How Does Air Speed Change in a Diffuser?

[Luchekv]

Homework Statement

Air enters a diffuser:
The ambient temp at the inlet is 300K.
The temp leaving the diffuser is 350K
The flow speed of air at the inlet is 400m/s

Assumptions:
Potential energy = 0
Kinetic energy at the diffuser exit is = 0
Air is an ideal gas

Q1: assuming air to be an ideal gas with variable spec heats, calculate the flow speed (m/s) for the outlet

Homework Equations

Ein=Eout
m*(h1+ (V₁²/2)=m*(h2+ (V₂²/2)

V₂=[(V₁²+2(h1-h2)]^0.5[/B]

The Attempt at a Solution

I looked up the enthapies for the given temps on the 'ideal-gas properties table'
and got:
h1=300.19
h2=350.49

then I sub into this formula:
V₂=[(V₁²+2(h1-h2)]^0.5

0=(350.49-300.19)kj/kg+((V₂²-(400m/s)²)/2)* (1kj/kg/1000m²/s²)
V₂=243.721

but apparently the answer is 237m/s...

 

[anathema]

Your solution is almost correct, but there is a small error in your calculation. The equation you used for kinetic energy (KE) is missing a multiplication by the mass of the air (m). The correct equation should be:

Ein = Eout
m*(h1+ (V12/2))=m*(h2+ (V22/2))

V2=[(V12+2(h1-h2))]*0.5

Substituting the values, we get:

0=(350.49-300.19)kJ/kg+((V22-(400m/s)2)/2)* (1kJ/kg/1000m2/s2)
V2=237.145 m/s

Therefore, the flow speed at the outlet is approximately 237 m/s.