Page 128 - Aerodynamics for Engineering Students

Potential flow 11 1

Fig. 3.6
(a datum) and taking the most convenient contour for integration as OQP where QP
is an arc of a circle of radius r,
$ = flow across OQ + flow across QP
= velocity across OQ x OQ + velocity across QP x QP
m
=O+-xrO
27rr
Therefore
$ = m13/27r
or putting e = tan-' b/x)

There is a limitation to the size of e here. 0 can have values only between 0 and 21r.
For $ = m13/27r where 8 is greater \ban 27r would mean that $, i.e. the amount of fluid
flowing, was greater than m m2 s- , which is impossible since m is the capacity of the
source and integrating a circuit round and round a source will not increase its strength.
Therefore 0 5 0 5 21r.
For a sink
$ = -(m/21r)e

To find the velocity potential # of a source
The velocity everywhere in the field is radial, i.e. the velocity at any point P(r, e) is given by
and
4 = dm 4 = 4n here, since 4t = 0. Integrating round OQP where Q is point (r, 0)
4 = 1 qcosPds + ipqcosBds
OQ
= S,, 4ndr + ipqtraQ= S,, 4n dr+ 0

But
m
4n =-
27rr
Therefore
m m r
4 = LGdr = T;;'n,,
where ro is the radius of the equipotential 4 = 0.