Question: Current Attempt in Progress table [ [ E , , S 5 It , , 0 5 , 8 ] , [ 0 ,

Current Attempt in Progress \table[[E,,S5It,,05,8],[0,6,SSIT,,000's,/_],[,S65,86,,000\deg I,9],[,860\tau ,SESE,,000\deg ZI,],[0,60T,,,05,AA],[980,8899,,,05,\epsi ],[,8899,IE,,000'5,],[,8899,9058,095,000'ZI,\tau ]]\table[[State,p(kPa),T(\deg C),h(kJ/kg),s(kJ/kg*K),x],[1,12,000,560,3506,6.683,],[2,5,000,,3221,6.683,],[3,50,2323,6.683,0.86,],[4,50,3412,1.093,0,],[5,12,000,353.5,1.093,,],[6,12,000,982.4,2.595,,],[7,5.000,1155,2.921,0,],[8,50,1155,3.387,0.353,]]
As indicated in the figure below, a power plant operates on a regenerative vapor power cycle with one closed
feedwater heater. Steam enters the first turbine stage at state 1 where pressure is p1=12 MPa and temperature is
560\deg C. Steam expands to state 2 where pressure is p2=5 MPa and some of the steam is extracted and diverted to the closed feedwater heater. Condensate exits the feedwater heater at state 7 as saturated liquid at a pressure of
p7=5 MPa, undergoes a throttling process through a trap to a pressure of pg =50 kPa at state 8, and then enters
the condenser. The remaining steam expands through the second turbine stage to a pressure of p3=50 kPa at state
3 and then enters the coldenser. Saturated liquid feedwater exiting the condenser at state 4 at a pressure of pa
=50 kPa enters a pump and exits the pump at a pressure of p5=12 MPa. The feedwater then flows through the
closed feedwater heater, exiting at state 6 with a pressure of p6=12 MPa. The net power output for the cycle is
330 MW.
For isentropic processes in each turbine stage and the pump, determine:
(a) the percent cycle thermal efficiency.
(b) the mass flow rate into the first turbine stage, in kg/s.
(c) the rate of entropy production in the closed feedwater heater, in kW/K.
(d) the rate of entropy production in the steam trap, in kW/K.
Current Attempt in Progress \ table [ [ E , < , S

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