Page 86 - Handbook of Energy Engineering Calculations

turbine; hence, its enthalpy is fully utilized. Thus, cogeneration schemes are
more efficient.
2
At 1000 lb/in (abs) (6890 kPa) and 900°F (482.2°C), the enthalpy, h =
1
1448 Btu/lb (3368 kJ/kg) from the steam tables. The entropy of steam at this
condition, from the steam tables, is 1.6121 Btu/lb°F (6.748 kJ/kg K). At 1
2
lb/in (abs) (6.89 kPa), the entropy of the saturated liquid, s , is 0.1326 Btu/lb
f
°F (0.555 kJ/kg K), and the entropy of the saturated vapor, s , is 1.9782
g
Btu/lb °F (8.28 kJ/kg K), again from the steam tables.
Now we must determine the quality of the steam, X, at the exhaust of the
2
steam turbine at 1 lb/in (abs) (6.89 kPa) from (entropy at turbine inlet
condition) = (entropy at outlet condition)(X) + (1 − X)(entropy of the
saturated fluid at the outlet condition); or X = 0.80. The enthalpy of steam
corresponding to this quality condition is h = [enthalpy of the saturated
2s
2
steam at 1 lb/in (abs)] (X) + [enthalpy of the saturated liquid at 1 lb/in 2
(abs)](l − X) = (1106)(0.80) + (1 − 0.80)(70) = 900 Btu/lb (2124 kJ/kg).

2. Compute the power output of the turbine

Use the equation P = (W )(e )(h − h )/34l3, where P = electrical power
2s
t
1
s
generated, kW; W = steam flow through the turbine, lb/h (kg/h); e = turbine
s
t
efficiency expressed as a decimal; h = enthalpy of the steam at the turbine
1
inlet, Btu/lb (kJ/kg); h = enthalpy of the steam after isentropic expansion
2s
through the turbine, Btu/lb (kJ/kg). Substituting, P = (60,000)(0.70)(1448 −
900)/3413 = 6743 kW = (6743)(3413) = 23 MM Btu/h (24.26 MM kJ).

3. Find the steam enthalpy after expansion in the cogeneration scheme

2
The steam is utilized for process heating after expansion to 200 lb/in (abs)
(1378 kPa) in the back-pressure turbine. We must compute the enthalpy of
the steam after expansion in order to find the energy available.

2
At 200 lb/in (abs) (1378 kPa), using the same procedure as in step 1
above, h = 1257.7 Btu/lb (2925.4 kJ/kg). Since we know the turbine
2s
efficiency we can use the equation, (e )(h − h ) = (h − h ); or (0.70)(1448 −
1
2
t
l
2s
1257.7) = (1448 − h ); h = 1315 Btu/lb (3058.7 kJ/kg), h = actual enthalpy
2
2
2
after expansion, Btu/lb (kJ/kg).