UFC 3-280-03
23 JULY 2003
For 30-d storage quantity of ferric chloride required = [(1150 kg/mo) / (0.40 x 1.45 kg/L] = 1990
L/mo (530 gal/mo)
Therefore, minimum storage should be provided for 2000 L or 2 m3 (550 gal-
lons).
5-2.3.3.5 Compute Size of the Sludge Pump Used Before the Conditioning Tank.
Assume that the conditioning tank is connected to all the filter processes, and that all
the operating filters are fed at one time.
Total sludge pumped per operating day = (37,850 L/d x 7 d/wk)/(5 d/wk operating) = 53,000 L/d =
(14,000 gal/d)
Pumping rate per cycle (filling time 100 min [6000 s]) = [53,000 L/d]/[(4 cycle/day) x (6000 s/cycle)]
= 2.2 L/s (35 gpm)
Therefore, the total minimum required pumping capacity is 2.2 L/s (35 gpm).
For this application a minimum of two pumps should be provided, one operating pump
and one standby pump. To size these pumps, the system's hydraulic (static and
dynamic) head requirements would need to be determined. This would be based on
specific system requirements, in addition to a safety factor of 10 to 25%, as described in
Subparagraph 2-7.2.2.1, to overcome any effects from the sludge such as its thixotropic
properties. Knowing the pump capacity and head, the designer should use
manufacturers' catalogs and select the appropriate pump.
5-2.3.3.6 Compute the Size of the Conditioning Tank. Assume use of an in-line
conditioning tank with a 10-minute (600-second) detention and mixing time.
Volume required for conditioning tank = 2.2 L/s x 600 s = 1320 L (350 gallons)
Therefore, a conditioning tank should be selected with a minimum capacity of
1320 L or 1.32 m3 (350 gallons). This conditioning tank should also be equipped with a
mixer and level switches to control the operation of the sludge and dilute conditioning
chemical metering pumps.
5-2.3.3.7 Compute the Size of the Sludge Feed Pumps to the Filter Presses. Each
filter press will be equipped with one operating sludge feed pump and one standby
pump. The total quantity of conditioned sludge plus chemical solution to the filter
presses follows:
Sludge quantity = 53,000 L/d
Water in lime at 10% solution =[106 kg/d] / [(0.1 x 1 kg/L)] = 1060 L/d
Water in ferric chloride at 10% solution =[53 kg/d] / [(0.1 x 1 kg/L)] = 530 L/d
Total quantity to be pumped =[53,000 + 1060 + 530] L/d = 54,590 L/d ~ 54,600 L/d (14,440 gal/d)
5-8
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