Chapter 10.9, Problem 85P

Reconsider Prob. 10–83. Determine which components of the combined cycle are the most wasteful of work potential.

10–83 A combined gas–steam power cycle uses a simple gas turbine for the topping cycle and simple Rankine cycle for the bottoming cycle. Atmospheric air enters the gas turbine at 101 kPa and 20°C, and the maximum gas cycle temperature is 1100°C. The compressor pressure ratio is 8; the compressor isentropic efficiency is 85 percent; and the gas turbine isentropic efficiency is 90 percent. The gas stream leaves the heat exchanger at the saturation temperature of the steam flowing through the heat exchanger. Steam flows through the heat exchanger with a pressure of 6000 kPa and leaves at 320°C. The steam-cycle condenser operates at 20 kPa, and the isentropic efficiency of the steam turbine is 90 percent. Determine the mass flow rate of air through the air compressor required for this system to produce 100 MW of power. Use constant specific heats for air at room temperature.

To determine

Which component of the combined cycle is the most wasteful of work potential.

Answer to Problem 85P

The combustor of the gas-steam cycle has largest exergy destruction $\left({\dot{X}}_{\text{destroyed,}\left(\text{6-7}\right)}=23,970\text{ kW}\right)$ of all other components in this cycle and that is the most wasteful of work potential.

Explanation of Solution

Show the $T-s$ diagram of the both cycle.

Refer Figure 1.

Consider the gas cycle (topping cycle) and their respective process states such as 5, 6, $6s$, 7, 8, $8s$ alone.

Write the temperature and pressure relation at isentropic state and for the process 5-6-$6s$.

$T_{6s}=T_5{\left(\frac{P_6}{P_5}\right)}^{\left(k-1\right)/k}$ (I)

Here, the temperature is $T$, the pressure is $P$, the ratio of specific heats is $k$, and the subscripts 5, 6, and $6s$ indicates the process states.

Write the formula for isentropic efficiency of compressor for the process 5-6-$6s$.

$\begin{array}{c}{\eta }_C=\frac{h_{6s}-h_5}{h_6-h_5}\\ =\frac{c_p\left(T_{6s}-T_5\right)}{c_p\left(T_6-T_5\right)}\\ =\frac{T_{6s}-T_5}{T_6-T_5}\end{array}$ (II)

Here, the enthalpy is $h$.

Rearrange and rewrite the equation (II) to obtain $T_6$.

$T_6=T_5+\frac{T_{6s}-T_5}{{\eta }_C}$ (III)

Write the temperature and pressure relation at isentropic state and for the process 7-8-$8s$.

$T_{8s}=T_7{\left(\frac{P_8}{P_7}\right)}^{\left(k-1\right)/k}$ (IV)

Write the formula for isentropic efficiency of gas turbine $\left({\eta }_{gT}\right)$ for the process 7-8-$8s$.

$\begin{array}{c}{\eta }_{gT}=\frac{h_7-h_8}{h_7-h_{8s}}\\ =\frac{c_p\left(T_7-T_8\right)}{c_p\left(T_7-T_{8s}\right)}\\ =\frac{T_7-T_8}{T_7-T_{8s}}\end{array}$ (V)

Rearrange and rewrite the equation (V) to obtain $T_8$.

$T_8=T_7-{\eta }_{gT}\left(T_7-T_{8s}\right)$ (VI)

At state 9: (heat exchanger)

The temperature $\left(T_9\right)$ is equal to the saturation temperature at the pressure of $6\text{MPa}\left(6000\text{kPa}\right)$.

$T_9=T_{\text{sat}\text{@}\text{6000}\text{kPa}}$

Refer Table A-5, “Saturated water-Pressure table”.

The saturation temperature corresponding to the pressure of $6\text{MPa}\left(6000\text{kPa}\right)$ is $275.59°\text{C}$ or $548.59\text{K}\simeq \text{548}\text{.6}\text{K}$.

Refer Figure 1.

Consider the steam cycle (bottoming cycle) and their respective process states such as 1, 2, 3, 4, $4s$ alone.

