Unformatted text preview: x 0.05/0.12
∆p= 2.88 Pascals.
u = {∆p/4Lµ}(R2  r2) which is made equal to the average velocity 0.05 m/s
0.05 = (2.88/4 x 1 x 0.018)(0.052  r2)
r = 0.035 m or 35.3 mm. © D.J.DUNN 12 2.6. FLOW BETWEEN FLAT PLATES Consider a small element of fluid moving at velocity u with a length dx and height dy at
distance y above a flat surface. The shear stress
acting on the element increases by dτ in the y
direction and the pressure decreases by dp in the x
dp
d 2u
=µ 2
direction. It was shown earlier that −
dx
dy
It is assumed that dp/dx does not vary with y so it
may be regarded as a fixed value in the following
work.
Fig.2.5
dp
du
=µ
+A
Integrating once  y
dy
dx
y 2 dp
= µu + Ay + B............(2.6 A)
2 dx
A and B are constants of integration. The solution of the equation now depends upon
the boundary conditions that will yield A and B.
Integrating again  WORKED EXAMPLE No.2.4 Derive the equation linking velocity u and height y at a given point in the x direction
when the flow is laminar between two stationary flat parallel plates distance h apart. Go
on to derive the volume flow rate and mean velocity.
SOLUTION
When a fluid touches a surface, it sticks to it and moves with it. The velocity at the flat
plates is the same as the plates and in this case is zero. The boundary conditions are
hence
u = 0 when y = 0
Substituting into equation 2.6A yields that B = 0
u=0 when y=h
Substituting into equation 2.6A yields that A = (dp/dx)h/2
Putting this into equation 2.6A yields
u = (dp/dx)(1/2µ){y2  hy} (The student should do the algebra for this). The result is a parabolic distribution similar
that given by Poiseuille's equation earlier only this time it is between two flat parallel
surfaces. © D.J.DUNN 13 FLOW RATE
To find the flow rate we consider flow through a small rectangular slit of width B and
height dy at height y. Fig.2.6
dQ = u Bdy =(dp/dx)(1/2µ){y2  hy} Bdy The flow through the slit is Integrating between y = 0 and y = h to find Q yields
The mean velocity is Q = B(dp/dx)(h3/12µ)
um = Q/Area = Q/Bh hence um = (dp/dx)(h2/12µ) (The student should do the algebra) 2.7 CONCENTRIC CYLINDERS
This could be a shaft rotating in a bush filled with oil or a rotational viscometer.
Consider a shaft rotating in a cylinder with the gap between filled with a Newtonian
liquid. There is no overall flow rate so equation 2.A does not apply. Due to the stickiness of the fluid, the liquid sticks to
both surfaces and has a velocity u = ωRi at the inner
layer and zero at the outer layer.
Fig 2.7 © D.J.DUNN 14 If the gap is small, it may be assumed that the change in the velocity across the gap
changes from u to zero linearly with radius r.
τ = µ du/dy
But since the change is linear du/dy = u/(RoRi) = ω Ri /(RoRi)
τ = µ ω Ri /(RoRi)
Shear force on cylinder F = shear stress x surface area
F = 2πRi hτ = 2πRi2 hµω
Ro − Ri Torque = F x R i
2πRi3 hµω
Ro − Ri
In the case of a rotational viscometer we rearrange so that
T ( Ro − R )
µ=
2πRi3 hω
In reality, it is unlikely that t...
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 Spring '14

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