Momentum-depth relationship in a rectangular channel

Introduction to Momentum

The basic momentum relationship in mechanics is that the sum of all forces in the s direction is equal to the change in momentum in the s direction where momentum is mass times the velocity in the s direction:

\[\sum F_s = \Delta (mV_s)\]

Where:

• For units: L = length, t = time, M = mass
• F_s = Force in the s direction (ML/t²)
• m = mass (M)
• V_s = velocity in the s direction (L/t)

The General Momentum Equation for Horizontal Channels

Momentum for one-dimensional flow in an ideal, frictionless, horizontal channel can be expressed by the following equation:

\[M = {Q^2 \over gA} + A \overline{y}\]

Where:

• For units: L = length, t = time
• A = area of flow (L²)
• g = gravitational constant (L/t²)
• M = momentum of the flow (L³)
• Q = flowrate or discharge in the channel (L³/t)
• ȳ = distance from the centroid of A to the water surface (L)

The Momentum Equation for Horizontal, Rectangular Channels

When considering an ideal, frictionless, horizontal, rectangular channel, this equation can be rewritten as:

\[M = {Q^2 \over g(by)} + {(by){y \over 2}}\]

Where:

• For units: L = length, t = time
• A = by = area of flow (L²)
• b = base width of the channel (L)
• g = gravitational constant (L/t²)
• M = momentum of the flow (L³)
• Q = flowrate or discharge in the channel (L³/t)
• ȳ = y/2 = distance from the centroid of A to the water surface (L)
• y = flow depth (L)

This can be further simplified when considering a rectangular channel of unit width by diving through by b and by noting that unit flowrate, q = Q/b. The resulting general equation for horizontal, rectangular channels of unit width is then:

\[M_{unit} = {q^2 \over gy} + {y^2 \over 2}\]

This equation is only valid in certain unique circumstances, such as in a laboratory flume, where the channel is truly rectangular and the channel slope is zero or assumed to be very small. When this is the case, it is possible to assume that a hydrostatic pressure distribution applies. With this equation, momentum, M, becomes unit momentum, M_unit, and now is associated with units of length² rather than length³. If the channel width is known, the full momentum (length³) at a point can be determined by multiplying M_unit by the base width, b.

Conservation of Momentum

For open-channel flow, there are certain conditions in which momentum can be assumed to be conserved from one point in the channel to another, such as in a hydraulic jump (discussed later). In these circumstances, it is possible to equate the momentum at an upstream point, M₁, to the momentum at a downstream point, M₂. In this way, the following relationship can be developed for an ideal, frictionless, horizontal, rectangular channel when considering a control volume over which momentum is conserved:

\[M_1=M_2 \quad \Rightarrow \quad {q^2 \over gy_1} + {y_1^2 \over 2} = {q^2 \over gy_2} + {y_2^2 \over 2}\]

Figure 1 depicts a hydraulic jump, a phenomenon in open-channel flow in which there is an abrupt change in water surface elevation caused when an upstream, shallow, fast-moving flow comes in contact with a downstream, deeper, slower-moving flow. A hydraulic jump is one example of a device or occurrence over which momentum is conserved. The law of conservation of momentum states that the total momentum of a closed system of objects (which has no interactions with external agents) is constant. In the case of a jump, this means that the momentum at a location downstream of the jump (y₂) is equal to the momentum at a location upstream of the jump (y₁) assuming that the jump is not caused by some outside influence.

File:Conservation of Momentum – Hydraulic Jump .png

Figure 1: Conservation of Momentum – Hydraulic Jump

The green box in the figure represents the control volume enclosing the jump system and shows the major pressure forces on the system (F_P1 and F_P2). As this system is considered to be horizontal (or nearly horizontal) and frictionless, the horizontal components of force that normally exist due to friction (F_f) and the weight of water from a sloping channel (F_w) are neglected.

