1. Volume: Descriptions, Calculations & Industrial Applications
Industrial Skills
Volume: Descriptions, Calculations & Industrial Applications

2. Description: Area A Two-Dimensional Quantity
Area measurements utilize linear measurements to calculate the number of unit squares within a given boundary.
Measurements of area are expressed in “square units” such as square inches, square feet, square meters, etc.
They have length and width but thickness is not considered in the calculations.
Never mix units in the same calculation.

3. Description: Volume A Three-Dimensional Quantity
Volume is the space an object occupies or the capacity of a container. Volume calculations combine three measurements: usually length, height, and depth or thickness.
Volume is always expressed in “cubic units” such as cubic inches, cubic feet, cubic meters, etc.
A variety of different volume formulas may be used, depending on the shape of the object.
Check your handout for exact formulas.
Never mix units in the same calculation.

4. Industrial Applications: Common Quantities and Conversions
1 square foot = 144 square inches
1 square yard = 9 square feet = 1,296 sq. in.
1 cubic foot = 1,728 cubic inches
1 cubic yard = 27 cubic feet = 46,656 cu. in.
1 cubic foot = 7.47 gallons of water*
1 gallon of water = pounds*
1 square centimeter = 100 square millimeters
1 square meter =10,000 sq.cm.=1,000,000 sq.mm
1 cubic meter = 1,000,000,000 cubic millimeters
1 cubic meter = 1,000 liters of water*
1 cubic meter = gallons of water*
1 liter of water = Kilogram*

5. Industrial Applications: Container Questions – Gallons & Weight
Calculate the volume of the inside of the container assigned to you and your partner(s).
Calculate how many gallons of water the container could theoretically hold.
1 cubic foot = 7.47 gallons of water*
Calculate how much the water in the container would weigh.
1 gallon of water = pounds*

6. Industrial Applications: Calculations
Scrap Rope Box:
Outside Dimensions: 36” x 32” x 18”
36” x 32” x 18” = 20,736 cu.in. divided by 1,728cu.in.
= 12 cubic feet
Inside Dimensions: 33¾” x 29¾” x 15¾”
33.75” x 29.75” x 15.75” = 15, cu.in. ÷ 1,728cu.in.
= cubic feet
9.152 cu.ft. x 7.47 gallons = gallons of water
gal. x 8.34 pounds = pounds

7. Other Calculations: Box A: Wood Box with Handles
Outside Dimensions: 12” x 16” x 32”
12” x 16” x 32” = 6,144 cu.in. divided by 1,728cu.in.
= cubic feet
Inside Dimensions: 9¾” x 13¾” x 29¾”
9.75” x 13.75” x 29.75” = 3, cu.in. ÷ 1,728cu.in.
= cubic feet
2.308 cu.ft. x 7.47 gallons = gallons of water
gal. x 8.34 pounds = pounds

8. Other Calculations: Box B: Small Wood Box
Outside Dimensions: 16” x 16” x 18¼”
16” x 16” x 18.25” = 4,672 cu.in. divided by 1,728cu.in.
= cubic feet
Inside Dimensions: 14” x 14” x 16½”
14” x 14” x 16. 5” = 3,234cu.in. ÷ 1,728cu.in.
= cubic feet
1.872 cu.ft. x 7.47 gallons = gallons of water
gal. x 8.34 pounds = pounds

9. Other Calculations: Box C: “Grainger” Cardboard Box
Outside Dimensions: 12½” x 12½” x 17¾”
12.5” x 12.5” x 17.75” = cu.in. divided by 1,728cu.in.
= cubic feet
Inside Dimensions: 12¼” x 12¼” x 17½”
12.25” x 12.25” x 17.5” = 2, cu.in. ÷ 1,728cu.in.
= cubic feet
1.520 cu.ft. x 7.47 gallons = gallons of water
gal. x 8.34 pounds = pounds

10. Other Calculations: Box D: Plain Cardboard Box
Outside Dimensions: 15¼” x 15¼” x 15¼”
15.25” x 15.25” x 15.25” = cu.in. divided by 1,728cu.in.
= cubic feet
Inside Dimensions: 15” x 15” x 15”
15” x 15” x 15” = 3,375 cu.in. ÷ 1,728cu.in.
= cubic feet
1.953 cu.ft. x 7.47 gallons = gallons of water
gal. x 8.34 pounds = pounds

11. Other Calculations: Stackable Fish Box:
Inside Dimensions (Rough): 11¼” x 16¾” x 29¾”
11.25” x 16.75” x 29.75” = cu.in. divided by 1,728cu.in.
= cubic feet
Subtract a total of cu. ft.
True Inside Volume* = cubic feet
2.996 cu.ft. x 7.47 gallons = gallons of water
gal. x 8.34 pounds = pounds

