Question

A turbine operates under steady flow conditions, receiving steam at the following state: Pressure

1.5 MPa, temperature 185 degree C, enthalpy 2785 kJ/kg, velocity 33.3 m/s and elevation 3 m.

The steam leaves the turbine at the following state: Pressure 22 kPa, enthalpy 2512 kJ/kg, velocity

100 m/s, and elevation 0 m. Heat is lost to the surroundings at the rate of 0.29 kJ/s. If the rate of

steam flow through the turbine is 0.42 kg/s, what is the power output of the turbine in kW?

Answer 1

Solution;

Applying the steady flow equation;

$\dot{m}[h_1+\frac{V_1^2}{2000}+\frac{gz_1}{1000}]+Q=\dot{m}[h_2+\frac{V_2^2}{2000}]+W$

By direct substitution;

$W=\dot{m}[(h_1-h_2)+(\frac{V_1^2-V_2^2}{2000})+\frac{gz_1}{1000}]+Q$

By direct substitution;

$W=0.42[(2785-2512)+\frac{33.3^2-100^2}{2000}+\frac{9.81×3}{1000}]+0.29$

$W=268.87kJ/s=268.87kW$