XiTqilC DQtmt r1000

mJ ult

Also, assume that all of the cations are removed.

£1 = 1412.8 gea = 1412.8 ^000^ = 1.96 ^ M ) uit m3 720( 1000) g g



Assume swell = 0.8

CatTBedVol = (1 + 0.8) = 0.13 m3 Ans 16.6.2 Quantity of Regenerant

Let CatRegenerant in kilogram equivalents be the quantity of regenerant required and R be the regenerant requirement in equivalents per equivalent of solute removed. The concentration of removable cations in gram equivalents per liter is [CatT]eq = X\Ilqi[Cs.']; therefore,

CatRegenerant = £ q[ ^ ]( R )( Q )( tmt ^J (16.13)

This equation represents the regenerant required between intervals of regeneration. The kilograms of regenerant depend upon the regenerant used. For example, if the regenerant is common table salt, then CatRegenerant should be multiplied by NaCl to obtain the kilograms of the salt.

By analogy with cation exchangers, the kilogram equivalents of regenerant, AnionRegenerant, used to regenerate anion exchangers is

'=m i 1 \ AnionRegenerant = X U[A^;](R)(Q)(U^J (16.14)

Example 16.4 Using a bed exchanger, 75 m of water per day is to be treated for hardness removal between regenerations having intervals of 8 hours. The raw water contains 80 mg/L of Ca2+ and 15 mg/L of Mg2+. The exchanger is a resin of exchange capacity of 1412.8 geq/m . Assume that the packed density of the resin is 720 kg/m . Calculate the kilograms of sodium chloride regenerant required assuming R = 2 and that all of the cations were removed.


i=m +q (i ^ CatRegenerant = Y qt[ C*' ](R)(Q)(tmt )i 24)

y q [ c+qi ] =_880_+_15_= 5 22 (10-3) -geq y* 1 (40.1/2)(1000) (24.3/2)(1000) V L


= 0.26 kg per interval of regeneration Ans

16.6.3 Wastewater Production

In the operation of ion exchangers wastewaters are produced. These come from the solvent water used to dissolve the regenerant and the backwash and rinse requirements. To illustrate the method for the production of wastewater from the solvent water, let us use an example calculation. In the sodium cycle, the concentration of NaCl is about 5 to 10% for an average of 7.5%. This means that, for example, if the quantity of regenerant required is 0.26 kg, the volume of wastewater produced from regeneration can be calculated as follows: the total mass of regenerant solution is 0.26/0.075 = 3.47 kilograms; the corresponding volume is 3.47/1000 = 0.0035 m . For an interval of regeneration of 8 h and assuming a rate of flow for the water treated of 75 m /d, the volume of water treated is 75/24(8) = 25 m . Thus, the wastewater produced is 0.0035/25 (100) = 0.014% by volume.

The other wastewater produced as a result of regeneration is the backwash and rinse waters. As soon as the bed is exhausted, it must be backwashed to remove debris accumulated during the service cycle. In addition, after regeneration, the bed should be rinsed to remove any residual regenerant. Backwash and rinse water requirements should be determined by experiment on an actual exchanger bed to be used in design.

The quantity of backwash and rinse requirement is best expressed as a function of bed volume. Let CatBackwashRinseVol be the m of backwash and rinse waters

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