12. Industrial Applications: Volume Question – Sea Salt Dilution
Calculate how much “synthetic sea salt” should be added to the water in “fish box” container. The mixing instructions stipulate: 7 pounds of Coralife Scientific Grade Marine Salt to 25 gallons of water.
Fish Box: Volume*: cu.ft.
Volume of Water: gallons

13. Industrial Applications: Volume Question – Sea Salt Dilution
7 pounds of “Salt” to 25 gallons of water.
Fish Box: Volume*: cu.ft.
Volume of Water: gallons
7 pounds = 112 ounces (7 x 16 ounces per pound)
112 oz. divided by 25 gal. = 4.48 oz. of salt per gallon of water
gal. x 4.48 oz. = oz. of salt
divided by 16 oz. = pounds of salt

14. Industrial Applications: Container Question – Space Utilization
A 5’x 8’ trailer has interior dimensions of 7ft.7in. long by 4ft.9in. wide by 5ft.8in. high. How many of the cardboard boxes shown earlier could we pack into the trailer?
Cardboard Box: Outside – 12.5” x 12.5” x 17.75”

15. Industrial Applications: Container Question – Space Utilization
ANSWER A:
Trailer: Inside - 7ft.7in. by 4ft.9in. by 5ft.8in.
cubic feet
Cardboard Box: Outside – 12.5” x 12.5” x 17.75”
1.605 cubic feet
20 Boxes on bottom. (5 boxes long x 4 boxes wide)
5 x 17.75” = inches (possible 91”)
4 x 12.5” = 50 inches (possible 57”)
5 Layers of Boxes High. (4 additional layers x 20 boxes)
5 x 12.5” = 62.5 inches (possible 68”)
Answer: 100 Boxes (100 x cu.ft. = cu.ft.)

16. Container Question – Space Utilization Answer A:
57”
57”
inside
inside
68”
Inside
height
91”
inside
5 Layers of Boxes High. (5 layers x 20 boxes)
5 x 12.5” = 62.5 inches (possible 68”)
Answer: 100 Boxes
(100 x cu.ft. = cu.ft.)
20 Boxes on bottom. (5 boxes long x 4 boxes wide)
5 x 17.75” = inches (possible 91”)
4 x 12.5” = 50 inches (possible 57”)

17. Industrial Applications: Container Question – Space Utilization
ANSWER B:
Trailer: Inside - 7ft.7in. by 4ft.9in. by 5ft.8in.
cubic feet
Cardboard Box: Outside – 12.5” x 12.5” x 17.75”
1.605 cubic feet
21 Boxes on bottom. (7 boxes long x 3 boxes wide)
7 x 12.5” = 87.5 inches (possible 91”)
3 x 17.75” = inches (possible 57”)
5 Layers of Boxes High. (4 additional layers x 21 boxes)
5 x 12.5” = 62.5 inches (possible 68”)
Answer: 105 Boxes (105 x cu.ft. = cu.ft.)

18. Container Question – Space Utilization Answer B:
57”
57”
inside
inside
68”
Inside
height
91”
inside
5 Layers of Boxes High. (5 layers x 21 boxes)
5 x 12.5” = 62.5 inches (possible 68”)
Answer: 105 Boxes
(105 x cu.ft. = cu.ft.)
21 Boxes on bottom. (7 boxes long x 3 boxes wide)
7 x 12.5” = 87.5 inches (possible 91”)
3 x 17.75” = inches (possible 57”)

19. Industrial Applications: Calculations: Volume of Cylinders (Pipe)
Volume = πR²L or AL
Volume is always expressed in cubic units.
6.031”
6.625”

20. Industrial Applications: Calculations: Volume of Cylinders (Pipe)
A section of the 6 inch clear PVC pipe used for the preservation of a large fish specimen is 42 inches long. What is the volume of the pipe?
6.031”
42”
Volume = πR²L or AL
*Always square radius first
6.031” Diameter = ” Radius (6.031”/2)
Area = πR² = π x ” ² = π x ”
= x ”
= sq.in.
Volume = πR²L = ” ² x 42” length
= cubic inches
”³ / 1728 ”³ = cubic feet

21. Industrial Applications: Calculations: Volume of Cylinders (Pipe)
What is the fluid volume of the pipe?
6.031”
Volume = ”³
= cubic feet
1 cubic foot = 7.47 gallons of water*
1 gallon of water = 231cu.in. (1728 ”³/7.47gal.)
1 gallon of water = pounds*
42”
Volume = ”³ divided by 231”³
= gal.

