# Fig. P4.94 ```364
Solutions Manual x Fluid Mechanics, Fifth Edition
resulting velocity distribution. The fluid is at rest
Fig. P4.94
far from the cylinder. [HINT: the cylinder does
Solution: We already have the useful hint that vr = 0. Continuity then tells us that
(1/r)wvT/wT = 0, hence vT does not vary with T. Navier-Stokes then yields the flow. From Eq.
D.6, the tangential momentum relation, with wp/wT 0 and vT = f(r), we obtain Eq. (4.139):
vT
1 d dvT
(r
)
, Solution : vT
dr
r dr
r2
As r o f , vT o 0 , hence C1
0
C1 r C2
r
C2
:R 2
; C 2 :R 2 ; Finally, vT
Ans.
R
r
Rotating a cylinder in a large expanse of fluid sets up (eventually) a potential vortex flow.
At r
R , vT
:R
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*P4.95
Two immiscible liquids of
equal thickness h are being sheared
V
h
U2, P2
y
between a fixed and a moving plate,
U1, P1
h
as in Fig. P4.95. Gravity is neglected,
x
Fixed
and there is no variation with x.
Fig. P4.95
Find an expression for (a) the velocity at the
interface; and (b) the shear stress in each fluid. Assume steady laminar flow.
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