Water flows through a horizontal nozzle in steady flow at the rate of 10m3/s. the inlet and outlet diameters are d1 = 0.5m and d2 = 0.25m. water exits the nozzle to the atmosphere. calculate the pressure at the entrance to the nozzle and the force required to hold the nozzle in place. neglect the weight of the nozzle. rho = 1000kg/m3, g = 10m/s2.

P₁ = 2.215 10⁷ Pa, F₁ = 4.3 106 N,

Explanation:

This problem of fluid mechanics let's start with the continuity equation to find the speed of water output

        Q = A v

        v = Q / A

The area of a circle is

       A = π r² = π d² / 4

Let's look at the speeds at each point

       v₁ = Q / A₁ = Q 4 /π d₁²

       v₁ = 10 4 /π 0.5²

       v₁ = 50.93 m / s

       v₂ = Q / A₂

       v₂ = 10 4 /π 0.25²

       v₂ = 203.72 m / s

Now we can use Bernoulli's equation in the colon

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

Since the tube is horizontal y₁ = y₂. The output pressure is P₂ = Patm = 1.013 10⁵ Pa, let's clear

       P₁ = P2 + ½ rho (v₂² - v₁²)

      P₁ = 1.013 10⁵ + ½ 1000 (203.72² - 50.93²)

      P₁ = 1.013 10⁵ + 2.205 10⁷

      P₁ = 2.215 10⁷ Pa

la definicion de presion es

      P₁ = F₁/A₁

     F₁ = P₁ A₁

     F₁ = 2.215 10⁷ pi d₁²/4

     F₁ = 2.215 10⁷ pi 0.5²/4

     F₁ = 4.3 106 N