Chapter 10.9, Problem 77P

The gas-turbine portion of a combined gas–steam power plant has a pressure ratio of 16. Air enters the compressor at 300 K at a rate of 14 kg/s and is heated to 1500 K in the combustion chamber. The combustion gases leaving the gas turbine are used to heat the steam to 400°C at 10 MPa in a heat exchanger. The combustion gases leave the heat exchanger at 420 K. The steam leaving the turbine is condensed at 15 kPa. Assuming all the compression and expansion processes to be isentropic, determine (a) the mass flow rate of the steam, (b) the net power output, and (c) the thermal efficiency of the combined cycle. For air, assume constant specific heats at room temperature.

To determine

The mass flow rate of the steam.

Answer to Problem 77P

The mass flow rate of the steam is $\underset{\_}{1.275\text{kg}/\text{s}}$.

Explanation of Solution

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

Determine the temperature of gas cycle at state 6.

$T_6=T_5\times {\left(\frac{P_6}{P_5}\right)}^{\left(k-1\right)/k}$ (I)

Here, the temperature of gas cycle at state 5 is $T_5$, the pressure of gas cycle at state 6 is $P_6$, the pressure of gas cycle at state 5 is $P_5$, and the ratio of specific heat is $k$.

Determine the rate of heat transfer into the gas turbine.

$\begin{array}{c}{\dot{Q}}_{\text{in}}={\dot{m}}_{\text{air}}\times \left(h_7-h_6\right)\\ ={\dot{m}}_{\text{air}}\times c_p\times \left(T_7-T_6\right)\end{array}$ (II)

Here, the mass flow rate of air is ${\dot{m}}_{\text{air}}$, the specific heat of constant pressure is $c_p$, the temperature of gas cycle at state 7 is $T_7$, and the temperature of gas cycle at state 6 is $T_6$.

Determine the power rate for compressor of gas turbine.

$\begin{array}{c}{\dot{W}}_{\text{C,gas}}={\dot{m}}_{\text{air}}\times \left(h_6-h_5\right)\\ ={\dot{m}}_{\text{air}}\times c_p\times \left(T_6-T_5\right)\end{array}$ (III)

Determine the temperature of gas cycle at state 8.

$T_8=T_7\times {\left(\frac{P_8}{P_7}\right)}^{\left(k-1\right)/k}$ (IV)

Here, the pressure of gas cycle at state 8 is $P_8$ and the pressure of gas cycle at state 7 is $P_7$.

Determine the power rate for gas turbine of gas turbine.

$\begin{array}{c}{\dot{W}}_{\text{T,gas}}={\dot{m}}_{\text{air}}\times \left(h_7-h_8\right)\\ ={\dot{m}}_{\text{air}}\times c_p\times \left(T_7-T_8\right)\end{array}$ (V)

Determine the net power output of the gas cycle.

${\dot{W}}_{\text{net,gas}}={\dot{W}}_{\text{T,gas}}-{\dot{W}}_{\text{C,gas}}$ (VI)

Determine input work done per unit mass of the isentropic process for the steam cycle.

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

Here, the specific volume of the steam is $v_1$, the pressure of steam cycle at state 2 is $P_2$, and the pressure of steam cycle at state 1 is $P_1$.

Determine the specific enthalpy at state 2 of the steam cycle.

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

Here, the specific enthalpy at the state 1 of the steam cycle is $h_1$.

Determine the quality at state 4 of the stream cycle.

$x_4=\frac{s_4-s_f}{s_{fg}}$ (IX)

Here, the specific entropy at state 4 is $s_4$, the specific entropy of saturated liquid is $s_f$, and the specific entropy change upon vaporization is $s_{fg}$.

Determine the specific enthalpy at state 4 of the steam cycle.

$h_4=h_f+x_4{h}_{fg}$ (X)

Here, the specific enthalpy of saturated liquid is $h_f$ and the specific enthalpy change upon vaporization is $h_{fg}$.

Write the expression for the steady-flow energy balance equation.

