A simple ideal steam power plant cycle with water as the working fluid operates

Question

A simple ideal steam power plant cycle with water as the working fluid operates between the pressure limits of 15 MPa in the boiler and 100 kPa in the condenser. Saturated vapor enters the isentropic turbine, and saturated liduid enters the pump.

1) What is the enthalpy at the exit of the turbine?

2) determine the work produced by the turbine?

A simple ideal steam power plant cycle with water as the working fluid operates between the pressure limits of 15 MPa in the boiler and 100 kPa in the condenser. Saturated vapor enters the isentropic turbine, and saturated liquid enters the pump. What is the enthalpy at the exit of the turbine? determine the work produced by the turbine?

Explanation / Answer

entropy at the nlet of turbine = 6.4430 Kj/Kg.K

so, quality at the outlet of turbine => 6.4430 = 1.3028+x*6.0562

so quality = .8487

enthalpy at the exit of turbine = 417.51 + .8487*2257.5 = 2333.563 Kj/Kg

enthalpy at the inlet of turbine = 2791.0 Kj/Kg

work done by turbine = 2791-2333.6 = 457.43 KJ/Kg

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