2
$\begingroup$

Wave drag is an especially high drag that affects supersonic vehicles above mach 1, but I've found it hard to find source materials to estimate wave drag. I'm looking at the transonic region till Mach 1.2 and so I'm looking to have an estimate about how I can design better for wave drag by estimating how it would affect my flight in the region mach 0.8 - 1.2.

$\endgroup$
  • 3
    $\begingroup$ Welcome to Aviation.SE. Speeds up to Mach 1 are subsonic by definition and do not have wave drag. You might want to simplify your question to focus on one aspect. It requires several answers in its current form, and this site tends to like one at a time. $\endgroup$ – Pilothead Sep 18 '18 at 18:15
  • $\begingroup$ Hey @Pilothead, thanks for the heads up. Thanks for welcoming me , and I have simplified the question and perhaps I'll add the other elements in another question. $\endgroup$ – Rajath Pai Sep 19 '18 at 1:28
1
$\begingroup$

Usually wave drag $c_{d_w}$ starts becoming relevant at the drag divergence Mach number $M_{DD}$, which is defined as

$$\frac{\partial c_{d_w}}{ \partial M}_{M=M_{DD}}=0.1 \tag{1}$$

Or in words, the Mach number at which the wave drag coefficient versus M has a gradient of 0.1

Lock$^1$ has derived an empirical formula that shows the development of $c_{d_w}$ after the critical Mach number, $M_{cr}$, which is the number at which the flow becomes locally supersonic on the wing:

$$c_{d_w} = 20(M-M_{cr})^4 \tag{2}$$

If we apply the definition of $(1)$ to $(2)$ we obtain:

$$M_{cr} = M_{DD} - \left(\frac{0.1}{80}\right)^{1/3} \tag{3}$$

As you can see, the value of $M_{DD}$, is a little bit higher than the value of $M_{cr}$. First you get local super sonic flow, but the drag does not increase rapidly, as you further increase the Mach number, the drag starts to increase significantly, and you reach $M_{DD}$. See this image, where $M_{DD}$ is denoted by $M_{CDR}$.

enter image description here

Source

The value of $M_{DD}$ can be estimated using:

$$ M_{DD} = \frac{K_A}{cos \Lambda} + \frac{t/c}{cos^2 \Lambda} +\frac{c_l}{cos ^3\Lambda} \tag{4}$$

With $\Lambda$ is the sweep angle, $t/c$ the relative thickness, $c_l$ the lift coefficient, and $K_A$ the airfoil technology factor (a correction factor denoting the quality of the airfoil).

So the workflow is as follows:

  1. Estimate $M_{DD}$ using $(4)$
  2. Derive $M_{cr}$ using $(3)$
  3. Determine the development of $c_{d_w}$ using $(2)$.

For more info, see this link and this link

$^1$ Hilton, W.F., High Speed Aerodynamics, Longmans, Green & Co., London, 1952, pp. 47-49

Note, this method is only for estimating the wave drag coefficient around the drag divergence Mach number, I'm not sure to what Mach number it provides an accurate estimate

$\endgroup$
  • $\begingroup$ thanks a lot for all the formulae and the info. You've been super helpful and I will look into the book as well. $\endgroup$ – Rajath Pai Sep 20 '18 at 0:17
  • $\begingroup$ @RajathPai You're welcome. As you can see your question has been 'on hold', because it's a very broad question. As such, other people can not answer to it anymore. Very broad questions are difficult to answer, please focus on one question per post. You can edit the question to make it more specific and then it can be reopened again. If you have any other doubts concerning how to improve your question let me know. If you want to ask another (different) question on an aviation topic, please make a new post. $\endgroup$ – ROIMaison Sep 20 '18 at 7:54
  • $\begingroup$ I already edited it and reduced it once. I will try to reword it and be more specific. Thank you for your advice. $\endgroup$ – Rajath Pai Sep 20 '18 at 10:07

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.