Answer to Question #100676 in Physics for Ahmed

Question #100676
Steam is the working fluid in a modified Rankine cycle where the turbine and pump each have an isentropic efficiency of 85%. Saturated vapor enters the turbine at 8.0 MPa and saturated liquid exits the condenser at a pressure of 0.008 MPa. The net power output of the cycle is 100 MW. Determine for the modified cycle: (a) the thermal efficiency, (b) the mass flow rate of the steam, in kg/h, (c) the rate of heat transfer,
1
Expert's answer
2019-12-23T11:28:04-0500

(a)

the thermal efficiency

k=(h1h2)(h4h3)(h4h1)(1)k=\frac {(h_1 – h_2) - (h_4 – h_3)}{(h_4 – h_1)} (1)

Using (1) we got

k=(27581939.5)(183.3173.9)(2758183.3)=0.314k=\frac {(2758 – 1939.5) - (183.3 – 173.9)}{(2758 – 183.3)}=0.314


(b)

the mass flow rate of the steam is

m=(100MV)(1000kV/MV)809.1kJ/kg)=4.449×105kg/hm=\frac {(100 MV)(1000 kV/MV)}{809.1 kJ/kg)}=4.449×10^5 kg/h


(с)

the rate of heat transfer to the steam

Q=m(h1h4)=3.182×105kWQ=m(h_1-h_4)=3.182×10^5 kW




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