Non-Newtonian Fluids Non-Newtonian Flow
Goals
• Describe key differences between a Newtonian and non-Newtonian fluid
• Identify examples of Bingham plastics (BP) and power law (PL) fluids
• Write basic equations describing shear stress and velocities of non-Newtonian fluids
• Calculate frictional losses in a non-Newtonian flow system Non-Newtonian Fluids Newtonian Fluid du t = -µ z rz dr
Non-Newtonian Fluid du t = -h z rz dr η is the apparent viscosity and is not constant for non-Newtonian fluids. η - Apparent Viscosity The shear rate dependence of η categorizes non-Newtonian fluids into several types. Power Law Fluids: Ø Pseudoplastic – η (viscosity) decreases as shear rate increases (shear rate thinning) Ø Dilatant – η (viscosity) increases as shear rate increases (shear rate thickening)
Bingham Plastics:
Ø η depends on a critical shear stress (t0) and then becomes constant Non-Newtonian Fluids
Bingham Plastic: sludge, paint, blood, ketchup
Pseudoplastic: latex, paper pulp, clay solns.
Newtonian
Dilatant: quicksand Modeling Power Law Fluids Oswald - de Waele n é n-1 ù æ du z ö æ du z ö æ du z ö t rz = Kç- ÷ = êKç ÷ ú ç- ÷ è dr ø ëê è dr ø ûú è dr ø
where: K = flow consistency index n = flow behavior index µeff
Note: Most non-Newtonian fluids are pseudoplastic n<1. Modeling Bingham Plastics
du t du t ³t t = -µ z ±t rz 0 rz ¥ dr 0 Frictional Losses Non-Newtonian Fluids Recall: L V 2 h = 4 f f D 2 Applies to any type of fluid under any flow conditions Laminar Flow Mechanical Energy Balance Dp DaV 2 + + g Dz + h =Wˆ r 2 f 0 0 0 MEB (contd) Combining: 1 æ D ö 2 æ Dp ö f = ç- ÷ ç ÷ 2 ç ÷ 4 è L øV è r ø Momentum Balance m! (b2V2 - b1V1 )= p1S1 - p2S2 - Fw - Fg 0 0 2 2p rLt rz = p r (- Dp) L - Dp = 2 t r rz Power Law Fluid n æ du z ö t rz = Kç- ÷ è dr ø 1 n du æ 1 Dp ö z = -ç- ÷ r1 n dr è 2 KL ø Boundary Condition r = R uz = 0 Velocity Profile of Power Law Fluid Circular Conduit Upon Integration and Applying BC 1 n é n+1 n+1 ù æ 1 Dp ö æ n ö n n uz = ç- ÷ ç ÷êR - r ú è 2 KL ø è n +1øë û Power Law (contd) Need bulk average velocity 1 1 V = udS = (2p ru z )dr S òS p R2 ò 1 n é 1 Dp ù æ n ö n +1 V = - ç ÷ R n ëê 2 KLûú è 3n + 1 ø Power Law Results (Laminar Flow) n n 2 æ 3n + 1ö n 2 + ç ÷ LKV n Dp = - è ø D n+1 ↑ Hagen-Poiseuille (laminar Flow) for Power Law Fluid ↑ Recall 1 æ D öæ 2 ö Dp f = - ç ÷ç ÷ 4 è L øèV 2 ø r Power Law Fluid Behavior Power Law Reynolds Number and Kinetic Energy Correction n 2-n n 3-n æ n ö V D r RePL = 2 ç ÷ è 3n +1ø K (4n + 2)(5n + 3) Re = 2100 PL,critical 3(3n +1)2 3(3n +1)2 a = (2n +1)(5n + 3) Laminar Flow Friction Factor Power Law Fluid n æ 3n +1ö 2 n+1 ç ÷ K n f = è ø V 2-n D n r 16 f = RePL Turbulent Flow Friction Factor Power Law Fluid (Smooth Pipe) Power Law Fluid Example A coal slurry is to be transported by horizontal pipeline. It has been determined that the slurry may be described by the power law model with a flow index of 0.4, an apparent viscosity of 50 cP at a shear rate of 100 /s, and a density