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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=18.5Re0.6
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)2πr2}δp
The shear force opposing is
{(τ+δτ)(2π)(r+dr)  τ2πr}δx
Equating, sim...
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 Spring '14

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