An isentropic steam turbine processes 2 kg/s of steam at 3 MPa, which is exhausted at 50 kPa and 100 oC. 5 percent of this flow is diverted for feedwater heating at 500 kPa. Determine the power produced by this turbine, in kW.

Power =2286.75W

Explanation:

Workdone from the energy balance relation is given by:

M1h1= M2h2 + M3h3 + W

First,determine the enthalpy of the exhausted steam from the given pressure and temperature tables A-6 and the corresponding entries.

h3= 2682.4kJ/kg

s= 76953kJ/kg

Since the enthalpy is constant, using entropy and adequate pressure values while interpolation and approximations with data from A-6.

W= M1( h1 - 0.05h2 - 0.95h3)

W= 2( 3854.85 - 0.05(3207.21) - 0.95(2652.4))

W= 2 (3854.85 - 160.36 - 2551.11)

W = 2 × 1143.38

W= 2286.75W

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explanation:

the answer is a 3.8cm

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