Chapter 9.12, Problem 152P

A gas-turbine power plant operates on the regenerative Brayton cycle between the pressure limits of 100 and 700 kPa. Air enters the compressor at 30°C at a rate of 12.6 kg/s and leaves at 260°C. It is then heated in a regenerator to 400°C by the hot combustion gases leaving the turbine. A diesel fuel with a heating value of 42,000 kJ/kg is burned in the combustion chamber with a combustion efficiency of 97 percent. The combustion gases leave the combustion chamber at 871°C and enter the turbine, whose isentropic efficiency is 85 percent. Treating combustion gases as air and using constant specific heats at 500°C, determine (a) the isentropic efficiency of the compressor, (b) the effectiveness of the regenerator, (c) the air–fuel ratio in the combustion chamber, (d) the net power output and the back work ratio, (e) the thermal efficiency, and (f) the second-law efficiency of the plant. Also determine (g) the second-law efficiencies of the compressor, the turbine, and the regenerator, and (h) the rate of the exergy flow with the combustion gases at the regenerator exit.

To determine

The isentropic efficiency of the compressor.

Answer to Problem 152P

The isentropic efficiency of the compressor is $88.1\%$.

Explanation of Solution

Draw the layout of the gas-turbine plant functioning on the regenerative Brayton cycle as shown in Figure (1).

Consider, the pressure is $P_i$ , the specific volume is $v_i$, the temperature is $T_i$, the entropy is $s_i$, the enthalpy is $h_i$ corresponding to $i^{th}$ state.

Write the expression to calculate the temperature and pressure relation ratio for the isentropic compression process 1-2s.

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

Here, the specific heat ratio is k.

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

${\eta }_C=\frac{\left(T_{2s}-T_1\right)}{\left(T_2-T_1\right)}$ (II)

Write the expression to calculate the temperature and pressure relation ratio for the expansion process 3-4s.

$T_{4s}=T_3{\left(\frac{P_4}{P_3}\right)}^{\left(k-1\right)/k}$ (III)

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

$\begin{array}{l}{\eta }_T=\frac{\left(T_3-T_4\right)}{\left(T_3-T_{4s}\right)}\\ T_4=T_3-{\eta }_T\left(T_3-T_{4s}\right)\end{array}$ (IV)

Conclusion:

From Table A-2b, “Ideal-gas specific heats of various common gases”, obtain the following values of air at $500°\text{C}\left(773\text{K}\right)$.

$\begin{array}{l}{c}_p=1.093\text{kJ}/\text{kg}\cdot \text{K}\\ c_v=0.806\text{kJ}/\text{kg}\cdot \text{K}\\ R=0.287\text{kJ}/\text{kg}\cdot \text{K}\\ k=1.357\end{array}$

Substitute 303 K for $T_1$, 700 kPa for $P_2$, 100 kPa for $P_1$ and 1.357 for k to find $T_{2s}$ in Equation (I).

$\begin{array}{c}{T}_{2s}=\left(303\text{K}\right){\left(\frac{700\text{kPa}}{100\text{kPa}}\right)}^{1.357-1/1.357}\\ =505.6\text{K}\end{array}$

Substitute 303 K for $T_1$, 505.6 K for $T_{2s}$, and 533 K for $T_2$ in Equation (II).

$\begin{array}{c}{\eta }_C=\frac{\left(505.6-303\right)\text{K}}{\left(533-303\right)\text{K}}\\ =0.881\times 100\%\\ =88.1\%\end{array}$

Thus, the isentropic efficiency of the compressor is $88.1\%$.

Substitute 1144 K for $T_3$, 100 kPa for $P_4$, 700 kPa for $P_3$, and 1.4 for k to find $T_{4s}$ in Equation (III).

$\begin{array}{c}{T}_{4s}=\left(1144\text{K}\right){\left(\frac{100\text{kPa}}{700\text{kPa}}\right)}^{1.357-1/1.357}\\ =685.6\text{K}\end{array}$

Substitute 1144 K for $T_3$, 685.6 K for $T_{4s}$, and 0.85 for ${\eta }_T$ in Equation (IV).

$\begin{array}{c}{T}_4=1144\text{K}-\left(0.85\right)\left(1144-685.6\right)\text{K}\\ =754.4\text{K}\end{array}$

To determine

The effectiveness of the regenerator for regenerative Brayton cycle.

