Borgnakke and Sonntag
11.43
A steam power plant has high and low pressures of 20 MPa and 10 kPa, and one
open feedwater heater operating at 1 MPa with the exit as saturated liquid. The
maximum temperature is 800°C and the turbine has a total power output of 5
MW. Find the fraction of the flow for extraction to the feedwater and the total
condenser heat transfer rate.
The physical components and the T-s diagram is as shown in Fig. 11.10 in the
main text for one open feedwater heater. The same state numbering is used. From
the Steam Tables:
State 5: (P, T) h5 = 4069.8 kJ/kg, s5 = 7.0544 kJ/kg K,
State 1: (P, x = 0) h1 = 191.81 kJ/kg, v1 = 0.00101 m3/kg
State 3: (P, x = 0) h3 = 762.8 kJ/kg, v3 = 0.001127 m3/kg
Pump P1: wP1 = v1(P2 - P1) = 0.00101 × 990 = 1 kJ/kg
h2 = h1 + wP1 = 192.81 kJ/kg
Turbine 5-6: s6 = s5 ⇒ h6 = 3013.7 kJ/kg
wT56 = h5 - h6 = 4069.8 – 3013.7 = 1056.1 kJ/kg
.
.
Feedwater Heater (mTOT = m5):
.
.
.
xm5h6 + (1 - x)m5h2 = m5h3
h3 - h2 762.8 - 192.81
⇒ x = h - h = 3013.7 - 192.81 = 0.2021
6
2
To get state 7 into condenser consider turbine.
s7 = s6 = s5
⇒
x7 = (7.0544 - 0.6493)/7.5009 = 0.85391
h7 = 191.81 + 0.85391 × 2392.82 = 2235.1 kJ/kg
Find specific turbine work to get total flow rate
.
.
.
.
WT = mTOTh5 - xmTOTh6 - (1 - x)mTOTh7 =
.
.
= mTOT × (h5 - xh6 - (1 - x)h7) = mTOT × 1677.3
.
mTOT = 5000/1677.3 = 2.98 kg/s
.
.
QL = mTOT (1-x) (h7-h1) = 2.98 × 0.7979(2235.1 - 191.81) = 4858 kW
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