Question

The diffuser in a jet engine is designed to decrease the kinetic energy of the air entering the engine compressor without any work or heat interactions. Calculate the velocity at the exit of a diffuser when air at 100 kPa and 30°C enters it with a velocity of 350 m/s and the exit state is 200 kPa and 90°C. (Page 216).

Answer 1

V₂=43.58 m/s

Step-by-step explanation:

Given that

P₂=200 KPa

T₂=90°C

V₂=? m/s ( exit speed)

T₁=30°C

P₁=100 KPa

V₁=350 m/s

We know that

Heat capacity for air Cp=1.005 KJ/kg.k

We know that for air change in enthalpy only depends only on temperature

Now from first law for open system

`h_1+(V_1^2)/(2000)=h_2+(V_2^2)/(2000)+W`

`1.005* 303+(350^2)/(2000)=1.005* 363+(V_2^2)/(2000)`

`1.005* 303+(350^2)/(2000)-1.005* 363=(V_2^2)/(2000)`

`(V_2^2)/(2000)=0.95`

V₂=43.58 m/s