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A hydraulic bore is a propagating abrupt depth change described as a moving hydraulic jump. In the shock frame, an admissible rightward through-flow goes from supercritical flow to subcritical flow, with the appropriate characteristic curves entering the jump.

In a stationary rectangular-channel hydraulic jump, orient the discharge $q=hu>0$ from side 1 to side 2. Momentum conservation gives $q^2=g{h}_1{h}_2(h_1+h_2)/2$ for unequal depths. The specific mechanical energy is $B=u^2/2+gh$, soDividing by $g$ gives the displayed loss of head, measured as a length. The mechanical energy flux lost is $\rho q(B_1-B_2)$, so a dissipative jump must have $h_2>h_1$. With $R=h_2/h_1>1$, the squared Froude numbers are $R(1+R)/2>1$ upstream and $(1+R)/(2R^2)<1$ downstream. Thus the shallow water equations connect supercritical flow to subcritical flow through the admissible jump.

Assume a steady inviscid light layer of constant reduced gravity $g^{\prime }$ under a roof at $z=H(x)$, with depth $h$, interface $z_i=H-h$, and slowly varying width $W(x)$. The ambient is deep and approximately motionless; neglect entrainment, friction and approach variations across the section. Let $p^{\prime }$ be pressure relative to the ambient hydrostatic pressure. Then $p_z^{\prime }={\rho }_0{g}^{\prime }$ in the light layer and $p^{\prime }={\rho }_0{g}^{\prime }(z-z_i)$ when the ambient pressure offset is chosen zero. The steady Bernoulli equation along a light-fluid streamline is $u^2/2+p^{\prime }/{\rho }_0-g^{\prime }z=\mathrm{constant}$. With $Q=Whu$, this becomes the hydraulic flow in an inverted channel energyAt fixed $x$ and fixed flux,The energy curve has a minimum at $h_c=[Q^2/(g^{\prime }{W}^2){]}^{1/3}$, with $E_c=3g^{\prime }{h}_c/2-g^{\prime }H$. Above that minimum there are a thin supercritical branch and a thick subcritical branch. A lowered roof increases $E_c$; it is the inverted counterpart of raising a bed crest.

For a single smooth obstruction in a constant-width channel, energy exceeding its largest $E_c$ permits entirely subcritical passage or entirely supercritical passage, depending on the boundary data. At the limiting head a subcritical upstream current can pass through critical flow at the roof minimum and emerge supercritical; this is the usual freely discharging control. A reverse transcritical branch would need suitable imposed downstream data and is not the usual free discharge. If the supplied head is below the required minimum, no smooth steady flow with that flux exists: upstream adjustment changes the discharge or head. A supercritical current encountering deep downstream warm fluid can undergo a hydraulic jump to the subcritical branch, with an energy loss rather than a second conservative branch switch.

For a rectangular section the jump conserves $h_1{u}_1=h_2{u}_2=q$ and $q^2/h+g^{\prime }{h}^2/2$. Hence $h_2/h_1=[\sqrt{1+8F_1^2}-1]/2$ for $F_1>1$, and the lost specific energy is $g^{\prime }(h_2-h_1{)}^3/(4h_1{h}_2)$. This remains positive after vertical reflection. The energy and profile sketches label the roof and the lower boundary of the light layer explicitly.

Let $W_1$ be the first doorway width and $W_2$ the conservatory's outer doorway width. For a hydraulic comparison assume the same reduced gravity and conserved volume flux through both, negligible losses except at any hydraulic jump, and roofs at the same doorway height $D$. Their critical depths and limiting energies areThe narrower doorway has the larger minimum required head. This gives the possible two successive hydraulic controls rather than a universal jump location.

If $W_2>W_1$, the first doorway can remain the discharge control. Its supercritical outflow may pass through the wider second doorway without being controlled there. Alternatively, if warm air accumulates in the conservatory, the first jet can undergo an inverted hydraulic jump, feed a subcritical intermediate reservoir, and accelerate to a second critical section at the outer doorway. The wider second throat has a lower critical energy, so the difference can accommodate the positive head loss at the jump. Whether this second configuration occurs, and where its jump stands, depends on conservatory dimensions and downstream head.

If $W_2<W_1$, an outflow set solely by critical flow at the first doorway lacks the energy to pass the narrower second doorway: $E_{c2}>E_{c1}$, and a jump cannot increase energy. Warm air backs up in the conservatory. The downstream doorway becomes the dominant control, and the first doorway is subcritical or submerged by the intermediate layer. During adjustment an upstream-moving bore or jump can propagate towards or through the first doorway. A permanently freely controlled first doorway followed by a narrower same-height control is impossible under the conserved-buoyancy assumptions.

