Lecture 12 - Non-Newtonian Pipe Flow

Pr 2l laminar flow in a pipe of nnf z0 ss w max p0 r

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Unformatted text preview: 3 Laminar Pipe Flow of Bingham Fluid Time independent non-Newtonian Fluid with yield stress; Substitute fluid constitutive model (Bingham), Solve Average pipe velocity Volumetric flow rate: vz , B s s y B 2 L Pr 2 s y s w s y P B 2L vz 4 dA R 4 P 4 s y 1 s y 1 A Q 8 L 3 s 3 s B w w Q A.vz v Rearrange equation for average velocity to get pressure drop in terms of velocity, pipe dimensions, etc.. Then obtain frictional losses to use in MFEE (hf ); 2 L v2 64.(6 Re B He) Re v D He s y D hf f ; f B 2 B B 2 D 2g 6 Re B Reynolds Number, Power Law fluid: Critical Value Critical value for transition from laminar to turbulent flow depends on power law index. Different approaches/experiments used. (note, other factors for NNF such as fluid elasticity can also affect this critical transition) Newtonian fluids Steffe, p 108. Example A Newtonian oil (specific gravity = 0.9, = 0.1 Pas) is being pumped into a mixer at v=1 m/s along a pipe of diameter of D=0.20 m. The process has been altered so that an oil-in-water emulsion is used in place of the oil. The scientist claims they have matched the viscosity so that flow conditions will be the same. What is the wall shear rate ? Hint: shear rate = 8v/D What is the Re ? What flow regime is it in? Re = VD/ You observe differences in the pressure drop for the emulsion, and later discover that the scientist matched the viscosity using a rheometer at a shear rate of 1s-1. But when you measure the emulsion rheology using a rheometer, you find it follows a power-law with K=0.1 Pas0.5 and n = 0.5. What flow regime is the emulsion in ? Determine apparent and actual wall shear rate, the apparent viscosity at the wall shear rate and Re. Reynolds Number: Power Law Fluid For Power Law Fluid: Re v D 4Q n n V w 34n1 R 34n1 8D Still valid, but now viscosity depends on shear rate 3 Kw n 1 Re PL v 2n D n 8n1 K 34nn1 n K=0.1 Pas0.5 and n = 0.5. D=0.20 m v=1 m/s The feed line is being altered to inject an oil-in-water emulsion at the same velocity (V=1 m/s , D= 0.20 m). The emulsion was measured on a rheometer to be described by a power-law with K=0.1 Pas0.5 and n = 0.5. What flow regime do you reckon it is in ? Estimate the Re ? 0.35 Viscosity (Pas) 0.3 0.25 oil 0.2 Emulsion 0.15 0.1 0.05 0 0.1 1 Shear rate 10 100 MFEE (energy equation): Power Law fluid MFEE can be applied to time independent non-Newtonian fluids. The friction term (hf ) can be calculated in a similar way 2 2 P P v v 1 1 z hq 2 2 z2 h f hs g g 2g 2g 1 hf For Laminar flow of Newtonian fluids: hf f L v2 ; D 2g P g f 64 / Re For Laminar flow of Power Law fluids: L v2 hf f ; D 2g f 64 / Re PL Note on nomenclature: I use: Steffe’ uses: s = shear stress sy = yield stress Re = Reynolds No. s = shear stress s0 = yield stress NRe = Reynolds No. Also used in various texts and/or tutorial problems: t = shear stress t0 = ty = yield stress I also use = viscosity of Newtonian fluid = viscosity of Non-Newtonian fluid (e.g. Shear rate dependent) MFEE – calculating friction factor Als...
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This note was uploaded on 04/17/2013 for the course CHEE 2003 taught by Professor Jasonstokes during the One '12 term at Queensland.

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