TM 5-814-3/AFM 88-11, Volume III
(2) Determine surface loading rate:
1,200 lb/day
' 96 gpd/sq ft.
Sludge flow =
(8.34 lb/gal)(0.01 solid/liquid)(150 sq ft)
(3) Check detention time:
DT = (150 12) cu ft 7.48 24 ' 22.5 hrs.
14,388 gal/day
C-18. Vacuum filtration. (Refer to para 16-5c.)
a. Design requirements and criteria. Design a vacuum filter to dewater 10,000 gpd digested sludge
containing 5 percent solids. Assume the following conditions:
Required vacuum = 25 in Hg;
Weight of filtered solids per unit volume of filtrate = 4 lb/cu ft;
Cycle time, 0 (from manufacturer)6 min;
Form time (from manufacturer)3 min.
b. Calculations and results.
25 14.7
' 12.3 psi.
Pressure, P = 25" =
29.9
3 107
r
'
' 3.0;
R=
7
7
10
10
Form time 3
x
' 0.5;
=
Cycle time 6
O = 6min;
= 0.896 centipoise.
Calculate filter yield:
0.5(12.3)(0.064)
L = 35.6
(eq C-20)
0.896(3.0)(6)
= 5.56 lb/sq ft/hr.
These values are typical values for digested sludge. For design purposes, it is preferred to use values based
on pilot or laboratory studies.
Compute the total solids in the sludge (S):
S = 0.0248.3450,000
= 10,000 lb/day
= 417 lb/hr.
Required filter area:
417
' 75 sq ft.
A=
5.56
Therefore:
Filter diameter = 4 ft;
Filter length = 6 ft.
If specific resistance is unknown for a specific application, use loading rates as follows:
3 lbs/sq ft/hr activated sludge;
10 lbs/sq ft/hr primary sludge;
5 lbs/sq ft/hr combined primary and activated sludge.
C-40
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