CHAPTER 9
CONTACT EQUILIBRIUM PROCESSES (cont'd)

SUMMARY

1. The equilibrium concentrations of components of mixtures often differ across the boundary between one phase and another. Such boundaries occur between liquid and solid, liquid and vapour, between immiscible liquids, and between liquids or gases separated by membranes.

2. These differences can be used to effect separations by the enrichment of one phase relative to the other, by differential transfer of mass of particular components across the phase boundary.

3. Rates of mass transfer across the phase boundaries are controlled by the differences between actual concentrations and equilibrium concentrations, which constitute the mass transfer driving force, and by resistances which impede transfer. Therefore the rate of mass transfer is very generally determined by a driving force and by a mass-transfer coefficient.

dw/dt = kA(c - c_equilibrium)

4. Analysis of mass transfer contact equilibrium systems is carried out by comparing the equilibrium conditions to the actual conditions in the system; and using the difference, together with material conservation relationships that describe the movements within and between the phases, to follow the transfer of mass.

PROBLEMS

1. The composition of air is 23 % oxygen, 77% nitrogen by weight, and the Henry's Law constant for oxygen in water is 3.64 x 10⁴ atm mole fraction^-1 at 20°C. Calculate the solubility of oxygen in water (a) as the mole fraction and (b) as a percentage by weight.
[ (a) 0.056 x 10^{-4 mole fraction ; (b) 0.00103% ]}

2. If in the deodorizer of worked Example 9.5, relative flow rates of cream and steam are altered to 1:1, what will be the final concentration of the taint in the cream coming from a plant with three contact stages?
[ 0.05 ppm ]

3. Casein is to be washed, in a multistage system, by water. The casein curd has initially a water content of 60% and between stages it is drained on an inclined screen to 80% water (both on a wet basis). The initial lactose content of the casein is 4.5 % on a wet basis , and it is necessary to produce casein with a lactose content of less than 1 % on a dry basis. How many washing steps would be needed if the wet casein is washed with twice its own weight of fresh water in each step?
[ 3 washing steps reduce to 0.56% on a dry basis ]

4. Estimate the osmotic pressure of a solution of sucrose in water containing 20% by weight of sucrose. The density of this solution is 1081 kg m^-3, and the temperature is 20°C
[ 1540 kPa ]

5. In a six-step sugar-boiling crystallization process, the proportions of the sucrose present removed in the successive crystallizations are 66.7%, 60%, 60%, 50%, 50% and 33%. If the original sugar was associazted with 0.3% of its weight of non-sucrose solids, calculate (a0 the percentage of non-sucrose material in the final molasses and (b) the proportion of the original sugar that remains in the molasses. Assume that after each crystallization, all of the impurities remain with the mother liquor.
[ (a) 25% ; (b) 0.89% ]

6. For a particular ultrafiltration plant concentrating skim milk, for a concentration ratio of 7:1 of protein relative to lactose, the plant capacity is 570 kg m^-2 h^-1 of skim milk. Assume that this is the flow rate through the membrane. Estimate (a) the plant capacity at a concentration ratio of 2:1 and (b) the percentage of the water in the skim milk removed by the ultrafiltration.
[ (a) 1600 kgh^{-1 ; (b) 50% ]}


CHAPTER 10: MECHANICAL SEPARATIONS

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