Derivation of Momentum Equation from Momentum-Force Balance

Definition of variables:

• For units: L = length, t = time, M = mass
• A = area of flow (L²)
• A₁ = area of flow at upstream of jump – supercritical flow area (L²)
• A₂ = area of flow at downstream of jump – subcritical flow area (L²)
• b = base width of the channel (L)
• Fr₁ = Froude number of the upstream flow – supercritical Froude number (Fr₁ > 1) (unitless)
• Fr₂ = Froude number of the downstream flow – subcritical Froude number (Fr₂ < 1) (unitless)
• F_f = force due to friction (ML/t²)
• F_P1 = upstream pressure force (ML/t²)
• F_P2 = downstream pressure force (ML/t²)
• F_w = gravitational force due to weight of water (ML/t²)
• g = gravitational constant (L/t²)
• M₁ = momentum of the flow upstream of jump – supercritical momentum (Length³)
• M₂ = momentum of the flow downstream of jump – subcritical momentum (Length³)
• Q = flowrate or discharge in the channel (L³/t)
• q = unit flowrate or discharge – for a rectangular channel, discharge per unit channel width (L²/t)
• ρ = fluid density (M/L³)
• V = flow velocity (L/t)
• V₁ = flow velocity upstream of jump – supercritical flow velocity (L/t)
• V₂ = flow velocity downstream of jump – subcritical flow velocity (L/t)
• y = flow depth (L)
• y₁ = flow depth upstream of jump – supercritical flow depth (L)
• y₂ = flow depth downstream of jump – subcritical flow depth (L)

The momentum-force balance of the normal jump can be generally represented as:

\[M_{1} + M_{2} = F_{w} + F_{P1} + F_{P2} + F_{f}\]

Neglecting \(F_f\) and \(F_w\):

\[M_{1x} + M_{2x} = F_{P1x} + F_{P2x}\]

Substituting in the components of momentum and of the pressure forces:

\[M_{1x} = \rho Q (V_1\cdot \mathbf{n}) = - \rho Q V_1 \quad and \quad F_{P1x} = \overline{P_1}A_1\]

\[M_{2x} = \rho Q (V_2\cdot \mathbf{n}) = \rho Q V_2 \quad and \quad F_{P2x} = \overline{P_2}A_2\]

The equation becomes:

\[- \rho Q V_1 + \rho Q V_2 = \overline{P_1}A_1 - \overline{P_2}A_2\]

For a rectangular channel, the pressure distribution forms a right triangle down from the water surface so the average pressure can be assumed to exist at half of the flow depth. Also, assuming continuity holds:

At \(h={y/2}\): \[P = P_{avg} = \overline{P} = {1 \over 2} \rho g y \quad and \quad Q = VA \quad \Rightarrow \quad V = {Q \over A} \quad \Rightarrow \quad V_1 = {Q \over A_1} \quad and \quad V_2 = {Q \over A_2}\]

Using these characteristics, the above conservation of momentum equation can be rewritten as:

\[-{\rho Q^2 \over A_1} +{\rho Q^2 \over A_2} = \left({1 \over 2} \rho g y_1 \right) A_1 - \left({1 \over 2} \rho g y_2 \right) A_2\]

Dividing through by \(\rho g\):

\[-{Q^2 \over gA_1} + {Q^2 \over gA_2} = \left({y_1 \over 2} \right) A_1 - \left({y_2 \over 2} \right) A_2\]

Separating the variables based on the sides of the jump that they occur on (sides 1 and 2):

\[{Q^2 \over gA_1} + \left({y_1 \over 2} \right) A_1 = {Q^2 \over gA_2} + \left({y_2 \over 2} \right) A_2\]

And by recognizing that ȳ\(= {y/2}\) for the case of the rectangular channel, we are brought back to the original general momentum equation:

\[{Q^2 \over gA_1} + \overline{y} A_1 = {Q^2 \over gA_2} + \overline{y} A_2\]

This derivation demonstrates how the momentum equation for an ideal, frictionless, horizontal, rectangular channel can be obtained and how it can be related to the general momentum equation for open channel flow.

The M-y Diagram

An M-y diagram is a plot of the depth of flow, y, versus momentum, M. This produces a specific momentum curve that is generated by calculating momentum for a range of depth values and plotting the results. Each M-y curve is unique for a specific flowrate, Q, or unit discharge, q. The momentum on the x-axis of the plot can either have units of length³ (when using the general M equation) or units of length² (when using the rectangular M_unit equation for unit widths). In a rectangular channel of unit width, an M-y curve is plotted using:

\[M_{unit} = {q^2 \over gy} + {y^2 \over 2}\]

Figure 2 depicts a sample M-y diagram showing the plots of four specific momentum curves for a horizontal, rectangular channel. Each of these curves corresponds to a specific q as noted in the figure. As unit discharge increases, the momentum curve shifts to the right and slightly upward (Figure 2).

