Chapter 9.12, Problem 151P

A gas turbine operates with a regenerator and two stages of reheating and intercooling. Air enters this engine at 14 psia and 60°F, the pressure ratio for each stage of compression is 3, the air temperature when entering a turbine is 940°F, the engine produces 1000 hp, and the regenerator operates perfectly. The isentropic efficiency of each compressor is 88 percent and that of each turbine is 93 percent. Which process of the cycle loses the greatest amount of work potential? The temperature of the heat source is the same as the maximum cycle temperature, and the temperature of the heat sink is the same as the minimum cycle temperature. Use constant specific heats at room temperature.

To determine

Which process of the cycle loses the greatest amount of work potential.

Answer to Problem 151P

The exergy destruction associated with process 1-2 and 3-4 is $4.50\text{Btu}/\text{lbm}$.

The exergy destruction associated with process 5-6 and 7-8 is $4.73\text{Btu}/\text{lbm}$.

The exergy destruction associated with process 6-7 and 8-9 is $3.14\text{Btu}/\text{lbm}$.

The exergy destruction associated with process 10-1 and 2-3 is $8.61\text{Btu}/\text{lbm}$.

The exergy destruction associated at regenerator is $0\text{Btu}/\text{lbm}$.

During the heat rejection process the highest energy destruction occurs.

Explanation of Solution

Draw the $T-s$ diagram for regenerative gas turbine cycle as shown in Figure (1).

Write the expression for the temperature and pressure relation for the isentropic process 1-2s.

$T_{2s}=T_1{r}_p^{\left(k-1\right)/k}$ (I)

Here, the pressure ratio is $r_p$, the specific heat ratio is k, temperature at state 1 is $T_1$, and temperature at state 2 for isentropic process is $T_{2s}$.

Write the expression for the efficiency of the compressor $\left({\eta }_C\right)$.

$\begin{array}{l}{\eta }_C=\frac{c_p\left(T_{2s}-T_1\right)}{c_p\left(T_2-T_1\right)}\\ T_2=T_1+\frac{T_{2s}-T_1}{{\eta }_C}\end{array}$ (II)

Here, the specific heat at constant pressure is $c_p$.

Write the expression for the temperature and pressure relation ratio for the expansion process 6-7s.

$T_{7s}=T_6{\left(\frac{1}{r_p}\right)}^{\left(k-1\right)/k}$ (III)

Here, temperature at state 7s for isentropic process is $T_{7s}$ and temperature at state 6 for is $T_6$.

Write the expression for the efficiency of the turbine $\left({\eta }_T\right)$.

$\begin{array}{c}{\eta }_T=\frac{c_p\left(T_6-T_7\right)}{c_p\left(T_6-T_{7s}\right)}\\ T=T_6+{\eta }_T\left(T_6-T_{7s}\right)\end{array}$ (IV)

Here, temperature at state 7 is $T_7$,

Write the expression to calculate the heat input for the two-stage gas turbine $\left(q_{\text{in,5-6}}\right)$.

$q_{\text{in,5-6}}=c_p\left(T_6-T_5\right)$ (V)

Here, the specific heat of air at constant pressure is $c_p$ and temperature at state 5 for is $T_5$.

Write the expression to calculate the heat output for the two-stage gas turbine $\left(q_{\text{in,10-1}}\right)$.

$q_{\text{in,10-1}}=c_p\left(T_{10}-T_1\right)$ (VI)

Write the expression for the exergy destruction during the process of as steam from an inlet to exit state.

$\begin{array}{l}{x}_{\text{destroyed}}=T_0{s}_{gen}\\ x_{\text{destroyed}}=T_0\left(s_e-s_i-\frac{q_{in}}{T_{source}}+\frac{q_{in}}{T_{\text{sink}}}\right)\end{array}$

Here, entropy generation is $s_{gen}$ , entropy at inlet is $s_e$ and entropy at exit is $s_i$.

