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In fluid mechanics we consider a fluid filling a region $D\subset \mathbb{R}^3$ together with a function $\rho : D \times \mathbb{R} \to \mathbb{R}$ called the mass density such that for any $3$-dimensional submanifold $W\subset D$ the total mass of fluid in $W$ is

$$m(W, t) = \int_W \rho(x , t) \ dV$$

That is fine, we also have $\mathbf{u} $ the spatial velocity field of the fluid, that is $\mathbf{u}(a, t)$ is the velocity of the particle of fluid at the point $a$ and time $t$. Suppose $\partial W$ is the boundary of $W$ and that $\mathbf{n}$ is the unit normal, then

$\mathbf{u}\cdot \mathbf{n}$ is the "volume flow rate per unit area" and $\rho \mathbf{u}\cdot \mathbf{n}$ is the "mass flow rate per unit area"

Now I simply can't grasp why. I've seem for many years people saying that for any vector field $\mathbf{V}$ and any surface with normal $\mathbf{n}$ the dot product of those is the flow of the field per unit area. This is already hard to swallow, and I never trully understood. How do we understand this in the context of fluid mechanics?

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2 Answers 2

up vote 5 down vote accepted

It can be shown via simple dimensional analysis. We know that $[u]=m/s$, so just multiply by 1 in terms of an area: $$ [u]=\frac{m}{s}\cdot\frac{m^2}{m^2}=\frac{m^3}{m^2\cdot s}=\color{red}{\frac{1}{m^2}}\cdot\color{blue}{\frac{m^3}{s}} $$ The blue term is the volumetric flow rate while the red term is the area, thus we have a volumetric flow rate per unit area. Multiplying this volumetric flow rate by the mass density gives a final unit of $$ \left[\rho u\right]=\color{red}{\frac1{m^2}}\cdot\color{blue}{\frac{kg}{s}} $$ which is the mass flow rate (the amount in mass that flows through the surface).

Since flows are generally three-dimensional, we are interested in how much of a fluid (either in terms of the volume or the mass) goes through an arbitrary surface (area) in some unit of time.

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You could express flow rate as a velocity. But if you want to have a quick measure of how much material flows through (for example) a pipe, you need to know both the velocity and the area - a quick diagram shows you that velocity x area = volume that passes through the area per unit time.

enter image description here

So if

$$v \cdot A = Vol/time$$

Then it follows that

$$v = \frac{Vol/time}{A}$$

and that is the units you see...

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