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

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:    \begin{aligned}h_{4}^{\prime} &=880.36 kJ / kg \quad\left(T_{4}^{\prime}=853 K \right) \\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}

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

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

\begin{aligned}\dot{E}_{\text {in }} &=\dot{E}_{\text {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}\right) &=\dot{m}_{g}\left(h_{4}^{\prime}-h_{5}^{\prime}\right) \\\dot{m}_{s}(3411.4-144.78) &=\dot{m}_{g}(880.36-451.80)\end{aligned}

Thus,

\frac{\dot{m}_{s}}{\dot{m}_{g}}=y=0.131

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

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

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

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

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.