Chapter 4.5, Problem 20P

A piston–cylinder device contains 0.15 kg of air initially at 2 MPa and 350°C. The air is first expanded isothermally to 500 kPa, then compressed polytropically with a polytropic exponent of 1.2 to the initial pressure, and finally compressed at the constant pressure to the initial state. Determine the boundary work for each process and the net work of the cycle.

To determine

The boundary work for the isothermal expansion process of a piston-cylinder device.

The boundary work for the polytropic compression process of a piston-cylinder device.

The boundary work for the constant pressure compression process of a piston-cylinder device.

The net-work for cycle is the sum of the works for each process of a piston-cylinder device.

Answer to Problem 20P

The boundary work for the isothermal expansion process of a piston-cylinder device is $\underset{\_}{37.18\text{kJ}}$.

The boundary work for the polytropic compression process of a piston-cylinder device is $\underset{\_}{-34.9\text{kJ}}$.

The boundary work for the constant pressure compression process of a piston-cylinder device is $\underset{\_}{-6.98\text{kJ}}$.

The net-work for cycle is the sum of the works for each process of a piston-cylinder device is $\underset{\_}{-4.69\text{kJ}}$.

Explanation of Solution

Show the free body diagram of the piston-cylinder device contains air.

Determine the process 1 volume of an ideal gas.

${\nu }_1=\frac{mRT}{P_1}$ (I)

Here, the mass of a piston-cylinder device is $m$, the process 1 pressure of a piston-cylinder device is $P_1$, the temperature is $T$, and the ideal-gas constant for nitrogen is $R$.

Determine the process 2 volume of an ideal gas.

${\nu }_2=\frac{mRT}{P_2}$ (II)

Here, the mass of a piston-cylinder device is $m$, the process 2 pressure of a piston-cylinder device is $P_2$, the temperature is $T$, and the ideal-gas constant for nitrogen is $R$.

Write the expression for the boundary work for isothermal expansion process (1-2) of an ideal gas.

$W_{\text{b,1-2}}=P_1{\nu }_1\ln \left(\frac{{\nu }_2}{{\nu }_1}\right)$ (III)

Determine an ideal gas for polytropic compression process.

$P_2{\nu }_2^n=P_3{\nu }_3^n$ (IV)

Here, the process 3 pressure of a piston-cylinder device is $P_3$ and the process 3 volume of a piston-cylinder device is ${\nu }_3$.

Write the expression for the boundary work for polytropic compression process of an ideal gas.

$W_{\text{b,2-3}}=\frac{P_3{\nu }_3-P_2{\nu }_2}{1-n}$ (V)

Write the expression for the boundary work for constant pressure compression process of an ideal gas.

$W_{\text{b,3-1}}=P_3\left({\nu }_1-{\nu }_3\right)$ (VI)

Determine the net-work for cycle is the sum of the works for each process of a piston-cylinder device.

$W_{net}=W_{b,1-2}+W_{b,2-3}+W_{b,3-1}$ (VII)

Conclusion:

From the Table 4-2a, “Ideal-gas specific heat of various common gases”, obtain the value gas constant for air as $0.287\text{kJ}/\text{kg}\cdot \text{K}$

Substitute $0.15\text{kg}$ for $m$, $0.287\text{kJ}/\text{kg}\cdot \text{K}$ for $R$, $2000\text{kPa}$ for $P_1$, and $350°\text{C}$ for $T$ in Equation (I).

$\begin{array}{c}{\nu }_1=\frac{\left(0.15\text{kg}\right)\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\left(350°\text{C}\right)}{2000\text{kPa}}\\ =\frac{\left(0.04305\text{kJ}/\text{K}\right)\left(350°\text{C}+273\text{K}\right)}{2000\text{kPa}}\\ =0.01341\text{kJ}/\text{kPa}\times \left(\frac{1{\text{m}}^3}{1\text{kJ}/\text{kPa}}\right)\\ =0.01341{\text{m}}^{\text{3}}\end{array}$

Substitute $0.15\text{kg}$ for $m$, $0.287\text{kJ}/\text{kg}\cdot \text{K}$ for $R$, $500\text{kPa}$ for $P_1$, and $350°\text{C}$ for $T$ in Equation (II).

$\begin{array}{c}{\nu }_2=\frac{\left(0.15\text{kg}\right)\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\left(350°\text{C}\right)}{500\text{kPa}}\\ =\frac{\left(0.04305\text{kJ}/\text{K}\right)\left(350°\text{C}+273\text{K}\right)}{500\text{kPa}}\\ =0.05364\text{kJ}/\text{kPa}\times \left(\frac{1{\text{m}}^3}{1\text{kJ}/\text{kPa}}\right)\\ =0.05364{\text{m}}^{\text{3}}\end{array}$

Substitute $2000\text{kPa}$ for $P_1$, $0.01341{\text{m}}^{\text{3}}$ for ${\nu }_1$, and $0.05364{\text{m}}^{\text{3}}$ for ${\nu }_2$ in Equation (III).

