Page 81 - Handbook of Energy Engineering Calculations

1188) + 0.1268(1377 − 1245) + (1 − 0.1268) (1377 − 1007)] = 6.52 lb/kWh
(2.96 kg/kWh).

4. Calculate the turbine thermal efficiency
(c) The thermal efficiency, E = [(H − H ) + x(H − H ) + (1 − x)(H −
3
1
t
x
3
2
H )]/[(H − H ) + (H − H )]. Or, E = [(1372 − 1188) + (0.1268)(1377 −
c
t
3
fx
1
2
1245) + (1 − 0.1268)(1377 − 1007)]/[(1377 − 1188) + (1372 − 219)] =
0.3903, or 39 percent. It is interesting to note that in an ideal cycle the
thermal efficiency of the turbine is the same as that of the cycle.
5. Determine the condition of the exhaust
(d) The engine efficiency of the turbine alone = (actual turbine combined
efficiency)/actual generator efficiency). Or, using the given data, engine
efficiency of the turbine alone = 0.72/0.94 = 0.765.
Using the computed engine efficiency of the turbine alone and the Mollier
chart, (H − H ) = 0.765(H − H ) = 283. Solving, H = H − 283 = 1094
3
c′
3
c′
3
c
Btu/lb (2549 kJ/kg). From the Mollier chart, the condition at H is 1.1
c′
percent moisture. The exhaust steam quality is therefore 100 − 1.1 = 98.9
percent.
Related Calculations. This procedure is valid for a variety of cycle
arrangements for industrial, central-station, commercial, and marine plants.
By using a combination of the steam tables, Mollier chart, and cycle diagram,

a full analysis of the plant can be quickly made.

ENERGY EFFICIENCY ANALYSIS FOR BINARY CYCLE

STEAM PLANT

A binary cycle steam and mercury plant is being considered by a public
utility. Steam and mercury temperature will be 1000°F (538°C). The mercury
2
is condensed in the steam boiler, Fig. 14a at 10 lb/in (abs) (68.9 kPa) and the
2
steam pressure is 1200 lb/in (abs) (8268 kPa). Condenser pressure is 1 lb/in 2
(abs) (6.89 kPa). Expansions in both turbines are assumed to be at constant

entropy. The steam cycle has superheat but no reheat. Find the efficiency of