Chapter 3.8, Problem 36P

To determine

The values of the final temperature and enthalpy.

Answer to Problem 36P

The values of the final temperature and enthalpy are $105°\text{F}$ and $125\text{Btu}/\text{lbm}$ respectively.

Explanation of Solution

Write the initial specific volume of the piston-cylinder device $\left(v_1\right)$.

$v_1=v_f+x\left(v_g-v_f\right)$ (I)

Here, specific volume of saturated liquid is $v_f$, quality is $x$, and specific volume of saturated vapor is $v_g$.

Calculate the initial volume of the piston-cylinder device $\left(V_1\right)$.

$V_1=m{v}_1$ (II)

Here, mass of the refrigerant-134a is m.

Write the final volume $\left(V_2\right)$ for $50\%$ increase in the volume.

$V_2=1.5V_1$ (III)

Write the area of the piston $\left(A_p\right)$.

$A_p=\frac{\pi D^2}{4}$ (IV)

Here, diameter of the piston is D.

Write the distance that the piston $\left(\Delta x\right)$ slides between the initial and final circumstances.

$\begin{array}{c}\Delta x=\frac{\Delta V}{A_p}\\ =\frac{V_2-V_1}{A_p}\end{array}$ (V)

Here, change in volume is $\Delta V$, and area of the piston is $A_p$.

Write the difference in pressure $\left(\Delta P\right)$.

$\begin{array}{c}\Delta P=\frac{\Delta F}{A_p}\\ =\frac{k\Delta x}{A_p}\end{array}$ (VI)

Here, change in force is $\Delta F$, and spring constant is k.

Write the final pressure of the piston-cylinder device $\left(P_2\right)$.

$P_2=P_1+\Delta P$ (VII)

Write the final specific volume of the piston-cylinder device $\left(v_2\right)$.

$v_2=\frac{V_2}{m}$ (VIII)

Conclusion:

Refer the table A-11E, “Saturated refrigerant-134a–Temperature table”, obtain the properties of refrigerant-134a at temperature of $-30°\text{F}$.

Specific volume of saturated liquid, $v_f\text{= 0}\text{.01143}{\text{ft}}^3/\text{lbm}$.

Specific volume of saturated vapor, $v_g\text{= 4}\text{.4286}{\text{ft}}^3/\text{lbm}$.

Saturation pressure, $P_1=P_{\text{sat}}=9.868\text{psia}$.

Substitute $\text{0}\text{.01143}{\text{ft}}^3/\text{lbm}$ for $v_f$, $\text{4}\text{.4286}{\text{ft}}^3/\text{lbm}$ for $v_g$, and 0.80 for x in Equation (I).

$\begin{array}{c}{v}_1=\text{0}\text{.01143}{\text{ft}}^3/\text{lbm}+\left(0.80\right)\left(\text{4}\text{.4286}{\text{ft}}^3/\text{lbm}-\text{0}\text{.01143}{\text{ft}}^3/\text{lbm}\right)\\ =\text{3}\text{.5452}{\text{ft}}^3/\text{lbm}\end{array}$

Substitute $0.13\text{lbm}$ for m, and $\text{3}\text{.5452}{\text{ft}}^3/\text{lbm}$ for $v_1$ in Equation (II).

$\begin{array}{c}{V}_1=\left(0.13\text{lbm}\right)\left(\text{3}\text{.5452}{\text{ft}}^3/\text{lbm}\right)\\ =0.4609{\text{ft}}^3\end{array}$

Substitute $0.4609{\text{ft}}^3$ for $V_1$ in Equation (III).

$\begin{array}{c}{V}_2=1.5\left(0.4609{\text{ft}}^3\right)\\ =0.6913{\text{ft}}^3\end{array}$

Substitute $12\text{in}\text{.}$ for D in Equation (IV).

$\begin{array}{c}{A}_p=\frac{\pi {\left(12\text{in}\text{.}\right)}^2}{4}\\ =\frac{\pi {\left(12\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}^2}{4}\\ =0.7854{\text{ft}}^2\end{array}$

Substitute $0.4609{\text{ft}}^3$ for $V_1$, $0.6913{\text{ft}}^3$ for $V_2$, and $0.7854{\text{ft}}^2$ for $A_p$ in Equation (V).

$\begin{array}{c}\Delta x=\frac{0.6913{\text{ft}}^3-0.4609{\text{ft}}^3}{0.7854{\text{ft}}^2}\\ =\frac{0.2304{\text{ft}}^3}{0.7854{\text{ft}}^2}\\ =0.2934\text{ft}\end{array}$

Substitute $37\text{lbf}/\text{in}\text{.}$ for k, $0.2934\text{ft}$ for $\Delta x$, and $0.7854{\text{ft}}^2$ for $A_p$ in Equation (VI).

$\begin{array}{c}\Delta P=\frac{\left(37\text{lbf}/\text{in}\text{.}\right)\left(0.2934\text{ft}\right)}{\left(0.7854{\text{ft}}^2\right)}\\ =\frac{\left(37\text{lbf}/\text{in}\text{.}\right)\left(0.2934\text{ft}\times \frac{12\text{in}\text{.}}{1\text{ft}}\right)}{\left(0.7854{\text{ft}}^2\times {\left(\frac{12\text{in}\text{.}}{12\text{ft}}\right)}^2\right)}\\ =1.152\text{lbf}/\text{in}{\text{.}}^2\\ =1.152\text{psia}\end{array}$

Substitute $9.868\text{psia}$ for $P_1$, and $1.152\text{psia}$ for $\Delta P$ in Equation (VII).

$\begin{array}{c}{P}_2=9.868\text{psia}+1.152\text{psia}\\ =\text{11}\text{.02}\text{psia}\end{array}$

Substitute $0.6913{\text{ft}}^3$ for $V_2$, and $0.13\text{lbm}$ for m in Equation (VIII).

$\begin{array}{c}{v}_2=\frac{0.6913{\text{ft}}^3}{0.13\text{lbm}}\\ =5.318{\text{ft}}^3/\text{lbm}\end{array}$

Refer the table A-13E, “Superheated refrigerant-134a table”, obtain the value of the final temperature and final enthalpy at final pressure of $\text{11}\text{.02}\text{psia}$ and final specific volume of $5.318{\text{ft}}^3/\text{lbm}$ as $105°\text{F}$ and $125\text{Btu}/\text{lbm}$.

Thus, the values of the final temperature and enthalpy are $105°\text{F}$ and $125\text{Btu}/\text{lbm}$ respectively.