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

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

Substitute $0$ for $\Delta {\dot{E}}_{\text{system}}$ in Equation (XI)

$\begin{array}{l}{\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=\Delta {\dot{E}}_{\text{system}}\\ {\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=0\\ {\dot{E}}_{\text{in}}={\dot{E}}_{\text{out}}\\ {\dot{m}}_s\left(h_3-h_2\right)={\dot{m}}_{air}\left(h_8-h_9\right)\end{array}$

$\begin{array}{c}{\dot{m}}_s=\frac{\left(h_8-h_9\right)}{\left(h_3-h_2\right)}\times {\dot{m}}_{air}\\ {\dot{m}}_s=\frac{c_p\left(T_8-T_9\right)}{\left(h_3-h_2\right)}\times {\dot{m}}_{air}\end{array}$ (XII)

Here, the temperature of gas cycle at state 8 is $T_8$, the temperature of gas cycle at state 9 is $T_9$, the specific enthalpy of steam cycle at the state 3 is $h_3$ and the specific enthalpy of steam cycle at the state 2 is $h_2$.

Determine the power rate for gas turbine of steam cycle.

${\dot{W}}_{\text{T,steam}}={\dot{m}}_s\times \left(h_3-h_4\right)$ (XIII)

Here, the mass flow rate of the steam is ${\dot{m}}_s$ and the specific enthalpy of steam cycle at the state 4 is $h_4$.

Determine the power rate of the isentropic process for the steam cycle.

${\dot{W}}_{\text{p,steam}}={\dot{m}}_s\times w_p$ (XIV)

Here, the mass flow rate of the steam is ${\dot{m}}_s$ and the specific enthalpy of steam cycle at the state 4 is $h_4$.

Determine the net power output of the steam cycle.

${\dot{W}}_{\text{net,steam}}={\dot{W}}_{\text{T,steam}}-{\dot{W}}_{\text{p,steam}}$ (XV)

Conclusion:

From the Table A-2, “Ideal-gas specific heats of various common gases”, obtain the value of specific heat of constant pressure and the ratio of specific heat at temperature of $300\text{K}$ as $1.005\text{kJ}/\text{kg}\cdot \text{K}$ and 1.4.

Substitute 300 K for $T_5$, 16 for $P_6/P_5$, and 1.4 for $k$ in Equation (I).

$\begin{array}{c}{T}_6=\left(300\text{K}\right)\times {\left(16\right)}^{\left(1.4-1\right)/1.4}\\ =\left(300\text{K}\right)\times \left(2.208179\right)\\ =662.45\text{K}\\ \cong 662.5\text{K}\end{array}$

Substitute $14\text{kg}/\text{s}$ for ${\dot{m}}_{\text{air}}$, $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, 1500 K for $T_5$, and 662.5 K for $T_6$ in Equation (II).

$\begin{array}{c}{\dot{Q}}_{\text{in}}=\left(14\text{kg}/\text{s}\right)\times \left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\times \left(1500-662.5\right)\text{K}\\ =\left(14\text{kg}/\text{s}\right)\times \left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\times \left(837.5\text{K}\right)\\ =11783.6\text{kJ}/\text{s}\\ \cong 11784\text{kJ}/\text{s}\times \left(\frac{1\text{kW}}{1\text{kJ}/\text{s}}\right)\end{array}$

     $\cong 11783.6\text{kW}$

Substitute $14\text{kg}/\text{s}$ for ${\dot{m}}_{\text{air}}$, $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, 1500 K for $T_5$, and 300 K for $T_5$ in Equation (III).

$\begin{array}{c}{\dot{W}}_{\text{C,gas}}=\left(14\text{kg}/\text{s}\right)\times \left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\times \left(662.5-300\right)\text{K}\\ =\left(14\text{kg}/\text{s}\right)\times \left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\times \left(362.5\text{K}\right)\\ =5100.375\text{kJ}/\text{s}\\ \cong 5100\text{kJ}/\text{s}\times \left(\frac{1\text{kW}}{1\text{kJ}/\text{s}}\right)\end{array}$

        $\cong 5100\text{kW}$

Substitute 1500 K for $T_7$, $1/16$ for $P_8/P_7$, and 1.4 for $k$ in Equation (IV).

$\begin{array}{c}{T}_8=\left(1500\text{K}\right)\times {\left(\frac{1}{16}\right)}^{\left(1.4-1\right)/1.4}\\ =\left(1500\text{K}\right)\times \left(0.452862\right)\\ =679.2927\text{K}\\ \cong 679.3\text{K}\end{array}$

Substitute $14\text{kg}/\text{s}$ for ${\dot{m}}_{\text{air}}$, $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, 1500 K for $T_7$, and 679.3 K for $T_8$ in Equation (V).

