Chapter 4.5, Problem 56P

To determine

The change in the internal energy of the air.

The change in the enthalpy of the air.

Answer to Problem 56P

The change in the internal energy of the air is $-78.9\text{Btu}/\text{lbm}$..

The change in the enthalpy of the air is $-110.7\text{Btu}/\text{lbm}$..

Explanation of Solution

Write the expression for the mass of the spring loaded piston cylinder device $\left(m\right)$.

$m=\frac{P_1{\nu }_1}{R{T}_1}$ (I)

Here, the initial volume of air is ${\nu }_1$, the universal gas constant is $R$, the initial temperature of the air gas is $T_1$, and the initial pressure of the air is $P_1$.

Write the final specific volume equation $\left({\nu }_2\right)$.

${\nu }_2=\frac{V_2}{m}$ (II)

Here, final volume is $V_2$.

Write the change in length of the spring $\left(\Delta x\right)$.

$\begin{array}{c}\Delta x=\frac{\Delta V}{A_p}\\ =\frac{\Delta V}{\frac{\pi }{4}{d}^2}\end{array}$ (III)

Here, change in volume is $\Delta V$, the spring constant is $k$, the area of the system is $A$, diameter of the piston is $d$, and change in displacement of the spring is $\Delta x$.

Write the linear P-v process for spring loaded piston cylinder device.

$\begin{array}{c}{P}_2=P_1-\Delta P\\ =P_1-\frac{\Delta F}{A}\end{array}$ (IV)

Substitute $k\Delta x$ for $\Delta F$ and $\frac{\pi }{4}{d}^2$ for $A$.

$P_2=P_1-\left(\frac{k\Delta x}{\frac{\pi }{4}{d}^2}\right)$ (V)

Here, the final pressure is $P_2$, the initial pressure of system is $P_1$.

Write the final temperature of the air.

$T_2=\frac{P_2{\nu }_2}{mR}$ (VI)

Here, the final pressure of the air is $P_2$, the mass of the spring loaded piston cylinder device is $m$, the universal gas constant is $R$, and the final volume of the spring loaded piston cylinder device is ${\nu }_2$.

Write the internal energy of the spring loaded piston-cylinder device.

$\begin{array}{c}\Delta u=c_{\nu }\Delta T\\ =c_{\nu }\left(T_2-T_1\right)\end{array}$ (VII)

Here, the specific heat of constant volume is $c_{\nu }$.

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

$\begin{array}{c}\Delta h=c_p\Delta T\\ =c_p\left(T_2-T_1\right)\end{array}$ (VIII)

Here, the specific heat of constant pressure is $c_p$.

Conclusion:

Refer Table A-2E(a), “Ideal-gas specific heats of various common gases” to obtain the value of gas constant, specific heat of constant volume and pressure for air gas is $0.3704\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}$, $0.171\text{Btu}/\text{lbm}\cdot \text{R}$, and $0.240\text{Btu}/\text{lbm}\cdot \text{R}$.

Substitute $0.3704\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}$ for $R$,$1{\text{ft}}^3$ for ${\nu }_1$, $460°\text{F}$ for $T_1$, and $250\text{psia}$ for $P_1$ in Equation (I).

$\begin{array}{c}m=\frac{250\text{psia}\times 1{\text{ft}}^3}{0.3704\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}\times 460°\text{F}}\\ =\frac{250\text{psia}\times 1{\text{ft}}^3}{0.3704\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}\times \left(460+460\right)\text{R}}\\ =0.7336\text{ lbm}\end{array}$

Substitute $0.7336\text{ lbm}$ for $m$ and $0.5{\text{ft}}^3$ for $V_2$ in Equation (II).

$\begin{array}{c}{\nu }_2=\frac{0.5{\text{ft}}^3}{0.7336\text{ lbm}}\\ =0.6816{\text{ft}}^3/\text{lbm}\end{array}$

Substitute $0.5{\text{ ft}}^3$ for $\Delta V$ and $10\text{in}\text{.}$ for $d$ in Equation (III).

$\begin{array}{c}\Delta x=\frac{\left(0.5{\text{ ft}}^3\right)}{\frac{\pi }{4}{\left(10\text{ in}\text{.}\right)}^2}\\ =\frac{\text{0}{\text{.5ft}}^3}{\frac{\pi }{4}{\left(10\text{ in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}^2}\\ =0.9167\text{ft}\times \frac{\text{12}\text{in}\text{.}}{1\text{ft}}\\ =11\text{in}\text{.}\end{array}$

Substitute $11\text{ in}\text{.}$ for $\Delta x$, $250\text{psia}$ for $P_2$, $5\text{lbf}/\text{in}\text{.}$ for $k$ and $0.3704\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}$ for $R$ in Equation (V).

$\begin{array}{c}{P}_2=250\text{psia}-\frac{5\text{lbf}/\text{in}\text{.}\times 11\text{ in}\text{.}}{\frac{\pi }{4}\times {\left(10\text{in}\text{.}\right)}^2}\\ =249.3\text{psia}\end{array}$

Substitute $249.3\text{psia}$ for $P_2$, $0.5{\text{ft}}^{\text{3}}$ for ${\nu }_2$,$0.7336\text{lbm}$ for $m$, and $0.3704\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}$ for $R$ in Equation (VI).

$\begin{array}{c}{T}_2=\frac{\left(249.3\text{psia}\right)\times 5\text{lbf}/\text{in}\text{.}\times 11\text{ in}\text{.}}{\left(0.7336\text{lbm}\right)\times 0.3704\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}}\\ =458.7\text{R}\end{array}$

Substitute $0.171\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_{\nu }$, $458.7\text{R}$ for $T_2$, $900\text{R}$ for $T_1$ in Equation (VII).

$\begin{array}{c}\Delta u=\left(0.171\text{Btu}/\text{lbm}\cdot \text{R}\right)\left(458.7\text{R}-920\text{R}\right)\\ =\left(0.171\text{Btu}/\text{lbm}\cdot \text{R}\right)\left(-461.3\text{ R}\right)\\ =-78.9\text{Btu}/\text{lbm}\end{array}$

Thus, the change in the internal energy of the air is $-78.9\text{Btu}/\text{lbm}$.

Substitute $1.039\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, 363 K for $T_2$, 300 K for $T_1$ in Equation (VII).

$\begin{array}{c}\Delta u=\left(0.240\text{Btu}/\text{lbm}\cdot \text{R}\right)\left(458.7\text{R}-920\text{R}\right)\\ =\left(0.240\text{Btu}/\text{lbm}\cdot \text{R}\right)\left(-461.3\text{ R}\right)\\ =-110.7\text{Btu}/\text{lbm}\end{array}$

Thus, the change in the enthalpy of the air is $-110.7\text{Btu}/\text{lbm}$.