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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 $rsin\theta$ about the x-axis.

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

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

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

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

Spherical polar coordinates for axisymmetric flow

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
$\psi ={\frac {-1}{2}}U_{\infty }r^{2}sin^{2}\theta$ $\phi =U_{\infty }rcos\theta$ (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

$\psi =mcos\theta$ $\phi =-{\frac {m}{r}}$ (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

$\psi _{doublet}=\lim _{a\to 0\ 2am=\lambda }(mcos\theta _{source}-mcos\theta _{sink})={\frac {\lambda sin^{2}\theta }{r}}$ (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.

$\psi ={\frac {-1}{2}}U_{\infty }r^{2}sin^{2}\theta +mcos\theta$ (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.

Fig 3:Streamlines for Rankine half-body of revolution.

At $\theta =70.5^{o}$ , $r=a{\sqrt {3}}$ , $V_{s}=1.155U_{\infty }$ , 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.

Fig 4:Streamlines and potential lines for inviscid flow past a 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)

References

Fluid Mechanics - Frank M. White
Fluid Mechanics and Hydraulic mechanics by R.K. Bansal.

Category:Mechanical engineering