A water contains 38.50 mg/L as CaCO3 of carbon dioxide, 288.00 mg/L as CaCO3 of

Question

A water contains 38.50 mg/L as CaCO3 of carbon dioxide, 288.00 mg/L as CaCO3 of Ca2+ and 53.00 mg/L as CaCO3 of Mg2+. All of the hardness is carbonate hardness. Using the stoichiometry of the lime soda ash softening equations, what is the daily sludge production (in dry weight, kg/day) if the plant treats water at a rate of 56.00 MGD? Assume that the effluent water contains no carbon dioxide, 30.0 mg/L as CaCO3 of Ca2+ and 10.0 mg/L as CaCO3 of Mg2+. Be sure to calculate the mass of CaCO3 and Mg(OH)2 sludge produced each day.

Explanation / Answer

Total input hardness =38.5 + 288+53 = 379.5 mg/l

outgoing hardness = 30 + 10 = 40 kg/l oc CaCO3

slug production equivlent to 379.5 - 40 =339.5 mg/L of CaCO3

sluge ag Mg(OH)2 = 53-10 = 43 mg/L

Sluge as CaCO3 = 339.5 - 43=296.5 mg/L

Total volume = 56MGD = 56*3.78541 = 211.983 =212 MLD

mass of CaCO3 = 296.5*212 kg = 62747 kg = 62.747 tonn

mass of Mh(OH)2 = 43*212 = 9116 kg = 9.116 tonn

total = 62.747 tonn + 9.116 tonn = 71.863 tonn

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