Ask Question
21 June, 01:38

Water vapor at 5 bar, 320°C enters a turbine operating at steady state with a volumetric flow rate of 0.65 m3/s and expands adiabatically to an exit state of 1 bar, 200°C. Kinetic and potential energy effects are negligible. Determine for the turbine: the power developed, in kW

+3
Answers (1)
  1. 21 June, 02:32
    0
    Power = 371.28 kW

    Explanation:

    Initial pressure, P1 = 5 bar

    Final pressure, P2 = 1 bar

    Initial temperature, T1 = 320°C

    Final temperature, T2 = 160°C

    Volume flow rate, V = 0.65m³/s

    From steam tables at state 1,

    h1 = 3105.6 kJ/kg, s1 = 7.5308 kJ/kgK

    v1 = 0.5416 m³/kg

    Mass flow rate, m = V/v1

    m = 1.2 kg/s

    From steam tables, at state 2

    h2 = 2796.2 kJ/kg, s2 = 7.6597 kJ/kgK

    Power developed, P = m (h1 - h2)

    P = 1.2 (3105.6-2796.2)

    P = 371.28 kW
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “Water vapor at 5 bar, 320°C enters a turbine operating at steady state with a volumetric flow rate of 0.65 m3/s and expands adiabatically ...” in 📘 Engineering if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers