Find the Steam rate Heat transfer in the boiler heat transfer in the condensor t

Question

Find the Steam rate

Heat transfer in the boiler

heat transfer in the condensor

thermal efficiency

For a Rankine cycle, the steam enters the boiler at P1=8,500kPa, and exits at T2=600 degree C, it then enters a turbine in which the pressure is reduced to P3=10kPa. The efficiency of the turbine and the pump are both 0.75. Assuming a power rate at 70,000 kW, sketch this Rankine cycle on a T-S diagram and determine: The steam rate The heat-transfer rate in the boiler The heat-transfer rate in the condenser The thermal efficiency of the plant

Explanation / Answer

W(pump)in=v(P1-P3)/0.75

where v is specific volume @Pressure 10kPa = 0.001 m^3/kg

So, W(pump)in = 0.001*(8500-10)/0.75 kJ/kg

=11.2 kJ/kg.

h@(P1,T2) (h1) = 3638.3 kJ/kg

h@P3(steam) (h2) = 2583.89 kJ/kg

h@(lliq. water) (h3) = 191.84 kJ/kg

h4 = 191.84+11.2 = 203.04 kJ

W(turbine)out =0.75(h1-h2)

=790.81 kJ/kg

So, net work(W) = W(turbine)out - W(pump)in = 790.81 - 11.2 = 779.61 kJ/kg

a)Now Power rate = 70000 kW = m*W (where m is steam rate)

So, m = 70000/779.61 kg/s = 89.79 kg/s.

b)Qin(boiler) = m(h1-h4) = 89.79(3638.3 - 203.04) = 308419 kW.

c)Qout(condensor) = m(h2-h3) = 89.79(2583.89 - 191.84) = 214782 kW

d)efficiency = work done/Qin = 70000/308419 = 22.7%

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