Question 12.7: FEEDING DILUTED POLYMER OR OTHER LIQUID CHEMICAL TO AWATER T...
FEEDING DILUTED POLYMER OR OTHER LIQUID CHEMICAL TO AWATER TREATMENT PLANT
For any liquid chemical, such as liquid polymer, with less than 2% concentration, the chemical feed rate can be calculated by a simple material balance equation,
(Q_{lc} + Q_{wtp})C_{wtp} = Q_{lc}C_{lc} ,
where Q_{wtp} = water treatment plant flow; C_{wtp} = liquid chemical dosage required for water treatment at WTP; Q_{lc} = liquid chemical feed rate; and C_{lc} = liquid chemical’s feed concentration. The units of Q_{wtp} and Q_{lc} should be identicals while the units of C_{wtp} and C_{lc} should be identical.
The desired polymer dose should be determined by jar testing, pilot filters, zeta meter, or colloid titration method (Wang et al., 1978). Once the dose is determined, calculations may be done to set the chemical feed pumps. Typically, the chemical feed pump rate is determined by pumping into or out of a graduated cylinder that has the gradations marked in mL. As a result, the chemical feed rate in mL/min may be calculated.
Determine the polymer solution feed rate Q_{lc} (mL/min), if the polymer solution concentration Q_{lc} = 0.5% = 5,000 mg/L, the WTP flow rate Q_{wtp} = 2.5 MGD = 9.46 MLD, and the required polymer dosage Q_{wtp} = 1 mg/L.
1 (US Customary System):
Since Q_{wtp} is much greater than Q_{lc}, the above material balance equation may be reduced to a simplified material balance equation,
Q_{wtp}C_{wtp} = Q_{lc}C_{lc}
(2.5 MGD)(1 mg∕L) = (Q_{lc} MGD)(5,000 mg∕L)
Q_{lc} = (2.5∕5,000) MGD = 0.0005 MGD
= (2.5∕5,000) MGD × (10^{6} gal∕MG)(3.784 × 1,000 mL∕gal)(1 d∕1,440 min)
= 1,314 mL∕min
Note: 1 MGD = 2,628,472 mL/min
2 (SI System):
(9.46 MLD)(1 mg∕L) = (Q_{lc} MLD)(5,000 mg∕L)
Q_{lc} = (9.46∕5,000) MLD = 0.001892 MLD
= (9.46∕5,000) MLD × (10^{6} L∕ML)(1,000 mL∕L)(1 d∕1,440 min)
= 1,314 mL∕min
Note: 1 MLD = 694,444 mL/min