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NPTEL IIT Kharagpur: Prof. K.P. Sinhamahapatra, Dept. of
Aerospace Engineering
2
Axial ly Symmetric Flow
No variation with , conditions are same in every meridian plane.
The potential equation is
( )2 2
22 21 1 0M
r r r x + + =
The incompressible case has the basic solution
2 2
A
x r=
+; either a source or sink
- A for source
If the source is at , 0x r= =
( )( )
2 2,
Ax r
x r
=
+
Since, the governing equation is linear, superposition is
allowed
( )( ) ( )
0 1 2
2 2 2 22 2
1 2
, ...A A A
x rx r x r x r
= + + + + + +
represents the flow due to a series of sources placed alongx
axis.
Introducing a source distribution, where ( )f is the source
strength per unit length, and
( ) ( )
( )20 2
,l f d
x r
x r
=
+
( )f is obtained by satisfying the boundary conditions.
Solution is most often obtained through numerical method with
finite number of sources and sinks.
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NPTEL IIT Kharagpur: Prof. K.P. Sinhamahapatra, Dept. of
Aerospace Engineering
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Subsonic fl ow
0122 >= M
The potential equation is
2 2
2 2 2 2
1 10
r r r x
+ + =
or0
112
2
22
2
22
2
=
+
+
zyx
Introducing a transformation (affine transformation)
,x x r r = =
or , ,x x y y z z = = =
2 2
2 2
10
r r r x
+ + =
or 02
2
2
2
2
2
=
+
+
zyx
incompressible flow equation
Two alternative solutions
1) Solution for the incompressible problem in the affinely
transformed domain. Velocity and pressure
fields to be transformed to the physical domain (Gothert
similarity rule).
2) Solution for the physical problem. Solution computed in the
physical domain.
Basic solution in the original coordinates is
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NPTEL IIT Kharagpur: Prof. K.P. Sinhamahapatra, Dept. of
Aerospace Engineering
4
( )( )
2 2 2
, ,A
x r
x r
= +
( ) ( )2 2 2 2 2, ,
A
x y z x y z =
+ +
and the general solution is
( ) ( )
( )20 2 2
,l f d
x r
x r
=
+
Supersonic flow
0122 >= M
The equation becomes
2 22
2 2
10
r r r x
+ =
(wave equation)
By analogy,
( )( )
2 2 2
, A
x r
x r
=
The equation is satisfied. Representing flows is problematic,
since denominator may become zero or
negative, resulting infinite or imaginary .
Potential may be represented as an integral over a distribution
of sources
( ) ( )
( )20 2 2
,x r f d
x r
x r
=
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NPTEL IIT Kharagpur: Prof. K.P. Sinhamahapatra, Dept. of
Aerospace Engineering
5
For suitable distribution of sources, solution may be obtained
that have no singularities off the axis.
Even though the sources are distributed along the x axis from 0
to 1, the value ofat ( ),x r include
only the sources up to x r = . The sources downstream of x r =
have no influence on the
conditions at ( ),x r
