TM 5-814-3/AFM 88-11, Volume III
b. Calculations and results.
(1) Calculate head loss, using Manning*s Equation, to give friction loss; F (for water):
4 f L V2
2 f V2 L
F'
'
(eq C-17)
D 2gc
D
gc
f = coefficient of friction;
L = length of pipe, ft;
D = diameter of pipe, ft;
V = mean velocity, fps;
gc = gravity constant.
(5 fps)2
4(0.01)(150 ft)
F=
(8 in 1/12 ft/in) 2(32.17 ft lb/lb force sec2)
= 5.04 ft of fluid.
Assuming friction losses for sludge 3 times that for water, the head loss from friction is:
ht = (5.04 ft)(3) = 15.1 ft, use 15 ft.
(2) Calculate total pumping head:
H = 10 ft + 15 ft = 25 ft of sludge.
Add 3 ft to this account for losses due to valves, elbows, etc.
Total H = 29 ft of sludge.
(3) Assuming a pump efficiency of 60 percent, calculate horsepower requirement:
QPW(sp gr)H
hp =
(eq C-18)
' (PW ' density of water)
550 eff
Q = (6 fps) ( (0.333)2) = 2.1 cfs;
(2.1 cu ft/sec)(62.4 lb/cu ft)(1.02)(28 ft)
hp =
550 (0.60)
= 11.3 up, use 15 hp.
(4) Determine discharge pump pressure:
(Total head)(Specific gravity)(weight H2O)
P=
144 sq in/sq ft
(28 ft)(1.02)(6.24 lb/cu ft)
=
144 sq in/sq ft
= 12.4 psi.
C-17. Gravity sludge thickener. (Refer to para 16-3.)
a. Design requirements and criteria. Size a gravity thickener tank to handle an activated sludge.
Assume the following conditions apply:
Amount of sludge to thickener = 1,200 lb/day;
Sludge solids content = 1 percent (i.e., 10,000 mg/L);
Solids loading for thickener = 8 lbs/sq ft/day;
Tank depth = 12 ft.
b. Calculations and results.
(1) Determine surface area:
1,200 lb/day
SA =
' 150 sq ft.
8 lb/sq ft/day
C-39
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