Chapter 3. Saturated Water Flow - LAWR

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SSC107 – Fall 2000 Chapter 3 Page 3-2 Pressure difference across tube = P 1 - P 2 (N/m 2 ). Do force balance on cylindrical water volume with r radius r < R. Pressure force across two ends = (P 1 - P 2 ).A = ∆P πr 2 Shear force on external area of water volume = τ(2πrL) Equating these 2 forces leads to: τ = ∆Pr/2L Earlier we found that τ = -υ dv/dr Equate and put terms with r together on left-hand side and integrate from r to R (v=0): rdr = - 2L υ ∆P dv R ∫ r rdr = -2L υ ∆P 0 ∫ v dv ∆ yields v(r)= P 4L ( R 2 - r 2 υ ) When plotting velocity as a function of radius r, its distribution is parabolic: R Radial position, r 0 Water velocity v(r ) v max ✪ For which value of r is the water velocity at its maximum ?
SSC107 – Fall 2000 Chapter 3 Page 3-3 At r = 0, where v = v max = R 2 ∆P/4Lν To find volumetric flow rate Q (volume per unit time), we must integrate v(r) over the area of the cylinder. Use cylindrical coordinates: Q = ∫ ∫ v(r)dA Yielding Q = 4 π R ∆P 8Lυ Using flow rate per unit area (water flux) J = R 2 2 ∆P R ρ g H K H w ∆ ∆ w [ = = ] 8Lυ 8Lν L NOTE: This equation is known as Poiseuille's law (Water flux is a function of pore radius) ✪ Would you expect differences in flow rates between sands and clays, and if so why ? In 1856, Darcy found for flow through saturated sand that Q = V ∆ H and is proportional to t ∆ X Q - discharge rate ( volume water per time ) V - volume of H O t - time X - distance or position ∆H ∆X - Total head gradient A gradient is the change of some parameter with position. 2
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Chapter 3. Saturated Water Flow - LAWR

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