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 356 m/s and the exit state is 200 kPa and 90°C. The specific heat of air at the average temperature of 60°C

Answer 1

Answer:

The exit velocity air is 77.56 m/s.

Explanation:

Specific heat of air at 60°C is about 1.006 KJ/kg-K (from standard table).

Now from first law for open system

$h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+w$

For ideal gas h=CT

Given that $T_1=30$°C

$T_2=90$°C

$V_1=356 m/s$

By putting the values

$1.006\times (273+30)+\dfrac{356^2}{2000}=1.006\times (273+90)+\dfrac{V_2^2}{2000}+$

$V_2=77.56 m/s$

So the exit velocity air is 77.56 m/s.