At some point the resistance balances the force of

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Unformatted text preview: he velocity varies linearly with radius and the bottom of the cylinder would have an affect on the torque. T = Fr = 2.8 FALLING SPHERES This theory may be applied to particle separation in tanks and to a falling sphere viscometer. When a sphere falls, it initially accelerates under the action of gravity. The resistance to motion is due to the shearing of the liquid passing around it. At some point, the resistance balances the force of gravity and the sphere falls at a constant velocity. This is the terminal velocity. For a body immersed in a liquid, the buoyant weight is W and this is equal to the viscous resistance R when the terminal velocity is reached. R = W = volume x density difference x gravity πd 3 g (ρ s − ρ f ) R =W = 6 ρs = density of the sphere material ρf = density of fluid d = sphere diameter The viscous resistance is much harder to derive from first principles and this will not be attempted here. In general, we use the concept of DRAG and define the DRAG COEFFICIENT as CD = Resistance force Dynamic pressure x projected Area © D.J.DUNN 15 The dynamic pressure of a flow stream is The projected area of a sphere is CD = πd ρu 2 2 2 4 8R ρu 2πd 2 Research shows the following relationship between CD and Re for a sphere. Fig. 2.8 For Re<0.2 the flow is called Stokes flow and Stokes showed that R = 3πdµu hence CD=24µ/ρfud = 24/Re For 0.2 < Re < 500 the flow is called Allen flow and CD=18.5Re-0.6 For 500 < Re < 105 CD is constant CD = 0.44 An empirical formula that covers the range 0.2 < Re < 105 is as follows. CD = 24 6 + + 0.4 R e 1 + Re For a falling sphere viscometer, Stokes flow applies. Equating the drag force and the buoyant weight we get 3πdµu = (πd3/6)(ρs - ρf) g µ = gd2(ρs - ρf)/18u for a falling sphere vicometer The terminal velocity for Stokes flow is u = d2g(ρs - ρf)18µ This formula assumes a fluid of infinite width but in a falling sphere viscometer, the liquid is squeezed between the sphere and the tube walls and additional viscous resistance is produced. The Faxen correction factor F is used to correct the result. © D.J.DUNN 16 2.9 THRUST BEARINGS Consider a round flat disc of radius R rotating at angular velocity ω rad/s on top of a flat surface and separated from it by an oil film of thickness t. Fig.2.9 Assume the velocity gradient is linear in which case du/dy = u/t = ωr/t at any radius r. ωr du =µ The shear stress on the ring is τ = µ dy t The shear force is dF = 2πr 2 drµ ω t The torque is dT = rdF = 2πr 3 drµ ω t The total torque is found by integrating with respect to r. R T = ∫ 2πr 3 drµ 0 ω t = πR 4 µ ω 2t In terms of diameter D this is T = µπωD 4 32t There are many variations on this theme that you should be prepared to handle. © D.J.DUNN 17 2.10 MORE ON FLOW THROUGH PIPES Consider an elementary thin cylindrical layer that makes an element of flow within a pipe. The length is δx , the inside radius is r and the radial thickness is dr. The pressure difference between the ends is δp and the shear stress on the surface increases by dτ from the inner to the outer surface. The velocity at any point is u and the dynamic viscosity is µ. Fig.2.10 The pressure force acting in the direction of flow is {π(r+dr)2-πr2}δp The shear force opposing is {(τ+δτ)(2π)(r+dr) - τ2πr}δx Equating, sim...
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This document was uploaded on 02/07/2014.

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