Answer :
Answer:
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 = [tex]\frac{A_1}{A_2} \ v_1[/tex]
we substitute
P₁ + ½ ρ v₁² = P₂ + ½ ρ v₁² (\frac{A_1}{A_2} \ v_1 )²
P₂ = P₁ + ½ ρ v₁² ( [tex]1 - \frac{A_1}{A_2}[/tex] )
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