Chapter 10.9, Problem 86P

Consider a combined gas–steam power plant that has a net power output of 280 MW. The pressure ratio of the gas-turbine cycle is 11. Air enters the compressor at 300 K and the turbine at 1100 K. The combustion gases leaving the gas turbine are used to heat the steam at 5 MPa to 350°C in a heat exchanger. The combustion gases leave the heat exchanger at 420 K. An open feedwater heater incorporated with the steam cycle operates at a pressure of 0.8 MPa. The condenser pressure is 10 kPa. Assuming isentropic efficiencies of 100 percent for the pump, 82 percent for the compressor, and 86 percent for the gas and steam turbines, determine (a) the mass flow rate ratio of air to steam, (b) the required rate of heat input in the combustion chamber, and (c) the thermal efficiency of the combined cycle.

To determine

The mass flow rate ratio of the air to the steam.

Answer to Problem 86P

The mass flow rate ratio of the air to the steam is $9.259$.

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 8, 9, $9s$, 10, 11, $11s$ alone.

At state 8:

The air enters the compressor at the temperature of $300\text{K}$. Since, the air is an ideal gas.

Refer Table A-17, “Ideal-gas properties of air”.

The enthalpy $\left(h_8\right)$ and the relative pressure $\left(P_{r8}\right)$ corresponding to the temperature of $300\text{K}$ is $300.19\text{kJ/kg}$ and $1.3860$.

Write the relative pressure and absolute pressure relation for the process 8-9-$9s$.

$\begin{array}{l}\frac{P_{r9s}}{P_{r8}}=\frac{P_9}{P_8}\\ P_{r9s}=\frac{P_9}{P_8}{P}_{r8}\end{array}$ (I)

Here, the relative pressure is $P_r$ , the pressure is $P$, and the subscripts 8, 9, and $9s$ indicates the process states.

Write the formula for isentropic efficiency of compressor for the process 8-9-$9s$.

$\begin{array}{c}{\eta }_C=\frac{h_{9s}-h_8}{h_9-h_8}\\ h_9=h_8+\frac{h_{9s}-h_8}{{\eta }_C}\end{array}$ (II)

Here, the enthalpy is $h$.

At state 10:

The air enters the turbine at the temperature of $1100\text{K}$.

Refer Table A-17, “Ideal-gas properties of air”.

The enthalpy $\left(h_{10}\right)$ and the relative pressure $\left(P_{r10}\right)$ corresponding to the temperature of $1100\text{K}$ is $1161.07\text{kJ/kg}$ and $167.1$.

Write the relative pressure and absolute pressure relation for the process 10-11-$11s$.

$\begin{array}{l}\frac{P_{r11s}}{P_{r10}}=\frac{P_{11}}{P_{10}}\\ P_{r11s}=\frac{P_{11}}{P_{10}}{P}_{r10}\end{array}$ (III)

Write the formula for isentropic efficiency of gas turbine $\left({\eta }_{gT}\right)$ for the process 10-11-$11s$.

$\begin{array}{c}{\eta }_C=\frac{h_{10}-h_{11}}{h_{10s}-h_{11}}\\ h_{11}=h_{10}-{\eta }_{gT}\left(h_{10s}-h_{11}\right)\end{array}$ (IV)

At state 12: (heat exchanger)

The enthalpy $\left(h_{12}\right)$ corresponding to the temperature of $420\text{K}$ is $421.26\text{kJ/kg}$.

Refer Figure 1.

Consider the steam cycle (bottoming cycle) and their respective process states such as 1, 2, 3, 4, 5, 6, $6s$, 7, $7s$ alone.

At state 1: (Pump I inlet)

The water exits the condenser as a saturated liquid at the pressure of $10\text{kPa}$. Hence, the enthalpy, and specific volume at state 1 is as follows.

$\begin{array}{l}{h}_1=h_{f@10\text{kPa}}\\ v_1=v_{f@10\text{kPa}}\end{array}$

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

The enthalpy $\left(h_1\right)$, specific volume $\left(v_1\right)$ and entropy $\left(s_1\right)$ at state 1 corresponding to the pressure of $10\text{kPa}$ is $191.81\text{kJ/kg}$, and $0.001010{\text{m}}^3\text{/kg}$ respectively.

At state 2:

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

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

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.

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

At state 3: (Pump II inlet)

The water exits the open feed water heater as a saturated liquid at the pressure of $0.8\text{MPa}\left(800\text{kPa}\right)$. Hence, the enthalpy, and specific volume at state 3 is as follows.

