Fluid Dynamics -- Use the Milne-Thomson circle theorem to show the complex potential for a fluid....

[zigggy23]

Homework Statement

Two equal line sources of strength k are located at x = 3a and x = −3a, near a circular cylinder of radius a with axis normal to the x, y plane and passing through the origin. The fluid is incompressible and the flow is irrotational (and inviscid). Use the Milne-Thomson circle theorem to show that the complex potential for this flow is:

w(z)=kln(a^4−(9a^2)*(z^2)−(9a)6z^2+81a^4).

I have done this the question then goes on to say:

Differentiate w(z) with respect to z, and hence show that the speed of the flow at the surface of the cylinder is:

v = |v1 − iv2| = 18k sin 2θ/a(41 − 9 cos 2θ)

and the final part is:

Determine the position of points on the surface of the cylinder at which the pressure is a minimum. Sketch (roughly) streamlines in the vicinity of the cylinder, and mark any stagnation points and the points of minimum pressure.

Homework Equations

I think dw/dz is the speed at the surface of the cylinder.

The Attempt at a Solution

I have differentiated w(z) wrt z to get:

18k(Z^4-a^4)/z(9z^4-82az^2+9a^4)

but don't know how to get it in the form required. I am also completely lost on the third part so any help would be appreciated

 

[KataKoniK]

.

First, let's review the Milne-Thomson circle theorem. This theorem states that the complex potential for a flow around a circular cylinder can be expressed as:

w(z) = k ln(z - z1) + k ln(z - z2)

where z1 and z2 are the locations of two equal line sources of strength k, located on the x-axis at x = 3a and x = -3a, respectively. This complex potential satisfies the Laplace equation, which is the governing equation for an inviscid and incompressible flow.

To show that the given complex potential w(z) satisfies the Laplace equation, we can take the derivative with respect to z and then show that the resulting expression is equal to zero. So, let's do that:

dw/dz = k/(z - z1) + k/(z - z2)

The derivative of a logarithm is given by:

d/dz ln(z) = 1/z

So, applying this to the above expression, we get:

dw/dz = k/(z - z1)^2 + k/(z - z2)^2

Now, using the fact that z1 = 3a and z2 = -3a, we can rewrite this as:

dw/dz = k/(z - 3a)^2 + k/(z + 3a)^2

Next, we can use the identity:

1/(z - a)^2 + 1/(z + a)^2 = 2/(z^2 - a^2)

to simplify the above expression as:

dw/dz = 2k/(z^2 - (3a)^2)

Now, using the identity:

z^2 - a^2 = (z - a)(z + a)

we can rewrite the above expression as:

dw/dz = 2k/(z - 3a)(z + 3a)

Finally, using the fact that a = 3a, we can simplify this expression further to get:

dw/dz = 2k/(z - a)^2 = 0

Thus, we have shown that the given complex potential satisfies the Laplace equation, as required.

Now, let's move on to the next part of the problem, which asks us to differentiate w(z) with respect to z to get the speed of the flow at the surface of the cylinder. This can be done by using the