Chemical Engineering 374

Transcription

1 Chemical Egieerig 374 Fluid Mechaics NoNewtoia Fluids Outlie 2 Types ad properties of o-newtoia Fluids Pipe flows for o-newtoia fluids Velocity profile / flow rate Pressure op Frictio factor Pump power Rheological Parameters Power Law Fluids

2 No-Newtoia Fluids 3 Newtoia Bigham Plastic Pseudoplastic Dilitat =* =* + y =κ* Bigham Plastic 4 3D elastic structures. Weak solid structures that must be broke Resists small shear, but structure breaks apart with large shear. The is ~ liear with du/ Some slurries (coal, grai slurries), sewage sludge. Toothpaste (o ip) Larger particles à weak solid structure à breaks Bigham Plastic =* + y 2

3 Pseudoplastic 5 Most commo Dissolved or dispersed particles, like dissolved log chai molecules. Have a radom orietatio i the fluid at rest, but lie up whe the fluid is sheared. decreases with strai rate ops as molecules alig Polymer melts, paper pulp suspesios, pigmet suspesios, hair gel, blood, muds, most slurries Shear-Thiig motor oil Pseudoplastic =K*(-) Dilitat 6 Rare Slurries of solid particles with barely eough liquid to keep apart. (cor starch, water squeezed out at high shear) At low strai rates, the fluid ca lubricate solids; at high strai rates, this lubricatio breaks dow. icreases with strai à icreases. Shear thickeig. Dilitat =K*(-) 3

4 No-Newtoia 7 Time depedece 8 Thixotropic Slurries/solutios of polymers May kow fluids Most are pseudoplastic Aligable particles/molecules with weak bods (H-bodig) Pait Rheopectic Rare Fewer kow examples Usually fluids oly show this behavior uder mild shearig Chages occur withi the first 60 sec. for most processes. Hard to describe Viscoelastic time time 4

5 Power Law Fluids 9 Goverig equatios are correct i terms of Expressio for is the model. Called a costitutive relatio Also have these for mass ad heat fluxes i heat ad mass trasfer. Newtoia flow = dv dy For dilitat ad pseudoplastic fluids (most commo) Power Law = K dv dy > à Dilitat < à Psuedoplastic =, K= à Newtoia K, are empirical costats May other forms Simpler oes have 3 parameters ad give a better fit, but are more complex tha power law form. See Hadout of Book Chapter o Webpage. Lamiar Pipe Flow 0 r+δr r x P x P x+δx r Force Balace: Pressure, stress (P x P x+ x )(2 r r)+(2 xr) r (2 x)(r + r) r+ r =0 Divide 2πΔrΔx r P x P x+ x x Limit Δx, Δr à 0 = d(r ) = C r + r r (r + r) r+ r r =0 Separate variables ad itegrate with =0 at r=0 = r 2 5

6 No-Newtoia Pipe Flow Most o-newtoia flows are lamiar. Key results: (remember, Q is just volumetric flow rate-vdot) Force balace: Power law costitutive relatio Itegrate with B.C. v=0 at r=r = r 2 dv = K v = 2K / R + r + + Q = Av avg Q is volumetric flow rate Q = 3 8(3 + ) / D 4K 2(3 + ) / D 4K Kietic Eergy Correctio Factor: Mometum Flux Correctio Factor: = 3(3 + )2 ) (5 + 3)(2 + ) = Pressure Drop Lamiar Flow 2 Defie f = 4 w 2 v2 avg Newtoia Force Balace = R 2 Z A 2 w R v(r)da 2 No-Newtoia v = R2 4 R2 8 r 2 R v = 2K / 2(3 + ) + R + r + / D 4K w = 4V avg R f = 64 Re Solve 4 for / ad isert ito 2 Isert ito for w w =(some complex expressio) f = 8K 2(3 + )Vavg Vavg 2 6

7 Turbulet Flow 3 Defie the frictio factor as before: (Lamiar or Turbulet) For turbulet flow we had f = f(re, ε/d) from dimesioal aalysis. Questio: Will this work for o-newtoia Flow? Questio: What is the Reyolds umber? No clear defiitio of Re sice is ot costat (depeds o the strai rate dv/, which depeds o V avg ) Use the same defiitio as the lamiar frictio factor: Re=64/f f = 8K V 2 avg f = 8 w V 2 avg 2(3 + )Vavg Re = 8 V 2 avg K = P D/L 2 V avg 2 2(3 + )V avg (Defiitio based o lamiar Newtoia, but used for all regimes) Plot frictio factor versus Re as for Newtoia flows, usig the red defiitio of Re. No-Newtoia Frictio Factor (Power Law) 4 f Faig = (/4) f Darcy Re = 8 V avg 2 K 2(3 + )V avg 7

8 Rheological Parameters (power law) 5 Problem: No-Newtoia fluid has: How to fid K, ad for a give fluid? You eed to measure somethig (what?) Try a pipe flow D, Q, / Here s what we kow: dv = K = r 2 dv w = K w = R 2 D, Q, / à V avg, w. The relate these to K, : Compute (-dv/) w from v(r) w v = 2K Q = / 3 8(3 + ) 2(3 + ) w = K D 4K dv w + / D 4K = K dv R + r + / Rheological Parameters (power law) 6 From v(r), we get: Now = K So a plot of l( w ) versus l(-dv/) w is liear with slope, ad itercept l(k). But, ote that (-dv/) w ivolves, which is ukow à what to do? Just rearrage: dv 2(3 + )Vavg = w dv l( w )=l(k)+ l( dv/) w l( w )=l(k)+ l(2(3 + )V avg /) l( w )=l(v avg )+{l(k)+l(2(3 + )/)} Now, a plot of l( w ) versus l(v avg ) is liear with slope. Oce is kow, K ca be computed from the itercept (term i {}), or just compute it aalytically from dv = K ad dv 2(3 + )Vavg = w which give w K = (2(3 + )V avg /) 8

9 Recap 7 To compute K, for a o-newtoia fluid Measure Q, D, / Compute V avg from Q ad D (area), that is, Q=A*V avg Compute w from Plot l( w ) versus l(v avg ) Fit a lie to the data (the liear part of the data) The slope is K is computed from the itercept, or from K = w = R 2 w (2(3 + )V avg /) Note, the uits o K are (kg*s -2 /m) Example 8 Give: Diameter Pressure Drop Flow Rate Compute: K, Re Power through a give pipe is as usual, Q*ΔP Note, here xi = 8*V avg /D, ad istead of plottig l(tau w ) versus l(v avg ), I m plottig l(tau w ) versus l(xi). The approach is the same, but the itercept has a differet formula for gettig K. By the way, xi = 8*V avg /D is (-dv/) w for Newtoia fluids, hece that choice here. 9