$\begin{array}{l}{h}_3=h_{f@800\text{kPa}}\\ v_3=v_{f@800\text{kPa}}\end{array}$

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

The enthalpy $\left(h_3\right)$, and specific volume $\left(v_3\right)$ at state 3 corresponding to the pressure of $0.8\text{MPa}\left(800\text{kPa}\right)$ is $720.87\text{kJ/kg}$ and $0.001115{\text{m}}^3\text{/kg}$ respectively.

At state 4:

Write the formula for work done by the pump during process 3-4.

$w_{\text{pII,in}}=v_3\left(P_4-P_3\right)$ (VII)

Here, the specific volume is $v$, the pressure is $P$, and the subscripts 3 and 4 indicates the process states.

Write the formula for enthalpy $\left(h\right)$ at state 4.

$h_4=h_3+w_{\text{pII,in}}$ (VIII)

At state 5:

The steam enters the turbine as superheated vapour.

Refer Table A-6, “Superheated water”.

The enthalpy $\left(h_5\right)$ and entropy $\left(s_5\right)$ at state 5 corresponding to the pressure of $5\text{MPa}\left(5000\text{kPa}\right)$ and the temperature of $500°\text{C}$ is as follows.

$\begin{array}{l}{h}_5=3069.3\text{kJ/kg}\\ s_5=6.4516\text{kJ/kg}\cdot \text{K}\end{array}$

At state $6s$:

The steam expanded to the pressure of $0.8\text{MPa}\left(800\text{kPa}\right)$ and in the state of saturated mixture.

The quality of water at state $6s$ is expressed as follows.

$x_{6s}=\frac{s_{6s}-s_{f,6s}}{s_{fg,6s}}$ (IX)

The enthalpy at state $6s$ is expressed as follows.

$h_{6s}=h_{f,6s}+x_{6s}{h}_{fg,6s}$ (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 $6s$ indicates the ideal process state $6s$.

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

Obtain the following properties corresponding to the pressure of $0.8\text{MPa}\left(800\text{kPa}\right)$.

$\begin{array}{l}{h}_{f,6s}=720.87\text{kJ/kg}\\ h_{fg,6s}=2047.5\text{kJ/kg}\\ s_{f,6s}=2.0457\text{kJ/kg}\cdot \text{K}\\ s_{fg,6s}=4.6160\text{kJ/kg}\cdot \text{K}\end{array}$

The isentropic efficiency of the steam turbine for the process 5-6-$6s$ is expressed as follows.

$\begin{array}{l}{\eta }_{sT}=\frac{h_5-h_6}{h_5-h_{6s}}\\ h_6=h_5-{\eta }_{sT}\left(h_5-h_{6s}\right)\end{array}$ (XI)

At state $7s$:

The steam enters the condenser at the pressure of $10\text{kPa}$ and at the state of saturated mixture.

The quality of water at state $7s$ is expressed as follows.

$x_{7s}=\frac{s_{7s}-s_{f,7s}}{s_{fg,7s}}$ (XII)

The enthalpy at state $7s$ is expressed as follows.

$h_{7s}=h_{f,7s}+x_{7s}{h}_{fg,7s}$ (XIII)

Here, the subscript $7s$ indicates the ideal process state $7s$.

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

Obtain the following properties corresponding to the pressure of $10\text{kPa}$.

$\begin{array}{l}{h}_{f,7s}=191.81\text{kJ/kg}\\ h_{fg,7s}=2392.1\text{kJ/kg}\\ s_{f,7s}=0.6492\text{kJ/kg}\cdot \text{K}\\ s_{fg,7s}=7.4996\text{kJ/kg}\cdot \text{K}\end{array}$

The isentropic efficiency of the steam turbine for the process 5-7-$7s$ is expressed as follows.

$\begin{array}{l}{\eta }_{sT}=\frac{h_5-h_7}{h_5-h_{7s}}\\ h_7=h_5-{\eta }_{sT}\left(h_5-h_{7s}\right)\end{array}$ (XIV)

Here, the subscript $6s$, and $7s$ indicates the ideal process states; all other subscripts such as 1, 2, 3, 4, 5, 6 and 7 indicates the actual process states.

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 (XVI) is reduced as follows for the heat exchanger.