of 90 lb/ft3. What horsepower would be required to pump the slurry at a rate of 900 GPM through an 8 in. Schedule 40 pipe that is 50 miles long ? P = 1atm P = 1atm L = 50 miles n æ ¶V ö æ ¶V ö t rz = K ç ÷ = µeff ç ÷ è ¶r ø è ¶r ø 1-0.4 æ100 ö kg K = 50cPç ÷ = 0.792 è s ø m s1.6 ~ æ 900gal ö æ 1ft3 ö æ1min ö æ 1 ö æ m ö m V = *ç ÷* *ç ÷*ç ÷ =1.759 ç ÷ ç ÷ ç ÷ ç 2 ÷ ç ÷ è min ø è 7.48gal ø è 60s ø è 0.3474 ft ø è 3.281ft ø s 1.6 é 0.4 æ kg öæ m ö ù 0.4 ê(0.202 m) ç1442 ÷ç1.759 ÷ ú æ 0.4 ö m3 s RE = 2(3-0.4)ç ÷ ê è øè ø ú = 7273 N ç ÷ ê kg ú è 3*(0.4) +1ø 0.792 ê 1.6 ú ë m s û Friction Factor (Power Law Fluid) DP DaV 2 gDZ Wp = + + + hf r 2gc gc æ L öV 2 Wp = hf = 4 f ç ÷ è D ø 2 f = 0.0048 Fig 5.11 2 æ m ö ç1.760 ÷ 2 æ 80460m ö è s ø m Wp = hf = 4(0.0048)ç ÷ =11,845 è 0.202m ø 2 s2 • m æ kg ö kg m =1.759 *(0.0323 m2 )*ç1442 ÷ = 81.9 s è m3 ø s kg æ m2 ö 81.9 ç11,845 ÷ s ç s2 ÷ Power = è ø = 970.1 kW =1300 Hp 1000 Bingham Plastics Bingham plastics exhibit Newtonian behavior after the shear stress exceeds to. For flow in circular conduits Bingham plastics behave in an interesting fashion. Bingham Plastics Unsheared Core t 0 2 r £ rc uz = uc = (R - rc ) 2µ¥rc Sheared Annular Region (R - r) ét rz æ r ö ù r > rc uz = ê ç1+ ÷ -t 0 ú µ¥ ë 2 è R ø û Laminar Bingham Plastic Flow 16 é He He4 ù f = 1+ - (Non-linear) ê 3 7 ú Re BP ë 6Re BP 3 f (Re BP ) û D2 rt 0 Hedstrom Number He = 2 µ¥ DrV Re BP = µ¥ Turbulent Bingham Plastic Flow a -0.193 f =10 Re BP -5 a = -1.378(1+ 0.146e-2.9x10 He ) Drilling Rig Fundamentals Bingham Plastic Example Drilling mud has to be pumped down into an oil well that is 8000 ft deep. The mud is to be pumped at a rate of 50 GPM to the bottom of the well and back to the surface through a pipe having an effective diameter of 4 in. The pressure at the bottom of the well is 4500 psi. What pump head is required to pump the mud to the bottom of the drill string ? The drilling mud has the properties of a Bingham plastic with a yield stress of 100 dyn/cm2, a limiting (plastic) viscosity of 35 cP, and a density of 1.2 g/cm3. P = 14.7 psi L = 8000 ft P = 4500 psi 4 D = ft = 0.3333 ft Area = 0.0873 ft2 12 gal æ min ö æ ft3 ö æ 1 ö ft V = 50 * *ç ÷*ç ÷ =1.276 ç ÷ ç ÷ ç 2 ÷ min è 60s ø è 7.48gal ø è 0.0873 ft ø s lb lb r =1.2*62.4 m = 74.88 m ft3 ft3 æ lb ö ç 6.7197 ´10 - 4 m ÷ ft s lb µ = 35 cP *ç ÷ = 0.0235 m ç cP ÷ ft s ç ÷ è ø æ ft ö æ lbm ö 0.3333 ft *ç1.276 ÷*ç74.88 ÷ è s ø ft3 N = è ø =1355 RE lb 0.0235 m ft s dyn g t =100 =100 o cm 2 s2cm 2 æ æ 2.54 cm öö æ g ö æ100g ö ç4in ç ÷÷ *ç1.2 3 ÷*ç 2 ÷ è è in øø è cm ø è s cm ø 5 N HE = 2 =1.01´10 æ g ö ç0.35 ÷ è cms ø f = 0.14 DP DaV 2 gDZ Wp = + + + hf r 2gc gc lb æ144 in2 ö æ 2 ö (4500 -14.7) f ç ÷ ç æ ft ö ÷ 2 ç 2 ÷ ç ç1.276 ÷ ÷ in è ft ø ft lbf 4*0.14 *(8000 ft) è s ø W = -8000 + ç ÷ p lb m lbm 0.3333 ft ç æ 32.2 ft lbm ö ÷ 74.88 3 2*ç ÷ ft ç ç lb s2 ÷ ÷ è è f ø ø ft lb Wp = (8626 -8000 + 339) = 965 f lbm 16 é He He4 ù f = 1+ - = 0.14 ê 3 7 ú Re BP ë 6Re BP 3 f (Re BP ) û