Chapter 7.13, Problem 199RP

To determine

The total work required to fill the tank.

The total heat transferred from the air in the tank.

Answer to Problem 199RP

The total work required to fill the tank is $\underset{\_}{-7.454\times {10}^7\text{Btu}}$.

The total heat transferred from the air in the tank is $\underset{\_}{4.990\times {10}^7\text{Btu}}$.

Explanation of Solution

Refer to Table A-2Ea, obtain the properties of air at room temperature.

$\text{Gas}\text{constant},R=0.3704\frac{\text{psia}\cdot {\text{ft}}^{\text{3}}}{\text{lbm}\cdot \text{R}}$

$\text{Constant}\text{pressure}\text{specific}\text{heat},c_p=0.240\text{Btu/lbm}\cdot \text{R}$

$\text{Constant}\text{volume}\text{specific}\text{heat},c_v=0.171\text{Btu/lbm}\cdot \text{R}$

$\text{specific}\text{heat}\text{ratio},k=1.4$

Calculate the initial mass of air in the tank.

$m_{\text{initial}}=\frac{P_{\text{initial}}V}{R{T}_{\text{initial}}}$ (I)

Here, pressure and temperature at initial state is $P_{\text{initial}}\text{and}{T}_{\text{initial}}$, and the tank volume is V.

Calculate the final mass of air in the tank.

$m_{\text{final}}=\frac{P_{\text{final}}V}{R{T}_{\text{final}}}$ (II)

Here, pressure and temperature at final state is $P_{\text{final}}\text{and}{T}_{\text{final}}$ respectively.

Since the compressor operates as an isentropic device, express the temperature at state 2.

$T_2=T_1{\left(\frac{P_2}{P_1}\right)}^{\frac{k-1}{k}}$

Here, specific heat ratio is k.

Express the conservation of mass applied to the tank.

$\frac{dm}{dt}={\dot{m}}_2$

Here, change of mass is $dm$, change in time is dt, and mass flow rate at state 2 is ${\dot{m}}_2$.

Write the equation of total power required to fill the tank using first law.

$\dot{Q}=\frac{d\left(mu\right)}{dt}-h_2\frac{dm}{dt}$ (III)

Here, specific enthalpy at state 2 is $h_2$.

Merge the ideal gas equation of state and constant specific heats in Equation (III).

$\dot{Q}=\frac{V{c}_v}{R}\frac{dP}{dt}-c_p{T}_2\frac{V}{RT}\frac{dP}{dt}$ (IV)

Use the temperature relation across the compressor and multiply by dt in Equation (IV).

$\begin{array}{c}\dot{Q}dt=\left(\frac{V{c}_v}{R}\frac{dP}{dt}-c_p{T}_2\frac{V}{RT}\frac{dP}{dt}\right)dt\\ =\frac{V{c}_v}{R}dP-c_p{T}_2\frac{V}{RT}dP\end{array}$ (V)

Here, change in pressure is dP.

Integrate Equation (V) for initial and final states.

$Q=\frac{V{c}_v}{R}\left(P_f-P_i\right)-\frac{k}{2k-1}\frac{c_{p}V}{R}\left[P_f{\left(\frac{P_f}{P_i}\right)}^{\frac{k-1}{k}}-P_i\right]$ (VI)

Here, pressures at initial and final states are $P_i,\text{and}{P}_f$ respectively.

Apply the first law to the tank and compressor.

$\left(\dot{Q}+\dot{W}\right)dt=d\left(mu\right)-h_{1}dm$ (VII)

Here, specific enthalpy at initial state is $h_1$ and

Integrate Equation (VII) to calculate the total heat transferred from the air in the tank.

$\begin{array}{l}Q+W=\left(m_f{u}_f-m_i{u}_i\right)-h_1\left(m_f-m_i\right)\\ W=\left(m_f{u}_f-m_i{u}_i\right)-h_1\left(m_f-m_i\right)-Q\\ W=-Q-\left(c_p-c_v\right)T\left(m_f-m_i\right)\end{array}$ (VIII)

Here, specific internal energy at initial and final state is $u_i,u_f$, mass at initial and final state is $m_i,m_f$.

