Handbook of Civil Engineering Calculations

SIZING A POLYMER DILUTION/

FEED SYSTEM

1.0 Mgd (3,785 m
3
/d) of equalized secondary effluent from a municipal wastewater treat-
ment facility is to undergo coagulation and flocculation in a direct filtration process. The
coagulant used will be an emulsion polymer with 30 percent active ingredient. Size the
polymer dilution/feed system including: the quantity of dilution water required, and the
amount of neat (as supplied) polymer required.

Calculation Procedure:

1. Determine the daily polymer requirements
   Depending on the quality of settled secondary effluent, organic polymer addition is often
   used to enhance the performance of tertiary effluent filters in a direct filtration process:
   see Design of a Rapid Mix Basin and Flocculation Basin. Because the chemistry of the
   wastewater has a significant effect on the performance of a polymer, the selection of a
   type of polymer for use as a filter aid generally requires experimental testing. Common
   test procedures for polymers involve adding an initial polymer dosage to the wastewater
   (usually 1 part per million, ppm) of a given polymer and observing the effects. Depending
   upon the effects observed, the polymer dosage should be increased or decreased by 0.5
   ppm increments to obtain an operating range. A polymer dosage of 2 ppm (2 parts poly-
   mer per 1 x 106 parts wastewater) will be used here.
   In general, the neat polymer is supplied with approximately 25 to 35 percent active
   polymer, the rest being oil and water. As stated above, a 30 percent active polymer will be
   used for this example. The neat polymer is first diluted to an extremely low concentration
   using dilution water, which consists of either potable water or treated effluent from the
   wastewater facility. The diluted polymer solution usually ranges from 0.005 to 0.5 per-
   cent solution. The diluted solution is injected into either a rapid mix basin or directly into
   a pipe. A 0.5 percent solution will be used here.
   The gallons per day (gal/day) (L/d) of active polymer required is calculated using the
   following:

Active polymer (gal/day) = (wastewater flow, Mgd)

x (active polymer dosage, ppm)

Using the values outlined above:

Active polymer = (1.0 Mgd)(2 ppm) = 2 gal/day active polymer (pure polymer)

= 0.083 gal/hr (gal/h) (0.31 L/h)

2. Find the quantity of dilution water required
   The quantity of dilution water required is calculated using the following:

^M. , ,„, active polymer, gal/h

Dilution water (gal/h) = ————^—^——*—.—-

% solution used (as a decimal)

Therefore, using values obtain above:

Dilution water = — —— = 16.6 gal/h (62.8 L/h)

0.005