Page 71 - Handbook of Energy Engineering Calculations

162,000 lb/h (73,548 kg/h). The no-load steam rate will then be = (half-load
rate) − (difference between full-load and half-load rates) = 194,000 − 162,000
= 32,000 lb/h (14,528 kg/h).

FIGURE 10 (a) Straight-line rating characteristic. (b) T-S diagram.

2. Determine the steam rate and heat rate at quarter-load points
(b) Using Fig. 10b, we see that the actual turbine efficiency, E = 3413/(w )
t
s
(H − H ) , w = steam flow, lb/kWh (kg/kWh); H = enthalpy of entering
s
f
1
1
steam, Btu/lb (kJ/kg): H = enthalpy of condensate at the exhaust pressure,
f
Btu/lb (kJ/kg).
Further, turbine heat rate = 3413/E Btu/kWh (kJ/kWh) = w (H − H ),
t
k
f
1
where w = the actual steam rate, lb/kWh (kg/kWh) = w = w /kW output,
s
k
where the symbols are as defined earlier.
Substituting w = 32,000 no-load throttle flow + (difference between full-
s
load and half-load throttle flow rate/kW output at half load)(kW output) =
32,000 + (162,000/20,000)(kW) = 32,000 + 8.1 kW for this turbine-generator
set. Also w = (32,000/kW) + 8.1.
k
Using the steam tables, we find H = 1398 Btu/lb (3257.3 kJ/kg); H = 83
f
1
Btu/lb (193.4 kJ/kg). Then, H − H = 1315 Btu/lb (3063.9 kJ/kg).
1
f
Substituting, heat rate = [(l3l5)(32,000)/(kW)] + (1316)(8.l) = 10.651.5 +
(42,080,000/kW).