The mass flow rate through that we saw was rho v.n dS. And then momentum is mass times velocity. So the momentum rate is going to be mass flow rate times velocity. So take the mass flow rate and multiply it by the velocity. So you have to do this. If you do this rigorously, you can say that that is right. So I am doing it by analogy here. And then if I integrate it over this entire control surface, I will get the net momentum outflow rate through that surface. In the mass flow case, mass going out should be equal to the mass coming in for steady flow. In the momentum conservation case, momentum going out minus momentum coming in is balanced by the forces.

So this gives me an aggregate view of that term. And now I need an aggregate view of the forces on the control volume. And the forces we are considering are pressure and the viscous forces. Let's consider the pressure first. On dS, the pressure is going to be along the invert normal. So that's going to be p. So the magnitude of the pressure force is going to be p dS. And the direction is going to be in minus n because its pressure acts along the invert normal. So if I write this that way, then that gives me the pressure force. The force is a vector on that elemental surface.

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- Summer '18