Volume Calculations
Volume of a Cylinder
3.14 x radius squared x height or (3.14r2 h)
Volume of a Cone
1.047 x radius squared x height or (1.047r2h)
Volume of a Rectangle
Length x Width x Height
or
(L x W x H)
Volume of a Roof or Triangle
Length x Width x Height
-----------------------------2
or
(L x W x H)
--------------2
Volume of Steel Grain Bin = Cylinder + Cone
Volume of a flat Storage or Warehouse = Rectangle + Roof
Diameter = Circumference
----------------3.14
Radius = Diameter
----------2
Story Problem. An inspector was called out to help an applicator figure out how many
phostoxin pellets to place in a grain silo. The silo has an 80’ circumference, 25’ height
from eaves to ground, and 5’ high cone. Temperature is 80 degrees F. This silo also
contains 500 bushels of grain.
1. Total Volume of Silo?
Diameter = 80 / 3.14  Diameter = 25.47  Radius = 25.47 / 2  Radius = 12.73
Cylinder = 3.14 x 12.732 x 25  Cylinder = 3.14 x 162.05 x 25  Cylinder = 12,721 ft3
Cone = 1.047 x 162.05 x 5  Cone = 848.33 cubic ft.
Total Volume = 848.33 + 12,721  13,569.33 cubic ft
2. How many pellets need to be used?
Page 10  450 – 900 pellets / 1,000 bu withtarp500 bushel  225 – 450 pellets
Page 10 150 – 700 pellets / 1,000 ft3  no tarp  2,035 – 9,498 pellets
* Ranges with severity of infestation, leaks, etc…
3. How long before aeration?
48 hours
Jordon’s Feed Mill Story Problem
Jordon’s Feed Mill has a grain silo that is 75’ in circumference and a vertical distance
from the eaves to the bottom of the bin of 15’. The cone top is 3’ high.
Larry’s Pest Control has been contracted to fumigate one of the silos for the lesser grain
borer. The silo is filled with approximately 500 bushels of spring wheat. Larry can
either fumigate on Tuesday July 21st in the evening, or Thursday July 23rd in the evening.
The weather for Tuesday is forecast to be partly cloudy with a 70% chance of rain and a
high of 95 degrees F. Thursday is forecast to be sunny with a temperature of 101 degrees
F. The peak grain temperature ranges from 70 – 80 degrees F. There are horses 50 yards
from grain silo.
Detail your fumigation management plan strategy to include at least the following details.
1. What is the volume of this grain bin?
Diameter = 75 / 3.14  Diameter = 23.88  Radius = diameter / 2  Radius = 11.94
cyl = 3.14 x 11.942 x 15  volume = 3.14 x 142.45 x 15  volume = 6,714.74 cubic ft
cone = 1.047 x 11.942 x 3  cone = 447.43 cubic ft
Volume = 447.43 + 6,714.74  Volume = 7,162.17 cubic ft
2. How many pellets will you need to use when fumigating?
Page 10  200 – 900 pellets / 1,000 bu  with tarp  500 bushel  100 – 450 pellets
Page 10 150 – 700 pellets / 1,000 ft3  no tarp  1,050 – 4,900 pellets
* Ranges with severity of infestation, leaks, etc…
3. Calculate the required fumigation period based on fumigant grain temperature?
Page 8  48 hr  60 – 80 degrees F grain temperature  if sealed properly
4. Which day would be the best one to start the fumigation?
Tuesday July 21st because of moisture that afternoon. Moisture will helps convert
aluminum or magnesium phosphide into phosphine gas.
5. When should ventilation begin?
Thursday evening or Friday morning, < 0.3 ppmillion
6. What personal protective equipment should you wear and how does this relate
phosphine gas concentrations?
Page 11 - wear dry cotton gloves, wash after, respiratory equipment (NIOSH /
phosphine gas approved)-up to 15 ppm
7. Please review the Fumigant Management Plan and fill in the blank fields (with ?).