Department of Mathematics Comsats Institute of Information Technology MTH- 486: Fluid Mechanics Instructor: Dr. Fahad Munir Abbasi Assistant Professor Department of Mathematics Comsats Institute of Information Technology Islamabad, Pakistan

Layout of lecture # 5 Summary of the previous lecture Newtonian and non-Newtonian fluids Types of non-Newtonian fluids Flow between two plates: Solved examples Summary

Previously we studied… Velocity of the fluid at a point: local and particle rate of change Viscosity of a fluid Real and Ideal fluids

MTH-486, Fluid Mechanics Lec. # 05 Newtonian fluids: Fluids in which the shear stress is directly proportional to the rate of deformation are called the “Newtonian fluids”. For such fluids, the Newton’s law of viscosity holds, i.e. Where the constant of proportionality “μ” is known as the viscosity of the fluid. 𝒔𝒉𝒆𝒂𝒓 𝒔𝒕𝒓𝒆𝒔𝒔∝ 𝒅𝒖 𝒅𝒚 τ 𝒙𝒚 =μ 𝒅𝒖 𝒅𝒚 (1)

Non-Newtonian fluids: MTH-486, Fluid Mechanics Lec. # 05 Non-Newtonian fluids: Fluid in which the shear stress is not directly proportional to the rate of deformation are called the “non-Newtonian fluid”. Examples: Toothpaste, blood, ketchups, paint, drilling mud, biological fluids, etc. For such fluids, the “Power law model” holds, i.e.: 𝒔𝒉𝒆𝒂𝒓 𝒔𝒕𝒓𝒆𝒔𝒔∝ 𝒅𝒖 𝒅𝒚 𝒏 , 𝒏≠𝟏. τ 𝒙𝒚 =𝒌 𝒅𝒖 𝒅𝒚 𝒏 (2)

MTH-486, Fluid Mechanics Lec. # 05 Where “n” is flow behavior index, “k” the consistency index. Note that Eq. (2) reduces to Newton’s law of viscosity for n = 1 and k = μ. Also from Eq. (2) τ 𝒙𝒚 =η 𝒅𝒖 𝒅𝒚 , η=𝒌 𝒅𝒖 𝒅𝒚 𝒏−𝟏 Is the apparent viscosity. (3) (4)

The rod climbing or the Weissenberg effect MTH-486, Fluid Mechanics Lec. # 05 The rod climbing or the Weissenberg effect When a rod is rotated in a beaker containing a polymer (non-Newtonian) solution. The solution moves in the opposite direction of the motion of the rod and climbs up the rod. This effect is known as the “rod climbing effect” or the “Weissenberg effect”. For a beginner, this is a simple test to differentiate between the Newtonian and non-Newtonian fluid.

Time independent fluids MTH-486, Fluid Mechanics Lec. # 05 Types of non-Newtonian fluids: The non-Newtonian fluids are divided into three broad groups: Non-Newtonian Fluids Time independent fluids Time dependent fluids Viscoelastic fluids

Time independent fluids MTH-486, Fluid Mechanics Lec. # 05 . Time independent fluids Pseudoplastic or shear thinning fluids (n < 1) Dilatant or shear thickening fluids (n >1) Ideal or Bingham plastic fluids

Time dependent fluids Thixotropic fluids Rheopectic fluids MTH-486, Fluid Mechanics Lec. # 05 . Time dependent fluids Thixotropic fluids Rheopectic fluids

Time independent non-Newtonian fluids: MTH-486, Fluid Mechanics Lec. # 05 Time independent non-Newtonian fluids: 1. Pseudoplastic (or shear thinning) fluids: Fluids in which the apparent viscosity decreases with increasing deformation rate, i.e. n < 1. Examples: Polymer solution such as rubber, Colloidal suspensions and paper pulp In water, blood, milk etc. 2. Dilatant (or shear thickening) fluids: Fluids in which the apparent viscosity increases with increasing deformation rate, i.e. n > 1. Examples: suspensions of starch and of sand, butter, printing ink, suger in water etc.

MTH-486, Fluid Mechanics Lec. # 05 3. Ideal or Bingham plastic fluids: Fluids that behave as solids until a minimum yield strength, τ 𝒚 is exceeded and subsequently exhibits a linear relation between stress and rate of deformation. Mathematically τ 𝒙𝒚 = τ 𝒚 + μ 𝒑 𝒅𝒖 𝒅𝒚 Examples: Drilling muds, toothpaste and clay suspensions, jellies etc.