Write the expression of exergy destruction for process 1-2 $\left(x_{\text{destroyed,1-2}}\right)$.

$x_{\text{destroyed,1-2}}=x_{\text{destroyed,3-4}}=T_0\left(c_p\ln \frac{T_2}{T_1}-R\ln \frac{P_2}{P_1}\right)$ (VII)

Here, pressure at state 2 is $P_2$, the pressure at state 1 is $P_1$ and surrounding temperature is $T_0$.

Write the expression of exergy destruction for process 5-6 $\left(x_{\text{destroyed,5-6}}\right)$.

$x_{\text{destroyed,5-6}}=x_{\text{destroyed,7-8}}=T_0\left(c_p\ln \frac{T_6}{T_5}-R\ln \frac{P_6}{P_5}-\frac{q_{in,5-6}}{T_{\text{source}}}\right)$ (VIII)

Here, pressure at state 5 is $P_5$ and the pressure at state 6 is $P_6$.

Write the expression of exergy destruction for process 6-7 $\left(x_{\text{destroyed,6-7}}\right)$.

$x_{\text{destroyed,6-7}}=x_{\text{destroyed,8-9}}=T_0\left(c_p\ln \frac{T_7}{1400\text{R}}-R\ln \frac{P_7}{P_6}\right)$ (IX)

Here, pressure at state 7 is $P_7$

Write the expression of exergy destruction for process 10-1 $\left(x_{\text{destroyed,10-1}}\right)$.

$x_{\text{destroyed,10-1}}=x_{\text{destroyed,2-3}}=T_0\left(c_p\ln \frac{T_1}{T_{10}}-R\ln \frac{P_1}{P_{10}}-\frac{q_{out}}{T_{\text{sink}}}\right)$ (X)

Here, pressure at state 10 is $P_{10}$ and temperature at state 10 is $T_{10}$.

Write the expression of exergy destruction for regenerator $\left(x_{\text{destroyed,regen}}\right)$.

$\begin{array}{l}{x}_{\text{destroyed,regen}}=T_0\left(\Delta s_{4-5}+\Delta s_{9-10}\right)\\ x_{\text{destroyed,regen}}=T_0\left(c_p\ln \frac{T_5}{T_4}+c_p\ln \frac{T_{10}}{T_9}\right)\end{array}$ (XI)

Conclusion:

Substitute $60°\text{F}$ for $T_1$, $3$ for $r_p$, and 1.4 for k in Equation (I).

$\begin{array}{c}{T}_{2s}=60°\text{F}{\left(3\right)}^{1.4-1/1.4}\\ T_{2s}=\left(60+460\right)\text{R}{\left(3\right)}^{0.4/1.4}\\ T_{2s}=T_{4s}=711.7\text{R}\end{array}$

Substitute $60°\text{F}$ for $T_1$, $711.7\text{R}$ for $T_{2s}$, and 0.88 for ${\eta }_C$ in Equation (II).

$\begin{array}{c}{T}_2=60°\text{F}+\frac{711.7\text{R}-60°\text{F}}{0.88}\\ =\left(60+460\right)\text{R}+\frac{711.7\text{R}-\left(60+460\right)\text{R}}{0.88}\\ =737.8\text{K}\end{array}$

Substitute $940°\text{F}$ for $T_6$, $3$  for $r_p$ and 1.4 for k in Equation (III).

$\begin{array}{c}{T}_{7s}=940°\text{F}{\left(\frac{1}{3}\right)}^{\left(1.4-1\right)/1.4}\\ =\left(940+460\right)\text{R}{\left(\frac{1}{3}\right)}^{\left(1.4-1\right)/1.4}\\ =1023\text{R}\end{array}$

Substitute $800°\text{C}$ for $T_4$, 592.3 K for $T_{5s}$, and 0.90 for ${\eta }_T$ in Equation (IV).

