Momentum–depth relationship in a rectangular channel

In classical physics, momentum is the product of mass and velocity and is a vector quantity, but in fluid mechanics it is treated as a longitudinal quantity (i.e. one dimension) evaluated in the direction of flow. Additionally, it is evaluated as momentum per unit time, corresponding to the product of mass flow rate and velocity, and therefore it has units of force. The momentum forces considered in open channel flow are dynamic force – dependent of depth and flow rate – and static force – dependent of depth – both affected by gravity.

The principle of conservation of momentum in open channel flow is applied in terms of specific force, or the momentum function; which has units of length cubed for any cross sectional shape, or can be treated as length squared in the case of rectangular channels. Although not being technically correct, the term momentum will be used to replace the concept of the momentum function. The conjugate depth equation, which describes the depths on either side of a hydraulic jump, can be derived from the conservation of momentum in rectangular channels, based upon the relationship between momentum and depth of flow. The concept of momentum can also be applied to evaluate the thrust force on a sluice gate, a device that conserves specific energy but loses momentum.

Derivation of the Momentum Function Equation

In fluid dynamics, the momentum-force balance over a control volume is given by:

$M_{1}+M_{2}=F_{w}+F_{f}+F_{P1}+F_{P2}$

Where:

• M = momentum on the upstream side per unit time (ML/t²)
• F_w = gravitational force due to weight of water (ML/t²)
• F_f = force due to friction (ML/t²)
• F_P = pressure force (ML/t²)
• subscripts 1 and 2 represent upstream and downstream locations, respectively
• Units: L = length, t = time, M = mass

Applying the momentum-force balance in the direction of flow, in a horizontal channel (i.e. F_w = 0) and neglecting the frictional force (smooth channel bed and walls):

$M_{1x}+M_{2x}=F_{P1x}+F_{P2x}$

Substituting the components of momentum per unit time and pressure force (with their respective positive or negative directions):

$M_{1x}={\dot {m}}V_{1x}=-\rho QV_{1}\quad and\quad F_{P1x}={\overline {P}}_{1}A_{1}$

$M_{2x}={\dot {m}}V_{2x}=-\rho QV_{2}\quad and\quad F_{P2x}={\overline {P}}_{2}A_{2}$

The equation becomes:

$-\rho QV_{1}+\rho QV_{2}={\overline {P}}_{1}A_{1}-{\overline {P}}_{2}A_{2}$

Where:

• ${\dot {m}}$ = mass flow rate (M/t)
• ρ = fluid density (M/L^3)
• Q = flow rate or discharge in the channel (L³/t)
• V = flow velocity (L/t)
• ${\overline {P}}$ = average pressure (M/Lt²)
• A = cross sectional area of flow (L²)
• subscripts 1 and 2 represent upstream and downstream locations, respectively
• Units: L (length); t (time); M (mass)

The hydrostatic pressure distribution has a triangular shape from the water surface to the bottom of the channel (Figure 1). The average pressure can be obtained from the integral of the pressure distribution:

${\overline {P}}={1 \over 2}\rho gy$

Where:

• y = flow depth (L)
• g = gravitational constant (L/t²)

hydrostatic pressure distribution

Applying the continuity equation:

$Q=V_{1}A_{1}=V_{2}A_{2}$

For the case of rectangular channels (i.e. constant width “b”) the flow rate, Q, can be replaced by the unit discharge q, where q = Q/b, which yields:

$q=V_{1}y_{1}=V_{2}y_{2}$

And therefore:

$V_{1}=q/y_{1}\quad and\quad V_{2}=q/y_{2}$

By dividing the left and right side of the momentum force equation by the channel's width, and substituting the above relationships:

$-\rho q{q \over y_{1}}+\rho q{q \over y_{2}}=({1 \over 2}\rho gy_{1})y_{1}-({1 \over 2}\rho gy_{2})y_{2}$

subscripts 1 and 2 represent upstream and downstream locations, respectively

Dividing through by ρg:

${-q^{2} \over {gy}_{1}}+{q^{2} \over {gy}_{2}}={{y_{1}}^{2} \over 2}-{{y_{2}}^{2} \over 2}$

Separating variables based on the sides of the jump:

${{y_{1}}^{2} \over 2}+{q^{2} \over {gy}_{1}}={{y_{2}}^{2} \over 2}+{q^{2} \over {gy}_{2}}$

In the above relationship both sides correspond to the specific force, or momentum function per channel width, also called M_unit.

