## M

Mo2Fe = ¡|{0)Lo[0)(e5]^ = 0.00014A[Fe ]mg ¥ (13.45)

M caOMn = 2l!^1i0{;00;0|([M145m)g ^^ = 0.0010A[ Mn ]mg ¥ (13.46)

MNaoHMn = 4x'000(Mn4mg)¥ = 0.0015 A[ Mn ]mg¥ (13.47)

1000(27.45)

Example 13.4 A raw water contains 2.5 mg/L Mn. Calculate the kilograms of lime per cubic meter needed to meet the limit concentration of 0.05 mg/L, if removal is to be done at the high pH range.

Solution:

MCaOMn = 0.0010A[Mn]mg¥ = 0.0010(2.5-0.05)( 1) = 2.45(10-3) kg/m3 Ans

Example 13.5 A raw water contains 40 mg/L iron. Calculate the kilograms of lime per cubic meter needed to meet the limit concentration of 0.3 mg/L, if removal is to be done at the high pH range.

Solution:

M CaOFe = 0.0010A[ Fe ]mg ¥ = 0.0010(40 - 0.3)(1) = 0.040 kg/m3 Ans

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