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Rheology ppt

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PRESENTATION ON RHEOLOGY

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Rheology ppt

  1. 1. ERT 142 Engineering Properties of Bio Materials Biosystems Engineering School of Bioprocess Engineering
  2. 2. Rheology  rheo – to flow  logos – science  ology – the study of  is the study of the flow of materials that behave in an interesting or unusual manner.  Unusual materials such as mayonnaise, peanut butter, chocolate, bread dough, paints, inks, road building aterials, cosmetics, dairy products, etc.
  3. 3. Rheology  The study of viscosity is of true liquids, solutions, dilute and concentrated colloidal systems is of much importance in this study  It is involved in the mixing and flow of materials, their packaging into containers, the pouring from the bottle, extrusion from a tube or a passage of the liquid to a syringe needle.
  4. 4. Rheology  Can affect the patient’s acceptability of the product, physical stability, biologic availability, absorption rate of drugs in the gastrointestinal tract  Influence the choice of processing equipments in the pharmaceutical system
  5. 5. What is a Fluid ? (Fluid vs. Solid) • A substance which deforms continuously under the action of a shearing stress. • A perfectly elastic solid can resist a shear stress by static deformation; a fluid cannot. • An elastic solid can behave like a fluid beyond its yield point, at which point it behaves as a "plastic". • Viscoelastic fluids behave like fluids and solids (i.e. egg whites, which have a small tendency to return to their original shape). Corollary: A fluid at rest must be in a state of zero shear stress.
  6. 6. Liquid vs. Gas • Gases typically expand to fill the shape of container. • Liquids assume shape of only part of container. • Equation of state for pressure • Gases typically obey equations of state for the pressure e.g. the ideal gas law p = ρ R T • Liquids are typically assumed to be incompressible and so p is a very weak function of ρ and T. • Sound speed in gases is typically smaller than in liquids (air ~ 343 m/s, water ~ 1484 m/s, iron 5120 m/s).
  7. 7. NEWTON’S LAW AND VISCOSITY  Some of the food substances are in the form of LIQUID.  One of the PHYSICAL PROPERTIES FOR LIQUID FOOD  VISCOSITY (Kelikatan).  REASON FOR APPLYING THE SCIENCE OF VISCOSITY: (1) Will influence the perception of consumer: (a) e.g.: ‘KAYA’  Cannot be too dilute, viscose is the best. (b) e.g.: SAUCE  Must be in certain value of viscosity. (c) e.g.: JUICES  Need to be diluted, not viscose. (2) Easy to transport: (a) e.g.: For transporting the liquid food to the other location in the factory. (3) Easy for gripping (adhesion) : (a) e.g.: Adhesion of coated flour to the chicken meat. (b) e.g.: Adhesion to the container or wrapper.
  8. 8. Load and Response  Stress force per unit area  Strain deformation ○amount of deformation divided by original length
  9. 9.  Movement of FLUID will occur if the STRESS (TEGASAN) is given.  STRESS/PRESSURE = Force (N)/Area (m2 )  If the FORCE (N) is perpendicular (at 90o angle) with surface  It is called NORMAL STRESS (Tegasan Normal).  Normally, it is called = PRESSURE (P)  If the FORCE is HORIZONTAL/PARALLEL (1800 ) with the surface  It is called SHEAR STRESS (Tegasan Ricihan).  Different materials will give different effect of the SHEAR STRESS. P3P2P1 (Force) F1 = N (V 0 to A) V 0 = m/s Fluid ∆ r (distance of the fluid movement), m Viscosity (µ) = Pa.s OR kg/m.s V A = m/s
  10. 10.  VISCOSITY, µ (Kelikatan) = “resistance (rintangan) of GASSES OR FLUIDS towards the flow of the SHEAR STRESS”.  When STRESS is applied  Fluids will move.  SO,  STRESS (F/A);  VELOCITY (m/s).  Fluids is assumed to be combination of FLUID LAYERS.  When the lower layer is given the STRESS, it will effect the upper layer.  USUALLY, viscosity of lower layer > upper layer.  Each layer will move at different velocity (m/s).  The far the layer from the source of STRESS, the  velocity (m/s) of that layer.  Upper layer has < velocity (m/s).  The subsequent layer has > velocity (m/s).
