SOLUTIONS TO CHAPTER 5: COLLOIDS AND FINE PARTICLES

SOLUTIONS TO CHAPTER 5: COLLOIDS AND FINE PARTICLES EXERCISE 5.1: Colloidal particles may be either dispersed or aggregated. (a) What causes the difference between these two cases? Answer in terms of interparticle interactions. (b) Name and describe at least two methods to create each type of colloidal dispersion. (c) Describe the differences in the behaviour of the two types of dispersions (including but not limited to rheological behaviour, settling rate, sediment bed properties.) SOLUTION TO EXERCISE 5.1: (a) Dispersed particles are those with repulsive forces between them. They remain in suspension as individuals. Aggregated particles have attractive forces between them. They exist as aggregates in suspension. (b) i) Colloidal particles can be dispersed by either 1) creating electrical double layer (EDL) repulsion by adjusting the ph of the suspension away from the isoelectric point and keeping the salt concentration low or 2) creating a steric repulsion by adsorbing a low moleculear weight polymer on to the surface in sufficient quantity to coat the surface with a layer of polymer. ii) Colloidal particles may be aggregated by either 1) adjusting the ph to the isoelectric point or adding high concentrations of salt in order to reduce the EDL repulsion so that van der Waals attraction dominates or 2) add a high molecular weight polymer which adsorbs to the particles and bridges between the particles holding them together. The concentration of polymer should not be so high as to completely coat all the particles surfaces. (c) Dispersed particles tend to produce suspensions with low viscosity relative to aggregated suspensions at the same volume fraction of solids. The sedimentation rate of dispersed colloidal particles is extremely low, because the mass of individual particles is so low that Brownian motion dominates and the dispersed colloids remain stable for extended periods. Once these particles eventually do settle or are consolidated by pressure to their equilibrium packing density, they form dense sediments with high packing fractions of solids. Aggregated particles have higher viscosity then dispersed particles at the same volume fraction. The behaviour is typically shear thinning. At volume fractions SOLUTIONS TO CHAPTER 5 EXERCISES: COLLOIDS AND FINE PARTICLES Page 5.1

above those where an attractive particle network is formed, aggregated particles have behaviour characteristic of solids such as viscoelasticity and yield stress. The sedimentation rates of aggregated colloidal suspensions is much greater than for dispersed suspensions and can be on order of between 10 and 100 m/s which can be useful for solid/liquid separation processes. The sediment beds formed from aggregated particles, pack to lower density than sediments from dispersed particles because of the attractive bond between particles. Low zeta potential (near iep) High salt (coagulation) Bridging polymers Attraction High zeta potential (away from iep) Low salt Polymer cushions (steric repulsion) Repulsion V V D Flocculated or Aggregated D Dispersed or Stabilised Aggregation Rapid Sedimentation Low Density Sediments (high moisture) High Viscosity and yield stress Individual Particles Slow Settling Dense Sediments (low moisture) Low Viscosity Figure 5.10 reproduced above is useful in illustrating the answer to Exercise 5.1 EXERCISE 5.2: (a)what forces are important for colloidal particles? What forces are important for non-colloidal particles? (b)what is the relationship between inter-particle potential energy and inter-particle force? (c) Which three types of rheological behaviour are characteristic of suspensions of attractive particles? SOLUTION TO EXERCISE 5.2: (a) Brownian and surface forces are important for colloidal particles because of the small mass and large surface area of colloidal particles. Body forces governed by F = ma are important for non-colloidal particles where the mass of the particle is large. Hydrodynamic forces can be important for both colloidal and non-colloidal particles. SOLUTIONS TO CHAPTER 5 EXERCISES: COLLOIDS AND FINE PARTICLES Page 5.2