At state 1:

The water exits the condenser as a saturated liquid at the pressure of $20\text{}\text{kPa}$. Hence, the enthalpy, specific volume and entropy at state 1 is as follows.

$\begin{array}{l}{h}_1=h_{f@20\text{}\text{kPa}}\\ v_1=v_{f@20\text{}\text{kPa}}\\ s_1=s_{f@20\text{}\text{kPa}}\end{array}$

Refer Table A-5, “Saturated water-Pressure table”.

The enthalpy $\left(h_1\right)$ and specific volume $\left(v_1\right)$ at state 1 corresponding to the pressure of $20\text{}\text{kPa}$ is $251.42\text{kJ/kg}$, $0.001017{\text{m}}^3\text{/kg}$, and $0.8320\text{kJ/kg}\cdot \text{K}$ respectively.

At state 2:

Write the formula for work done by the pump during process 1-2.

$w_{\text{p,in}}=v_1\left(P_2-P_1\right)$ (VII)

Here, the specific volume is $v$, the pressure is $P$ and the subscripts 1 and 2 indicates the process states.

Write the formula for enthalpy $\left(h\right)$ at state 2.

$h_2=h_1+w_{\text{p,in}}$ (VIII)

At state 3: (Turbine inlet)

The steam enters the turbine as superheated vapour.

Refer Table A-6, “Superheated water”.

The enthalpy $\left(h_3\right)$ and entropy $\left(s_3\right)$ at state 3 corresponding to the pressure of $6\text{MPa}\left(6000\text{kPa}\right)$ and the temperature of $320°\text{C}$ is as follows.

$\begin{array}{l}{h}_3=2953.6\text{kJ/kg}\\ s_3=6.1871\text{kJ/kg}\cdot \text{K}\end{array}$

From Figure 1,

$s_3=s_{4s}=6.1871\text{kJ/kg}\cdot \text{K}$

At state 4: (Turbine exit or condenser inlet)

The steam exits the condenser as a saturated liquid at the pressure of $20\text{}\text{kPa}$.

The quality of water at the exit of the turbine is expressed as follows.

$x_{4s}=\frac{s_{4s}-s_{f,4s}}{s_{fg,4s}}$ (IX)

The enthalpy at state $4s$ is expressed as follows.

$h_{4s}=h_{f,4s}+x_{4s}{h}_{fg,4s}$ (X)

Here, the enthalpy is $h$, the entropy is $s$, the quality of the water is $x$, the suffix $f$ indicates the fluid condition, the suffix $fg$ indicates the change of vaporization phase; the subscript $4s$ indicates the ideal process and the subscript 4 indicates the actual process.

Refer Table A-5, “Saturated water-Pressure table”.

Obtain the following properties corresponding to the pressure of $20\text{}\text{kPa}$.

$\begin{array}{l}{h}_{f,4s}=251.42\text{kJ/kg}\\ h_{fg,4s}=2357.5\text{kJ/kg}\\ s_{f,4s}=0.8320\text{kJ/kg}\cdot \text{K}\\ s_{fg,4s}=7.0752\text{kJ/kg}\cdot \text{K}\end{array}$

Write the formula for isentropic efficiency of the steam turbine $\left({\eta }_{sT}\right)$.

${\eta }_{sT}=\frac{h_3-h_4}{h_3-h_{4s}}$ (XI)

Rearrange the Equation (XI) to obtain the enthalpy $h_4$.

$h_4=h_3-{\eta }_{sT}\left(h_3-h_{4s}\right)$ (XII)

Write the formula for net work output of the gas cycle.

$\begin{array}{c}{w}_{\text{net,gas cycle}}=w_{\text{gT,out}}-w_{\text{C,in}}\\ =\left(h_7-h_8\right)-\left(h_6-h_5\right)\\ =c_p\left(T_7-T_8\right)-c_p\left(T_6-T_5\right)\\ =c_p\left(T_7-T_8-T_6+T_5\right)\end{array}$ (XIII)

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

Write the formula for net work output of the steam cycle.

$\begin{array}{c}{w}_{\text{net,steam cycle}}=w_{\text{sT,out}}-w_{\text{P,in}}\\ =\left(h_3-h_4\right)-w_{\text{P,in}}\end{array}$ (XIV)

Write the general energy rate balance equation.

