Estimate the amount of time that each of the following suspensions will remain stabilized against sedimentation due to Brownian motion at room temperature (300K). (a) 200 nm diameter alumina ðr ¼ 3980 kg=m3Þ in water (typical ceramic processing
suspension);
(b) 200 nm diameter latex particlesðr ¼ 1060 kg=m3Þ in water (typical paint formulation);
(c) 150 nm diameter fat globulesðr ¼ 780 kg=m3Þ in water (homogenized milk); (d) 1000 nm diameter fat globulesðr ¼ 780 kg=m3Þ in water (non-homogenized milk). Solution
The time that the suspension remains stable against gravity can be approximated by equating the average distance moved by a particle due to Brownian motion to the distance settled due to gravity. This time is presented in Equation (5.13) as follows:
t ¼ 216kTm
pg2ðr p rfÞ
2x5
wherek ¼ 1:381 1023J=K; mwater¼ 0:001 Pa s; g ¼ 9:8 m=s2; rwater¼ 1000 kg=m3.
then t ¼216ð1:381 10 23J=KÞ300 Kð0:001 Pa sÞ pð9:8 m=s2Þ2ðr p 1000 kg=m3Þ 2x5 t ¼ 2:96 1024kg2s=m ðrpkg=m3 1000 kg=m3Þ 2x5 m5 WORKED EXAMPLES 145
(a) For the alumina suspension
t ¼ 2:96 1024kg2s=m
ð3980 kg=m3 1000 kg=m3Þ2ð200 109Þ5m5¼ 1042 s ¼ 17:4 min
(b) For the latex particles in paint
t ¼ 2:96 1024kg2s=m
ð1060 kg=m3 1000 kg=m3Þ2ð200 109Þ5
m5¼ 2:57 10
6s ¼ 30 days
(c) For the homogenized milk
t ¼ 2:96 1024kg2s=m
ð780 kg=m3 1000 kg=m3Þ2ð150 109Þ5
m5¼ 8:06 10
5s¼ 9:3 days
(d) For the non-homogenized milk
t ¼ 2:96 1024kg2s=m
ð780 kg=m3 1000 kg=m3Þ2ð1000 109Þ5
m5¼ 61 s
These characteristic times correspond best with the time for the first particle to settle out. The time for all the particles to settle out depends upon the height of the container. Nonetheless, one can understand the reasons why the alumina suspension needs to be mixed to keep all of the particles suspended for extended periods of time, why latex paints must be stirred if kept for a month and why milk is homogenized to prevent cream from forming while the milk is in your fridge for a week or two.
WORKED EXAMPLE 5.2
van der Waals and EDL Forces
Use the DLVO equation, FT¼ peeox2okekDðAx=24 D2Þ, to plot the total interparticle
force (FT) versus interparticle separation distance (D) for two alumina particles and for two oil droplets under the following conditions. The particles are spherical, 1mm in diameter and suspended in water that contains 0.01 M NaCl. Plot three conditions for each material: (a) at the IEP; (b) with ¼ 30 mV; and (c) with ¼ 60 mV. Comment on the differences in the behaviour of the two different materials. Which particles are easier to disperse and why?
Solution
Assume the surface potential equals the zeta potentialðo¼ Þ.
Calculate the inverse Debye length (k) with Equation (5.10).
k ¼ 3:29pffiffiffiffiffi½cðnm1Þ ¼ k ¼ 3:29p0:01ffiffiffiffiffiffiffiffiffiðnm1Þ ¼ 0:329 ðnm1Þ ¼ 3:29 108m1 The relative permittivity of water ðeÞ is 80 and the permittivity of free space ðeoÞ is
8:854 1012C2=J=m.
The diameter of the particles is 1 106m. Then FT¼ p80ð8:854 1012C2=J=mÞð1 106mÞ2oð3:29 108m1Þeð3:2910 8m1ÞD Að1 106mÞ 24D2 FT¼ ð7:32 107C2=J=mÞ2oeð3:2x10 8m1ÞD ð1 106mÞA 24D2 whereois in volts andD in meters.
From Table 5.1 the Hamaker constants (A) are: For alumina A ¼ 5:0 1020J
For oil A ¼ 0:4 1020J Then for alumina
FT¼ ð7:32 107C2=J=mÞ2oeð3:2910 8m1ÞD ð1 106mÞð5 1020JÞ 24D2 FT¼ ð7:32 107C2=J=mÞ2oeð3:2910 8m1ÞD ð5 1026J mÞ 24D2 and for oil
FT ¼ ð7:32 107C2=J=mÞ2oeð3:2910 8m1ÞD ð1 106mÞð0:4 1024D2 20JÞ FT ¼ ð7:32 107C2=J=mÞ2oeð3:2910 8 m1ÞDð0:4 1026J mÞ 24D2
These equations can be plotted by a standard data plotting software such as Excel, KG, or Sigmaplot, but first the units and typical values must be checked by hand to insure no errors are made while writing the equation to the spreadsheet. Unit analysis FT¼ C2=J=m V2em 1 m J m m2 where V¼ J=C FT ¼ C2=J=m J2=C2em 1 m J m m2 so FT¼ m 1J J m¼ J mand J¼ N m so FT is in newtons so the units are OK:
Figures 5W2.1 and 5W2.2 are the plotted results.
