# Getting the Jump on Hydraulic Jump

Hydraulic jump is one of the more interesting phenomena in open channel flow.  It is used, for example, in stilling basins, where it is necessary to reduce the fluid velocity coming out from a dam by dissipating the kinetic energy in the flow.  An example of what it looks like is below.

This is a very basic treatment of the subject.  It is based on Streeter (1966) and also Dodge and Thompson (1937), where many of the graphics come from.  To keep things simple we will assume that the jump takes place in an open channel with two vertical parallel walls.  But first we need to introduce a couple of concepts.

## Froude’s Number

In James Warrington’s monograph about propulsive power, he mentions “the investigations of the elder Froude.”  The Froude Number, which is named after him, is basically the ratio of the inertia force to the gravity force, and is defined in Wind Tunnel Testing as

$Fr = \frac{u^2}{g_cD}$

Although it’s confusing, the Froude Number is frequently the square root of this value, and for this study will be designated by the equation

$Fr = \frac{u}{\sqrt{g_cy}}$

In this case $u$ is the average velocity of the fluid in a one-dimensional channel; the fact that it’s average is very important.  The constant $g_c$ is the gravitational constant and $y$ is the depth of the fluid (usually water) in the channel.  This is the way we will use the Froude Number in this study.

Since we’re restricting ourselves to vertically-walled channels with rectangular cross-sections, we can make another modification to the Froude Number.  If we define the flow as

$Q = u w y$

where $Q$ is the flow in volume per unit time, $w$ is the width of the channel and $y$ is defined as before, the Froude Number can be stated as

$Fr = \frac{Q}{w{g_c}^{\frac{1}{2}}y^{\frac{3}{2}}}$

Keep in mind this is only for channels with vertical walls and rectangular flow cross-sections.

## Specific Energy

Let us consider a vertical cross section of a one-dimensional horizontal channel flow.  Bernoulli’s Equation states that the fluid energy/head at that cross section should be

$E = y + \frac{u^2}{2 g_c}$

What we have done here is to eliminate the pressure term in Bernoulli’s Equation, which does not apply for a surface.  We are left with the potential term (the depth of the fluid in the channel) and the kinetic term (due to the movement of the fluid.)  The energy $E$ is thus in units of length, or is a head term.

Again assuming vertical, parallel walls and substituting the flow term, this equation can be stated as

$E = y + \frac{Q^2}{2w^2 y^2 g_c}$

The resulting curve looks like this:

Obviously this figure allows for non-parallel sides, but the basic theory is the same.  The depth which we call $y$ is called $d$ in this diagram.  At a depth of zero the specific energy is infinite, decreasing to a minimum at the point where the curve in (c) comes closest to the ordinate, and then decreases to the point where the second term on the right hand side vanishes and the specific energy equals to the depth.  We can find the location of this minimum by differentiating with respect to depth as follows:

$\frac{dE}{dy}=0=1-\frac{Q^2}{w^2y^3g_c}$

Solving this equation for $y$ yields a cubic equation with one real and two complex roots.  Using the real root only, the depth at which the energy is minimised is given by the equation

$y_c = \frac{Q\frac{2}{3}}{w^\frac{2}{3}g_c^\frac{1}{3}}$

This is referred to as the critical depth, and flow at this point is referred to as critical flow.  For depths shallower than this, the flow is referred to as rapid or supercritical.  For depths above this, the flow is referred to as subcritical or tranquil.

Some algebra will also reveal the following:

1. At critical flow, the Froude Number is unity.
2. At critical flow, the specific energy is $\frac{3}{2}y_c$.
3. Since the Froude Number is inversely proportional to $y^\frac{3}{2}$, for values of y greater than critical (subcritical/tranquil flow) the Froude Number is less than unity.
4. Conversely, for values of y less than critical (supercritical/rapid flow) the Froude Number is greater than unity.

## Solving for Hydraulic Jump

Now that we’ve laid the groundwork for some basics of flow in channels, it’s time to consider hydraulic jump itself, which is illustrated below.

The first thing worth noting is that the incoming flow (from the left) must be supercritical for hydraulic jump to take place.  From that the flow on the right must be subcritical.  In the process of the jump energy is dissipated, which is a big purpose of inducing hydraulic jump in, say, a stilling basin.  But this also means that conservation of energy does not apply here.  What does apply is a) conservation of momentum and b) conservation of mass flow.

