Question

A manometer consists of an inclined glass tube which communicates with a metal cylinder standing upright; liquid fills the apparatus to a fixed zero mark on the tube when both the cylinder and the tube are open to atmosphere. The upper end of the cylinder is then connected to a gas supply at a pressure p and manometer liquid rises through a distance l in the tube. Establish the relation;
h=Sl [ sin(alpha) + (d/D)^2 ]



for the pressure head h of water column in terms of inclination alpha of the tube, specific gravity S of the fluid, and ratio of diameter d of the to the diameter D of the cylinder.

Also determine the value of so that the error disregarding the change in level in the cylinder will not exceed 0.1 percent when = .

Answer 1

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