I know that an hydraulic jump can occur only when the Froude number is higher than 1 (supercritical flow), and I am familiar with the Bélanger equation. However, I feel like I have a fundamental misunderstanding about hydraulic jumps: if a jump can occur when Froude number is larger then 1, what prevents the flow in rivers from "jumping" immediately whenever its Froude number becomes slightly more than 1?

I mean how can a fluid develop a very high Froude number without making a hydraulic jump? According to what I read, in industrial applications such jumps are specially induced on the fluid to dissipate its energy. So I want to understand how it's done.

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    $\begingroup$ It's not clear what you want to know. The pressure edge and the resultant rise in surface height just plain happens when those critical parameters are exceeded. To induce a jump, design your spillway accordingly. Perhaps you can find what you want to know at the many technical article links in the Wikipedia article titled "hydraulic jump" $\endgroup$ Commented Apr 13, 2016 at 18:07
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    $\begingroup$ Just because supercritical flow is a requirement for hydraulic jumps to occur doesn't mean that it's the fact of supercritical flow that causes the jump. The Froude number simply represents a limiting condition on the flow; if the Froude number is less than 1, jumps won't occur regardless of the geometry of the channel. If you read the WP article carefully, you should start to see that it's the geometry of the channel that actually causes the jump. $\endgroup$
    – Air
    Commented Apr 13, 2016 at 18:21
  • $\begingroup$ That seems like a reasonable answer to the question. Q: "What prevents jumps?" A: "The geometry." $\endgroup$ Commented Apr 13, 2016 at 22:14
  • $\begingroup$ Can you describe in details how the geometry of the channel causes the jump? Can a growing cross section (and thus lower flow velocity) of the channel induce a jump? What is the condition on the gradient of cross section that is needed to induce the jump?. I ask all these questions because there must be some conditions - otherwise a jump can occur everywhere. $\endgroup$
    – user2554
    Commented Apr 14, 2016 at 10:41
  • $\begingroup$ I'm not sure I can give a detailed and accurate description of how the jump occurs. It's not a simple concept, it's been years since I studied hydraulics and even then I struggled (we all struggled) to understand. I'm mainly reluctant to answer because I don't remember how to predict the location of the jump, except in a very general conceptual sense - and so I would rather you get an answer from a hydraulics expert on that point. $\endgroup$
    – Air
    Commented Apr 15, 2016 at 17:38

1 Answer 1


First a hydraulic jump is where fluid will transition from supercritical (having a Froude Number greater than one, to subcritical (having a Froude Number less than one). When the Froude number is near one, the jump is very weak and somewhat gradual such that you might not even be able to tell it is "jumping".

To figure out where the jump occurs you need to know when the Froude Number will transition from less than one to more than one. To help with this is an extremely useful equation for modeling open channel flow:


Where $y$ represents the depth of flow in a riverbed, x is the distance along the flow, $S$ is the slope of the riverbed (positive being flowing downhill, negative, flowing uphill), $y_c$ is the depth of flow corresponding to $Fr=1$ and $y_n$ is the normal depth, the depth of flow corresponding to flow that is balancing the frictional losses with gravitation gains. $$y_c=\sqrt[3]{\frac{q^2}g}$$ $$y_n=\sqrt[3]{\frac{q^2}{C^2S}}$$

Where $q$ is the flow rate per width of river, $C$ is Chezy's coefficient (approximated as constant), and $g$ is acceleration due to gravity.

A depth greater than $y_c$ corresponds to a slower velocity, and thus subcritical flow (having a Froude number less than 1). When this is the case, the bottom of the fraction is positive between 0 and 1.

Similarly, when the depth is less than $y_c$ the flow is supercritical and the bottom of the fraction is negative.

In this range therefore, if the top of the fraction is also negative, then the flow will be getting deeper, but if it is positive, the flow will be getting shallower. This corresponds to a depth of flow less than and greater than $y_n$ respectively. This means that while the flow is supercritical, the depth of flow will always be moving towards $y_n$

This means that if you have a steep slope such that $y_n$ is less than $y_c$ the depth of flow will move to and then stay at $y_n$. If you would like the flow to have a hydraulic jump simply decrease the slope till $y_n$ is greater than $y_c$ then the flow will deepen at progressively faster rates until you reach the hydraulic jump.

