Aerodynamics: Optimization of an aircraft engine

  • Thread starter Thread starter dd5139
  • Start date Start date
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 777 views
dd5139
Messages
1
Reaction score
0
New user has been reminded to always show their work on schoolwork problems.
I am writing an essay about optimization of an aircraft engine. I have to show the derivation of aerodynamic drag eq. I need help about it because i couldn't find where it came from.
 
Physics news on Phys.org
Hi there, welcome to the forum! :welcome:

It is probably better to start a new topic on this next time instead of replying to an old thread (last post almost a year ago). We try to keep threads limited to the original topic.

But do you mean by 'the aerodynamic drag eq.' this one:
$$
F_d = \frac{1}{2}\rho u^2 A C_d
$$
If so then you ought to know that this is just the inverse of the definition of the drag coefficient (##C_d = \frac{2 F_d}{\rho u^2 A}##). The drag coefficient comes from dimensional analysis (The Buckingham ##\pi## Theorem). This drag coefficient is found to be similar for similarly shaped object. There are whole books written that give (a range of) ##C_d## values for differently shaped objects (most notably the historic 'Aerodynamic Drag' by Hoerner). But please note that ##C_d## is not always constant for even the same object in different flow conditions (like for different Reynolds numbers or different Mach numbers). Also, the million dollar question is how ##C_d## varies for small changes in the geometry.
 
dd5139 said:
I am writing an essay about optimization of an aircraft engine. I have to show the derivation of aerodynamic drag eq. I need help about it because i couldn't find where it came from.
Welcome to PF.

Please show us your Google search terms that you have been using so far. This seems like a pretty easy search, so we would like to help you learn how to do searches like this better in the future. :smile: