Respuesta :
Answer:
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
The pressure when the pipe is horizontal is 22psi.
How to calculate the pressure
Firstly, we'll apply the equation of continuity for the system. This will be:
A₁V₁ = A₂V₂
(π × 7²/4) × 12 = (π × 3²/4) × V₂
V₂ = 65.33 fps
For the horizontal pipe, the pressure will be calculated thus:
50 × 12² + 0.5 × 1.95 × 12² + pgh = P₂ + 0.5 × 1.95 × 65.33² + pgh
P₂ = 3179 lb/ft²
P₂ = 22psi
For the vertical pipe, the pressure will be:
= 50 × 12² + 1/2 × 1.95 × 12² + 62.4 × 40 = P₂ + 1/2 × 1.95 × 65.33²
P₂ = 5675lb/ft²
P₂ = 39.41 psi
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