Answer to Problem 152P

The effectiveness of the regenerator for regenerative Brayton cycle is $0.632$.

Explanation of Solution

Write the expression to calculate the effectiveness of the regenerator $\left({\epsilon }_{\text{regen}}\right)$.

${\epsilon }_{\text{regen}}=\frac{T_5-T_2}{T_4-T_2}$ (V)

Conclusion:

Substitute 673 K for $T_5$, 533 K for $T_2$, and 754.4 K for $T_4$ in Equation (V).

$\begin{array}{c}{\epsilon }_{\text{regen}}=\frac{\left(673-533\right)\text{K}}{\left(754.4-533\right)\text{K}}\\ =0.632\end{array}$

Thus, the effectiveness of the regenerator for regenerative Brayton cycle is $0.632$.

To determine

The air-fuel ratio in the combustion chamber.

Answer to Problem 152P

The air-fuel ratio in the combustion chamber is $78.16$.

Explanation of Solution

Write the expression for the heat input for the regenerative Brayton cycle $\left({\dot{Q}}_{\text{in}}\right)$.

${\dot{Q}}_{\text{in}}={\dot{m}}_f{q}_{\text{HV}}{\eta }_c=\left({\dot{m}}_f+{\dot{m}}_a\right)c_p\left(T_3-T_5\right)$ (VI)

Here, the specific heat at constant pressure is $c_p$, the mass flow rate of diesel fuel is ${\dot{m}}_f$, heating value of diesel fuel is $q_{\text{HV}}$, combustion efficiency is ${\eta }_c$, and mass flow rate of air is ${\dot{m}}_a$.

Write the expression to calculate the air-fuel ratio in the combustion chamber (AF).

$\text{AF =}\frac{{\dot{m}}_a}{{\dot{m}}_f}$ (VII)

Write the expression to calculate the total mass of the air-fuel mixture $\left(\dot{m}\right)$.

$\dot{m}={\dot{m}}_f+{\dot{m}}_a$ (VIII)

Write the expression to calculate the heat input for the regenerative cycle $\left({\dot{Q}}_{\text{in}}\right)$.

${\dot{Q}}_{\text{in}}={\dot{m}}_f{q}_{\text{HV}}{\eta }_c$ (IX)

Conclusion:

Substitute $42,000\text{kJ}/\text{kg}$ for $q_{\text{HV}}$, 0.97 for ${\eta }_c$, $1.093\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, $12.6\text{kg}/\text{s}$ for ${\dot{m}}_a$, 1144 K for $T_3$, and 673 K for $T_5$ in Equation (VI).

$\begin{array}{c}{\dot{m}}_f\left(42,000\text{kJ}/\text{kg}\right)\left(0.97\right)=\left({\dot{m}}_f+12.6\right)\text{kg}/\text{s}\left(1.093\text{kJ}/\text{kg}\cdot \text{K}\right)\left(1144-673\right)\text{K}\\ \text{79}\text{.14}{\dot{m}}_f={\dot{m}}_f+12.6\\ \text{78}\text{.14}{\dot{m}}_f=12.6\\ {\dot{m}}_f=0.1612\text{kg}/\text{s}\end{array}$

Substitute $12.6\text{kg}/\text{s}$ for ${\dot{m}}_a$, and $0.1612\text{kg}/\text{s}$ for ${\dot{m}}_f$ in Equation (VII).

$\begin{array}{c}\text{AF=}\frac{12.6\text{kg}/\text{s}}{0.1612\text{kg}/\text{s}}\\ =78.16\end{array}$

Thus, the air-fuel ratio in the combustion chamber is $78.16$.

Substitute $12.6\text{kg}/\text{s}$ for ${\dot{m}}_a$, and $0.1612\text{kg}/\text{s}$ for ${\dot{m}}_f$ in Equation (VIII).

$\begin{array}{c}\dot{m}=\left(12.6+0.1612\right)\text{kg}/\text{s}\\ =12.76\text{kg}/\text{s}\end{array}$

Substitute $0.1612\text{kg}/\text{s}$ for ${\dot{m}}_f$, $42,000\text{kJ}/\text{kg}$ for $q_{\text{HV}}$, and 0.97 for ${\eta }_c$ in Equation (IX).

