Question

when the atmospheric pressure is 1 atm a water fountain ejects a stream of water that rises to a height of 5 m. There is a 1.9-cm-radius pipe that leads from a pressurized tank to the opening that ejects the water. 1. What would happen if the fountain were operating when the eye of a hurricane passes through? 2. Assume that the atmospheric pressure in the eye is 0.877 atm and the tank's pressure remains the same. What will be the height?

Answer 1

Answer:

1) When we are in the eye of a hurricane the atmospheric pressure decreases so the water must rise more         2) y = 25.3 m

Explanation:

In this exercise we must use the concepts and mechanics of fluids and kinematics to solve them. Let's start by finding the speed (v₁) so the atmosphere left the pipe, for this we use the vertical launch equations

           v² = v₀² - 2 g y

At the highest point the speed is zero

          v₀ = v₁ = √ 2 g y

          v₁ = √ 2 9.8 5

          v₁1 = 9.90 m / s

Now let's use the continuity equation to find the velocity (v₂) in the wide part of the tank

         A₁ v₁ = A₂ v₂

They indicate that the radius of the pipe r₂/r₁ = 1.9, in area of ​​a circle is

        A = π R²

       A₂ / A₁ = π r₂² /π r₁²

        A₂ / A₁ = (r₂ / r₁)²

       v₂ = A₁ /A₂   v₁

       v₂ = (1 / 1.9)² 9.9

       v₂ = 2.74 m / s

Now let's use Bernoulli's equation, to find the pressure, point 1 is at the exit and point 2 is in the tank,

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

In the problem it indicates that P₁ = 1 atm, they do not give explicit indications about the height of the pipe, suppose that it is horizontal (the most common) y₁ = y₂

      P₂ - $P_{atm}$ = ½ ρ (v₁² - v₂²)

      P₂ = $P_{atm}$ + ½ ρ (v₁² -v₂²)

Let's calculate

      P₂ = 1.013 10⁵ + ½ 1000 (9.90² - 2.74²)

      P₂ = 1.013 10⁵ + 4.525 10⁴

      P₂ = 1.47 10⁵ Pa

1) When we are in the eye of a hurricane the atmospheric pressure decreases so the water must rise more

2) Let's use Bernoulli's equation to find the output velocity (v₁)

       ½ ρ v₁² = P₂- P₁ + ½ ρ v₂²)

Let's calculate

       P₂ = 0.877 1.013 10⁵ Pa

       P₂ = 0.888 10⁵ Pa

       ½ 1000 v₁² = 1.47 10⁵ - 0.888 10⁵ + ½ 1000 2.74²

      500 v₁² = 0.582 10⁵ + 3753.8

      v₁ = √ (61953.8 / 500)

      v₁ = 11.13 m / s

now we use kinematics

     v₀ = 11.13 m/s

     v² = v₀² - 2 g y

    y = v₀² / 2g

    y = 11.13 2/2 9.8

    y = 25.3 m