$\begin{array}{c}{T}_7=\left(940°\text{F}\right)-\left(0.93\right)\left(940°\text{F}-1023\text{R}\right)\\ T_7=\left(940+460\right)\text{R}-\left(0.93\right)\left[\left(940+460\right)\text{R}-1023\text{R}\right]\\ T_7=T_9=1049\text{K}\end{array}$

The regenerator is ideal, the effectiveness is 100% and therefore, $\left(T_5=T_7\right)$ and $\left(T_{10}=T_2\right)$.

Substitute $0.240\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_p$, $1400\text{R}$ for $T_6$, and $1049\text{R}$ for $T_5$ in Equation (V).

$\begin{array}{c}{q}_{\text{in,5-6}}=0.240\text{Btu}/\text{lbm}\cdot \text{R}\left(1400\text{R}-1049\text{R}\right)\\ q_{\text{in,5-6}}=q_{\text{in,7-8}}=84.24\text{Btu}/\text{lbm}\end{array}$

Substitute $0.240\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_p$, $737.8\text{R}$ for $T_{10}$, and $520\text{R}$ for $T_1$ in Equation (VI).

$\begin{array}{c}{q}_{\text{in,10-1}}=0.240\text{Btu}/\text{lbm}\cdot \text{R}\left(737.8\text{R}-520\text{R}\right)\\ q_{\text{in,10-1}}=q_{\text{in,2-3}}=52.27\text{Btu}/\text{lbm}\end{array}$

Substitute $520\text{R}$ for $T_0$, $0.240\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_p$ , $737.8\text{R}$ for $T_2$ , $520\text{R}$ for $T_1$ , $0.0685\text{Btu}/\text{lbm}\cdot \text{R}$ for $R$ and $3$ for $\left(\frac{P_2}{P_1}\right)$ in Equation (VII).

$\begin{array}{c}{x}_{\text{destroyed,1-2}}=x_{\text{destroyed,3-4}}=520\text{R}\left(\begin{array}{l}0.240\text{Btu}/\text{lbm}\cdot \text{R}\ln \frac{737.8\text{R}}{520\text{R}}\\ -\left(0.0685\text{Btu}/\text{lbm}\cdot \text{R}\right)\ln \left(3\right)\end{array}\right)\\ x_{\text{destroyed,1-2}}=x_{\text{destroyed,3-4}}=4.50\text{Btu}/\text{lbm}\end{array}$

Thus, the exergy destruction associated with process 1-2 and 3-4 is $4.50\text{Btu}/\text{lbm}$.

Substitute $520\text{R}$ for $T_0$, $0.240\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_p$ , $1400\text{R}$ for $T_6$ , $1049\text{R}$ for $T_5$ , $0.0685\text{Btu}/\text{lbm}\cdot \text{R}$ for $R$ , $84.24\text{Btu}/\text{lbm}$ for $q_{in,5-6}$ and $1400\text{R}$ for $T_{\text{source}}$ in Equation (VIII).

$\begin{array}{l}{x}_{\text{destroyed,5-6}}=x_{\text{destroyed,7-8}}=T_0\left(\begin{array}{l}\left(0.240\text{Btu}/\text{lbm}\cdot \text{R}\right)\ln \frac{1400\text{R}}{1049\text{R}}\\ -R\ln \frac{P_6}{P_5}-\frac{84.24\text{Btu}/\text{lbm}}{1400\text{R}}\end{array}\right)\\ x_{\text{destroyed,5-6}}=x_{\text{destroyed,7-8}}=T_0\left(\begin{array}{l}\left(0.240\text{Btu}/\text{lbm}\cdot \text{R}\right)\ln \frac{1400\text{R}}{1049\text{R}}\\ -\left(0\right)-\frac{84.24\text{Btu}/\text{lbm}}{1400\text{R}}\end{array}\right)\\ x_{\text{destroyed,5-6}}=x_{\text{destroyed,7-8}}=4.73\text{Btu}/\text{lbm}\end{array}$

Thus, the exergy destruction associated with process 5-6 and 7-8 is $4.73\text{Btu}/\text{lbm}$.

