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

Answer:

V₂=43.58 m/s

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

$h_1+\dfrac{V_1^2}{2000}=h_2+\dfrac{V_2^2}{2000}+W$

$1.005\times 303+\dfrac{350^2}{2000}=1.005\times 363+\dfrac{V_2^2}{2000}$

$1.005\times 303+\dfrac{350^2}{2000}-1.005\times 363=\dfrac{V_2^2}{2000}$

$\dfrac{V_2^2}{2000}=0.95$

V₂=43.58 m/s