{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Problem 3.31 - The mercury in the right leg is displaced...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Problem 3.31 [Difficulty: 2] Given: A U-tube manometer is connected to the open tank filled with water as shown (manometer fluid is mercury). The tank is sealed and pressurized. D 1 2.5 m = D 2 0.7 m = d 0.2 m = p o 0.5 atm = SG Hg 13.55 = (From Table A.1, App. A) Find: The manometer deflection, l Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh ρ g = (Hydrostatic Pressure - h is positive downwards) ρ SG ρ water = (Definition of Specific Gravity) Assumptions: (1) Static liquid (2) Incompressible liquid D 1 D 2 d c d Integrating the hydrostatic pressure equation we get: Δ p ρ g Δ h = When the tank is filled with water and pressurized, the mercury in the left leg of the manometer is displaced downward by l/2. The mercury in the right leg is displaced
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: The mercury in the right leg is displaced upward by the same distance, l/2. Beginning at the free surface of the tank, and accounting for the changes in pressure with elevation: p atm p o + ρ water g ⋅ D 1 D 2 − d + l 2 + ⎛ ⎝ ⎞ ⎠ ⋅ + ρ Hg g ⋅ l ⋅ − p atm = Upon simplification: p o ρ water g ⋅ D 1 D 2 − d + l 2 + ⎛ ⎝ ⎞ ⎠ ⋅ + ρ Hg g ⋅ l ⋅ = l p o ρ water g ⋅ D 1 + D 2 − d + SG Hg 1 2 − = Substituting values into the expression: l 0.5 atm ⋅ 1.013 10 5 × N ⋅ m 2 atm ⋅ × 1 1000 × m 3 kg ⋅ 1 9.8 × s 2 m ⋅ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2.5 m ⋅ + 0.7 m ⋅ − 0.2 m ⋅ + 13.55 1 2 − = l 0.549m =...
View Full Document

{[ snackBarMessage ]}