Problem Statement:

Air enters an adiabatic nozzle at 400 kPa and 900 K with negligible velocity. If the flow is ideal and exit pressure is 100 kPa, the exit temperature in K and the exit velocity in m/s are respectively?


Solution:

Given,

Inlet pressure (P1) = 400 kPa

Inlet temperature (T1) = 900 K

Exit Pressure (P2) = 100 kPa

Velocity at the inlet (V1) = 0 m/s


Assumption:

The flow is ideal and adiabatic.


Formula:

The temperature at the exit of the nozzle (T2) can be calculated using the isentropic relation,

T2 = T1*(P2/P1)^((γ-1)/γ)

where γ = Cp/Cv is the ratio of specific heats.


The velocity at the exit of the nozzle (V2) can be calculated using the Bernoulli's equation,

P1 + ½ρV1^2 = P2 + ½ρV2^2

where ρ is the density of air.


Calculation:

Given, Cp/Cv = γ = 1.4 for air.

Using the isentropic relation,

T2 = T1*(P2/P1)^((γ-1)/γ)

T2 = 900*(100/400)^((1.4-1)/1.4)

T2 = 536.24 K


Using the Bernoulli's equation,

P1 + ½ρV1^2 = P2 + ½ρV2^2

Since the velocity at the inlet (V1) is negligible, V1^2 can be neglected.

Hence, P1 = P2 + ½ρV2^2

The density of air (ρ) can be calculated using the ideal gas law,

P1V1/T1 = P2V2/T2

ρ = P1/(RT1)

where R = 287 J/kg.K is the gas constant for air.

Substituting the values, we get,

ρ = 400000/(287*900)

ρ = 1.627 kg/m^3

Substituting the values in the equation P1 = P2 + ½ρV2^2

400000 = 100000 + ½*1.627*V2^2

V2 = 365.26 m/s


Answer:

The exit temperature of air (T2) is 536.24 K and the exit velocity of air (V2) is 365.26 m/s.


Conclusion:

The exit temperature and velocity of air in an adiabatic nozzle can be calculated using the isentropic relation and Bernoulli's equation respectively.