Page 96 - Handbook of Energy Engineering Calculations

For heater 1, energy in = energy out, or H m + H (l − m ) = H , where
1
13
12
1
3′
m = bleed-steam flow to the feedwater heater, lb/lb of throttle steam flow.
(The subscript refers to the heater under consideration.) Then H m + (H +
3′ 1
11
W ) (1 – m ) = H , where W = work done by pump 2, Fig. 18, in Btu/lb
p2
13
1
p2
per pound of throttle flow. Then 1372.2m + (355.4 + 1.7) (1 – m ) = 500.8;
1
1
m = 0.1416 lb/lb (0.064 kg/kg) throttle flow; H = H + W = 500.8 + 4.7 =
p1
1
l3
1
505.5 Btu/lb (1175.8 kJ/kg), where W = work done by pump 1, Fig. 18. For
p1
each pump, find the work from the chart accompanying the compressed-
liquid table in Keenan and Keyes—Steam Tables by entering the chart at the
heater inlet pressure and projecting vertically at constant entropy to the heater
outlet pressure, which equals the next heater inlet pressure. Read the enthalpy
values at the respective pressures, and subtract the smaller from the larger to
obtain the pump work during passage of the feedwater through the pump
from the lower to the higher pressure. Thus, W = 1.7 − 0.0 = 1.7 Btu/lb (4.0
p2
2
kJ/kg), from enthalpy values for 200 lb/in (abs) (1379.0 kPa) and 750 lb/in 2
(abs) (5171.3 kPa), the heater inlet and discharge pressures, respectively.
For heater 2, energy in = energy out, or H m + H (1 – m − m ) = H (1
4′ 2
1
11
2
10
– m ) H m + (H + W ) (1 – m − m ) = H (l – m )1258.4m + (184.4 +
1
p3
1
11
2
2
9
4′ 2
1
0.5)(0.8584 − m ) = 355.4(0.8584) m = 0.1365 lb/lb (0.0619 kg/kg) throttle
2
2
flow.
For heater 3, energy in = energy out, or H m + H (1 − m – m – m ) =
5′ 3
8
2
3
1
H (1 – m − m ) H m + (H + W ) (1 – m – m – m ) = H (1 – m −
5′ 3
9
1
1
1
p4
2
7
9
2
3
m )1099.2m + (47.1 + 0.1)(0.7210 − m ) = 184.4(0.7219)m = 0.0942 lb/lb
3
3
3
2
(0.0427 kg/kg) throttle flow.
4. Compute the turbine work output
The work output per section W Btu is W = H – H = 1474.5 − 1372.1 =
3′
2
1
102.3 Btu (107.9 kJ), from the previously computed enthalpy values. Also W 2
= (H – H ) (1 – m ) = (1372.2 − 1258.4) (1 − 0.1416) = 97.7 Btu (103.1 kJ);
3′
4′
1
W = (H − H ) (1 − m − m ) = (1258.4 − 1099.2) (1 − 0.1416 − 0.1365) =
4′
5′
3
1
2
115.0 Btu (121.3 kJ); W = (H – H ) (1 – m – m – m ) = (1099.2 − 938.4)
5′
4
3
2
6′
1
(1 − 0.1416 − 0.1365 − 0.0942) = 100.9 Btu (106.5 kJ). The total work output