$\begin{array}{c}{\dot{Q}}_{\text{in}}=\left(0.1612\text{kg}/\text{s}\right)\left(42,000\text{kJ}/\text{kg}\right)\left(0.97\right)\\ =6567\text{kW}\end{array}$

To determine

The net power developed by the gas-turbine plant and the back work ratio for the gas-turbine plant.

Answer to Problem 152P

The net power developed by the gas-turbine plant is $\text{2266 kW}$.

The back work ratio for the gas-turbine plant is $0.583$.

Explanation of Solution

Write the expression to calculate the power given to the compressor $\left({\dot{W}}_{\text{C,in}}\right)$.

${\dot{W}}_{\text{C,in}}={\dot{m}}_a{c}_p\left(T_2-T_1\right)$ (X)

Write the expression to calculate the power developed by the turbine $\left({\dot{W}}_{\text{T,out}}\right)$.

${\dot{W}}_{\text{T,out}}=\dot{m}{c}_p\left(T_3-T_4\right)$ (XI)

Write the expression to calculate the net power developed by the gas-turbine plant $\left({\dot{W}}_{\text{net}}\right)$.

${\dot{W}}_{\text{net}}={\dot{W}}_{\text{T,out}}-{\dot{W}}_{\text{C,in}}$ (XII)

Write the expression to calculate the back work ratio for the gas-turbine plant $\left(r_{\text{bw}}\right)$.

$r_{\text{bw}}=\frac{{\dot{W}}_{\text{C,in}}}{{\dot{W}}_{\text{T,out}}}$ (XIII)

Conclusion:

Substitute $12.6\text{kg}/\text{s}$ for ${\dot{m}}_a$, $1.093\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, 533 K for $T_2$, and 303 K for $T_1$ in Equation (X).

$\begin{array}{c}{\dot{W}}_{\text{C,in}}=\left(12.6\text{kg}/\text{s}\right)\left(1.093\text{kJ}/\text{kg}\cdot \text{K}\right)\left(533-303\right)\text{K}\\ \text{= 3168}\text{kW}\end{array}$

Substitute $12.76\text{kg}/\text{s}$ for $m_a$, $1.093\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, 1,144 K for $T_3$, and 754.4 K for $T_4$ in Equation (XI).

$\begin{array}{c}{\dot{W}}_{\text{T,out}}=\left(12.76\text{kg}/\text{s}\right)\left(1.093\text{kJ}/\text{kg}\cdot \text{K}\right)\left(1144-754.4\right)\text{K}\\ \text{= 5434}\text{kW}\end{array}$

Substitute 3168 kW for ${\dot{W}}_{\text{T,out}}$, and 5434 kW for ${\dot{W}}_{\text{C,in}}$ in Equation (XII).

$\begin{array}{c}{\dot{W}}_{\text{net}}=\left(5434-3168\right)\text{kW}\\ =\text{2266 kW}\end{array}$

Thus, the net power developed by the gas-turbine plant is $\text{2266 kW}$.

Substitute 3168 kW for ${\dot{W}}_{\text{T,out}}$, and 5434 kW for ${\dot{W}}_{\text{C,in}}$ in Equation (XIII).

$\begin{array}{c}{r}_{\text{bw}}=\frac{3168\text{kW}}{5434\text{kW}}\\ =0.583\end{array}$

Thus, the back work ratio for the gas-turbine plant is $0.583$.

To determine

The thermal efficiency of the gas-turbine plant.

Answer to Problem 152P

The thermal efficiency of the gas-turbine plant is $34.5\%$.

Explanation of Solution

Write the expression to calculate the thermal efficiency of the gas-turbine plant $\left({\eta }_{\text{th}}\right)$.

${\eta }_{\text{th}}=\frac{{\dot{W}}_{\text{net}}}{{\dot{Q}}_{\text{in}}}$ (XIV)

Conclusion:

Substitute 2266 kW for ${\dot{W}}_{\text{net}}$, and 6567 kW for ${\dot{Q}}_{\text{in}}$ in Equation (XIV).

$\begin{array}{c}{\eta }_{\text{th}}=\frac{2266\text{kW}}{6567\text{kW}}\\ =0.345\times 100\%\\ =34.5\%\end{array}$

Thus, the thermal efficiency of the gas-turbine plant is $34.5\%$.

