Chapter 8.8, Problem 120RP

To determine

The change in the work potential of the air stored in the tank.

The change in exergy of the air stored in the tank.

Answer to Problem 120RP

The change in the work potential of the air stored in the tank is $-\text{3}\text{.516}\times {\text{10}}^8\text{kJ}$.

The change in exergy of the air stored in the tank is $7.875\times {\text{10}}^8\text{kJ}$.

Explanation of Solution

Write the formula to calculate initial mass of air in the tank $\left(m_i\right)$.

$m_i=\frac{P_{i}V}{R{T}_i}$ (I)

Here, initial pressure of air is $P_i$, volume of the tank is $V$, initial temperature of air is $T$ and gas constant of air is $R$.

Write the formula to calculate final mass of air in the tank $\left(m_f\right)$.

$m_f=\frac{P_{f}V}{R{T}_f}$ (II)

Here, final pressure of air is $P_f$ and final temperature of air is $T_f$.

Write the formula to calculate temperature of air at state 2 using isentropic relation $\left(T_{\text{2}}\right)$.

$T_{\text{2}}=T_1{\left(\frac{P_2}{P_1}\right)}^{\frac{k-1}{k}}$

Here, temperature of air at state 1 is $T_1$, specific heat ratio of air is $k$, pressure of air at state 1 is $P_1$ and pressure of air at state 2 is $P_2$.

Apply the conservation of mass to the tank which gives the following relation.

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

Here, rate of change in mass of air is $\frac{dm}{dt}$ and initial mass flow rate of air is ${\dot{m}}_1$.

Write the equation for the rate of heat transfer using the first law of thermodynamics $\left(\dot{Q}\right)$.

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

Here, enthalpy of air is h, internal energy of air is u, volume of tank is V, specific heat capacities at constant pressure and constant volume are $c_p$ and $c_v$ respectively and change in pressure is $dP$.

From the final temperature equation and multiplying the above relation by $dt$, it gives,

$\dot{Q}dt=\frac{V{c}_v}{R}dP-c_p\left(T_1{\left(\frac{P_2}{P_1}\right)}^{\frac{k-1}{k}}\right)\frac{V}{RT}dP$

Integrate the above relation.

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

Apply the first law to the tank and compressor.

$\left(\dot{Q}-{\dot{W}}_{\text{out}}\right)dt=d\left(mu\right)-h_{1}dm$

Here, rate of work potential of the air stored in the tank is ${\dot{W}}_{\text{out}}$.

Integrate the above relation.

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

Here, change in the work potential of the air stored in the tank is $W_{\text{out}}$ and initial and final internal energies of air are $u_i$ and $u_f$ respectively.

Apply the first law and second law to the tank and compressor and the mass balance incorporated. It gives,

${\dot{W}}_{\text{rev}}=\dot{Q}\left(1-\frac{T_0}{T_R}\right)-\frac{d\left(U-T_{0}ds\right)}{dt}+\left(h-T_{0}s\right)\frac{dm}{dt}$

Here, dead state temperature is $T_0$, rate of reversible work done on the system is ${\dot{W}}_{\text{rev}}$ and change in entropy is $ds$.

Integrate the above relation.

$\begin{array}{c}{W}_{\text{rev}}=Q\left(1-\frac{T_0}{T}\right)+m_i\left[\left(u_i-h_1\right)-T_0\left(s_i-s_1\right)\right]-m_f\left[\left(u_f-h_1\right)-T_0\left(s_f-s_1\right)\right]\\ =Q\left(1-\frac{T_0}{T}\right)+m_i\left[T_i\left(c_v-c_p\right)\right]-m_f\left[T_f\left(c_v-c_p\right)-T_{0}R\ln \frac{P_f}{P_i}\right]\end{array}$ (V)

Here, reversible work done on the system is $W_{\text{rev}}$.

Conclusion:

Refer Table A-2, "Ideal-gas specific heats of various common gases", obtain the properties of air at the room temperature.

$\begin{array}{l}R=0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}\\ c_p=1.005\text{kJ}/\text{kg}\cdot \text{K}\\ c_v=0.718\text{kJ}/\text{kg}\cdot \text{K}\\ k=1.4\end{array}$

Substitute $\text{100}\text{kPa}$ for $P_i$, $\text{500,000}{\text{m}}^{\text{3}}$ for $V$, $0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}$ for $R$ and $20°\text{C}$ for $T_i$ in Equation (I).

$\begin{array}{c}{m}_i=\frac{\left(\text{100}\text{kPa}\right)\left(\text{500,000}{\text{m}}^{\text{3}}\right)}{\left(0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}\right)\left(20°\text{C}\right)}\\ =\frac{\left(\text{100}\text{kPa}\right)\left(\text{500,000}{\text{m}}^{\text{3}}\right)}{\left(0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}\right)\left(20+273\text{K}\right)}\\ =0.5946\times {10}^6\text{kg}\end{array}$

Substitute $\text{600}\text{kPa}$ for $P_f$, $\text{500,000}{\text{m}}^{\text{3}}$ for $V$, $0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}$ for $R$ and $20°\text{C}$ for $T_f$ in Equation (II).

