Respuesta :
Answer:
ΔP = - 0.262 ρ v₁²
Explanation:
This is a fluid mechanics problem, let's use the subscript for region I and the subscript for region II. Let's write Bernoulli's equation
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
suppose the pipe is horizontal, therefore
y₁ = y₂
P₁ -P₂ = ½ ρ (v₁² -v₂²) 1
Now let's use the continuity equation
v₁ A₁ = v₂ A₂ 2
The area of a circle is
A = π r²
also the radius is half the diameter r = d /2
A₁ = π d₁² / 4
let's substitute in equation 2
v1 π d₁² / 4 = v2 π d₂² / 4
v₁ d₁² = v₂ d₂²
v₂ = v₁ d₁² / d₂²
let's substitute in equation 1
P₁ - P₂ = ½ ρ (v₁² - v₁² (d₁² / d₂²)² )
Now let's use that the diameter d₂ = d is 10% smaller than the larger diameter d₁ = D, we assume that it is in region 1
d₂ = D -0.1D = 0.9 D
we substitute in the previous equation
P₁ - P₂ = ½ ρ v₁² (1 - (D / 0,9D)⁴)
P₁- P₂ = ½ ρ v₁² (1 - 1 / 0,9⁴ )
P1 - P2 = ½ ρ v12 (- 0.524)
ΔP = - 0.262 ρ v₁²
in this solution we assume that the data in zone I is known
