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| 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]].
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| 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.
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| ==Derivation of the Momentum Function Equation from Momentum-Force Balance==
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| In fluid dynamics, the momentum-force balance over a control volume is given by:
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| :<math>M_1 + M_2 = F_w + F_f + F_{P1} + F_{P2}</math> | |
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| Where:
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| * M = momentum per unit time (ML/t<sup>2</sup>)
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| * F<sub>w</sub> = [[gravitational force]] due to weight of water (ML/t<sup>2</sup>)
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| * F<sub>f</sub> = force due to [[friction]] (ML/t<sup>2</sup>)
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| * F<sub>P</sub> = [[pressure]] force (ML/t<sup>2</sup>)
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| * subscripts 1 and 2 represent upstream and downstream locations, respectively
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| * Units: L = length, t = time, M = mass
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| Applying the momentum-force balance in the direction of flow, in a horizontal channel (i.e. F<sub>w</sub> = 0) and neglecting the frictional force (smooth channel bed and walls):
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| :<math>M_{1x} + M_{2x} = F_{P1x} + F_{P2x}</math>
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| Substituting the components of momentum per unit time and pressure force (with their respective positive or negative directions):
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| :<math>M_{1x} = \dot{m}V_{1x} = - \rho QV_1 \quad and \quad F_{P1x} = \overline{P}_1A_1</math>
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| :<math>M_{2x} = \dot{m}V_{2x} = - \rho QV_2 \quad and \quad F_{P2x} = \overline{P}_2A_2</math>
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| The equation becomes:
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| :<math>-\rho QV_1 + \rho QV_2 = \overline{P}_1A_1 - \overline{P}_2A_2</math>
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| Where:
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| * <math>\dot{m}</math> = mass flow rate (M/t)
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| * ρ = fluid density (M/L<sup>3</sup>)
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| * Q = flow rate or discharge in the channel (L<sup>3</sup>/t)
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| * V = flow velocity (L/t)
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| * <math>\overline{P}</math> = average pressure (M/Lt<sup>2</sup>)
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| * A = cross sectional area of flow (L<sup>2</sup>)
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| * subscripts 1 and 2 represent upstream and downstream locations, respectively
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| * Units: L (length); t (time); M (mass)
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| 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:
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| :<math> \overline{P} = {1 \over 2} \rho gy</math>
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| Where:
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| * y = flow depth (L)
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| * g = gravitational constant (L/t<sup>2</sup>)
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| [[File:Figure 1 Static Pressure Distribution.png|thumb|center|600px|alt=Figure 1: hydrostatic pressure distribution|'''Figure 1: Static Pressure Distribution''']]
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| Applying the continuity equation:
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| :<math>Q = V_1A_1 = V_2A_2</math>
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| 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:
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| :<math>q = V_1y_1 = V_2y_2</math>
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| And therefore:
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| :<math>V_1 = q/y_1 \quad and \quad V_2 = q/y_2</math>
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| By dividing the left and right side of the momentum-force equation by the channel’s width, and substituting the above relationships:
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| :<math>-{\rho q \left({q \over {y_1}}\right)} + {\rho q \left({q \over {y_2}}\right)} = \left({1 \over 2} \rho g{y_1} \right) {y_1} -{\left({1 \over 2} \rho g{y_2} \right) {y_2}} </math> | |
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| * subscripts 1 and 2 represent upstream and downstream locations, respectively.
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| Dividing through by ρg:
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| :<math>-{{q^2 \over g{y_1}}} + {q^2 \over g{y_2}} = {{y_1} ^2 \over 2}-{{y_2} ^2 \over 2}</math>
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| Separating the variables based on the sides of the jump:
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| :<math>{{y_1} ^2 \over 2}+{q^2 \over g{y_1}} = {{y_2} ^2 \over 2}+{q^2 \over g{y_2}}</math>
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| In the above relationship both sides correspond to the [[specific force]], or momentum function per channel width, also called M<sub>unit</sub>.
