# How should energy loss in a hydraulic jump be calculated?

The energy loss in a hydraulic jump is still calculated with the old equation of Bresse from year 1860; (I.e., equation 7 in this paper from 2017)

$$\frac{\Delta E}{E_1} = \frac{(\sqrt{1+8Fr^2}-3)^3}{16(\sqrt{1+8Fr^2}-1)(1+\frac{1}{2}Fr^2)}$$

Here Fr = Froude number, E = Energy. This equation has a long known discrepancy to experimental data, i.e. figure 15.3. in the book of Chow 1959 shows an impossible difference; There is no measured energy loss when $$Fr<\sqrt3$$, though this equation predicts at $$Fr=\sqrt3$$ a loss of; $$\frac{\Delta E}{E_1} = \frac{(\sqrt{1+8*3}-3)^3}{16(\sqrt{1+8*3}-1)(1+\frac{1}{2}*3)}=\frac{2^3}{16(4)(2\frac{1}{2})}=\frac{8}{160}=5\%$$

This is obviously wrong, as it violates badly the conservation of energy, which must mean that the whole equation of Bresse is simply wrong.

Is there a better way to calculate this loss, where the logic is rigorously derived from the fundamentals?

Equation 15-1 from the book of Chow 1959 gives of course the same result for $$Fr=\sqrt3$$, as it's just another presentation of the same equation of Bresse 1860; $$\frac{E_2}{E_1} = \frac{(1+8Fr^2)^{3/2}-4Fr^2+1}{8Fr^2(2+Fr^2)}=0.95$$

If the previosly ignored roots are considered, (ie. p. 59 in the book of Chanson) then an energy loss equation can be created, which fits exactly to the experience. This means that the negative root of this quadratic equation is also considered;

$$\frac{d_2}{d_1} = \frac{\pm\sqrt{1+8Fr^2}-1}{2}$$

To keep this answer readible I get to the point, the full derivation is shown in the paper Navier-Stokes existency and smoothness problem, -The Answer.

The Energy factor (named as $$\Delta S$$ in the paper) can be created as follows;

$$\Delta S = d_2+ \frac{d_1}{2}$$

Through Froude definition can be written; $$d_1 = \frac{1}{1+\frac{1}{2}Fr^2}$$ Thus;

$$\Delta S = \frac{\sqrt{1+8Fr^2}}{2+Fr^2}$$

The energy loss calculated this manner produces perfect fit to the experimental data. This calculation idea can also be expanded to pipe flows, where it also produces exact loss factors ie. for the losses in the case of flow exit from tank, like shown in this question; Air core Vortex; Physical explanation of the "air Entrainment Hook" at $F_{co}=0.7$

Based on V.T. Chow classification in Open-Channel Hydraulics (1973), H. Chanson says in his book (p. 60) that there is no hydraulic jump for Froude number between 1 and 1.7 and only a weak one for Froude number between 1.7 and 2.5. He adds that the energy loss is negligible for the first one and weak for the second one. As $$\textit{Fr} =\sqrt(3)\simeq 1.7$$, this equation may not be relevant in this case to evaluate the energy loss. Also the equation may still be valid for strong hydraulic jump.

This equation is found in a rectangular cross-section for a steady flow. But hydraulic jumps are quite unsteady flows. The assumptions made to get this equation are probably not valid for a Froude number that low. The applications that Chanson gives in the book use a larger Froude number, greater than 4.

Maybe you can still use it to have a first idea of the possible energy loss but a proper study of the problem may be needed in complex cases. You talk about a rigorous derivation from the fundamentals, but as often in fluid mechanics, it is not that easy.

• I just got time take my copy of this book in my hand. Yes, your write this wrong. "No Hydraulic jump" is just on exactly Fr = 1. There is Undular Jump between 1...1.7 and here there is said to be negligible energy losses. (but non-zero) as you write. Mar 29, 2019 at 11:28