Chapter 4.5, Problem 41P

A piston–cylinder device initially contains 0.6 m³ of saturated water vapor at 250 kPa. At this state, the piston is resting on a set of stops, and the mass of the piston is such that a pressure of 300 kPa is required to move it. Heat is now slowly transferred to the steam until the volume doubles. Show the process on a P-V diagram with respect to saturation lines and determine (a) the final temperature, (b) the work done during this process, and (c) the total heat transfer.

To determine

The final temperature of the piston cylinder device.

Answer to Problem 41P

The final temperature of the piston cylinder device is $\underset{\_}{662\text{°C}}$.

Explanation of Solution

Write the expression for the energy balance equation.

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

Here, the total energy entering the system is $E_{\text{in}}$, the total energy leaving the system is $E_{\text{out}}$, and the change in the total energy of the system is $\Delta E_{\text{system}}$.

Substitute $Q_{\text{in}}$ for $E_{\text{in}}$, $W_{b,out}$ for $E_{\text{out}}$, and $\Delta U$ for $\Delta E_{\text{system}}$ in Equation (I).

$\begin{array}{l}{Q}_{in}-W_{b,out}=\Delta U\\ Q_{in}=m\left(u_3-u_1\right)+W_{b,out}\end{array}$ (II)

Here, the mass of the piston cylinder device is $m$, the final specific internal energy is $u_3$, the initial specific internal energy is $u_1$, and the work done during the process is $W_{b,out}$.

Write the expression for the mass of the system.

$m=\frac{{\nu }_1}{v_1}$ (III)

Here, the initial volume of the system is ${\nu }_1$ and the specific volume of the system is $v_1$.

Determine the final specific volume of the piston cylinder device.

$v_3=\frac{{\nu }_3}{m}$ (IV)

The final volume of the piston cylinder device is ${\nu }_3$.

Conclusion:

From the Table (A-4 through A-6), obtain the value of initial specific volume, the specific internal energy at initial pressure of $250\text{kPa}$.

$\begin{array}{l}{\nu }_1={\nu }_{g@250\text{kPa}}=0.71873{\text{m}}^{\text{3}}/\text{kg}\\ u_1=u_{g@250\text{kPa}}=2536.8\text{kJ}/\text{kg}\end{array}$

Substitute $0.6{\text{m}}^{\text{3}}$ for ${\nu }_1$ and $0.71873{\text{m}}^{\text{3}}/\text{kg}$ for $v_1$ in Equation (III)

$\begin{array}{c}m=\frac{0.6{\text{m}}^{\text{3}}}{0.71873{\text{m}}^{\text{3}}/\text{kg}}\\ =0.8348\text{kg}\end{array}$

Substitute $1.2{\text{m}}^{\text{3}}$ for ${\nu }_3$ and $0.8348\text{kg}$ for $m$ in the Equation (IV).

$\begin{array}{c}{v}_3=\frac{\left(1.2{\text{m}}^{\text{3}}\right)}{\left(0.8348\text{kg}\right)}\\ =1.43747{\text{m}}^{\text{3}}/\text{kg}\\ \cong 1.4375{\text{m}}^{\text{3}}/\text{kg}\end{array}$

Unit conversion of final pressure from kPa to MPa.

$\begin{array}{c}{P}_3=300\text{kPa}\times \left(\frac{\text{1}\text{MPa}}{1000\text{kPa}}\right)\\ =0.30\text{MPa}\end{array}$

Refer to Table A-6, “Superheated water”, obtain the below properties at the final pressure of 0.30 MPa using interpolation method of two variables.

Write the formula of interpolation method of two variables.

$y_2=\frac{\left(x_2-x_1\right)\left(y_3-y_1\right)}{\left(x_3-x_1\right)}+y_1$ (V)

Here, the variables denote by x and y are temperature and specific volume.

Show the temperature at $600°\text{C}$ and $700°\text{C}$ as in Table (1).

