This question is incomplete, the complete question is;
Air flow is measured in a Venturi meter that has a large diameter of 1.5 m and a small diameter of 0.9 m. A 1:12 scale model with water as the fluid is used to calibrate the meter. For the model, it is determined that when the volume flowrate is 0.07 m3 /s, the pressure drop from the large diameter portion (Section 1) to the small diameter portion (Section 2) is 172 kPa.
Calculate The corresponding flowrate in the prototype.
Assume a water temperature of 15°C and standard properties of air
Answer: The corresponding flowrate in the prototype is 10.21 m³/s
Step-by-step explanation:
Given that;
Lm = Lp/12 and lp = 12Lm, Qm = 0.07 m³/s, ΔPm = 172 Kpa
properties of water at 15°C ---- Vm = 1.2015 × 10⁻⁶ m²/s, Sm = 1001.2 kg/m₂
Also for Air---- Vp = 1.46041 × 10⁻⁵, Sp = 1.225 kg/m³
Now by Using Reynold's model law; (Vm × Lm)/Vm = (Vp × Lp)/Vp
(Vm × Lm) / 1.2015 × 10⁻⁶ = (Vp ×12 × Lm) / 1.46041 × 10⁻⁵
Vm/Vp = 0.9872
we know that
Discharge = Area × Velocity
Qm/Qp = Lm²/Lpl × Vm/Vp
= (1/12)² × 0.9872
= 6.856 × 10⁻³
so Qp = (0.07 / 6.856 × 10⁻³) = 10.21 m³/s
Therefore The corresponding flowrate in the prototype is 10.21 m³/s