The water pressure in the pipes going into an office building is 4 atm at street level. in other words, the waterpressure at street level is 3 atm above. the water enters the 5.0 cm diameter street level pipe at a speedof 1 m/s. the pipe tapers down to a diameter of 3.0 cm at the top floor, 15 m above street level where afaucet is left on. what is the flow velocity and pressure in the pipe at the top floor. assume no branch pipes and ignore viscosity.

v₂ = 2.78 m / s ,   P₂ = 2.55 10⁵ Pa

Explanation:

For this exercise in fluid mechanics, we use Bernoulli's equation, which is an expression of energy conservation.

         P + ½ ρ v² +ρg y = cte.

And the continuity equation

        Q = A v

Let's define two points the street level, point 1 and the level of the third floor point 2, let's see what they give us

At point 1 we have the pressure (P₁ = 4 atm), the diameter of the pipe (d₁ = 5.0 cm) the speed with which it arrives (v₁ = 1.0 m / s)

At point 2 they give the pressure (P₂ = 1 atm), pipe diameter (d₂ = 3.0 cm) and height (y₂ = 15m)

Ask for speed, let's use the continuity equation at the two points

        A₁ v₁ = A₂ v₂

The area and the pipe are

       A = π r² = π (d/2)²

       π / 4 d₁² v₁ = π / 4 d₂² v²

       v₂ = v₁ d₁² / d₂²

       v₂ = 1 5.0² / 3.0²

       v₂ = 2.78 m / s

Let's reduce the magnitudes to the SI System

       P₁ = 4 atm (1,013 10⁵ Pa / 1 atm) = 4,052 10⁵ Pa

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

       P₂ = P₁ + ½ ρ (v₁² - v₂²) + ρ g (y₁ - y₂)

calculate

       P₂ = 4,052 10⁵+ ½ 1000 (1² - 2.78²) + 1000 9.8 (0 - 15)

       P₂ = 4.052 10⁵– 3364.2 - 147000

       P₂ = 254835.8 Pa

       P₂ = 2.55 10⁵ Pa

no. dot product   =0 when vecs at right angles

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