Water with pressure of 190 kPa flows down hill through a pipe with a 3.90 cm diameter hose at a speed of 1.2 m/s. It then exits through a smaller pipe m of diameter 1.50 cm which is 5.00 m below the upper pipe.

a. What is the speed of the water exiting at the lower pipe opening?

b. What is the water pressure at the exit of the lower pipe.?

a) v₂ = 8,112 m / s, b) P₂ = 206.8 10³ Pa

Explanation:

This is a fluid mechanics exercise, let's use index 1 for thick pipe and index 2 for thinner pipe

a) with the continuity equation it allows to find the speed of the water

v₁ A₁ = v₂ A₂

the area of a circle is

A₁ = π r₁² = π d₁² / 4

v₂ = v₁ d₁² / d₂²

let's calculate

v₂ = 1.2 3.90² / 1.50²

v₂ = 8,112 m / s

b) We use Bernoulli's equation

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

P2 = P1 + ½ ρ (v₁² - v₂²) + ρ g (y₁ - y₂)

let's calculate

P₂ = 190 10³ + ½ 1000 (1.2² - 8.112²) + 1000 9.8 5

P₂ = 206.8 10³ Pa