ABE313 Topic 7 Rheological Properties PDF

Topic 7 RHEOLOGICAL PROPERTIES INTRODUCTION TO RHEOLOGY Rheology is the science that studies the deformation of materials including flow. Rheological data are required in product quality evaluation, engineering calculations, and process design. An understanding of flow behavior is necessary...

Topic 7 RHEOLOGICAL PROPERTIES INTRODUCTION TO RHEOLOGY Rheology is the science that studies the deformation of materials including flow. Rheological data are required in product quality evaluation, engineering calculations, and process design. An understanding of flow behavior is necessary to determine the size of the pump and pipe and the energy requirements. The rheological models obtained from the experimental measurements can be useful in design of food engineering processes if used together with momentum, energy, and mass balances. Effects of processing on rheological properties must be known for process control. Rheology can be classified into different groups as shown in Fig. 7.1. Figure 7. 1. Classification of rheology. FLOW OF MATERIAL Newton’s Law of Viscosity Consider a fluid between two large parallel plates of area 𝐴, separated by a very small distance 𝑌. The system is initially at rest but at time 𝑡 = 0, the lower plate is set in motion in the 𝑧-direction at a constant velocity 𝑉 by applying a force 𝐹 in the 𝑧-direction while the upper plate is kept stationary. At 𝑡 = 0, the velocity is zero everywhere except at the lower plate, which has a velocity 𝑉 (Fig. 7.2). Then, the velocity distribution starts to develop as a function of time. Finally, steady state is achieved and a linear velocity distribution is obtained. The velocity of the fluid is experimentally found to vary linearly from zero at the upper plate to velocity 𝑉 at the lower plate, corresponding to no-slip conditions at each plate.  Figure 7. 2. Velocity profile in steady flow for a Newtonian fluid between two parallel plates. Experimental results show that the force required to maintain the motion of the lower plate per unit area is proportional to the velocity gradient, and the proportionality constant, 𝜇, is the viscosity of the fluid: 𝐹 𝑉 =𝜇 7.1 𝐴 𝑌 The microscopic form of this equation is known as Newton’s law of viscosity: 𝑑v𝑧 𝜏𝑦𝑧 = −𝜇 = −𝜇𝛾̇𝑦𝑧 7.2 𝑑𝑦 where 𝜏𝑦𝑧 = shear stress (N/m2), 𝜇 = viscosity (Pa·s), 𝛾̇𝑦𝑧 = shear rate (1/s). Shear stress and shear rate have two subscripts: z represents the direction of force and 𝑦 represents the direction of normal to the surface on which the force is acting. A negative sign is introduced into the equation because the velocity gradient is negative, that is, velocity decreases in the direction of transfer of momentum. Example 7.1. Two parallel plates are 0.1 m apart. The bottom plate is stationary while the upper one is moving with a velocity 𝑉 (Fig E.7.1). The fluid between the plates is water, which has a viscosity of 1 cp. (a) Calculate the momentum flux necessary to maintain the top plate in motion at a velocity of 0.30 m/s. (b) If water is replaced with a fluid of viscosity 100 cp, and momentum flux remains constant, find the new velocity of the top plate. Solution: (a) 𝜇𝑤 = 1 cp= 1 × 10−3 Pa · s  Newton’s law of viscosity is used to determine shear stress: 𝑑v𝑧 𝜏𝑦𝑧 = −𝜇 7.2 𝑑𝑦 (0 − 0.3) m⁄s 𝜏𝑦𝑧 = −1 × 10−3 Pa · s = 0.003 Pa (0.1 − 0) m⁄s Figure E.7. 1. Illustration of Example 7.1. (b) 𝜇 = 100 cp = 0.1 Pa · s (0 − 𝑉) m⁄s 0.003 = −0.1 Pa · s ⇒ 𝑉 = 0.003 m⁄s (0.1 − 0) m⁄s Viscosity is defined as the resistance of a fluid to flow. The unit of dynamic viscosity is (Pa · s) in the SI system and poise (g/cm · s) in the CGS system. Viscosity varies with temperature. The difference in the effect of temperature on viscosity of liquids and gases is related to the difference in their molecular structure. Viscosity of most of the liquids decreases with increasing temperature. Theories have been proposed regarding the effect of temperature on viscosity of liquids. According to the Eyring theory, there are vacancies in liquid. Molecules continuously move into these vacancies. This process permits flow but requires energy. This activation energy is more readily available at higher temperatures and the fluid flows easily. The temperature effect on viscosity can be described by an Arrhenius type equation: 𝐸𝑎 𝜇 = 𝜇∞ exp ( ) 7.3 𝑅𝑇 where 𝐸𝑎 = activation energy (J/kg mol), 𝑅 = gas constant (8314.34 J/kg mol K), 𝑇 = absolute temperature (K), 𝜇∞ = constant (Pa · s). Liquid molecules are closely spaced with strong cohesive forces between them. The temperature dependence of viscosity can also be explained by cohesive forces between the molecules. As temperature increases, these cohesive forces between the molecules decrease and flow becomes freer. As a result, viscosities of liquids