Water flows in a pipeline. At a point in the line where the diameter is 7 in., the velocity is 12 fps and the pressure is 50 psi. At a point 40 ft away, the diameter reduces to 3 in. Calculate the pressure here when the pipe is (a) horizontal, (b) vertical with flow downward, and (c) vertical with the flow upward. Explain why there is a difference in the pressure for the different situations.

a)   P₂ = 3219.11 lbf / ft² , b)    P₂ = 721.91 lbf / ft² , c)  P₂ = 5707.31 lbf / ft²

Explanation:

For this exercise we can use the fluid mechanics equations, let's start with the continuity equation, index 1 is for the starting point and index 2 for the end point of the reduction

     A₁ v₁ = A₂ v₂

     v₂ = v₁ A₁ / A₂

The area of a circle is

    A = π r² = π/4  d²

     v₂ = v₁ (d₁ / d₂)²

Let's calculate

    v₂ = 12 (7/3)²

    v₂ = 65 feet / s

Now let's use Bernoulli's equation

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

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

Case 1. The pipe is horizontal, so

      y₁ = y₂

      P₁ - P₂ = ½ ρ  (v₂² –v₁²)

      P₂ = P₁ - ½ ρ (v₂² –v₁²)

     ρ = 62.43 lbf / ft³

     P₁ = 50 psi (144 lbf/ ft² / psi) = 7200 lbf / ft²

    P₂ = 7200 - ½ 62.43 / 32 (65² -12²)

    P₂ = 7200 - 3980.89

    P₂ = 3219.11 lbf / ft²

Case 2 vertical pipe with water flow up

        y₂ –y₁ = 40 ft

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

        7200 - P₂ = 62.43 (40) + ½ 62.43 / 32 (65 2 - 12 2) =

        P₂ = 7200 - 2497.2 - 3980.89

         P₂ = 721.91 lbf / ft²

Case 3. Vertical water pipe flows down

         y₂ –y₁ = -40

         P₂ = 7200 + 2497.2 - 3980.89

         P₂ = 5707.31 lbf / ft²

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