  11. 11.  The English unit for VISCOSITY  are called “Poise” or “centipoise” (cp) = g/cm.s  The SI UNIT for VISCOSITY  Pa.s ( N.s/m2 OR kg/m.s) 1 cp = 1 × 10-3 kg/m.s = 1 × 10-3 Pa.s = 1 × 10-3 N.s/m2 (SI Unit) 1 cp = 0.01 poise = 0.01 g/cm.s (English Unit)
  12. 12. (1) For ELASTIC SOLID  when the shear stress is applied, it will change accordingly to the flow of the stress.  The material will RETURN BACK to its normal size when the stress is removed.  e.g.: rubber materials. (2) For SOLID that have a plastic properties  when the SHEAR STRESS is applied, it will change accordingly to the flow of the stress.  BUT = it will not come back to its original shape right after the stress is removed.  e.g.: jelly (agar-agar). (3) For FLUIDS  when the SHEAR STRESS is applied, it will change accordingly to the flow of the stress.  This material will not come back to its original shape & will moving towards the flow of the stress.
  13. 13. A shear stress, is applied to the top of the square while the bottom is held in place. This stress results in a strain, or deformation, changing the square into a parallelogram.
  14. 14. Generic Stress-Strain Relationship Strain (deformation) Stress(load) ElasticRegion Elastic Limit (Yield Point) Plastic Region
  15. 15. Mechanical Strength  The strength of a material has to do with the maximum stress (or strain) the material is able to withstand before failure.
  16. 16. Strain  Strain is the quantification of the deformation of a material
  17. 17.  Linear Strain Occurs as a result of a change in the object’s length.  Shear Strain Occurs with a change in orientation of adjacent molecules as a result of these molecules slipping past each other.
  18. 18. Instron Measuring stress and strain in biological materials
  19. 19. Two Categories of Flow & Deformation Newtonian (Newtonian Law of Flow)Newtonian (Newtonian Law of Flow)  “the higher the viscosity of a liquid, the greater is the force per unit area (shearing stress) required to produce a certain rate of shear”  Shear – as a stress which is applied parallel or tangential to a face of a material, as opposed to a normal stress which is applied perpendicularly.  Shear stress  Measured in (SI unit): pascal  Commonly used symbols: τ  Expressed in other quantities: τ = F / A
  20. 20. Two Categories of Flow & Deformation Newtonian (Newtonian Law of Flow)Newtonian (Newtonian Law of Flow) A Newtonian fluid (named for Isaac Newton) is a fluid whose stress versus rate of strain curve is linear and passes through the origin. The constant of proportionality is known as the viscosity. A simple equation to describe Newtonian fluid behavior is where • τ is the shear stress exerted by the fluid ("drag") [Pa] • μ is the fluid viscosity - a constant of proportionality [Pa·s] • du is the velocity gradient perpendicular to the direction dy of shear [s−1]
  21. 21. Two Categories of Flow & Deformation Newtonian (Newtonian Law of Flow)Newtonian (Newtonian Law of Flow) In common terms, this means the fluid continues to flow, regardless of the forces acting on it. For example, water is Newtonian, because it continues to exemplify fluid properties no matter how fast it is stirred or mixed. For a Newtonian fluid, the viscosity, by definition, depends only on temperature and pressure (and also the chemical composition of the fluid if the fluid is not a pure substance), not on the forces acting upon it.
