ}/\text{kg}\cdot \text{K}\ln \frac{282.6\text{kPa}}{500\text{kPa}}\end{array}\right)\\ -\left(1.3\text{kg}\right)\left(\begin{array}{l}\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \frac{\left(30+273\right)\text{K}}{\left(70+273\right)\text{K}}\\ -0.287\text{kJ}/\text{kg}\cdot \text{K}\ln \frac{282.6\text{kPa}}{500\text{kPa}}\end{array}\right)\end{array}\right\}\\ =0.09714\text{kJ}/\text{K}\end{array}$

Thus, the entropy generated for the process is $0.09714\text{kJ}/\text{K}$.