Non-Newtonian Fluids CBE 150A – Transport Spring Semester 2014 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 CBE 150A – Transport Spring Semester 2014 Non-Newtonian Fluids Newtonian Fluid du z τ rz = − µ dr Non-Newtonian Fluid du z τ rz = −η dr η is the apparent viscosity and is not constant for non-Newtonian fluids. CBE 150A – Transport Spring Semester 2014 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 (τ0) and then becomes constant CBE 150A – Transport Spring Semester 2014 Non-Newtonian Fluids Bingham Plastic: sludge, paint, blood, ketchup Pseudoplastic: latex, paper pulp, clay solns. Newtonian Dilatant: quicksand CBE 150A – Transport Spring Semester 2014 Modeling Power Law Fluids Oswald - de Waele n n −1 ⎡ ⎛ du z ⎞ ⎛ du z ⎞ ⎤ ⎛ du z ⎞ τ rz = K ⎜ − ⎟ = ⎢K ⎜ ⎟ ⎥ ⎜− ⎟ ⎝ dr ⎠ ⎢⎣ ⎝ dr ⎠ ⎥⎦ ⎝ dr ⎠ where: K = flow consistency index n = flow behavior index µ eff Note: Most non-Newtonian fluids are pseudoplastic n<1. CBE 150A – Transport Spring Semester 2014 Modeling Bingham Plastics τ rz < τ 0 du z =0 dr τ rz ≥ τ 0 du z τ rz = − µ ∞ ±τ 0 dr CBE 150A – Transport Rigid Spring Semester 2014 Frictional Losses Non-Newtonian Fluids Recall: LV hf = 4 f D 2 2 Applies to any type of fluid under any flow conditions CBE 150A – Transport Spring Semester 2014 Laminar Flow Mechanical Energy Balance Δp 2 ΔαV + + g Δ z + h f = Wˆ ρ 2 0 CBE 150A – Transport 0 0 Spring Semester 2014 MEB (contd) Combining: 1 ⎛ D ⎞ 2 ⎛ Δp ⎞ ⎟⎟ f = ⎜ ⎟ 2 ⎜⎜ − 4 ⎝ L ⎠V ⎝ ρ ⎠ CBE 150A – Transport Spring Semester 2014 Momentum Balance m! (β 2V2 − β1V1 ) = p1S1 − p2 S2 − Fw − Fg 0 0 2 2π rLτ rz = π r (− Δp ) L − Δp = 2 τ rz r CBE 150A – Transport Spring Semester 2014 Power Law Fluid ⎛ du z ⎞ τ rz = K ⎜ − ⎟ ⎝ dr ⎠ n 1n du z ⎛ 1 Δp ⎞ 1 n = −⎜ − ⎟ r dr ⎝ 2 KL ⎠ Boundary Condition r=R CBE 150A – Transport uz = 0 Spring Semester 2014 Velocity Profile of Power Law Fluid Circular Conduit Upon Integration and Applying BC 1n ⎛ 1 Δp ⎞ ⎛ n ⎞ ⎡ uz = ⎜ − ⎟ ⎜ ⎟⎢ R ⎝ 2 KL ⎠ ⎝ n + 1 ⎠ ⎣ CBE 