# Velocity as a function of area in pipe flow

1. Apr 7, 2016

### mech-eng

There is an equation which perplexes me and it is about calculation of average velocity in a pipe but over area.

In the image, the velocity is already function of diameter, i.e, u=u(r) so how can we think velocity as a function of area?

Source: Fluid Mechanics, Fundamentals and Applications by Çengel/Cimbala

Thank you.

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Last edited: Apr 7, 2016
2. Apr 7, 2016

You've sort of answered your own question. It's the difference between the velocity at a point, which is a function of $r$, and the average velocity across the cross section, which averages that $r$-dependence away and depends on $A$.

3. Apr 7, 2016

### Staff: Mentor

u(r) is the local axial velocity at radial coordinate r. The axial velocity is maximum at the center of the tube u(0), and is zero at the wall, u(R) = 0. If you want to get the average axial velocity, you calculate the volumetric flow rate and divide by the total cross sectional area of the tube.

4. Apr 7, 2016

### mech-eng

Can thinking of u(Area) instead of U(D) be the same? Because Area is function of diamter as well. This looks like a differential equation problem and while integrating over area it becomes integration over diameter because dA=2*pi*r*dr

Thank you.

5. Apr 7, 2016

In general, the velocity varies over the range of $r$. Doing the integration so that it is $u_{avg}(A)$ loses all that information about the $r$ dependence. You could certainly cast it in terms of $u_{avg}(D)$ as long as you are just referring to $D$ as the inner diameter of the pipe and not some kind of surrogate for $r$, but you couldn't get $u(D)$ just like you couldn't get $u(A)$. It is less intuitive though, since using $A$ comes from conservation of mass.