Note since $q$ is based on your width it's also possible to modify the width rather than slope to cause $y_n$ to be greater than $y_c$.

But wait! After the hydraulic jump the flow will be subcritical, which means the flow downstream can now influence the water level and push the hydraulic jump upstream. How far upstream? Depends on the flow downstream.

In subcritical flows it is easiest to work upstream. If you try to work downstream from an initial guess, you'll end up with non-physical flows unless you happen to guess perfectly. This is because in subcritical flows, as you move downstream, the depth moves away from the normal depth. If you're less than the normal depth, you can quickly end up with a negative depth, and if you're more than the normal depth you can end up with a depth the would overflow your walls quick quickly. However, if you work backwards, going upstream then the depth always moves towards the normal depth, which should always give a valid depth.

So then the question is where to get your initial condition? Downstream of subcritical flow, and upstream on subscrital flow. The opposite transition of hydraulic jump. This transition is always smooth and thus according to the equation always happens when $y=y_n=y_c$ with $y_n$ decreasing downstream. An extreme example would be the lip of a waterfall or edge of an overflowing dam. For given flow rate, the depth will always be easy to predict at these choke points. As an added bonus the slope at these choke points is largely flow rate independent:


$$\sqrt[3]{\frac{q^2}g}=\sqrt[3]{\frac{q^2}{C^2S}}$$ $$\frac1g=\frac1{C^2S}$$ $$S=\frac{g}{C^2}$$

So anytime the slope goes from shallower than $\frac{g}{C^2}$ to steeper there might be a transition to supercritical flow right there, and anytime the slope goes from steeper than $\frac{g}{C^2}$ to shallower there might be a hydraulic jump a bit upstream.

I say might because it is possible that a choke point further downstream is causing the depth to be too high to cause any transitions (you could be at the bottom of a lake for example.)

However, if the slope is constant and shallower than $\frac{g}{C^2}$ for a long time (many miles), then odds are, near the top of this section the depth will have converged to $y_n$. So by this shortcut you don't have to start all the way back from a choke point.

In either case, you can model the flow going upstream from the subcritical section and downstream from the supercritical section. Then having modeled the supercritical section you can plot vs position the depth of flow that would result from a hydraulic jump at that location. Any time that plot intersects the subcritical plot is a location where a hydraulic jump could be stable.

Equations are derived in full in MIT's open courseware chapter on open channels. My analysis of those equations was also guided by the analysis presented there.

  • $\begingroup$ How are these equations derived? $\endgroup$
    – Hans
    Commented Apr 30, 2019 at 8:28
  • $\begingroup$ @Hans I added attribution for the equations. Thanks for heads up that they were missing. $\endgroup$
    – Rick
    Commented Apr 30, 2019 at 12:04
  • $\begingroup$ +1. Thank you, Rick. I like the justification for using a single number to approximate the speeds over the whole cross section as well as approximating the pressure by the hydrostatic pressure in Paragraph 22 on page 165 of the MIT open courseware. I have been wondering about the proper justification for the two simplifications. $\endgroup$
    – Hans
    Commented May 1, 2019 at 8:25
  • $\begingroup$ @Hans I believe the reason they only used a single number for flow velocity is that they are looking for equations that work for slow changes in mostly steady state flow. Under these conditions the shape of the flow profile doesn't change much, so you get information about the entire flow from just one number. I could see that if the stream split that you'd need a number for each branch. Or if you have significant vertical or cross flow, then the flow would not have an approximately similar flow profile across it's cross section. In this case these equations would poorly approximate the flow. $\endgroup$
    – Rick
    Commented May 1, 2019 at 14:38
  • $\begingroup$ As for the pressure, a similar statement holds: if you assume that the flow profile shape is constant, then all the pressures can be predicted based on the single velocity and the depth. So the assumption/approximation is that the flow profile is always constant. $\endgroup$
    – Rick
    Commented May 1, 2019 at 14:42

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