In a jet engine a diffuser decreases the

Answered question

2022-04-14

In a jet engine a diffuser decreases the kinetic energy of the air entering the engine.  The air enters a the diffuser at 132 kPa and 25°C with a velocity of 502 m/s and leaves at 229 kPa and 64°C.  The heat lost from the diffuser is estimated to be 6 kJ/kg.  Determine the exit velocity in m/s to 1 decimal place.  Take the specific heat at constant pressure of air to be 1.009 kJ/(kgK).

Answer & Explanation

user_27qwe

user_27qwe

Skilled2023-04-27Added 375 answers

First, we can find the change in temperature of the air:
ΔT=T2T1=64°C25°C=39°C=312 K
Next, we can find the change in pressure of the air:
ΔP=P2P1=229 kPa132 kPa=97 kPa
To find the change in specific enthalpy, we can use the equation:
Δh=CpΔT
where Cp is the specific heat at constant pressure of air, given as 1.009 kJ/(kgK).
Δh=1.009 kJ/(kgK)·312 K=315.408 kJ/kg
Since the diffuser decreases the kinetic energy of the air, the change in specific enthalpy should be negative. Therefore, we can write:
Δh=12v22+12v12
where v1 is the initial velocity of the air (502 m/s) and v2 is the exit velocity of the air, which we want to find.
Rearranging this equation, we can solve for v2:
v2=2Δhρ+v1
where ρ is the density of air. To find the density, we can use the ideal gas law:
PV=mRT
where P is the pressure, V is the volume, m is the mass, R is the specific gas constant for air, and T is the temperature in Kelvin.
We can assume that the volume of the air is constant, so we can write:
P1ρ1RT1=P2ρ2RT2
Solving for ρ2, we get:
ρ2=P2RT2·ρ1RT1P1=229 kPa(287 J/(kgK))·(64+273) K·(132 kPa)(287 J/(kgK))·(25+273) K=1.685 kg/m3
Substituting the values we found into the equation for v2, we get:
v2=2(315.408 kJ/kg)1.685 kg/m3+502 m/s=283.4 m/s
Therefore, the exit velocity of the air is approximately 283.4 m/s.

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