${\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=\Delta {\dot{E}}_{\text{system}}$ (XV)

Here, the rate of energy in is ${\dot{E}}_{\text{in}}$, the rate of energy out is ${\dot{E}}_{\text{out}}$, and rate of change in net energy of the system is $\Delta {\dot{E}}_{\text{system}}$.

Consider the heat exchanger operates on steady state. Hence, the rate of change in net energy of the system is zero.

$\Delta {\dot{E}}_{\text{system}}=0$

The Equation (XV) is reduced as follows for the heat exchanger.

$\begin{array}{l}{\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=0\\ {\dot{E}}_{\text{in}}={\dot{E}}_{\text{out}}\\ {\displaystyle \sum {\dot{m}}_{\text{in}}{h}_{\text{in}}}={\displaystyle \sum {\dot{m}}_{\text{out}}{h}_{\text{out}}}\\ {\dot{m}}_a\left(h_8-h_9\right)={\dot{m}}_w\left(h_3-h_2\right)\end{array}$

$\begin{array}{l}{\dot{m}}_a{c}_p\left(T_8-T_9\right)={\dot{m}}_w\left(h_3-h_2\right)\\ {\dot{m}}_w=\frac{{\dot{m}}_a{c}_p\left(T_8-T_9\right)}{h_3-h_2}\end{array}$ (XVI)

Here, the mass flow rate of air is ${\dot{m}}_a$, the mass flow rate of water is ${\dot{m}}_w$, and the specific heat of air at constant pressure is $c_p$.

Write the formula for mass flow rate of air through the compressor.

${\dot{m}}_a=\frac{{\dot{W}}_{\text{net}}}{w_{\text{net}}}$ (XVII)

Write the formula for the exergy destruction for the process 3-4 (turbine).

${\dot{X}}_{\text{destroyed}\left(3-4\right)}={\dot{m}}_w{T}_0\left(s_4-s_3\right)$ (XVIII)

Write the formula for the exergy destruction for the process 4-1 (condenser).

${\dot{X}}_{\text{destroyed}\left(\text{4-1}\right)}={\dot{m}}_w{T}_0\left(s_1-s_4+\frac{q_{\text{out}}}{T_{\text{sink}}}\right)$ (XIX)

Write the formula for the exergy destruction for heat exchanger.

$\begin{array}{c}{\dot{X}}_{\text{destroyed, heat exchanger}}={\dot{m}}_a{T}_0\Delta s_{89}+{\dot{m}}_w{T}_0\Delta s_{23}\\ ={\dot{m}}_a{T}_0\left(c_p\ln \frac{T_9}{T_8}\right)+{\dot{m}}_w{T}_0\left(s_3-s_2\right)\end{array}$ (XX)

Write the formula for the exergy destruction for the process 5-6 (compressor).

${\dot{X}}_{\text{destroyed}\left(5-6\right)}={\dot{m}}_a{T}_0\left(c_p\ln \frac{T_6}{T_5}-R\ln \frac{P_6}{P_5}\right)$ (XXI)

Write the formula for the exergy destruction for the process 6-7 (combustion chamber).

${\dot{X}}_{\text{destroyed}\left(6-7\right)}={\dot{m}}_a{T}_0\left(c_p\ln \frac{T_7}{T_6}-\frac{q_{\text{in,6-7}}}{T_{\text{source}}}\right)$ (XXII)

Write the formula for the exergy destruction for the process 7-8 (gas turbine).

${\dot{X}}_{\text{destroyed}\left(7-8\right)}={\dot{m}}_a{T}_0\left(c_p\ln \frac{T_8}{T_7}-R\ln \frac{P_8}{P_7}\right)$ (XXIII)

Here, the specific heat at constant pressure of air is $c_p$, the gas constant of air is $R$, the surrounding temperature is $T_0$, the source temperature is $T_{\text{source}}$, the sink temperature is $T_{\text{sink}}$,

Refer Table A-2, “Ideal-gas specific heats of various common gases”.