$\begin{array}{l}{\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=0\\ {\dot{E}}_{\text{in}}={\dot{E}}_{\text{out}}\\ {\displaystyle \sum {\dot{m}}_{\text{in}}{h}_{\text{in}}}={\displaystyle \sum {\dot{m}}_{\text{out}}{h}_{\text{out}}}\\ {\dot{m}}_a\left(h_{11}-h_{12}\right)={\dot{m}}_w\left(h_5-h_4\right)\end{array}$

$\frac{{\dot{m}}_a}{{\dot{m}}_w}=\frac{h_5-h_4}{h_{11}-h_{12}}$ (XVI)

Here, the mass flow rate of air is ${\dot{m}}_a$ and the mass flow rate of water is ${\dot{m}}_w$.

Conclusion:

Substitute $11$ for $P_9/P_8$, and $1.3860$ for $P_{r8}$ in Equation (I).

$\begin{array}{c}{P}_{r9s}=\left(11\right)\left(1.3860\right)\\ =15.246\\ \simeq 15.25\end{array}$

Refer Table A-17, “Ideal-gas properties of air”.

The enthalpy $\left(h_{9s}\right)$ corresponding to the relative pressure of $P_{r9s}=15.25$ is $595.84\text{kJ/kg}$ (interpolation method).

Substitute $300.19\text{kJ/kg}$ for $h_8$, $595.84\text{kJ/kg}$ for $h_{9s}$, and $0.82$ for ${\eta }_C$ in Equation (II).

$\begin{array}{c}{h}_9=300.19\text{kJ/kg}+\frac{595.84\text{kJ/kg}-300.19\text{kJ/kg}}{0.82}\\ =300.19\text{kJ/kg}+360.5487\text{kJ/kg}\\ =660.7388\text{kJ/kg}\\ \simeq 660.74\text{kJ/kg}\end{array}$

Substitute $1/11$ for $P_{11}/P_{10}$, and $167.1$ for $P_{r10}$ in Equation (III).

$\begin{array}{c}{P}_{r11s}=\left(\frac{1}{11}\right)\left(167.1\right)\\ =15.1909\\ \simeq 15.19\end{array}$

Refer Table A-17, “Ideal-gas properties of air”.

The enthalpy $\left(h_{11s}\right)$ corresponding to the relative pressure of $P_{r11s}=15.19$ is $595.18\text{kJ/kg}$ (interpolation method).

Substitute $1161.07\text{kJ/kg}$ for $h_{10}$, $595.18\text{kJ/kg}$ for $h_{11s}$, and $0.86$ for ${\eta }_{gT}$ in Equation (IV).

$\begin{array}{c}{h}_{11}=1161.07\text{kJ/kg}-0.86\left(1161.07\text{kJ/kg}-595.18\text{kJ/kg}\right)\\ =1161.07\text{kJ/kg}-486.6654\text{kJ/kg}\\ =674.4046\text{kJ/kg}\\ \simeq 674.40\text{kJ/kg}\end{array}$

Substitute $0.001010{\text{m}}^3\text{/kg}$ for $v_1$, $10\text{kPa}$ for $P_1$, and $800\text{kPa}$ for $P_2$ in Equation (V).

$\begin{array}{c}{w}_{\text{pI,in}}=\left(0.001010{\text{m}}^3/\text{kg}\right)\left(800\text{kPa}-10\text{kPa}\right)\\ =0.7979\text{kPa}\cdot {\text{m}}^3/\text{kg}\times \frac{1\text{kJ}}{1\text{kPa}\cdot {\text{m}}^3}\\ =0.7979\text{kJ}/\text{kg}\\ \simeq 0.80\text{kJ}/\text{kg}\end{array}$

Substitute $191.81\text{kJ/kg}$ for $h_1$, and $0.80\text{kJ}/\text{kg}$ for $w_{\text{pI,in}}$ in Equation (VI).

$\begin{array}{c}{h}_2=191.81\text{kJ/kg}+0.80\text{kJ}/\text{kg}\\ =192.61\text{kJ}/\text{kg}\\ \simeq 192.60\text{kJ}/\text{kg}\end{array}$

Substitute $0.001115{\text{m}}^3\text{/kg}$ for $v_3$, $800\text{kPa}$ for $P_3$, and $5000\text{kPa}$ for $P_4$ in

Equation (VII).

