Question 2.14: Calculate rW, ηTH and SSC for a Rankine-superheat cycle work...
Calculate r_{W}, η_{TH} and SSC for a Rankine-superheat cycle working between 100 bar and 0.3 bar, with 139°C of superheat.
The calculation of the liquid work is the same as in the previous example. At 100 bar T_{SAT} = 311 °C, therefore superheated temperature is 450°C. The isentropic work out of the turbine requires enthalpy in:
h_{100bar,450°C} = 3241 kJ/kgand the enthalpy at exhaust.
Calculate the exhaust enthalpy using the mixture fraction at constant entropy. Entropy at 100 bar and 450°C is 6.419 kJ/kgK. Using h_{f} and h_{g} at 0.3 bar as 289 and 2625 kJ/kg respectively, results in dryness fraction:
6.149 = 0.944(1 – x) + 7.767x
x = 0.762
and hence exhaust enthalpy:
h_{3} = 2625x + 289(1 – x)
h_{3} = 2069 kJ/kg
hence work out is 3,241 – 2,069 = 1172 kJ/kg.
r_{W} = \frac{(1172 – 10)}{1172} = 99.1%
External heat supplied is
h_{100bar,450°C} – (289 + 10) = 3241 – 299 = 2942 kJ/kg\eta _{TH} = \frac{(1172 – 10)}{2842} = 39.5%
SSC = \frac{3600}{(1172 – 10)} = 3.098 kg/kWh