Question

Steam is the working fluid in an ideal Rankine cycle with superheat and reheat. Steam
enters the high-pressure turbine at 8.0MPa, 480̊C, and expands to 0.6MPa. It is then reheatedto 450̊C before entering the low-pressure turbine where it expands to the condenser pressure of 10kPa. The net power output is 110MW.

A closed feedwater heater (CFWH) which uses steam extracted fromthe low-pressure turbine at 0.5MP. The extracted steam leaves the CFWH as a saturated liquid and is then pumped up to the boiler pressure before mixing with the feedwater. The feedwater leaves the CFWH at the same temperature as the now condensed extracted steam leaving the CFWH.
i.Plot the cycle on a T-s diagram, labeling all states.
ii.Thermal efficiency of the cycle
iii.The mass flow rate of steam, in kg/s
iv.The rate of heat transfer from the condensing steam as it passes through the condenser, in MW

Answer 1

P1= 8MPa, T1=480 C

h1=3348.4 KJ/Kg, S1=6.6586 KJ/Kg K.

At state , we have P2= 0.6 MPa = 600 kPa, and S2=S1=6.6586 kJ/Kg/K.

This is a two-phase mixture, x2=s2-sf/sfg = (6.6586- 1.9922)/4.7158 = 0.9895

Then for h2 we get,

h2= hf +x2hfg = 697.20 + (0.9895)(2066.30) = 2741.85 kJ/Kg.

state 3,

T3 = 450 C, P3= 600 kpa

The thermal efficiency is ηt = wnet/ qin = 1523.22 /3777.91 = 0.403

The net power per unit mass flow is wnet = wturbine 1 + wturbine 2 − wpump, wnet = (h1 − h2) + (h3 − h4) + (h6 − h5), wnet = (3358.4 − 2741.8) + (3353.3 − 2428.57) − (181.94 − 173.88), wnet = 1523.22 kJ kg

Then heat that is paid for is qin = (h1 − h6) + (h3 − h2), qin = (3348.4 − 181.94) + (3353.3 − 2741.85), qin = 3777.91 kJ kg .