TM 5-814-3/AFM 88-11, Volume III
(2) Bed depth required at 30 min contact time:
(1,042 gpm)(30 min)
Volume required '
' 2,090 cu ft/unit;
(7.48 gpm)(2 units)
2,090 cu ft/unit
' 18.5 ft per unit, use d ' 20 ft;
Bed depth '
113 sq ft/unit
(20 ft)(226 sq ft)(7.48 gal/cu ft)
Actual contact time '
' 32.4 minutes.
1,042 gpm
Supply 40 percent expansion room at top for backwashing and 3.0 ft freeboard at bottom:
Actual depth = .4(20*) + 3.0 ft + 20 ft = 31 ft.
(3) Regeneration of carbon:
BOD removal required = 25 mg/L 10 mg/L = 15 mg/L;
lb BOD removed
Carbon exhaustion rate =
0.2
;
lb carbon
BOD removed/day = 15 mg/L (8.34 lb/gal)(l.5 mgd) = 188 lb/day.
188 lb BOD removed/day
Carbon required '
' 940 lb carbon/day.
lb BOD removed
0.2
lb carbon
940
lb/column
Lbs carbon required per column per day '
' 470
2 units
day
Carbon in column at 26 lb/cu ft = (20 ft)(ll3 sq ft)(26 lb/cu ft) = 58,760 lb/column.
58 760 lb/column
Regeneration required at interval of '
' 125 days.
lb/column
470
day
C-12. Phosphorus removal. (Refer to para 15-7.)
a. Design requirements and criteria. Determine what post-secondary mineral treatment for phosphorus
removal will meet an effluent requirement of 1 mg/L as P. Assume the following conditions apply:
Wastewater flow = 0.5 mgd;
Influent phosphorus concentration = 12 mg/L;
Treatment will be post secondary;
Influent alkalinity = 300 mg/L as CaCO3.
b. Calculations and results.
(1) Determine lb/day of phosphorus incoming:
lb/mil gal
lb/day ' (12 mg/L 8.34
(0.5 mgd)
mg/L
= 50.0 lb/day phosphorus as P.
C-34
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