Problem 8.09 Water is the working fluid in an ideal Rankine cycle. The pressure
ID: 703404 • Letter: P
Question
Problem 8.09 Water is the working fluid in an ideal Rankine cycle. The pressure and temperature at the turbine inlet are 1600 Ibf /in.2 and 1100°F, respectively, and the condenser pressure is 250 Ibf/in.2 The mass flow rate of steam entering the turbine is 1.4 x 106 lb/h. The cooling water experiences a temperature increase from 60 to 80°F, with negligible pressure drop, as it passes through the condenser. Determine for the cycle (a) the net power developed, in Btu/h (b) the thermal efficiency (c) the mass flow rate of cooling water, in lb/h Btu/h lb/hExplanation / Answer
At state 1 (superheated steam)
Pressure P1 = 1600 lbf/in2
Temperature T1 = 1100 °F
Specific enthalpy h1 = 1548.26 Btu/lb
Specific entropy S1 = 1.6319 Btu/lb-°R
At state 2 (saturated steam)
P2 = 250 lbf/in2
S2 = S1
Quality of steam from entropy balance
x2 = (S2 - Sf)/(Sg - Sf)
= (1.6319 - 0.5679)/(1.527 - 0.5679)
= 1.10937
Enthalpy at state 2
h2 = hf + x2 (hg - hf)
= 376.235 + 1.10937 (1201.81 - 376.235 )
= 1292.103 Btu/lb
At state 3 (saturated liquid)
P3 = P2 = 250 lbf/in2
h3 = hf3 = 376.235 Btu/lb
Specific volume v3 = vf3 = 0.018650 ft3/lb
At state 4
h4 = h3 + v3 (P4 - P3)
= 376.235 Btu/lb + 0.018650 ft3/lb x (1600 - 250) lbf/in2 x 144 in2/ft2 x 1 Btu/778 ft-lbf
= 380.895 Btu/lb
Part a
Power developed by turbine Wt = m (h1 - h2)
Power developed by pump Wp = m (h4 - h3)
Net power developed W = Wt - Wp
= m [(h1 - h2) - (h4 - h3)]
= 1.4 x 10^6 lb/h [ (1548.26 - 1292.103) - (380.895 - 376.235)] Btu/lb
W = 3.521 x 10^8 Btu/h
Part b
Rate of heat transfer in
Q = m (h1 - h4)
= 1.4 x 10^6 lb/h (1548.26 - 380.895) Btu/lb
= 1.634 x 10^9 Btu/h
Thermal efficiency = W/Q = 3.521 x 10^8 / 1.634 x 10^9
= 0.2154 = 21.54%
Part C
Apply steady state mass and energy balance around condenser
m2 = m3 = m
ma = mb = mcw
0 = m (h2 - h3) + mcw (ha - hb)
mcw = m(h2 - h3) / (hb - ha)
ha (at 250 lbf/in2 and 60 °F) = 28.785 Btu/lb
hb (at 250 lbf/in2 and 80 °F) = 48.75 Btu/lb
mcw = 1.4 x 10^6 lb/h x (1292.103 - 376.235) / (48.75 - 28.785)
= 6.42 x 10^7 lb/h