Loss in a hydraulic jump.

It is possible to make an estimate of the loss of energy in a hydraulic jump in a horizontal rectangular channel using the same method as was used to find an expression for the loss of energy in a sudden enlargement in a pipe. I cannot imagine it having any great value for mechanical engineers but the method used is likely to be useful when trying to analyse other flows that are important in mechanical engineering. It turns out that we must first find the depth after the jump in terms of the depth before the jump and then substitute in the total head equation.

 

This can be done for a channel of rectangular section if we isolate, between two cross sections, a part of the water containing the hydraulic jump. I have shown this diagrammatically in figure 10-13 which is a redraft of 10-9. The jump is now between two sections 1 and 2. There are two forces that are not in line and, as I have said earlier, these forces reduce the speed of flow and impart angular momentum to the water as it goes through the jump. The normal decision taken in order to get a result is first to treat the flow as being one-dimensional and then to ignore the fact that the forces are out of line and equate the net force on the section of water in the direction of motion and the resulting change in momentum per second. Whether this is justified depends not on validity of the mathematics but on whether it is any value when it is used in ordinary engineering analysis. My experience is that it is useful.

 

The force in direction of motion =  = .

Now, if we regard the velocities in sections 1 and 2 as uniform, we can say that the resulting change in momentum per second is equal to .

 It follows that,  and that :-

                                      

 We can also use continuity of flow to write  and, from this,

          and then  .

This is an equation in the flow per unit width and the depths before and after the jump. It must be possible to manipulate it to plot a graph of  against  for a given value of q.[1] It is common to leave the manipulation to the reader but I will lay it out in full.

 from which,  .

This can be rearranged to give a quadratic equation in  and other quantities that are known.

 

      and this is the required quadratic equation in .

It can be manipulated to give a solution and a graph plotted from it.

                                or 

.

This gives the final form:-   which is quite elegant.

 

Mathcad can be used us to explore the physics of the hydraulic jump in a rectangular channel. The outcome is in graph 10-6. I have used a value of  of  as before and plotted everything against values of the depth before the jump. The depth after the jump comes from the equation directly above. Once the final depth is known the loss can be evaluated from  and the loss as a proportion of  by division. We have expressions for the critical velocity and for the critical depth and these are constants for a given value of flow per unit width of  equal to 4.43 m/s and 0.74 m. These can be used to plot the values of Fr before and after the jump.

 

The graphs have no meaning above the critical depth of 0.74 m.

 

In assessing this graph one should remember that nothing has been said about the depth required before the gate to create the rapid flow initially. At the small depth of 0.1 m the velocity before the jump is about 20 m/s and this would be associated with a kinetic head of 20.4 m and the loss is 17.63 m. If this comes from a transition from tranquil flow then the upstream depth would be about 20 m, a depth that is most unlikely to occur.

 

Not surprisingly this physics tells us that, as the upstream Froude number increases, the loss in the jump as a proportion of the upstream total energy also increases. The strong jump is useful as an energy dissipation device.

 

This graph and its implications result from delving into a very simple mathematical model of the flow through a hydraulic jump.  It is worth going into MathCad and plotting these graphs for other rates of flow.