Quantcast Section C -- Calculating Aquatic Herbicide Requirements

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64
AFM 91-19 / TM 5-629 / NAVFAC MO-314
24 May 1989
(4) Adjust the feed mechanism as necessary
9-10. Applications Based on Water Volume.
and repeat the procedure until the desired rate is
For herbicide treatments based on pounds of
achieved.
herbicide per acre-foot of water, it is necessary
to know the surface area of the body of water in
b. Hand and Mechanical Spreaders Without
acres and the average depth of the water. The
volume of water to be treated can then be
(1) Begin with a weighed amount of herbi-
determined in acre-feet. An acre-foot of water is
cide formulation. Apply the material to a mea-
the amount of water that can cover an acre to a
sured area, preferably 435.6 ft2 or 1/100 acre.
depth of 1 foot. The amount of herbicide to be
Weigh the material remaining in the spreader,
applied is determined by multiplying the applica-
and subtract the weight from the initial weight
tion rate by the acre-feet of water to be treated.
to determine how much was applied. A shop
Example:
vacuum-cleaner facilitates recovering the unused
Water surface area = 2.5 acres
herbicide formulation. Make appropriate adjust-
Average water depth = 3 ft
ments in the spreader, and repeat the procedure
Total volume of water = 2.5 x 3 = 7.5
until the desired rate can be approximated in
acre-feet
repeated trials.
Application rate = 4 lb per acre-foot
(2) If the treatment period extends over
Herbicide required = 7.5 x 4 = 30 lb
several hours or days, occasionally check the
rate being applied by weighing the amount to a
9-11. Applications Based on Concentration. T o
measured area, and make any necessary adjust-
determine the amount of herbicide required to
ments.
treat a body of water when the application rate
is expressed in parts per million (p/m) by
Section C -- Calculating Aquatic Herbicide Re-
weight, use the following calculations:
quirements
1 acre-foot of water weighs 2,722,500 lb
1 p/m by weight of an acre-foot of water =
9-9. Applications Based on Water Surface Area.
2,722,500 lb = 2.7 lb
When the application rate is expressed in pounds
1,000,000
or other units per surface acre, it is necessary to
know the surface area of the body of water in
Herbicide required = acre-feet of water to
acres. The amount of herbicide to be applied is
be treated x p/m by weight desired x 2.7 lb
determined by multiplying the application rate
Example:
by the surface area.
Water surface area = 5 acres
a. The surface area of a rectangular body of
Average water depth = 4 ft
water can be calculated by multiplying the length
Water to be treated = 5 x 4 = 20 acre-feet
in feet by the width in feet and dividing the
Application rate = 2 p/m by weight
result by the number of square feet in an acre
Herbicide required = 20 x 2 x 2.7 = 108 lb
(43,560).
9-12. Applications in Flowing Water. The vol-
Area in acres = length in ft x width in ft
ume of waterflow must be accurately determined
43,560 ft2 per acre
before treating flowing water to control aquatic
b. The surface area of a circular body of
weeds. Herbicide requirements are based on the
water is calculated by multiplying the square of
volume of flow.
the radius in feet by 3.1416 and dividing the
a. Weirs and other devices can provide accu-
result by the number of square feet in an acre.
rate measurements of waterflow, or, in the
absence of such devices, the approximate volume
of flow can be obtained by determining the
Area in acres = 3.1416 x square of radius in ft
cross-sectional area of the channel in square feet
43,560 ft2 per acre
(average water depth in feet x average width of
c. The radius is one-half the diameter. For
the channel in feet) and the velocity of flow. A
bodies of water that are somewhat round, but
convenient and fairly accurate measurement of
not perfectly so, an approximate radius can be
velocity of flow may be made by placing a dense
determined by measuring the perimeter in feet
floating object, such as a waterlogged stick, in
and dividing it by 6.28. The area in acres can
the center of the channel and measuring the
then be determined from the formula above.
length of time in seconds that are required for





 


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