  22. 22. Two Categories of Flow & Deformation Newtonian (Newtonian Law of Flow)Newtonian (Newtonian Law of Flow) For a Newtonian fluid, the viscosity, by definition, depends only on temperature and pressure (and also the chemical composition of the fluid if the fluid is not a pure substance), not on the forces acting upon it. If the fluid is incompressible and viscosity is constant across the fluid, the equation governing the shear stress is expressed in the Cartesian coordinate system
  23. 23. Two Categories of Flow & Deformation Newtonian (Newtonian Law of Flow)Newtonian (Newtonian Law of Flow) Cartesian coordinate system where, by the convention of tensor notation, • τij is the shear stress on the ith face of a fluid element in the jth direction • ui is the velocity in the ith direction • xj is the jth direction coordinate
  24. 24. Two Categories of Flow & Deformation Newtonian (Newtonian Law of Flow)Newtonian (Newtonian Law of Flow) Cartesian coordinate system Tensor - are geometrical entities introduced into mathematics and physics to extend the notion of scalars, (geometric) vectors, and matrices - Components of stress, a second-order tensor, in three dimensions. The tensor in the image is the row vector, of the forces acting on the X, Y, and Z faces of the cube. Those forces are represented by column vectors. The row and column vectors that make up the tensor can be represented together by a matrix.
  25. 25. Two Categories of Flow & Deformation Non-NewtonianNon-Newtonian  AA non-Newtonian fluidnon-Newtonian fluid is ais a fluidfluid whose flowwhose flow properties are not described by a single constantproperties are not described by a single constant value ofvalue of viscosityviscosity..  Many polymer solutions and molten polymers are non-Newtonian fluids, as are many commonly found substances such as ketchup, starch suspensions, paint, blood and shampoo.  In a Newtonian fluid, the relation between the shear stress and the strain rate is linear (and if one were to plot this relationship, it would pass through the origin), the constant of proportionality being the coefficient of viscosity.
  26. 26. Two Categories of Flow & Deformation  Non-NewtonianNon-Newtonian  A In a non-Newtonian fluid, the relation between theA In a non-Newtonian fluid, the relation between the shear stressshear stress and theand the strain ratestrain rate is nonlinear, and canis nonlinear, and can even be time-dependent. Therefore a constanteven be time-dependent. Therefore a constant coefficient of viscosity cannot be defined.coefficient of viscosity cannot be defined.  A ratio between shear stress and rate of strain (or shear-dependent viscosity) can be defined, this concept being more useful for fluids without time- dependent behavior.
  27. 27. (A) (B) (C) SLOPE: (A) > (B) > (C) = APPARENT VISCOSITY (µapp) ----  µapp with  SHEAR RATE (γ) (n > 1.0) (n < 1.0) (n = 1.0)
  28. 28. Two Categories of Flow & Deformation  Non-Newtonian ExamplesNon-Newtonian Examples  An inexpensive, non-toxic example of a non- Newtonian fluid is a suspension of starch (e.g. cornflour) in water, sometimes called oobleck (uncooked imitation custard, being a suspension of primarily cornflour, has the same properties). The sudden application of force — for example:  by stabbing the surface with a finger, or rapidly inverting the container holding it — leads to the fluid behaving like a solid rather than a liquid.  This is the "shear thickening" property of this non-Newtonian fluid. More gentle treatment, such as slowly inserting a spoon, will leave it in its liquid state.  Trying to jerk the spoon back out again, however, will trigger the return of the temporary solid state.  A person moving quickly and applying sufficient force with their feet can literally walk across such a liquid.
  29. 29. Two Categories of Flow & Deformation  Non-Newtonian ExamplesNon-Newtonian Examples  There are fluids which have a linear shear stress/shear strain relationship which require a finite yield stress before they begin to flow. That is the shear stress, shear strain curve doesn't pass through the origin. These fluids are called 1. Bingham plastics.  clay suspensions, drilling mud, toothpaste, mayonnaise, chocolate, and mustard. The classic case is ketchup which will not come out of the bottle until you stress it by shaking.
