Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

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

Poiseuille's Law relies on the fact that velocity is not constant throughout a cross-section of the pipe (it is zero at the boundary due to the no-slip condition and maximum in the center). By Bernoulli's Law, this means that pressure is maximum at the boundary and minimum at the center. But in the book I have it is assumed that the pressure gradient is independent of radius (distance from the center of the pipe), and the pressure gradient is thus extricated from a radius-integral. Can anyone justify this?

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

First of all, Bernoulli's law is applicable only to inviscid flow, while Poiseuille's flow is for the viscous fluid. The fact that pressure is constant along the orthogonal cross section of the pipe could be derived from the assumption that the flow is parallel, that is everywhere inside the pipe the velocity field has only z-component (assuming the cylindrical coordinate system, with pipe oriented along the z-axis). Then the r-component of Navier-Stokes equation is then reduced to $0 =- \frac1{\rho}\frac{\partial p}{\partial r}$ (all terms containing velocity components are equal to zero here), which gives the pressure independent of radial coordinate.

The assumption that the flow in the pipe would be parallel in derivations of Poiseuille's flow is just an ansatz, compatible with the symmetries of the problem, that is later justified by producing correct solution of Navier-Stokes equation. However one should remember that such flow would be stable only for relatively large viscosity of the fluid (that is for Reynolds number not exceeding the certain critical value).

share|cite|improve this answer
Why couldn't there be parallel flow if the pressure was not constant along the orthogonal cross section? I can see that it would be true, but I don't really see why. Is there a way to explain this without using the Navier-Stokes equation (I haven't studied this yet). – Joshua Meyers Jun 8 '13 at 23:42
Without the NS equations you could simply consider the forces acting on the fluid element. The acceleration of such element is zero for stationary parallel flow, so the total sum of forces acting on it must be zero. We have only two sources of such forces: the viscous friction and pressure. For parallel (shearing) flow the friction force will only have component parallel to velocity. That means that the pressure gradient force must also have only the components parallel to velocity. Which means constant pressure on orthogonal cross section. – user23660 Jun 9 '13 at 4:59
I understand, thanks. – Joshua Meyers Jun 9 '13 at 10:54

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.