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Published bySucianty Setiabudi Modified over 6 years ago
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An Alternative Derivation of the Hagen-Poiseuille Equation
Group 14 Jill Brown Sabine Knedlik Jonathan Lok
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Alternative Derivation of the Hagen-Poiseuille Equation
z Control volume r+r Po PL Looking at the geometry of this system, we choose cylindrical coordinates. z
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0 = - r [P/z] + r [rz/r]
We can factor out r and simplify to get: 0 = - r [P/z] + r [rz/r] Now we shrink the control volume element to infinitesimal size by taking the limit as r and z The pressure and shear stress are functions of only z and r. When flow is fully developed, the P is constant = dP/dz. 0 = - r dP/dz + r drz/dz This becomes a separable differential equation: r dP/dz = r drz/dr ∫ r dP/dz dr = ∫ r drz rz = (dP/dz) r/2 + C/r
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The constant C can be evaluated if the value of rz is known at some radius r; we can take r=0 at the center (axis). Since rz must be a finite value at r=0, C must be equal to zero. rz = (dP/dz) r/2 Shear stress varies linearly in pipe from r=0 to r=R. We assume a Newtonian viscosity relationship for laminar flow. rz = dvz/dr dvz/dr = (dP/dz) r/2
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Rearranging, we get: Integrate: We have a no-slip boundary condition , r = R By plugging in C value and factoring, we get:
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The velocity is at a maximum when r=o (center of pipe)
Shear stress is greatest on wall! Velocity profile = parabolic So if we rewrite, The average
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For a parabolic profile:
Rearranging we get: if This only applies for a Newtonian fluid, steady laminar flow, fully developed and incompressible!!!! Then:
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THE HAGEN- POISEUILLE EQUATION!!!!
The velocity profile across the pipe is parabolic. We have to determine the mass flow rate Q of fluid passing in each second through any cross section of the pipe. A mass of passes each second through element of the cross-sectional area. where THE HAGEN- POISEUILLE EQUATION!!!! Hence:
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