# Homework Help: Volume Flux for a Hydraulic Jump

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1. Sep 12, 2017

1. The problem statement, all variables and given/known data
Problem is given in this image,
https://gyazo.com/454370ff9549dcd7c53604ebfe5df105

2. Relevant equations

Continuity or conservation of mass equation:
$$\frac{\partial u}{\partial x} + \frac{\partial w}{\partial z} = 0$$
Where u is the horizontal velocity and w is the vertical velocity

3. The attempt at a solution

Firstly I integrated the conservation of mass equation with respect to x between the two points:

$$\int_ {x_1} ^ {x_2} \frac{\partial u}{\partial x} \mathrm{d}x + \int_ {x_1} ^ {x_2} \frac{\partial w}{\partial z} \mathrm{d}x = 0$$
Which after evaluating I get,
$$u(x_2) - u(x_1) + x_2 \frac{\partial w}{\partial z} - x_1 \frac{\partial w}{\partial z} = 0$$
Firstly here I am not 100% sure I can assume w is just a function of z only, but I have yet to see it as a function of anything else other than t in 2D flow?

I then integrated the mass equation vertically first from $z = 0$ to $z = h_1$ and then from $z=0$ to $z= h_2$

$$\int_{0}^{h_1} \frac{\partial u}{\partial x} dz + \int_{0}^{h_1} \frac{\partial w}{\partial z} dz = 0$$

which yields,
$$h_1 \frac{\partial u}{\partial x} - w(0) = 0$$
Since $$w(h_1) = 0$$

Similarly for $z=0$ to $z= h_2$ I get,
$$h_2 \frac{\partial u}{\partial x} - w(0) = 0$$

This is where I am kind of stuck, from these two equations it appears $h_1 = h_2$ which doesn't make sense since this is a hydraulic jump and the nature of such is to increase the surface of the liquid.

Any hints where I may be going wrong here or missing something is appreciated.

Thanks.

2. Sep 12, 2017

### Staff: Mentor

First integrate with respect to z, using the Leibnitz rule for differentiation under the integral sign on the u term.

3. Sep 13, 2017

I try to integrate with respect to z first, I haven't used the Leibniz rule before but from what I can gather I can treat my bounds as constant thus I get this for the u term.

$$\int_{0}^{h_1} \frac {\partial u}{\partial x} dz = \frac {\partial}{\partial x} \int_{0}^{h_1} u dz$$

After evaluating this I get,
$$h_1 \frac {\partial u}{\partial x}$$

I'm not sure if I did this correct but following this doesn't seem to get me much further?
I suspect my math is off I haven't done any multi variable calc for a while.

4. Sep 13, 2017

### Staff: Mentor

$$\frac{\partial [\int_0^{h(x)}udz]}{\partial x}=\int_0^{h(x)}{\frac{\partial u}{\partial x}dz}+u(x,h)\frac{dh}{dx}$$