ME 367 HW 3 Solutions

ME 367 HW 3 Solutions - Problem 3.32[3 Given Find Solution...

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Problem 3.32 [3] Given: Data on inclined manometer Find: Angle θ for given data; find sensitivity Solution: Basic equation dp dy ρ g = or, for constant ρ Δ p ρ g Δ h = where Δ h is height difference Under applied pressure Δ pS G Mer ρ g L sin θ () x + = (1) From Table A.1 SG Mer 0.827 = and Δ p = 1 in. of water, or Δ p ρ g h = where h2 5 m m = h 0.025m = Δ p 1000 kg m 3 9.81 × m s 2 0.025 × m Ns 2 kg m × = Δ p 245Pa = The volume of liquid must remain constant, so xA res LA tube = xL A tube A res = L d D 2 = (2) Combining Eqs 1 and 2 Δ G Mer ρ g L sin θ L d D 2 + = Solving for θ sin θ Δ p SG Mer ρ g L d D 2 = sin θ ( ) 245 N m 2 1 0.827 × 1 1000 × m 3 kg 1 9.81 × s 2 m 1 0.15 × 1 m kg m s 2 N × 8 76 2 = 0.186 = θ 11 deg = The sensitivity is the ratio of manometer deflection to a vertical water manometer s L h = 0.15 m 0.025 m = s6 =
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Problem 3.51 [3] F A H = 25 ft y R = 10 ft h A B z x y Given: Geometry of gate Find: Force F A for equilibrium Solution: Basic equation F R A p d = dp dh ρ g = Σ M z 0 = or, use computing equations F R p c A = y' y c I xx Ay c + = where y would be measured from the free surface Assumptions: static fluid; ρ = constant; p atm on other side; door is in equilibrium Instead of using either of these approaches, we note the following, using y as in the sketch Σ M z 0 = F A R A yp d = with p ρ g h = (Gage pressure, since p = p atm on other side) F A 1 R A y ρ g h d = with dA r dr d θ = and y r sin θ () = hHy = Hence F A 1 R 0 π θ 0 R r ρ g r sin θ H r sin θ r
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This note was uploaded on 09/21/2009 for the course CHEM 220 taught by Professor Cumming during the Spring '09 term at Nevada.

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ME 367 HW 3 Solutions - Problem 3.32[3 Given Find Solution...

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