Mechanics of Fluids

216 Laminar flow between solid boundaries
higher orders of small quantities being neglected. Dividing by τr 2 we obtain
which may be integrated to give
δτ
τ =−2δr
r
τ = A/r 2
(6.29)
where A is a constant.
We now have to evaluate τ in terms of the dynamic viscosity µ. Since
we are here concerned with the viscous stress over an area perpendicular
to the radius, the velocity gradient must be calculated along the radius.
The tangential velocity at radius r is given by ωr where ω represents the
angular velocity of the fluid, and the full velocity gradient is therefore
∂u
∂r
∂
= (ωr) = ω + r∂ω
∂r ∂r
In this expression, however, only the second term contributes to relative
motion between particles. Suppose that the angular velocity of the fluid
does not vary with the radius. Then ∂ω/∂r is zero and ∂u/∂r reduces to ω.
In this case there is no relative motion between the particles of fluid, even
though ∂u/∂r is not zero; the entire quantity of fluid rotates as if it were a
solid block. (One may imagine a cylinder of liquid placed on a gramophone
turntable: when conditions are steady the liquid will have the same angular
velocity as the turntable and there will be no relative motion between
particles at different radii, even though the peripheral velocity increases
with radius.) Therefore the rate of shear, which represents relative motion
between particles, is simply r∂ω/∂r and so the stress τ is given by µr∂ω/∂r.
∴ µ ∂ω
∂r
= τ
r
= A
r 3
(6.30)
(from eqn 6.29). Since ω is here a function of r alone, eqn 6.30 may be
integrated to give
µω =− A
+ B where B is a constant (6.31)
2r2 Now if the rotating cylinder has radius a and angular velocity � and the
stationary cylinder has radius b, the condition of no slip at a boundary
requires ω to be zero when r = b. Substituting these simultaneous values in
eqn 6.31 gives B = A/2b2 . The same no slip requirement makes ω = � at
r = a, and these values substituted in eqn 6.31 give
whence
µ� =− A A
+
2a2 2b2 = A(a2 − b2 )
2a2b2 A = 2a2 b 2 µ�
a 2 − b 2