Page 409 - Introduction to Continuum Mechanics

Newtonian Viscous Fluid 393

therefore,

d<f>
(b) From v/ = —jr-, we have
1

(c) We have, at (0,0,0), v 1 = 0, v 2 = 0, v 3 = 0, p = p 0, and Q = 0
therefore, from the Bernoulli's equation, [Eq. (6.21.9)]

we have

and

or

2
(d) On the plane y - 0, vj = -3x and v 2 = 0. Now, v^ - 0 means that the normal com-
ponents of velocity are zero on the plane, which is what it should be if y = 0 is a solid fixed
2
boundary. Since vi = -3* , the tangential components of velocity are not zero on the
plane, that is, the fluid slips on the boundary. In inviscid theory, consistent with the assumption
of zero viscosity, the slipping of fluid on a solid boundary is allowed. More discussion on this
point will be given in the next section.

Example 6.21.2

A liquid is being drained through a small opening as shown. Neglect viscosity and assume
that the falling of the free surface is so slow that the flow can be treated as a steady one. Find
the exit speed of the liquid jet as a function of h.