ProblemSet5_soln_18.pdf - ECH 141 Problem Set#5 1 The...

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ECH 141 Problem Set #5 1. The generalized Newtonian fluid model provides a simple way to describe non-Newtonian fluids in steady unidirectional flow. Here you will derive a generalized form of the Hagen-Poiseuille equation for fully developed flow of a power-law fluid through a long, cylindrical tube. As the flow is steady and fully developed, you may assume that v z = v z r ( ) only, v r = v θ = 0 and consequently that η = m dv z dr n 1 and τ rz = m dv z dr n 1 dv z dr a) If the dynamic pressure at z=0 is P 0 and at z=L is P L , derive an expression for the velocity profile v Z (r) of the fluid in the tube. b) Use your result from part (a) to derive an expression for the flow rate Q. Show that your result simplifies to the Newtonian, Hagen-Poiseuille equation in the limit n=1 and m= μ . Solution ,
By the same reasoning as is used for a Newtonian fluid, one therefore finds that dP dz is constant or dP dz = P 0 P L L . Integrate the z-component stress equation once to obtain τ rz = dP dz r 2 + C 1 r . One can proceed and use the constitutive equation from this point, but it is easier to require a finite viscous stress at r=0, which means it must be that C 1 = 0 . Now use the constitutive equation, but from physical intuition note that in tube flow dv z dr = dv z dr so that m dv z dr n 1 dv z dr = dP dz r 2 or m dv z dr n = dP dz r 2 . Then dv z dr n = 1 m dP dz r 2 or dv z dr = − − 1 2m dP dz 1/n r 1/n . Now integrate to find v z r ( ) = n n + 1 1 2m dP dz 1/n r n + 1 n + C 2 .
Imposing the no-slip condition at the wall, v z = 0 at r=R then yields v z r ( ) = n n + 1 1 2m dP dz 1/n R n + 1 n n n + 1 1 2m dP dz 1/n r n + 1 n or v z r ( ) = n n + 1 1 2m dP dz 1/n R n + 1 n 1 r R n + 1 n . For a Newtonian fluid, n=1 and m= μ so we recover the result (2.8-18 in Whitaker, derived via shell balances) v z r ( ) = R 2 4 μ dP dz 1 r R 2 . b) To obtain the volumetric flow rate, we must integrate the velocity over the circular cross-section of the tube, or Q = n n + 1 1 2m dP dz 1/n R n + 1 n 1 r R n + 1 n 0 R 0 2 π rdrd θ . The angular integration is trivial, so Q = 2 π n n + 1 1 2m dP dz 1/n R n + 1 n 1 r R
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