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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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BlueSky06

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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