  30. 30. Two Categories of Flow & Deformation  Non-Newtonian ExamplesNon-Newtonian Examples These fluids are called  1. Pseudoplastic Flow  Polymers in solutions such as tragacant, sodium alginate, methylcellulose  Viscosity decreases with an increase in shear thinning  Caused by the re-alignment of polymer and/or the release of solvents associated with the polymers  2. Dilatant Flow  Volume increases when sheared  Shear thickening  Suspension containing high-concentration of small deflocculated particles
  31. 31. Two Categories of Flow & Deformation  Non-Newtonian ExamplesNon-Newtonian Examples  There are also fluids whose strain rate is a function of time. Fluids that require a gradually increasing shear stress to maintain a constant strain rate are referred to as rheopectic.  An opposite case of this, is a fluid that thins out with time and requires a decreasing stress to maintain a constant strain rate (thixotropic).
  32. 32. THIXOTROPY  is the property of some non-Newtonian pseudoplastic fluids to show a time-dependent change in viscosity; the longer the fluid undergoes shear stress, the lower its viscosity.  A thixotropic fluid is a fluid which takes a finite time to attain equilibrium viscosity when introduced to a step change in shear rate.  the term is sometimes applied to pseudoplastic fluids without a viscosity/time component. Many gels and colloids are thixotropic materials, exhibiting a stable form at rest but becoming fluid when agitated.
  33. 33. THIXOTROPY  pseudoplastic fluids ○ Shear thinning is an effect where viscosity decreases with increasing rate of shear stress. Materials that exhibit shear thinning are called pseudoplastic. ○ This property is found in certain complex solutions, such as lava, ketchup, whipped cream, blood, paint, and nail polish. ○ Pseudoplasticity can be demonstrated by the manner in which shaking a bottle of ketchup causes the contents to undergo an unpredictable change in viscosity. The force causes it to go from being thick like honey to flowing like water. ○ thixotropic fluid viscosity decreases over time at a constant shear rate.
  34. 34. THIXOTROPY The distinction between a thixotropic fluid and a shear thinning fluid: ○ A thixotropic fluid displays a decrease in viscosity over time at a constant shear rate. ○ A shear thinning fluid displays decreasing viscosity with increasing shear rate. ○ Some fluids are anti-thixotropic: constant shear stress for a time causes an increase in viscosity or even solidification. Constant shear stress can be applied by shaking or mixing. Fluids which exhibit this property are usually called rheopectic. They are much less common.
  35. 35. IDENTIFICATION OF FLUIDS RHEOLOGY USING VISCOMETER  RHEOLOGY  “The science of the flow and deformation (pembentukan) of fluids”.  2 types of measurement equipment : (1) Rotational Type Viscometer (Viskometer Jenis Putaran):  Concentric cylinder (silinder konsentrik/berpusat sama). UKM: known as ---- “Wide-gap rotational viscometer with spindle cylinder”  Parallel plate (plat selari).  Cone & plate (kon & plat).  Mixer (pengacau). (2) Tube Type Viscometer (Viskometer Jenis Tiub):  Glass capillary (kapillari kaca).  Pipe (paip).  High pressure capillary (kapilari tekanan tinggi).