If $W_2=W_1$, the equal-energy idealization is marginal. Both throats can reach critical conditions in a lossless configuration, but any positive interposed jump loss prevents two independent critical sections at the same flux and buoyancy. Small losses or differences in effective discharge geometry select the downstream restriction and can drown the first. A constriction should not be assumed controlled simply because it exists.

The sketches show these admissible regimes. Mixing with cool conservatory air, heat loss, finite return-flow inertia, different roof elevations or aperture losses change the comparison and need additional data. In particular relative widths alone do not determine the actual conservative-room temperature or the position of a jump.

The shallow-water approximation requires the depth to be small compared with the horizontal scale of variation, small vertical accelerations and nearly hydrostatic pressure. Take cross-sectionally uniform lower-layer velocity and a sufficiently deep, effectively stationary upper layer. Under the Boussinesq approximation, $g^{\prime }=g({\rho }_1-{\rho }_0)/{\rho }_0$ is the reduced gravity. Conservation of lower-layer volume gives constant discharge $Q=bhu$.

Dividing the Bernoulli equation by $g^{\prime }$ gives the shallow-water specific energyAt fixed $Q,b$, the energy curve has its minimum at the critical depth $h_c=(Q^2/(g^{\prime }{b}^2){)}^{1/3}$, where $E_c=3h_c/2$. Above this minimum there are two depth branches: deeper subcritical flow has $F<1$, and shallower supercritical flow has $F>1$.

The two long-wave speeds relative to the bed are $u\pm \sqrt{g^{\prime }h}$. In subcritical flow one family can propagate upstream and one downstream; in supercritical flow both travel downstream for $u>0$. At hydraulic control, $u-\sqrt{g^{\prime }h}=0$, and a smooth transcritical solution must also satisfy the geometric regularity condition. The flow may remain entirely on either branch, cross smoothly through the control from subcritical to supercritical, or take the reverse smooth branch under appropriate boundary conditions. A downstream hydraulic jump can return supercritical flow to the subcritical branch while losing energy.

The sketch distinguishes smooth branch changes at a critical section from a jump: a jump conserves mass and momentum, not the original Bernoulli head.

Assume the jump is stationary, short compared with the scale of width variation, hydrostatic on either side, and has negligible bed stress integrated over its short length. Treat its width $b_j=\sqrt{32/27}$ as constant. It must occur on the downstream supercritical branch, at $x_j=L\sqrt{b_j-1}>0$. The discharge from the control gives the local unit-width dischargeBefore the jump the energy is still $H$. Writing $y=h/H$, its equation is $y+1/(8y^2)=1$. Its positive roots are $1/2$ and $(1+\sqrt{5})/4$; the former is supercritical. ThusMass conservation gives $h_1{u}_1=h_2{u}_2=q_j$. The hydraulic jump momentum balance conserves $q_j^2/h+g^{\prime }{h}^2/2$. Factoring out the non-jump solution $h_2=h_1$ gives the conjugate-depth relationConsequently the conditions immediately after the jump areIts Froude number is below one. The specific-energy loss isSelecting the deeper root of the unchanged energy equation would be wrong: that would conserve the energy across a turbulent jump instead of the momentum.

Bed and sidewall stress, turbulent mixing at the interface, wave radiation and hydraulic jumps convert or remove mechanical energy. In the specified local head-loss model, differentiate $B=u^2/2+g^{\prime }h$ while keeping $Q=bhu$ constant:At a regular hydraulic control the coefficient of $h^{\prime }$ vanishes. Therefore $B^{\prime }=-\alpha$ requiresThus the control moves to the widening downstream side, $x_c>0$. This is the exact local compatibility condition; it is the head-loss correction to hydraulic control with loss expressed as Bernoulli potential per unit length.

For a small loss and a control remaining near the throat, use the lossless depth $h_c=2H/3+O(\alpha )$ and $b^{\prime }/b=2x_c/L^2+O(x_c^3/L^4)$. The leading displacement isFor a finite loss, the local derivative alone does not specify the accumulated loss from the far-upstream reservoir, so it does not fix a unique revised discharge or numerical control depth. Given $Q$, the exact condition can instead be solved implicitly for $x_c$; if the inviscid discharge is imposed, it becomes $\alpha =(4g^{\prime }H/3L)(x_c/L)[1+(x_c/L{)}^2{]}^{-5/3}$. The small-displacement root is the one continuous from the throat. This distinction avoids treating the original upstream energy as unchanged after an unspecified finite upstream loss.