150A – Transport n +1 n −r n +1 n ⎤ ⎥ ⎦ Spring Semester 2014 Power Law (contd) Need bulk average velocity 1 1 V = ∫ u dS = 2 S S πR 1n ∫ (2π ru )dr z ⎡ 1 Δp ⎤ ⎛ n ⎞ V = ⎢− R ⎜ ⎟ ⎥ ⎣ 2 K L ⎦ ⎝ 3n + 1 ⎠ CBE 150A – Transport n +1 n Spring Semester 2014 Power Law Results (Laminar Flow) n 2 Δp = − ↑ n+2 ⎛ 3n + 1 ⎞ ⎜ ⎟ LK V ⎝ n ⎠ n +1 D n Hagen-Poiseuille (laminar Flow) for Power Law Fluid ↑ Recall 1 ⎛ D ⎞⎛ 2 ⎞ Δp f = − ⎜ ⎟⎜ 2 ⎟ 4 ⎝ L ⎠⎝ V ⎠ ρ CBE 150A – Transport Spring Semester 2014 Power Law Fluid Behavior Power Law Reynolds Number and Kinetic Energy Correction n RePL 2−n n n V D ρ ⎛ ⎞ 3− n =2 ⎜ ⎟ K ⎝ 3n + 1 ⎠ ( 4n + 2)(5n + 3) Re PL ,critical = 2100 2 3(3n + 1) 3(3n + 1) 2 α= (2n + 1)(5n + 3) CBE 150A – Transport Spring Semester 2014 Laminar Flow Friction Factor Power Law Fluid n ⎛ 3n + 1 ⎞ 2 ⎜ ⎟ K n ⎠ ⎝ f = V 2−n D n ρ n +1 16 f = RePL CBE 150A – Transport Spring Semester 2014 Turbulent Flow Friction Factor Power Law Fluid (Smooth Pipe) CBE 150A – Transport Spring Semester 2014 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 CBE 150A – Transport Spring Semester 2014 n ⎛ ∂V ⎞ ⎛ ∂V ⎞ τ rz = K ⎜ ⎟ = µ eff ⎜ ⎟ ⎝ ∂r ⎠ ⎝ ∂r ⎠ 1− 0.4 ⎛ 100 ⎞ K = 50cP ⎜ ⎟ ⎝ s ⎠ = 0.792 kg m s1.6 3 ⎞ ⎛ m ⎞ 1 m ⎛ 900 gal ⎞ ⎛ 1 ft ⎞ ⎛ 1 min ⎞ ⎛ ⎟⎟ ∗ ⎜ ⎟ ⎜ ⎟ V =⎜ ∗ = 1 . 759 ⎟ ∗ ⎜⎜ ⎟ ∗ ⎜⎜ 2 ⎟ ⎜ ⎟ s ⎝ min ⎠ ⎝ 7.48 gal ⎠ ⎝ 60 s ⎠ ⎝ 0.3474 ft ⎠ ⎝ 3.281 ft ⎠ ~ 1.6 ⎡ kg ⎞⎛ m⎞ ⎤ 0.4 ⎛ 0.4 ⎢ (0.202 m ) ⎜1442 ⎟⎜1.759 ⎟ ⎥ ⎞ ⎢ 0.4 m 3 ⎠⎝ s⎠ ⎥ ⎝ (3− 0.4 ) ⎛ ⎜⎜ ⎟⎟ RE N = 2 = 7273 kg ⎢ ⎥ 3 ∗ ( 0 . 4 ) + 1 ⎝ ⎠ 0.792 ⎢ ⎥ m s1.6 ⎣ ⎦ CBE 150A – Transport Spring Semester 2014 Friction Factor (Power Law Fluid) CBE 150A – Transport Spring Semester 2014 ΔαV 2 gΔZ Wp = + + + hf ρ 2gc gc ΔP 2 ⎛ L ⎞V Wp = h f = 4 f ⎜ ⎟ ⎝ D⎠ 2 f = 0.0048 Fig 5.11 2 m⎞ ⎛ ⎜1.760 ⎟ m2 s⎠ ⎛ 80460 m ⎞ ⎝ W p = h f = 4(0.0048)⎜ = 11,845 2 ⎟ 2 s ⎝ 0.202m ⎠ • m = 1.759 m kg ⎞ kg ⎛ ∗ (0.0323 m 2 )∗ ⎜1442 3 ⎟ = 81.9 s m ⎠ s ⎝ kg ⎛ m2 ⎞ 81.9 ⎜⎜11,845 2 ⎟⎟ s ⎝ s ⎠ Power = = 970.1 kW = 1300 Hp 1000 CBE 150A – Transport Spring