decrease as temperature increases. In liquids, the intermolecular (cohesive) forces play an important role. Viscosities of liquids show little dependence on density, molecular velocity, or mean free path. In most liquids, viscosity is constant up to a pressure of 10.134MPa, but at higher pressures viscosity increases as pressure increases. In gases, in contrast to liquids, molecules are widely spaced and intermolecular forces are negligible. In most gases, viscosity increases with increasing temperature, which can be expressed by the kinetic theory. The first analysis of viscosity by kinetic theory was made by Maxwell in 1860. Resistance to relative motion is the result of the exchange of momentum of gas molecules between adjacent layers. As molecules are transported by random motion from a region of low bulk velocity to mix with molecules in a higher velocity region (and vice versa), there is a momentum exchange that resists the relative motion between the layers. As temperature increases the random molecular activity increases, which corresponds to an increase in viscosity. Consider a pure gas composed of rigid, nonattracting spherical molecules of diameter 𝑑 and mass 𝑚 present in a concentration of 𝑁 molecules per unit volume. It is considered that 𝑁 is small enough so that the average distance between molecules is many times their diameter 𝑑. According to kinetic theory, it is assumed that an average molecule traverses a distance equal to the mean free path between impacts. If mean free path is 𝜆, one may consider the length of this path is the thickness of the layer of gas in which viscous action takes place. On the two sides of a gas layer having a thickness of 𝜆, the difference of streaming velocity in the gas is expressed 𝑑v 𝑑v as 𝜆 , for the velocity gradient normal to the motion of gas,. Molecules coming from upper to 𝑑𝑧 𝑑𝑧 𝑑v the lower layer carry an excess momentum of 𝑚𝜆 from the upper to the lower side. It can be 𝑑𝑧 said that on the average one third of the molecules are moving with paths that are up or down. Thus, the number of molecules of speed (𝑐̅) going up or down per unit area per second will be one third of 𝑁𝑐̅. The momentum transferred across this layer up and down by the molecules can be expressed as: 1 𝑑v 𝐹 = 𝑁𝑐̅𝑚𝜆 7.4 3 𝑑𝑧 From Newton’s law of viscosity: 𝑑v 𝐹=𝜇 7.5 𝑑𝑧 From Eqs. (7.4) and (7.5): 1 𝜇 = 𝑁𝑐̅𝑚𝜆 7.6 3 The mean free path is given by the following equation: 1 𝜆= 7.7 √2𝜋𝑑 2 𝑁 Inserting Eq. (7.7) into (7.6) gives: 𝑚𝑐̅ 𝜇= 7.8 3√2𝜋𝑑 2 According to kinetic theory, molecular velocities relative to fluid velocity have an average magnitude given by the following equation: 8𝑅𝑇 𝑐̅ = √ 7.9 𝜋𝑁𝐴 𝑚 where 𝑁𝐴 is the Avogadro number, 𝑚 is the mass of the molecule, 𝑅 is the gas constant, and 𝑇 is the absolute temperature. Thus, 2 𝜇= √𝑚𝐾𝑇 7.10 3𝜋 3⁄2 𝑑 2 where 𝐾 is the Boltzmann constant (𝐾 = 𝑅/𝑁𝐴 ). Equation (7.10) predicts that viscosity should increase with the square root of temperature. Experimental results showed that viscosity increased with temperature more rapidly. Gases have the lowest viscosity values. Viscosities of gases are constant up to 1 MPa pressure but increase as pressure increases above this level. Momentum diffusivity or kinematic viscosity, which has the same units as thermal diffusivity (𝛼 = 𝑘⁄𝜌 𝑐𝑝 ) in heat transfer and mass diffusivity (𝐷𝐴𝐵 ) in mass transfer, is defined to make the transport properties analogous. Its unit is m 2/s in SI and stoke (cm2/s) in CGS. It is the ratio of dynamic viscosity to density of fluid: 𝜇 𝑣= 7.11 𝜌 Viscous Fluids Viscous fluids tend to deform continuously under the effect of an applied stress. They can be categorized as Newtonian or non-Newtonian fluids. Newtonian Fluids. Fluids that follow Newton’s law of viscosity (Eq. 7.2) are called Newtonian fluids. The slope of the shear stress versus shear rate graph, which is viscosity, is constant and independent of shear rate in Newtonian fluids (Figs. 7.3 and 7.4). Gases; oils; water; and most liquids that contain more than 90% water such as tea, coffee, beer, carbonated beverages, fruit juices, and milk show Newtonian behavior.  