Basics of Fluid Mechanics, 2014a

356 CHAPTER 10. POTENTIAL FLOW
U θ = 1 ∂φ
(10.147b)
r ∂θ
Thus the relationships that were obtained before for Cartesian coordinates is written in
cylindrical coordinates as
∂φ
∂r = 1 ∂ψ
r ∂θ
(10.148a)
1 ∂φ
r ∂θ = −∂ψ ∂r
(10.148b)
In the case of the dipole, the knowledge of the potential function is used to obtain
the stream function. The derivative of the potential function as respect to the radius is
And
∂φ
∂r = Q 0 cos θ
2 π
1
r 2 r
1 ∂φ
r ∂θ = Q 0 sin θ
2 π r 2 − ∂ψ
∂r
From equation (10.149) after integration with respect to θ one can obtain
∂ψ
∂θ
(10.149)
(10.150)
ψ = Q 0
sin θ + f(r) (10.151)
2 πr
and from equation (10.150) one can obtain that
− ∂ψ
∂r = Q 0
2 πr 2 sin θ + f ′ (r) (10.152)
The only way that these conditions co–exist is f(r) to be constant and thus f ′ (r) is
zero. The general solution of the stream function is then
ψ = Q 0 sin θ
2 πr
(10.153)
Caution: mathematical details which can be skipped
The potential function and stream function describe the circles as following: In
equation (10.153) it can be recognized that r = √ x 2 + y 2 Thus, multiply equation
(10.153) by r and some rearrangement yield
2 πψ
Q 0
⎛ ⎞
r
{ }} 2
{
⎜
⎝x 2 + y 2 ⎟
⎠ =
r sinθ
{}}{
y (10.154)