Chapter 12.6, Problem 81P

To determine

The exergy destruction associate with the process.

Answer to Problem 81P

The exergy destruction associate with the process is $23.3916\text{kJ/kg}$.

Explanation of Solution

Write formula for enthalpy departure factor $\left(Z_h\right)$.

$Z_h=\frac{{\left(h_{\text{ideal}}-h\right)}_{T,P}}{R{T}_{\text{cr}}}$ (I)

Here, the enthalpy at ideal gas state is $h_{\text{ideal}}$, the enthalpy and normal state is $h$, the gas constant of propane is $R$, and the critical temperature is $T_{\text{cr}}$; The subscripts $T,P$ indicates the correspondence of given temperature and pressure.

Rearrange the Equation (I) to obtain $h$.

$h=h_{\text{ideal}}-Z_{h}R{T}_{\text{cr}}$ (II)

Refer Equation (II) express as two states of enthalpy difference (final – initial).

$h_2-h_1={\left(h_2-h_1\right)}_{\text{ideal}}-\left(Z_{h2}-Z_h{}_1\right)R{T}_{\text{cr}}$ (III)

Consider the propane as the real gas and the express the equation of state.

$P\nu =ZRT$ (IV)

Here, the compressibility factor is $Z$.

Here the compressibility factor is,

$Z=Z_{\text{avg}}=\frac{Z_1+Z_2}{2}$

Rewrite the Equation (IV) as follows.

$P\nu =Z_{\text{avg}}RT$

Consider the term $Z_{\text{avg}}RT$ as constant $\left(C\right)$.

$Z_{\text{avg}}RT=C$

The internal energy is expressed as follows.

$u=h-P\nu$

Here, the enthalpy is $h$, the pressure is $P$, and the volume is $\nu$.

Write the formula for change in internal energy.

$u_2-u_1=\left(h_2-h_1\right)-R\left(Z_2{T}_2-Z_1{T}_1\right)$ (V)

While compression the boundary work is done on the system (piston-cylinder).

Write the formula for boundary work input.

$\begin{array}{c}{w}_{b,\text{in}}=-{\displaystyle {\int }_1^{2}Pd\nu }\\ =-{\displaystyle {\int }_1^2\frac{C}{\nu }d\nu }\\ =-C{\displaystyle {\int }_1^2\frac{1}{\nu }d\nu }\\ =-C\ln \frac{{\nu }_2}{{\nu }_1}\end{array}$

$\begin{array}{c}=-\left(Z_{\text{avg}}RT\right)\ln \frac{Z_{2}RT/P_2}{Z_{1}RT/P_1}\\ =-\left(Z_{\text{avg}}RT\right)\ln \frac{Z_2/P_2}{Z_1/P_1}\end{array}$ (VI)

Here, the negative sign indicates the work done on the system.

Write the energy balance equation for the system (piston-cylinder).

$\begin{array}{l}{E}_{\text{in}}-E_{\text{out}}=\Delta E_{\text{system}}\\ \left(q_{\text{in}}+w_{b,\text{in}}\right)-0=\Delta u\\ q_{\text{in}}+w_{b,\text{in}}=u_2-u_1\end{array}$ (VII)

Here, the net energy in is $E_{\text{in}}$, the net energy out is $E_{\text{out}}$ and the change in net energy of the system is $\Delta E_{\text{system}}$.

Refer Table A-1, “Molar mass, gas constant, and critical-point properties”.

The critical temperature and pressure of propane gas is as follows.

$\begin{array}{l}{T}_{\text{cr}}=370\text{K}\\ P_{\text{cr}}=4.26\text{MPa}\end{array}$

The gas constant $\left(R\right)$ of propane is $0.1885\text{kJ/kg}\cdot \text{K}$.

The specific heat at constant pressure $\left(c_p\right)$ of propane is $1.6794\text{kJ/kg}\cdot \text{K}$.

