# Diluting 20,000-Fold with a 30 Gallon Remaining Volume in a 1,200 Gallon Tank

Our goal in this example is to dilute by a factor of 20,000.

The maximum amount of dilution possible with a 1,200 gallon tank and a 30 gallon remainder is 1200/30=40.

The formulae:

Dilution per Rinse = final dilution ^(1/# of rinses)

Rinse Volume = (dilution per rinse * remaining volume) – remaining volume

• One rinse diluting by 20,000 – impossible with a 1,200 gallon tank (max achievable is 40-fold);
• Two sequential rinses, each diluting by a factor of 20,000^(1/2) = 141. Also impossible with a 1,200 gallon tank;
• Three sequential rinses, each diluting by a factor of 20,000^(1/3) = 27. A volume of 780 gallons can do this  (27*30)-30=780 gallons. For three rinses, the total volume is 2,340 gallons.
• Four sequential rinses, each diluting by a factor of 20,000^(1/4) = 12. A volume of 330 gallons can do this, for a total volume of 1,320 gallons;
• Five sequential rinses, each diluting by a factor of 20,000^(1/5) = 7. A volume of 180 gallons can do this, for a total volume of 900 gallons;
• Six sequential rinses, each diluting by a factor of 20,000^(1/6) = 5.2. A volume of 126 gallons can do this, for a total volume of 757 gallons.

Second, let’s assume the operator is prepared to prime the boom where it doesn’t harm soybeans. Now the first new product tank takes care of the last dilution, lowering the cleanout dilution requirement by 1,200/30 = a factor of 40. Now the cleanout dilution requirement is only 20,000/40 = 500.

• One 1,200 gallon tank rinse can only achieve 40-fold dilution.
• Two rinses, each diluting by 500^(1/2) = 22. Rinse volumes of 640 gallons are sufficient, for a total of 1,280 gallons.
• Three sequential rinses, each diluting by a factor of 500^(1/3) = 7.9. A volume of 210 gallons can do this, for a total volume of 630 gallons;
• Four sequential rinses, each diluting by a factor of 500^(1/4) = 4.7. A volume of 112 gallons can do this, for a total volume of 448 gallons.
Sending
5 (1 vote)