## Info

M CaO — M CaOCaHCO3 + M CaOMgHCO3 + M CaOMgCa + M CaOCO2 + M CaO Excess M CaOCaHCO3 — 0'35 f CaHCO3 MT CaHCO3 ■

Removal of calcium bicarbonate takes precedence. Therefore, meq/L of Ca(HCO3)2—2.5 2 5-03

— 36.88 (20.05 ) Ca(HCO3)2 — 36.88(20.05 )( -1 j

Therefore, MCaOCaHCO — 0.35 (0.88 )(2989.12) — 920.65 kg/d

MCaOMgHCO3 — 0.77 fMgHCO3MTMgHCO3 — 0 M CaOMgCa — 2.30 f MgCa MT Mg:

f MgCa — ^-j0.32 — 0.68; MTMg — 1.0 (-0ô)( 14,750) — 14.75 kgeq/d

Therefore,

MCaOMgCa — 2.30(0.68 )( 179.21 ) — 280.28 kg/d MCaOCO2 — 1.27 ( Mco2) — 1.27 (22.0)(^)( 14,750) — 412.12 kg/d M CaO Excess — 0.028 ¥ kg = 0.028( 14,750) = 413 kg/d

M Ca0 = 920.65 + 0 + 280.28 + 412.12 + 413 = 2026.05 kg/d of pure lime = 20026005 = 2251.17 kg/d of lime Ans

Soda ash will not be used in the first stage.

Example 10.5 In Example 10.4, calculate the chemical requirements in the second stage.

Solution:

M CaO = M Ca0CaHC03 + M Ca0MgHC03 + M CaOMgCa + M Ca0C02 + M CaOExcess

First, we have to find the influent hardness concentrations to the second stage. Calcium introduced to the second stage from the first stage is equal to the limit of technology of 15.0 mg/L CaC03, plus the calcium from the lime used to remove the noncarbonate hardness of magnesium, which is 50 mg/L CaC03, plus the noncarbonate calcium from the raw water, which is 199.5 - 125 = 74.5 mg/L CaC03. This will produce a total outflow of calcium from the first stage of 50 + 15 + 74.5 = 139.5 mg/L as CaC03 from a flow of 14,750 m /d. As this outflow mixes with the bypass flow, the influent calcium concentration to the second stage then becomes (139.5(14,750) + 199.5(10,250))/25,000 = 164.1 mg/L as CaC03 = 3.28 meq/L. Calcium to remain in the effluent is 120 - 30 = 90 mg/L as CaC03 = 1.8 meq/L. Because of the limit of technology of 0.3 meq/L, pseudo calcium concentration to remain in effluent is 1.8 -0.3 = 1.5 meq/L. Thus, calcium to be removed is 3.28 - 1.5 = 1.78 meq/L.

Considering the bypass, the influent HCO- concentration to the second stage is = (0(14,750) + 2.5(10,250))/25,000 = 1.025 meq/L. Because we have 3.28 meq/L of influent calcium but only 1.025 meq/L of bicarbonate, by the order of removal, influent Ca(HC03)2 = 1.025 meq/L. An amount of 1.78 meq/L of calcium is to be removed. Thus, noncarbonate calcium to be removed = 1.78 - 1.025 = 0.755 meq/L. Noncarbonate calcium = 3.28 - 1.025 = 2.255 meq/L.

The limit of technology for magnesium hydroxide is 16 mg/L as CaC03 = 0.32 meq/L. Considering the bypass flow, Mg influent concentration to the second stage = (0.32(14,750) + 1.0(10,250))/25,000 = 0.60 meq/L.

MCa0CaHC03 = °.35 f CaHC03 MTCaHC03

M caoco2 = 1.27 (M co2) = 1.27 (22.0)(-°°)( 10,250) = 286.39 kg/d

Therefore,

MCa0 = 514.16 + 0 + 0 + 286.39 - 413 = 387.55 kg/d of pure lime 387.55

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