To determine

The second-law efficiency of the gas-turbine plant.

Answer to Problem 152P

The second-law efficiency of the gas-turbine plant is $46.9\%$.

Explanation of Solution

Write the expression to calculate the second-law efficiency of the gas-turbine plant $\left({\eta }_{{\rm I}{\rm I}}\right)$.

${\eta }_{{\rm I}{\rm I}}=\frac{{\eta }_{\text{th}}}{{\eta }_{\max }}$ (XV)

Here, the maximum possible efficiency of the gas-turbine plant is ${\eta }_{\max }$.

Write the expression to calculate the maximum possible efficiency of the gas-turbine plant.

${\eta }_{\max }=1-\frac{T_1}{T_3}$ (XVI)

Conclusion:

Substitute 303 K for $T_1$, and 1144 K for $T_3$ in Equation (XV).

$\begin{array}{c}{\eta }_{\max }=1-\frac{303\text{K}}{1144\text{K}}\\ =0.735\end{array}$

Substitute 0.735 for ${\eta }_{\max }$, and 0.345 for ${\eta }_{\text{th}}$ in Equation (4) to find ${\eta }_{{\rm I}{\rm I}}$ in Equation (XVI).

$\begin{array}{c}{\eta }_{{\rm I}{\rm I}}=\frac{0.345}{0.735}\\ =0.469\\ =46.9\%\end{array}$

Thus, the second-law efficiency of the gas-turbine plant is $46.9\%$.

To determine

The exergy efficiency for compressor , turbine and regenerator.

Answer to Problem 152P

The exergy efficiency for the compressor is $92.9\%$.

The exergy efficiency for the turbine is $93.2\%$.

The exergy efficiency for the regenerator is $90.12\%$.

Explanation of Solution

Write the expression to calculate the stream exergy difference between the inlet and exit of the compressor $\left(\Delta {\dot{X}}_{\text{C}}\right)$.

$\begin{array}{c}\Delta {\dot{X}}_{\text{C}}={\dot{m}}_a\left[h_2-h_1-T_0\left(s_2-s_1\right)\right]\\ ={\dot{m}}_a\left\{c_p\left(T_2-T_1\right)-T_0\left[c_p\ln \frac{T_2}{T_1}-R\ln \frac{P_2}{P_1}\right]\right\}\end{array}$ (XVII)

Here, the temperature of the surroundings is $T_0$.

Write the expression to calculate the exergy efficiency for the compressor $\left({\eta }_{\text{ΙΙ,C}}\right)$.

${\eta }_{\text{ΙΙ,C}}=\frac{\Delta {\dot{X}}_{\text{C}}}{{\dot{W}}_{\text{C,in}}}$ (XVIII)

Write the expression to calculate the stream exergy difference between the inlet and exit of the turbine $\left(\Delta {\dot{X}}_{\text{T}}\right)$.

$\Delta {\dot{X}}_{\text{T}}=\dot{m}\left\{c_p\left(T_3-T_4\right)-T_0\left[c_p\ln \frac{T_3}{T_4}-R\ln \frac{P_3}{P_4}\right]\right\}$ (XIX)

Write the expression to calculate the exergy efficiency for the turbine $\left({\eta }_{\text{ΙΙ,T}}\right)$.

${\eta }_{\text{ΙΙ,T}}=\frac{{\dot{W}}_{\text{T,out}}}{\Delta {\dot{X}}_{\text{T}}}$ (XX)

Applying energy balance for the regenerator process.

${\dot{m}}_a{c}_p\left(T_5-T_2\right)=\dot{m}{c}_p\left(T_4-T_6\right)$ (XXI)

Write the expression to calculate the exergy increase of the cold fluid for the regenerator $\left(\Delta {\dot{X}}_{\text{regen,hot}}\right)$.

$\Delta {\dot{X}}_{\text{regen,hot}}=\dot{m}\left\{c_p\left(T_4-T_6\right)-T_0\left[c_p\ln \frac{T_4}{T_6}-0\right]\right\}$ (XXII)

Write the expression to calculate the exergy decrease of the cold fluid for the regenerator $\left(\Delta {\dot{X}}_{\text{regen,cold}}\right)$.

