(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.

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.

(1) Determine surface area:

1,200 lb/day

SA =

' 150 sq ft.

8 lb/sq ft/day

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