Semester 2014 Bingham Plastics Bingham plastics exhibit Newtonian behavior after the shear stress exceeds τo. For flow in circular conduits Bingham plastics behave in an interesting fashion. CBE 150A – Transport Spring Semester 2014 Bingham Plastics Unsheared Core τ0 2 (R − rc ) u z = uc = 2µ ∞ rc r ≤ rc Sheared Annular Region r > rc CBE 150A – Transport uz ( R − r ) ⎡τ rz ⎛ r ⎞ = 1+ −τ µ∞ ⎢2 ⎜ ⎣ ⎝ ⎟ R⎠ ⎤ 0⎥ ⎦ Spring Semester 2014 Laminar Bingham Plastic Flow 16 f = Re BP ⎡ ⎤ He He 4 − 3 ⎢1 + 7⎥ ⎣ 6 Re BP 3 f (Re BP ) ⎦ (Non-linear) 2 He = D ρτ 0 Re BP = CBE 150A – Transport µ ∞2 Hedstrom Number DρV µ∞ Spring Semester 2014 Turbulent Bingham Plastic Flow a −0.193 f = 10 Re BP ( a = −1.378 1 + 0.146e CBE 150A – Transport − 2.9 x10−5 He ) Spring Semester 2014 Drilling Rig Fundamentals CBE 150A – Transport Spring Semester 2014 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 CBE 150A – Transport Spring Semester 2014 D= 4 ft = 0.3333 ft 12 Area = 0.0873 ft 2 ⎞ gal ⎛ min ⎞ ⎛ ft 3 ⎞ ⎛ 1 ft ⎟⎟ ∗ ⎜⎜ ⎟ = 1.276 V = 50 ∗⎜ ⎟ ∗ ⎜⎜ 2 ⎟ min ⎝ 60 s ⎠ ⎝ 7.48 gal ⎠ ⎝ 0.0873 ft ⎠ s ρ = 1.2 ∗ 62.4 lbm lb = 74.88 m3 3 ft ft lb ⎞ ⎛ ⎜ 6.7197 × 10 − 4 m ⎟ lb ft s ⎟ µ = 35 cP ∗ ⎜ = 0.0235 m ⎜ ⎟ cP ft s ⎜ ⎟ ⎝ ⎠ N RE lb ⎞ ft ⎞ ⎛ ⎛ 0.3333 ft ∗ ⎜1.276 ⎟ ∗ ⎜⎜ 74.88 m ⎟⎟ s⎠ ⎝ ft 3 ⎠ ⎝ = = 1355 lbm 0.0235 ft s τ o = 100 CBE 150A – Transport dyn g = 100 2 2 cm s cm Spring Semester 2014 2 N HE ⎛ g ⎞ ⎛ 100 g ⎞ ⎛ 2.54 cm ⎞ ⎞ ⎛ ⎜ 4in ⎜ ⎟ ⎟ ∗ ⎜1.2 3 ⎟ ∗ ⎜ 2 ⎟ ⎝ in ⎠ ⎠ ⎝ cm ⎠ ⎝ s cm ⎠ =⎝ = 1.01 ×10 5 2 g ⎞ ⎛ ⎜ 0.35 ⎟ cms ⎠ ⎝ f = 0.14 Wp = ΔP ρ + ΔαV 2 gΔZ + + hf 2gc gc 2 ⎛ ⎞ lb f ⎛ 144 in 2 ⎞ ft ⎞ ⎜ ⎛ ⎟ ⎜ ⎟ 1.276 ⎟ ⎜ 2 ⎜ 2 ⎟ ⎜ ⎟ ft lb f 4 ∗ 0.14 ∗ (8000 ft ) in ⎝ ft ⎠ s⎠ ⎝ − 8000 + ⎜ ⎟ lbm lb 0 . 3333 ft ⎛ ⎞ 32 . 2 ft lbm ⎜ ⎜ m 74.88 3 ⎟⎟ ⎜ 2 ∗ ⎜ lb s 2 ⎟ ⎟ ft f ⎠⎠ ⎝ ⎝ (4500 − 14.7 ) Wp = Wp = (8626 − 8000 + 339) = 965 CBE 150A – Transport ft lb f lbm Spring Semester 2014 4 ⎡ ⎤ 16 He He f = − 3 = 0.14 ⎢1 + 7⎥ Re BP ⎣ 6 Re BP 3 f (Re BP ) ⎦ CBE 150A – Transport Spring Semester 2014