Question

Steam enters a turbine at 30 bar, 300°C (u = 2750 kJ/kg, h = 2993 kJ/kg) and exits the turbine as saturated liquid at 15 kPa (u = 225 kJ/kg, h = 226 kJ/kg). Heat loss to the surrounding is 50 kJ/kg of steam flowing through the turbine. Neglecting changes in kinetic energy and potential energy, the work output of the turbine (in kJ/kg of steam) is

Answer 1

Concept:

Applying SFEE (Steady Flow Energy Equation)

\({h_1} + \frac{{V_1^2}}{2} + g{Z_1} + \frac{{dQ}}{{dm}} = {h_2} + \frac{{V_2^2}}{2} + g{Z_2} + \frac{{dW}}{{dm}}\)

1 – Entry point

2 – Exit point

Calculation:

Given:

h₁ = 2993 kJ/kg, h₂ = 226 kJ/kg

dQ/dm = - 50 kJ/kg

Neglecting the change in potential and velocity head

\({h_1} + \frac{{dQ}}{{dm}} = {h_2} + \frac{{dW}}{{dm}}\)

2993 - 50 = 226 + dW/dm

dW/dm = (2993 - 226 - 50) = 2717 kJ/kg

Work output = 2717 kJ/kg