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A horizontal steady-state turbine is being designed to serve as an energy source for a small...

A horizontal steady-state turbine is being designed to serve as an energy source for a small electrical generator. (assume its adiabatic and frictionless) The power required from the turbine is 1037 kW. The inlet to the turbine is steam at T1 = 600 0 C and P1 = 10 bar, with a mass flow rate of 2.5 kg/s. The inlet pipe is 10 cm in diameter. The conditions at the turbine exit are T2 = 400 0 C and P2 = 1 bar. From this information, recommend a diameter for the outlet pipe.

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Answer #1

Approximate calculation:

Assuming ideal behavior, V2/V1 = P1*T2/(P2*T1) = 10*673/(1*873) =7.709

Ratio of areas A2/A1= 7.709

(D2/D1)^2 = 7.709

D2/D1 = 2.776

Hence outlet diameter = 2.776*10 cm = 27.76 cm

Detailed Calculation :

Here work required by turbine, W = mx(h1-h2)

m is mass flowrate, x is steam fraction,h1 & h2 are Enthalpies at inlet (P1=10 bar,T1=873K) and outlet(P2=1bar,T2=673K)

x = 1.0 for pure steam

w= 1037 kW

m= 2.5 kg/s

From steam temperature table ratio of specific volumes V2/V1 = 6.014

D2= 24.52 cm

From steam pressure table ,ratio of specific volume ,V2/V1= 8.613

D2= 29.35 cm

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