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The potential flow which is symmetric to the axis of a plane flow is known as axisymmetric potential flow. It can be determined using superposition technique of plane flows. Some of the examples are as follows:

Spherical Polar Coordinates
Spherical polar coordinates are used to express axisymmetric potential flows. Only two coordinates (r,θ) are used and the flow properties are constant on a circle of radius $$r sin \theta$$ about the x-axis.

The equation of continuity for incompressible flow in spherical polar coordinates is:

$$\frac{\partial}{\partial x} (r^2 v_r sin\theta) $$ +$$\frac{\partial}{\partial \theta} (r v_\theta sin\theta)= 0 $$                               (1)

where $$v_r$$ and $$v_\theta$$ are radial and tangential velocities. Therefore a spherical polar stream function exists such that

$$v_r$$ = $$ \frac  {-1}{ r^2  sin\theta}   \frac {\partial \psi}{\partial \theta} $$ $$v_\theta $$ = $$ \frac {1}{ r  sin \theta}   \frac {\partial \psi}{\partial r} $$                    (2)

Similarly a velocity potential $$ \phi(r, \theta)$$ exists such that

$$ v_r$$= $$\frac{\partial \phi}{\partial r} $$ $$v_\theta$$ = $$\frac{1}{r}\frac{\partial \phi}{\partial x} $$                                                (3)



These formulas help to deduce the $$\psi$$ and $$\phi$$ functions for various elementary axisymmetric potential flows.

Uniform stream in the x Direction
The components of a stream $$U_\infty$$ in x direction are:

$$v_r = U_\infty cos\theta $$                     $$v_\theta = - U_\infty  sin\theta$$                            (4)

Substituting in eqs. 2 and and 3 and integrating them gives $$ \psi= \frac{-1}{2} U_\infty r^2 sin^2 \theta$$                      $$\phi= U_\infty  r cos\theta$$                   (5)

The arbitrary constants of integration have been neglected.

Point Source or Sink
Consider a volume flux Q emanating from a point source. The flow will spread out radially and at radius r, it will be equal to $$ \frac{Q}{4\pi r^2}$$. Thus

$$ v_r = \frac {Q}{4 \pi r^2} = \frac {m}{r^2}$$ $$ v_\theta = 0 $$                    (6)

With $$m=\frac{Q}{4\pi}$$ for convenience. Integrating Eqs. 2 and 3 gives

$$ \psi = m cos\theta $$ $$ \phi = -\frac{m}{r}$$                        (7)

For a point sink, change m to –m in Eq. 6.

Point Doublet
A source can be placed at $$(x,y) = (-a, 0)$$ and an equal sink at $$(+a, 0)$$. On taking a limit when $$a$$ tends to 'zero' with product $$2am = \lambda$$ being constant

$$\psi_{doublet} =\lim_{a\to 0\ 2am = \lambda}( m cos \theta_{source} - m cos\theta_{sink})= \frac {\lambda sin^2 \theta} {r} $$                        (8)

The velocity potential of point doublet can be given by:

$$\phi_{doublet} =\lim_{a\to 0\ 2am = \lambda}$$ $$(\frac {-m}{r_{source}} + \frac{m}{r_{sink}}) = \frac{\lambda cos\theta}{r^2 } $$                                                                    (9)

Uniform Stream plus a point source
On combination of Eqns. (5) and (7) we get the stream function for a uniform stream and a point source at the origin.

$$\psi= \frac{-1}{2} U_\infty r^2 sin^2\theta +m cos\theta $$                                                                                    (10)

From Eqn. (2), the velocity components can be written after differentiation as:

$$v_r$$= $$U_\infty cos\theta + \frac {m}{r^2}$$                     $$v_\theta= -U_\infty   sin\theta $$                        (11)



Equating these equations with zero gives a stagnation point at $$\theta = 180^o$$ and at $$r=a=( \frac {m} {U_\infty})^{1/2}$$, as shown in the Fig. Suppose m = $$U_\infty a^2$$, we can write the stream function as:

$$\frac{\psi}{U_\infty a^2} = cos\theta-  \frac{1}{2} (\frac{r}{a^2})^2  sin^2\theta$$                                               (12)

The value of stream surface passing through the stagnation point $$(r, \theta) = (a, \pi)$$ is $$\psi = -U_\infty a^2$$ which forms a half body of revolution enclosing a point source, as shown in Fig. Using this half body, a pitot tube can be simulated. The half body approaches the constant radius $$R = 2a$$ about the x-axis far down the stream.



At $$\theta = 70.5^o$$,$$r = a\sqrt3$$, $$V_s = 1.155U_\infty$$, there occurs the maximum velocity and minimum pressure along the half body surface. There exists an adverse gradient downstream of this point because Vs slowly decelerates to $$U_\infty$$, but no flow separation is indicated by boundary layer theory. Thus for a real half body flow, Eqn. (12) proves to be a realistic simulation. But if one adds the uniform stream to a sink to form a half body rear surface, the separation will be predictable and inviscid pattern would not be realistic.

Uniform Stream plus Point Doublet
From Eqns. (5) and (8), if we combine a uniform steam and a point doublet at the origin, we get

$$\psi = \frac{-1}{2} U_\infty r^2 sin^2\theta + \frac{\lambda}{r}sin^2\theta $$                      	          (13)

On examining this relation, the steam surface $$\psi = 0$$ corresponds to the sphere of radius:

$$ r = a = (\frac{2\lambda}{U_\infty})^\frac{1}{3}$$                                                         (14)

Taking$$ \lambda = \frac{1}{2} U_\infty a^3$$ for convenience, we rewrite Eqn. (13) as

$$ \frac{\psi}{\frac{1}{2}U_\infty a^2} $$= $$- sin^2\theta (\frac {r^2}{a^2}- \frac {a}{r}) $$                                         (15)

Below is the plot of streamlines for this sphere. Differentiating Eqn. (2), we get the velocity components as

$$v_r = U_\infty cos\theta (1- \frac{a^3}{r^3})$$                                            (16)

$$v_\theta = -\frac{1}{2} U_\infty sin\theta (2+ \frac{a^3}{r^3})$$                                     (17)

The radial velocity vanishes at the surface of sphere r = a, as expected. A stagnation point exists at the front $$(a, \pi)$$ and the rear $$(a, 0)$$ of the sphere.



At the shoulder $$(a, \pm \frac{1}{2 \pi})$$, there is maximum velocity where $$v_r = 0$$ and $$v_\theta -\frac{3}{2} U_\infty$$. The surface velocity distribution is

$$V_s = -v_{\theta|r=a} = \frac {3}{2} U_\infty sin\theta$$               (18)