Thermofluids: Air Flow Measurement (Pitot & Venturi)

AI Thread Summary
The experiment measured air flow using a pitot static tube and a venturi meter, yielding different volumetric flow rates: 0.55386 m³/s for the pitot tube and 0.34575 m³/s for the venturi meter. The discrepancy raised questions about whether the flow rates should theoretically be the same, suggesting a potential need for correction factors due to non-uniform velocity across the duct's cross-section. The pitot tube was positioned in the middle for maximum velocity pressure, but access to the equipment for further testing was unavailable after the lab session. The discussion emphasizes the importance of understanding flow measurement principles and potential factors affecting accuracy. Clarifying these concepts could enhance experimental outcomes in future studies.
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Homework Statement


I did an experiment to measure air flow using a pitot static tube and a venturi meter.
First attempt: I connected both long and short manometers to the pitot static tube. Then I switch on the fan. The difference in manometer reading is 0.59kPa, which is the pressure. Then I found the volumetric flow rate to be 0.55386.

Second attempt: I connected both long and short manometers to the venturi meter. Then I switch on the fan. The difference in manometer reading is 1.24kPa, which is the pressure. Then I found the volumetric flow rate to be 0.34575.

What I don't understand is that, should theoretically the volumetric flow rate of both of them the same? Or it is supposed to be different? Why is it different? Thank you.

Homework Equations


Pitot tube:
Q = a*vm
a = area of duct (duct diameter 0.15m)
vm = 1.291*sqrt Pv

Venturi Meter:
Q = Cd*a2 〖[2∆P/ {ρ(1-〖(a_2/a_1 )〗^2)}]〗^0.5

Cd= 0.98
d1= duct diameter 0.15m, a1= duct area
d2= Throat diameter (0.095) a2= throat area
ρ= 1.2 kg/m^3

The Attempt at a Solution


Pitot-static tube:
vm = 1.291 √(P_v )
vm = 1.291 √(0.59×〖10〗^3 )
vm = 31.36 m2/s

Q = avm
Q = π/4×〖0.15〗^2×31.36
Q = 0.554 m3/s

Venturi meter:
Q = Cda2 〖[2∆P/ {ρ(1-〖(a_2/a_1 )〗^2)}]〗^0.5 (m3/s)
Q = 0.98 × π/4 ×〖0.095〗^2 〖[2×1.23×〖10〗^3/ {1.2(1-〖(〖0.095〗^2/〖0.15〗^2 )〗^2)}]〗^0.5
Q = 6.943×〖10〗^(-3) 〖[2460/(1.2 × 0.839)]〗^0.5
Q = 0.3436 m3/s
 
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Can you try the experiment again but move the pitot tube to different positions within the duct cross section?
 
paisiello2 said:
Can you try the experiment again but move the pitot tube to different positions within the duct cross section?

Unfortunately, I can't. I am not given the access to the pitot and venturi meter after the lab session was finished.

I remembered I put it in the middle to obtain max velocity pressure though.
 
So should there be a correction factor to account for the fact that the velocity is not uniform along the cross section of the duct? This might explain why the pitot value is higher than the venturi value.
 
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The truth is I don't know. Do you mean a correction factor for the pitot static tube?
Does this mean that, theoretically, the value of the flow rate should have been the same?
Thank you very much.
 
Yes, that's what I mean.
 
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