A steam power plant operates with with a high pressure of 5 MPa and has a boiler

Question

A steam power plant operates with with a high pressure of 5 MPa and has a boiler exit temperature of of 600°C receiving heat from a 700°C source. The ambient at 20°C provides cooling for the condenser so it can maintain 45°C inside. All the components are ideal except for the turbine which has an exit state with a quality of 97%. Find the work and heat transfer in all components per kg water and the turbine isentropic efficiency. Find the rate of entropy generation per kg water in the boiler/heat source setup?

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Explanation / Answer

Solution:
Take CV around each component steady state in standard Rankine Cycle.
1: v = 0.00101; h = 188.42, s = 0.6386 (saturated liquid at 45°C).

3: h = 3666.5 kJ/kg, s = 7.2588 kJ/kg K superheated vapor

4ac: h = 188.42 + 0.97 × 2394.8 = 2511.4 kJ/kg
CV Turbine: no heat transfer q = 0

wac = h3 – h4ac = 3666.5 – 2511.4 = 1155.1 kJ/kg

Ideal turbine: s4 = s3 = 7.2588 => x4s = 0.88, h4s = 2295 kJ/kg
ws = h3 – h4s = 3666.5 – 2295 = 1371.5 kJ/kg,

Eff = wac / ws = 1155.1 / 1371.5 = 0.842
CV Condenser: no shaft work w = 0

qout = h4ac – h1 = 2511.4 – 188.42 = 2323 kJ/k

g
CV Pump: no heat transfer, q = 0 incompressible flow so v = constant
w = v(P2 – P1) = 0.00101(5000 – 9.59) = 5.04 kJ/kg

CV Boiler: no shaft work, w = 0

qH = h3 - h2 = h3 - h1 - wP = 3666.5 - 188.42 - 5.04 = 3473 kJ/kg
s2 + (qH/ TH) + sGen = s3 and s2 = s1 (from pump analysis)

sgen = 7.2588 – 0.6386 – 3473700 + 273 = 3.05 kJ/kg K

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