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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク Momentum-depth relationship in a rectangular channel ： ウィキペディア英語版 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 from Momentum-Force Balance== In fluid dynamics, the momentum-force balance over a control volume is given by: :$M\_1\; +\; M\_2\; =\; F\_w\; +\; F\_f\; +\; F\_\; +\; F\_$ Where: * M = momentum per unit time (ML/t2) * Fw = gravitational force due to weight of water (ML/t2) * Ff = force due to friction (ML/t2) * FP = 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. Fw = 0) and neglecting the frictional force (smooth channel bed and walls): :$M\_\; +\; M\_\; =\; F\_\; +\; F\_$ Substituting the components of momentum per unit time and pressure force (with their respective positive or negative directions): :$M\_\; =\; \backslash dotV\_\; =\; -\; \backslash rho\; QV\_1\; \backslash quad\; and\; \backslash quad\; F\_\; =\; \backslash overline\_1A\_1$ :$M\_\; =\; \backslash dotV\_\; =\; -\; \backslash rho\; QV\_2\; \backslash quad\; and\; \backslash quad\; F\_\; =\; \backslash overline\_2A\_2$ The equation becomes: :$-\backslash rho\; QV\_1\; +\; \backslash rho\; QV\_2\; =\; \backslash overline\_1A\_1\; -\; \backslash overline\_2A\_2$ Where: * $\backslash dot$ = mass flow rate (M/t) * ρ = fluid density (M/L3) * Q = flow rate or discharge in the channel (L3/t) * V = flow velocity (L/t) * $\backslash overline$ = 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: :$\backslash overline\; =\; \backslash rho\; gy$ Where: * y = flow depth (L) * g = gravitational constant (L/t2) Applying the continuity equation: :$Q\; =\; V\_1A\_1\; =\; V\_2A\_2$ 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: :$q\; =\; V\_1y\_1\; =\; V\_2y\_2$ And therefore: :$V\_1\; =\; q/y\_1\; \backslash quad\; and\; \backslash quad\; V\_2\; =\; q/y\_2$ By dividing the left and right side of the momentum-force equation by the channel’s width, and substituting the above relationships: :$-\; +\; =\; \backslash left(\; \backslash rho\; g\; \backslash right)\; -\; \backslash right)\; \}$ * subscripts 1 and 2 represent upstream and downstream locations, respectively. Dividing through by ρg: :$-\}\; +\; ^2\; \backslash over\; 2\}+\; =\; +$ 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 small. When this is the case, it is possible to assume that a hydrostatic pressure distribution applies. Munit is expressed in units of L2. If the channel width is known, the full specific force (L3) at a point can be determined by multiplying Munit by the width, b. 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Momentum-depth relationship in a rectangular channel」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.