Answer:
48.61
Explanation:
See attached diagram.
The level rise in the tube is l sin α.
The level drop in the cylinder (let's call it y) is:
π/4 D² y = π/4 d² l
D² y = d² l
y = l (d/D)²
The elevation difference is the sum:
l sin α + l (d/D)²
l (sin α + (d/D)²)
From Bernoulli's principle:
P = ρgl (sin α + (d/D)²)
Divide both sides by density of water (ρw) and gravity:
P/(ρw g) = (ρ/ρw) l (sin α + (d/D)²)
h = S l (sin α + (d/D)²)
If we disregard the level change in the cylinder:
h = S l (sin α)
We want the percent error between these two expressions for h to be 0.1% when α = 25°.
[ S l (sin α + (d/D)²) − S l (sin α) ] / [ S l (sin α + (d/D)²) ] = 0.001
[ S l sin α + S l (d/D)² − S l sin α ] / [ S l (sin α + (d/D)²) ] = 0.001
[ S l (d/D)² ] / [ S l (sin α + (d/D)²)] = 0.001
(d/D)² / (sin α + (d/D)²) = 0.001
(d/D)² = 0.001 (sin α + (d/D)²)
(d/D)² = 0.001 sin α + 0.001 (d/D)²
0.999 (d/D)² = 0.001 sin α
d/D = √(0.001 sin α / 0.999)
When α = 25°:
d/D ≈ 0.02057
D/d ≈ 48.61
In this exercise we have to use the knowledge of pressure to be able to determine the value that has inside the cylinder:
48.61
Using the formula for the area of the cylinder we find that:
[tex]\pi/4 D^2 y = \pi/4 d^2 l\\D^2 y = d^2 l\\y = l (d/D)^2[/tex]
The elevation difference is the sum:
[tex]l sin \theta + l (d/D)^2\\l (sin \theta + (d/D)^2)[/tex]
From Bernoulli's principle:
[tex]P = \rho gl (sin \theta + (d/D)^2)\\P/(\rho w g) = (\rho /\rho w) l (sin \theta + (d/D)^2)\\h = S l (sin \theta + (d/D)^2)[/tex]
We want the percent error between these two expressions for h to be 0.1% when α = 25°.
[tex][ S l (sin \theta + (d/D)^2) − S l (sin \theta) ] / [ S l (sin \theta + (d/D)^2) ] = 0.001\\S l (d/D)^2 / S l (sin \theta + (d/D)^2) = 0.001\\(d/D)^2 = 0.001 (sin \theta + (d/D)^2)\\(d/D)^2 = 0.001 sin \theta + 0.001 (d/D)^2\\d/D = 0.02057\\D/d = 48.61[/tex]
See more about pressure at brainly.com/question/356585