$\begin{array}{c}{w}_{\text{pII,in}}=\left(0.001115{\text{m}}^3\text{/kg}\right)\left(5000\text{kPa}-800\text{kPa}\right)\\ =4.683\text{kPa}\cdot {\text{m}}^3/\text{kg}\times \frac{1\text{kJ}}{1\text{kPa}\cdot {\text{m}}^3}\\ =4.683\text{kJ}/\text{kg}\\ \simeq 4.68\text{kJ}/\text{kg}\end{array}$

Substitute $720.87\text{kJ/kg}$ for $h_3$, and $4.68\text{kJ}/\text{kg}$ for $w_{\text{pII,in}}$ in Equation (VIII).

$\begin{array}{c}{h}_4=720.87\text{kJ/kg}+4.68\text{kJ}/\text{kg}\\ =725.55\text{kJ}/\text{kg}\end{array}$

From Figure 1.

$s_5=s_{6s}=s_{7s}=6.4516\text{kJ/kg}\cdot \text{K}$

Substitute $6.4516\text{kJ/kg}\cdot \text{K}$ for $s_{6s}$, $2.0457\text{kJ/kg}\cdot \text{K}$ for $s_{f,6s}$, and $4.6160\text{kJ/kg}\cdot \text{K}$ for $s_{fg,6s}$ in Equation (IX).

$\begin{array}{c}{x}_{6s}=\frac{6.4516\text{kJ/kg}\cdot \text{K}-2.0457\text{kJ/kg}\cdot \text{K}}{4.6160\text{kJ/kg}\cdot \text{K}}\\ =0.9545\end{array}$

Substitute $720.87\text{kJ/kg}$ for $h_{f,6s}$, $2047.5\text{kJ/kg}$ for $h_{fg,6s}$, and $0.9545$ for $x_{6s}$ in

Equation (X).

$\begin{array}{c}{h}_{6s}=720.87\text{kJ/kg}+0.9545\left(2047.5\text{kJ/kg}\right)\\ =720.87\text{kJ/kg}+1954.3387\text{kJ}/\text{kg}\\ =2675.2087\text{kJ}/\text{kg}\\ \simeq 2675.2\text{kJ/kg}\end{array}$

Substitute $3069.3\text{kJ/kg}$ for $h_5$, $0.86$ for ${\eta }_{sT}$, and $2675.2\text{kJ/kg}$ for $h_{6s}$ in Equation (XI).

$\begin{array}{c}{h}_6=3069.3\text{kJ/kg}-0.86\left(3069.3\text{kJ/kg}-2675.2\text{kJ/kg}\right)\\ =3069.3\text{kJ/kg}-338.926\text{kJ/kg}\\ =2730.374\text{kJ/kg}\\ \simeq 2730.3\text{kJ/kg}\end{array}$

Substitute $6.4516\text{kJ/kg}\cdot \text{K}$ for $s_{7s}$, $0.6492\text{kJ/kg}\cdot \text{K}$ for $s_{f,7s}$, and $7.4996\text{kJ/kg}\cdot \text{K}$ for $s_{fg,7s}$ in Equation (XII).

$\begin{array}{c}{x}_{7s}=\frac{6.4516\text{kJ/kg}\cdot \text{K}-0.6492\text{kJ/kg}\cdot \text{K}}{7.4996\text{kJ/kg}\cdot \text{K}}\\ =0.7737\end{array}$

Substitute $191.81\text{kJ/kg}$ for $h_{f,7s}$, $2392.1\text{kJ/kg}$ for $h_{fg,7s}$, and $0.7737$ for $x_{7s}$ in

Equation (XIII).

$\begin{array}{c}{h}_{7s}=191.81\text{kJ/kg}+0.7737\left(2392.1\text{kJ/kg}\right)\\ =191.81\text{kJ/kg}+1850.7678\text{kJ}/\text{kg}\\ =2042.5778\text{kJ}/\text{kg}\\ \simeq 2042.6\text{kJ}/\text{kg}\end{array}$

Substitute $3069.3\text{kJ/kg}$ for $h_5$, $0.86$ for ${\eta }_{sT}$, and $2042.6\text{kJ}/\text{kg}$ for $h_{7s}$ in Equation (XIV).

$\begin{array}{c}{h}_7=3069.3\text{kJ/kg}-0.86\left(3069.3\text{kJ/kg}-2042.6\text{kJ}/\text{kg}\right)\\ =3069.3\text{kJ/kg}-882.962\text{kJ/kg}\\ =2186.338\text{kJ/kg}\\ \simeq 2186.3\text{kJ/kg}\end{array}$

Substitute $3069.3\text{kJ/kg}$ for $h_5$, $725.55\text{kJ}/\text{kg}$ for $h_4$, $674.40\text{kJ/kg}$ for $h_{11}$, and $421.26\text{kJ/kg}$ for $h_{12}$ in Equation (XVI).

