Answer to Question #100676 in Physics for Ahmed

Question

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,

Answer 1

(a)

the thermal efficiency

$k=\frac {(h_1 – h_2) - (h_4 – h_3)}{(h_4 – h_1)} (1)$

Using (1) we got

$k=\frac {(2758 – 1939.5) - (183.3 – 173.9)}{(2758 – 183.3)}=0.314$

(b)

the mass flow rate of the steam is

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

(с)

the rate of heat transfer to the steam

$Q=m(h_1-h_4)=3.182×10^5 kW$