Geoscience Reference
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Solution: To calculate the total alkalinity required, we must first calculate the alkalinity that will
react with 52 mg/L alum:
0.45 mg/L alkalinity
1mg/Lalum
= x
mg/L alka
linity
mg/L alum
52
04552
. ×= x
23.4 mg/L alkal
inity = x
The total alkalinity requirement can now be determined:
Total alkalinity required = Alkalinity reacting with alum + Residual alkalinity
= 23.4 mg/L + 30 mg/L = 53.4 mg/L
Next calculate how much alkalinity must be added to the water:
Alkalinity to be added = Total alkalinity required - Alkalinity present in water
= 53.4 mg/L - 36 mg/L = 17.4 mg/L
Finally, calculate the lime required to provide this additional alkalinity:
0.45 mg/L alkalinity
0.35 mg/L lime
= 17 .
mg/
Lalkalinity
mg/L alum
x
045 74
.
x
.
035
.
4035
045
13 5
17
×
.
x
=
.
x
=
.
mg/L lime
23.5.8 d etermining l lime d osage ( lb / day )
After the lime dose has been determined in terms of mg/L, it is a fairly simple matter to calculate
the lime dose in lb/day, which is one of the most common calculations in water and wastewater treat-
ment. To convert from mg/L to lb/day lime dose, we use the following equation:
Lime (lb/day) = Lime (mg/L) × Flow (MGD) × 8.34 lb/gal
(23.57)
EXAMPLE 23.61
Problem: The lime dose for a raw water has been calculated to be 15.2 mg/L. If the flow to be
treated is 2.4 MGD, how many lb/day lime will be required?
Solution:
Lime (lb/day) = Lime (mg/L) × Flow (MGD) × 8.34 lb/gal
= 15.2 mg/L × 2.4 MGD × 8.34 lb/gal = 304 lb/day lime
 
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