SIMPLE EXAMPLES

To evaluate the complex potential of two-dimensional flowfields, we shall apply Eq. (6.13) to the results of some basic flows that were treated ifl Chapter 3.

6.3.1 Uniform Stream and Singular Solutions

The complex potential for the flow of a uniform stream of speed Qx in the x direction is obtained by substituting the results for the velocity potential and stream function into Eq. (6.13) to get

F = Ф + /Ф = Qx(x + iz) = G»Y

Now, consider the stream to be at an angle a to the x axis and repeat the process. The complex potential becomes

F = Qx(x cos a + z sin a) + iQ^(-x sin a + z cos a)

= Q Jcos a – і sin a)(x + iz) = Q„Ye~ia (6.15a)

This illustrates the general result that the complex potential for one flowfield
can be made to represent the same flowfield rotated counterclockwise by or if У is replaced by Ye~’a.

Consider a source of strength a at the origin. Its complex potential can be obtained similarly and using polar coordinates we get

F = ~(nr + ів) = ~nY (6.16)

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Note that it is easy to demonstrate that for a source at У = Y0 = x0 + tz,,, the complex potential is

F = ^ln(y-y0) (6.16a)

and in general a flowfield can be translated by У0 by replacing У by У – У0 in the complex potential. The complex potential for a vortex with clockwise circulation Г at У = У0 is

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The complex potential of a doublet at the origin whose axis is in the x direction is

JLi

2л Y

Using the above rules, the complex potential for a doublet at Y=Y0 with an axis at an angle a to the x direction is given by