A steam turbine receive a steam flow of 13kg/s and the power output is 500kw calculate
a. The chance specific enthalpy across the turbine when velocity at entrance and exist and the difference in elevation negliable
b. The chance specific enthalpy when the velocity of entrance is 60m/s, The velocity of exist is 36m/s and inlet pipe is 3m above the exhausted pipe.
a) v1 = v2, Z1 = Z2, Q = 0
Find the work done
W = P/m
W = 500 kW / 1.35 kg/s = 370 kJ/kg
Since ∆KE = 0 and ∆PE = 0
We get
∆h = W
∆h = 370 kJ/kg
b) v1 = 60 m/s, v2 = 360 m/s, Z1 = 0, Z2 = 3 m, Q = 0
W= ∆h + ∆KE + ∆PE
∆KE = (v22 - v12) / 2 = ((360 m/s)2 – (60 m/s)2) / 2 = 63 kJ/kg
∆PE = g × (Z2 - Z1) = 9.81 × 3 = 29.43 J/kg = 29.43 ×10-3 kJ/kg
We find ∆h
∆h = W – (∆KE + ∆PE)
Finally
∆h = 370 kJ/kg – (63 kJ/kg + 29.43 ×10-3 kJ/kg) = 307 kJ/kg
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