  36. 36. CONTINUE: Tube Type Viscometer: High pressure capillary (kapilari tekanan tinggi) NON-NEWTONIAN SUBTANCES: •Lubricating oil •Polymer solutions •Fuel oil •Emulsions •Fat melts •Suspensions •Printing inks •Liquid detergents •Latex •Adhesives •Lacquers •Glues
  37. 37. Rotational Type Viscometer Concentric cylinder (silinder konsentrik) NON-NEWTONIAN SUBSTANCES: •suspensions •diary products •lacquers or varnishes •printing inks •emulsion paints •lubricants •latex •polymer solutions •coal suspensions •glues or adhesives •resins •coating slips •chocolate suspensions •sealing compounds •fruit mash or preparations •vegetable mash •coating colours •cosmetics •solutions •emulsions •mud
  38. 38. Rotational Type Viscometer: Mixer (pengacau)
  39. 39.  The most common type of equipment  concentric cylinder viscometer (viskometer silinder konsentrik).  Consisted with 2 concentric cylinder (silinder berpusat sama).  Fluid substances is put between 2 cylinders (INTERNAL & EXTERNAL CYLINDER) .  INTERNAL CYLINDER (spindle)  will spin & gives SHEAR FORCE to the fluid substances.  There are particular sizes of SPINDLE can be found depending on the viscosity of the fluid substances (make a visual interpretation before selecting the sizes of spindle).  Resistance towards the flow will be experienced by SPINDLE & it will measured the resistance.  The measured values will be multiply with CERTAIN CONVERSION FACTOR for obtaining the  “ACTUAL VISCOSITY”.
  40. 40.  The CONVERSION FACTOR  is different according to the SIZE of the SPINDLE & SPIN SPEED.  The example of viscometer brand: “Brookefield Viscometer”  CALCULATION EXAMPLES:  Spindle = No. 2  Spindle speed (N) = 60 rpm  CF = 5  Reading from the device = 64 ∴ Viscosity (µapp:Non-newtonian OR µNewtonian) : 64 x 5 = 320 cp  Spindle : No. 3; Spindle speed (N) = 6 rpm; CF = 200  Reading from the device = 64.4 ∴ (µapp:Non-newtonian OR µNewtonian) : 64.4 x 200 = 12,880 cp
  41. 41. NEWTONIAN FLUIDS: ROTATIONAL VISCOMETER  For NEWTONIAN FLUID  it will give actual value of viscosity (µ).  n (flow behavior index) for NEWTONIAN = 1.0 and  Plot graph Log τ vs. Log γ.  PLOT FROM ORIGIN (0,0) ------ SLOPE = Actual viscosity (µ) SHEAR STRESS; τ (N.m2 ) α torque (N.m) ------- (1) SHEAR RATE; γ (s-1 ) α N (spindle speed, rpm) -- (2) (µ) = slope (τ) (γ)
  42. 42.  The SHEAR RATE (γ) at the surface of the spindle for NEWTONIAN FLUIDS is as follows with n = 1: 4πN/1 - (Rb/Rc)2 Where: (a) Rb - radius of the spindle, m (b) Rc - radius of the outer cylinder or container, m (c) ω - angular velocity of the spindle, rad/s ω = 2πN/60 ,when N is the RPM  The SHEAR STRESS (τ) at the wall of the spindle: τ = A/2πLRb 2 γ = Where: A = Torque (N.m) L = Height of the spindle, m Rb = radius of the spindle, m =
  43. 43. VARIOUS SPECIAL CASES CAN BE DERIVED FROM EQUATION (1): (1) VERY LARGE GAP (Rb/Rc < 0.1) = This is the case of a spindle immersed in a large beaker of test fluid. [ω = 2πN/60 ,when N is the RPM] Test fluid Rb Rc Spindle Container or cylinder γ = = 4πN/n