$\begin{array}{c}\frac{{\dot{m}}_a}{{\dot{m}}_w}=\frac{3069.3\text{kJ/kg}-725.55\text{kJ}/\text{kg}}{674.40\text{kJ/kg}-421.26\text{kJ/kg}}\\ =\frac{2343.75}{253.14}\\ =9.2587\\ \simeq 9.259\end{array}$

Thus, the mass flow rate ratio of the air to the steam is $9.259$.

To determine

The required rate of heat input in the combustion chamber.

Answer to Problem 86P

The required rate of heat input in the combustion chamber is $671300\text{kW}$.

Explanation of Solution

Refer Equation (XV).

Consider the open feed water heater operates on steady state. Hence, the rate of change in net energy of the system is zero.

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

Write the energy rate balance equation for open feed water heater.

$\begin{array}{l}{\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=0\\ {\dot{E}}_{\text{in}}={\dot{E}}_{\text{out}}\\ {\displaystyle \sum {\dot{m}}_{\text{in}}{h}_{\text{in}}}={\displaystyle \sum {\dot{m}}_{\text{out}}{h}_{\text{out}}}\\ {\dot{m}}_6{h}_6+{\dot{m}}_2{h}_2={\dot{m}}_3{h}_3\end{array}$ (XVII)

Rewrite the Equation (XVII) in terms of mass fraction $y$.

$\begin{array}{l}y{h}_6+\left(1-y\right)h_2=1h_3\\ y{h}_6+h_2-y{h}_2=h_3\\ y\left(h_6-h_2\right)=h_3-h_2\\ y=\frac{h_3-h_2}{h_6-h_2}\end{array}$ (XVIII)

Here, the mass fraction steam extracted from the turbine to the inlet mass of the boiler $\left({\dot{m}}_6/{\dot{m}}_3\right)$ is $y$.

Write the formula for work output of the steam turbine.

$w_{sT}=\left(h_5-h_6\right)+\left(1-y\right)\left(h_6-h_7\right)$ (XIX)

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

$\begin{array}{c}{w}_{\text{net,steam cycle}}=w_{sT}-w_{\text{P,in}}\\ =w_{sT}-\left[\left(1-y\right)w_{\text{pI,in}}-w_{\text{pII,in}}\right]\end{array}$ (XX)

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

$\begin{array}{c}{w}_{\text{net,gas cycle}}=w_{gT}-w_{\text{C,in}}\\ =\left(h_{10}-h_{11}\right)-\left(h_9-h_8\right)\\ =h_{10}-h_{11}-h_9+h_8\end{array}$ (XXI)

Write the formula for the net work output of the gas-steam cycle per unit mass of gas.

$\begin{array}{c}{w}_{\text{net}}=\left(1\text{kg}\times w_{\text{net,gas cycle}}\right)+\left(\frac{1}{9.259}\times w_{\text{net,steam cycle}}\right)\\ =w_{\text{net,gas cycle}}+0.108\left(w_{\text{net,steam cycle}}\right)\end{array}$ (XXII)

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

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

Write the formula for rate of heat input to the combustion chamber.

${\dot{Q}}_{\text{in}}={\dot{m}}_a\left(h_{10}-h_9\right)$ (XXIV)

Conclusion:

Substitute $720.87\text{kJ/kg}$ for $h_3$, $192.60\text{kJ}/\text{kg}$ for $h_2$, and $2730.3\text{kJ/kg}$ for $h_6$ in

Equation (XVIII).

$\begin{array}{c}y=\frac{720.87\text{kJ/kg}-192.60\text{kJ}/\text{kg}}{2730.3\text{kJ/kg}-192.60\text{kJ}/\text{kg}}\\ =\frac{528.27}{2537.7}\\ =0.2082\end{array}$

Substitute $0.86$ for ${\eta }_{sT}$, $3069.3\text{kJ/kg}$ for $h_5$, $2730.3\text{kJ/kg}$ for $h_6$, $0.2082$ for $y$, and $2186.3\text{kJ/kg}$ for $h_7$ in Equation (XIX).