The specific heat at constant pressure $\left(c_p\right)$ and the specific heat ratio $\left(k\right)$ of air is $1.005\text{kJ/kg}\cdot \text{K}$ and $1.4$ respectively.

Conclusion:

Substitute $20°\text{C}$ for $T_5$, $8$ for $P_6/P_5$, and 1.4 for $k$ in Equation (I).

$\begin{array}{c}{T}_{6s}=\left(20°\text{C}\right){\left(8\right)}^{\left(1.4-1\right)/1.4}\\ =\left[\left(20+273\right)\text{K}\right]{\left(8\right)}^{0.2857}\\ =530.754\text{K}\\ \simeq 530.8\text{K}\end{array}$

Substitute $20°\text{C}$ for $T_5$, $530.8\text{K}$ for $T_{6s}$, and $0.85$ for ${\eta }_C$ in Equation (III).

$\begin{array}{c}{T}_6=20°\text{C}+\frac{530.8\text{K}-20°\text{C}}{0.85}\\ =\left(20+273\right)\text{K}+\frac{530.8\text{K}-\left(20+273\right)\text{K}}{0.85}\\ =293\text{K}+279.7647\text{K}\\ =572.7647\text{K}\end{array}$

$\simeq 572.8\text{K}$

Substitute $1100°\text{C}$ for $T_7$, $\frac{1}{8}$ for $\frac{P_8}{P_7}$, and 1.4 for $k$ in Equation (IV).

$\begin{array}{c}{T}_{8s}=\left(1100°\text{C}\right){\left(\frac{1}{8}\right)}^{\left(1.4-1\right)/1.4}\\ =\left[\left(1100+273\right)\text{K}\right]{\left(0.125\right)}^{0.2857}\\ =757.9799\text{K}\\ \simeq 758\text{K}\end{array}$

Substitute $1100°\text{C}$ for $T_7$, $758\text{K}$ for $T_{8s}$, and $0.90$ for ${\eta }_{gT}$ in Equation (VI).

$\begin{array}{c}{T}_8=1100°\text{C}-0.90\left(1100°\text{C}-758\text{K}\right)\\ =\left(1100+273\right)\text{K}-0.90\left[\left(1100+273\right)\text{K}-758\text{K}\right]\\ =1373\text{K}+553.5\text{K}\\ =819.5\text{K}\end{array}$

Substitute $0.001017{\text{m}}^3\text{/kg}$ for $v_1$, $6000\text{kPa}$ for $P_2$ and $20\text{kPa}$ for $P_1$ in Equation (VII).

$\begin{array}{c}{w}_{\text{p,in}}=\left(0.001017{\text{m}}^3\text{/kg}\right)\left(6000\text{kPa}-20\text{kPa}\right)\\ =6.0816\text{kPa}\cdot {\text{m}}^3\text{/kg}\times \frac{1\text{kJ}}{\text{kPa}\cdot {\text{m}}^3}\\ =6.0816\text{kJ/kg}\\ \simeq 6.08\text{kJ/kg}\end{array}$

Substitute $251.42\text{kJ/kg}$ for $h_1$ and $6.08\text{kJ/kg}$ for $w_{\text{p,in}}$ in Equation (VIII).

$\begin{array}{c}{h}_2=251.42\text{kJ/kg}+6.08\text{kJ/kg}\\ =257.5\text{kJ/kg}\end{array}$

Here, $s_3=s_{4s}=6.1871\text{kJ/kg}\cdot \text{K}$.

Substitute $6.1871\text{kJ/kg}\cdot \text{K}$ for $s_{4s}$, $0.8320\text{kJ/kg}\cdot \text{K}$ for $s_{f,4s}$ and $7.0752\text{kJ/kg}\cdot \text{K}$ for $s_{fg,4s}$ in Equation (IX).

$\begin{array}{c}{x}_{4s}=\frac{6.1871\text{kJ/kg}\cdot \text{K}-0.8320\text{kJ/kg}\cdot \text{K}}{7.0752\text{kJ/kg}\cdot \text{K}}\\ =0.7569\end{array}$

Substitute $251.42\text{kJ/kg}$ for $h_{f,4s}$, $0.7569$ for $x_{4s}$ and $2357.5\text{kJ/kg}$ for $h_{fg,4s}$ in

Equation (X).

