Velocity as a function of area in pipe flow

[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.

 

[boneh3ad]

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 $A$.

 

[Chestermiller]

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.

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.

 

[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.

 

[boneh3ad]

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.