Question

Calculate the quantity of the pure lime and soda required for softening of 1,00,000 litres of water containing the following impurities in ppm. The purity of lime is 70% and soda is BS% Ca(HCO₃) = 30.2 , Mg(HCO₃) = 20.8 ,CaCI₂ = 28.1 ,MgCl₂ = 8.78, CaSO₄ = 35, MgSO₄ = 6.7 .

Answer 1

 

impurities( mg/lit)	Multification fector	CaC03 equivalent Mg/lif	Requirement
Ca(Hco3)=30.2	`100/162`	`30.2 xx 100/162 = 18.64`	L
Mg(Hco3) =20.8	`100/146`	`20.8 xx 100/146 = 14.24`	2L
CaCl2 = 28.1	`100/111`	`28.1 xx 100/111=25.32`	S
CaS04=35	`100/136`	`35 xx 100/136 = 25.74`	S
MgS04 = 6.7	`100/120`	`6.7xx 100/120=5.58`	L+S
MgCl2 = 8.7	`100/95`	`8.7 xx 100/95 = 9.16`	L+S

`"Lime" =74/100 [CaCO_3  "equivalent of"  Ca (HCO_3)+ 2 xx Mg (HCO_3) + MgCI_2 xx MgSO_4 ]xx ("Volume of water")/(1000) xx 100/(%"Purity").`

`=74/100 [18.64 + 2 xx 14.25 + 9.16 + 5.58] xx (100000)/1000 xx 100/70.`

= 6541.6 gms

`"Soda" = 106/100 [CaCO_3  "equivalent of " MgCI_2 + MgSO_4 + CaSO_4 ] xx "Volum of water "/1000 xx 100/(%"Purity")`

`= 106/100 [25.31 + 9.16 +25.73 + 5.58 ] xx (100000)/1000 xx 100/85`.

= 8203 .15gms.

The lime requirement is 6541.6 gms and soda requirement is 8203. 15 gms.