Conclusion:

Substitute 1 atm for $P_{\text{initial}}$, $0.3704\frac{\text{psia}\cdot {\text{ft}}^{\text{3}}}{\text{lbm}\cdot \text{R}}$ for R, $1\times {10}^6{\text{ft}}^{\text{3}}$ for V, and $70°\text{F}$ for $T_{\text{initial}}$ in Equation (I).

$\begin{array}{c}{m}_{\text{initial}}=\frac{1\text{atm}\left(1\times {10}^6{\text{ft}}^{\text{3}}\right)}{\left(0.3704\frac{\text{psia}\cdot {\text{ft}}^{\text{3}}}{\text{lbm}\cdot \text{R}}\right)\left(70°\text{F}\right)}\\ =\frac{1\text{atm}\times \text{14}\text{.696}\frac{\text{psia}}{1\text{atm}}\left(1\times {10}^6{\text{ft}}^{\text{3}}\right)}{\left(0.3704\frac{\text{psia}\cdot {\text{ft}}^{\text{3}}}{\text{lbm}\cdot \text{R}}\right)\left(70+459.67\right)\text{R}}\\ =\frac{14.696\text{psia}\left(1\times {10}^6{\text{ft}}^{\text{3}}\right)}{\left(0.3704\frac{\text{psia}\cdot {\text{ft}}^{\text{3}}}{\text{lbm}\cdot \text{R}}\right)\left(530\right)\text{R}}\\ =74,860\text{lbm}\end{array}$

Substitute 10 atm for $P_{\text{final}}$, $0.3704\frac{\text{psia}\cdot {\text{ft}}^{\text{3}}}{\text{lbm}\cdot \text{R}}$ for R, $1\times {10}^6{\text{ft}}^{\text{3}}$ for V, and $70°\text{F}$ for $T_{\text{final}}$ in Equation (II).

$\begin{array}{c}{m}_{\text{final}}=\frac{10\text{atm}\left(1\times {10}^6{\text{ft}}^{\text{3}}\right)}{\left(0.3704\frac{\text{psia}\cdot {\text{ft}}^{\text{3}}}{\text{lbm}\cdot \text{R}}\right)\left(70°\text{F}\right)}\\ =\frac{10\text{atm}\times \text{14}\text{.696}\frac{\text{psia}}{1\text{atm}}\left(1\times {10}^6{\text{ft}}^{\text{3}}\right)}{\left(0.3704\frac{\text{psia}\cdot {\text{ft}}^{\text{3}}}{\text{lbm}\cdot \text{R}}\right)\left(70+459.67\right)\text{R}}\\ =\frac{146.96\text{psia}\left(1\times {10}^6{\text{ft}}^{\text{3}}\right)}{\left(0.3704\frac{\text{psia}\cdot {\text{ft}}^{\text{3}}}{\text{lbm}\cdot \text{R}}\right)\left(530\right)\text{R}}\\ =748,600\text{lbm}\end{array}$

Substitute $1\times {10}^6{\text{ft}}^{\text{3}}$ for V, $0.3704\frac{\text{psia}\cdot {\text{ft}}^{\text{3}}}{\text{lbm}\cdot \text{R}}$ for R, 10 atm for $P_{\text{final}}$, 1 atm for $P_{\text{initial}}$, $0.171\text{Btu/lbm}\cdot \text{R}$ for $c_v$, $0.240\text{Btu/lbm}\cdot \text{R}$ for $c_p$, and 1.4 for k in Equation (VI).