$\begin{array}{c}{m}_f=\frac{\left(\text{600}\text{kPa}\right)\left(\text{500,000}{\text{m}}^{\text{3}}\right)}{\left(0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}\right)\left(20°\text{C}\right)}\\ =\frac{\left(\text{600}\text{kPa}\right)\left(\text{500,000}{\text{m}}^{\text{3}}\right)}{\left(0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}\right)\left(20+273\text{K}\right)}\\ =3.568\times {10}^6\text{kg}\end{array}$

Substitute $\text{100}\text{kPa}$ for $P_i$, $\text{600}\text{kPa}$ for $P_f$, $\text{500,000}{\text{m}}^{\text{3}}$ for $V$, $0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}$ for $R$, $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, $0.718\text{kJ}/\text{kg}\cdot \text{K}$ for $c_v$ and $1.4$ for $k$ in Equation (III).

$\begin{array}{c}Q=\left\{\begin{array}{l}\frac{\left(\text{500,000}{\text{m}}^{\text{3}}\right)\left(0.718\text{kJ}/\text{kg}\cdot \text{K}\right)}{0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}}\left(\text{600}\text{kPa}-\text{100}\text{kPa}\right)\\ -\frac{1.4}{2\left(1.4\right)-1}\frac{\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\left(\text{500,000}{\text{m}}^{\text{3}}\right)}{0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}}\\ \left(\left(\text{600}\text{kPa}\right){\left(\frac{\text{600}\text{kPa}}{\text{100}\text{kPa}}\right)}^{\frac{1.4-1}{1.4}}-\text{100}\text{kPa}\right)\end{array}\right\}\\ =-6.017\times {10}^8\text{kJ}\end{array}$

Substitute $-6.017\times {10}^8\text{kJ}$ for $Q$, $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, $0.718\text{kJ}/\text{kg}\cdot \text{K}$ for $c_v$, $20°\text{C}$ for $T$, $0.5946\times {10}^6\text{kg}$ for $m_i$ and $3.568\times {10}^6\text{kg}$ for $m_f$ in Equation (IV).

$\begin{array}{c}{W}_{\text{out}}=\left\{\begin{array}{c}-6.017\times {10}^8\text{kJ}+\left(1.005\text{kJ}/\text{kg}\cdot \text{K}-0.718\text{kJ}/\text{kg}\cdot \text{K}\right)\left(20°\text{C}\right)\\ \left(3.568\times {10}^6\text{kg}-0.5946\times {10}^6\text{kg}\right)\end{array}\right\}\\ =\left\{\begin{array}{c}-6.017\times {10}^8\text{kJ}+\left(1.005\text{kJ}/\text{kg}\cdot \text{K}-0.718\text{kJ}/\text{kg}\cdot \text{K}\right)\left(20+273\text{K}\right)\\ \left(3.568\times {10}^6\text{kg}-0.5946\times {10}^6\text{kg}\right)\end{array}\right\}\\ =-3.516\times {10}^8\text{kJ}\end{array}$

The negative sign shows that the work is done on the compressor.

Thus, the change in the work potential of the air stored in the tank is $-\text{3}\text{.516}\times {\text{10}}^8\text{kJ}$.

Substitute $-6.017\times {10}^8\text{kJ}$ for $Q$, $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, $0.718\text{kJ}/\text{kg}\cdot \text{K}$ for $c_v$, $20°\text{C}$ for $T_i$,$20°\text{C}$ for $T_f$,$20°\text{C}$ for $T_0$,$20°\text{C}$ for $T$, $0.5946\times {10}^6\text{kg}$ for $m_i$,$3.568\times {10}^6\text{kg}$ for $m_f$ $\text{100}\text{kPa}$ for $P_i$, $\text{600}\text{kPa}$ for $P_f$ and $0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}$ for $R$ in Equation (V).

$\begin{array}{c}{W}_{\text{rev}}=\left\{\begin{array}{l}-6.017\times {10}^8\text{kJ}\left(1-\frac{20°\text{C}}{20°\text{C}}\right)\\ +\left(0.5946\times {10}^6\text{kg}\right)\left[\left(20°\text{C}\right)\left(0.718\text{kJ}/\text{kg}\cdot \text{K}-1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\right]\\ -\left(3.568\times {10}^6\text{kg}\right)\left[\begin{array}{l}\left(20°\text{C}\right)\left(0.718\text{kJ}/\text{kg}\cdot \text{K}-1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\\ -\left(20°\text{C}\right)\left(0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}\right)\ln \frac{\text{600}\text{kPa}}{\text{100}\text{kPa}}\end{array}\right]\end{array}\right\}\\ =\left\{\begin{array}{l}-6.017\times {10}^8\text{kJ}\left(1-\frac{20+273\text{K}}{20+273\text{K}}\right)\\ +\left(0.5946\times {10}^6\text{kg}\right)\left[\left(20+273\text{K}\right)\left(0.718\text{kJ}/\text{kg}\cdot \text{K}-1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\right]\\ -\left(3.568\times {10}^6\text{kg}\right)\left[\begin{array}{l}\left(20+273\text{K}\right)\left(0.718\text{kJ}/\text{kg}\cdot \text{K}-1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\\ -\left(20+273\text{K}\right)\left(0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}\right)\ln \frac{\text{600}\text{kPa}}{\text{100}\text{kPa}}\end{array}\right]\end{array}\right\}\\ =7.875\times {10}^8\text{kJ}\end{array}$

This is the exergy change of the air stored in the tank.

Thus, the change in exergy of the air stored in the tank is $7.875\times {\text{10}}^8\text{kJ}$.