TM 5-814-3/AFM 88-11, Volume III
Convert this to horsepower:
1 hp
140.4
'
' 0.26; use 0.3hp.
Hp ' P
550 ft lb/sec 550
(5) Determine the paddle area from the following formula:
2P
A'
CDv 3
where
CD = dimensionless coefficient of drag (= 1.8);
2
= mass fluid density, lb/cu ft/g (1.94 lb&sec 2at 20EC);
ft 4
v
= relative relocity of paddles in fluid, fps (assume to be 0.75 times paddle-tip speed)
= 0.75 1.2 fps;
= 0.9 fps, with paddle-tip speed of 1.2 fps;
in this case,
2(160.3 ft/lb sec)
A'
1.8(1.94lb.&sec2/ft 4)(0.9ft/sec)3
= 125.9 sq ft.
C-3. Sedimentation. (Refer to para 11-2.)
a. Design requirements and criteria. Design a sedimentation unit to provide settling for a sewage flow
rate of 4 mgd, with suspended solids concentration of 300 mg/L. The following conditions apply:
Surface loading rate = 600 gpd/sq ft;
Suspended solids removal = 60%;
Sludge solids content = 4%;
Sludge specific density = 1.02.
b. Calculations and results.
(1) Calculate total tank surface area:
Flow Rate
4,000,000 gpd
Surface Area '
'
' 6,666.7; use 6,670 sq ft.
Surface Loading .Rate
600 gpd/sq ft
(3) Using a depth of 8 ft, calculate total volume:
V
=
8 6,670 = 53,360 cu ft.
(3) This volume can be divided among three rectangular tanks (in parallel), 20 ft wide and 120 ft long,
with a satisfactory length-to-width ratio of 6:1. Two circular tanks (in parallel), 35 ft in diameter, would also
be suitable. This will provide flexibility of operation during routine or emergency maintenance.
(4) Calculate weir length requirement, assuming 3 rectangular tanks and allowable weir loading rate
of 15,000 gpd/linear ft.
Design flow/tank = Total Flow ' 4,000,000 gpd ' 1,333,333 gpd;
3
3
1,333,333 gpd
' 89 lin ft.
Weir lenth/tank =
15,000 gpd/lin ft
(5) Comnplete weight of solids removed, assuming 60% removal:
Weight removed = 4 mgd 300 mg/L 0.60 = 6,000 lb/day;
therefore, 1,500 lbs are removed per 1 mpd flow.
C-15
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