(b) The force (F) is simply the negative of the derivative of the potential energy (V) dv versus particle surface to surface separation distance (D). F = dd (c) Shear thinning, yield stress and viscoelasticity. EXERCISE 5.3: (a) Describe the mechanism responsible for shear thinning behaviour observed for concentrated suspensions of micron sized hard sphere suspensions. (b) Consider the same suspension as in a) except instead of hard sphere interactions, the particles are interacting with a strong attraction such as when they are at their isoelectric point. In this case describe the mechanism for the shear thinning behaviour observed. (c) Draw a schematic plot (log-log) of the relative viscosity as a function of shear rate comparing the behaviour of the two suspensions described in a) and b). Be sure to indicate the relative magnitude of the low shear rate viscosites. (d) Consider two suspensions of particles. All factors are the same except for the particle shape. One suspension has spherical particles and the other rod-shaped particles like grains of rice. i) Which suspension will have a higher viscosity? ii) What two physical parameters does the shape of the particles influence that affect the suspension viscosity. SOLUTION TO EXERCISE 5.3: (a) Brownian motion dominates the behaviour of concentrated suspensions at rest and at low shear rate such that a random particle structure results that produces a viscosity dependent upon the particle volume fraction. As shown in Figure 5.14 there is typically a range of low shear rates over which the viscosity is independent of shear rate. This region is commonly referred to as the low shear rate Newtonian plateau. At high shear rates, hydrodynamic interactions are more significant than Brownian motion and preferred flow structures such as sheets and strings of particles develop as in Figure 5.14. The viscosity of suspensions with such preferred flow structures is much lower than the viscosity of the same volume fraction suspension with randomized structure. The preferred flow structure that minimizes the particleparticle hydrodynamic interaction develops naturally as the shear rate is increased. There is typically a range of high shear rates where the viscosity reaches a plateau. The shear thinning behaviour observed in concentrated hard sphere suspensions is due to the transition from the randomized structure of the low shear rate Newtonian SOLUTIONS TO CHAPTER 5 EXERCISES: COLLOIDS AND FINE PARTICLES Page 5.3

plateau to the fully developed flow structure of the high shear rate Newtonian plateau as illustrated in Figure 5.14. 100 random structure Suspension Viscosity (Pa.s) 10 1 shear thinning behaviour of typical concentrated suspension preferred flow structure 0.1 0.01 0.1 1 10 100 1000 Shear Rate (s -1 ) Figure 5.14. The transition from Brownian dominated random structures to preferred flow structures as shear rate is increased is the mechanism for the shear thinning behaviour of concentrated suspensions of hard sphere colloids. (b) The shear thinning of an attractive particle network is more pronounced than for hard sphere suspensions of the same particles at the same volume fraction and is caused by a different mechanism. The mechanism for shear thinning is illustrated in Figure 5.18. At rest, the particle network spans the entire volume of the container and resists flow. At low shear rates, the particle network is broken up in to large clusters that flow as units. A large amount of liquid is trapped within the particle clusters and the viscosity is high. As shear rate is increased and hydrodynamic forces overcome inter-particle attraction, the particle clusters are broken down into smaller and smaller flow units releasing more and more liquid and reducing the viscosity. At very high shear rates the particle network is completely broken down and the particles flow as individuals again almost as if they were non-interacting. SOLUTIONS TO CHAPTER 5 EXERCISES: COLLOIDS AND FINE PARTICLES Page 5.4

