# Thermofluids: Air Flow Measurement (Pitot & Venturi)

**1. The problem statement, all variables and given/known data**

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.

**2. Relevant 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

**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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