Question

SP-25 Consider a regenerative steam power plant with one open feedwater heater and one closed feedwater heater. Superheated steam enters the turbine with a mass flow rate of 120 kg/s at 16 MPa and 560°C (State 1). Some fraction of the steam is extracted at 40 bar (State 2) and is supplied to the closed feedwater heater. The remaining steam expands to a pressure of 3 bar (State 3), another fraction is extracted at this pressure, and is supplied to the open feedwater heater. Steam then expands to the condenser pressure of 8 kPa (State 4). The isentropic efficiency of the turbine is 85%. Assume that saturated liquid water exits the condenser (State 5) and compressed liquid water exits the low-pressure pump (State 6). The low-pressure pump has an isentropic efficiency of 85%. Saturated liquid at 3 bar (State 7) exits the open feedwater heater, is compressed to 160 bar (State 8) in the high-pressure pump, and flows into the closed feedwater heater. The high-pressure pump has an isentropic efficiency of 85%. Compressed liquid leaves the closed feedwater heater at 160 bar with a temperature equal to the saturation temperature at 40 bar (State 9) and enters the boiler. Saturated liquid condensate at 40 bar (State 10) drains from the closed feedwater heater and is throttled to 3 bar (State 11) into the open feedwater heater Boiler l Turbine osed Open liquid pl 10 Sat liguid Pump 2 Pump 1 (a) Determine the net power output (kW) of the cycle (b) Find the rate of heat transfer (kW) to the steam passing through the boiler (c) Calculate thermal efficiency (%) of the cycle (d) Compare the thermal efficiency of this cycle with the value calculated in SP-23(c) and comment about regeneration. Answer: (c) 4 1.3%

Answer 1

2 (D 3 6

Sen 2sーか 65132 - 2L 13s F 91 = 0.9 228 ahove pro (73. ge- hp9 2 2403-1

vem 85 · 0.95 233. ol kal 0.85 42251

Pump ik ーケ 0.85 -

Yo EinEout /0 2. /0 2

nound う)+ (l.yc. 중 しは1.296 4 균 40.9%

138,135 kh elead 338,560.8 kW