Question

A turbine, operating under steady-flow conditions, receives 4500 kg of steam per hour. The steam enters the turbine at a velocity of 2800 m/min, an elevation of 5.5 m and a specific enthalpy of 2800 kJ/kg. It leaves the turbine at a velocity of 5600 m/min, an elevation of 1.5 m and a specific enthalpy of 2300 kJ/kg. Heat losses from the turbine to the surroundings amout to 16000 kJ/h. 

Determine the power output of the turbine

Answer 1

Quantity of steam supplied to the turbine, m = 4500 kg/h 

Steam velocity at the entrance to the turbine, C₁ = 2800 m/min 

Elevation at the entrance, Z₁ = 5.5 m 

Specific enthalpy at the entrance, h₁ = 2800 kJ/g 

Steam velocity at the exit, C₂ = 5600 m/min 

Elevation at the exit, Z₂ = 1.5 m 

Specific enthalpy at the exit, h₂ = 2300 kJ/kg 

Heat losses from the turbine to the surroundings, Q = – 16000 kJ/h 

Applying the steady flow energy equation at entry (1) and exit (2)

– 4.44 – W = 1.25 (500 + 3.26 – 0.039) or W = 633.44 kJ/s 

∴ Power output of the turbine = 633.44 kW.

Comments

Why have you taken heat loss as Q in this problem,but subtracted it from heat transfer in the last sum? Please Explain