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 100kPa and 20oC enters it with a velocity of 500 m/s and the exit is 200kPa and 90oC.

330.18m/s

Explanation:

Given data:

T1 = 20 °C

T2 = 90 °C

V = 500m/s

P1 = 100kpa

P2 = 200kpa.

Cp = 1.007kj/KgK

Solution:

Energy balance

Ein - Eout = change in Esystem.

We know

Exit velocity

V2 = (V1^2 + 2(h1 - h2)^0.5 = [V1^2 + 2Cp (T1-T2)]^0.5

input given data into the formula

V2 = [500^2 + 2 (1.007) * (20 - 90) * 1000/1 ]^0.5

V2 = 109020^0.5

V2 = 330.18m/s

Velocity at the exit is 330.18m/s

heats surface water up