Question

A turbine operating under steady-flow conditions receives steam at the following state: pressure 13.8 bar, specific volume 0.143 m3/kg, specific internal energy 2590 kJ/kg, velocity 30 m/s. The state of the steam leaving the turbine is as follows: pressure 0.35bar,specific volume 4.37 m3/kg, specific internal energy 2360 kJ/kg, velocity 90 m/s Heat is rejected to the surroundings at the rate of 0.25 kW and the rate of steam flow through the turbine is 0.38 kg/s. Calculate the power developed by the turbine.

Answer 1

From the question the value of :

• velocity is in m/s
• the mass flow is in kg/s
• pressure in pa
• Q is in J/s
• internal energy in j/k
• specific volume m³/kg

$\therefore$ the unit of W will be J/s or W

$m[u{_1}+p{_1}v{_1}\frac{C{_1}^{2}}{2}+Z{_1}g]±Q=m[u{_2}+p{_2}v{_2}\frac{C{_2}^{2}}{2}+Z{_2}g]+W$

$W=m[(u{_1}-u{_2}+(p{_1}v{_1}-p{_2}v{_2})\frac{C{_1}^{2}-C{_2}^{2}}{2}]-Q$

$Z{_1}=Z{_2}=0$

$W=0.38kJ/kg[(2590\frac{kJ}{kg}-2360\frac{kJ}{kg})+(13.5.10^{5}pa\times0.143\frac{m^{3}}{kg}-0.35.10^{5}pa\times4.37\frac{m^{3}}{kg})+(\frac{30^{2}\frac{m}{s}-90^{2}\frac{m}{s}}{2})]-0.25kJ/s$

$=1.029.10^{5}J/s or 102.9kW$

$=102.9kW$