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5 March, 17:06

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.

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  1. 5 March, 20:16
    0
    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
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