Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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'm having trouble with understanding the meaning of the following sentence:

Let $n$ be the outward normal defined at points of the boundary of a region $W$ and let $dA$ denote the area element on this boundary. The volume flow rate across $\partial W$ per unit area is $$u \cdot n$$

Can someone please explain to me why does this scalar product interpretation is the volume flow rate? Is there any way I can see this? [edit: $u$ is the velocity field of the fluid ]

Thanks !!!

share|cite|improve this question
What does $u$ represent in your problem? – JohnD Jan 27 '13 at 19:49
I'm sorry. $u$ is the velocity vector field of a fluid. – fluidon Jan 27 '13 at 22:03
See this page: Flux (Surface Integrals of Vectors Fields) – Rahul Jan 27 '13 at 22:09
Thanks a lot !!!!! – fluidon Jan 29 '13 at 21:55
up vote 1 down vote accepted

At time $t$ the fluid (moving at velocity $v$) moves a distance $v\Delta t$. If the cross-section of the pipe has area $A$, then the volume that moves past a given flat surface is $\Delta V = Av\Delta t$. The flow rate is the volume per time, ${\Delta V\over \Delta t}=Av$.

enter image description here

However, if the cross-section isn't perpendicular to the fluid flow (shown below),

enter image description here

we adjust our calculations accordingly. The fluid still moves a distance $v\Delta t$, and the volume that moves through the cross section is the area $A$ times $v\Delta t$. The area of a parallelogram is the length of one side times the perpendicular distance $h$ from that side to its opposite side. (See below.)

enter image description here

Similarly the volume of a parallelepiped is the area of one side times the perpendicular distance from that side to the side opposite. The perpendicular distance is $v\Delta t\cos\alpha$. It can be described by the angle that the normal to the plane makes with the direction of the fluid velocity, which yields $$ \Delta V = Ah = A(v\Delta t)\cos\alpha. $$ The flow rate is then $${\Delta V\over \Delta t}=A v\cos\alpha.\tag{1}$$ If we introduce the normal vector $\hat{n}$, then $(1)$ can be rewritten in terms of a dot product involving $\hat{n}$: $$ {\Delta V\over \Delta t}=A v\cos\alpha=A\vec{v}\cdot\hat{n}. $$

Finally, the volume flow rate per unit area that you seek is given by $$ {\Delta V/\Delta t\over A}=\vec{v}\cdot\hat{n}, $$ where $\vec{v}$ is the fluid velocity.

Source for diagrams (and similar explanation).

share|cite|improve this answer
Thanks a lot !!!!!! – fluidon Jan 29 '13 at 21:56

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.