𝑛 = flow behavior index. For shear thinning (pseudoplastic) fluids n < 1, For shear thickening fluids 𝑛 > 1. Newtonian fluids can be considered as a special case of this model in which 𝑛 = 1 and 𝑘 = μ. The slope of shear stress versus shear rate graph is not constant for non-Newtonian fluids (Fig. 7.3). For different shear rates, different viscosities are observed. Therefore, apparent viscosity or a consistency term is used for non-Newtonian fluids. The variation of apparent viscosities with shear rates for different types of non-Newtonian fluids is shown in Fig. 7.4. The symbol 𝜂 is often used to represent the apparent viscosity to distinguish it from a purely Newtonian viscosity, 𝜇. The ratio of shear stress to the corresponding shear rate is therefore called apparent viscosity at that shear rate: 𝜏 𝜂(𝛾̇ ) = 7.13 𝛾̇ The apparent viscosity and the Newtonian viscosity are identical for Newtonian fluids but apparent viscosity for a power law fluid is: 𝑘(𝛾̇ )𝑛 𝜂(𝛾̇ ) = = 𝑘(𝛾̇ )𝑛−1 7.14 𝛾̇ (a) Shear Thinning (Pseudoplastic) Fluids. In these types of fluids, as shear rate increases friction between layers decreases. Shearing causes entangled, long-chain molecules to straighten out and become aligned with the flow, reducing viscosity. A typical example for shear thinning fluids is paint. When paint is on the surface but brushing is not applied, its viscosity increases and prevents it from flowing under the action of gravity. When paint is applied to a surface by brushing which shears the paint, its viscosity decreases. Another example for a pseudoplastic fluid is the ink in a ballpoint pen. When the pen is not in use, the ink is so viscous that it does not flow. When we begin to write, the small ball on its point rolls and the turning of ball creates the shearing movement. As a result, the viscosity of ink decreases and it flows on the paper. Fruit and vegetable products such as applesauce, banana puree, and concentrated fruit juices are good examples of pseudoplastic fluids in food systems. Many researchers analyzed rheological properties of fruit and vegetable products. The consistency coefficient 𝑘 increased exponentially while the flow behavior index 𝑛 decreased slightly with concentration. Flow behavior index was close to 0.5 for pulpy products and close to 1.0 for clear juices. While the flow behavior index was assumed to be relatively constant with temperature, the effect of temperature on both apparent viscosity, 𝜂 and consistency coefficient of the power law model, 𝑘 was explained by an Arrhenius-type equation. The flow properties of peach pulp–granular activated carbon mixtures at temperatures of 15 to 40◦C and at granular activated carbon concentrations of 0.5 to 5.0 kg/m 3 exhibited a shear thinning behavior. The flow behavior index and the consistency coefficient of peach pulp–granular activated carbon mixtures were in the ranges of 0.328 to 0.512 and 2.17 to 6.18 Pa · s n, respectively. Both the consistency coefficient and the flow behavior index decreased with increasing temperature. Rheological behavior of foods may change depending on concentration. The rheological behavior of concentrated grape juice with a Brix value of 82.1 showed shear thinning behavior. However, diluted samples with a Brix value of 52.1 to 72.9 were found to be Newtonian. The rheological behavior of sesame paste–concentrated grape juice blends, which is a traditional food product in Turkish breakfasts, was studied at 35 to 65◦C and at 20% to 32% sesame paste concentration. All the blends showed shear thinning behavior with a flow behavior index of 0.70 to 0.85. The consistency coefficient was described by an Arrhenius-type equation. Researchers studied the rheological behavior of cake batter with different fat concentrations and emulsifier types and found that cake batter exhibited shear thinning behavior. The increase in fat content and addition of emulsifier caused a decrease in the apparent viscosity. Flow behavior index was found to be independent of composition of cake batter. (b) Shear Thickening Fluids. In these types of fluids, as shear rate increases, the internal friction and apparent viscosity increase. A person falling in a swamp tries to escape as soon as possible. However, as he tries to move in panic, sudden shearing is created and the more he tries to escape, the greater the force is required for his movement. Walking on wet sand on a beach is another example of shear thickening fluids. If a sand–water suspension has settled for some time, the void fraction occupied by water is minimal. Any shear will disturb close packing and the void fraction will increase. Water will no longer fill the space between the sand granules and the lack of lubrication will cause an increased resistance to flow. In food systems, corn starch suspension is an example of shear thickening fluids. A study showed the shear thickening phenomenon in unmodified starches (waxymaize, waxyrice, waxybarley, waxypotato, wheat, rice, maize) that had been dissolved and dispersed at 3.0% concentrations in 0.2 𝑁 NaOH. Waxy starches (maize, rice, barley, and potato) showed this behavior to a greater extent than did normal wheat, rice, or maize starches. The amylopectin component was responsible for shear thickening properties.  