Water vapor at 5 bar, 3208C enters a turbine operating at steady state with a volumetric flow rate of 0.65 m3/s and expands adiabatically to an exit state of 1 bar, 1608C. Kinetic and potential energy effects are negligible. Determine for the turbine (a) the power developed, in kW, (b) the rate of entropy production, in kW/K, and (c) the isentropic turbine efficiency.

Answer:

A) 371.28 kW

b) 0.1547 Kw/K

c) 85%

Explanation:

pressure (p1) = 5 bar

exit pressure ( p2 ) = 1 bar

Initial Temperature ( T1 ) = 320°C

Final temp ( T2 ) = 160°C

Volume ( V ) = 0.65 m^3/s

A) Calculate power developed ( kW )

P = m( h1 - h2 ) = 1.2 ( 3105.6 - 2796.2 ) = 371.28 kW

B) Calculate the rate of entropy production

Δs = m ( S2 - S1 ) = 1.2 ( 7.6597 - 7.5308 ) = 0.1547 Kw/K

c) Calculate the isentropic turbine efficiency

For an isentropic condition  : S2s = S1

therefore at state , value of h2 at isentropic condition

attached below is the remaining part of the solution

Note : values of [ h1,  h2,  s1,  s2 , v1 and m ]   are gotten from the steam tables at state 1 and state 2

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