# Meaning of scalar product with normal in fluid flow

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 !!!

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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

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, ${\Delta V\over \Delta t}=Av$.

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

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.)

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).

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Thanks a lot !!!!!! – fluidon Jan 29 '13 at 21:56