Conjugate depth

In fluid dynamics, the conjugate depths refer to the depth (y₁) upstream and the depth (y₂) downstream of the hydraulic jump whose momentum fluxes are equal for a given discharge (volume flux) q. The depth upstream of a hydraulic jump is always supercritical. It is important to note that the conjugate depth is different from the alternate depths for flow which are used in energy conservation calculations.

Mathematical derivation

M–y diagram.

Beginning with an equal momentum flux M and discharge q upstream and downstream of the hydraulic jump:

$M=\frac{y_1^2}{2}+\frac{q^2}{g y_1}=\frac{y_2^2}{2}+\frac{q^2}{g y_2}.$

Rearranging terms gives:

$\frac{q^2}{g}\left(\frac{1}{y_1}-\frac{1}{y_2}\right)=\frac{1}{2}\left(y_z^2-y_1^2\right).$

Multiply to get a common denominator on the left-hand side and factor the right-hand side:

$\frac{q^2}{g}\left(\frac{y_2-y_1}{y_1 y_2}\right)=\frac{1}{2}(y_2-y_1)(y_2+y_1).$

The (y₂−y₁) term cancels out:

$\frac{q^2}{g}\left(\frac{1}{y_1 y_2}\right)=\frac{1}{2}(y_2+y_1)\qquad\text{where }q_1^2=y_1^2 v_1^2=y_2^2 v_2^2.$

Divide by y₁²

$\frac{v_1^2}{g}\left(\frac{1}{y_1 y_2}\right)=\frac{1}{2 y_1^2}(y_2+y_1)\qquad\text{recall }F r_1^2=\frac{v_1^2}{g y_1}.$

Thereafter multiply by y₂ and expand the right hand side:

$F r_2^2=\frac{y_2^2}{2 y_1^2}+\frac{y_2}{2 y_1}.$

Substitute x for the constant y₂/y₁:

$F r_1^2=\frac{x^2}{2}+\frac{x}{2} \Rightarrow 0=\frac{x^2}{2}+\frac{x}{2}-F r_1^2.$

Solving the quadratic equation and multiplying it by $\tfrac{\sqrt{4}}{2}$ gives:

$x=\frac{-\tfrac{1}{2}\pm\sqrt{(1/2)^2-4(1/2)(F r_1^2)}}{2(1/2)}=-\frac{1}{2}\sqrt{(1/4)-8(F r_1^2)}.$

Substitute the constant y₂/y₁ back in for x to get the conjugate depth equation

$\frac{y_2}{y_1}=\frac{1}{2}\left(\sqrt{1+8 F r_1^2} - 1\right).$