$\begin{array}{c}{\dot{W}}_{\text{T,gas}}=\left(14\text{kg}/\text{s}\right)\times \left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\times \left(1500-679.3\right)\text{K}\\ =\left(14\text{kg}/\text{s}\right)\times \left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\times \left(820.7\text{K}\right)\\ =11547.25\text{kJ}/\text{s}\\ \cong 11547\text{kJ}/\text{s}\times \left(\frac{1\text{kW}}{1\text{kJ}/\text{s}}\right)\end{array}$

        $\cong 11547\text{kW}$

Substitute 11547 kW for ${\dot{W}}_{\text{T,gas}}$ and 5100 kW for ${\dot{W}}_{\text{C,gas}}$ in Equation (VI).

$\begin{array}{c}{\dot{W}}_{\text{net,gas}}=\left(11547-5100\right)\text{kW}\\ =6447\text{kW}\end{array}$

From the Table A-4, “Saturated water-Pressure table”, obtain the value of the initial specific enthalpy at liquid state, specific volume at the liquid state, the specific entropy at liquid state, the specific enthalpy change upon vaporization at pressure, and the specific entropy change upon vaporization at pressure of 15 kPa as:

$\begin{array}{l}{h}_1=225.94\text{kJ}/\text{kg}\\ h_{fg}=2372.3\text{kJ}/\text{kg}\\ v_1=0.001014{\text{m}}^{\text{3}}/\text{kg}\\ s_f=0.7549\text{kJ}/\text{kg}\cdot \text{K}\\ s_{fg}=7.2522\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Substitute $0.001014{\text{m}}^{\text{3}}/\text{kg}$ for $v_1$, $10\text{MPa}$ for $P_2$ and $15\text{kPa}$ for $P_1$ in Equation (VII).

$\begin{array}{c}{w}_{\text{pI,in}}=\left(0.001014{\text{m}}^{\text{3}}/\text{kg}\right)\left(10\text{MPa}-15\text{kPa}\right)\\ =\left(0.001014{\text{m}}^{\text{3}}/\text{kg}\right)\left(10\text{MPa}\times \left(\frac{1000\text{kPa}}{1\text{MPa}}\right)-15\text{kPa}\right)\\ =10.12\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\\ =10.12\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\times \left(\frac{1\text{kJ}}{1\text{kPa}\cdot {\text{m}}^{\text{3}}}\right)\end{array}$

        $=10.12\text{kJ}$

Substitute $225.94\text{kJ}/\text{kg}$ for $h_1$ and $10.12\text{kJ}/\text{kg}$ for $w_{\text{pI,in}}$ in Equation (VIII).

$\begin{array}{c}{h}_2=225.94\text{kJ}/\text{kg}+10.12\text{kJ}/\text{kg}\\ =236.06\text{kJ}/\text{kg}\end{array}$

From the Table A-6, “Superheated water”, obtain the value of the specific enthalpy at state 3 and the specific entropy at state 3 at pressure of 10 MPa and temperature of $400°\text{C}$ is $3097.0\text{kJ}/\text{kg}$ and $6.2141\text{kJ}/\text{kg}\cdot \text{K}$.

Substitute $6.2141\text{kJ}/\text{kg}\cdot \text{K}$ for $s_4$, $0.7549\text{kJ}/\text{kg}\cdot \text{K}$ for $s_f$, and $7.2522\text{kJ}/\text{kg}\cdot \text{K}$ for $s_{fg}$ in Equation (IX).

$\begin{array}{c}{x}_4=\frac{\left(6.2141\text{kJ}/\text{kg}\cdot \text{K}\right)-\left(0.7549\text{kJ}/\text{kg}\cdot \text{K}\right)}{\left(7.2522\text{kJ}/\text{kg}\cdot \text{K}\right)}\\ =\frac{\left(0.7549\text{kJ}/\text{kg}\cdot \text{K}\right)}{\left(7.2522\text{kJ}/\text{kg}\cdot \text{K}\right)}\\ =0.752765\\ =0.7528\end{array}$

Substitute 0.7528 for $x_4$, $225.94\text{kJ}/\text{kg}$ for $h_1$, and $2372.3\text{kJ}/\text{kg}$ for $h_{fg}$ in Equation (X).

