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:

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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

 The corresponding flowrate in the prototype is 10.21 m³/s

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

energy can cross the boundaries of a closed system in the form of heat and mass

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