$\begin{array}{c}{h}_{4s}=251.42\text{kJ/kg}+0.7569\left(2357.5\text{kJ/kg}\right)\\ =251.42\text{kJ/kg}+1784.3917\text{kJ/kg}\\ =2035.8117\text{kJ/kg}\\ \simeq \text{2035}\text{.8}\text{kJ/kg}\end{array}$

Substitute $2953.6\text{kJ/kg}$ for $h_3$, $0.90$ for ${\eta }_{sT}$, and $\text{2035}\text{.8}\text{kJ/kg}$ for $h_{4s}$ in Equation (XII).

$\begin{array}{c}{h}_4=2953.6\text{kJ/kg}-0.90\left(2953.6\text{kJ/kg}-\text{2035}\text{.8}\text{kJ/kg}\right)\\ =2953.6\text{kJ/kg}-826.02\text{kJ/kg}\\ =2127.58\text{kJ/kg}\\ \simeq 2127.6\text{kJ/kg}\end{array}$

Substitute $1.005\text{kJ/kg}\cdot \text{K}$ for $c_p$, $1373\text{K}$ for $T_7$, $819.5\text{K}$ for $T_8$, $572.8\text{K}$ for $T_6$, and $293\text{K}$ for $T_5$ in Equation (XIII).

$\begin{array}{c}{w}_{\text{net,gas cycle}}=1.005\text{kJ/kg}\cdot \text{K}\left(1373\text{K}-819.5\text{K}-572.8\text{K}+293\text{K}\right)\\ =1.005\text{kJ/kg}\cdot \text{K}\left(273.7\text{K}\right)\\ =275.0685\text{kJ/kg}\\ \simeq 275.1\text{kJ/kg}\end{array}$

Substitute $2953.6\text{kJ/kg}$ for $h_3$, $2127.6\text{kJ/kg}$ for $h_4$, and $6.08\text{kJ/kg}$ for $w_{\text{p,in}}$ in

Equation (XIV).

$\begin{array}{c}{w}_{\text{net,steam cycle}}=\left(2953.6\text{kJ/kg}-2127.6\text{kJ/kg}\right)-6.08\text{kJ/kg}\\ =826\text{kJ/kg}-6.08\text{kJ/kg}\\ =819.92\text{kJ/kg}\\ \simeq 819.9\text{kJ/kg}\end{array}$

Substitute $1.005\text{kJ/kg}\cdot \text{K}$ for $c_p$, $819.5\text{K}$ for $T_8$, $\text{548}\text{.6}\text{K}$ for $T_9$, $2953.6\text{kJ/kg}$ for $h_3$, and $257.5\text{kJ/kg}$ for $h_2$ in Equation (XVI).

$\begin{array}{c}{\dot{m}}_w=\frac{{\dot{m}}_a\left(1.005\text{kJ/kg}\cdot \text{K}\right)\left(819.5\text{K}-\text{548}\text{.6}\text{K}\right)}{2953.6\text{kJ/kg}-257.5\text{kJ/kg}}\\ =\left({\dot{m}}_a\right)\frac{272.2545\text{kJ/kg}}{2696.1\text{kJ/kg}}\\ =0.10098{\dot{m}}_a\\ \simeq 0.1010{\dot{m}}_a\end{array}$ (XXIV)

When, the mass flow rate of air is $1\text{kg/s}$.

Substitute $1\text{kg/s}$ for ${\dot{m}}_a$ in Equation (XXIV).

$\begin{array}{c}{\dot{m}}_w=0.1010\left(1\text{kg/s}\right)\\ =0.1010\text{kg/s}\end{array}$

Thus, the Equation (XXIV) describes that $1\text{kg}$ of exhaust gas can heat only the $0.1010\text{kg}$ of water. Hence, the net work output of the gas steam cycle is expressed as follows.

