Question

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.

Answer 1

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, $C_{p} = 1.005 KJ/kgK$

specific heat at constant volume, $C_{p}$ = 1.005 KJ/kgK

k = $\frac{C_{p}}{C_{V}}$

(a) To calculate temperature at exit, T'

Using steady flow Eqn:

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

where

h = enthalpy = $C_{p}T$

h'= $C_{p}T'$

Now, from eqn (1)-

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

$C_{p}T + \frac{v^{2}}{2} = C_{p}T' + \frac{v'^{2}}{2}$

$1005\times 500 + \frac{40^{2}}{2} = 1005\times T' + \frac{v'^{2}}{2}$

             

T' = 469.70 K

(b) To calculate % isentropic nozzle efficiency:

Using the relation:

$\frac{T_{2s}}{T} = (\frac{P'}{P})^\frac{k - 1}{k}$

⇒ $\frac{T_{2s}}{500} = (\frac{0.85}{1.3})^\frac{1.4 - 1}{1.4}$

${T_{2s} = 0.88 \times 500 = 442.84 K$

Now,

% isentropic nozzle efficiency, $\eta =\frac{T - T' }{T - T_{2s}}\times 100$

%  $\eta =\frac{500 - 469.70 }{500 - 442.84}\times100 = 53.00$ %

$\eta = 53.00$ %