Question

The exhaust nozzle of a jet engine expands air at 300 kPa and 180°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.

Answer 1

Answer:

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

Given:

Temperature, T = $180^{\circ} = 273 + 180^{\circ} = 453 K$

Pressure, P = 300 kPa

Pressure, P' = 100 kPa

isentropic efficiency, $\eta_{s} = 96%$

Solution:

In case of isentropic process:

$\frac{T_{s}}{T} = (\frac{P'}{P})^{\frac{k - 1}{k}}$

${T_{s} = 453\times (\frac{100}{300})^{\frac{1.4 - 1}{1.4}}$

${T_{s} = 453\times 0.0233 = 330.96 K$

Using steady flow energy eqn:

$h + \frac{v^{2}}{2} = h' + \frac{v'^{2}}{2}$

$h - h' = \frac{v'^{2}}{2} - \frac{v^{2}}{2}$

$mC_{p}(T - T') = \frac{1}{2}m(v'^{2} - v^{2})$

$C_{p}(T - T') = \frac{1}{2}(v'^{2} - v^{2})$

$\Delta KE = C_{p}(T - T')$                       (1)

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

$\Delta KE = C_{p}(T - T_{s})$

where

$C_{p} = 1.005 kJ/kgK$

Now,

$\Delta KE = 1.005\times (453 - 330.96) = 122.04 kJ/kg$

(a) For actual kinetic energy :

$\eta_{s} = \frac{\Delta KE_{actual}}{\Delta KE}$

$0.96 = \frac{\Delta KE_{actual}}{122.04}$

$\Delta KE_{actual} = 117.16 kJ/kg$

Also,

$\Delta KE_{actual} = \frac{1}{2}(v'^{2} - v^{2})$              (2)

Now,

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

Using eqn (2):

$117.16 = \frac{1}{2}(v'^{2})$

v' = 15.3 m/s