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
