Chapter 8.8, Problem 29P

Which is a more valuable resource for work production in a closed system −15 ft³ of air at 100 psia and 250°F or 20 ft³ of helium at 60 psia and 200°F? Take T₀ = 77°F and P₀ = 14.7 psia.

To determine

The more valuable resource for work production in a closed system; air or helium.

Answer to Problem 29P

The air has the more valuable resource for work production in a closed system.

Explanation of Solution

Express the mass in the system.

$m=\frac{P\nu }{RT}$ (I)

Here, mass is $m$, pressure is $P$, volume is $\nu$, gas constant is $R$ and temperature is $T$.

Express the entropy change between the given state and dead state.

$s-s_0=c_p\ln \frac{T}{T_0}-R\ln \frac{P}{P_0}$ (II)

Here, entropy change between the given state and dead state is $s-s_0$, specific heat at constant pressure is $c_p$, surrounding temperature is $T_0$ and surrounding pressure is $P_0$.

Express the specific volume at the given state.

$v=\frac{RT}{P}$ (III)

Express the specific volume at dead state.

$v_0=\frac{R{T}_0}{P_0}$ (IV)

Express the specific closed system exergy.

$\begin{array}{c}\varphi =u-u_0+P_0\left(v-v_0\right)-T_0\left(s-s_0\right)\\ =c_v\left(T-T_0\right)+P_0\left(v-v_0\right)-T_0\left(s-s_0\right)\end{array}$ (V)

Here, specific internal energy is $u$, specific internal energy at dead state is $u_0$ and specific heat at constant volume is $c_v$.

Express the total exergy available for the production of work.

$\Phi =m\varphi$ (VI)

Here, mass is $m$.

Conclusion:

Solve for AIR:

Refer Table A-1E, “molar mass, gas constant and critical point properties”, and write the gas constant of air.

$\begin{array}{c}R=0.06855\text{Btu}/\text{lbm}\cdot \text{R}\\ =\text{0}\text{.3704}\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}\end{array}$

Refer Table A-2E, “ideal gas specific heats of various common gases”, and write the properties of air.

$\begin{array}{l}{c}_p=0.240\text{Btu}/\text{lbm}\cdot \text{R}\\ c_v=0.171\text{Btu}/\text{lbm}\cdot \text{R}\\ k=1.4\end{array}$

Substitute $\text{100}\text{psia}$ for $P$, $\text{15}{\text{ft}}^{\text{3}}$ for $\nu$, $0.3704\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}$ for $R$ and $\text{250°F}$ for $T$ in Equation (I).

$\begin{array}{c}m=\frac{\left(\text{100}\text{psia}\right)\left(\text{15}{\text{ft}}^{\text{3}}\right)}{\left(0.3704\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}\right)\left(\text{250°F}\right)}\\ =\frac{\left(\text{100}\text{psia}\right)\left(\text{15}{\text{ft}}^{\text{3}}\right)}{\left(0.3704\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}\right)\left[\left(\text{250}+460\right)\text{R}\right]}\\ =\frac{\left(\text{100}\text{psia}\right)\left(\text{15}{\text{ft}}^{\text{3}}\right)}{\left(0.3704\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}\right)\left(\text{710}\text{R}\right)}\\ =5.704\text{lbm}\end{array}$

Substitute $0.240\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_p$, $\text{250°F}$ for $T$, $\text{77°F}$ for $T_0$, $0.06855\text{Btu}/\text{lbm}\cdot \text{R}$ for $R$, $\text{100}\text{psia}$ for $P$ and $\text{14}\text{.7}\text{psia}$ for $P_0$ in Equation (II).

