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.

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.

(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.

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