Question

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Steam at 0.3 kg/s enters an adiabatic turbine at 2 MPa and 500°C. It exits at 100 kPa. If the isentropic efficiency is 95%, how much power is produced, in units of kW?

Correct Answer is 514.1 m/s

Thank you.

Answer 1

solution: - Given deita: P = 2MPa =20bar Ti 500°c P = 100 kPa = 1 bar Tsentropic Officiency, l = 0.35 = 0.3 kg/s maus of St Steam, ma flow rate P = 2 - 2 MPa Ti = sood P = lookfa Refer. Thermodynamic tables steam for super heated conditions we find At = 20 bar and Ti=500 hi 3467.55 Iling Si = 7.4316 KJ kg-K С. KJ

where of team at inlet h = enthal dondition Si= entropy of the am aut inlet Si condition 1 hi Isentropic prowe actual process h hal La h25 25 Si=S2s S for isentropic process Si = S25 At B = 1 bar and Sas=5,= 7.4316 KJ we tables find from the smodynamic of steam

has 2702.43 kg Now the sentropic officiency Can be given by NT hi-hra hinhas 0.35 (3467.55 ha) (3467.55-2702:43 Bo (3467.55-620) = 0.95(3467.55– 2702543) 3467.55- ha = 726.864 2740.686 thorbine hza Now the power produced by the can be given by : P (hi-hea) P= 0.3(3467.55 – 2740.686) m P=218.1 kW Any