Understanding Conservation of Momentum for Fluids

In Fluid Dynamics, the equation $$\sum F = \dfrac{d}{dt} \int_{CV} \mathbf{U} \rho dV + \int_{CS} \mathbf{U} \rho \mathbf{U} . d\mathbf{A}$$ was introduced, where is $\mathbf{U}$ and $V$ is volume. It makes sense in a way knowing that the force on a body equals the rate change of momentum and net momentum influx.

What doesn't make sense is integrating a vector over a different element that isn't a vector.

In vector calculus, one learns about integrating vector fields across surfaces, in which you can intuitively understand the formula $\int_S \mathbf{F} . \mathbf{n} dS$, where we project the force vector onto the normal vector pointing out of the surface.

But, for example, the first term of the equation presented in fluid dynamics integrates a vector field $\mathbf{U}$ over $dV$, which isn't a vector.

Can someone explain the equation in an intuitive way to me?

Imagine a ball made of lots of atoms. For each atom, calculate that atoms mass times it’s (vector) velocity: its momentum. Add all those up and you get the total momentum. That’s a vector quantity.

This is analogous. Each bit of fluid has a (normalized) momentum, which the integral sums up to a total momentum in the volume. That’s a vector.

So the first term is the rate of change (the derivative) of the momentum of the volume of fluid.

The second term captures the amount of momentum entering or leaving that volume. Imagine a bit $dA$ of the surface of our volume. Per unit time, the depth next to that given by the local velocity $U$ will go through the $dA$: that’s a volume flux of $U dA$. As before, the total momentum in that volume is $m U = \rho dV U = U \rho U dA$. Integrating over the entire surface, that’s the amount of momentum leaving or entering our original volume per unit time.

Finally, the equation says “momentum leaving or entering is what caused the momentum contained to change”. Which seems reasonable.

• That makes sense. So to extend to the formula, say for the first term, I imagine taking a small volume and density (equals mass) and then multiply by velocity? Then I add all these up (the integral) to get that value? – 14tim4 Apr 12 '18 at 3:18
• @14tim4 Right. Each $dV$ holds a bit of fluid with mass $\rho dV$, and multiplied by velocity you gets its momentum. – Bob Jacobsen Apr 12 '18 at 3:21
• So, when calculating the first term, we would be splitting up the velocity vector into its $x,y$ and $z$ components, then integrating over each component. Explicitly, something like $\int_{CV} U_i \rho dv + \int_{CV} U_j \rho + \int_{CV} U_k\rho$. Also, could you also explain briefly an intuitive way to think about the second term (like you did for the first term)? – 14tim4 Apr 12 '18 at 5:14
• Added a bit more to the answer. – Bob Jacobsen Apr 12 '18 at 5:27