S. No	specific volume ${\text{m}}^{\text{3}}/\text{kg}$ $\left(x\right)$	Temperature, $°\text{C}$ $\left(y\right)$
1	$1.34139{\text{m}}^{\text{3}}/\text{kg}$	$600°\text{C}$
2	$1.4375{\text{m}}^{\text{3}}/\text{kg}$	$y_2=?$
3	$1.4958{\text{m}}^{\text{3}}/\text{kg}$	$700°\text{C}$

Calculate final temperature at final pressure of 0.30 MPa for liquid phase using interpolation method.

Substitute $1.34139{\text{m}}^{\text{3}}/\text{kg}$ for $x_1$, $1.4375{\text{m}}^{\text{3}}/\text{kg}$ for $x_2$, $1.4958{\text{m}}^{\text{3}}/\text{kg}$ for $x_3$, $600°\text{C}$ for $y_1$, and $700°\text{C}$ for $y_3$ in Equation (V).

$\begin{array}{c}{y}_2=\frac{\left(1.4375{\text{m}}^{\text{3}}/\text{kg}-1.34139{\text{m}}^{\text{3}}/\text{kg}\right)\left(700°\text{C}-600°\text{C}\right)}{\left(1.4958{\text{m}}^{\text{3}}/\text{kg}-1.34139{\text{m}}^{\text{3}}/\text{kg}\right)}+600°\text{C}\\ =662.24°\text{C}\\ =\text{662}°\text{C}\end{array}$

From above calculation the final temperature is $\text{662}°\text{C}$ final pressure of 0.30 MPa.

Repeat the above statement for the final specific internal energy.

$u_3=3411.4\text{kJ}/\text{kg}$

Thus, the final temperature of the piston cylinder device is $\underset{\_}{662\text{°C}}$.

To determine

The work done during the piston-cylinder process.

Answer to Problem 41P

The work done during the piston-cylinder process is $\underset{\_}{180\text{kJ}}$.

Explanation of Solution

Determine the work done during the constant pressure process.

$\begin{array}{c}{W}_{b,out}={\displaystyle {\int }_2^{3}Pd\nu }\\ =P\left({\nu }_3-{\nu }_2\right)\end{array}$ (V)

Here, the final pressure is $P$.

Conclusion:

Substitute $300\text{kPa}$ for $P$, $1.2{\text{m}}^{\text{3}}$ for ${\nu }_3$, and $0.6{\text{m}}^{\text{3}}$ for ${\nu }_2$ in Equation (V).

$\begin{array}{c}{W}_{b,out}=\left(300\text{kPa}\right)\times \left(1.2-0.6\right){\text{m}}^{\text{3}}\\ =\left(300\text{kPa}\right)\times \left(0.6{\text{m}}^{\text{3}}\right)\\ =180\text{kPa}\cdot {\text{m}}^{\text{3}}\times \left(\frac{1\text{kJ}}{1\text{kPa}\cdot {\text{m}}^{\text{3}}}\right)\\ =180\text{kJ}\end{array}$

Thus, the work done during the piston-cylinder process is $\underset{\_}{180\text{kJ}}$.

To determine

The heat transfer during the piston-cylinder process.

Answer to Problem 41P

The heat transfer during the piston-cylinder process is $\underset{\_}{910\text{kJ}}$.

Explanation of Solution

Conclusion:

Substitute $0.8348\text{kg}$ for $m$, $180\text{kJ}$ for $W_{b,out}$, $3411.4\text{kJ}/\text{kg}$ for $u_2$, and $2536.8\text{kJ}/\text{kg}$ for $u_1$ in Equation (II).

$\begin{array}{c}{Q}_{\text{in}}=\left(0.8348\text{kg}\right)\left(3411.4\text{kJ}/\text{kg}-2536.8\text{kJ}/\text{kg}\right)+180\text{kJ}\\ =910\text{kJ}\end{array}$

Thus, the heat transfer during the piston-cylinder process is $\underset{\_}{910\text{kJ}}$.