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19 January, 00:52

Hot combustion gases enter the nozzle of a turbojet engine at 260 kpa, 747oc, and 80 m/s. the gases exit at a pressure of 85 kpa. assuming an isentropic efficiency of 92 percent and treating the combustion gases as air, determine

a. the exit velocity

b. the exit temperature

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  1. 19 January, 04:10
    0
    Hot combustion gases are accelerated in a 92% efficient adiabatic nozzle from low velocity to a specified velocity. The exit velocity and the exit temp are to be determined.

    Given:

    T1 = 1020 K à h1 = 1068.89 kJ/kg, Pr1 = 123.4

    P1 = 260 kPa

    T1 = 747 degrees Celsius

    V1 = 80 m/s - >nN = 92% - > P2 = 85 kPa

    Solution:

    From the isentropic relation,

    Pr2 = (P2 / P1) PR1 = (85 kPa / 260 kPa) (123.4) = 40.34 = h2s = 783.92 kJ/kg

    There is only one inlet and one exit, and thus, m1 = m2 = m3. We take the nozzle as the system, which is a control volume since mass crosses the boundary.

    h2a = 1068.89 kJ/kg - (((728.2 m/s) 2 - (80 m/s) 2) / 2) (1 kJ/kg / 1000 m2/s2) = 806.95 kJ/kg/

    From the air table, we read T2a = 786.3 K
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