SSC JE Mechanical Engineering 25 Sep 2019 Shift 1 Paper

Question : 136 of 200

Marks: +1, -0

0.5 kg/s of steam of enthalpy 3140 kJ/kg enters a steam turbine. It leaves the turbine with an enthalpy of 2640 kJ/kg. The turbine’s power output is:

Solution:

Given:
ṁ = 0.5 kg/s,

h_1

= 3140 kJ/kg,

h_2

= 2640 kJ/kg
P = 0.5 × (3140 – 2640) = 250 kW

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