A horizontally oriented pipe has a diameter of 5.8 cm and is filled with water. The pipe draws water from a reservoir that is initially at rest. A manually operated plunger provides a force of 397 N in the pipe. Assuming that the other end of the pipe is open to the air, with what speed does the water emerge from the pipe?

v₂ = 97.4 m / s

Explanation:

Let's write the Bernoulli equation

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

Index 1 is for tank and index 2 for exit

We can calculate the pressure in the tank with the equation

P = F / A

Where the area of a circle is

A = π r²

E radius is half the diameter

r = d / 2

A = π d² / 4

We replace

P = F 4 / π d²2

P₁ = 397 4 / π 0.058²

P₁ = 1.50 10⁵ Pa

The water velocity in the tank is zero because it is at rest (v1 = 0)

The outlet pressure, being open to the atmosphere is P1 = 1.13 105 Pa

Since the pipe is horizontal y₁ = y₂

We replace on the first occasion

P₁ = P₂ + ½ ρ v₂²

v₂ = √ (P1-P2) 2 / ρ

v₂ = √ [ (1.50-1.013) 10⁵ 2/1000]

v₂ = 97.4 m / s