Chapter 3.8, Problem 63P

The spring-loaded piston–cylinder device shown in Fig. P3–63 is filled with 0.5 kg of water vapor that is initially at 4 MPa and 400°C. Initially, the spring exerts no force against the piston. The spring constant in the spring force relation F = kx is k = 0.9 kN/cm and the piston diameter is D = 20 cm. The water now undergoes a process until its volume is one-half of the original volume. Calculate the final temperature and the specific enthalpy of the water.

FIGURE P3–63

To determine

The final temperature of the spring-loaded piston-cylinder device.

The enthalpy of the spring-loaded piston-cylinder device.

Answer to Problem 63P

The final temperature of the spring-loaded piston-cylinder device is $\underset{\_}{220°\text{C}}$.

The enthalpy of the spring-loaded piston-cylinder device is $\underset{\_}{1721\text{kJ}/\text{kg}}$

Explanation of Solution

Write the final temperature of the spring-loaded piston-cylinder device using linear $P-\nu$ process.

$\begin{array}{c}{P}_2-P_1=c\left(v_2-v_1\right)\\ =\left(\frac{km}{A^2}\right)\left(\frac{v_1}{2}-v_1\right)\\ =\left(\frac{km}{{\left(\frac{\pi D^2}{4}\right)}^2}\right)\left(\frac{v_1}{2}-v_1\right)\\ =-\frac{\left(\frac{km}{{\left(\frac{\pi D^2}{4}\right)}^2}\right)}{2}\left(v_1\right)\end{array}$ (I)

Here, the spring constant is $k$, the mass of the water is $m$, the diameter of the piston device is $D$, and the initial specific volume is $v_1$.

Determine the final specific volume of the spring-loaded piston-cylinder device.

$v_2=\frac{v_1}{2}$ (II)

Determine the quality of final state for the spring-loaded piston-cylinder device.

$x_2=\frac{v_2-v_f}{\left(v_g-v_f\right)}$ (III)

Here, the specific volume of saturated liquid is $v_f$ and the specific volume of saturated vapour is $v_g$.

Determine the enthalpy at the final state of the spring-loaded piston-cylinder device.

$h_2=h_f+x_2{h}_{fg}$ (IV)

Here, the specific enthalpy of saturated liquid is $h_f$ and the specific volume change upon vapourization is $h_{fg}$.

Conclusion:

From the Table A-6, “Superheated water” to obtain the value of the specific volume of steam at 4 MPa of pressure and $400°\text{C}$ of temperature as $0.07343{\text{m}}^{\text{3}}/\text{kg}$.

Substitute $90\text{kN}/\text{m}$ for $k$, $0.5\text{kg}$ for $m$, $0.2\text{m}$ for $D$, $4\text{MPa}$ for $P_1$, and $0.07343{\text{m}}^{\text{3}}/\text{kg}$ for $v_1$ in Equation (I)

$\begin{array}{l}{P}_2-\left(4\text{MPa}\right)=-\frac{\left(\frac{\left(90\text{kN}/\text{m}\right)\left(0.5\text{kg}\right)}{{\left(\frac{\pi {\left(20\text{cm}\right)}^2}{4}\right)}^2}\right)}{2}\left(0.07343{\text{m}}^{\text{3}}/\text{kg}\right)\\ P_2-\left(4\text{MPa}\right)\times \left(\frac{1000\text{kPa}}{1\text{MPa}}\right)=-\frac{\left(\frac{\left(90\text{kN}/\text{m}\right)\left(0.5\text{kg}\right)}{{\left(\frac{\pi {\left(20\text{cm}\right)}^2\times \left(\frac{1\text{m}}{100\text{cm}}\right)}{4}\right)}^2}\right)}{2}\left(0.07343{\text{m}}^{\text{3}}/\text{kg}\right)\\ P_2-\left(4000\text{kPa}\right)=-\frac{\left(45,594.5\text{kN}\cdot \text{kg}/{\text{m}}^{\text{5}}\right)}{2}\left(0.07343{\text{m}}^{\text{3}}/\text{kg}\right)\\ P_2=\left(4000\text{kPa}\right)-\frac{\left(45,594.5\text{kN}\cdot \text{kg}/{\text{m}}^{\text{5}}\right)}{2}\left(0.07343{\text{m}}^{\text{3}}/\text{kg}\right)\end{array}$

$\begin{array}{c}{P}_2=\left(4000\text{kPa}\right)-\frac{\left(45,594.5\text{kN}\cdot \text{kg}/{\text{m}}^{\text{5}}\right)}{2}\left(0.07343{\text{m}}^{\text{3}}/\text{kg}\right)\\ =\left(4000\text{kPa}\right)-\left(22797.27\text{kN}\cdot \text{kg}/{\text{m}}^{\text{5}}\right)\left(0.07343{\text{m}}^{\text{3}}/\text{kg}\right)\\ =2325.9\text{kPa}\\ \cong \text{2326}\text{kPa}\end{array}$

Substitute $0.07343{\text{m}}^{\text{3}}/\text{kg}$ for $v_1$ in Equation (II).