10000 Suspension Viscosity (Pa.s) 1000 100 10 1 shear thinning of attractive particle network shear thinning of hard sphere suspension 0.1 0.01 0.1 1 10 100 1000 Shear Rate (s -1 ) Figure 5.18. Comparison of typical shear thinning behaviour of attractive particle network to less pronounced shear thinning of hard sphere suspensions. The attractive particle network is broken down into smaller flow units as the shear rate is increased. (c) Figure 5.18 above is a comparison of the shear thinning behaviour observed for hard sphere and attractive particle suspensions. The low shear rate viscosity of the attractive particle network is higher than for the hard sphere suspension. (d) i) The rod shaped particles will produce suspensions with higher viscosities than the spherical particles at the same volume fraction and equivalent size. The reason is described below. ii) The shape of the particles influences the maximum packing density (φ max ) and the intrinsic viscosity ([η]). The maximum packing density decreases as particle shape deviates from spherical. The intrinsic viscosity increases as the particles become nonspherical. Examination of the Kreiger-Dougherty relationship μ * s φ μ 1 φmax = l [ η ] φ max indicates that both factors influence the value of the exponent. Typically the intrinsic viscosity increases at a faster rate than the maximum packing fraction decreases such that the product [η]φ max increases for non-spherical particles. This increase and the SOLUTIONS TO CHAPTER 5 EXERCISES: COLLOIDS AND FINE PARTICLES Page 5.5

reduced maximum packing factor result in suspension viscosities which increase as the particle shape deviates from sphericity. High aspect ratio particles such as needles and thin plates produce suspensions with very high viscosities. EXERCISE 5.4: (a) Explain why the permeability of the sediment from a flocculated mineral suspension (less than 5 microns) is greater than the permeability of the sediment of the same mineral suspension that settles while dispersed. (b) Fine clay particles (approximately 0.15 microns diameter) wash from a farmer s soil into a river due to rain. i) Explain why the particles will remain suspended and be carried down stream in the fast flowing fresh water. ii) Explain what happens to the clay when the river empties into the ocean. SOLUTION TO EXERCISE 5.4: (a) The flocculated suspension has particles which have attractive forces between them. They form low density sediments relative to the dispersed particles which pack to high densities due to the repulsive forces between particles. The higher void fraction in the flocculated sediments results in greater permeability. The relation ship between packed bed permeability and voidage is described in Chapter 6. (b) i) The clay particles develop a negative surface charge in the fresh (low salt) water producing repulsion between them so they tend not to aggregate. Brownian motion and the hydrodynamic forces and turbulence of the fast flowing water keep the particles suspended in the flow and wash the particles down stream. ii) When the river empties into the ocean, the high salt concentration of the ocean reduces the charge on the particles and van der Waals attraction dominates. The attraction causes the particles to aggregate. Relative to the fast flowing river, the ocean (away form the breaking waves) is calm. The mass of the aggregates is large enough to cause them to settle out onto the ocean floor. This is how river deltas are created. EXERCISE 5.5: (a) Calculate the effective volume fraction for a suspension of 150 nm silica particles at 40 volume percent solids in a solution of 0.005 M NaCl. SOLUTION TO EXERCISE 5.5: (a) The effective volume fraction of the particles is the fraction of the total volume of the suspension occupied by the volume of both the solid and the excluded volume: SOLUTIONS TO CHAPTER 5 EXERCISES: COLLOIDS AND FINE PARTICLES Page 5.6

φ eff = volume of solid + excluded volume total volume For particles stablised by electrical double layer repulsion, the excluded volume can be estimated using the range of the repulsion as estimated by the Debye length (κ -1 ). The inverse Debye length (κ) is a function of the salt concentration ([c]) for monovalent salts. κ = 3.29 [ c] nm -1 for 0.005 M NaCl, κ = 3.29 [0.005] = 0.233 nm -1 so the Debye length is 4.3 nm. The volume of one particle corresponds to φ = 0.40 so then the volume of one particle plus its excluded volume is equal to the effective volume fraction of the suspension. For the particle alone, π 3 π 3 6 Vol = x = (150nm) = 1.77x10 nm 6 6 3 For the particle and excluded volume, π 3 π 3 6 Vol = x = (150 + 4.3 + 4.3nm) = 2.09x10 nm 6 6 φ 1.77 Then = 2. 09, so that φ eff = 0.472 φ eff 3 EXERCISE 5.6: You are a sales engineer working for a polymer supply company selling poly acrylic acid (PAA). PAA is a water soluble anionic (negatively charged polymer) that comes in different molecular weights: 10,000, 100,000, 1 million and 10 million. You have two customers. The first customer is using 0.8 micron alumina to produce ceramics. This customer would like to reduce the viscosity of the suspension of 40 volume percent solids suspensions. The second customer is trying to remove 0.8 micron alumina from wastewater. There is about 2 volume percent alumina in the water and he wants to remove it by settling. What would you recommend to each customer? SOLUTIONS TO CHAPTER 5 EXERCISES: COLLOIDS AND FINE PARTICLES Page 5.7