If the increase in viscosity is accompanied by volume expansion, shear thickening fluids are called dilatant fluids. All the dilatant fluids are shear thickening but not all shear thickening fluids are dilatant. Plastic Fluids Bingham Plastic Fluids. In these types of fluids, fluid remains rigid when the magnitude of shear stress is smaller than the yield stress (𝜏0 ) but flows like a Newtonian fluid when the shear stress exceeds 𝜏0. Toothpaste is a typical example of Bingham plastic fluid. It does not flow unless the tube is squeezed. In food systems, mayonnaise, tomato paste, and ketchup are examples of this type of fluid. Equation (7.15) shows the behavior of Bingham plastic fluids. 𝑑v𝑧 𝜏𝑦𝑧 = 𝜏0 + 𝑘 ( ) 7.15 𝑑𝑦 The apparent viscosities for Bingham plastic fluids can be determined by taking the ratio of shear stress to the corresponding shear rate: 𝜏0 + 𝑘(𝛾̇ ) 𝜏0 𝜂(𝛾̇ ) = = +𝑘 7.16 𝛾̇ 𝛾̇ Non-Bingham Plastic Fluids. In these types of fluids, a minimum shear stress known as yield stress must be exceeded before flow begins, as in the case of Bingham plastic fluids. However, the graph of shear stress versus shear rate is not linear. Fluids of this type are either shear thinning or shear thickening with yield stress. Fluids that obey the Herschel-Bulkley model are characterized by the presence of a yield stress term (𝜏0 ) in the power law equation: 𝑛 𝜏𝑦𝑧 = 𝜏0 + 𝑘(𝛾̇𝑦𝑧 ) 7.17 Minced fish paste and raisin paste obey Herschel-Bulkley model. Flow behavior of rice flour-based batter used in fried products was found to obey the Herschel-Bulkley model. The Casson model is expressed as: 0.5 0.5 (𝜏𝑦𝑧 ) = (𝜏0 )0.5 + 𝑘(𝛾̇𝑦𝑧 ) 7.18 Molten milk chocolate obeys the Casson model. When the effect of particle size distribution of nonfat solids on the flow characteristic of molten milk chocolate was investigated, Casson yield stress value was correlated with diameter and specific surface area of non-fat solids. Time Dependency When some fluids are subjected to a constant shear rate, they become thinner (or thicker) with time (Fig. 7.5). Fluids that exhibit decreasing shear stress and apparent viscosity with respect to time at a fixed shear rate are called thixotropic fluids (shear thinning with time). This phenomenon is  Figure 7. 7. Shear stress versus shear rate curve showing hysteresis. In rheopectic fluids (shear thickening with time), shear stress and apparent viscosity increase with time, that is, the structure builds up as shearing continues (Fig. 7.5). Bentonite–clay suspensions show this type of flow behavior. It is rarely observed in food systems. Starch–milk–sugar pastes showed a time dependent flow behavior. If the pasting process was done at 85 and 95◦C, starch–milk–sugar pastes exhibited a thixotropic behavior, while pastes processed at 75◦Cbehaved like a rheopectic fluid. It was noted that the thixotropy occurred at high shear stress (above 50 Pa), and the rheopexy occurred at low shear stress (below45 Pa). When soy protein was added to tomato juices, thixotropic behavior was observed at low shear rate but this was followed by a transition to rheopectic behavior at higher shear rates. Solution Viscosity In the case of solutions, emulsions, or suspensions, viscosity is often measured in comparative terms, that is, the viscosity of the solution, emulsion, or suspension is compared with the viscosity of a pure solvent. Solution viscosities are useful in understanding the behavior of some biopolymers including aqueous solutions of locust bean gum, guar gum, and carboxymethyl cellulose. Viscosities of pure solvents and suspensions can be measured and various values can be calculated from the resulting data. The relative viscosity, 𝜂rel is expressed as: 𝜂suspension 𝜂rel = = 1 + 𝑘𝑋𝑑𝑣 7.19 𝜂solvent where 𝑋𝑑𝑣 = volume fraction occupied by the dispersed phase,  𝑘 = constant. The specific viscosity, 𝜂sp is: 𝜂sp = 𝜂rel − 1 7.20 The reduced viscosity, 𝜂red is: 𝜂sp 𝜂red = 7.21 𝐶 where 𝐶 is the mass concentration of the solution in g/100 mL. Inherent viscosity, 𝜂inh is: ln 𝜂rel 𝜂inh = 7.22 𝐶 Intrinsic viscosity, 𝜂int , can be determined from dilute solution viscosity data: 𝜂sp 𝜂int = lim ( ) 7.23 𝐶→0 𝐶 In dilute solutions, polymer chains are separate and intrinsic viscosity of a polymer in solution depends only on dimensions of the polymer chain. The equations commonly used for determination of intrinsic viscosity of food gums are Huggins (7.24) and Kramer (7.25) equations. 