Question

A turbine operates under steady flow conditions, receiving steam at the following state: pressure 1200 kPa, temperature 188 degrees C, enthalpy 2785 kJ/kg, speed 33.3 m/s and elevation 3 m. The steam leaves the turbine at the following state; pressure 20 kPa, enthalpy 2512 kJ/kg, speed 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;

Given;

At Inlet of Turbine;

$P_i=1200kPa$

$T_i=188°c$

$h_i=2785kJ/kg$

$v_i=33.3m/s$

$Q_r=0.29kW$

$\displaystyle\dot m=0.42kg/s$

At outlet of the turbine;

$P_o=20kPa$

$h_o=2512kJ/kg$

$v_o=100m/s$

$z_o=0$

From steady state equation for a Turbine;

$m[h_i+\frac{v_i^2}{2000}+\frac{gz_i}{1000}]+Q=m[h_o+\frac{v_o^2}{2000}+\frac{gz_o}{1000}]+W$

Hence;

$0.42[2785+\frac{33.3^2}{2000}+\frac{9.81×3}{1000}]-0.29=0.42[2512+\frac{100^2}{2000}+0]+W$

Factorise;

$1169.95-0.29=1057.14+W$

$W=1169.66-1057.14$

$W=112.52kW$