  44. 44. VARIOUS SPECIAL CASES CAN BE DERIVED FROM EQUATION (1): (1) VERY NARROW GAP (Rb/Rc > 0.99) = This is similar to flow parallel plates. Taking the shear rate at radius (Rb + Rc)/2. [ω = 2πN/60 ,when N is the RPM] Test fluid Rb RcSpindle Container or cylinder γ =
  45. 45. EXAMPLE (NON-NEWTONIAN): Rotational Type Viscometer Concentric cylinder (Wide-gap rotational viscometer with spindle): A wide-gap rotational viscometer with a spring constant of 7,187 dynes.cm are used to measure a viscosity (cps; %) of a tomato sauce. The outer radius (Rc), spindle radius(Rb)and height (L) of cylindrical spindle are 2.70 cm, 0.15 cm and 5.0 cm respectively. Determine the K and n? Calculate also the apparent viscosity (µapp) of tomato sauce in N.s/m2 at spindle speed of 35 rpm. Measurement values are given below: Spindle speed (N) Device reading; cps (% scale) 20 rpm 29 50 rpm 44 100 rpm 60
  46. 46. Spindle cylinder Test fluid - Tomato sauce Beaker “Wide-gap rotational viscometer with spindle cylinder” L RcRb
  47. 47. ANS: Rb/Rc = 0.15 cm/2.7 cm = 0.055 < 0.1 --- So, equation very large gap is used n value determination: (1) Sauce tomato = NON-NEWTONIAN FLUIDS (2) Firstly, we have to construct a graph of ---- Log torque vs. Log N (3) Convert: Device reading; cps (%)  Torque (dynes.cm) Torque (dynes.cm) = Device reading; cps (%) × spring constant (7,187 dynes.cm) e.g.: Torque (dynes.cm) = 29/100 × 7187 = 2,084.23 (4) Convert: {dynes.cm}  {N.m} by using the conversion factor: CONVERSION FACTOR: (1 dynes = 10-5 Newton) e.g.: 2084.23 dynes.cm 10-5 N 1 m = 0.000208 N.m 1 dynes 100 cm (5) Log (N) rpm ---- “see table”
  48. 48. (6) Plot the graph: N (rpm) Log N Cps (%) Torque (dynes.cm) Torque (N.m) Log Torque 20 rpm 1.30 29 2084.23 0.000208 -3.68 50 rpm 1.70 44 3162.28 0.000316 -3.50 100 rpm 2.00 60 4312.20 0.000431 -3.37
  49. 49. (7) So, we have the LINEAR equation: y = 0.451x - 4.268 -------- n value is the SLOPE ∴n = 0.451 (n < 1.0 = PSEUDOPLASTICS FLUIDS) K value determination: (1) Convert: Torque (N.m)  τ (N/m2 ) by using the formula: τ = A/2πLRb 2 ; A = Torque (N.m) e.g.: τ = 0.000208/[2 × 3.14 × 5.0/100 × (0.15/100)2 ] = 294.62 N/m2 (2) Log τ ------- ‘see table’ (3) Convert: N (rpm)  γ (s-1 ) by using the formula: γ = 4πN/n e.g.: γ = [4 × 3.14 × (20 rpm/60 s)]/0.451 = 9.26 s-1
  50. 50. (3) Plot the graph: N (rpm) γ (s-1 ) Log γ Torque (N.m) τ (N/m2 ) Log τ 20 rpm 9.26 0.97 0.000208 294.62 2.47 50 rpm 23.16 1.36 0.000316 447.60 2.65 100 rpm 46.32 1.67 0.000431 610.36 2.79
  51. 51. (4) So, we have the LINEAR equation: y = 0.451x + 2.033 ---------- intercept (c) = K ∴c = Log τ ---- Antilog (c): Antilog [2.033] = K = 107.89 N.s0.451 /m2 Apparent viscosity (µapp) in N.s/m2 at spindle speed of 35 rpm: Convert: N (rpm)  γ (s-1 ) by using the formula: γ = 4πN/n γ = [4 × 3.14 × (35 rpm/60 s)]/0.451 = 16.25 s-1 Log [16.25] = 1.21 Based on the equation: y = 0.451x +2.033; y = 0.451( 1.21) + 2.033 = 2.58 Antilog (y) = τ = 380.19 N/m2 ---- SO; ∴(µapp) = τ/ γ = 380.19/16.25 = 23.40 N.s/m2 OR ALTERNATIVELY using equation: µapp = τ/γ = K(γ)n-1 = 107.89[(16.25)0.451-1 ] µapp = 107.89 × 16.25-0.549 = 23.40 N.s/m2

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