Water Treatment Processes - Mathematics and Statistics for the Environment - page 629

In Step 2, we make a comparison between required alkalinity and alkalinity already in the raw
water to determine how many mg/L alkalinity should be added to the water. The equation used to
make this calculation is shown below:

Alkalinity to be added = Total alkalinity required - Alkalinity present in water (23.56)

■ EXAMPLE 23.58
Problem: A total of 44-mg/L alkalinity is required to react with alum and ensure proper precipita-
tion. If the raw water has an alkalinity of 30 mg/L as bicarbonate, how much mg/L alkalinity should
be added to the water?

Solution:

Alkalinity to be added = Total alkalinity required - Alkalinity present in water

= 44 mg/L - 30 mg/L = 14 mg/L

In Step 3, after determining the amount of alkalinity to be added to the water, we determine how
much lime (the source of alkalinity) must be added. We accomplish this by using the ratio shown
in Example 23.59.

■ EXAMPLE 23.59
Problem: It has been calculated that 16 mg/L alkalinity must be added to a raw water. How much
mg/L lime will be required to provide this amount of alkalinity? (1 mg/L alum reacts with 0.45
mg/L alkalinity and 1 mg/L alum reacts with 0.35 mg/L lime.)

Solution: First determine the mg/L lime required by using a proportion that relates bicarbonate
alkalinity to lime:

0.45 mg/L alkalinity
0.35 mg/L lime

= 16

mg/L alkalinity

mg/L lime

x

Then cross-multiply:

0.45 x =× 16 0.35

16 0.35
0.45
12.4 mg/L lime

×

x =

x =

In Example 23.60, we use all three steps to determine the lime dosage (mg/L) required.

■ EXAMPLE 23.60
Problem: Given the following data, calculate the lime dose required, in mg/L:

Alum dose required (determined by jar tests) = 52 mg/L
Residual alkalinity required for precipitation = 30 mg/L
1 mg/L alum reacts with 0.35 mg/L lime
1 mg/L alum reacts with 0.45 mg/L alkalinity
Raw water alkalinity = 36 mg/L