The exhaust nozzle of a jet engine expands air at 300 kpa and 180â°c adiabatically to 100 kpa. determine: (a) the air velocity at the exit when the inlet velocity is low, and (b) the nozzle isentropic efficiency is 96 percent.

air velocity at the exit, v' = 15.3 m/s

Given:

Temperature, T =

Pressure, P = 300 kPa

Pressure, P' = 100 kPa

isentropic efficiency,

Solution:

In case of isentropic process:

Using steady flow energy eqn:

                      (1)

Now, by using eqn (1), we can calculate the isentropic kinetic energy:

where

Now,

(a) For actual kinetic energy :

Also,

             (2)

Now,

(a) air velocity at the exit, v = 0:

Using eqn (2):

v' = 15.3 m/s

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answer:

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