$\begin{array}{c}{w}_{\text{net}}=\left(1\times w_{\text{net,gas cycle}}\right)+\left(0.1010\times w_{\text{net,steam cycle}}\right)\\ =\left(1\times 275.1\text{kJ/kg}\right)+\left(0.1010\times 819.9\text{kJ/kg}\right)\\ =275.1\text{kJ/kg}+82.8099\text{kJ/kg}\\ =357.9099\text{kJ/kg}\end{array}$

Substitute $100\text{MW}$ for ${\dot{W}}_{\text{net}}$ and $357.9099\text{kJ/kg}$ for $w_{\text{net}}$ in Equation (XVII).

$\begin{array}{c}{\dot{m}}_a=\frac{100\text{MW}}{357.9099\text{kJ/kg}}\\ =\frac{100\text{MW}\times \frac{{10}^3\text{kJ/s}}{1\text{MW}}}{357.9099\text{kJ/kg}}\\ =279.3999\text{kg/s}\\ =279.4\text{kg/s}\end{array}$

$\simeq 279\text{kg/s}$

Thus, the mass flow rate of the air through the air compressor required for this system to produce $100\text{MW}$ of power is $279\text{kg/s}$.

Substitute $279.4\text{kg/s}$ for ${\dot{m}}_a$ in Equation (XXIV).

$\begin{array}{c}{\dot{m}}_w=0.1010\left(279.4\text{kg/s}\right)\\ =28.2194\text{kg/s}\\ \simeq \text{28}\text{.22}\text{kg/s}\end{array}$

Consider the process 1 to 2 (Pump).

Here, the pump is isentropic. Hence the exergy destruction during the process 1-2 is zero.

${\dot{X}}_{\text{destroyed,}\left(\text{1-2}\right)}=0$

Consider the process 3 to 4 (steam turbine).

Here,

$T_0=20°\text{C}=\text{293}\text{K}$

$s_4=6.4627\text{kJ/kg}\cdot \text{K}$

Substitute $28.22\text{kg/s}$ for ${\dot{m}}_w$, $293\text{}\text{K}$ for $T_0$, $6.4627\text{kJ/kg}\cdot \text{K}$ for $s_4$, and $6.1871\text{kJ/kg}\cdot \text{K}$ for $s_3$, in Equation (XVIII).

$\begin{array}{c}{\dot{X}}_{\text{destroyed, }\left(\text{3-4}\right)}=\left(28.22\text{kg/s}\right)\left(293\text{}\text{K}\right)\left(6.4627\text{kJ/kg}\cdot \text{K}-6.1871\text{kJ/kg}\cdot \text{K}\right)\\ =\left(28.22\text{kg/s}\right)\left(293\text{}\text{K}\right)\left(0.2756\text{kJ}/\text{kg}\cdot \text{K}\right)\\ =2278.7875\text{kJ/s}\times \frac{1\text{}\text{kW}}{1\text{kJ/s}}\\ \simeq 2278.78\text{kW}\end{array}$

Thus, the exergy destruction during process 3-4 is $2278.78\text{kW}$.

Substitute $28.21\text{kg/s}$ for ${\dot{m}}_w$, $293\text{}\text{K}$ for $T_0$, $0.8320\text{kJ/kg}\cdot \text{K}$ for $s_1$, $6.1871\text{kJ/kg}\cdot \text{K}$ for $s_4$, $1876.2\text{kJ}/\text{kg}$ for $q_{\text{out}}$, and $293\text{ K}$ for $T_{\text{sink}}$ in equation (XIX)

$\begin{array}{c}{\dot{X}}_{\text{destroyed,}\left(\text{4-1}\right)}=\left(28.21\text{kg/s}\right)\left(293\text{}\text{K}\right)\left(0.8320\text{kJ/kg}\cdot \text{K}-6.1871\text{kJ/kg}\cdot \text{K+}\frac{1876.2\text{kJ}/\text{kg}}{293\text{ K}}\right)\\ =8665\text{kJ/s}\times \frac{1\text{}\text{kW}}{1\text{kJ/s}}\\ \simeq 8665\text{kW}\end{array}$

Substitute $279.3\text{ kg/s}$ for ${\dot{m}}_a$, 293 K for $T_0$, $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, $548.6\text{ K}$ for $T_9$, $819.5\text{ K}$ for $T_8$, $28.21\text{kg}/\text{s}$ for ${\dot{m}}_w$, $293\text{ K}$ for $T_0$, $6.1871\text{kJ}/\text{kg}\cdot \text{K}$ for $s_3$, and $0.8320\text{kJ}/\text{kg}\cdot \text{K}$ for $s_2$ in equation (XX).