It is given that the propane is compressed isothermally. At ideal gas state the enthalpy is solely depends on temperature. Since, the process is isothermal (temperature is constant). Hence the change in enthalpy at ideal gas state becomes zero.

${\left(h_2-h_1\right)}_{\text{ideal}}=0$

The reduced pressure $\left(P_{R1}\right)$ and temperature $\left(T_{R1}\right)$ at initial state is expressed as follows.

$\begin{array}{c}{T}_{R1}=\frac{T_1}{T_{\text{cr}}}\\ =\frac{\left(100+273\right)\text{K}}{370\text{K}}\\ =1.008\end{array}$

$\begin{array}{c}{P}_{R1}=\frac{P_1}{P_{\text{cr}}}\\ =\frac{1\text{MPa}}{4.26\text{MPa}}\\ =0.235\end{array}$

The reduced pressure $\left(P_{R2}\right)$ and temperature $\left(T_{R2}\right)$ at final state is expressed as follows.

$\begin{array}{c}{T}_{R2}=\frac{T_2}{T_{\text{cr}}}\\ =\frac{\left(100+273\right)\text{K}}{370\text{K}}\\ =1.008\end{array}$

$\begin{array}{c}{P}_{R2}=\frac{P_2}{P_{\text{cr}}}\\ =\frac{4\text{MPa}}{4.26\text{MPa}}\\ =0.939\end{array}$

Write the entropy balance equation for closed system.

$S_{in}+S_{gen}-S_{out}=\Delta S_{system}$ (VIII)

Here, the entropy input is $S_{in}$, the entropy output is $S_{out}$, the entropy generation in the system is $S_{gen}$ and the change in entropy of system is $\Delta S_{system}$.

Rewrite the Equation (VII) as follows by substituting 0 for $S_{in}$, $\frac{Q_{out}}{T_{b,out}}$ for $S_{out}$ and $m\left(s_2-s_1\right)$ for $\Delta S_{system}$, for this case.

$\begin{array}{l}{S}_{gen}-\frac{Q_{out}}{T_{b,out}}=m\left(s_2-s_1\right)\\ S_{gen}=m\left(s_2-s_1\right)+\frac{Q_{out}}{T_{b,out}}\\ m{s}_{gen}=m\left(s_2-s_1\right)+\frac{m{q}_{out}}{T_{b,out}}\\ s_{gen}=\left(s_2-s_1\right)+\frac{q_{out}}{T_{b,out}}\end{array}$ (IX)

Here, mass flow rate is $\dot{m}$, amount of heat transfer is $q_{out}$ and surrounding temperature is $T_{b,out}$.

Write the formula for change in entropy $\left({\left(s_2-s_1\right)}_{ideal}\right)$ for ideal gas.

${\left(s_2-s_1\right)}_{ideal}=c_p\ln \frac{T_2}{T_1}-R\ln \frac{P_2}{P_1}$ (X)

Here, the gas constant is R, the specific heat at constant pressure is $c_p$, the initial pressure is $P_1$, the final pressure is $P_2$, the final temperature is $T_2$ and the initial temperature is $T_1$.

Write the formula for change in entropy $\left(s_2-s_1\right)$ using generalized entropy departure chart relation.

$s_2-s_1=R\left(Z_{s1}-Z_{s2}\right)+{\left(s_2-s_1\right)}_{ideal}$ (XI)

Here, the entropy departure factor is $Z_s$.

Write the formula for exergy destruction associate with process.

$x_{\text{destruction}}=T_{b,out}{s}_{gen}$

Substitute $m\left(s_2-s_1\right)+\frac{q_{out}}{T_{b,out}}$ for $s_{gen}$.

$x_{\text{destruction}}=T_{b,out}\left(s_2-s_1+\frac{q_{out}}{T_{b,out}}\right)$ (XII)

At initial:

Refer Figure A-29, “Generalized enthalpy departure chart”.