$\begin{array}{c}s-s_0=\left(0.240\text{Btu}/\text{lbm}\cdot \text{R}\right)\left[\ln \frac{\text{250°F}}{\text{77°F}}\right]-\left(0.06855\text{Btu}/\text{lbm}\cdot \text{R}\right)\left[\ln \frac{\text{100}\text{psia}}{\text{14}\text{.7}\text{psia}}\right]\\ =\left(0.240\text{Btu}/\text{lbm}\cdot \text{R}\right)\left\{\begin{array}{c}\left[\ln \frac{\left(\text{250}+460\right)\text{R}}{\left(\text{77}+460\right)\text{R}}\right]-\left(0.06855\text{Btu}/\text{lbm}\cdot \text{R}\right)\\ \left[\ln \frac{\text{100}\text{psia}}{\text{14}\text{.7}\text{psia}}\right]\end{array}\right\}\\ =\left(0.240\text{Btu}/\text{lbm}\cdot \text{R}\right)\left[\ln \frac{710\text{R}}{537\text{R}}\right]-\left(0.06855\text{Btu}/\text{lbm}\cdot \text{R}\right)\left[\ln \frac{\text{100}\text{psia}}{\text{14}\text{.7}\text{psia}}\right]\\ =-0.06441\text{Btu}/\text{lbm}\cdot \text{R}\end{array}$

Substitute $\text{0}\text{.3704}\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}$ for $R$, $\text{250°F}$ for $T$, and $\text{100}\text{psia}$ for $P$ in Equation (III).

$\begin{array}{c}v=\frac{\left(\text{0}\text{.3704}\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}\right)\left(\text{250°F}\right)}{\text{100}\text{psia}}\\ =\frac{\left(\text{0}\text{.3704}\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}\right)\left(\text{250}+460\right)\text{R}}{\text{100}\text{psia}}\\ =\frac{\left(\text{0}\text{.3704}\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}\right)\left(710\text{R}\right)}{\text{100}\text{psia}}\\ =2.630{\text{ft}}^{\text{3}}/\text{lbm}\end{array}$

Substitute $\text{0}\text{.3704}\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}$ for $R$, $\text{77°F}$ for $T_0$, and $\text{14}\text{.7}\text{psia}$ for $P_0$ in Equation (IV).

$\begin{array}{c}{v}_0=\frac{\left(\text{0}\text{.3704}\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}\right)\left(\text{77°F}\right)}{\text{14}\text{.7}\text{psia}}\\ =\frac{\left(\text{0}\text{.3704}\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}\right)\left(\text{77}+460\right)\text{R}}{\text{14}\text{.7}\text{psia}}\\ =\frac{\left(\text{0}\text{.3704}\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}\right)\left(537\text{R}\right)}{\text{14}\text{.7}\text{psia}}\\ =13.53{\text{ft}}^{\text{3}}/\text{lbm}\end{array}$

Substitute $0.171\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_v$, $\text{250°F}$ for $T$, $\text{77°F}$ for $T_0$, $\text{14}\text{.7}\text{psia}$ for $P_0$, $2.630{\text{ft}}^{\text{3}}/\text{lbm}$ for $v$, $13.53{\text{ft}}^{\text{3}}/\text{lbm}$ for $v_0$, and $-0.06441\text{Btu}/\text{lbm}\cdot \text{R}$ for $s-s_0$ in Equation (V).

$\begin{array}{c}\varphi =\left\{\begin{array}{l}\left(0.171\text{Btu}/\text{lbm}\cdot \text{R}\right)\left(\text{250°F}-\text{77°F}\right)+\left(\text{14}\text{.7}\text{psia}\right)\left[\left(2.630-13.53\right){\text{ft}}^{\text{3}}/\text{lbm}\right]\\ -\text{77°F}\left(-0.06441\text{Btu}/\text{lbm}\cdot \text{R}\right)\end{array}\right\}\\ =\left\{\begin{array}{l}\left(0.171\text{Btu}/\text{lbm}\cdot \text{R}\right)\left(710\text{R}-\text{537}\text{R}\right)+\left(\text{14}\text{.7}\text{psia}\right)\left[\left(2.630-13.53\right){\text{ft}}^{\text{3}}/\text{lbm}\right]\\ -\left(\text{537}\text{R}\right)\left(-0.06441\text{Btu}/\text{lbm}\cdot \text{R}\right)\end{array}\right\}\\ =34.52\text{Btu}/\text{lbm}\end{array}$

Substitute $34.52\text{Btu}/\text{lbm}$ for $\varphi$ and $5.704\text{lbm}$ for $m$ in Equation (VI).

