Steam is the working fluid in a simple, ideal Rankine cycle. Saturated vapor ent
ID: 2996524 • Letter: S
Question
Steam is the working fluid in a simple, ideal Rankine cycle. Saturated vapor enters the turbine at 8 MPa and saturated liquid exits the condenser at a pressure of 8 kPa. The net power output of the cvcle is 100 MW. Determine for the cycle: Thermal efficiency Back work ratio Mass flow-rate of the steam in kg/h Rate of heat transfer to the working fluid as it passes through the boiler in MW Rate of heat transfer from the working fluid as is passes through the condenser in MW The mass flow rate of the condenser cooling water in kg/h if the cooling water enters the condenser at 15 degree C and exits at 35 degree C.Explanation / Answer
P1 = 8Mpa P2 = 8kPa power = 100MW
from steam table
h1 = 2757.9 s1 = 5.7431 hf4 = 1316.61
h3 = hf2 = 721.20 hfg2 = 2046.1 sf2 = 2.0458 sfg2 = 4.6044
as isentropic process 1 to 2
s1 = s2 = sf2 +x*sfg2
5.7431 = 2.0458 + x*4.6044
x = 0.80
h2 = hf2 +x*hfg2 = 721.20+ 0.8*2046.1 = 2358.08
1) Thermal effeciency = h1-h2 / h1-hf4 = 2757.9-2358.08 / 2757.9-1316.61 = 0.2774 = 27.74%
2) Back Work Ratio = power generated in Turbine / Power dissipated in Pump = 345.26/409.22 = 0.84
(Calculated in 4) and 5))
3) Power = mass flow rate * ( h1-h2)
100*10^3 = m*(2757.9-2358.08)
m = 250kg/s = 900000kg/h
4) rate of heat transfer to fluid in boiler = m(h1-hf4) = 250(2757.9-1316.61) = 345255kW = 345.26MW
5) rate of heat transfer from fluid in condenser = m(h2-hf3) = 250(2358.08-721.20) = 409220kW = 409.22MW
6) Q = m*cp*(deltaT)
100*10^6 = m*4.18*(35-15)
m = 4.3*10^9 kg/h