File:M-y Curves for Various Unit Flowrates.png

Figure 2: M-y Curves for Various Unit Flowrates

M-y diagrams can provide a few key pieces of information about the characteristics and behavior of a certain discharge in a channel. Primarily, an M-y diagram will show which flow depths correspond to supercritical or subcritical flow for a given discharge, as well as defining the critical depth and critical momentum of a flow. In addition, M-y diagrams can aid in finding conjugate depths of flow that have the same specific momentum, as in the case of flow depths on either side of a hydraulic jump.

Critical Flow

A flow is termed critical if the bulk velocity of the flow \(V\) is equal to the propagation velocity of a shallow gravity wave \(\sqrt {gy}\). At critical flow, the specific energy and the specific momentum (force) are at a minimum for a given discharge. Figure 3 shows this relationship by showing a specific energy curve (E-y diagram) side-by-side to its corresponding specific momentum curve (M-y diagram) for a unit discharge q = 10 ft²/s. The green line on these figures intersects the curves at the minimum x-axis value that each curve exhibits. As noted, both of these intersections occur at a depth of approximately 1.46 ft, which is the critical flow depth for the specific conditions in the given channel. This critical depth represents the transition depth in the channel where the flow switches from supercritical flow to subcritical flow or vice versa.

File:E-y & M-y Curves for a Rectangular Channel.png

Figure 3: E-y & M-y Curves for a Rectangular Channel with q = 10 ft²/s

In a rectangular channel, critical depth (y_c) can also be found mathematically using the following equation:

\[y_c = \left({q^2 \over g} \right)^{1 \over 3}\]

Where:

• For units: L = length, t = time
• g = gravitational constant (L/t²)
• q = unit flowrate or discharge – for a rectangular channel, discharge per unit channel width (L²/t)

Supercritical Flow versus Subcritical Flow on an M-y Diagram

As mentioned before, an M-y diagram can provide an indication of flow classification for a given depth and a given discharge. When flow is not moving critically, it can either be classified as subcritical or supercritical. This distinction is based on the Froude number of the flow, which relates the bulk velocity (V) to the propagation velocity of a shallow wave ([g*y]^0.5) as follows:

\[F_{r_1} = {V_1 \over {\sqrt {gy_1}}} \quad or \quad F_{r_2} = {V_2 \over {\sqrt {gy_2}}}\]

When this ratio is greater than one, the flow is deemed supercritical, whereas a Froude number less than one represents a subcritical flow. In general, supercritical flows are shallower and have a high velocity and subcritical flows are deeper and have a low velocity. These different flow classifications are also represented on M-y diagrams, on which different regions of the plot represent different flow types. Figure 4 can be used to show these regions, with a specific momentum curve corresponding to a q = 10 ft²/s. As stated previously, critical flow is represented by the minimum momentum that exists on the curve (green line). Supercritical flows correspond to any point on the momentum curve that has a depth less that the critical depth and subcritical flows exist above the critical depth. The supercritical branch of the curve asymptotically approaches the horizontal axis while the subcritical branch extends upward and indefinitely to the right.

File:M-y Curve for a Horizontal, Rectangular Channel.png

Figure 4: M-y Curve Corresponding to a q = 10 ft²/s in a Horizontal, Rectangular Channel

A Hydraulic Jump

One of the most common applications of the momentum equation in open channel flow is the analysis of a hydraulic jump. A hydraulic jump is a region of rapidly varied flow and is formed in a channel whenever a supercritical flow transitions into a subcritical flow. This change in flow type is manifested as an abrupt change in the flow depth from the shallower, faster-moving supercritical flow to the deeper, slower-moving subcritical flow. In natural systems, the energy of a flow is dissipated as it travels along a channel due to frictional resistance. This resistance results in a decrease in velocity and therefore an increase in depth in the direction of flow. This difference in flow depth is essentially what initiates the jump conditions.

A jump causes the water surface to rise abruptly, and as a result, surface rollers are formed, intense mixing occurs, air is entrained, and usually a large amount of energy is dissipated. For these reasons, a hydraulic jump is sometimes forced in an attempt to dissipate flow energy, to mix chemicals, or to act as an aeration device.

Despite the fact that there is an energy loss, momentum across a hydraulic jump is still conserved. This means that the flow depth on either side of the jump will have the same momentum, and in this way, if the momentum and flow depth at either side of the jump is known, it is possible to determine the depth on the other side of the jump. These paired depths are known as sequent depths, or conjugate depths. Figure 5 shows a typical profile of a jump in a horizontal, rectangular channel with y₁ and y₂ representing conjugate depths.