$\begin{array}{c}Q=\left\{\begin{array}{l}\left[\frac{\left(1\times {10}^6{\text{ft}}^{\text{3}}\right)0.171\text{Btu/lbm}\cdot \text{R}}{0.3704\frac{\text{psia}\cdot {\text{ft}}^{\text{3}}}{\text{lbm}\cdot \text{R}}}\left(10\text{atm}-1\text{atm}\right)\right]-\\ \left\{\begin{array}{l}\left(\frac{1.4}{2\left(1.4\right)-1}\right)\\ \frac{\left(0.240\text{Btu/lbm}\cdot \text{R}\right)1\times {10}^6{\text{ft}}^{\text{3}}}{\left(0.3704\frac{\text{psia}\cdot {\text{ft}}^{\text{3}}}{\text{lbm}\cdot \text{R}}\right)}\\ \left[\begin{array}{l}10\text{atm}{\left(\frac{10\text{atm}}{1\text{atm}}\right)}^{\frac{1.4-1}{1.4}}-\\ 1\text{atm}\end{array}\right]\end{array}\right\}\end{array}\right\}\\ =\left\{\begin{array}{l}\left[\frac{\left(1\times {10}^6{\text{ft}}^{\text{3}}\right)0.171\text{Btu/lbm}\cdot \text{R}}{0.3704\frac{\text{psia}\cdot {\text{ft}}^{\text{3}}}{\text{lbm}\cdot \text{R}}}\left(10\text{atm}\times \text{14}\text{.696}\frac{\text{psia}}{1\text{atm}}-1\text{atm}\times \text{14}\text{.696}\frac{\text{psia}}{1\text{atm}}\right)\right]-\\ \left\{\begin{array}{l}\left(\frac{1.4}{2\left(1.4\right)-1}\right)\\ \frac{\left(0.240\text{Btu/lbm}\cdot \text{R}\right)1\times {10}^6{\text{ft}}^{\text{3}}}{\left(0.3704\frac{\text{psia}\cdot {\text{ft}}^{\text{3}}}{\text{lbm}\cdot \text{R}}\right)}\\ \left[\begin{array}{l}10\text{atm}\times \text{14}\text{.696}\frac{\text{psia}}{1\text{atm}}{\left(\frac{10\text{atm}\times \text{14}\text{.696}\frac{\text{psia}}{1\text{atm}}}{1\text{atm}\times \text{14}\text{.696}\frac{\text{psia}}{1\text{atm}}}\right)}^{\frac{1.4-1}{1.4}}-\\ 1\text{atm}\times \text{14}\text{.696}\frac{\text{psia}}{1\text{atm}}\end{array}\right]\end{array}\right\}\end{array}\right\}\\ =-7.454\times {10}^7\text{Btu}\end{array}$

Thus, the total work required to fill the tank is $\underset{\_}{-7.454\times {10}^7\text{Btu}}$.

Substitute $70°\text{F}$ for T, $0.171\text{Btu/lbm}\cdot \text{R}$ for $c_v$, $0.240\text{Btu/lbm}\cdot \text{R}$ for $c_p$, $748,600\text{lbm}$ for $m_f$, $74,860\text{lbm}$ for $m_i$, and $-7.454\times {10}^7\text{Btu}$ for Q in Equation (VIII).

$\begin{array}{c}W=-\left(-7.454\times {10}^7\text{Btu}\right)-\left(\begin{array}{l}0.240\text{Btu/lbm}\cdot \text{R}-\\ 0.171\text{Btu/lbm}\cdot \text{R}\end{array}\right)70°\text{F}\left(\begin{array}{l}748,600\text{lbm}-\\ 74,860\text{lbm}\end{array}\right)\\ =\left(7.454\times {10}^7\text{Btu}\right)-\left(\begin{array}{l}0.240\text{Btu/lbm}\cdot \text{R}-\\ 0.171\text{Btu/lbm}\cdot \text{R}\end{array}\right)\left(70+459.67\right)\text{R}\left(\begin{array}{l}748,600\text{lbm}-\\ 74,860\text{lbm}\end{array}\right)\\ =4.990\times {10}^7\text{Btu}\end{array}$

Thus, the total heat transferred from the air in the tank is $\underset{\_}{4.990\times {10}^7\text{Btu}}$.