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2 answers:
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Answer:
The exit temperature is 293.74 K.
Explanation:
Given that
At inlet condition(1)
P =80 KPa
V=150 m/s
T=10 C
Exit area is 5 times the inlet area
Now
If consider that density of air is not changing from inlet to exit then by using continuity equation
So
m/s
Now from first law for open system
Here Q=0 and w=0
When air is treating as ideal gas
Noe by putting the values
So the exit temperature is 293.74 K.
Answer
diameter of parking lot = 18 in
flowrate = 10 ft³/s
pressure drop = 100 ft
using general equation
taking f = 0.0185
at Z₁ = Z₂
ΔP = 0.266 psi
b) when flow is uphill z₂-z₁ = 2
c) When flow is downhill z₂-z₁ = -2