Chapter 5.5, Problem 183RP

A piston–cylinder device initially contains 1.2 kg of air at 700 kPa and 200°C. At this state, the piston is touching on a pair of stops. The mass of the piston is such that 600-kPa pressure is required to move it. A valve at the bottom of the tank is opened, and air is withdrawn from the cylinder. The valve is closed when the volume of the cylinder decreases to 80 percent of the initial volume. If it is estimated that 40 kJ of heat is lost from the cylinder, determine (a) the final temperature of the air in the cylinder, (b) the amount of mass that has escaped from the cylinder, and (c) the work done. Use constant specific heats at the average temperature.

FIGURE P5–183

To determine

The final temperature of air in the cylinder.

Answer to Problem 183RP

The final temperature of air in the cylinder is $415\text{K}$.

Explanation of Solution

Write the equation of mass balance.

$m_{\text{in}}-m_{\text{e}}=\Delta m_{\text{system}}$ (I)

Here, the inlet mass is $m_{\text{in}}$, the exit mass is $m_{\text{e}}$ and the change in mass of the system is $\Delta m_{\text{system}}$.

The change in mass of the system for the control volume is expressed as,

$\Delta m_{\text{system}}={\left(m_2-m_1\right)}_{\text{cv}}$

Here, the suffixes 1 and 2 indicates the initial and final states of the system.

Consider the piston-cylinder as the control volume. Initially the cylinder is filled with air and the valve is in closed position, further no other mass is allowed to enter the cylinder. Hence, the inlet mass is neglected i.e. $m_{\text{in}}=0$.

Rewrite the Equation (I) as follows.

$\begin{array}{l}0-m_{\text{e}}={\left(m_2-m_1\right)}_{\text{cv}}\\ -m_{\text{e}}=m_2-m_1\\ m_{\text{e}}=m_1-m_2\end{array}$ (II)

Write the formula for initial volume of air present in the cylinder.

${\nu }_{\text{1}}=\frac{m_{1}R{T}_1}{P_1}$ (III)

Here, the mass of air is $m$, the pressure is $P$, the volume is $\nu$, the gas constant of is $R$, the temperature is $T$ and the subscript 1 indicates the initial state.

Write the formula for mass of air present in the cylinder at final state.

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

Here, the subscript 2 indicates the final state.

Write the energy balance equation.

$\begin{array}{l}{E}_{in}-E_{out}=\Delta E_{system}\\ \left\{\begin{array}{l}\left[Q_{\text{in}}+W_{\text{in}}+m_{\text{in}}{\left(h+\text{ke}+\text{pe}\right)}_{\text{in}}\right]\\ -\left[Q_{\text{e}}+W_{\text{e}}+m_{\text{e}}{\left(h+\text{ke}+\text{pe}\right)}_{\text{e}}\right]\end{array}\right\}={\left[\begin{array}{l}{m}_2{\left(u+\text{ke}+\text{pe}\right)}_2\\ -m_1{\left(u+\text{ke}+\text{pe}\right)}_1\end{array}\right]}_{\text{system}}\end{array}$ (V)

Here, the heat transfer is $Q$, the work transfer is $W$, the enthalpy is $h$, the internal energy is $u$, the kinetic energy is $\text{ke}$, the potential energy is $\text{pe}$ and the change in net energy of the system is $\Delta E_{\text{system}}$; the suffixes 1 and 2 indicates the inlet and outlet of the system.

The pressure of $600\text{kPa}$ is required to move the piston i.e. boundary work required to move the piston. When the valve is opened and air is withdrawn at the bottom of the cylinder. The boundary work is done on the system. i.e. $W_{\text{in}}=W_{\text{b,in}},W_{\text{e}}=0$. The heat transfer occurs while air escapes the cylinder. Neglect the kinetic and potential energy changes i.e. $\left(\Delta \text{ke}=\Delta \text{pe}=0\right)$.

The Equation (V) reduced as follows.

$W_{\text{b,in}}-Q_{\text{e}}-m_{\text{e}}{h}_{\text{e}}=m_2{u}_2-m_1{u}_1$ (VI)

Write the formula for boundary work done on the cylinder.