𝜂sp 2 = 𝜂int + 𝑘1 𝜂int 𝐶 7.24 𝐶 ln 𝜂rel 2 = 𝜂int + 𝑘2 𝜂int 𝐶 7.25 𝐶 where 𝑘1 and 𝑘2 are the Huggins and Kramer constants, respectively and they are theoretically related as: 𝑘1 = 𝑘2 + 0.5 7.26 VISCOSITY MEASUREMENT The most commonly used viscosity measurement devices are capillary flow viscometers, orifice type viscometers, falling ball viscometers, and rotational viscometers. Capillary Flow Viscometers Capillary flow viscometers are generally in the form of a U-tube. These types of viscometers are very simple, inexpensive, and suitable for low-viscosity fluids. There are different designs of capillary viscometers. A typical design of capillary viscometer is shown in Fig. 7.8. In capillary flow viscometers, the time for a standard volume of fluid to pass through a known length of capillary tubing is measured. The flow rate of material due to a known pressure gradient is determined. The driving pressure is usually generated by the force of gravity acting on a column of the liquid although it can be generated by the application of compressed air or by mechanical means. Gravity-operated glass capillaries are suitable only for Newtonian fluids having viscosities in the range of 0.4 to 20,000 mPa·s. To measure the viscosities of more viscous fluids, external pressure may be applied. For non-Newtonian fluids, this device is less suitable because the measurement cannot be done at a constant shear rate. Capillary viscometers can be used only for non-Newtonian fluids if the applied external pressure is more significant than static pressure. Figure 7. 8. Cannon-Fenske capillary flow viscometer. The diameter of a capillary viscometer should be small enough to provide laminar flow. Capillary viscometers are calibrated with Newtonian oils of known viscosities since the flow rate depends on the capillary radius, which is difficult to measure. For the viscosity measurement, the viscometer is accurately filled with an accurately known volume of test fluid and the apparatus is immersed in a constant temperature bath until equilibrium is reached. Then, fluid is sucked up from the other limb through the capillary tube until it is above the marked level (A) (Fig. 7.8). Then, suction is removed and fluid flows through the capillary tube under the influence of gravity or the induced pressure head and the time for the fluid to flow from mark A to B is recorded. This time is a direct measure of the kinematic viscosity since it depends on both viscosity and density of fluid. This can be written as: 𝑣 = 𝐶𝑡 7.27 where 𝐶 is the calibration constant. Assuming that the flow is laminar, fluid is incompressible, velocity of the fluid at the wall is zero (no-slip condition), and end effects are negligible, making a macroscopic force balance for a fluid flowing through a horizontal cylindrical tube of length (𝐿) and inner radius (𝑟), the following equation is obtained: ∆𝑃𝜋𝑟 2 = 𝜏2𝜋𝑟𝐿 7.28 where ∆𝑃 is the pressure drop causing flow and 𝜏 is the shear stress resisting flow. This equation can be solved for shear stress: ∆𝑃𝑟 𝜏= 7.29 2𝐿 For a Newtonian fluid, both shear stress and shear rate vary linearly from zero at the center (𝑟 = 0) of the capillary to a maximum at the wall (𝑟 = 𝑅). For a Newtonian fluid, this results in the parabolic velocity profile. Then, the shear stress on the fluid at the wall (𝜏𝑤 ) is related to the pressure drop along the length of the tube: ∆𝑃𝑅 𝜏𝑤 = 7.30 2𝐿 The flow in capillary viscometers is described by the Hagen Poiseuille equation: 8𝜇v𝐿 ∆𝑃 = 7.31 𝑅2 Substituting Eq. (7.31) into Eq. (7.30), shear stress can also be expressed as: 4v 𝜏𝑤 = 𝜇 7.32 𝑅 Then, shear rate at the wall (𝛾̇𝑤 ) for a Newtonian fluid is given by: 4v 4𝑄 𝛾̇𝑤 = = 7.33 𝑅 𝜋𝑅 3 where 𝑄 is the volumetric flow rate. Newton’s law of viscosity can be written in terms of pressure gradient and volumetric flow rate as: ∆𝑃𝑅 4𝑄 = 𝜇 ( 3) 7.34 2𝐿 𝜋𝑅 and viscosity of the fluid can be determined from the pressure drop and volumetric flow rate or velocity data. For non-Newtonian fluids, the relation between shear stress and shear rate has to be known to derive these equations. Compared to the parabolic profile for a Newtonian fluid, the profile for a shear thinning fluid is more blunted. The shear rate at the wall can be determined from the Rabinowitsch-Mooney equation: 3𝑄 𝑑(𝑄 ⁄𝜋𝑅 3 ) 𝛾̇𝑤 = ( 3 ) + 𝜏𝑤 [ ] 7.35 𝜋𝑅 𝑑𝜏𝑤 This equation can also be expressed in terms of the apparent wall shear rate, 𝛾̇𝑎𝑝𝑝 = 4𝑄/𝜋𝑅 3 :  3 𝜏𝑤 𝑑𝛾̇app 𝛾̇𝑤 = ( ) 𝛾̇app + ( ) ( ) 7.36 4 4 𝑑𝜏𝑤 Equation (7.36) can also be written as: 3 1 𝑑(ln 𝛾̇app ) 𝛾̇𝑤 = [( ) + ( ) ( )] 𝛾̇app 7.37 4 4 𝑑(ln 𝜏𝑤 ) Equation (7.37) can be written in the following simplified form: 3𝑛′ + 1 𝛾̇𝑤 = ( ) 𝛾̇app 7.38 4𝑛′ where 𝑛’ is the point slope of the ln 𝜏𝑤 versus ln 𝛾̇app. That is: 𝑑(ln 𝜏𝑤 ) 𝑛’ = 7.39 𝑑(ln 𝛾̇app ) If the fluid behaves as a power law fluid, the slope of the derivative is a straight line and 𝑛’ = 𝑛. Different shear rates can be achieved by varying the flow rate for a single capillary section or by using several capillary sections of different diameters in series. Example 7.2. The pressure drop versus volumetric flow rate data is obtained for chocolate melt using a capillary viscometer with a pipe diameter of 1 cm and a length of 60 cm (Table E.7.2). (a) Show that chocolate melt is not a Newtonian fluid. (b) Determine the rheological model constants of the power law, Herschel-Bulkley, and