Water Source and Storage Calculations - Basic Mathematics for Water and Wastewater Operators - page 229

■ Example 15.18
Problem: A small lake has an average length of 300 ft, an average width of 95
ft, and a maximum depth of 22 ft. What is the estimated volume of the lake in
gallons?

Note: For small ponds and lakes, the average depth is generally about 0.4 times
the greatest depth; therefore, to estimate the average depth, measure the greatest
depth and then multiply that number by 0.4.

Solution : First, the average depth of the lake must be estimated:

Estimatedaverage depth (ft)Greatestdepth = (ft) × 0.4 ft = 22 ft × 0.4 ft = 8.8 ft

Then the lake volume can be determined:

3

Volume = Avg. length (ft)Avg. width (ft)Avg × × .. depth (ft) × 7.48 gal/ft

3

= 300 ft × 95 ft × 8.8 ft × 7.48 gal/ft = 1,875,984 gal

COPPER SULFATE DOSING
Algal control by applying copper sulfate is perhaps the most common in situ treat-
ment of lakes, ponds, and reservoirs; the copper ions in the water kill the algae.
Copper sulfate application methods and dosages will vary depending on the spe-
cific surface water body being treated. The desired copper sulfate dosage may be
expressed in mg/L copper, lb/ac-ft copper sulfate, or lb/ac copper sulfate.
For a dose expressed as mg/L copper, the following equation is used to calculate
the pounds of copper sulfate required:

Copper(mg/L)Volume(M × G) 8.34 lb/gal

%Availablecopper/100

×

Coppersulfate (lb) = (15.16)

■ Example 15.19
Problem: For algae control in a small pond, a dosage of 0.5 mg/L copper is desired.
The pond has a volume of 15 MG. How many pounds of copper sulfate will be
required? (Copper sulfate contains 25% available copper.)

Solution:

Copper(mg/L)Volume(MG)8. × × 34 lb/gal

%Availablecopper/100

Coppersulfate =

0.5 mg/L × 55MG 1 × 8.34 lb/gal

= = 250 lb

25/100