A venturi meter is measuring the flow of water; it has a main diameter of 3.5 cm tapering down to a throat diameter of 1.0 cm. If the pressure difference is measured to be 18 mm-Hg, what is the speed of the water entering the venturi throat? (0.18 m/s) 


Given:

Main diameter  = d₁ = 3.5 cm = 0.035 m
Throat diameter = d₂ = 1.0 cm = 0.010 m
Pressure difference = p₁ - p₂ = 18 mm - hg = 18 ｘ 133.3 Pa = 2399.4 Pa
Density of water = ⍴ = 1000 kg m⁻³


To Find:

Speed of water entering the venturi throat = v₁ = ?

Solution:


The speed of water v₁ entering the venturi throat is given by the relation:

v₁ = A₂ `\sqrt frac {2(P₁ - P₂)}{⍴(A₂^2 - A₁^2)}` ------ (1)

So, we have to calculate first the areas A₁ and A₂ which are:

A₁ = 𝜋 r₁² = 𝜋 (d₁/2)² = 𝜋 (0.035 m/2)² = 0.00096 m²

and 

A₂ = 𝜋 r₂² = 𝜋 (d₂/2)² = 𝜋 (0.010 m/2)² = 0.000078 m²

Putting values in equation (1) we have

v₁ = 0.00096 m²ｘ`\sqrt frac {2(2399.4 Pa)}{1000 kg m⁻³((0.000078 m²)^2 - (0.00096 m²)^2)}`

After simplifying and rounding off the number we get

v₁ = 0.18 m s⁻¹  -------------Ans


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