Respuesta :
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, [tex]C_{p} = 1.005 KJ/kgK[/tex]
specific heat at constant volume, [tex]C_{p}[/tex] = 1.005 KJ/kgK
k = [tex]\frac{C_{p}}{C_{V}}[/tex]
(a) To calculate temperature at exit, T'
Using steady flow Eqn:
[tex]h + \frac{v^{2}}{2} = h' + \frac{v'^{2}}{2}[/tex] (1)
where
h = enthalpy = [tex]C_{p}T[/tex]
h'= [tex]C_{p}T'[/tex]
Now, from eqn (1)-
[tex]h + \frac{v^{2}}{2} = h' + \frac{v'^{2}}{2}[/tex]
[tex]C_{p}T + \frac{v^{2}}{2} = C_{p}T' + \frac{v'^{2}}{2}[/tex]
[tex]1005\times 500 + \frac{40^{2}}{2} = 1005\times T' + \frac{v'^{2}}{2}[/tex]
T' = 469.70 K
(b) To calculate % isentropic nozzle efficiency:
Using the relation:
[tex]\frac{T_{2s}}{T} = (\frac{P'}{P})^\frac{k - 1}{k}[/tex]
⇒ [tex]\frac{T_{2s}}{500} = (\frac{0.85}{1.3})^\frac{1.4 - 1}{1.4}[/tex]
[tex]{T_{2s} = 0.88 \times 500 = 442.84 K[/tex]
Now,
% isentropic nozzle efficiency, [tex]\eta =\frac{T - T' }{T - T_{2s}}\times 100 [/tex]
% [tex]\eta =\frac{500 - 469.70 }{500 - 442.84}\times100 = 53.00[/tex] %
[tex]\eta = 53.00[/tex] %
