Info

MFeOH3FeIIIchl _ 35.6 : nnn,^ ^ + -n^ V (using FeCl3)

1000( 54.1 ) 1000 _ 0.00066 [FeII opt ]mg ¥ + [[-1p0]f ¥ (using FeCl3 )

MFeOH3FeIIIsul _ 35.6^000(56.(6l5l + lp00Lg ^ (using Fe2(SO4)3)

_ 0.00053 [FeIIopt]mg ¥ + [-p-00Lg ¥ (using Fe2( SO4)3 )

([spJS]mg/1000)¥ is the kilograms of solids produced from the suspended solids of the raw water. The corresponding volumes of the sludges may be obtained by using the percent solids and the mass density of sludge.

Example 12.7 A raw water containing 100 mg/L of suspended solids is subjected to a coagulation treatment using Fe2(SO4)3. The current acidity and pH are, respectively, 30 mg/L as CaCO3 and 5.9. Calculate the amount of sludge produced if the optimum dose is 40 mg/L of Fe2(SO4)3.

Solution:

MFeOH3FeIIIsul _ 0.00053 [FeIIIopt]mg V + -100f V

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