Question

Steam flows through an adiabatic turbine at the rate of 100 lb/min with KE = 0

and Q = 0. At entry , its pressure is 175 psia , its volume is 3.16 ft3

/ lb and its internal

energy is 1166.7 BTU / lb. At the exit, its pressure is 0.813 psia, its volume is 328 ft3

/lb

and its internal energy is 854.6 BTU / lb. What horsepower is developed

Answer 1

$h_1 =u_1 +p_1v_1 \\ h_1 = 1166.7 + \frac{(175 \times 3.16)}{5.40395} = 1269.03 \; BTU/lb \\ h_2 =u_2+p_2v_2 \\ h_2 = 854.6 + \frac{0.813 \times 328}{5.40395} = 903.946 \; BTU/lb \\ Power = m(h_1 -h_2) = 100(1269.03- 903.946) = 36508.64 \;BTU/lb = 860.91 \;HP$