The Ideal Reheat–Regenerative Rankine Cycle Consider a steam power plant that operates on an ideal reheat–regenerative Rankine cycle with one open feedwater heater, one closed feedwater heater, and one reheater. Steam enters the turbine at 15 MPa and 600°C and is condensed in the condenser at

Question

The Ideal Reheat–Regenerative Rankine Cycle

Consider a steam power plant that operates on an ideal reheat–regenerative Rankine cycle with one open feedwater heater, one closed feedwater heater, and one reheated. Steam enters the turbine at 15 MPa and 600°C and is condensed in the condenser at a pressure of 10 kPa. Some steam is extracted from the turbine at 4 MPa for the closed feedwater heater, and the remaining steam is reheated at the same pressure to 600°C. The extracted steam is completely condensed in the heater and is pumped to 15 MPa before it mixes with the feedwater at the same pressure. Steam for the open feedwater heater is extracted from the low-pressure turbine at a pressure of 0.5 MPa. Determine the fractions of steam extracted from the turbine as well as the thermal efficiency of the cycle.

Accepted Answer

A steam power plant operates on the ideal reheat–regenerative Rankine cycle with one open feedwater heater, one closed feedwater heater, and one reheated. The fractions of steam extracted from the turbine and the thermal efficiency are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 In both open and closed feedwater heaters, feedwater is heated to the saturation temperature at the feedwater heater pressure. (Note that this is a conservative assumption since extracted steam enters the closed feedwater heater at 376°C and the saturation temperature at the closed feedwater pressure of 4 MPa is 250°C).
Analysis The schematic of the power plant and the T-s diagram of the cycle are shown in Fig. 10–19. The power plant operates on the ideal reheat– regenerative Rankine cycle and thus the pumps and the turbines are isentropic; there are no pressure drops in the boiler, reheated, condenser, and feedwater heaters; and steam leaves the condenser and the feedwater heaters as saturated liquid.
The enthalpies at the various states and the pump work per unit mass of fluid flowing through them are

[latex]\begin{array}{lrl}h_{1} & =191.81 kJ / kg & h_{9} & =3155.0 kJ / kg \h_{2} & =192.30 kJ / kg & h_{10} & =3155.0 kJ / kg \h_{3} & =640.09 kJ / kg & h_{11} & =3674.9 kJ / kg \h_{4} & =643.92 kJ / kg & h_{12} & =3014.8 kJ / kg \h_{5} & =1087.4 kJ / kg & h_{13} & =2335.7 kJ / kg \h_{6} & =1087.4 kJ / kg & w_{\text {pump I,in }} & =0.49 kJ / kg \h_{7} & =1101.2 kJ / kg & w_{\text {pump II,in }} & =3.83 kJ / kg \h_{8} & =1089.8 kJ / kg & w_{\text {pump III,in }} & =13.77 kJ / kg\end{array}[/latex]

The fractions of steam extracted are determined from the mass and energy balances of the feedwater heaters:
Closed feedwater heater:

[latex]\begin{gathered}\dot{E}_{\text {in }}=\dot{E}_{\text {out }} \y h_{10}+(1-y) h_{4}=(1-y) h_{5}+y h_{6} \y=\frac{h_{5}-h_{4}}{\left(h_{10}-h_{6}\right)+\left(h_{5}-_{4}\right)}=\frac{1087.4-643.92}{(3155.0-1087.4)+(1087.4-643.92)}=0.1766\end{gathered}[/latex]

Open feedwater heater:

[latex]\begin{aligned}\dot{E}_{\text {in }} &=\dot{E}_{\text {out }} \z h_{12}+(1-y-z) h_{2} &=(1-y) h_{3} \z=\frac{(1-y)\left(h_{3}-h_{2}\right)}{h_{12}-h_{2}} &=\frac{(1-0.1766)(640.09-192.30)}{3014.8-192.30}=0.1306\end{aligned}[/latex]

The enthalpy at state 8 is determined by applying the mass and energy equations to the mixing chamber, which is assumed to be insulated:

[latex]\begin{aligned}\dot{E}_{\text {in }} &=\dot{E}_{\text {out }} \(1) h_{8} &=(1-y) h_{5}+y h_{7} \h_{8} &=(1-0.1766)(1087.4) kJ / kg +0.1766(1101.2) kJ / kg \&=1089.8 kJ / kg\end{aligned}[/latex]

Thus,

[latex]\begin{aligned}q_{\text {in }} &=\left(h_{9}-h_{8}\right)+(1-y)\left(h_{11}-h_{10}\right) \&=(3583.1-1089.8) kJ / kg +(1-0.1766)(3674.9-3155.0) kJ / kg \&=2921.4 kJ / kg \q_{\text {out }} &=(1-y-z)\left(h_{13}-h_{1}\right) \&=(1-0.1766-0.1306)(2335.7-191.81) kJ / kg \&=1485.3 kJ / kg\end{aligned}[/latex]

and

[latex] \eta_{ th }=1-\frac{q_{\text {out }}}{q_{\text {in }}}=1-\frac{1485.3 kJ / kg }{2921.4 kJ / kg }= 0 . 4 9 2 \text { or } 4 9 . 2 \% [/latex]

Discussion This problem was worked out in Example 10–4 for the same pressure and temperature limits with reheat but without the regeneration process.
A comparison of the two results reveals that the thermal efficiency of the cycle has increased from 45.0 to 49.2 percent as a result of regeneration.
The thermal efficiency of this cycle could also be determined from

[latex] \eta_{ th }=\frac{w_{ net }}{q_{ in }}=\frac{w_{ turb , out }-w_{ pump , in }}{q_{ in }} [/latex]

where

[latex] \begin{aligned} &w_{\text {turb,out }}=\left(h_{9}-h_{10}\right)+(1-y)\left(h_{11}-h_{12}\right)+(1-y-z)\left(h_{12}-h_{13}\right) \ &w_{\text {pump,in }}=(1-y-z) w_{\text {pump I,in }}+(1-y) w_{\text {pump II,in }}+(y) w_{\text {pump III,in }} \end{aligned} [/latex]

Also, if we assume that the feedwater leaves the closed FWH as a saturated liquid at 15 MPa (and thus at [latex] T_{5}=342^{\circ} C \text { and } h_{5}=1610.3 kJ / kg [/latex]), it can be shown that the thermal efficiency would be 50.6.

Question 10.6: The Ideal Reheat–Regenerative Rankine Cycle Consider a steam...

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