At the crest the two depth branches meet at $h=h_c=h_1{F}^{1/3}$. For a smooth nondegenerate maximum of the bed, $D^{\ast }-D$ is proportional to $(x-x_c{)}^2$, while expansion of the depth relation gives $D^{\ast }-D\sim 3(h-h_c{)}^2/(2h_c)$. A differentiable flow through the crest therefore switches branches rather than following the cusp obtained by staying on the same branch. Subcritical upstream flow becomes supercritical downstream; supercritical upstream flow becomes subcritical downstream.

When $F<1$ and the bed returns to zero, put $y=h_{\mathrm{down}}/h_1$. The head equation becomes $2y^3-(2+F)y^2+F=0$, orThe root $y=1$ is the original subcritical branch. The other positive root is smaller and lies on the supercritical branch, givingThis is the smooth transcritical, shock-free solution. A downstream hydraulic jump would require further downstream conditions and energy loss. A degenerate flat crest can also require a branch-selection condition; the standard critical-control interpretation uses a nondegenerate smooth crest. For $F=1$, $D^{\ast }=0$, so a positive hump is incompatible with the prescribed upstream critical head. If $F=0$, the head equation is the still-water relation $h+D=h_1$; a crest of height $h_1$ dries out, so the wet transcritical-flow interpretation requires $F>0$.

Use the steady inviscid, hydrostatic, slowly varying channel approximation, without a hydraulic jump. Conservation of volume flux per unit width gives $Q=u(x)h(x)=u_1{h}_1$. The free-surface Bernoulli equation isDividing by $h_1$ yieldsThe parameter $F$ here is the squared upstream Froude number. For nonzero discharge, regarding $D$ as a function of $h$, its derivative is $dD/dh=-1+Q^2/(g{h}^3)$. It has its unique maximum at $h_c=(Q^2/g{)}^{1/3}=h_1{F}^{1/3}$, where the local Froude number is one. Substitution gives $D^{\ast }=h_1(1-3F^{1/3}/2+F/2)$. This maximum is the hydraulic control over a smooth hump.

Let states $1$ and $2$ be just left and right of a hydraulic bore moving at speed $s$, and let $[f]=f_2-f_1$. Integrating the conservative prismatic-channel shallow water equations across it gives the Rankine-Hugoniot conditionEquivalently the discharge in the shock frame is common on the two sides,Thus, for unequal positive depths,Momentum is conserved through the bore in this ideal model, but mechanical energy is dissipated; an energy-conservation jump must not be imposed as well.

For a stationary bore with rightward flow, $s=0$, $A_1{u}_1=A_2{u}_2>0$, and the momentum jump must still hold. A compressive hydraulic jump has $h_2>h_1$ andThe $u-c$ characteristic curves approach the stationary bore from both sides: upstream flow is supercritical flow, while downstream flow is subcritical flow using the local wave speed $c_i=\sqrt{2g{h}_i/3}$. The $u+c$ family travels downstream on both sides. These characteristic directions accompany, rather than replace, the two stationary conservation jumps.

Orient the flow so $q>0$, and let subscripts 1 and 2 denote upstream and downstream. The mechanical energy flux is $\rho qB$, where $B=u^2/2+gh$. Inserting the relation for $q^2$ in i gives the hydraulic jump head loss,An admissible jump dissipates mechanical energy, so $\rho q(B_1-B_2)>0$. The denominator is positive, henceThe zero depth-change case is continuous flow. A depth-decreasing discontinuity would require an energy input at the jump; it is not the dissipative hydraulic jump selected by this model.

Take a unit-width control volume spanning the stationary hydraulic jump. Mass conservation makes the discharge $q=hu$ constant. The horizontal momentum flux is $\rho h{u}^2$, and the integrated hydrostatic pressure force on a vertical cross-section is ${\int }_0^h\rho g(h-z)dz=\rho g{h}^2/2$. Momentum conservation therefore givesThese are the Rankine-Hugoniot conditions for the shallow water equations in the stationary shock frame. With depths $h_1\ne h_2$ on the two sides, eliminating the velocity gives

For two same-height throats carrying the same light-layer volume flux and reduced gravity, the narrower throat has the larger critical minimum energy. A narrower downstream throat can back up and submerge an upstream throat. If the first throat is narrower, it can control discharge and feed a supercritical current through the second; a hydraulic jump followed by a second control is also possible when the available energy drop and intermediate geometry permit it. Equal throat widths are a marginal lossless case. Relative width alone does not fix jump position or mixing in the intervening reservoir.