A Combined Gas–Steam Power Cycle Consider the combined gas–steam power cycle shown in Fig. 10–25. The topping cycle is a gas-turbine cycle that has a pressure ratio of 8. Air enters the compressor at 300 K and the turbine at 1300 K. The isentropic efficiency of the compressor is 80 percent,

Question

A Combined Gas–Steam Power Cycle

Consider the combined gas–steam power cycle shown in Fig. 10–25. The topping cycle is a gas-turbine cycle that has a pressure ratio of 8. Air enters the compressor at 300 K and the turbine at 1300 K. The isentropic efficiency of the compressor is 80 percent, and that of the gas turbine is 85 percent. The bottoming cycle is a simple ideal Rankine cycle operating between the pressure limits of 7 MPa and 5 kPa. Steam is heated in a heat exchanger by the exhaust gases to a temperature of 500°C. The exhaust gases leave the heat exchanger at 450 K. Determine (a) the ratio of the mass flow rates of the steam and the combustion gases and (b) the thermal efficiency of the combined cycle.

Accepted Answer

A combined gas–steam cycle is considered. The ratio of the mass flow rates of the steam and the combustion gases and the thermal efficiency are to be determined.
Analysis The T-s diagrams of both cycles are given in Fig. 10–25. The gasturbine cycle alone was analyzed in Example 9–6, and the steam cycle in Example 10–8b, with the following results:

Gas cycle: [latex]\begin{aligned}h_{4}^{\prime} &=880.36 kJ / kg \quad\left(T_{4}^{\prime}=853 K ight) \q_{ in } &=790.58 kJ / kg \quad w_{ net }=210.41 kJ / kg \quad \eta_{ th }=26.6 \% \h_{5}^{\prime} &=h_{@ 450 K }=451.80 kJ / kg\end{aligned}[/latex]

Steam cycle: [latex]\begin{aligned}h_{2} &=144.78 kJ / kg & &\left(T_{2}=33^{\circ} C ight) \h_{3} &=3411.4 kJ / kg & &\left(T_{3}=500^{\circ} C ight) \w_{ ext {net }} &=1331.4 kJ / kg & & \eta_{ ext {th }}=40.8 \%\end{aligned}[/latex]

(a) The ratio of mass flow rates is determined from an energy balance on the heat exchanger:

[latex]\begin{aligned}\dot{E}_{ ext {in }} &=\dot{E}_{ ext {out }} \\dot{m}_{g} h_{5}^{\prime}+\dot{m}_{s} h_{3} &=\dot{m}_{g} h_{4}^{\prime}+\dot{m}_{s} h_{2} \\dot{m}_{s}\left(h_{3}-h_{2} ight) &=\dot{m}_{g}\left(h_{4}^{\prime}-h_{5}^{\prime} ight) \\dot{m}_{s}(3411.4-144.78) &=\dot{m}_{g}(880.36-451.80)\end{aligned}[/latex]

Thus,

[latex]\frac{\dot{m}_{s}}{\dot{m}_{g}}=y=0.131[/latex]

That is, 1 kg of exhaust gases can heat only 0.131 kg of steam from 33 to 500°C as they are cooled from 853 to 450 K. Then the total net work output per kilogram of combustion gases becomes

[latex]\begin{aligned}w_{ ext {net }} &=w_{ ext {net,gas }}+y w_{ ext {net,steam }} \&=(210.41 kJ / kg ext { gas })+(0.131 kg ext { steam} / kg ext { gas })(1331.4 kJ / kg ext { steam }) \&=384.8 kJ / kg ext { gas }\end{aligned}[/latex]

Therefore, for each kg of combustion gases produced, the combined plant will deliver 384.8 kJ of work. The net power output of the plant is determined by multiplying this value by the mass flow rate of the working fluid in the gas-turbine cycle.
(b) The thermal efficiency of the combined cycle is determined from

[latex] \eta_{ ext {th }}=\frac{w_{ ext {net }}}{q_{ ext {in }}}=\frac{384.8 kJ / kg ext { gas }}{790.6 kJ / kg ext { gas }}=0.487 ext { or } 48.7 \%[/latex]

Discussion Note that this combined cycle converts to useful work 48.7 percent of the energy supplied to the gas in the combustion chamber. This value is considerably higher than the thermal efficiency of the gas-turbine cycle (26.6 percent) or the steam-turbine cycle (40.8 percent) operating alone.

Question 10.9: A Combined Gas–Steam Power Cycle Consider the combined gas–s...

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