Question 2.12: Find the SSC, work ratio and thermal efficiency of the Carno...
Find the SSC, work ratio and thermal efficiency of the Carnot cycle operating between 100 bar and 0.3 bar.
For SSC, we require the work done to compress the fluid, and the net work done. For the Carnot cycle, we need the enthalpy change between 4 and 1 on the diagram in Figure 2.64, as follows:
h_{f,100bar} = 1,408 kJ/kg
s_{f,100bar} = 3.36 kJ/kgK
Dryness fraction at 4 is from entropy h_{4} = h_{1}:
s_{4f} (1 – x) + s_{4g}x = s_{1}0.944(1 – x) + 7.767x = 3.36
x = 0.35
Therefore enthalpy at 4 is:
h_{4} = 0.35h_{g0.3bar} + 0.65h_{f0.3bar}h_{4} = 0.35 \times 2.625 + 0.65 \times 289 = 1.106 kJ/kg
Therefore
W_{4-1} = 1408 – 1106 = 302 kJ/kgThe work out of the turbine between 2 and 3 is:
\Delta h = h_{g100bar} – h_{3}h_{3} is found by first getting the dryness fraction as above for constant entropy, which is 0.685, and then using that to get the mixture enthalpy, which is 1,889 kJ/kg. Therefore
\Delta h = 2725 – 1889 = 836 kJ/kg.
SSC for Carnot is
\frac{1 \times 3600}{(836 – 302)} = 6.742 kg/kWh.
r_{W} for Carnot is
\frac{836 – 302}{836} = 64\%.
For thermal efficiency, the external heat added is required. For the Carnot cycle, this is
h_{fg,100bar} = 1317 kJ/kg Therefore \eta _{TH,Carnot} = \frac{836 – 302}{1317} = 40.5\%.
(Using data from tables in Rogers and Mayhew, 1995.)