Casson models for the given data. (c) Which model best represents the rheological behavior of the chocolate melt? Table E.7.2 Pressure Drop Versus Volumetric Flow Rate Data for Chocolate Melt in Capillary Viscometer Pressure Drop (Pa) Flow Rate (cm3/s) 3840 0.01 4646 0.06 5762 0.13 6742 0.24 7798 0.37 10,454 0.72 11,760 0.94 Solution: (a) Newtonian fluids follow Newton’s law of viscosity (Eq. 7.2): 𝑑v𝑧 𝜏𝑦𝑧 = −𝜇 7.2 𝑑𝑦 Using pressure drop data, the shear stress at the wall is calculated from Eq. (7.30): ∆𝑃𝑅 𝜏𝑤 = 7.30 2𝐿  The shear rate is calculated from Eq. (7.33) for different flow rates: 4𝑄 𝛾̇𝑤 = 7.33 𝜋𝑅 3 The plot of shear stress versus shear rate is shown in Fig. E.7.2.1. For a fluid to be Newtonian, variation of shear stress versus shear rate should be linear and the intercept must be zero. Since this is not the case, it can be concluded that chocolate melt is not a Newtonian fluid. Figure E.7.2.1 Shear stress versus shear rate plot for a chocolate melt. (b) The power law equation is: 𝑑v𝑧 𝑛 𝑛 𝜏𝑦𝑧 = 𝑘 ( ) = 𝑘(𝛾̇𝑦𝑧 ) 7.12 𝑑𝑦 The power law equation can be linearized by taking the natural logarithm of both sides to determine the flow behavior index (𝑛) and the consistency coefficient (𝑘). ln 𝜏𝑦𝑧 = ln 𝑘 + 𝑛 ln 𝛾̇𝑦𝑧 A logarithmic plot of shear stress versus shear rate (ln 𝜏𝑦𝑧 versus ln 𝛾̇𝑦𝑧 ) yields a straight line with a slope of 𝑛 and intercept of ln 𝑘 (Fig. E.7.2.2). ln 𝜏𝑤 = 3.173 + 0.273 ln 𝛾̇𝑤 Being the flow behavior index n to be different than 1, also show that the fluid is not Newtonian. For a power law fluid: 𝑛’ = 𝑛 3𝑛′ + 1 𝛾̇𝑤 =( ) 𝛾̇app 7.38 4𝑛′  ln 𝜏𝑤 = 0.273 ln 𝛾̇w + 0.034 The slope of the model gives 𝑛, which is 0.273. From the intercept, which is ln 𝑘 = 3.034, the value of 𝑘 is calculated as 20.78 Pa · sn. The coefficient of determination (𝑟 2 ) for the model is 0.953. Thus, the power law expression is: 𝜏𝑤 = 20.78𝛾̇𝑤0.273 The Herschel-Bulkley expression is given in Eq. (7.17): 𝜏𝑤 = 𝜏0 + 𝑘(𝛾̇𝑤 )𝑛 7.17 To determine the value for 𝜏0 , shear stress at the wall (𝜏𝑤 ) is plotted with respect to shear rate at the wall (𝛾̇𝑤 ) and 𝜏0 is found to be 13 Pa by extrapolation. Figure E.7.2.4. Herschel Bulkley model for a chocolate melt. To find the model constants Herschel-Bulkley expression is linearized as: ln(𝜏𝑤 − 𝜏0 ) = ln 𝑘 + 𝑛 ln 𝛾̇𝑤 The plot of ln(𝜏𝑤 − 𝜏0 ) versus ln 𝛾̇𝑤 is shown in Fig. E.7.2.4. When linear regression is used, the equation of ln(𝜏𝑤 − 𝜏0 ) = 1.897 + 0.601 ln 𝛾̇𝑤 is determined with a coefficient of determination (𝑟 2 ) of 0.999. From the intercept, which is ln 𝑘 = 1.897, the value of 𝑘 is calculated as 6.66 Pa · sn. Thus, the Herschel-Bulkley expression is: 𝜏𝑤 = 13 + 6.66𝛾̇𝑤0.601 The Casson model is given in Eq. (7.18):  (𝜏𝑤 )0.5 = (𝜏0 )0.5 + 𝑘(𝛾̇𝑤 )0.5 7.18 The plot of (𝜏𝑤 )0.5 versus (𝛾̇𝑤 )0.5 is given in Fig. E.7.2.5 and linear regression was performed. Figure E.7.2.5. Casson model for a chocolate melt. 0.5 = 3.570 + 0.861𝛾̇ 0.5 (𝑟 2 = 1.000) As a result, 𝜏𝑤 𝑤 Then, yield stress is; 𝜏0 = (3.570)1/0.5 = 12.74 Pa Thus, the Casson expression is: 0.5 = 12.741/0.5 + 0.861𝛾̇ 0.5 𝜏𝑤 𝑤 (c) The Casson model is the best model defining the flow behavior of chocolate melt since it has the highest coefficient of determination (𝑟 2 = 1.000) compared to others. Example 7.3. Viscosity of refined sunflower oil was measured at different temperatures by a glass capillary viscometer. Table E.7.3.1 shows the density values and the timing results at different temperatures of sunflower oil. As the reference liquid for calibration of the viscometer, a 50% sucrose solution was used. The density and the viscosity of the reference liquid are known to be 1227.4 kg/m3 and 0.0126 Pa · s, respectively at 25◦C. It took the reference liquid 100 s to fall from one mark to the other of capillary viscometer. Show that the temperature effect on the viscosity of sunflower oil can be expressed by an Arrhenius-type equation. Determine the activation energy and Arrhenius equation constant. Solution: Kinematic viscosity is correlated with time as shown in Eq. (7.27):  𝑣 = 𝐶𝑡 7.27 where 𝐶 is the calibration constant. Then, the following equation can be written: 𝜇 𝜌∙𝑡 = 𝜇ref 𝜌ref 𝑡ref where 𝜇ref and 𝜌ref are the viscosity and density of reference liquid, respectively. Inserting the data given in the question for reference fluid, viscosities of sunflower oil at different temperatures are calculated (Table E.7.3.2). Table E.7.3.1 Density of Sunflower Oil and Timing Results in Capillary Viscometer at Different Temperatures Temperature (◦C) Time (s) Density (kg/m3) 25 521 916 35 361 899 45 262 883 55 198 867 Table E.7.3.2 Viscosity of Sunflower Oil at Different Temperatures Temperature (◦C) 𝜇 (Pa·s) 25 0.049 35 0.033 45 0.024 55 0.018 The Arrhenius type equation is: 𝐸𝑎 𝜇 = 𝜇∞ exp ( ) 7.3 𝑅𝑇 Taking the natural logarithm of both sides: 𝐸𝑎 ln 𝜇 = ln 𝜇∞ + ( ) 𝑅𝑇 The plot of (ln 𝜇) versus (1/𝑇) is shown in Fig. E.7.3.1. The model equation is found as: 3332.4 ln 𝜇 = −14.209 + 𝑇 From the intercept of this equation, the Arrhenius constant is found to be 6.75 × 10 −7 Ns/m2. The slope of the equation 𝐸𝑎/𝑅 is 3332.4. The gas constant, 𝑅, is 8.314 × 10−3 kJ/mole · K. Thus, activation energy, 𝐸𝑎, is calculated as 27.705 kJ/mole.  