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| :<math>M_{unit} = {y^2 \over 2}+{q^2 \over gy} </math>
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| 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 small. When this is the case, it is possible to assume that a [[hydrostatic pressure]] distribution applies. M<sub>unit</sub> is expressed in units of L<sup>2</sup>. If the channel width is known, the full specific force (L<sup>3</sup>) at a point can be determined by multiplying M<sub>unit</sub> by the width, b.
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| == Hydraulic Jumps and Conservation of Momentum ==
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| Figure 2 depicts a [[hydraulic jump]]. A hydraulic jump is a region of rapidly varied flow and is formed in a channel when a [[supercritical flow]] transitions into a [[subcritical flow]].<ref name=" Henderson ">Henderson, F. M. (1966). Open Channel Flow, MacMillan Publishing Co., Inc, New York, NY.</ref> 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. Assuming no additional drag forces, momentum is conserved.
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| A jump causes the water surface to rise abruptly, and as a result, surface rollers are formed, intense mixing occurs, [[Air entrainment|air is entrained]], and usually a large amount of energy is dissipated. For these reasons, in engineered systems a hydraulic jump is sometimes forced in an attempt to dissipate flow energy, to mix chemicals, or to act as an [[aeration]] device.<ref name=" Chaudhry ">Chaudhry, M. H. (2008). Open-Channel Flow, Springer Science+Business Media, LLC, New York, NY.</ref><ref name="Sturm">Sturm, T. W. (2010). Open Channel Hydraulics, McGraw-Hill, New York, NY.</ref>
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| The law of [[Momentum|conservation of momentum]] states that the total momentum of a [[closed system]] of objects (which has no interactions with external agents) is constant.<ref name="Finnemore">Finnemore, E. J., and Franzini, J. B. (2002). Fluid Mechanics with Engineering Applications, McGraw-Hill, New York, NY.</ref> 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 on 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 ''[[Hydraulic jumps in rectangular channels#Conjugate depths relationships|conjugate depths]]''. The latter is valid unless the jump is caused by an external force or outside influence.
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| [[File:Conservation of Momentum – Hydraulic Jump .png|thumb|center|600px|alt=Figure 1: Conservation of Momentum – Hydraulic Jump|'''Figure 2: Conservation of Momentum – Hydraulic Jump''']]
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| The green box in Figure 2 represents the [[control volume]] enclosing the jump system and shows the major [[pressure]] forces on the system (F<sub>P1</sub> and F<sub>P2</sub>). 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<sub>f</sub>) and the weight of water from a sloping channel (F<sub>w</sub>) are neglected. It is worth to noting that the slope of the triangular hydrostatic pressure distributions on each location corresponds to the specific weight of water (γ), which has units of (m/L<sup>2</sup>t<sup>2</sup>)
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| == The M-y Diagram ==
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| An M-y diagram is a plot of the depth of flow (y) versus momentum (M). It should be noted that in this case M does not refer to momentum (M/Lt<sup>2</sup>), but to the momentum function (L<sup>3</sup> or L<sup>2</sup>). This produces a specific momentum curve that is generated by calculating momentum for a range of depth values and graphing 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 length<sup>3</sup> (when using the general momentum function equation) or units of length<sup>2</sup> (when using the rectangular form M<sub>unit</sub> equation). In a rectangular channel of unit width, an M-y curve is plotted using:
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| :<math>M_{unit} = {q^2 \over gy} + {y^2 \over 2}</math>
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| Figure 3 depicts a sample M-y diagram showing the plots of four specific momentum curves. Each of these curves correspond to a specific ''q'' as noted in the figure. As unit discharge increases, the curve shifts to the right.
