Question

Steam enters a turbine operating at steady state at 600°F and 200 lbf/in^2 with a velocity of 80 ft/s and leaves as saturated vapor at 5 lbf/in^2 with a velocity of 300 ft/s. The power developed by the turbine is 200 horsepower. Heat transfer from the turbine to the surroundings occurs at a rate of 50,000 Btu/h.
a. Neglecting potential energy effects, determine the mass flow rate of steam, in lb/s.

Answer 1

Answer:

49.05 lb / s

Explanation:

Given:

Inlet 1

P_1 = 200 psia

T_1 = 600 F

V_1 = 80 ft / s

Exit 2

P_2 = 5 psia

saturated vapor

V_2 = 300 ft / s

Power W = 200-hp = 509000 Btu/h = 8483.333 Btu / s

Heat Loss Q = 50,000 Btu / h = 833.3333 Btu / s

Solution:

From Table A-6E for steam inlet conditions:

h_1 = 1322.3 Btu / lbm

From Table A-5E for steam inlet conditions:

h_2 = 1130.7 Btu / lbm

Energy Balance:

Q - W = m_flow *( (h_2 - h_1) + (V_2^2 - V_1^2) / 2*25037)

m_flow = ( Q - W ) / ((h_2 - h_1) + (V_2^2 - V_1^2) / 2*25037)

m_flow = ( - 833.333 - 8483.33) / ((1130.7-1322.3) + (300^2-80^2)/50074)

m_flow = -9316.66663 / -189.9305

m_flow = 49.05 lb/s