Figure E.7.3.1. Arrhenius plot for variation of viscosity with temperature. Orifice Type Viscometers In orifice type viscometers, the time for a standard volume of fluid to flow through an orifice is measured. They are used for Newtonian or near-Newtonian fluids when extreme accuracy is not required. In the food industry, the most commonly used one is a dipping type Zahn viscometer that consists of a 44-mL capacity stainless steel cup with a handle and with a calibrated circular hole in the bottom. The cup is filled by dipping it into the fluid and withdrawing it. The time from the start of withdrawing to the first break occurring in the issuing stream is recorded. Falling Ball Viscometers These types of viscometers involve a vertical tube where a ball is allowed to fall under the influence of gravity. It operates on the principle of measuring the time for a ball to fall through a liquid under the influence of gravity. When the ball falls through the fluid, it is subjected to gravitational force, drag force, and buoyancy force (Fig. 7.9). Making a force balance: Net force (𝐹Net ) = Gravitational force (𝐹G)− Buoyancy force (𝐹B )− Drag force (𝐹D) 𝜋𝐷𝑝3 𝜌𝑝 𝑑v 𝜋𝐷𝑝3 𝜌𝑝 𝑔 𝜋𝐷𝑝3 𝜌𝑓 𝑔 𝑐𝐷 𝜋𝐷𝑝2 𝜌𝑓 v 2 = − − 7.40 6 dt 6 6 8 where 𝐷𝑝 = diameter of the ball (m),  𝜌𝑝 = density of the ball (kg/m3), 𝜌𝑓 = density of the fluid (kg/m3), 𝑐𝐷 = drag coefficient, v = velocity of the ball (m/s). Figure 7. 9. Forces acting on a ball in falling ball viscometer. When equilibrium is attained, the upward and downward forces are balanced and the ball moves at a constant velocity. That is, the falling ball reaches a terminal velocity (vt ) when the acceleration due to the force of gravity is exactly compensated by the friction of the fluid on the ball. 𝑑v =0 7.41 𝑑𝑡 In the Stoke’s region, the drag coefficient is: 24 𝑐𝐷 = 7.42 Re Substituting Eqs. (7.41) and (7.42) into Eq. (7.40) gives: 𝜋𝐷𝑝3 𝜌𝑝 𝑔 𝜋𝐷𝑝3 𝜌𝑓 𝑔 6𝜋𝐷𝑝 𝜇v𝑡 = + 7.43 6 6 2 𝜋𝐷2𝑝 (𝜌𝑝 − 𝜌𝑓 ) 𝑔 ⇒𝜇= 7.44 18v𝑡 If the terminal velocity of the ball is calculated, it is possible to determine the dynamic viscosity of the fluid. Falling ball viscometers are more suitable for viscous fluids where the terminal velocity is low. Stoke’s law applies when the diameter of the ball is so much smaller than the diameter of the tube through which it is falling. Thus, there is no effect of the wall on the rate of fall of the ball. If the tube diameter is 10 times the ball diameter, wall effects can be neglected. The larger the ball, the faster it falls. Therefore, it is necessary to select the diameter of ball small enough to fall at a rate that can be measured with some degree of accuracy. This method is not suitable for opaque fluids.  The rising bubble viscometer operates on the same principle as the falling ball viscometer. In this case, a bubble of air is allowed to rise through a column of liquid. Rising time for a certain distance is correlated to viscosity. Example 7.4. To determine the viscosity of sunflower oil, a falling ball viscometer was used. Viscometer has a tube length of 10 cm and its ball has a diameter of 0.68 mm. Oil and the ball have densities of 921 kg/m3 and 2420 kg/m3, respectively. If it takes 44.5 s for the ball to fall from the top of the tube, calculate the viscosity of the oil. Solution: Terminal velocity is: 𝐿 v𝑡 = 𝑡 0.1 m = 44.5 s = 0.0022 m/s Then, viscosity can be calculated using Eq. (7.44): 𝜋𝐷2𝑝 (𝜌𝑝 − 𝜌𝑓 ) 𝑔 ⇒𝜇= 7.44 18v𝑡 −3 2 (0.68 × 10 m) (2420 kg/m3 − 921 kg/m3 )(9.81 m/s2 ) ⇒𝜇= 18(0.0022 m/s) Rotational Viscometers In rotational viscometers, the sample is sheared between the two parts of the measuring device by means of rotation. In agitation, the shear rate is proportional to the rotational speed. It is possible to measure the shear stress as the shear rate is changed. In addition, a sample can be sheared for as long as desired. Therefore, rotational viscometers are the best for characterization of non-Newtonian and time-dependent behavior. There are different forms of these viscometers, as described in the following sections. Concentric Cylinder (Coaxial Rotational) Viscometers. This type of viscometer consists of two annular cylinders with a narrow gap between them (Fig. 7.10). The fluid to be measured is placed in the gap. Either the inner (Searle system) or the outer (Couette system) cylinder is rotated. Although the Searle system is more common, the word Couette is mostly used to refer to any kind of concentric cylinder system. For both Couette and Searle systems, the equations relating rotation to torsion (𝑀) are the same. By changing the shear rate or shear stress, it is possible to obtain viscosity measurements over a range of shearing conditions on the same sample. It can be used for both Newtonian and non-Newtonian foods.  