File:Hydraulic Jump Showing Conjugate Depths (y1 and y2).png

Figure 5: Hydraulic Jump Showing Conjugate Depths (y₁ and y₂)

Types of Jumps

A hydraulic jump can assume several distinct forms depending on the approach Froude number, Fr₁. Each of these types has unique flow patterns and flow characteristics, such as the strength and formation of rollers and eddies, that help to determine the amount of energy dissipation that will occur in the jump. The following descriptions of jump types are based on specific ranges of Froude numbers, but it should be noted that these ranges are not precise and that overlap can occur near the endpoints.

Weak Jump (1 < Fr₁ < 2.5)

For the case when 1 < Fr₁ < 1.7, y₁ and y₂ are approximately equal and only a very small jump occurs. In this range, the water surface shows slight undulations and because of this, jumps in this range are sometimes known as undular jumps. These surface riffles generally result in very little energy dissipation. As Fr₁ approaches 1.7, a number of small rollers begin to form at the water surface at the jump location, but in general, the downstream water surface remains relatively smooth. Between 1.7 < Fr₁ < 2.5, the velocity remains fairly uniform on either side of the jump and energy loss is low.

Oscillating Jump (2.5 < Fr₁ < 4.5)

An oscillating jump can occur when 2.5 < Fr₁ < 4.5. During this jump, the jet of water at the entrance of the jump (supercritical) fluctuates from the bottom of the channel to the top of the channel at an irregular period. Turbulence created from this jet can be near the channel bottom at one instant and then suddenly transition to the water surface. This oscillation of the jet causes irregular waves to form, which can propagate for long distances downstream of the jump, potentially causing damage and degradation of the channel banks.

Steady Jump (4.5 < Fr₁ < 9)

When the Froude number falls into this range, the jump forms steadily and at the same location. In a steady jump, turbulence is confined within the jump and the location of the jump is the least susceptible to downstream flow conditions out of the four major types of jumps. Steady jumps are generally well-balanced and the energy dissipation is usually considerable (45-70%).

Strong Jump (Fr₁ > 9)

There is a large difference in conjugate depths in a strong jump. Strong jumps are characterized by a jump action that is very rough resulting in a high energy dissipation rate. At irregular intervals, slugs of water can be seen rolling down the front of the jump face. These slugs enter the high-velocity, supercritical jet and cause the formation of additional waves in the jump. Energy dissipation in strong jumps can reach up to 85%.

Jump Location

In general, a hydraulic jump is formed at a location where the upstream and downstream flow depths satisfy the conjugate depth equation. However, there can be conditions in a channel, such as downstream controls, that can alter where the conjugate depths form. Tailwater depth can play a very influential role on where the jump will occur in the channel, and changes in this depth can shift the jump either upstream or downstream. Figure 6 contains three scenarios of tailwater elevations (y_d): y_d is equal to the conjugate depth (y₂) of the upstream flow depth (y₁), y_d is less than the conjugate depth (y₂) of the upstream flow depth (y₁), and y_d is greater than the conjugate depth (y₂) of the upstream flow depth (y₁). The upstream depth (y₁) in all three cases is controlled by a sluice gate and remains constant. Its corresponding conjugate depth (y₂) is shown by the dashed line in each of the scenarios.

File:Effect of Tailwater Elevation on Jump Location.png

Figure 6: Effect of Tailwater Elevation on Jump Location

In the first situation (Scenario A), the jump is formed right at the apron, as it would if there was no downstream control. However, in the next scenario (Scenario B), the downstream tailwater depth has some control imposed on it such that it is less than the conjugate to y₁. In this case, the jump travels downstream and initiates at a point where the upstream flow depth (y₁’) has risen to the conjugate of the new downstream tailwater depth (y_d). This rise from y₁ to y₁’ is caused by frictional resistance in the channel; and velocity decrease, the depth increase. In this image, y₁’ and y₂’ represent the conjugate depths of the hydraulic jump where y₂’ assumes the depth of y_d. In contrast, in the third setup (Scenario C), there is a downstream control that forces the tailwater elevation to a depth above the original conjugate depth. Here, y_d is greater than the required depth so the jump is pushed upstream. In this scenario, the sluice gate inhibits the movement of the jump upstream so that the upstream conjugate cannot be attained. This leads to a situation known as a submerged or drowned hydraulic jump. These scenarios demonstrate how influential the role of tailwater is to jump formation and location.