22. Industrial Applications: Calculations: Volume of Pipe
A section of white 3 inch PVC pipe is 20¾ inches long. With the cap on one end the inside distance of the pipe is actually 21¼ in. What is the solid and the fluid volume (water) of this pipe?
3.042”
Volume = πR²L or AL
3.042” Diameter = 1.521” Radius (3.042”/2)
Area = πR² = π x 1.521” ² = π x ”
= x ”
= sq.in.
Volume = πR²L = ” ² x 21.25” length
= cubic inches (solid volume)
”³ / 1728 ”³ = cubic feet (solid volume)
21.25”
Fluid Volume: ”³ divided by 231”³ = gal.
or cubic feet x 7.47 gal. = gal.
Which can then be converted to fluid ounces by: x 128oz. = fl. oz.

23. Industrial Applications: Calculations: Volume of Pipe
A section of white 2 inch PVC pipe is in. long. What is the solid and the fluid volume (water) of this pipe?
2.049”
Volume = πR²L or AL
2.049” Diameter = ” Radius (2.049”/2)
Area = πR² = π x ”² = π x ”
= x ”
= sq.in.
Volume = πR²L = ” ² x ” length
= cubic inches (solid volume)
”³ / 1728 ”³ = cubic feet (solid volume)
61.375”
Fluid Volume: ”³ divided by 231”³ = gal.
or cubic feet x 7.47 gal. = gal.
Which can then be converted to fluid ounces by: x 128oz. = fl. oz.

24. Industrial Applications: Calculations: Volume of Pipe
1.033”
A section of white 1 inch PVC pipe is in. long. What is the solid and the fluid volume (water) of this pipe?
Volume = πR²L or AL
1.033” Diameter = ” Radius (1.033”/2)
Area = πR² = π x ”² = π x ”
= x ”
= sq.in.
Volume = πR²L = ”² x ” length
= cubic inches (solid volume)
”³ / 1728 ”³ = cubic feet (solid volume)
47.125”
Fluid Volume: ”³ divided by 231”³ = gal.
or cubic feet x 7.47 gal. = gal.
Which can then be converted to fluid ounces by: x 128oz. = fl. oz.

25. Industrial Applications: Calculations: Volume of Pipe
0.810”
A section of white ¾ inch PVC pipe is 47in. long. What is the solid and the fluid volume (water) of this pipe?
Volume = πR²L or AL
0.810” Diameter = 0.405” Radius (.810”/2)
Area = πR² = π x 0.405”² = π x ”
= x ”
= sq.in.
Volume = πR²L = ”² x 47.0” length
= cubic inches (solid volume)
”³ / 1728 ”³ = cubic feet (solid volume)
47”
Fluid Volume: ”³ divided by 231”³ = gal.
or cubic feet x 7.47 gal. = gal.
Which can then be converted to fluid ounces by: x 128oz. = fl. oz.

26. Industrial Applications: Calculations: Volume of Pipe
0.810”
A section of white ¾ inch PVC pipe is 37in. long. What is the solid and the fluid volume (water) of this pipe?
Volume = πR²L or AL
0.810” Diameter = 0.405” Radius (.810”/2)
Area = πR² = π x 0.405”² = π x ”
= x ”
= sq.in.
Volume = πR²L = ”² x 37.0” length
= cubic inches (solid volume)
”³ / 1728 ”³ = cubic feet (solid volume)
37”
Fluid Volume: ”³ divided by 231”³ = gal.
or cubic feet x 7.47 gal. = gal.
Which can then be converted to fluid ounces by: x 128oz. = fl. oz.

27. Industrial Applications: Calculations: Volume of Pipe
0.528”
A section of gray, schedule 80 - ½ inch PVC pipe is in. long. What is the solid and the fluid volume (water) of this pipe?
Volume = πR²L or AL
0.528” Diameter = 0.264” Radius (.528”/2)
Area = πR² = π x 0.264”² = π x ”
= x ”
= sq.in.
Volume = πR²L = ”² x ” length
= cubic inches (solid volume)
”³ / 1728 ”³ = cubic feet (solid volume)
46.375”
Fluid Volume: ”³ divided by 231”³ = gal.
or cubic feet x 7.47 gal. = gal.
Which can then be converted to fluid ounces by: x 128oz. = 5.62 fl. oz.

28. Industrial Applications: Calculations: Volume of Pipe
0.528”
A section of gray, schedule 80 - ½ inch PVC pipe is in. long. What is the solid and the fluid volume (water) of this pipe?
Volume = πR²L or AL
0.528” Diameter = 0.264” Radius (.528”/2)
Area = πR² = π x 0.264”² = π x ”
= x ”
= sq.in.
Volume = πR²L = ”² x ” length
= cubic inches (solid volume)
”³ / 1728 ”³ = cubic feet (solid volume)
51.375”
Fluid Volume: ”³ divided by 231”³ = gal.
or cubic feet x 7.47 gal. = gal.
Which can then be converted to fluid ounces by: x 128oz. = 6.23 fl. oz.