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| [[File:Figure 3 Different M y Curves.png|thumb|center|600px|alt=Figure 3: different m-y curves for different flows|'''Figure 3: M-y Curves for Various Unit Flowrates''']]
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| M-y diagrams can provide 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 flow|supercritical]] or [[subcritical flow]] for a given [[Flowrate|discharge]], as well as defining the critical depth and critical momentum of a flow. In addition, M-y diagrams can aid in finding [[Hydraulic jumps in rectangular channels#Conjugate depths relationships|conjugate depths]] of flow that have the same specific force or momentum function, as in the case of flow depths on either side of a [[hydraulic jump]]. A[[Dimensionless momentum-depth relationship in open-channel flow|dimensionless form of the M-y diagram]] representing any unit discharge can be created and utilized in place of the particular M-y curves discussed here and referred to in Figure 3.
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| == Critical Flow ==
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| A flow is termed ''critical'' if the bulk velocity of the flow <math>V</math> is equal to the propagation velocity of a shallow gravity wave <math>\sqrt {gy}</math>.<ref name=" Henderson ">Henderson, F. M. (1966). Open Channel Flow, MacMillan Publishing Co., Inc., New York, NY.</ref><ref name="Chow">Chow, V. T. (1959). Open-Channel Hydraulics, McGraw-Hill, New York, NY.</ref> At critical flow, the specific energy and the specific momentum (force) are at a minimum for a given discharge.<ref name=" Henderson "/> Figure 3 shows this relationship by showing a [[Energy-depth relationship in a rectangular channel|specific energy curve]] (E-y diagram) side-by-side to its corresponding specific momentum curve (M-y diagram) for a unit discharge q = 10 ft<sup>2</sup>/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.
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| [[File:Figure 4 E y vs M y.png|thumb|center|600px|alt=Figure 4: comparison of energy and momentum function curves|'''Figure 4: E-y & M-y Curves for a Rectangular Channel with q = 10 ft<sup>2</sup>/s''']]
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| In a rectangular channel, critical depth (y<sub>c</sub>) can also be found mathematically using the following equation:
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| :<math>y_c = \left({q^2 \over g} \right)^{1 \over 3}</math>
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| Where:
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| * g = [[Standard gravity|gravitational constant]] (L/t<sup>2</sup>)
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| * q = unit flowrate or discharge – for a rectangular channel, [[Flowrate|discharge]] per unit channel width (L<sup>2</sup>/t)
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| == Supercritical Flow versus Subcritical Flow on an M-y Diagram ==
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| As mentioned before, an M-y diagram can provide an indication of flow classification for a given depth and discharge. When flow is not critical it is classified as either ''subcritical'' or ''supercritical''. This distinction is based on the [[Froude number]] of the flow, which is the ratio of the bulk velocity (V) to the propagation velocity of a shallow wave :<math>\left({\sqrt {gy}}\right)</math>.<ref name="Chow"/> The generic equation of the Froude number is expressed in terms of gravity (g), the flow’s velocity (V) and the hydraulic depth (A/B), where (A) represents the cross sectional area and (B) the top width. For rectangular channels, this ratio is equal to the depth of flow (y).
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| :<math>F_r={{V}\over {\sqrt{g \left({A \over B} \right)}}}={{V}\over {\sqrt {gy}}} </math>
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| A Froude number greater than one is [[Supercritical flow|supercritical]], and a [[Froude number]] less than one is [[subcritical flow|subcritical]]. In general, supercritical flows are shallow and fast and subcritical flows are deep and slow. These different flow classifications are also represented on M-y diagrams where different regions of the graph represent different flow types. Figure 5 shows these regions, with a specific momentum curve corresponding to a q = 10 ft<sup>2</sup>/s. As stated previously, critical flow is represented by the minimum momentum that exists on the curve (green line). [[Supercritical flow]]s correspond to any point on the momentum curve that has a depth less that the critical depth with [[subcritical flow]]s having a depth greater than critical depth.<ref name="French">French, R. H. (1985). Open-Channel Hydraulics, McGraw-Hill New York, NY.</ref>
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| [[File:Figure 5 M y Curve. Subcritical and Supercritical Regions.png|thumb|center|600px|alt=Figure 5: regions of flow in a M-y curve|'''Figure 5: M-y Curve. Subcritical and Supercritical Regions for q = 10 ft<sup>2</sup>/s''']]
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| == Conjugate Depths in a Rectangular Channel ==
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| 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).