$\begin{array}{c}{W}_{\text{b,in}}=P_2\Delta \nu \\ =P_2\left({\nu }_1-{\nu }_2\right)\end{array}$ (VII)

Here, the pressure required to move the piston is $P_2$.

The enthalpy and internal energy in terms of temperature and specific heats are expressed as follows.

$\begin{array}{c}h=c_{p}T\\ u=c_{v}T\end{array}$

Rewrite the Equation (VI) as follows.

$W_{\text{b,in}}-Q_{\text{e}}-m_{\text{e}}{c}_p{T}_e=m_2{c}_v{T}_2-m_1{c}_v{T}_1$ (VIII)

The temperature of the air while exiting the cylinder is considered as the average temperature of initial and final temperatures.

$\begin{array}{c}{T}_e=\frac{T_1+T_2}{2}\\ =\frac{\left(200+273\right)\text{K}+T_2}{2}\\ =\frac{473\text{K}+T_2}{2}\\ =\frac{473+T_2}{2}\end{array}$

Refer Table A-1, “Molar mass, gas constant, and critical-point properties”.

The gas constant $\left(R\right)$ of air is $0.287\text{kPa}{\text{.m}}^3\text{/kg}\cdot \text{K}$.

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

The specific heat at constant pressure $\left(c_p\right)$ of air corresponding to the anticipated average temperature of $450\text{K}$ is $1.020\text{kJ/kg}\cdot \text{K}$ and the specific heat at constant volume $\left(c_v\right)$ is $0.733\text{kJ/kg}\cdot \text{K}$.

Conclusion:

Substitute $1.2\text{kg}$ for $m_1$, $0.287\text{kPa}{\text{.m}}^3\text{/kg}\cdot \text{K}$ for $R$, $200\text{°C}$ for $T_1$ , and  $700\text{kPa}$ for $P_1$ in Equation (III).

$\begin{array}{c}{\nu }_{\text{1}}=\frac{\left(1.2\text{kg}\right)\left(0.287\text{kPa}{\text{.m}}^3\text{/kg}\cdot \text{K}\right)\left(200°\text{C}\right)}{700\text{kPa}}\\ =\frac{\left(1.2\text{kg}\right)\left(0.287\text{kPa}{\text{.m}}^3\text{/kg}\cdot \text{K}\right)\left(20+273\right)\text{K}}{700\text{kPa}}\\ =\frac{162.9012\text{kPa}{\text{.m}}^3}{700\text{kPa}}\\ =0.2327{\text{m}}^3\end{array}$

It is given that the final volume is 80 % of initial volume.

$\begin{array}{c}{\nu }_2=\frac{80}{100}{\nu }_1\\ =0.8{\nu }_1\\ =0.8\left(0.2327{\text{m}}^3\right)\\ =0.1862{\text{m}}^3\end{array}$

Substitute $600\text{kPa}$ for $P$, $0.2327{\text{m}}^3$ for ${\nu }_1$, and $0.1862{\text{m}}^3$ for ${\nu }_2$ in Equation (VII).

$\begin{array}{c}{W}_{\text{b,in}}=600\text{kPa}\left(0.2327{\text{m}}^3-0.1862{\text{m}}^3\right)\\ =600\text{kPa}\left(0.0465{\text{m}}^3\right)\\ =27.9\text{kPa}\cdot {\text{m}}^3\times \frac{1\text{kJ}}{1\text{kPa}\cdot {\text{m}}^3}\\ =27.9\text{kJ}\end{array}$

Substitute $600\text{kPa}$ for $P_2$, $0.1862{\text{m}}^3$ for ${\nu }_2$, and $0.287\text{kPa}{\text{.m}}^3\text{/kg}\cdot \text{K}$ for $R$ in Equation (IV).

$\begin{array}{c}{m}_2=\frac{\left(600\text{kPa}\right)\left(0.1862{\text{m}}^3\right)}{\left(0.287\text{kPa}{\text{.m}}^3\text{/kg}\cdot \text{K}\right)T_2}\\ =\frac{389.27}{T_2}\text{kg}\end{array}$

Substitute $1.2\text{kg}$ for $m_1$ and $\frac{389.27}{T_2}\text{kg}$ for $m_2$ in Equation (I).

