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Legend: A Cation © Anion a Anlon-permeable membrane C - Cation-permeable membrane

FIGURE 8.1 Cation filtering membrane (a); the electrodialysis process (b).

ions in the membranes. As shown in the figure, two compartments become "cleaned" of ions and one compartment (the middle) becomes "dirty" of ions. The two compartments are diluting compartments; the middle compartment is a concentrating compartment. The water in the diluting compartments is withdrawn as the product water, and is the filtered water. The concentrated solution in the concentrating compartment is discharged to waste.

8.1.1 Power Requirement of Electrodialysis Units

The filtering membranes in Figure 8.1b are arranged as CACA from left to right, where C stands for cation and A stands for anion. In compartments CA, the water is deionized, while in compartment AC, the water is not deionized. The number of deionizing compartments is equal to two. Also, note that the membranes are always arranged in pairs (i.e., cation membrane C is always paired with anion membrane A). Thus, the number of membranes in a unit is always even. If the number of membranes is increased from four to six, the number of deionizing compartments will increase from two to three; if increased from six to eight, the number of deionizing membranes will increase from three to four; and so on. Thus, if m is the number of membranes in a unit, the number of deionizing compartments is equal to m/2.

As shown in the figure, a deionizing compartment pairs with a concentrating compartment in both directions; this pairing forms a cell. For example, deionizing compartment CA pairs with concentrating compartment AC in the left direction and with the concentrating compartment AC in the right direction of CA. In this paring (in both directions), however, only one cell is formed equal to the one deionizing compartment. Thus, the number of cells formed in an electrodialysis unit can be determined by counting only the number of deionizing compartments. The number of deionizing compartments in a unit is m/2, so the number of cells in a unit is also equal to m/2.

Because one equivalent of a substance is equal to one equivalent of electricity, in electrodialysis calculations, concentrations are conveniently expressed in terms of equivalents per unit volume. Let the flow to the electrodialysis unit be Qo. The flow per deionizing compartment or cell is then equal to Qo /(m/2). If the influent ion concentration (positive or negative) is [Co] equivalents per unit volume, the total rate of inflow of ions is [Co]Qo/(m/2) equivalents per unit time per cell. One equivalent is also equal to one Faraday. Because a Faraday or equivalent is equal to 96,494 coulombs, assuming a coulomb efficiency of n, the amount of electricity needed to remove the ions in one cell is equal to 96,494[Co]Qon/(m /2) coulombs per unit time. Coulomb efficiency is the fraction of the input number of equivalents of an ionized substance that is actually acted upon by an input of electricity.

If time is expressed in seconds, coulomb per second is amperes. Therefore, for time in seconds, 96,494[Co]Qon/(m/2) amperes of current must be impressed upon the membranes of the cell to effect the removal of the ions. The cells are connected in series, so the same current must pass through all of the cells in the electrodialysis unit, and the same 96,494[Co]Qon/(m/2) amperes of current would be responsible for removing the ions in the whole unit. To repeat, not only is the amperage impressed in one cell but in all of the cells in the unit.

In electrodialysis calculations, a term called current density (CD) is often used. Current density is the current in milliamperes that flows through a square centimeter of membrane perpendicular to the current direction: CD = mA/Acm, where mA is the milliamperes of electricity and Acm is the square centimeters of perpendicular area. A ratio called current density to normality (CD/N) is also used, where N is the normality. A high value of this ratio means that there is insufficient charge to carry the current away. When this occurs, a localized deficiency of ions on the membrane surfaces may occur. This occurrence is called polarization. In commercial electrodialysis units CD/N of up to 1,000 are utilized.

The electric current I that is impressed at the electrodes is not necessarily the same current that passes through the cells or deionizing compartments. The actual current that successfully passes through is a function of the current efficiency which varies with the nature of the electrolyte, its concentration in solution, and the membrane system. Call M the current efficiency. The amperes passing through the solution is equal to the amperes required to remove the ions. Thus,

The emf E across the electrodes is given by Ohm's law as shown below, where R is the resistance across the unit.

If I is in amperes and R is in ohms, then E is in volts.

From basic electricity, the power P is EI = 12R. Thus,

If I is in amperes, E is in volts, and R is in ohms, P in is in watts. Of course, the combined units of N and Qo must be in corresponding consistent units.

Example 8.1 A brackish water of 378.51 m /day containing 4,000 mg/L of ions expressed as Nacl is to be de-ionized using an electrodialysis unit. There are 400 membranes in the unit each measuring 45.72 cm by 50.8 cm. Resistance across the unit is 6 ohms and the current efficiency is 90%. CD IN to avoid polarization is 700. Estimate the impressed current and voltage, the coulomb efficiency, and the power requirement.

Solution:

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