Geoscience Reference
In-Depth Information
Copper(mg/L)Volume(M
× G) 8.34 lb/gal
%Available copper100
×
Coppersulfate (lb)
=
(23.17)
EXAMPLE 23.19
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
%Availablecopper100
Coppersulfate
=
0.5 mg/L
×
15
MG
×
8.34 lb/gal
=
=
250 lb
25 100
For calculating lb copper sulfate per ac-ft, use the following equation (assuming a desired copper
sulfate dose of 0.9 lb/ac-ft):
Coppersulfate (lb)= 0.9 lb Coppersulfate
× c-ft
(23.18)
1 ac-ft
EXAMPLE 23.20
Problem: A pond has a volume of 35 ac-ft. If the desired copper sulfate dose is 0.9 lb/ac-ft, how
many lb of copper sulfate will be required?
Solution:
0.9 lb Coppersulfate
× c-ft
a
Coppersulfate (lb)
=
1 ac-ft
0.9 lb Coppersulfate
1 ac-ft
= x lb Coppersulfate
35 ac-ft
Then solve for x :
0.9 × 35 = x = 31.5 lb copper sulfate
The desired copper sulfate dosage may also be expressed in terms of lb copper sulfate per acre. The
following equation is used to determine lb copper sulfate (assuming a desired copper sulfate dose
of 5.2 lb/ac):
Copper sulfate (lb) = (5.2 lb copper sulfate × acres)/1 ac
(23.19)
EXAMPLE 23.21
Problem: A small lake has a surface area of 6.0 ac. If the desired copper sulfate dose is 5.2 lb/ac,
how many pounds of copper sulfate are required?
Solution:
Copper sulfate (lb) = (5.2 lb copper sulfate × 6.0 ac)/1 ac = 31.2 lb copper sulfate
 
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