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

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. When this is the case, it is possible to assume that a hydrostatic pressure distribution applies. M_unit is expressed in units of L² . If the channel width is known, the full specific force (length3) at a point can be determined by multiplying M_unit by the base width, b.

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.

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 flow 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:

$M_{1x}=\rho Q(V_{1}\cdot \mathbf {n} )=-\rho QV_{1}\quad and\quad F_{P1x}={\overline {P_{1}}}A_{1}$

$M_{2x}=\rho Q(V_{2}\cdot \mathbf {n} )=\rho QV_{2}\quad and\quad F_{P2x}={\overline {P_{2}}}A_{2}$

The equation becomes:

$-\rho QV_{1}+\rho QV_{2}={\overline {P_{1}}}A_{1}-{\overline {P_{2}}}A_{2}$

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 $h={y/2}$:

$P=P_{avg}={\overline {P}}={1 \over 2}\rho gy\quad and\quad Q=VA\quad \Rightarrow \quad V={Q \over A}\quad \Rightarrow \quad V_{1}={Q \over A_{1}}\quad and\quad V_{2}={Q \over A_{2}}$

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

$-{\rho Q^{2} \over A_{1}}+{\rho Q^{2} \over A_{2}}=\left({1 \over 2}\rho gy_{1}\right)A_{1}-\left({1 \over 2}\rho gy_{2}\right)A_{2}$

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).

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. By dividing both sides of the M-y equation with critical depth, y_c, we can also obtain the dimensionless form of M-y diagram, which is valid for different values of q with only one curve in the diagram.

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.

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.

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.

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.

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.

check after this

Conjugate Depths in a Rectangular Channel

Conjugate, or sequent, depths are the paired depths that result upstream and downstream of a hydraulic jump, with the upstream flow being supercritical and downstream flow being subcritical. Conjugate depths can be found either graphically using a specific momentum curve or algebraically with a set of equations. Because momentum is conserved over a hydraulic jump conjugate depths have equivalent momentum, and given a discharge, the conjugate to any flow depth can be determined with an M-y diagram (Figure 6).

A vertical line that crosses the M-y curve twice (i.e. non-critical flow conditions) represents the depths on opposite sides of a hydraulic jump. Given sufficient momentum (momentum greater than critical flow), a conjugate depth pair exists at each point where the vertical line intersects the M-y curve. Figure 6 exemplifies this behavior with a momentum of 10 ft² for a unit discharge of 10 ft²/s. This momentum line crosses the M-y curve at depths of 0.312 (y₁) and 4.31 feet (y₂). Depth y₁ corresponds to the supercritical depth upstream of the jump and depth y₂ corresponds to the subcritical depth downstream of the jump.

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

$y_{2}={y_{1} \over 2}\left(-1+{\sqrt {(1+8{F_{r_{1}}}^{2})}}\right)\quad or\quad y_{1}={y_{2} \over 2}\left(-1+{\sqrt {(1+8{F_{r_{2}}}^{2})}}\right)$

Figure 6: M-y Diagram for q = 10 ft²/s Showing Conjugate Depths Corresponding to a Momentum of 10 ft²

Derivation of Conjugate Depth Equation for a Rectangular Channel

Start with the conservation of momentum function $M_{1}=M_{2}$, for rectangular channels:

$\left({q^{2} \over gy_{1}}+{y_{1}^{2} \over 2}\right)=\left({q^{2} \over gy_{2}}+{y_{2}^{2} \over 2}\right)$

Where:

• q = discharge per unit channel width (L²/t)
• g = gravitational constant (L/t²)
• y = flow depth (L)
• subscripts 1 and 2 represent upstream and downstream locations, respectively.

Isolate the q² terms on one side of the equal sign with the $y^{2}$ terms on the other side:

${q^{2} \over gy_{1}}-{q^{2} \over gy_{2}}=-{y_{1}^{2} \over 2}+{y_{2}^{2} \over 2}$

Factor 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 depth terms on the left side and expand the quadratic on the right side:

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

Divide by $(y_{2}-y_{1})$:

${q^{2} \over g}\left({1 \over {y_{1}y_{2}}}\right)={1 \over 2}({y_{2}+y_{1}})$

Recall from Continuity in a rectangular channel that:

$q=V_{1}y_{1}=V_{2}y_{2}$

Substitute $V_{1}y_{1}$ into the left side of the equation for q:

