Consider a venturi with a small hole drilled in the side of the throat. This hole is connected via a tube to a closed reservoir. The purpose of the venturi is to create a vacuumi in the reservoir when the venturi is placed in an airstream. The venturi has a throat-to-inlet area ratio of 0.85. Calculate the maximum vacuum obtainable in the reservoir when the venturi is placed in an airstream of 80 m/s at standard sea level conditions.

ΔP = - 689.92 Pa

Explanation:

This is a fluid mechanics exercise, let's use Bernoulli's equation, where subscript 1 is for the wide part and subscript 2 is for the narrow part

          P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

suppose the venturi tube is level, y₁ = y₂

Now let's use the continuity equation

         A₁ v₁ = A₂ v₂

         v₂ =

we substitute

         P₁ + ½ ρ v₁² = P₂ + ½ ρ v₁² (\frac{A_1}{A_2} \ v_1 )²

         P₂ = P₁ + ½ ρ v₁² ( )

we have assumed that the density of the air does not change,

         ρ (air) = 1,225 kg / m³

         P₁ = P_{atm}

the relation of the entry and restriction area is

         A₂ / A₁ = 0.85

Thus

         A₁ / A₂ = 1,176

let's calculate

         P₂ - P₁ =  ½ 1,225 80² (1- 1,176)

         ΔP = - 689.92 Pa

The pressure in the reservoir will spread the amount of 689.92 Pa of its initial pressure

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c)as speed changes, wavelength and frequency change.

everything changes!