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| 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 ft<sup>2</sup> for a unit discharge of 10 ft<sup>2</sup>/s. This momentum line crosses the M-y curve at depths of 0.312 (y<sub>1</sub>) and 4.31 feet (y<sub>2</sub>). Depth y<sub>1</sub> corresponds to the supercritical depth upstream of the jump and depth y<sub>2</sub> corresponds to the subcritical depth downstream of the jump.
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| Conjugate depths can also be calculated using the [[Froude number]] and depth of either the [[Supercritical flow|supercritical]] or [[Subcritical flow|subcritical]] flow. The following equations can be used to determine the [[Hydraulic jumps in rectangular channels#Conjugate depths relationships|conjugate depth]] to a known depth in a rectangular channel:
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| :<math>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)</math>
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| [[File:Figure 6 M y Diagram. Showing Conjugate Depths.png||thumb|center|600px|alt=Figure 6: Conjugate depths for a specific momentum force value|'''Figure 6: M-y Diagram for q = 10 ft<sup>2</sup>/s Showing Conjugate Depths Corresponding to a Momentum of 10 ft<sup>2</sup>''']]
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| == Derivation of Conjugate Depth Equation for a Rectangular Channel ==
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| Start with the [[conservation of momentum]] function <math>M_1 = M_2</math>, for rectangular channels:
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| :<math>\left({q^2 \over gy_1} + {y_1^2 \over 2} \right) = \left({q^2 \over gy_2} + {y_2^2 \over 2}\right)</math>
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| Where:
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| * q = discharge per unit channel width (L<sup>2</sup>/t)
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| * g = [[Standard gravity|gravitational constant]] (L/t<sup>2</sup>)
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| * y = flow depth (L)
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| * subscripts 1 and 2 represent upstream and downstream locations, respectively.
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| Isolate the q<sup>2</sup> terms on one side of the equal sign with the <math>y^2</math> terms on the other side:
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| :<math>{q^2 \over gy_1} - {q^2 \over gy_2} = -{y_1^2 \over 2} + {y_2^2 \over 2}</math>
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| Factor the constant terms q<sup>2</sup>/g and 1/2:
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| :<math>{q^2 \over g} \left({1 \over y_1} - {1 \over y_2}\right) = {1 \over 2} ({y_2^2} - {y_1^2})</math>
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| Combine the depth terms on the left side and expand the quadratic on the right side:
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| :<math>{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})</math>
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| Divide by <math>(y_2 - y_1)</math>:
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| :<math>{q^2 \over g} \left({1 \over {y_1y_2}} \right) = {1 \over 2}({y_2 + y_1})</math>
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| Recall from Continuity in a rectangular channel that:
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| :<math>q = V_1y_1 = V_2y_2</math>
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| Substitute <math>V_1y_1</math> into the left side of the equation for q:
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| :<math>{{V_1^2y_1^2} \over {gy_1y_2}} = {{V_1^2y_1} \over {gy_2}} = {{y_2 + y_1} \over 2}</math>
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| Divide by <math>{y_1/y_2}</math>:
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| :<math>{{V_1^2} \over {g}} = {{y_2({y_2 + y_1})} \over {2y_1}}</math>
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| Divide by <math>y_1</math> and recognize that the left hand side is now equal to F<sub>r1</sub><sup>2</sup>:
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| :<math>{{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}</math>
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| Rearrange and set the equation equal to zero:
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| :<math>0 = {y_2^2 \over {2y_1^2}} + {{y_2} \over 2y_1} - F_{r_1}^2</math>
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| To facilitate the next step, let <math>x = {y_2/y_1}</math>, and the above equation becomes:
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| :<math>0 = {1 \over {2}}x^2 + {1 \over {2}}x - F_{r_1}^2</math>
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| Solve for <math>x</math> using the [[quadratic equation]] with <math>a = {1/2}</math>, <math>b = {1/2}</math>, and <math>c = </math>-F<sub>r1</sub><sup>2</sup>:
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| :<math>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}</math>
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| Pull out the 1/4 inside of the square root:
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| :<math>{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)}</math>
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| Focus on the root with the positive second term:
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| :<math> {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)}</math>
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| Factor the (y<sub>1</sub>/2) terms:
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| :<math>y_2 = {y_1 \over 2} \left(-1 + \sqrt {\left(1+8{F_{r_1}^2}\right)}\right)</math>
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| 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 (y<sub>1</sub>, F<sub>r1</sub>) or downstream (y<sub>2</sub>, F<sub>r2</sub>).
