Question

Air enters a 200 mm diameter adiabatic nozzle at 195 deg C, 500 kPa and 100 m/s. It exits at 85 kPa. If the exit diameter is 158 mm, what are the temperature and velocity at the exit?

Answer 1

`v_2 = 160.23 m/s`

`T_2 = 475.797 k`

Step-by-step explanation:

given data:

Diameter =
`d_1 = 200mm`

`t_1 =195 degree`

`p_1 =500 kPa`

`v_1 = 100m/s`

`p_2 = 85kPa`

`d_2 = 158mm`

from continuity equation

`A_1v_1 = A_2v_2`

`v_2 = ((\pi)/(4)d_1^2 v_1^2)/((\pi)/(4)d_2^2)`

`v_2 = (d_2v_1)/(d_2^2)`

`v_2 = [(d_1)/(d_2)]^2 v_1`

`= [(0.200)/(0.158)]^2 * 100`

`v_2 = 160.23 m/s`

by energy flow equation

`h_1 + (v_1^2)/(2) +gz_1 +q =h_2 + (v_2^2)/(2) +gz_2 +w`

`z_1 =z_2` and q =0, w =0 for nozzle

therefore we have

`h_1 -h_2 =(v_1^2)/(2) -(v_2^2)/(2)`

`dh = (1)/(2) (v_1^2 -v_2^2)`

but we know dh = Cp dt

hence our equation become

`Cp(T_2 -T_1) = (1)/(2) (v_1^2 -v_2^2)`

`Cp (T_2 -T_1) = 7836.94`

`(T_2 -T_1) = (7836.94)/(1.005*10^3)`

`(T_2 -T_1) = 7.797`

`T_2 = 7.797 +468 = 475.797 k`