$\begin{array}{c}{W}_{\text{b,1-2}}=\left(2000\text{kPa}\right)\left(0.01341{\text{m}}^{\text{3}}\right)\ln \left(\frac{0.05364{\text{m}}^{\text{3}}}{0.01341{\text{m}}^{\text{3}}}\right)\\ =37.18\text{kPa}\cdot {\text{m}}^{\text{3}}\times \left(\frac{1\text{kJ}}{1\text{kPa}\cdot {\text{m}}^{\text{3}}}\right)\\ =37.18\text{kJ}\end{array}$

Thus, the boundary work for the isothermal expansion process of a piston-cylinder device is $\underset{\_}{37.18\text{kJ}}$.

Substitute $500\text{kPa}$ for $P_2$, $0.05364{\text{m}}^{\text{3}}$ for ${\nu }_2$, and $2000\text{kPa}$ for $P_3$ in Equation (IV).

$\begin{array}{l}\left(500\text{kPa}\right){\left(0.05364{\text{m}}^{\text{3}}\right)}^{1.2}=\left(2000\text{kPa}\right){\nu }_3^{1.2}\\ \left(14.94\text{kPa}\cdot {\text{m}}^{\text{3}}\right)=\left(2000\text{kPa}\right){\nu }_3^{1.2}\\ {\nu }_3=0.01690m^3\end{array}$

Substitute $2000\text{kPa}$ for $P_3$, $0.01690{\text{m}}^{\text{3}}$ for ${\nu }_3$, $500\text{kPa}$ for $P_2$, $0.05364{\text{m}}^{\text{3}}$ for ${\nu }_2$, and $1.2$ for $n$ in Equation (V).

$\begin{array}{c}{W}_{b,2-3}=\frac{\left(2000\text{kPa}\right)\left(0.01690{\text{m}}^{\text{3}}\right)-\left(500\text{kPa}\right)\left(0.05364{\text{m}}^{\text{3}}\right)}{1-1.2}\\ =\frac{\left(33.8\text{kPa}\cdot {\text{m}}^{\text{3}}\right)-\left(26.82\text{kPa}\cdot {\text{m}}^{\text{3}}\right)}{\left(-0.2\right)}\\ =\frac{\left(6.98\text{kPa}\cdot {\text{m}}^{\text{3}}\right)}{\left(-0.2\right)}\\ =-34.9\text{kPa}\cdot {\text{m}}^{\text{3}}\end{array}$

          $\begin{array}{l}=-34.9\text{kPa}\cdot {\text{m}}^{\text{3}}\times \left(\frac{1\text{kJ}}{\text{1}\text{kPa}\cdot {\text{m}}^{\text{3}}}\right)\\ =-34.9\text{kJ}\end{array}$

Thus, the boundary work for the polytropic compression process of a piston-cylinder device is $\underset{\_}{-34.9\text{kJ}}$.

Substitute $2000\text{kPa}$ for $P_3$, $0.01341{\text{m}}^{\text{3}}$ for ${\nu }_1$, and $0.01690{\text{m}}^{\text{3}}$ for ${\nu }_3$ in Equation (VI).

$\begin{array}{c}{W}_{\text{b,3-1}}=\left(2000\text{kPa}\right)\left[\left(0.01341{\text{m}}^{\text{3}}\right)-\left(0.01690{\text{m}}^{\text{3}}\right)\right]\\ =\left(2000\text{kPa}\right)\left(-0.00349{\text{m}}^{\text{3}}\right)\\ =-6.98\text{kPa}\cdot {\text{m}}^{\text{3}}\end{array}$

Thus, the boundary work for the constant pressure compression process of a piston-cylinder device is $\underset{\_}{-6.98\text{kJ}}$.

Substitute $37.18\text{kJ}$ for $W_{b,1-2}$, $-34.9\text{kJ}$ for $W_{b,2-3}$, and $-6.97\text{kJ}$ for $W_{b,3-1}$ in Equation (VII).

$\begin{array}{c}{W}_{net}=\left(37.18\text{kJ}\right)+\left(-34.9\text{kJ}\right)+\left(-6.97\text{kJ}\right)\\ =-4.69\text{kJ}\end{array}$

Thus, the net-work for cycle is the sum of the works for each process of a piston-cylinder device is $\underset{\_}{-4.69\text{kJ}}$.