Question

Steam flows through a turbine at the rate of 50 kg/min with ΔKE= 0 and Q = 0. At entry, its pressure is 1200kPaa, its specific volume = 0.2 m3 /kg and its internal energy is 1230 KJ/kg. At exit, its pressure is 5.6 kPaa, its specific volume is 20.5 m3 /kg and its internal energy is 902 KJ/kg. What horsepower is developed?, (b) The same as (a) except that the heat loss from the turbine is 10 KJ/kg of steam

Answer 1

A)

$w = m(n1-n2)$

$w = m[ ( u1-u2)+(p1v1 -p2v2)$

$Wout = \frac{50}{60}[ ( 1230 - 902) +(1200\times0.2 - 5.6\times20.5)]$

$W out = 377.667 kw = 506.256 hp$

B)

W out = power output in the previous case - Q_loss

$w = 377.667 - \frac{50}{60}\times10$

$Wout = 369.33 kw = 495.085 hp$