$\begin{array}{c}{\dot{X}}_{\text{destroyed,}\text{heat exchanger}}=\left[\begin{array}{l}\left(279.3\text{ kg/s}\right)\left(293\text{ K}\right)\times \\ \left[1.005\text{kJ}/\text{kg}\cdot \text{K}\ln \left(\frac{548.6\text{ K}}{819.5\text{ K}}\right)\right]\end{array}\right]+\left[\begin{array}{l}\left(28.21\text{kg}/\text{s}\right)\left(293\text{ K}\right)\\ \left(6.1871\text{kJ}/\text{kg}\cdot \text{K}-0.8320\text{kJ}/\text{kg}\cdot \text{K}\right)\end{array}\right]\\ =11260\text{ kW}\end{array}$

Substitute $279.3\text{ kg/s}$ for ${\dot{m}}_a$, 293 K for $T_0$, $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, $\text{572}\text{.7 K}$ for $T_6$, $293\text{ K}$ for $T_5$, $\text{0}\text{.287 }\text{kJ}/\text{kg}$ for $R$, $8$ for $\frac{P_6}{P_5}$,  in equation (XXI).

$\begin{array}{c}{\dot{X}}_{\text{destroyed,}\left(\text{5-6}\right)}=\left[\begin{array}{l}\left(279.3\text{ kg/s}\right)\left(293\text{ K}\right)\times \\ \left[1.005\text{kJ}/\text{kg}\cdot \text{K}\ln \left(\frac{572.7\text{ K}}{293\text{ K}}-0.287\text{kJ}/\text{kg}\cdot \text{K}\times \text{ ln}\left(8\right)\right)\right]\end{array}\right]\\ =6280\text{ kW}\end{array}$

Substitute $279.3\text{ kg/s}$ for ${\dot{m}}_a$, 293 K for $T_0$, $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, $\text{1373 K}$ for $T_7$, $572.7\text{ K}$ for $T_6$, $\text{804}\text{.3 }\text{kJ}/\text{kg}$ for $q_{in}$, 1373 K for $T_{source}$,  in equation (XXII).

$\begin{array}{c}{\dot{X}}_{\text{destroyed,}\left(\text{6-7}\right)}=\left[\begin{array}{l}\left(279.3\text{ kg/s}\right)\left(293\text{ K}\right)\times \\ \left[1.005\text{kJ}/\text{kg}\cdot \text{K}\ln \left(\frac{1373\text{ K}}{572.7\text{ K}}-\frac{804.3\text{ kJ/kg}}{1373\text{ K}}\right)\right]\end{array}\right]\\ =23,970\text{ kW}\end{array}$

Substitute $279.3\text{ kg/s}$ for ${\dot{m}}_a$, 293 K for $T_0$, $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, $\text{1373 K}$ for $T_7$, $0.287\text{kJ}/\text{kg}\cdot \text{K}$ for $R$, $\left(\text{1/8}\right)$ for in equation (XXIII).

$\begin{array}{c}{\dot{X}}_{\text{destroyed,}\left(\text{7-8}\right)}=\left[\begin{array}{l}\left(279.3\text{ kg/s}\right)\left(293\text{ K}\right)\times \\ \left[1.005\text{kJ}/\text{kg}\cdot \text{K}\ln \left(\frac{819.5\text{ K}}{1373\text{ K}}-0.287\text{kJ}/\text{kg}\cdot \text{K}\ln \left(\frac{1}{8}\right)\right)\right]\end{array}\right]\\ =6396\text{ kW}\end{array}$

The calculated exergy destruction value is greater for component combustor that is $23,970\text{ kW}$.

Hence, the combustor of the gas-steam cycle has largest exergy destruction of all other components and that is the most wasteful of work potential.