The difference between the two materials is that the Hamaker constant for the alumina is much greater than for the oil and thus the attraction between alumina particles is much stronger between alumina than between oil droplets. Thus, it is possible to just stabilize the oil droplets with 30 mV zeta potential, whereas when the alumina has 30 mV zeta potential, there is still attraction between the particles. Hence 60 mV is needed to stabilize the alumina to the same extent that 30 mV was able to stabilize oil droplets.
Figure 5W2.1 Force versus separation distance curves for alumina particles
Figure 5W2.2 Force versus separation distance curves for oil droplets
TEST YOURSELF
5.1 What is the typical size range of colloidal particles?
5.2 What two influences are more important for colloidal particles than body forces? 5.3 What is the influence of Brownian motion on a suspension of colloidal particles? 5.4 What is the relationship between surface forces and the potential energy between a
pair of particles?
5.5 Under what conditions can you expect van der Waals interactions to be attractive and under what conditions can you expect van der Waals interactions to be repulsive? Which set of conditions is more commonly encountered?
5.6 What are surface hydroxyl groups? What are surface ionization reactions? What is the isoelectric point?
5.7 What is the physical basis for the electrical double layer repulsion between similarly charged particles?
5.8 What is bridging flocculation? What type of polymers are most suitable to induce bridging attraction? What relative surface coverage of the polymer on the particles surface is typically optimum for flocculation?
5.9 What is steric repulsion? What type of polymers are most suitable to induce steric repulsion? What relative surface coverage of the polymer on the particles surface is typically optimum for steric stabilization?
5.10 What is meant by the DLVO theory?
5.11 Why are fine particles in typical atmospheric conditions cohesive? What would happen if all humidity were removed from the air?
5.12 What can be done to particles suspended in a liquid which typically cannot be done to particles suspended in a gas?
5.13 Explain why suspension of repulsive colloidal particles cannot be economically separated from liquid by sedimentation. What is the important parameter that is changed when the particles are flocculated that allows the particles to be ecomo- mically separated from the liquid by sedimentation?
5.14 How do interparticle interaction forces influence suspension consolidation?
5.15 Why does Einstein’s prediction of suspension rheology break down as solids
concentration increases above about 7 vol %?
5.16 What is the mechanism for shear thinning of hard sphere suspensions?
5.17 What happens to suspension viscosity as the volume fraction of solids is
increased?
5.18 How do repulsive forces influence suspension rheology? 5.19 How do attractive forces influence suspension rheology?
5.20 What three types of rheological behaviour are typical of attractive particle networks? 5.21 What is the mechanism for shear thinning in attractive particle networks?
5.22 Describe the influence of particle size on rheological properties of attractive particle networks.
5.23 How do you think the concepts presented in this chapter will be important for producing products from nano-particles?
EXERCISES
5.1 Colloidal particles may be either ‘dispersed’ or ‘aggregated’.
(a) What causes the difference between these two cases? Answer in terms of inter- particle 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.) 5.2
(a) What forces are important for colloidal particles? What forces are important for non- colloidal particles?
(b) What is the relationship between interparticle potential energy and interparticle force?
(c) Which three types of rheological behaviour are characteristic of suspensions of attractive particles?
5.3
(a) Describe the mechanism responsible for shear thinning behaviour observed for concentrated suspensions of micrometre 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.
5.4
(a) Explain why the permeability of the sediment from a flocculated mineral suspen- sion (less than 5mm) is greater than the permeability of the sediment of the same mineral suspension that settles while dispersed.
(b) Fine clay particles (approximately 0.15mm 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. 5.5 Calculate the effective volume fraction for a suspension of 150 nm silica particles at 40 vol % solids in a solution of 0.005 M NaCl.
(Answer: 0.473.)
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: 10000, 100000, 1 million and 10 million. You have two
0 20 40 60 80 100 1 0.8 0.6 0.4 0.2 0
Percent of surface covered by polymer.
Concentration of PAA (wt %) 1 x107
1 x106
1 x105
1 x104
Figure 5E6.1 Adsorption isotherms for PAA of various molecular weights
customers. The first customer is using 0.8mm alumina to produce ceramics. This customer would like to reduce the viscosity of the suspension of 40 vol % solids suspensions. The second customer is trying to remove 0.8mm alumina from wastewater. There is about 2 vol % alumina in the water and he wants to remove it by settling. What would you recommend to each customer? Consider if PAA is the right material to use, what molecular weight should be used and how much should be used. Figure 5E6.1 shows the adsorption isotherms for PAA with different molecular weights.
5.7
(a) Draw the typical log m versus log _g plot for suspensions of hard spheres of approximately micrometre sized particles at 40, 45, 50 and 55 vol % solids. (b) Draw the relative viscosityðms=mlÞ versus volume fraction curve for the low shear
viscosities of a typical hard sphere suspension.