Substitute $1049\text{R}$ for $T_7$, $0.240\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_p$ , $1400\text{R}$ for $T_6$ , $520\text{R}$ for $T_1$ , $0.0685\text{Btu}/\text{lbm}\cdot \text{R}$ for $R$ and $\frac{1}{3}$ for $\left(\frac{P_7}{P_6}\right)$ in Equation (IX).

$\begin{array}{l}{x}_{\text{destroyed,6-7}}=x_{\text{destroyed,8-9}}=T_0\left(\begin{array}{l}\left(0.240\text{Btu}/\text{lbm}\cdot \text{R}\right)\ln \frac{1049\text{R}}{1400\text{R}}\\ -\left(0.0685\text{Btu}/\text{lbm}\cdot \text{R}\right)\ln \left(\frac{1}{3}\right)\end{array}\right)\\ x_{\text{destroyed,6-7}}=x_{\text{destroyed,8-9}}=3.14\text{Btu}/\text{lbm}\end{array}$

Thus, the exergy destruction associated with process 6-7 and 8-9 is $3.14\text{Btu}/\text{lbm}$.

Substitute $520\text{R}$ for $T_1$, $0.240\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_p$ , $737.8\text{R}$ for $T_{10}$ , $520\text{R}$ for $T_0$ , $0.0685\text{Btu}/\text{lbm}\cdot \text{R}$ for $R$ , $52.27\text{Btu}/\text{lbm}$ for $q_{out}$ and $520\text{R}$ for $T_{\text{sink}}$ in Equation (X).

$\begin{array}{l}{x}_{\text{destroyed,10-1}}=x_{\text{destroyed,2-3}}=T_0\left(\begin{array}{l}\left(0.240\text{Btu}/\text{lbm}\cdot \text{R}\right)\ln \frac{520\text{R}}{737.8\text{R}}\\ -R\ln \frac{P_1}{P_{10}}-\frac{52.27\text{Btu}/\text{lbm}}{520\text{R}}\end{array}\right)\\ x_{\text{destroyed,10-1}}=x_{\text{destroyed,2-3}}=520\text{R}\left(\begin{array}{l}\left(0.240\text{Btu}/\text{lbm}\cdot \text{R}\right)\ln \frac{520\text{R}}{737.8\text{R}}\\ -\left(0\right)-\frac{52.27\text{Btu}/\text{lbm}}{520\text{R}}\end{array}\right)\\ x_{\text{destroyed,10-1}}=x_{\text{destroyed,2-3}}=8.61\text{Btu}/\text{lbm}\end{array}$

Thus, the exergy destruction associated with process 10-1 and 2-3 is $8.61\text{Btu}/\text{lbm}$.

Substitute $1049\text{R}$ for $T_5$, $0.240\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_p$ , $1400\text{R}$ for $T_4$ , $737.8\text{R}$ for $T_{10}$ , $1049\text{R}$ for $T_9$, and $520\text{R}$ for $T_0$ in Equation (XI).

$\begin{array}{c}{x}_{\text{destroyed,regen}}=520\text{R}\left(\begin{array}{l}\left(0.240\text{Btu}/\text{lbm}\cdot \text{R}\right)\ln \frac{1049\text{R}}{1400\text{R}}\\ +\left(0.240\text{Btu}/\text{lbm}\cdot \text{R}\right)\ln \frac{737.8\text{R}}{1049\text{R}}\end{array}\right)\\ x_{\text{destroyed,regen}}=0\text{Btu}/\text{lbm}\end{array}$

Thus, the exergy destruction associated at regenerator is $0\text{Btu}/\text{lbm}$.

During the heat rejection process the highest energy destruction occurs.