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3 years ago

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A water contains 50.40 mg/L as CaCO3 of carbon dioxide, 190.00 mg/L as CaCO3 of Ca2 and 55.00 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 2.935 m^3/s. Assume that the effluent water contains no carbon dioxide, 30.0 mg. L^-1 as CaCO3 of Ca2+ and 10.0 mg.L^1 as CaCO3 of Mg2+. Be sure to calculate the mass of CaC03 and Mg(OH)2 sludge produced each day.

2 answers:

Galina-37 [17]3 years ago

6 0

Answer:

Total sludge = 123426kg/d

Explanation:

The reaction is given as;

H2Co3 + Ca(OH)2 ⇆ CaCo3 + 2H20

1 1 1 2 moles

Calculating the concentration of C02, we have

Concentration of C02 = concentration of CaCo3/Molecular weight of Caco3

= 50.4/100.09

= 0.5035mol/L

Sludge of Co2 = Conc. of Co2 * Q * MW of CaCo3 *10^-6

= 0.5035 * 253.6 *10^6 * 100.09 * 10^-6

= 12780kg/d

From the equation Ca2+ + 2HCo3- + Ca(OH)2 ⇄ 2CaCo3 + 2H2O

1 mole of calcium yields 2 moles of CaCo3

Therefore, Concentration of Ca2+ = Conc. of CaCo3/Mw of CaCO3

= 190-30/100.09

=1.599mol/L

Calculating sludge of calcium:

Sludge of Ca = 2 * Conc. of ca * Q * mw of CaCO3 * 10^-6

= 2 * 1.599 *253.6*10^6* 100.09 * 10^-6

= 811742kg/d

From the equation,

Mg2+ +2HCO3- + Ca(OH)2 ⇄ MgCO3 + 2CaCO3 + 2H2O

1 mole of mg yields 2 moles CaCO3 and 1 mole of Mg(OH)2

Concentration of Mg2+ = Conc, of CaCO3 /Mw of CaCo3

= 55- 10/100.09

= 0.4496mol/L

Sludge of Mg = 2 * Conc. of Mg * Q * mw of CaCO3 * 10^-6 +* Conc. of Mg * Q * mw of Mg(OH)2 * 10^-6

= 2 * 0.4496 * 253.5*10^6 * 100.09 * 10^-6 + 0.4996* 253.5*10^6 58.3 * 10^-6

= 29472kg/d

Total Sludge = Sludge of CO2 + Sludge of Ca + Sludge of Mg

12780+ 81174 + 29472

= 123426kg/d

Simora [160]3 years ago

4 0

Answer:

Total Sludge = 123426 kg / d

Explanation:

Find attached the solution

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