Consider if PAA is the right material to use, what molecular weight should be used and how much should be used. Below in Figure 5E6.1 are the adsorption isotherms for each MW polymer. Percent of surface covered by polymer. 1 x10 7 1 x10 5 100 80 1 x10 6 1 x10 4 60 40 20 0 0 0.2 0.4 0.6 0.8 1 Concentration of PAA (wt %) Figure 5E6.1. adsorption isotherms for PAA of various molecular weights. SOLUTION TO EXERCISE 5.6: The first customer should use steric repulsion to reduce the viscosity of the alumina suspension for ceramic processing. PAA is a good polymer to use since it is negative and will adsorb to the positively charged surface of alumina over a wide range of ph. The low molecular weight polymer (1 x 10 4 Da) will be most effective in producing steric repulsion since it will not tend to bridge between particles. The entire surface should be covered with polymer to produce good steric repulsion. The adsorption isotherm indicates that about 1wt% of the polymer will be sufficient to cover the surface of the particles and produce good steric repulsion. The second customer should flocculate the particles by bridging attraction to cause them at aggregate and settle rapidly so they can be removed from the water. PAA is a good polymer to use since it is negative and will adsorb to the positively charged surface of alumina over a wide range of ph. The high molecular weight polymer (1 x 10 7 Da) will be most effective in producing bridging flocculation since the long chains will aid in bridging between particles. Good bridging typically occurs when half of the total surface area of the particles is covered by polymer so that the polymer can adsorb onto more than one particle s surface. The adsorption isotherm for the high molecular weight PAA indicate that about 0.15 wt % will be sufficient to cover about half of the particles surface area. EXERCISE 5.7: SOLUTIONS TO CHAPTER 5 EXERCISES: COLLOIDS AND FINE PARTICLES Page 5.8

(a) Draw the typical log μ verses log γ& plot for suspensions of hard spheres of approximately micron sized particles at 40, 45, 50, and 55 volume percent solids. (b) Draw the relative viscosity (μ S /μ L ) verses volume fraction curve for the low shear viscosities of a typical hard sphere suspension. SOLUTION TO EXERCISE 5.7: (a). 1000 Suspension Viscosity (Pa.s) 100 10 1 0.1 low shear rate Newtonian plateau shear thinning region shear thickening region high shear rate Newtonian plateau 55 v% 50 v% 45 v% < 40 v% 0.01 0.01 0.1 1 10 100 1000 10000 Shear Rate (s -1 ) Figure 5.16. Map of typical rheological behaviour of hard sphere suspensions as a function of shear rate for suspensions with volume fractions between about 40 and 55 volume percent solid particles. The dashed lines indicate the approximate location of the boundaries between Newtonian and non-newtonian behaviour. (b) 100000 5 Relative Viscosity 10000 1000 100 Relative Viscosity 4 3 2 1 0 0.1 0.2 0.3 0.4 Volume Fraction 10 1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Volume Fraction SOLUTIONS TO CHAPTER 5 EXERCISES: COLLOIDS AND FINE PARTICLES Page 5.9

Figure 5.15. Relative viscosity (μ s /μ l ) at low shear rate of hard sphere silica suspensions (circles). Quemada s model (solid line) with φ max = 0.631; Batchelor s model (dashed line) and Einstein s model (dotted line). Data from Jones et al., 1991. SOLUTIONS TO CHAPTER 5 EXERCISES: COLLOIDS AND FINE PARTICLES Page 5.10