1916] ON THE FLOW OF COMPRESSIBLE FLUID PAST AN OBSTACLE 403
where q is the resultant velocity, so that
(4)
reduces in this case to
or log p = 0 ^qz,
or a2 log(p//>o) = - %q\ ....
if pa correspond to ij = 0. From (2) and (5) we get
(5)
.(6)
__ __-t . _j____i_ J I
r 2a2 [dsc dx dy dy dz dz
When <f is small in comparison with a2, this equation may be employed to estimate the effects of compressibility. Taking a known solution for an incompressible fluid, we calculate the value of the right-hand member and by integration obtain a second approximation to the solution in the actual case. The operation may be repeated, and if the integrations can be effected, we obtain a solution in series proceeding by descending powers of a'-. It may be presumed that this series will be convergent so long as </ is less than a2.
There is no difficulty in the first steps for obstacles in the form of spheres or cylinders, and I will detail especially the treatment in the latter case. If U, parallel to 6 = 0, denote the uniform velocity of the stream at a distance, the velocity-potential for the motion of incompressible fluid is known to be
0 = U(r + c2/r) cos 0, ...........................(7)
the origin of polar coordinates (r, 6} being at the centre of the cylinder. At the surface of the cylinder r = c, d$ldr = 0, for all values of 6.
On the right hand of (6)
dq- _
and from (7)
d$ df dcj> d<f =
dai dx dy dy dr dr r* dO ~dQ '
drj
.(8)
2/9
Also
Accordingly
O v
"U dr
I ^ = _4c; U"~a dr r5
Urde
4c2
+ r COS
A-tO
~
in
rdv
ar8
The terms on the right of (10) are all of the form rŪ cos n0, so that for the present purpose we have to solve
- w r dr
f^-.. (11)
262<n