3
$\begingroup$

I am reading "Fundamental of Aerodynamics " by J.D.Anderson Fifth edition. Now I am at the chapter 3. If you got the book, please go to the page 286, the part that says: "when j = i, the contribution to the derivative is simply $\lambda/2$ ". How could it be $\lambda/2$ ? I thought it would be zero.

I will summarize the problem for those who don't have the book:

We have a uniform flow and a source sheet cover the the surface of a known body. We have to find out the $\lambda=\lambda(s)$ along the source sheet that when we superimpose the uniform flow with this source sheet, we obtain the streamline over surface of the body.

enter image description here

A source sheet is an infinite number of line sources side by side, where the strength of each line source is infinitesimally small

enter image description here

We approximate the source sheet by a series of straight panel over the body surface (if we take n panels, then we have to find n values of $\lambda$) and calculate the velocity at each panel and set the normal component of it equal zero.

enter image description here

Take a point $P$ at the middle of $i$th panel and calculate the velocity at $P$ induced by the panels, or just calculate the normal component of the velocity at P. The book said that the normal component of the velocity contributed by only $i$th panel at the mid point $P$ of this panel is $\lambda_i/2$. I don't know how to calculate this, I have tried and thought it would be zero. enter image description here

@Peter: Look at the image, you can see the velocity induced at P by the panel i is in the ab-line direction (I mean the horizontal direction). Then, the normal component (vertical direction) of it will be zero

$\endgroup$
  • $\begingroup$ Perhaps you could share how you tried and obtained zero? $\endgroup$ – Peter Schilling Aug 16 '17 at 11:38
  • $\begingroup$ A source panel will only produce flow normal to the panel. The horizontal component is zero, not the vertical. $\endgroup$ – Peter Schilling Aug 17 '17 at 1:01
  • $\begingroup$ normal at every point ?? I agree with this but it's only true for some point. At the control point or the midpoint of the panel, due to the symmetric, the velocity is zero $\endgroup$ – Dat Aug 17 '17 at 5:05
  • $\begingroup$ Yes, at every point. We define the normal component of the overall flow (freestream plus all panels) to be zero only at the control points, but you're specifically asking about one panel's contribution to itself. We only need to worry about the flow from the panel in question, which is always normal to it. $\endgroup$ – Peter Schilling Aug 17 '17 at 11:34
1
$\begingroup$

You can find a mathematical derivation of the $\lambda_i/2$ result in Katz and Plotkin, chapter 10. In this PDF, the result you are looking for is on page 234 (250 of the file itself). Anderson approaches things a bit differently, but you should be able to see at a high level how the approaches are analogous. The math gets a bit complicated and it's really not necessary to comprehend every step in detail to understand the application to aerodynamics. However, a good exercise if you really want to dive into the math is to translate Katz and Plotkin's approach into Anderson's notation, or vice versa. For me, though, it's good enough to see that it can be done and move on with the more interesting applications.

$\endgroup$
  • $\begingroup$ Thank you for the PDF file, I understand the $ \lambda / 2 $ result. I still do not agree with you that the source panel produces only normal velocity at every point (yes, the flow is produced by only one straight panel). In the PDF file, on page 250, equation (10.20) is the horizontal component of the velocity at P, it may not be zero at every point. It only equals zero when 2x = x1 + x2, or P is on the mid line of the panel. $\endgroup$ – Dat Aug 18 '17 at 7:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.