Question

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.

Answer 1

Answer:

a) T₂ = 376.905 K

b) $\eta_n$ = 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

$h_1+\frac{v_1^2}{2}=h_2+\frac{v_2^2}{2}$

where, h is the enthalpy

on rearranging we get

$\frac{v_2^2-v_1^2}{2}=h_1-h_2$

also

h = Cp × T

thus,

$\frac{v_2^2-v_1^2}{2}=C_p(T_1-T_2)$

on substituting the values, we get

$\frac{307^2-40^2}{2}=1.005\times 10^3(423-T_2)$

or

T₂ = 376.905 K

b) The  isentropic nozzle efficiency is given as:

$\eta_n=\frac{V_2^2/2}{V_1^2/2}=\frac{Actual\ expansion}{Ideal\ Expansion}$

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

$\frac{T_{2s}}{T_1}=(\frac{P_2}{P_1})^{\frac{k-1}{k}}$

on substituting the values and solving, we get

$T_{2s}$ = 374.65 K

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

$V_{2s}=\sqrt{2C_p(T_2-T_1)+V_1^2}$

on substituting the values, we get

$V_{2s}=\sqrt{2\times1.005\times10^3(423-374.65)+40^2}$

or

$V_{2s}=314.29\ m/s$

thus,

substituting in the formula for efficiency, we get

$\eta_n=\frac{307^2/2}{314.29^2/2}$

or

$\eta_n=0.954$ = 95.4%