The enthalpy departure factor $\left(Z_{h1}\right)$ corresponding the reduced pressure $\left(P_{R1}\right)$ and reduced temperature $\left(T_{R1}\right)$ is $0.28$.

Refer Figure A-30, “Generalized entropy departure chart”.

The entropy departure factor $\left(Z_{s1}\right)$ corresponding the reduced pressure $\left(P_{R1}\right)$ and reduced temperature $\left(T_{R1}\right)$ is $0.21$.

Refer Figure A-15, “Nelson–Obert generalized compressibility chart”.

The compressibility factor $\left(Z_1\right)$ corresponding the reduced pressure $\left(P_{R1}\right)$ and reduced temperature $\left(T_{R1}\right)$ is $0.92$.

At final:

Refer Figure A-29, “Generalized enthalpy departure chart”.

The enthalpy departure factor $\left(Z_{h2}\right)$ corresponding the reduced pressure $\left(P_{R2}\right)$ and reduced temperature $\left(T_{R2}\right)$ is $1.8$.

Refer Figure A-30, “Generalized entropy departure chart”.

The entropy departure factor $\left(Z_{s2}\right)$ corresponding the reduced pressure $\left(P_{R2}\right)$ and reduced temperature $\left(T_{R2}\right)$ is $1.5$.

Refer Figure A-15, “Nelson–Obert generalized compressibility chart”.

The compressibility factor $\left(Z_2\right)$ corresponding the reduced pressure $\left(P_{R2}\right)$ and reduced temperature $\left(T_{R2}\right)$ is $0.50$.

Conclusion:

The average compressibility factor is,

$\begin{array}{c}{Z}_{\text{avg}}=\frac{0.92+0.50}{2}\\ =\frac{1.42}{2}\\ =0.71\end{array}$

Substitute $0$ for ${\left(h_2-h_1\right)}_{\text{ideal}}$, $1.8$ for $Z_{h2}$, $0.28$ for $Z_{h1}$, $0.1885\text{kJ/kg}\cdot \text{K}$ for $R$, and $370\text{K}$ for $T_{\text{cr}}$ in Equation (III).

$\begin{array}{c}{h}_2-h_1=\left\{\begin{array}{l}0\\ -\left[\left(1.8-0.28\right)\left(0.1885\text{kJ/kg}\cdot \text{K}\right)\left(370\text{K}\right)\right]\end{array}\right\}\\ =0-106.0124\text{kJ/kg}\\ =-106.0124\text{kJ/kg}\end{array}$

Substitute $-106.0124\text{kJ/kg}$ for $h_2-h_1$, $0.1885\text{kJ/kg}\cdot \text{K}$ for $R$, $0.5$ for $Z_2$, $373\text{K}$ for $T_2$, $0.92$ for $Z_1$, and $373\text{K}$ for $T_1$ in Equation (V).

$\begin{array}{c}{u}_2-u_1=\left\{\begin{array}{l}-106.0124\text{kJ/kg}\\ -\left(0.1885\text{kJ/kg}\cdot \text{K}\right)\left[\left(0.5\times 373\text{K}\right)-\left(0.92\times 373\text{K}\right)\right]\end{array}\right\}\\ =-106.0124\text{kJ/kg}+29.5304\text{kJ/kg}\\ =-76.482\text{kJ/kg}\\ \simeq -76.5\text{kJ/kg}\end{array}$

Substitute $0.71$ for $Z_{\text{avg}}$, $0.1885\text{kJ/kg}\cdot \text{K}$ for $R$, $373\text{K}$ for $T$, $0.5$ for $Z_2$, $4\text{MPa}$ for $P_2$, $0.92$ for $Z_1$, and $1\text{MPa}$ for $P_1$ in Equation (VI).

