Problem 3. Th e power output of a Rankin cycle is 500 MW with a turbine inlet te
ID: 1767247 • Letter: P
Question
Problem 3. Th e power output of a Rankin cycle is 500 MW with a turbine inlet temperature of 700°C and pressure of 100 bar. The water leaves the turbine as saturated vapor and a pressure of 0.1 bar. The isentropic efficie t 28 C.For these conditions determine the following: (1) The mass flow rate of water in the Rankine cvcle; (2) The isentropic efficiency of the turbine; (3) The power input to the pump; (4) The thermal efficiency of the Rankine cycle; and (5) The mass flow rate of cooling water.Explanation / Answer
The power output of Rankine cycle = 500000000 Watts.
Turbine inlet temperature =T1= 700 C
Turbine inlet pressure = P1 = 100 bar
Inlet specific enthalpy at turbine for 700 c and 100 bar =h1= 3870 KJ/Kg.
Outlet condition is saturated vapour with 0.1 bar pressure
Outlet specific enthalpy at 0.1 bar = h2 = 2583 KJ/kg.
Power developed by rankine cycle = 500 MW = 500000 = (h1 - h2) - (h4 - h3).
Where (h1-h2) = enthalphy drop in turbine
(h4 - h3) = enthalpy expended for pressure rise in pump from 0.1 br to 100 bar.
Specific pump work = (h4 - h3) = Specific volume *(Pressure rise)
Specific pump work = (h4 - h3)= 0.0011*(100*105 - 0.1*105) = 10101 W = 10.10 KW.
Pump efficiency = 0.8
Pump work needed = 10.10/0.8 = 12.625 KW = (h4 - h3)
a)
Power = P = 500000 = m ( (h1 - h2) - (h4 - h3) )
Where m = mass flow rate of water through turbine in rankine cycle.
500000 = m((3870-2583) - 12.625 )
Mass flowrate = m = 392.50 kg/sec.
b)
Specific pump work = (h4 - h3)= 0.0011*(100*105 - 0.1*105) = 10101 W = 10.10 KW.
Pump efficiency = 0.8
Specific Pump work needed = 10.10/0.8 = 12.625 KW = (h4 - h3).
The mass flow rate of water = 392.5 kg/sec.
pump work = 12.625*392.5 = 4955.3 W.
c)
Thermal efficiency = Network done / work input = 500000000 / 505147500 = 0.98.
work input in turbine = mass flow rate * enthalphy.
d)
heat removed in cooling = m*4.18*(28-20).
Boiling temperature at 0.1 bar = 45.8 C.
So the vapour at 45.8 C converted into the water at 45.8 C by giving up the latent heat.
The mass flow rate of water = 392.5 kg.
Latent heat of water = 2.230 KJ.
Heat took out by the cooling water = 392.5*2.230 = 875.275 KJ.
Cooling water flow rate = 875.275 = m (28-20)
The mass flow rate of cooling water = m = 109.40 Kg/sec.
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