Steam enters the turbine of a basic Rankine power cycle at a pressure of 10 MPa and a temperature T2, and expands adiabatically to 6 kPa. The isentropic turbine efficiency is 85%. Saturated liquid water leaves the condenser at 6 kPa and the isentropic pump efficiency is 82%.
a.) For T2 = 580oC, determine the quality of the turbine effluent and the thermal efficiency of the cycle.
b.) Plot the quality of the turbine effluent and the thermal efficiency of the cycle for values of T2 ranging from 580oC to 700oC at 10oC increments.
11 comments:
Is there a trick to doing part B without TFT?
I got S1s=3.94 then i cannot find H3 from subcooled table of P=10MPa since 3.94 is too big...any hint?
thermo:
isn't state 2 superheated since T2 is at 580c?
and I got S1s=.516228 since S1s=S4 and S4 I just got by interpolation since for state for we have quality and pressure.
do I need mdot to find the thermal efficiency?
if the turbine affluent is superheated and entropy for a real turbine always increases, how can the turbine effluent have a quality?
Because it can still cross the two phase barrier. I think...
Parwiz:
No. Doing part (b) without TFT will be tedious because you have to lookup many numbers and do several interpolations. Please try the TFT.
Thermo:
Your value for S1S is not correct. S1S = S4 = Ssat liq at 6 kPa or 0.006 MPa = 0.519 kJ/kg-K. I don't know how you got 3.94 !
Chewbacca:
Yes, state 2 is superheated. Your S1S matches mine pretty well. I used TFT. You interpolated. All good. In part (b), you will want to use the TFT plug-in.
You do not need to know mdot to calculate thermal efficiency because it cancels as long as the mass flow rate is the same through every process (no splitters or mixers). So, thermal efficiency = W_hat_cycle / Q_H.
Y:
Effluent, with an E.
It is superheated.
The tail on the 2-phase envelope really slopes WAY out to the right. As a result, the turbine effluent can lie in the 2-phase region even though S_hat increases in the real turbine.
Good question !
Chewbacca:
Yes, exactly. Good job !
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