Starting with conservation of momentum, after a little algebra and using our y notation for depth (as opposed to d in the figures) we have

$g_cy_1^2 + u_1^2y_1 = g_cy_2^2+u^2_2y_2$

After a great deal more algebra and application of the Froude Number,

$\frac{1+2Fr_{1}^2}{Fr_1^\frac{4}{3}} = \frac{1+2Fr_{2}^2}{Fr_2^\frac{4}{3}}$

Conservation of mass flow requires that

$Q_1 = Q_2$

or

$Fr_1y_1^\frac{3}{2}=Fr_2y_2^\frac{3}{2}$

There is more than one way to combine the conservation of momentum and mass flow equations.  One way results in determining the relationship between the two Froude Numbers, thus

$Fr_2 = \frac{2\sqrt{2}Fr_1}{\left( \sqrt{1+8Fr_1^2}-1 \right)^\frac{3}{2}}$

Plot of the Froude Number before the jump (Fr1) and after (Fr2)

We can also reapply the conservation of mass flow equation and eliminate $Fr_2$ in this way:

$\frac{y_1}{y_2} = \frac{2}{ \sqrt{1+8Fr_1^2}-1 }$

We now have a definite relationship between a) the two Froude Numbers and b) the ratio of the water depths on both sides of the jump to the first Froude Number (and thus the second.)

## Determining the Flow

At this point we have the basics of hydraulic jump.  There are many ways we can apply this theory.  In this monograph we’ll look at two: using the results to estimate the flow, and using the results to estimate the energy loss.

We’ll start with presenting this graphic overview of hydraulic jump and the effects of varying the upstream Froude Number (and thus the downstream one, as we saw earlier) from Research Studies on Stilling Basins, Energy Dissapators and Associated Appertuances.

Let’s consider a worked example, taken from Streeter (1966).  A hydraulic jump takes place where the upstream depth is 5′ and the downstream depth is 31′.  The sluice is 50′ wide.  Determine the Froude Numbers on either side of the jump and the flow through the jump in cubic feet per second.

We start by rewriting the equation that relates the ratio of depths to the upstream Froude Number (1) as follows:

$\frac{y_1}{y_2} - \frac{2}{ \sqrt{1+8Fr_1^2}-1 } = 0$

We then note that $\frac{y_1}{y_2} = \frac{5}{31} = 0.161.$  We could go through more algebra and solve for $Fr_1$, but it’s simpler to use goal seek and compute it by finding the zero of the above expression; it comes out to $Fr_1 = 4.72$.

We then substitute this result into this equation

$Fr_2 = \frac{2\sqrt{2}Fr_1}{\left( \sqrt{1+8Fr_1^2}-1 \right)^\frac{3}{2}}$

and compute $Fr_2 = 0.306$.

Since we now know the Froude Numbers for both sides, we can compute the flow.  We defined the Froude Number earlier in terms of flow; rearranging that equation yields

$Q = Fr w{g_c}^{\frac{1}{2}}y^{\frac{3}{2}}$

Taking either the upstream Froude Number and depth or the downstream Froude Number and depth and substituting these, the sluice width (50′) and the gravitational constant, the result $Q = 14,986.4 \frac{ft^3}{sec}$.

## Computing the Energy Loss

One of the main purposes for inducing hydraulic jump is to dissipate kinetic energy in a fluid flow.  Earlier we saw that at any point in the flow there is a specific energy, or head, in the flow.  If we subtract the specific energy at the upstream (Point 1) from the downstream (Point 2) the difference between the two heads is

$\Delta E = \frac{\left( y_2 - y_1 \right)^3}{4y_1y_2}$

or strictly a function of the two water depths.  If we substitute the two water depths from the last example, this yields an energy/head change of 28.35′.  Although the units are of length, it is more accurate to say that this is an energy per unit weight of water, or ft-lb/lb.  The power dissipated during the hydraulic jump is thus

$P = Q\gamma\Delta E$

Substituting the head change we just computed with the flow we computed earlier and the unit weight of water yields P = 26,510,196 ft-lb/sec = 48,200 hp.

You are urged to set these equations up in a spreadsheet and substitute the values given to confirm the validity of your spreadsheet before using them for another case.

## References (other than ones on this site)

• Dodge, R.A., and Thompson, M.J. (1937) Fluid Mechanics. First Edition.  New York: McGraw-Hill Book Company
• Streeter, V.L.  (1966) Fluid Mechanics.  Fourth Edition.  New York: McGraw-Hill Book Company.