$\begin{array}{c}{m}_e=1.2\text{kg}-\frac{389.27}{T_2}\text{kg}\\ =1.2-\frac{389.27}{T_2}\end{array}$ (IX)

Substitute $27.9\text{kJ}$ for $W_{\text{b,in}}$, $40\text{kJ}$ for $Q_{\text{e}}$, $1.2-\frac{389.27}{T_2}$ for $m_e$, $1.020\text{kJ/kg}\cdot \text{K}$ for $c_p$, $\frac{473+T_2}{2}$ for $T_{\text{e}}$, $\frac{389.27}{T_2}\text{kg}$ for $m_2$, $0.733\text{kJ/kg}\cdot \text{K}$ for $c_v$, $1.2\text{kg}$ for $m_1$, $473\text{K}$ for $T_1$ in Equation (VIII).

$\begin{array}{l}\left\{\begin{array}{l}27.9\text{kJ}-40\text{kJ}\\ -\left[\begin{array}{l}\left(1.2-\frac{389.27}{T_2}\right)\left(1.020\text{kJ/kg}\cdot \text{K}\right)\\ \left(\frac{473+T_2}{2}\right)\end{array}\right]\end{array}\right\}=\left[\begin{array}{l}\left(\frac{389.27}{T_2}\text{kg}\right)\left(0.733\text{kJ/kg}\cdot \text{K}\right)T_2\\ -\left(1.2\text{kg}\right)\left(0.733\text{kJ/kg}\cdot \text{K}\right)\left(473\text{K}\right)\end{array}\right]\\ -12.1\text{kJ}-\left[\begin{array}{l}\left(\frac{1.2T_2-389.27}{T_2}\right)\left(1.020\text{kJ/kg}\cdot \text{K}\right)\\ \left(\frac{473+T_2}{2}\right)\end{array}\right]=285.3349\text{kJ}-416.0508\text{kJ}\\ -\left(\frac{1.2T_2-389.27}{T_2}\right)\left(1.020\text{kJ/kg}\cdot \text{K}\right)\left(\frac{473+T_2}{2}\right)=\left(\begin{array}{l}285.3349\text{kJ}-416.0508\text{kJ}\\ +12.1\text{kJ}\end{array}\right)\\ -\left(\frac{1.2T_2-389.27}{T_2}\right)\left(1.020\text{kJ/kg}\cdot \text{K}\right)\left(\frac{473+T_2}{2}\right)=-118.6159\text{kJ}\end{array}$

$\left(\frac{1.2T_2-389.27}{T_2}\right)\left(1.020\text{kJ/kg}\cdot \text{K}\right)\left(\frac{473+T_2}{2}\right)=118.6159\text{kJ}$ (X)

Use Engineering Equation Solver (EES) or online calculator to solve the Equation (X) and obtain the value of $T_2$ and consider the positive root alone.

$\begin{array}{c}{T}_2=414.96659\text{K}\\ \simeq 415\text{K}\end{array}$

Thus, the final temperature of air in the cylinder is $415\text{K}$.

To determine

The amount of mass escaped from the cylinder.

Answer to Problem 183RP

The amount of mass escaped from the cylinder is $0.262\text{kg}$.

Explanation of Solution

The amount of mass escaped from the cylinder is nothing but the mass of air vented out until final state i.e. $m_{\text{e}}$.

Refer Equation (II) and (IX).

Conclusion:

Substitute $415\text{K}$ for $T_2$ in Equation (IX).

$\begin{array}{c}{m}_e=1.2\text{kg}-\frac{389.27}{415\text{K}}\text{kg}\\ =1.2\text{kg}-0.938\text{kg}\\ =0.262\text{kg}\end{array}$

Thus, the amount of mass escaped from the cylinder is $0.262\text{kg}$.

To determine

The work done.

Answer to Problem 183RP

The amount of mass escaped from the cylinder is $0.262\text{kg}$.

Explanation of Solution

The work done is nothing but the work done on the piston to move it i.e. boundary work $\left(W_{\text{b,in}}\right)$.

Refer part (a).

$W_{\text{b,in}}=27.9\text{kJ}$

Thus, the work done is $27.9\text{kJ}$.