Problem 3.32 - ⎠ 2 L ⋅ = Therefore p 1 p 2 − ρ g ⋅...

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Problem 3.32 [Difficulty: 3] Given: Inclined manometer as shown. D 96 mm = d 8 mm = Angle θ is such that the liquid deflection L is five times that of a regular U-tube manometer. Find: Angle θ and manometer sensitivity. Solution: We will apply the hydrostatics equations to this system. Governing Equation: dp dz ρ g = (Hydrostatic Pressure - z is positive upwards) Assumptions: (1) Static liquid (2) Incompressible liquid x Integrating the hydrostatic pressure equation we get: Δ p ρ g Δ z = Applying this equation from point 1 to point 2: p 1 ρ g x L sin θ ( ) + ( ) p 2 = Upon simplification: p 1 p 2 ρ g x L sin θ ( ) + ( ) = Since the volume of the fluid must remain constant: π 4 D 2 x π 4 d 2 L = x d D 2 L = Therefore:
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Unformatted text preview: ⎠ 2 L ⋅ = Therefore: p 1 p 2 − ρ g ⋅ L ⋅ d D ⎛ ⎝ ⎞ ⎠ 2 sin θ ( ) + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅ = Now for a U-tube manometer: p 1 p 2 − ρ g ⋅ h ⋅ = Hence: p 1incl p 2incl − p 1U p 2U − ρ g ⋅ L ⋅ d D ⎛ ⎝ ⎞ ⎠ 2 sin θ ( ) + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅ ρ g ⋅ h ⋅ = For equal applied pressures: L d D ⎛ ⎝ ⎞ ⎠ 2 sin θ ( ) + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅ h = Since L/h = 5: sin θ ( ) h L d D ⎛ ⎝ ⎞ ⎠ 2 − = 1 5 8 mm ⋅ 96 mm ⋅ ⎛ ⎝ ⎞ ⎠ 2 − = θ 11.13 deg ⋅ = The sensitivity of the manometer: s L Δ h e = L SG h ⋅ = s 5 SG =...
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