Question

A steam turbine operating under steady state flow conditions, receives 3600Kg of steam per hour. The steam enters the turbine at a velocity of 80m/sec, an elevation of 10m and specific enthalpy of 3276KJ/kg. It leaves the turbine at a velocity of 150m/sec. An elevation of 3m and a specific enthalpy of 2465 KJ/kg. Heat losses from the turbine to the surroundings amount to 36MJ/hr. Estimate the power output of the turbine.

Answer 1

Steam flow rate = 3600Kg/hr = 3600/3600 = 1 Kg/sec

Steam velocity at inlet V1 = 80m/sec

Steam velocity at exit V2 = 150m/sec

Elevation at inlet Z1 = 10m

Elevation at exit Z2 = 3m

Sp. Enthalpy at inlet h1 = 3276KJ/kg Sp.

Enthalpy at exit h2 = 2465KJ/kg

Heat losses from the turbine to surrounding Q = 36MJ/hr = 36 x 106/3600 = 10KJ/sec

Turbine operates under steady flow condition,

so apply SFEE For unit mass basis

Q – Ws = (h2 – h1) +1/2( V22 –V12) + g(Z2 –Z1) J/Kg-sec

– 10 – Ws = [(2465 – 3276) + (1502 – 802)/2 × 1000 + 9.81(3 –10)/1000]

Ws = 793 KJ/Kg-sec = 793 KW