Conjugate depth

M–y diagram.

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

${\displaystyle M={\frac {y_{1}^{2}}{2}}+{\frac {q^{2}}{gy_{1}}}={\frac {y_{2}^{2}}{2}}+{\frac {q^{2}}{gy_{2}}}.}$

Rearranging terms gives:

${\displaystyle {\frac {q^{2}}{g}}\left({\frac {1}{y_{1}}}-{\frac {1}{y_{2}}}\right)={\frac {1}{2}}\left(y_{z}^{2}-y_{1}^{2}\right).}$

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

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

The (y₂−y₁) term cancels out:

${\displaystyle {\frac {q^{2}}{g}}\left({\frac {1}{y_{1}y_{2}}}\right)={\frac {1}{2}}(y_{2}+y_{1})\qquad {\text{where }}q_{1}^{2}=y_{1}^{2}v_{1}^{2}=y_{2}^{2}v_{2}^{2}.}$

Divide by y₁²

${\displaystyle {\frac {v_{1}^{2}}{g}}\left({\frac {1}{y_{1}y_{2}}}\right)={\frac {1}{2y_{1}^{2}}}(y_{2}+y_{1})\qquad {\text{recall }}Fr_{1}^{2}={\frac {v_{1}^{2}}{gy_{1}}}.}$

Thereafter multiply by y₂ and expand the right hand side:

${\displaystyle Fr_{2}^{2}={\frac {y_{2}^{2}}{2y_{1}^{2}}}+{\frac {y_{2}}{2y_{1}}}.}$

Substitute x for the constant y₂/y₁:

${\displaystyle Fr_{1}^{2}={\frac {x^{2}}{2}}+{\frac {x}{2}}\Rightarrow 0={\frac {x^{2}}{2}}+{\frac {x}{2}}-Fr_{1}^{2}.}$

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

${\displaystyle x={\frac {-{\tfrac {1}{2}}\pm {\sqrt {(1/2)^{2}+4(1/2)(Fr_{1}^{2})}}}{2(1/2)}}.}$

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

${\displaystyle {\frac {y_{2}}{y_{1}}}={\frac {1}{2}}\left({\sqrt {1+8Fr_{1}^{2}}}-1\right)}$

Note that this equation is only applicable to hydraulic jumps over flat beds.