Thursday, 11 December 2014

fluid dynamics - 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; Figure 15.3. in the book of Chow 1959


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; $$ \frac{E_2}{E_1} = \frac{(1+8Fr^2)^{3/2}-4Fr^2+1}{8Fr^2(2+Fr^2)}=0.95 $$



Answer



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 $F_{co}=0.7$


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