Chapter 7.13, Problem 200RP

To determine

The total work produced by the turbine.

The total heat transferred to the air in the tank during the discharge.

Answer to Problem 200RP

The total work produced by the turbine is $\underset{\_}{2.999\times {10}^7\text{Btu}}$.

The total heat transferred to the air in the tank during the discharge is $\underset{\_}{-1.468\times {10}^8\text{Btu}}$.

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.

Express the mass at any time.

$m_t=\frac{PV}{RT}$ (III)

Differentiate Equation (III) with respect to temperature.

$\frac{d{m}_t}{dt}=\frac{V}{RT}\frac{dP}{dt}$

Since the compressor operates as an isentropic device, express the temperature at state 4.

$T_4=T_3{\left(\frac{P_4}{P_3}\right)}^{\frac{k-1}{k}}$

Here, temperature at state 3 is $T_3$, specific heat ratio is k, constant atmospheric pressure is $P_4$, pressure in the tank is $P_3$.

Express the conservation of mass applied to the tank.

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

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

Calculate the total power produced by the turbine after applied the first law.

$\begin{array}{c}\dot{W}=\dot{m}\left(h_3-h_4\right)\\ =\dot{m}{c}_p\left(T_3-T_4\right)\\ =-\frac{dm}{dt}{c}_p\left(T_3-T_4\right)\end{array}$ (IV)

Here, specific enthalpy at state 3 and 4 is $h_3,h_4$.

Substitute $T_3{\left(\frac{P_4}{P_4}\right)}^{\frac{k-1}{k}}$ for $T_4$ and $\frac{V}{RT}\frac{dP}{dt}$ for $\frac{d{m}_t}{dt}$ in Equation (IV).

$\begin{array}{c}\dot{W}=-\left(\frac{V}{RT}\frac{dP}{dt}\right)c_p\left(T_3-T_3{\left(\frac{P_4}{P_3}\right)}^{\frac{k-1}{k}}\right)\\ =c_p{T}_3\left[{\left(\frac{P_4}{P_3}\right)}^{\frac{k-1}{k}}-1\right]\frac{V}{R{T}_3}\frac{dP}{dt}\end{array}$ (V)

Integrate Equation (V).

$\begin{array}{c}W=\frac{c_{p}V}{R}{\displaystyle \underset{i}{\overset{f}{\int }}\left[{\left(\frac{P_2}{P_1}\right)}^{\frac{k-1}{k}}-1\right]}dP\\ =\frac{c_{p}V}{R}\left\{k{P}_f\left[1-{\left(\frac{P_1}{P_2}\right)}^{1/k}\right]-\left(P_f-P_i\right)\right\}\end{array}$ (VI)

Calculate the total heat transferred to the air in the tank during the discharge.

$\begin{array}{c}Q=\left(m_f{u}_f-m_i{u}_i\right)+h\left(m_f-m_i\right)\\ =\left(m_f-m_i\right)T\left(c_p+c_v\right)\end{array}$ (VII)

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

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).

$\begin{array}{c}W=\frac{\left(0.240\text{Btu/lbm}\cdot \text{R}\right)1\times {10}^6{\text{ft}}^{\text{3}}}{0.3704\frac{\text{psia}\cdot {\text{ft}}^{\text{3}}}{\text{lbm}\cdot \text{R}}}\left\{\left(1.4\right)10\text{atm}\left[1-{\left(\frac{1\text{atm}}{10\text{atm}}\right)}^{1/1.4}\right]-\left(10\text{atm}-1\text{atm}\right)\right\}\\ =\frac{\left(0.240\text{Btu/lbm}\cdot \text{R}\right)1\times {10}^6{\text{ft}}^{\text{3}}}{0.3704\frac{\text{psia}\cdot {\text{ft}}^{\text{3}}}{\text{lbm}\cdot \text{R}}}\left\{\begin{array}{l}\left(1.4\right)10\text{atm}\times \text{14}\text{.696}\frac{\text{psia}}{1\text{atm}}\left[1-{\left(\frac{\begin{array}{l}1\text{atm}\times \\ \text{14}\text{.696}\frac{\text{psia}}{1\text{atm}}\end{array}}{\begin{array}{l}10\text{atm}\times \\ \text{14}\text{.696}\frac{\text{psia}}{1\text{atm}}\end{array}}\right)}^{1/1.4}\right]-\\ \left(10\text{atm}\times \text{14}\text{.696}\frac{\text{psia}}{1\text{atm}}-1\text{atm}\times \text{14}\text{.696}\frac{\text{psia}}{1\text{atm}}\right)\end{array}\right\}\\ =2.999\times {10}^7\text{Btu}\end{array}$

Thus, the total work produced by the turbine is $\underset{\_}{2.999\times {10}^7\text{Btu}}$.

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 $m_f$, and $74,860\text{lbm}$ for $m_i$ in Equation (VII).

$\begin{array}{c}Q=\left(748,600\text{lbm}-74,860\text{lbm}\right)70°\text{F}\left(\begin{array}{l}0.240\text{Btu/lbm}\cdot \text{R+}\\ 0.171\text{Btu/lbm}\cdot \text{R}\end{array}\right)\\ =\left(748,600\text{lbm}-74,860\text{lbm}\right)\left(70+459.67\right)\text{R}\left(\begin{array}{l}0.240\text{Btu/lbm}\cdot \text{R+}\\ 0.171\text{Btu/lbm}\cdot \text{R}\end{array}\right)\\ =-1.468\times {10}^8\text{Btu}\end{array}$

Thus, the total heat transferred to the air in the tank during the discharge is $\underset{\_}{-1.468\times {10}^8\text{Btu}}$.