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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