$\begin{array}{c}{v}_2=\frac{0.07343{\text{m}}^{\text{3}}/\text{kg}}{2}\\ =0.036715{\text{m}}^{\text{3}}/\text{kg}\\ \cong 0.03672{\text{m}}^{\text{3}}/\text{kg}\end{array}$

Refer to Table A-6, “Saturated water”, obtain the below properties at the final pressure $2326\text{kPa}$ 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 pressure and temperature.

Show the pressure at $2250\text{kPa}$ and $2500\text{kPa}$ as in Table (1).

S. No	Pressure, kPa $\left(x\right)$	Temperature, $°\text{C}$ $\left(y\right)$
1	$2250\text{kPa}$	218.41
2	$2326\text{kPa}$	$y_2=?$
3	$2500\text{kPa}$	223.95

Calculate final temperature at the pressure $2326\text{kPa}$ for liquid phase using interpolation method.

Substitute $2250\text{kPa}$ for $x_1$, $2326\text{kPa}$ for $x_2$, $2500\text{kPa}$ for $x_3$, $218.41°\text{C}$ for $y_1$, and $223.95°\text{C}$ for $y_3$ in Equation (V).

$\begin{array}{c}{y}_2=\frac{\left(2326\text{kPa}-2250\text{kPa}\right)\left(223.95°\text{C}-218.41°\text{C}\right)}{\left(2500\text{kPa}-2250\text{kPa}\right)}+\left(218.41°\text{C}\right)\\ =220.09°\text{C}\\ \cong 220°\text{C}\end{array}$

From above calculation the final temperature at the pressure $2326\text{kPa}$ is $220°\text{C}$

Thus, the final temperature of the spring-loaded piston-cylinder device is $\underset{\_}{220°\text{C}}$.

Repeat the above Equation (V), to obtain the value of specific volume of saturated liquid, the specific volume of saturated vapour, specific enthalpy of saturated liquid and the specific enthalpy of saturated vapour at the final pressure $2326\text{kPa}$ as:

$\begin{array}{l}{v}_f=0.001190{\text{m}}^{\text{3}}/\text{kg}\\ v_g=0.086094{\text{m}}^{\text{3}}/\text{kg}\\ h_f=943.55\text{kJ}/\text{kg}\\ h_{fg}=1857.4\text{kJ}/\text{kg}\end{array}$

Substitute $0.001190{\text{m}}^{\text{3}}/\text{kg}$ for $v_f$, $0.086094{\text{m}}^{\text{3}}/\text{kg}$ for $v_g$, and $0.03672{\text{m}}^{\text{3}}/\text{kg}$ for $v_2$ in Equation (III).

$\begin{array}{c}{x}_2=\frac{\left(0.03672{\text{m}}^{\text{3}}/\text{kg}\right)-\left(0.001190{\text{m}}^{\text{3}}/\text{kg}\right)}{\left(0.086094{\text{m}}^{\text{3}}/\text{kg}-0.001190{\text{m}}^{\text{3}}/\text{kg}\right)}\\ =\frac{\left(0.03553{\text{m}}^{\text{3}}/\text{kg}\right)}{\left(0.086904{\text{m}}^{\text{3}}/\text{kg}\right)}\\ =0.41847\\ \cong 0.4185\end{array}$

Substitute $943.55\text{kJ}/\text{kg}$ for $h_f$, 0.4185 for $x_2$, and $1857.4\text{kJ}/\text{kg}$ for $h_{fg}$ in Equation (IV).

$\begin{array}{c}{h}_2=\left(943.55\text{kJ}/\text{kg}\right)+\left(0.4185\right)\times \left(1857.4\text{kJ}/\text{kg}\right)\\ =\left(943.55\text{kJ}/\text{kg}\right)+\left(1857.4\text{kJ}/\text{kg}\right)\\ =1720.87\text{kJ}/\text{kg}\\ \cong 1721\text{kJ}/\text{kg}\end{array}$

Thus, the enthalpy of the spring-loaded piston-cylinder device is $\underset{\_}{1721\text{kJ}/\text{kg}}$