$\begin{array}{c}{w}_{b,\text{in}}=-\left(0.71\right)\left(0.1885\text{kJ/kg}\cdot \text{K}\right)\left(373\text{K}\right)\ln \left(\frac{0.5/4\text{MPa}}{0.92/1\text{MPa}}\right)\\ =-49.9204\text{kJ/kg}\left(-1.9960\right)\\ =99.6441\text{kJ/kg}\\ \simeq 99.6\text{kJ/kg}\end{array}$

Substitute $99.6\text{kJ/kg}$ for $w_{b,\text{in}}$ and $-76.5\text{kJ/kg}$ for $u_2-u_1$ in Equation (VII).

$\begin{array}{l}{q}_{\text{in}}+99.6\text{kJ/kg}=-76.5\text{kJ/kg}\\ q_{\text{in}}=-76.5\text{kJ/kg}-99.6\text{kJ/kg}\\ q_{\text{in}}=-176.1\text{kJ/kg}\end{array}$

Here, the negative sign indicates the heat is transferred out from the system.

$q_{\text{out}}=176.1\text{kJ/kg}$

Substitute $1.6794\text{kJ/kg}\cdot \text{K}$ for $c_p$, $373\text{K}$ for $T_2$, $373\text{K}$ for $T_1$, $0.1885\text{kJ/kg}\cdot \text{K}$ for $R$, $4\text{MPa}$ for $P_2$, and $1\text{MPa}$ for $P_1$ in Equation (IX).

$\begin{array}{c}{\left(s_2-s_1\right)}_{ideal}=\left(\begin{array}{l}\left(1.6794\text{kJ/kg}\cdot \text{K}\right)\ln \left(\frac{373\text{K}}{373\text{K}}\right)\\ -\left(0.1885\text{kJ/kg}\cdot \text{K}\right)\ln \left(\frac{4\text{MPa}}{1\text{MPa}}\right)\end{array}\right)\\ =\left(1.6794\text{kJ/kg}\cdot \text{K}\right)\left(0\right)-0.2613\text{kJ/kg}\cdot \text{K}\\ =-0.2613\text{kJ/kg}\cdot \text{K}\end{array}$

Substitute $-0.2613\text{kJ/kg}\cdot \text{K}$ for ${\left(s_2-s_1\right)}_{ideal}$, 0.21 for $Z_{s2}$, 1.5 for $Z_{s1}$ and $0.1885\text{kJ/kg}\cdot \text{K}$ for $R$ in Equation (X).

$\begin{array}{c}{s}_2-s_1=\left(0.1885\text{kJ/kg}\cdot \text{K}\right)\left(0.21-1.5\right)+\left(-0.2613\text{kJ/kg}\cdot \text{K}\right)\\ =-0.2432\text{kJ/kg}\cdot \text{K}-0.2613\text{kJ/kg}\cdot \text{K}\\ =-0.5044\text{kJ/kg}\cdot \text{K}\\ \simeq -0.504\text{kJ/kg}\cdot \text{K}\end{array}$

Substitute $30\text{°C}$ for $T_{b,out}$, $-0.504\text{kJ/kg}\cdot \text{K}$ for $s_2-s_1$, and $176.1\text{kJ/kg}$ for $q_{\text{out}}$ in Equation (XII).

$\begin{array}{c}{x}_{\text{destruction}}=\left(30\text{°C}\right)\left(-0.504\text{kJ/kg}\cdot \text{K}+\frac{176.1\text{kJ/kg}}{30\text{°C}}\right)\\ =\left(30+273\right)\text{K}\left(-0.504\text{kJ/kg}\cdot \text{K}+\frac{176.1\text{kJ/kg}}{\left(30+273\right)\text{K}}\right)\\ =303\text{K}\left(-0.504\text{kJ/kg}\cdot \text{K}+0.5812\text{kJ/kg}\cdot \text{K}\right)\\ =303\text{K}\left(0.0772\text{kJ/kg}\cdot \text{K}\right)\end{array}$

$=23.3916\text{kJ/kg}$

Thus, the exergy destruction associate with the process is $23.3916\text{kJ/kg}$.