Conjugate Depths in a Rectangular Channel

As stated previously, conjugate or sequent depths are the paired depths that result before and after a hydraulic jump. The first conjugate depth (y₁) corresponds to the supercritical flow depth before the jump whereas the second depth (y₂) corresponds to the subcritical flow depth after the jump. Conjugate depths can be found either graphically with the aid of an M-y diagram or algebraically with a set of equations.

By definition, conjugate depths have the same momentum, as momentum is conserved over a hydraulic jump. Taking a graphical approach, the conjugate to any flow depth can be determined with an M-y diagram corresponding to a specific unit flowrate. Figure 7 shows an M-y diagram corresponding to a unit discharge, q, equal to 10 ft²/s in a rectangular channel. A vertical line can be drawn on an M-y diagram representing a line of constant momentum. This line will either cross the M-y curve once (critical flow), twice (conjugate depth flow), or zero times (below critical flow). Given sufficient momentum (above critical momentum), a conjugate depth exists at each point where the vertical line intersects the M-y curve. Figure 7 exemplifies this behavior and shows a vertical line corresponding to a momentum of 25 ft². This line crosses the M-y curve at a depth of about 0.12 ft (y₁) and again at a depth of about 7.01 ft (y₂). The y₁ depth corresponds to the supercritical depth before the jump and the y₂ corresponds to the subcritical depth after the jump. In this way, an M-y diagram can be used to determine either the upstream or downstream depth of a jump, knowing either the flow depth or the momentum at one of sides.

File:M-y Diagram Showing Conjugate Depths Corresponding to a Momentum = 25 ft2.png

Figure 7: M-y Diagram for q = 10 ft²/s Showing Conjugate Depths Corresponding to a Momentum = 25 ft²

Conjugate depths can also be calculated algebraically if the Froude number and flow depth of either the supercritical or subcritical flow are known or can be determined. The following equations can be used to determine the conjugate depth to a known depth in a horizontal, rectangular channel:

\[y_2 = {y_1 \over 2} \left(-1 + \sqrt{(1 + 8{F_{r_1}}^2)}\right) \quad or \quad y_1 = {y_2 \over 2} \left(-1 + \sqrt{(1 + 8{F_{r_2}}^2)} \right)\]

These two equations were developed from the momentum conservation principle M₁ = M₂. Their derivation is as follows:

Derivation of Conjugate Depth Equation for a Rectangular Horizontal Channel

Definition of variables:

• For units: L = length, t = time
• A = area of flow (L²)
• b = base width of the channel (L)
• Fr₁ = Froude number of the upstream flow – supercritical Froude number (Fr₁ > 1) (unitless)
• Fr₂ = Froude number of the downstream flow – subcritical Froude number (Fr₂ < 1) (unitless)
• g = gravitational constant (L/t²)
• Q = flowrate or discharge in the channel (L³/t)
• q = unit flowrate or discharge – for a rectangular channel, discharge per unit channel width (L²/t)
• V = flow velocity (L/t)
• V₁ = flow velocity upstream of jump – supercritical flow velocity (L/t)
• V₂ = flow velocity downstream of jump – subcritical flow velocity (L/t)
• y = flow depth (L)
• y₁ = flow depth upstream of jump – supercritical flow depth (L)
• y₂ = flow depth downstream of jump – subcritical flow depth (L)

Start with the conservation of momentum, \(M_1 = M_2\):

\[\left({q^2 \over gy_1} + {y_1^2 \over 2} \right) = \left({q^2 \over gy_2} + {y_2^2 \over 2}\right)\]

Isolate the q terms on one side of the equals sign with the \(y^2\) terms on the other side:

\[{q^2 \over gy_1} - {q^2 \over gy_2} = -{y_1^2 \over 2} + {y_2^2 \over 2}\]

Pull out the constant terms q²/g and 1/2:

\[{q^2 \over g} \left({1 \over y_1} - {1 \over y_2}\right) = {1 \over 2} ({y_2^2} - {y_1^2})\]

Combine the terms on the left side and expand the quadratic on the right side:

\[{q^2 \over g} \left({{y_2 - y_1} \over {y_1y_2}} \right) = {1 \over 2} ({y_2 - y_1})({y_2 + y_1})\]

Cancel out the \((y_2 - y_1)\) term on each side of the equation:

\[{q^2 \over g} \left({1 \over {y_1y_2}} \right) = {1 \over 2}({y_2 + y_1})\]