${{V_{1}^{2}y_{1}^{2}} \over {gy_{1}y_{2}}}={{V_{1}^{2}y_{1}} \over {gy_{2}}}={{y_{2}+y_{1}} \over 2}$

Divide by ${y_{1}/y_{2}}$:

${{V_{1}^{2}} \over {g}}={{y_{2}({y_{2}+y_{1}})} \over {2y_{1}}}$

Divide by $y_{1}$ and recognize that the left hand side is now equal to F_r1²:

${{V_{1}^{2}} \over {gy_{1}}}={{y_{2}({y_{2}+y_{1}})} \over {2y_{1}^{2}}}\Rightarrow {V_{1}^{2} \over gy_{1}}=F_{r_{1}}^{2}={y_{2}^{2} \over {2y_{1}^{2}}}+{{y_{2}y_{1}} \over 2y_{1}^{2}}={y_{2}^{2} \over {2y_{1}^{2}}}+{{y_{2}} \over 2y_{1}}$

Rearrange and set the equation equal to zero:

$0={y_{2}^{2} \over {2y_{1}^{2}}}+{{y_{2}} \over 2y_{1}}-F_{r_{1}}^{2}$

To facilitate the next step, let $x={y_{2}/y_{1}}$, and 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:

${y_{2} \over y_{1}}=-{1 \over 2}\pm {\sqrt {{1 \over 4}\left(1+8{F_{r_{1}}^{2}}\right)}}=-{1 \over 2}\pm {1 \over 2}{\sqrt {\left(1+8{F_{r_{1}}^{2}}\right)}}$

Focus on the root with the positive second term:

${y_{2} \over y_{1}}=-{1 \over 2}+{1 \over 2}{\sqrt {\left(1+8{F_{r_{1}}^{2}}\right)}}\quad \Rightarrow \quad y_{2}=-{y_{1} \over 2}+{y_{1} \over 2}{\sqrt {\left(1+8{F_{r_{1}}^{2}}\right)}}$

Factor the (y₁/2) terms:

$y_{2}={y_{1} \over 2}\left(-1+{\sqrt {\left(1+8{F_{r_{1}}^{2}}\right)}}\right)$

The above is the conjugate depth equation in a rectangular channel and can be used to find the subcritical or supercritical depth from known conditions, either upstream (y₁, F_r1) or downstream (y₂, F_r2).

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.

Application of the momentum function equation to evaluate the thrust force on a sluice gate

The momentum equation can be applied to determine the force exerted by water on a sluice gate (Figure 7). Contrary to the conservation of fluid energy when a flow encounters a sluice gate, the momentum upstream and downstream of the gate is not conserved. The thrust force exerted by water on a gate placed in a rectangular channel can be obtained from the following equation, which can be derived in the same way as the conservation of momentum equation for rectangular channels:

$F_{thrust-gate}=\gamma b(\Delta M)=\gamma b(M_{unit,1}-M_{unit,2})$

Where:

• F_thrust-gate = force exerted by water on the sluice gate (ML/t²)
• γ = specific weight of water (M/L²t²)
• ΔM_unit = difference in momentum per unit width between the upstream and downstream sides of the sluice gate (L²).

Figure 7: Force on a Sluice Gate (simplified version)

Example

Water is flowing through a smooth, frictionless, rectangular channel at a rate of 100.0 cfs. The width of the channel is 10.0 ft. The flow depth upstream of the sluice gate was measured to be 16.3 ft with a corresponding alternate depths of 0.312 ft. The water temperature was measured to be 70 °F. What is the thrust force on the gate?

Applying the momentum per unit width equation to the upstream and downstream locations respectively:

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

$M_{unit,1}={{{y_{1}}^{2}} \over 2}+{{q^{2}} \over gy_{1}}={{{16.3}^{2}} \over 2}+{{{\left({100 \over 10}\right)}^{2}} \over {(32.2)(16.3)}}$

$M_{unit,1}=132\quad ft^{2}$

and

$M_{unit,2}=10\quad ft^{2}$

The specific weight of water at 70°F is 62.30 ${lb/ft^{3}}$. The resulting net thrust force on the sluice gate is:

$F_{thrust-gate}=\gamma b(M_{unit,1}-M_{unit,2})$

$F_{thrust-gate}=(62.30)(10)(132-10)$

$F_{thrust-gate}=76,142\quad lbs$