# Fluid Dynamics: Conservation of momentum equation

1. Mar 6, 2015

### Feodalherren

1. The problem statement, all variables and given/known data

2. Relevant equations
Conservation of linear momentum for fluids

3. The attempt at a solution
This seemingly simple problem has me confused. First of all, I want to set up the sum of the foces as:

-Min + Mout -Fx = 0

So first of all, not only does my Fx term have the wrong sign, I'm missing an entire term! How did he arrive at the final equation and where does the last term come from, what does it symbolize?!

2. Mar 6, 2015

### Staff: Mentor

I don't like the way they wrote the equation. I would write it as:
$$F_{AX}=\frac{1}{2}γ_wh_1A_1+v_1ρv_1A_1-v_2sin20ρv_2A_2$$
The first term on the right hand side represents the pressure force (hydrostatic) acting on section 1 of the control volume. The second term represents the horizontal momentum entering the control volume at section 1. The third term represents the horizontal momentum exiting the control volume at section 2.

Hope this helps.

Chet

3. Mar 6, 2015

### Feodalherren

That looks a lot more sensible to me, thanks. Can you elaborate on the hydrostatic pressure force? It seems to me like this is the only problem where that pops up and we did that material months ago.

4. Mar 6, 2015

### Staff: Mentor

You have an open channel, so, at the top of the fluid layer, the pressure is atmospheric pressure. At depth z, the gauge pressure is γw z. If you integrate this gauge pressure variation over section A, you get the first term on the right hand side of the equation. Atmospheric pressure contributes on all the surfaces of the control volume, so it cancels out. The hydrostatic pressure force at section 2 is negligible, because the pressures on both sides of the layer are atmospheric, and the gauge pressures are zero.

Chet