$\begin{array}{c}\Phi =\left(5.704\text{lbm}\right)\left(34.52\text{Btu}/\text{lbm}\right)\\ =197\text{Btu}\end{array}$ (VII)

Solve for HELIUM:

Refer Table A-1E, “molar mass, gas constant and critical point properties”, and write the gas constant of helium.

$\begin{array}{c}R=0.4961\text{Btu}/\text{lbm}\cdot \text{R}\\ =\text{2}\text{.6809}\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}\end{array}$

Refer Table A-2E, “ideal gas specific heats of various common gases”, and write the properties of helium.

$\begin{array}{l}{c}_p=1.25\text{Btu}/\text{lbm}\cdot \text{R}\\ c_v=0.753\text{Btu}/\text{lbm}\cdot \text{R}\\ k=1.667\end{array}$

Substitute $\text{60}\text{psia}$ for $P$, $\text{20}{\text{ft}}^{\text{3}}$ for $\nu$, $2.6809\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}$ for $R$ and $\text{200°F}$ for $T$ in Equation (I).

$\begin{array}{c}m=\frac{\left(\text{60}\text{psia}\right)\left(\text{20}{\text{ft}}^{\text{3}}\right)}{\left(2.6809\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}\right)\left(\text{200°F}\right)}\\ =\frac{\left(\text{60}\text{psia}\right)\left(20{\text{ft}}^{\text{3}}\right)}{\left(2.6809\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}\right)\left[\left(\text{200}+460\right)\text{R}\right]}\\ =\frac{\left(\text{60}\text{psia}\right)\left(20{\text{ft}}^{\text{3}}\right)}{\left(2.6809\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}\right)\left(\text{660}\text{R}\right)}\\ =0.6782\text{lbm}\end{array}$

Substitute $1.25\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_p$, $\text{200°F}$ for $T$, $\text{77°F}$ for $T_0$, $0.4961\text{Btu}/\text{lbm}\cdot \text{R}$ for $R$, $\text{60}\text{psia}$ for $P$ and $\text{14}\text{.7}\text{psia}$ for $P_0$ in Equation (II).

$\begin{array}{c}s-s_0=\left(1.25\text{Btu}/\text{lbm}\cdot \text{R}\right)\left[\ln \frac{\text{200°F}}{\text{77°F}}\right]-\left(0.4961\text{Btu}/\text{lbm}\cdot \text{R}\right)\left[\ln \frac{\text{60}\text{psia}}{\text{14}\text{.7}\text{psia}}\right]\\ =\left(1.25\text{Btu}/\text{lbm}\cdot \text{R}\right)\left\{\begin{array}{c}\left[\ln \frac{\left(\text{200}+460\right)\text{R}}{\left(\text{77}+460\right)\text{R}}\right]-\left(0.4961\text{Btu}/\text{lbm}\cdot \text{R}\right)\\ \left[\ln \frac{\text{60}\text{psia}}{\text{14}\text{.7}\text{psia}}\right]\end{array}\right\}\\ =\left(1.25\text{Btu}/\text{lbm}\cdot \text{R}\right)\left[\ln \frac{660\text{R}}{537\text{R}}\right]-\left(0.4961\text{Btu}/\text{lbm}\cdot \text{R}\right)\left[\ln \frac{\text{60}\text{psia}}{\text{14}\text{.7}\text{psia}}\right]\\ =-0.4400\text{Btu}/\text{lbm}\cdot \text{R}\end{array}$

Substitute $\text{2}\text{.6809}\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}$ for $R$, $\text{200°F}$ for $T$, and $\text{60}\text{psia}$ for $P$ in Equation (III).