$\begin{array}{c}{w}_{sT}=\left(3069.3\text{kJ/kg}-2730.3\text{kJ/kg}\right)+\left(1-0.2082\right)\left(2730.3\text{kJ/kg}-2186.3\text{kJ/kg}\right)\\ =339\text{kJ/kg}+430.7392\text{kJ/kg}\\ =769.7392\text{kJ/kg}\\ \simeq 769.74\text{kJ/kg}\end{array}$

Substitute $769.74\text{kJ/kg}$ for $w_{sT}$, $0.2082$ for $y$, $0.80\text{kJ}/\text{kg}$ for $w_{\text{pI,in}}$, and $4.68\text{kJ}/\text{kg}$ for $w_{\text{pII,in}}$ in Equation (XX).

$\begin{array}{c}{w}_{\text{net,steam cycle}}=769.74\text{kJ/kg}-\left(1-0.2082\right)\left(0.80\text{kJ/kg}\right)-4.68\text{kJ/kg}\\ =769.74\text{kJ/kg}-0.6334\text{kJ/kg}-4.68\text{kJ/kg}\\ =764.4266\text{kJ/kg}\\ \simeq 764.43\text{kJ/kg}\end{array}$

Substitute $1161.07\text{kJ/kg}$ for $h_{10}$, $674.40\text{kJ/kg}$ for $h_{11}$, $660.74\text{kJ/kg}$ for $h_9$, and $300.19\text{kJ/kg}$ for $h_8$ in Equation (XXI).

$\begin{array}{c}{w}_{\text{net,gas cycle}}=1161.07\text{kJ/kg}-674.40\text{kJ/kg}-660.74\text{kJ/kg}+300.19\text{kJ/kg}\\ =126.12\text{kJ/kg}\end{array}$

Substitute $126.12\text{kJ/kg}$ for $w_{\text{net,gas cycle}}$ and $764.43\text{kJ/kg}$ for $w_{\text{net,steam cycle}}$ in Equation (XXII).

$\begin{array}{c}{w}_{\text{net}}=126.12\text{kJ/kg}+0.108\left(764.43\text{kJ/kg}\right)\\ =126.12\text{kJ/kg}+82.5584\text{kJ/kg}\\ =208.6784\text{kJ/kg}\\ \simeq 208.69\text{kJ/kg}\end{array}$

Substitute $280\text{MW}$ for ${\dot{W}}_{\text{net}}$ and $208.69\text{kJ/kg}$ for $w_{\text{net}}$ in Equation (XXIII).

$\begin{array}{c}{\dot{m}}_a=\frac{280\text{MW}}{208.69\text{kJ/kg}}\\ =\frac{280\text{MW}\times \frac{{10}^3\text{kJ/s}}{1\text{MW}}}{208.69\text{kJ/kg}}\\ =1341.7030\text{kg/s}\\ \simeq 1341.7\text{kg/s}\end{array}$

Substitute $1341.7\text{kg/s}$ for ${\dot{m}}_a$, $1161.07\text{kJ/kg}$ for $h_{10}$, and $660.74\text{kJ/kg}$ for $h_9$ in

Equation (XXIV).

$\begin{array}{c}{\dot{Q}}_{\text{in}}=1341.7\text{kg/s}\left(1161.07\text{kJ/kg}-660.74\text{kJ/kg}\right)\\ =1341.7\text{kg/s}\left(500.33\text{kJ/kg}\right)\\ =671292.761\text{kJ/s}\times \frac{1\text{kW}}{1\text{kJ/s}}\\ \simeq 671300\text{kW}\end{array}$

Thus, the required rate of heat input in the combustion chamber is $671300\text{kW}$.

To determine

The thermal efficiency of the combined cycle.

Answer to Problem 86P

The thermal efficiency of the combined cycle is $41.7\%$.

Explanation of Solution

Write the formula for thermal efficiency.

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

Conclusion:

Substitute $280\text{MW}$ for ${\dot{W}}_{\text{net}}$ and $671300\text{kJ/s}$ for ${\dot{Q}}_{\text{in}}$ in Equation (XXV).

$\begin{array}{c}{\eta }_{\text{th}}=\frac{280\text{MW}}{671300\text{kJ/s}}\\ =\frac{280\text{MW}\times \frac{{10}^3\text{kJ/s}}{1\text{MW}}}{671300\text{kJ/s}}\\ =0.4171\times 100\\ \simeq 41.7\%\end{array}$

Thus, the thermal efficiency of the combined cycle is $41.7\%$.