Air at 1.3 bar, 500 k and a velocity of 40 m/s enters a nozzle operating at steady state and expands adiabatically to the exit, where the pressure is 0.85 bar and velocity is 250 m/s. for air modeled as an ideal gas with k = 1.4, determine for the nozzle: (a) the temperature at the exit, in k, and (b) the percent isentropic nozzle efficiency.

Given:

Pressure, P = 1.3 bar

Temperature, T = 500 K

velocity, v = 40 m/s

Pressure, P' = 0.85 bar

velocity, v' = 250 m/s

k = 1.4

Solution:

Now, we know that:

specific heat at constant pressure,

specific heat at constant volume, = 1.005 KJ/kgK

k =

(a) To calculate temperature at exit, T'

Using steady flow Eqn:

                 (1)

where

h = enthalpy =

h'=

Now, from eqn (1)-

T' = 469.70 K

(b) To calculate % isentropic nozzle efficiency:

Using the relation:

⇒

Now,

% isentropic nozzle efficiency,

%   %

%

answer: what are you talking about?

explanation:

v 154.9= 0.9 length length s little over 154.9cm