Steam enters a turbine at 120 bar, 508oC. At the exit, the pressure and quality are 50 kPa and 0.912, respectively. Determine the power (kW) produced by the turbine if the mass flow rate is 1.31 kg/s and the heat loss from the turbine is 225 kW.

The turbine produces 955.53 KW power.

Explanation:

Taking the turbine as a system. Applying Law of Conservation of Energy:

m(h₁ - h₂) - Heat Loss = P

where,

m = mass flow rate of steam = 1.31 kg/s

h₁ = enthalpy at state 1 (120 bar and 508°C)

h₂ = enthalpy at state 2 (50 KPa and x = 0.912)

Heat Loss = 225 KW

P = Power generated by turbine = ?

First, we find h₁ from super steam tables:

At,

T = 508°C

P = 120 bar = 12000 KPa = 12 MPa

we find that steam is in super-heated state with enthalpy:

Due to unavailibility of values in chart we approximate the state to 500° C and 12.5 MPa:

h₁ = 3343.6 KJ/kg

Now, for state 2, we have:

P = 50 KPa and x = 0.912

From saturated steam table:

h₂ = hf₂ + x(hfg₂) = 340.54 KJ/kg + (0.912)(2304.7 KJ/kg)

h₂ = 2442.4 KJ/kg

Now, using values in the conservation equation:

(1.31 kg/s)(3343.6 KJ/kg - 2442.4 KJ/kg) - 225 KW = P

P = 955.53 KW

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