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| == A Note on Conjugate Depths vs. Alternative Depths ==
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| It is important not to confuse [[Hydraulic jumps in rectangular channels#Conjugate depths relationships|conjugate depths]] (between which momentum is conserved) with [[Energy-depth relationship in a rectangular channel#Alternate depths|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. [[Energy-depth relationship in a rectangular channel#Alternate depths|Alternate depths]] are valid over energy conserving devices such as [[sluice gate]]s and [[Hydraulic jumps in rectangular channels#Conjugate depths relationships|conjugate depths]] are valid over momentum conserving devices such as [[hydraulic jump]]s.
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| ==Application of the momentum function equation to evaluate the thrust force on a sluice gate==
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| 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:
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| :<math>F_{thrust-gate} = \gamma b(\Delta M) = \gamma b(M_{unit,1} - M_{unit,2}) </math>
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| Where:
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| * F<sub>thrust-gate</sub> = force exerted by water on the sluice gate (ML/t<sup>2</sup>)
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| * γ = specific weight of water (M/L<sup>2</sup>t<sup>2</sup>)
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| *'''Δ'''M<sub>unit</sub> = difference in momentum per unit width between the upstream and downstream sides of the sluice gate (L<sup>2</sup>).
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| [[File:Figure 7 force on a sluice gate.png|thumb|center|600px|alt= dynamic force acting on a sluice gate|''' Figure 7: Force on a Sluice Gate (simplified version)]]
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| === Example===
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| 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 [[Energy-depth relationship in a rectangular channel#Alternate depths|alternate depths]] of 0.312 ft. The water temperature was measured to be 70 °F. What is the thrust force on the gate?
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| Applying the momentum per unit width equation to the upstream and downstream locations respectively:
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| :<math>M_{unit} = {{y^2}\over 2}+{{q^2}\over gy} </math>
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| :<math>M_{unit,1}= {{{y_1}^2}\over 2}+{{q^2}\over gy_1} = {{{16.3}^2}\over 2} +{{{\left({100\over10}\right)}^2}\over{(32.2)(16.3)}}</math>
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| :<math>M_{unit,1}= 132 \quad ft^2</math>
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| and
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| :<math>M_{unit,2}= 10 \quad ft^2</math>
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| The specific weight of water at 70°F is 62.30 <math>{lb/ft^3}</math>. The resulting net thrust force on the sluice gate is:
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| :<math>F_{thrust-gate} = \gamma b(M_{unit,1} - M_{unit,2}) </math>
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| :<math>F_{thrust-gate} = (62.30)(10)(132-10) </math>
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| :<math>F_{thrust-gate} = 76,142 \quad lbs </math>
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| ==References==
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| {{reflist}}
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| <!--- After listing your sources please cite them using inline citations and place them after the information they cite. Please see http://en.wikipedia.org/wiki/Wikipedia:REFB for instructions on how to add citations. --->
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| [[Category:Fluid dynamics]]
| |
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