$\Delta {\dot{X}}_{\text{regen,cold}}=\dot{m}\left\{c_p\left(T_5-T_2\right)-T_0\left[c_p\ln \frac{T_5}{T_2}-0\right]\right\}$ (XXIII)

Write the expression to calculate the exergy efficiency for the regenerator $\left({\eta }_{\text{ΙΙ,regen}}\right)$.

${\eta }_{\text{ΙΙ,regen}}=\frac{\Delta {\dot{X}}_{\text{regen,cold}}}{\Delta {\dot{X}}_{\text{regen,hot}}}$ (XXIV)

Conclusion:

Substitute $12.6\text{kg}/\text{s}$ for ${\dot{m}}_a$, $1.093\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, 533 K for $T_2$, 303 K for $T_1$, 303 K for $T_0$, 700 kPa for $P_2$, 100 kPa for $P_1$, and $0.287\text{kJ}/\text{kg}\cdot \text{K}$ for $R$ in Equation (XVII).

$\begin{array}{c}\Delta {\dot{X}}_{\text{C}}=\left(12.6\text{kg}/\text{s}\right)\left\{\begin{array}{l}\left(1.093\text{kJ}/\text{kg}\cdot \text{K}\right)\left(533\text{K}-303\text{}\text{K}\right)\\ -\left(303\text{K}\right)\left[\begin{array}{l}\left(1.093\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \frac{533\text{K}}{303\text{K}}\\ -\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \frac{700\text{kPa}}{100\text{kPa}}\end{array}\right]\end{array}\right\}\\ =2,943\text{kW}\end{array}$

Substitute $2943\text{kW}$ for $\Delta {\dot{X}}_{\text{C}}$, and $\text{3168}\text{kW}$ for ${\dot{W}}_{\text{C,in}}$ in Equation (XVIII).

$\begin{array}{c}{\eta }_{\text{ΙΙ,C}}=\frac{2943\text{kW}}{3168\text{kW}}\\ =0.929\times 100\%\\ =92.9\%\end{array}$

Thus, the exergy efficiency for the compressor is $92.9\%$.

Substitute $12.76\text{kg}/\text{s}$ for $\dot{m}$, $1.093\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, 1,144 K for $T_3$, 754.4 K for $T_4$, 303 K for $T_0$, 700 kPa for $P_3$, 100 kPa for $P_4$, and $0.287\text{kJ}/\text{kg}\cdot \text{K}$ for $R$ in Equation (XIX).

$\begin{array}{c}\Delta {\dot{X}}_{\text{T}}=\left(12.6\text{kg}/\text{s}\right)\left\{\begin{array}{l}\left(1.093\text{kJ}/\text{kg}\cdot \text{K}\right)\left(1144\text{K}-754.4\text{K}\right)\\ -\left(303\text{K}\right)\left[\begin{array}{l}\left(1.093\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \frac{1,144\text{K}}{754.4\text{K}}\\ -\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \frac{700\text{kPa}}{100\text{kPa}}\end{array}\right]\end{array}\right\}\\ =5,833\text{kW}\end{array}$

Substitute $5833\text{kW}$ for $\Delta {\dot{X}}_{\text{T}}$, and $\text{5434}\text{kW}$ for ${\dot{W}}_{\text{T,out}}$ in Equation (XX).

$\begin{array}{c}{\eta }_{\text{ΙΙ,T}}=\frac{5434\text{kW}}{5833\text{kW}}\\ =0.932\times 100\%\\ =93.2\%\end{array}$

Thus, the exergy efficiency for the turbine is $93.2\%$.

substitute $12.6\text{kg}/\text{s}$ for ${\dot{m}}_a$, $12.76\text{kg}/\text{s}$ for $\dot{m}$, $1.093\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, 673 K for $T_5$, 533 K for $T_2$, and 754.4 K for $T_4$ to find $T_6$ in Equation (XXI).