Figure 7. 10. Concentric cylinder viscometer. The following assumptions should be made in developing the mathematical relationships 1. Flow is laminar and steady. 2. Radial and axial velocity components are zero. 3. The test fluid is incompressible. 4. The temperature is constant. 5. End effects are negligible. 6. There is no slip at the wall of the instrument. Consider a test fluid placed in Searle type viscometer. When the inner cylinder rotates at a constant angular velocity (Ω) and the outer cylinder is stationary, the instrument measures the torque (𝑀) required to maintain this constant angular velocity of the inner cylinder. The opposing torque comes from the shear stress exerted on the inner cylinder by the fluid. Making a force balance: 𝑀 = 2𝜋𝑟ℎ𝑟𝜏 = 2𝜋ℎ𝑟 2 𝜏 7.45 where 𝑟 is any location in the fluid and ℎ is the height of the cylinder. Solving Eq. (7.45) for the shear stress: 𝑀 𝜏= 7.46 2𝜋ℎ𝑟 2 This equation shows that the shear stress is not constant over the gap between the two concentric cylinders but decreases in moving from the inner cylinder of radius, 𝑅𝑖 , to the outer cylinder of radius, 𝑅0. The shear stress at the inner cylinder can be written as:  𝑀 𝜏𝑖 = 7.47 2𝜋ℎ𝑅𝑖2 To determine the shear rate, the linear velocity (v) is expressed in terms of angular velocity (𝜔) at 𝑟 : 𝑑v 𝑑𝜔 𝛾̇ = − = −𝑟 7.48 𝑑𝑟 𝑑𝑟 The shear rate is a function of shear stress, and the functional relationship between shear stress and shear rate is shown by: 𝑑𝜔 𝛾̇ = −𝑟 = 𝑓(𝜏) 7.49 𝑑𝑟 An expression for the differential of the angular velocity yields: 𝑑𝑟 𝑑𝜔 = − 𝑓(𝜏) 7.50 𝑟 An expression for 𝑟 can be determined from Eq. (7.45): 𝑀 1/2 𝑟=( ) 7.51 2𝜋ℎ𝜏 Equation (7.51) is differentiated with respect to 𝜏 and substituted into Eq. (7.50) after substituting the value of torque from Eq. (7.45): 1 𝑑𝜏 𝑑𝜔 = 𝑓(𝜏) 7.52 2 𝜏 Integrating Eq. (7.52) over the fluid in the annulus gives the general expression for the angular velocity of the inner cylinder as a function of the shear stress in the gap: 0 𝜏0 1 𝑑𝜏 ∫ 𝑑𝜔 = ∫ 𝑓(𝜏) 7.53 2 𝜏 Ω 𝜏𝑖 However, the solution of Eq. (7.53) depends on 𝑓(𝜏), which depends on the behavior of the fluid. If the fluid is Newtonian: 𝜏 𝛾̇ = 𝑓(𝜏) = 7.54 𝜇 Substituting Eq. (7.54) into Eq. (7.53) gives: 𝜏0 1 1 Ω=− ∫ 𝑑𝜏 = (𝜏 − 𝜏0 ) 7.55 2𝜇 2𝜇 𝑖 𝜏𝑖 Substituting Eq. (7.46) into Eq. (7.55) and rearrangement gives Margules equation: 𝑀 1 1 Ω= ( 2 − 2) 7.56 4𝜋𝜇ℎ 𝑅𝑖 𝑅0 For a power law fluid:  𝜏 1/𝑛 𝛾̇ = 𝑓(𝜏) = ( ) 7.57 𝑘 Substituting Eq. (7.57) into Eq. (7.53) gives: 𝜏0 1 𝜏 1/𝑛 𝑑𝜏 1 Ω=− ∫( ) = 1/𝑛 [(𝜏𝑖 )1/𝑛 − (𝜏0 )1/𝑛 ] 7.58 2 𝑘 𝜏 2𝑘 𝜏𝑖 Substituting Eq. (7.46) into Eq. (7.58), an expression for the power law fluid is obtained: 1/𝑛 𝑛 𝑀 𝑅𝑖 1/𝑛 Ω = 1/𝑛 ( ) [1 − ( ) ] 7.59 2𝑘 2𝜋ℎ𝑅𝑖2 𝑅𝑜 When studying liquid foods, the simple shear, Newtonian, or power law approximations are often used. (a) Simple Shear Approximation. If there is a very small gap between the cylinders compared to the radius, shear stress can be taken as constant. Assuming a uniform shear rate across the gap: Ω𝑅𝑖 Ω 𝛾̇ 𝑖 = = 7.60 𝑅𝑜 − 𝑅𝑖 𝛼 − 1 𝑅𝑜 where 𝛼 = ⁄𝑅 𝑖 When calculating shear rates with Eq. (7.60), a corresponding average shear stress should be used: 1 𝑀(1 + 𝛼 2 ) 𝜏ave = (𝜏𝑖 + 𝜏0 ) = 7.61 2 4𝜋ℎ𝑅𝑜2 (b) Newtonian Approximation. Shear stress at the inner cylinder can be calculated using Eq. (7.47). An equation for the shear rate at the inner cylinder can be derived by substituting Eq. (7.45) into the Margules Eq. (7.56) and expressing shear stress in terms of shear rate using Newton’s law of viscosity: 𝛼2 𝛾̇ 𝑖 = 2Ω ( ) 7.62 𝛼2 − 1 (c) Power Law Approximation. Shear stress at the inner cylinder can be calculated using Eq. (7.47). An equation for the shear rate at the inner cylinder can be derived by substituting Eq. (7.45) into Eq. (7.59) and expressing shear stress in terms of shear rate using power law equation: 2Ω 𝛼 2/𝑛 𝛾̇ 𝑖 = ( ) ( 2/𝑛 ) 7.63 n 𝛼 −1 where the flow behavior index 𝑛 is: 𝑑(ln 𝜏𝑖 ) 𝑑(ln 𝑀) 𝑛= = 7.64 𝑑(ln Ω) 𝑑(ln Ω)  Cone and Plate Viscometers. The operating principle for cone and plate viscometers is similar to that for concentric cylinder viscometers. The system consists of a circular plate and a cone with radius R with its axis perpendicular to the plate and its vertex in the plane of the surface of the plate (Fig. 7.11). Usually, the cone is rotated at a known angular velocity (Ω). The fluid is placed in the gap between the cone and plate and transmits torque to the plate. If the angle 𝜃 between the cone and plate is small (