Question

A fireman standing on a 15 m high ladder operates a water hose with a round nozzle of diameter 2.02 inch. The lower end of the hose (15 m below the nozzle) is connected to the pump outlet of diameter 3.37 inch. The gauge pressure of the water at the pump is P (gauge) pump = P (abs) pump − Patm = 65.2 PSI = 449.538 kPa . Calculate the speed of the water jet emerging from the nozzle. Assume that water is an incompressible liquid of density 1000 kg/m3 and negligible viscosity. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m/s.

Answer 1

v₁ = 1,606 10⁴ m / s

Step-by-step explanation:

For this exercise we must use Bernoulli's equation, let's use index 1 for the nozzle on the stairs and index 2 the pump on the street

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

The pressure when the water comes out is the atmospheric pressure

P₁ = P_atm

The difference in height between the street and the nozzle on the stairs is

y₂-y₁ = 15 m

Now let's use the continuity equation

v₁ A₁ = v₂ A₁

The area of ​​a circle is

A = π r² = π (d/2)²

v₁ π d₁²/ 4 = v₂ π d₂²/ 4

v₂ = v₁ d₁² / d₂²

Let's replace

P₂-P_atm + ½ ρ [ (v₁ d₁² / d₂²)²- v₁² ] + ρ g (y2-y1) = 0

P₂- P_Atm + ρ g (y₂-y₁) = ½ ρ v₁² [1- (d₁/d₂)⁴]

v₁² [1- (d₁/d₂)⁴] = (P₂-P_atm) ρ / 2 + g (y₂-y₁) / 2

Let's reduce the magnitudes to the SI system

d₂ = 3.37 in (2.54 10⁻² m / 1 in) = 8.56 10⁻² m

d₁ = 2.02 in = 5.13 10⁻² m

Let's calculate

v₁² [1- (5.13 / 8.56) 4] = 449.538 10³ 10³/2 -9.8 15/2

v₁² [0.8710] = 2.2477 10⁸ - 73.5 = 2.2477 10⁸

v₁ = √ 2.2477 10⁸ /0.8710

v₁ = 1,606 10⁴ m / s