Problem 3. pressure of 100 bar. The water leaves the turbine as saturated vapor
ID: 1767243 • Letter: P
Question
Problem 3. pressure of 100 bar. The water leaves the turbine as saturated vapor an isentropic efficiency of the pump is 0.8, an at 28°C. For these conditions determine Th t temperature of 700°C and d a pressure of 0.1 bar. The e power output of a Rankin cycle is 500 MW with a turbine inle d cooling water is supplied to the condenser at 20°C and leaves the following: (1) The mass flow rate of water in the Rankine thermal cycle: (2) The isentropic efficiency of the turbine; (3) The power input to the pump:(4) The 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 = m*((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
Specific pump work needed = 10.10/0.8 = 12.625 KW = (h4 - h3)
a)mass flow rate
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) pump work needed
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 KW
c)Isentropic efficiency of turbine
Total work developed by rankine cycle = 500000 KW
Work consumed for pump = 4955.3 KW.
Work developed by trubine alone = Totalwork + work consumed in pump = 500000+4955.3 = 504955.3 KW.
Actual enthalphy drop in turbine for producing work of 504955.3 KW = 504955.3 / mass flow rate
Actual enthalphy drop in turbine = 504955.3/392.5 = 1286.5 KJ.
Ideal enthalpy drop = h1 - h2 = (3870-2583) = 1287 KJ
Isentropic efficiency of turbine= 1286.5/1287 = 0.996.
d) thermal efficiency of cycle
Thermal efficiency = Network done / work input at turbine= 500000 / 504955.3 = 0.99.
work input in turbine = mass flow rate * enthalphy.
e)
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 in condensor
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.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.