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In fluid dynamics , the conjugate depth s refer to the depth (y 1 ) upstream and the depth (y 2 ) 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
=
y
1
2
2
+
q
2
g
y
1
=
y
2
2
2
+
q
2
g
y
2
.
{\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:
q
2
g
(
1
y
1
−
1
y
2
)
=
1
2
(
y
z
2
−
y
1
2
)
.
{\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:
q
2
g
(
y
2
−
y
1
y
1
y
2
)
=
1
2
(
y
2
−
y
1
)
(
y
2
+
y
1
)
.
{\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 (y 2 −y 1 ) term cancels out:
q
2
g
(
1
y
1
y
2
)
=
1
2
(
y
2
+
y
1
)
where
q
1
2
=
y
1
2
v
1
2
=
y
2
2
v
2
2
.
{\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 y 1 2
v
1
2
g
(
1
y
1
y
2
)
=
1
2
y
1
2
(
y
2
+
y
1
)
recall
F
r
1
2
=
v
1
2
g
y
1
.
{\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 y 2 and expand the right hand side :
F
r
2
2
=
y
2
2
2
y
1
2
+
y
2
2
y
1
.
{\displaystyle Fr_{2}^{2}={\frac {y_{2}^{2}}{2y_{1}^{2}}}+{\frac {y_{2}}{2y_{1}}}.}
Substitute x for the constant y 2 /y 1 :
F
r
1
2
=
x
2
2
+
x
2
⇒
0
=
x
2
2
+
x
2
−
F
r
1
2
.
{\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
4
2
{\displaystyle {\tfrac {\sqrt {4}}{2}}}
gives:
x
=
−
1
2
±
(
1
/
2
)
2
−
4
(
1
/
2
)
(
F
r
1
2
)
2
(
1
/
2
)
=
−
1
2
(
1
/
4
)
−
8
(
F
r
1
2
)
.
{\displaystyle x={\frac {-{\tfrac {1}{2}}\pm {\sqrt {(1/2)^{2}-4(1/2)(Fr_{1}^{2})}}}{2(1/2)}}=-{\frac {1}{2}}{\sqrt {(1/4)-8(Fr_{1}^{2})}}.}
Substitute the constant y 2 /y 1 back in for x to get the conjugate depth equation
y
2
y
1
=
1
2
(
1
+
8
F
r
1
2
−
1
)
.
{\displaystyle {\frac {y_{2}}{y_{1}}}={\frac {1}{2}}\left({\sqrt {1+8Fr_{1}^{2}}}-1\right).}