$\begin{array}{l}\left\{\begin{array}{c}\left(12.6\text{kg}/\text{s}\right)\left(1.093\text{kJ}/\text{kg}\cdot \text{K}\right)\\ \left(673\text{K}-533\text{K}\right)\end{array}\right\}=\left\{\begin{array}{c}\left(12.76\text{kg}/\text{s}\right)\\ \left(1.093\text{kJ}/\text{kg}\cdot \text{K}\right)\left(754.4\text{K}-T_6\right)\end{array}\right\}\\ \left(754.4\text{K}-T_6\right)=138.24\text{K}\\ T_6=616.2\text{K}\end{array}$

Substitute $12.76\text{kg}/\text{s}$ for $\dot{m}$, $1.093\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, 754.4 K for $T_4$, 616.2 K for $T_6$, and 303 K for $T_0$ in Equation (XXII).

$\begin{array}{c}\Delta {\dot{X}}_{\text{regen,hot}}=\left(12.76\text{kg}/\text{s}\right)\left\{\begin{array}{l}\left(1.093\text{kJ}/\text{kg}\cdot \text{K}\right)\left(754.4\text{K}-616.2\text{}\text{K}\right)\\ -\left(303\text{K}\right)\left[\left(1.093\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \frac{754.4\text{K}}{616.2\text{K}}-0\right]\end{array}\right\}\\ =1073\text{kW}\end{array}$

Substitute $12.76\text{kg}/\text{s}$ for $\dot{m}$, $1.093\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, 673 K for $T_5$, 533 K for $T_2$, and 303 K for $T_0$ in Equation (XXIII).

$\begin{array}{c}\Delta {\dot{X}}_{\text{regen,cold}}=\left(12.76\text{kg}/\text{s}\right)\left\{\begin{array}{l}\left(1.093\text{kJ}/\text{kg}\cdot \text{K}\right)\left(673\text{}\text{K}-533\text{K}\right)-\\ \left(303\text{K}\right)\left[\left(1.093\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \frac{673\text{K}}{533\text{K}}-0\right]\end{array}\right\}\\ =\left(12.76\text{kg}/\text{s}\right)\left\{153.02\text{kJ}/\text{kg}-77.23\text{kJ}/\text{kg}\right\}\\ =967.08\text{kJ}/\text{s}\left(\frac{1\text{kW}}{1\text{kJ}/\text{s}}\right)\\ =967.08\text{kW}\end{array}$

Substitute $967.08\text{kW}$ for $\Delta {\dot{X}}_{\text{regen,cold}}$, and $1073\text{kW}$ for $\Delta {\dot{X}}_{\text{regen,hot}}$ in Equation (XXIV).

$\begin{array}{c}{\eta }_{\text{ΙΙ,regen}}=\frac{967.08\text{kW}}{1073\text{kW}}\\ =0.9012\times 100\%\\ =90.12\%\end{array}$

Thus, the exergy efficiency for the regenerator is $90.12\%$.

To determine

The rate of exergy of the combustion gases at the regenerator exit.

Answer to Problem 152P

The rate of exergy of the combustion gases at the regenerator exit is $1368.33\text{kW}$.

Explanation of Solution

Write the expression to calculate the rate of exergy of the combustion gases at the regenerator exit $\left({\dot{X}}_6\right)$.

${\dot{X}}_6=\dot{m}\left\{c_p\left(T_6-T_0\right)-T_0\left[c_p\ln \frac{T_6}{T_0}-0\right]\right\}$ (XXV)

Conclusion:

Substitute $12.76\text{kg}/\text{s}$ for $\dot{m}$, $1.093\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, 616.2 K for $T_6$, and 303 K for $T_0$ in Equation (XXV).

$\begin{array}{c}{\dot{X}}_6=\left(12.76\text{kg}/\text{s}\right)\left\{\begin{array}{l}\left(1.093\text{kJ}/\text{kg}\cdot \text{K}\right)\left(616.2\text{}\text{K}-303\text{K}\right)\\ -\left(303\text{K}\right)\left[\left(1.093\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \frac{616.2\text{K}}{303\text{K}}-0\right]\end{array}\right\}\\ =\left(12.76\text{kg}/\text{s}\right)\left[342.32\text{kJ}/\text{kg}-235.08\text{kJ}/\text{kg}\right]\\ =1368.33\text{kJ}/\text{s}\left(\frac{1\text{kW}}{1\text{kJ}/\text{s}}\right)\\ =1368.33\text{kW}\end{array}$

Thus, the rate of exergy of the combustion gases at the regenerator exit is $1368.33\text{kW}$.