# Turbine efficiency



## sislam012 (Sep 7, 2015)

Is it possible to calculate turbine efficiency given:

*Exit Condition*

Power (actual workout) = 8 MW = 8,000 kJ/sec

P2 = 20 KPa

*Inlet Condition* 

P1 = 1.5 MPa

T1 = 3000C

This is as far as I got:

nt = (Wactual/Wideal) = (h1 - h2') / (h1 - h2) = [ m_dotactual (8,000 kJ/sec) ] / [3,038 kJ/kg - 2,283 kJ/kg]

seems like need mass flow rate and not sure how to determine that

steam turbine problem.pdf


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## JHW 3d (Sep 10, 2015)

Is this a problem from a book or for your work? I'd say you seem to be on the right track as far as I can tell. I agree... You need to know the mass flow rate.


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## Audi Driver P.E. (Oct 12, 2015)

You don't need the mdot because it cancels out in the efficiency equation due to conservation of mass. You have the actual exit pressure. If it operated ideally, your exit entropy would be equal to the input condition. In your attachment you correctly calculated the actual exit entropy to obtain the quality for the given actual exit pressure. In the ideal, it would be the saturation temp and pressure and enthalpy that corresponds to the initial entropy. You simply need to find h exit for the ideal and make the appropriate comparison. (Hsubi - Hsube,act)/(Hsubi-Hsube, ideal). Remember Power =mdot(deltaH) for both the actual and the ideal. So if you are taking ratios of deltaH, the mdot (being equal) will cancel out.

Make sense?


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## Audi Driver P.E. (Oct 12, 2015)

And to add... the out put Power that is given is sort of a distractor for this problem, as you don't really need it since they give you the outlet presssure. If they'd given you the mdot instead, then you would have needed the output power.


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## JHW 3d (Nov 4, 2015)

Audi_driver said:


> You don't need the mdot because it cancels out in the efficiency equation due to conservation of mass. You have the actual exit pressure. If it operated ideally, your exit entropy would be equal to the input condition. In your attachment you correctly calculated the actual exit entropy to obtain the quality for the given actual exit pressure. In the ideal, it would be the saturation temp and pressure and enthalpy that corresponds to the initial entropy. You simply need to find h exit for the ideal and make the appropriate comparison. (Hsubi - Hsube,act)/(Hsubi-Hsube, ideal). Remember Power =mdot(deltaH) for both the actual and the ideal. So if you are taking ratios of deltaH, the mdot (being equal) will cancel out.
> 
> Make sense?


You can find H_ideal(out) using the quality calculated from the entropy. How do you find H_actual(out)?

(Rehashing this because in about a month I will have forgotten everything and won't be able to....)


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## Audi Driver P.E. (Nov 12, 2015)

Sorry, when I first read your problem, I thought it had the exit temperature as well.  You need that, mass flow rate or the quality at the exit.


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