$\begin{array}{c}{h}_4=225.94\text{kJ}/\text{kg}+\left(0.7528\right)\times \left(2372.3\text{kJ}/\text{kg}\right)\\ =225.94\text{kJ}/\text{kg}+\left(1785.867\text{kJ}/\text{kg}\right)\\ =2011.8\text{kJ}/\text{kg}\end{array}$

Substitute $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, 420 K for $T_9$, 679.3 K for $T_8$, $3097.0\text{kJ}/\text{kg}$ for $h_3$, $14\text{kg}/\text{s}$ for ${\dot{m}}_{\text{air}}$, and $236.06\text{kJ}/\text{kg}$ for $h_2$ in Equation (XII)

$\begin{array}{c}{\dot{m}}_s=\frac{\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\left(679.3-420\right)\text{K}}{\left(3097.0-236.06\right)\text{kJ}/\text{kg}}\times \left(14\text{kg}/\text{s}\right)\\ =\frac{\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\left(259.3\text{K}\right)}{\left(2860.94\text{kJ}/\text{kg}\right)}\times \left(14\text{kg}/\text{s}\right)\\ =\frac{\left(260.5965\text{kJ}/\text{kg}\right)}{\left(2860.94\text{kJ}/\text{kg}\right)}\times \left(14\text{kg}/\text{s}\right)\\ =1.275\text{kg}/\text{s}\end{array}$

Thus, the mass flow rate of the steam is $\underset{\_}{1.275\text{kg}/\text{s}}$.

Substitute $1.275\text{kg}/\text{s}$ for ${\dot{m}}_s$, $3097.0\text{kJ}/\text{kg}$ for $h_3$, and $2011.5\text{kJ}/\text{kg}$ for $h_4$ in Equation (XIII).

$\begin{array}{c}{\dot{W}}_{\text{T,steam}}=\left(1.275\text{kg}/\text{s}\right)\times \left(3097.0\text{kJ}/\text{kg}-2011.5\text{kJ}/\text{kg}\right)\\ =\left(1.275\text{kg}/\text{s}\right)\times \left(1085.5\text{kJ}/\text{kg}\right)\\ =1384.013\text{kJ}/\text{s}\\ =1384.013\text{kJ}/\text{s}\times \left(\frac{1\text{kW}}{1\text{kJ}/\text{s}}\right)\end{array}$

          $=1384.013\text{kW}$

Substitute $1.275\text{kg}/\text{s}$ for ${\dot{m}}_s$ and $10.12\text{kJ}/\text{kg}$ for $w_p$ in Equation (XIV).

$\begin{array}{c}{\dot{W}}_{\text{p,steam}}=\left(1.275\text{kg}/\text{s}\right)\times \left(10.12\text{kJ}/\text{kg}\right)\\ =12.903\text{kJ}/\text{s}\\ =12.903\text{kJ}/\text{s}\times \left(\frac{1\text{kW}}{1\text{kJ}/\text{s}}\right)\\ =12.903\text{kW}\end{array}$

Substitute 1384.013 kW for ${\dot{W}}_{\text{T,steam}}$ and 12.903 kW for ${\dot{W}}_{\text{T,steam}}$ in Equation (XV).

$\begin{array}{c}{\dot{W}}_{\text{net,steam}}=\left(1384.013-12.903\right)\text{kW}\\ =\text{1371}\text{.11}\text{kW}\\ \cong \text{1371}\text{kW}\end{array}$

To determine

The net work output of the combined cycle.

Answer to Problem 77P

The net work output of the combined cycle is $\underset{\_}{7819\text{kW}}$.

Explanation of Solution

Determine the net power output of combined cycle.

${\dot{W}}_{\text{net}}={\dot{W}}_{\text{net,steam}}+{\dot{W}}_{\text{net,gas}}$ (XVI)

Conclusion:

Substitute 1371 kW for ${\dot{W}}_{\text{net,steam}}$ and 6448 kW for ${\dot{W}}_{\text{net,gas}}$ in Equation (XVI).

$\begin{array}{c}{\dot{W}}_{\text{net}}=\left(1371+6448\right)\text{kW}\\ =7819\text{kW}\end{array}$

Thus, the net work output of the combined cycle is $\underset{\_}{7819\text{kW}}$.

To determine

The thermal efficiency of the combined cycle.

Answer to Problem 77P

The thermal efficiency of the combined cycle is $\underset{\_}{66.4\%}$.

Explanation of Solution

Determine the thermal efficiency of the combined cycle.

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

Conclusion:

Substitute 7819 kW for ${\dot{W}}_{\text{net}}$ and 11784 kW for ${\dot{Q}}_{\text{in}}$ in Equation (XVII).

$\begin{array}{c}{\eta }_{\text{th}}=\frac{7819\text{kW}}{11784\text{kW}}\\ =0.6635\\ \cong 0.664\times \left(1000\right)\%\\ \cong 66.4\%\end{array}$

Thus, the thermal efficiency of the combined cycle is $\underset{\_}{66.4\%}$.