$\begin{array}{c}v=\frac{\left(\text{2}\text{.6809}\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}\right)\left(\text{200°F}\right)}{\text{60}\text{psia}}\\ =\frac{\left(\text{2}\text{.6809}\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}\right)\left(\text{200}+460\right)\text{R}}{\text{60}\text{psia}}\\ =\frac{\left(2.6809\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}\right)\left(660\text{R}\right)}{\text{60}\text{psia}}\\ =29.49{\text{ft}}^{\text{3}}/\text{lbm}\end{array}$

Substitute $\text{2}\text{.6809}\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}$ for $R$, $\text{77°F}$ for $T_0$, and $\text{14}\text{.7}\text{psia}$ for $P_0$ in Equation (IV).

$\begin{array}{c}{v}_0=\frac{\left(\text{2}\text{.6809}\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}\right)\left(\text{77°F}\right)}{\text{14}\text{.7}\text{psia}}\\ =\frac{\left(\text{2}\text{.6809}\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}\right)\left(\text{77}+460\right)\text{R}}{\text{14}\text{.7}\text{psia}}\\ =\frac{\left(\text{2}\text{.6809}\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}\right)\left(537\text{R}\right)}{\text{14}\text{.7}\text{psia}}\\ =97.93{\text{ft}}^{\text{3}}/\text{lbm}\end{array}$

Substitute $0.753\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_v$, $\text{200°F}$ for $T$, $\text{77°F}$ for $T_0$, $\text{14}\text{.7}\text{psia}$ for $P_0$, $29.49{\text{ft}}^{\text{3}}/\text{lbm}$ for $v$, $97.93{\text{ft}}^{\text{3}}/\text{lbm}$ for $v_0$, and $-0.4400\text{Btu}/\text{lbm}\cdot \text{R}$ for $s-s_0$ in Equation (V).

$\begin{array}{c}\varphi =\left\{\begin{array}{l}\left(0.753\text{Btu}/\text{lbm}\cdot \text{R}\right)\left(\text{200°F}-\text{77°F}\right)+\left(\text{14}\text{.7}\text{psia}\right)\left[\left(29.49-97.93\right){\text{ft}}^{\text{3}}/\text{lbm}\right]\\ -\text{77°F}\left(-0.4400\text{Btu}/\text{lbm}\cdot \text{R}\right)\end{array}\right\}\\ =\left\{\begin{array}{l}\left(0.753\text{Btu}/\text{lbm}\cdot \text{R}\right)\left(660\text{R}-\text{537}\text{R}\right)+\left(\text{14}\text{.7}\text{psia}\right)\\ \left[\left(29.49-97.93\right){\text{ft}}^{\text{3}}/\text{lbm}\right]\frac{\text{Btu}}{5.404\text{psia}\cdot {\text{ft}}^{\text{3}}}-\left(\text{537}\text{R}\right)\left(-0.4400\text{Btu}/\text{lbm}\cdot \text{R}\right)\end{array}\right\}\\ =142.7\text{Btu}/\text{lbm}\end{array}$

Substitute $142.7\text{Btu}/\text{lbm}$ for $\varphi$ and $0.6782\text{lbm}$ for $m$ in Equation (VI).

$\begin{array}{c}\Phi =\left(0.6782\text{lbm}\right)\left(142.7\text{Btu}/\text{lbm}\right)\\ =96.8\text{Btu}\end{array}$ (VIII)

On comparing the total exergy available for the potential of work from Equations (VII) and (VIII), it is obtained that air has more potential.

$\begin{array}{l}\text{air}>\text{helium}\\ \Phi =197\text{Btu}>\Phi =96.8\text{Btu}\end{array}$

Hence, the air has the more valuable resource for work production in a closed system.