Time dependent non-Newtonian fluids: MTH-486, Fluid Mechanics Lec. # 05 Time dependent non-Newtonian fluids: 1. Thixotropic fluids: Fluids that show a decrease in η with time under a constant applied shear stress. Examples: Lipstick, some paints and enamels, etc. 2. Rheopectic fluids: Fluids that show an increase in η with time under a constant applied shear stress. Examples: gypsum suspension in water and bentonite solution, etc.

Viscoelastic non-Newtonian fluids: MTH-486, Fluid Mechanics Lec. # 05 Viscoelastic non-Newtonian fluids: Some fluid after deformation partially return to their original shape when the applied stress is released. Such fluids are named as “viscoelastic fluids.” Viscoelastic fluids have two major types: linear viscoelastic fluids e.g. the Maxwell and Jeffery’s fluids, and non-linear viscoelastic fluids e.g. Walter’s A and B fluid, Oldroyed A and B etc.

Graphical representation: MTH-486, Fluid Mechanics Lec. # 05 Graphical representation: Fig. 1: Flow curves for time independent fluids.

Graphical representation: MTH-486, Fluid Mechanics Lec. # 05 Graphical representation: Fig. 2: Flow curves for time dependent fluids

Graphical representation: MTH-486, Fluid Mechanics Lec. # 05 Graphical representation: Fig. 3: Flow curves for time dependent fluids

Flow between two plates: MTH-486, Fluid Mechanics Lec. # 05 Flow between two plates: An infinite plate is moved over a second plate on a layer of liquid. For small gap width, h = 0.3mm, we assume a linear velocity distribution in the liquid, U = 0.3 m/sec. The liquid viscosity is 0.65 X 𝟏𝟎 −𝟑 𝑲𝒈 𝒎.𝒔𝒆𝒄 and its specific gravity is 0.88. Find: The kinematic viscosity of the fluid. The shear stress on lower plate. Indicate the direction of shear stress.

Flow between two plates: MTH-486, Fluid Mechanics Lec. # 05 Flow between two plates:

MTH-486, Fluid Mechanics Lec. # 05  222222226622222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222 density

τ 𝒚𝒙 = τ 𝒍𝒐𝒘𝒆𝒓 =μ 𝒅𝒖 𝒅𝒚 =μ 𝑼 𝒉 =𝟎.𝟔𝟓 𝑲𝒈 𝒎. 𝒔𝒆𝒄 𝟐 MTH-486, Fluid Mechanics Lec. # 05 Now, ν= μ ρ = 𝟎.𝟔𝟓 𝑿 𝟏𝟎 −𝟑 𝟎.𝟖𝟖 𝑿 𝟏𝟎 𝟑 =−−− 𝒎 𝟐 𝒔𝒆𝒄 τ 𝒚𝒙 = τ 𝒍𝒐𝒘𝒆𝒓 =μ 𝒅𝒖 𝒅𝒚 =μ 𝑼 𝒉 =𝟎.𝟔𝟓 𝑲𝒈 𝒎. 𝒔𝒆𝒄 𝟐 Since the τ 𝒚𝒙 is positive so the direction of shear stress is along positive x-axis.

MTH-486, Fluid Mechanics Lec. # 05 Example # 02: Suppose that the fluid being sheared between to plates is SAE 30 oil (μ = 0.29 𝑲𝒈 𝒎.𝒔𝒆𝒄 ) at 20°C. Compute the shear stress in the oil if V=3 m/s and h = 2 cm. Solution: τ=μ 𝒅𝒖 𝒅𝒚 =μ 𝑽 𝒉 = 𝟎.𝟐𝟗 𝑲𝒈 𝒎.𝒔𝒆𝒄 𝑿 𝟑.𝟎 𝒎 𝒔𝒆𝒄 𝟎.𝟎𝟐 𝒎 =𝟎.𝟒𝟑 𝑲𝒈 𝒎. 𝒔𝒆𝒄 𝟐 =𝟒𝟑 𝑵 𝒎 𝟐 =𝟒𝟑𝑷𝒂

Summary Newtonian and non-Newtonian fluids Types of non-Newtonian fluids Solved examples

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