From Continuity in a rectangular channel and using a unit width:

\[Q = VA = Vby \qquad and \qquad q = {Q \over b} \qquad \Rightarrow \qquad q = V_1y_1 = V_2y_2\]

Substitute \(V_1y_1\) into the right side of the equation for q:

\[{{V_1^2y_1^2} \over {gy_1y_2}} = {{V_1^2y_1} \over {gy_2}} = {{y_2 + y_1} \over 2}\]

Divide each side of the equation by \({y_1/y_2}\):

\[{{V_1^2} \over {g}} = {{y_2({y_2 + y_1})} \over {2y_1}}\]

Divide each side of the equation by \(y_1\) and recognize that the left hand side is now equal to F_r1²:

\[{{V_1^2} \over {gy_1}} = {{y_2({y_2 + y_1})} \over {2y_1^2}} \Rightarrow {V_1^2 \over gy_1} = F_{r_1}^2 = {y_2^2 \over {2y_1^2}} + {{y_2y_1} \over 2y_1^2} = {y_2^2 \over {2y_1^2}} + {{y_2} \over 2y_1}\]

Set the equation equal to zero:

\[0 = {y_2^2 \over {2y_1^2}} + {{y_2} \over 2y_1} - F_{r_1}^2\]

If \(x = {y_2/y_1}\), the above equation becomes:

\[0 = {1 \over {2}}x^2 + {1 \over {2}}x - F_{r_1}^2\]

Solve for \(x\) using the quadratic equation with \(a = {1/2}\), \(b = {1/2}\), and \(c = \)F_r1²:

\[x=\frac{-b \pm \sqrt {b^2-4ac}}{2a} \quad \Rightarrow \quad x = {y_2 \over y_1} = \frac{-{1 \over 2} \pm \sqrt {{1 \over 2}^2-4\left({1 \over 2}\right)\left(-F_{r_1}^2\right)}}{2\left({1 \over 2}\right)} = \frac{-{1 \over 2} \pm \sqrt {{1 \over 4}+2{F_{r_1}^2}}}{1}\]

Pull out the 1/4 inside of the square root:

\[x = {y_2 \over y_1} = -{1 \over 2} \pm \sqrt {{1 \over 4} \left(1+8{F_{r_1}^2}\right)} = -{1 \over 2} \pm {1 \over 2} \sqrt {\left(1+8{F_{r_1}^2}\right)}\]

Solutions for \(x = {y_2/y_1}\):

\[x_1 = {y_2 \over y_1} = -{1 \over 2} + {1 \over 2} \sqrt {\left(1+8{F_{r_1}^2}\right)} \quad \Rightarrow \quad y_2 = -{y_1 \over 2} + {y_1 \over 2} \sqrt {\left(1+8{F_{r_1}^2}\right)}\]

\[x_2 = {y_2 \over y_1} = -{1 \over 2} - {1 \over 2} \sqrt {\left(1+8{F_{r_1}^2}\right)} \quad \Rightarrow \quad y_2 = -{y_1 \over 2} - {y_1 \over 2} \sqrt {\left(1+8{F_{r_1}^2}\right)}\]

Pull out the \({y_1/2}\) terms:

\[y_2 = {y_1 \over 2} \left(-1 + 1 \sqrt {\left(1+8{F_{r_1}^2}\right)}\right)\]

\[y_2 = {y_1 \over 2} \left(-1 - 1 \sqrt {\left(1+8{F_{r_1}^2}\right)}\right)\]

Negative answers do not yield meaningful physical results so neglect the negative solution:

\[y_2 = {y_1 \over 2} \left(-1 + \sqrt {\left(1+8{F_{r_1}^2}\right)}\right)\]

Conjugate Depth Equations: \[y_2 = {y_1 \over 2} \left(-1 + \sqrt {\left(1+8{F_{r_1}^2}\right)}\right) \quad or \quad y_1 = {y_2 \over 2} \left(-1 + \sqrt {\left(1+8{F_{r_2}^2}\right)}\right)\]

A Note on Conjugate Depths vs. Alternative Depths

It is important not to confuse conjugate depths (between which momentum is conserved) with alternate depths (between which energy is conserved). In the case of a hydraulic jump, the flow experiences a certain amount of energy headloss so that the subcritical flow downstream of the jump contains less energy than the supercritical flow upstream of the jump. Alternate depths are valid over energy conserving devices such as sluice gates and conjugate depths are valid over momentum conserving devices such as hydraulic jumps.