Question

Water, with a density of 1000 kg/m3, flows out of a spigot, through a hose, and out a nozzle into the air. The hose has an inner diameter of 2.25 cm. The opening in the nozzle that the water comes out of has a diameter of 2.00 mm. The water coming out of the nozzle, which is held at a height of 7.25 meters above the height of the spigot, has a velocity of 11.2 m/s. Neglecting viscosity and assuming that the water flow is laminar (not necessarily good assumptions, but let's not make this any harder than it already is), what is the pressure of the water in the hose right after it comes out of the spigot where the water enters the hose (to three significant digits)? Assume that ????=9.80 m/s2 and that the surrounding air is at a pressure of 1.013×105 N/m2

Answer 1

P₁ = 2.3506 10⁵ Pa

Step-by-step explanation:

For this exercise we use Bernoulli's equation and continuity, where point 1 is in the hose and point 2 in the nozzle

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

A₁ v₁ = A₂ v₂

Let's look for the areas

r₁ = d₁ / 2 = 2.25 / 2 = 1,125 cm

r₂ = d₂ / 2 = 0.2 / 2 = 0.100 cm

A₁ = π r₁²

A₁ = π 1.125²

A₁ = 3,976 cm²

A₂ = π r₂²

A₂ = π 0.1²

A₂ = 0.0452 cm²

Now with the continuity equation we can look for the speed of water inside the hose

v₁ = v₂ A₂ / A₁

v₁ = 11.2 0.0452 / 3.976

v₁ = 0.1273 m / s

Now we can use Bernoulli's equation, pa pressure at the nozzle is the air pressure (P₂ = Patm) the hose must be on the floor so the height is zero (y₁ = 0)

P₁ + ½ ρ v₁² = Patm + ½ ρ v₂² + ρ g y₂

P₁ = Patm + ½ ρ (v₂² - v₁²) + ρ g y₂

Let's calculate

P₁ = 1.013 10⁵ + ½ 1000 (11.2² - 0.1273²) + 1000 9.8 7.25

P₁ = 1.013 10⁵ + 6.271 10⁴ + 7.105 10⁴

P₁ = 2.3506 10⁵ Pa