Problem 2.73 - Problem 2.73 [Difficulty: 4] Given: Conical...

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Problem 2.73 [Difficulty: 4] ds AA U= ω r dz z r a Section AA Given: Conical bearing geometry Find: Expression for shear stress; Viscous torque on shaft Solution: Basic equation τμ du dy = dT r τ dA = Infinitesimal shear torque Assumptions: Newtonian fluid, linear velocity profile (in narrow clearance gap), no slip condition tan θ () r z = so r z tan θ = Then du dy = μ u y = μ ω r 0 a0 = μω z tan θ a = As we move up the device, shear stress increases linearly (because rate of shear strain does) But from the sketch dz ds cos θ = dA 2 π r ds = 2 π r dz cos θ = The viscous torque on the element of area is dT r τ dA = r z tan θ a 2 π r dz cos θ
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