Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I haven't ever studied fluid dynamics before and may mix something here, so please, be patient :).

Given flow potential of the form (homogeneus flow over a dipole):

$$ \phi = u_\infty x -\frac{M}{2\pi} \frac{x}{x^2 + y^2} $$

how do I calculate pressure field? I know the answer is like below, and it's derived using Bernoulli's principle. I just don't know how to get there analytically.

enter image description here

In my materials the answer is:

$$ C_p = \frac{p-p_\infty}{\frac{1}{2} \rho u_\infty^2} = 1 - \left( \frac{u}{u_\infty}\right)^2$$

but I don't understand where it comes from.

share|cite|improve this question
up vote 1 down vote accepted

The pressure coefficient at a certain point (at which the value of the pressure is $p$) is defined as

$C_p=\frac{p-p_\infty}{\frac12\rho_\infty u_\infty^2},$

where the $\infty$-Symbol denotes freestream quantities.

For an incompressible and steady fluid and assuming zero viscosity, Bernoulli's equation is given by

$p+\frac12\rho u^2=p_\infty+\frac12\rho u_\infty^2,$

which we can rearrange as

$\frac{p-p_\infty}{\frac12\rho u_\infty^2}=1-\frac{u^2}{u_\infty^2}.$

In order for this expression to be valid for given fluid, we have to show that it is incompressible and steady. Incompressibility means that the Laplacian of the flow potential $\phi$ vanishes, this can be shown to be true for the problem at hand. Furthermore, a fluid is steady if its flow does not depend explicitely on time, which is also the case.

share|cite|improve this answer
Thank you @Frederic Brünner. And how do I calculate pressure field for $\phi$ as given above using this formula? I'd like to know, how I can obtain the image above. – mmm Feb 18 '13 at 17:45
You can express the pressure from that equation in terms of the velocity, which is given by the gradient of $\phi$. – Frederic Brünner Feb 18 '13 at 17:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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