Air at 1.3 bar, 423 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 307 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 isentropic nozzle efficiency.

a) T₂ = 376.905 K

b) = 95.4%

Explanation:

Given:

Initial pressure, P₁ = 1.3 bar

Initial temperature, T₁ = 423 K

Initial velocity of air, v₁ = 40 m/s

Exit pressure, P₂ = 0.85 bar

Exit velocity of the air, v₂ = 307 m/s

gas constant for air, k = 1.4

also,

the specific heat for the air, Cp = 1.005 KJ/kg.K =1.005 × 10³ J/kg.K   (standard)

a) applying the energy rate balance, we have

where, h is the enthalpy

on rearranging we get

also

h = Cp × T

thus,

on substituting the values, we get

or

T₂ = 376.905 K

b) The  isentropic nozzle efficiency is given as:

for an isentropic process,we have the isentropic relation as:

on substituting the values and solving, we get

= 374.65 K

by applying the energy equation, we get the ideal exit velocity as:

on substituting the values, we get

or

thus,

substituting in the formula for efficiency, we get

or

= 95.4%

490 n

f=125n a=2.5m/s2