Lecture 12 - Non-Newtonian Pipe Flow

4 f re v d 1 04 4 1 n 075 log re pl f 2

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Average Velocity v v Recall from last lecture for shell (annulus) of fluid: vz dA A dA= 2 r dr A = R2 n1 2 n1 vz constant dA 2 R n r r n dr A R2 R n1 r 2 r n 1 R n 2 2 nn1 1 0 2 n1 L v P 2LK P 2LK 1 n 1 n n n 1 2 R2 n1 R 2 R n 1 R n 2 2 nn1 1 2 n1 n n (3n 1) R 2nR (3n 1) 3n 1 n1 n n 1 n 1 (3n 1) 2n (3n 1) 1 n1 n P 2LK n R n 3n 1 P 2LK nn1 R 1 n n1 n Laminar Pipe Flow of Power Law Fluid 4Q 3 R n w 34 n1 s K n s w K w n This puts the wall shear stress and shear rate as a direct function of the measurable pressure drop and flow rate. 4Q For a Newtonian fluid, i.e. n=1, this reduces to Newtonian 3 R Therefore, to simplify equations for Power law fluid, this shear rate is used as a reference, i.e. apparent shear rate is : 4Q 8v 3 R D n w 34 n1 . Laminar Pipe Flow of Power Law Fluid Average Velocity for Power Law Fluid: P v 2LK n 3nn1 R 1 n1 n Volumetric Flow rate , Q: P Q v.A R 2 2LK n 3nn1 R 1 n1 n P Q 2LK n 3nn1 R 1 3 n1 n Note, wall shear stress is directly proportional to the pressure drop: sw PR n K w 2L Power law model and the wall shear rate can be determined by subbing into equation for Q: n 4Q w 34 n1 3 R 1 s n w w K Laminar Pipe Flow of NNF: General Approach General expression can be used for any fluid model : (see derivation in Steffe) 1 Q 3 R 3 s w tw 0 s 2 f (s )dt Use Leibnitz’ rule to change variables to differentiate as function of sw and solve for shear rate at the wall, gives: d ds w w f (s w ) 3 1 s w 4 4 x dy d (ln y ) y dx d (ln x) d (ln ) w 3 1 4 4 d (ln s w ) d (ln ) 1 3n'1 w n' d (ln s w ) 4n' NB : 4Q 8v 3 R D Rabinowitsch-Mooney equation n’ can be obtained directly from P and Q data ! For Power Law, n’ = n...
View Full Document

This note was uploaded on 04/17/2013 for the course CHEE 2003 taught by Professor Jasonstokes during the One '12 term at Queensland.

Ask a homework question - tutors are online