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| In [[fluid dynamics]], the derivation of the '''[[Hagen–Poiseuille flow]] from the Navier–Stokes equations''' shows how this [[fluid flow|flow]] is an exact solution to the [[Navier–Stokes equations]].<ref>{{cite book|last=White|first=Frank M.|title=Fluid Mechanics|edition=5|year=2003|publisher=[[McGraw-Hill]]|chapter=6}}</ref><ref name=Bird>{{cite book | author=Bird, Stewart, Lightfoot | title=Transport Phenomena | year=1960}}</ref>
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| ==Derivation==
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| The [[laminar flow]] through a pipe of uniform (circular) cross-section is known as Hagen–Poiseuille flow. The equations governing the Hagen–Poiseuille flow can be derived directly from the [[Navier–Stokes equations]] in cylindrical coordinates by making the following set of assumptions:
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| # The flow is steady ( <math> \partial(...)/\partial t = 0 </math> ).
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| # The radial and swirl components of the fluid velocity are zero ( <math> u_r = u_\theta = 0 </math> ).
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| # The flow is axisymmetric ( <math> \partial(...)/\partial \theta = 0 </math> ) and fully developed (<math> \partial u_z/\partial z = 0 </math> ).
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| Then the second of the [[Navier%E2%80%93Stokes_equations#Cylindrical_coordinates|three Navier–Stokes momentum equations]] and the [[continuity equation]] are identically satisfied. The first momentum equation reduces to <math> \partial p/\partial r = 0 </math>, i.e., the [[pressure]] <math> p </math> is a function of the axial coordinate <math> z </math> only. The third momentum equation reduces to:
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| :<math> \frac{1}{r}\frac{\partial}{\partial r}\left(r \frac{\partial u_z}{\partial r}\right)= \frac{1}{\mu} \frac{\partial p}{\partial z}</math> where <math>\mu</math> is the dynamic viscosity of the fluid.
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| :The solution is
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| :<math> u_z = \frac{1}{4\mu} \frac{\partial p}{\partial z}r^2 + c_1 \ln r + c_2 </math>
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| Since <math> u_z </math> needs to be finite at <math> r = 0 </math>, <math> c_1 = 0 </math>. The no slip [[boundary condition]] at the pipe wall requires that <math> u_z = 0</math> at <math> r = R </math> (radius of the pipe), which yields
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| :<math> c_2 = -\frac{1}{4\mu} \frac{\partial p}{\partial z}R^2.</math>
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| Thus we have finally the following [[Parabola|parabolic]] [[velocity]] profile:
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| :<math> u_z = -\frac{1}{4\mu} \frac{\partial p}{\partial z} (R^2 - r^2). </math>
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| The maximum velocity occurs at the pipe centerline (<math> r=0 </math>):
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| :<math> {u_z}_{max}=\frac{R^2}{4\mu} \left(-\frac{\partial p}{\partial z}\right). </math>
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| The average velocity can be obtained by integrating over the pipe [[Cross section (geometry)|cross section]]:
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| : <math> {u_z}_\mathrm{avg}=\frac{1}{\pi R^2} \int_0^R u_z \cdot 2\pi r dr = 0.5 {u_z}_\mathrm{max}. </math>
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| The Hagen–Poiseuille equation relates the pressure drop <math> \Delta p</math> across a circular pipe of length <math> L </math> to the
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| average flow velocity in the pipe <math> {u_z}_\mathrm{avg} </math> and other parameters. Assuming that the pressure decreases linearly across the length
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| of the pipe, we have <math> - \frac{\partial p}{\partial z} = \frac{\Delta p}{L} </math> (constant). Substituting this and the expression for <math> {u_z}_\mathrm{max} </math> into the expression for <math> {u_z}_\mathrm{avg} </math>, and noting that the pipe diameter <math> D = 2R </math>, we get:
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| : <math> {u_z}_{avg} = \frac{D^2}{32 \mu} \frac{\Delta P}{L}. </math>
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| Rearrangement of this gives the Hagen–Poiseuille equation:
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| : <math> \Delta P = \frac{32 \mu L ~{u_z}_\mathrm{avg}}{D^2}. </math>
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| ==References==
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| {{reflist}}
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| ==See also==
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| *[[Poiseuille's Law]]
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| *[[Couette flow]]
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| *[[Pipe flow]]
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| {{DEFAULTSORT:Hagen-Poiseuille flow from the Navier-Stokes equations}}
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| [[Category:Fluid dynamics]]
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| [[Category:Fluid mechanics]]
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