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pro00020 - Additional Examples Related to Chapter II M.S...

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Unformatted text preview: Additional Examples Related to Chapter II M.S. Ghidaoui, Spring 2002 CIVL252 Note: you do not learn by simply reading a problem and its solution! You must try to solve the problems by yourself, and discuss the difficulties encountered with your classmates, the teaching assistants and myself. Problem 1: Determine the velocity distribution of a uniform flow in an open channel. The uniform depth in the channel is H and the channel slope is S . You may assume that the channel has an infinite width. Determine the flowrate per unit width q . Solution: The only non-zero velocity is u(x,y,z,t), implying that the flow is one-dimensional. That is, v=w= 0. In addition, the flow is steady, which means that all partial derivatives with time are zero (i.e., / = ∂ ∂ t ) and the velocity does not depend on time. Since the channel is of infinite width, the velocity does not depend on the coordinate perpendicular to the plane of the page. Therefore, u = u(x,y ) and the governing equations become as follows: Continuity: = ∂ ∂ x u x-Momentum: ρ ρ μ ρ x f z u y u x u x p x u u + ∂ ∂ + ∂ ∂ + ∂ ∂ + ∂ ∂- = ∂ ∂ ) ( 1 2 2 2 2 2 2 y-Momentum: ρ ρ y f y p + ∂ ∂- = 1 Since x on depend not does u x u ⇒ = ∂ ∂ ) ( y u u = ⇒ and 2 2 = ∂ ∂ x u . Therefore, the x- momentum equation becomes: ρ ρ μ ρ x f z u y u x p + ∂ ∂ + ∂ ∂ + ∂ ∂- = ) ( 1 2 2 2 2 Since the pipe has a slope of θ with respect to the horizontal, then the forces per unit volume due to gravity are: f x = ρ g S and f y =- ρ g cos θ . Therefore, 2 2 1 gS dy u d x p momentum x + + ∂ ∂- =- ρ μ ρ θ ρ cos 1 g y p momentum y- ∂ ∂- =- Integrating the y-momentum equation with respect to y from the water surface y=h to a depth y, where y=0 is at the channel bottom, and noticing that the slope is constant gives: ) ( cos ) ( ) ( y h g y p h p-- =- θ ρ or ) ( cos y h g p- = θ ρ . Plugging this result into the x-momentum equation gives: 2 2 ) ( cos gS dy u d x y h g momentum x + + ∂- ∂- =- ρ μ θ Since the water depth is uniform, the partial derivative of h with x is zero and since y and x are independent of one another the partial derivative of y with x is zero. Thus, 2 2 gS dy u d momentum x + =- ρ μ Integrating the above equation with respect to y from bottom of the channel y=0 to a depth y gives: D By y gS y u + +- = 2 2 ) ( μ ρ The condition that u=0 at y=0 gives D =0. In addition, condition the shear at the water surface is zero gives B = ρ S h/ μ . Therefore, ( 29 y h y gS y u- = 2 2 ) ( μ ρ The flowrate per unit width is: ( 29 3 3 3 3 6 2 2 2 ) ( h gS h gS h gS dy y h y gS dy y u q h h μ ρ μ ρ μ ρ μ ρ =- =- = = ∫ ∫ Note: the flow is driven by the channel slope!...
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This note was uploaded on 03/19/2012 for the course CIVL 000 taught by Professor Kk during the Spring '10 term at HKUST.

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pro00020 - Additional Examples Related to Chapter II M.S...

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