Industrial Skills Volume: Descriptions, Calculations & Industrial Applications

Industrial Skills

Volume: Descriptions, Calculations & Industrial Applications

Description: AreaA 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.

Description: VolumeA 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.

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 = 8.3453 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 = 264.2 gallons of water*• 1 liter of water = 1.000 Kilogram*

Industrial Applications:Container Questions – Gallons & Weight

1 Calculate the volume of the inside of the container assigned to you and your partner(s).

2 Calculate how many gallons of water the container could theoretically hold.• 1 cubic foot = 7.47 gallons of water*

3 Calculate how much the water in the container would weigh.• 1 gallon of water = 8.3453 pounds*

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” = 15,813.984cu.in. ÷ 1,728cu.in.

= 9.152 cubic feet

9.152 cu.ft. x 7.47 gallons = 68.365 gallons of water

68.365 gal. x 8.34 pounds = 570.164 pounds

Other Calculations:

Box A: Wood Box with Handles



= 3.556 cubic feet

Inside Dimensions: 9¾” x 13¾” x 29¾”

9.75” x 13.75” x 29.75” = 3,988.359cu.in. ÷ 1,728cu.in.

= 2.308 cubic feet



Other Calculations:

Box B: Small Wood Box



= 2.963 cubic feet

Inside Dimensions: 14” x 14” x 18”

14” x 14” x 18” = 3,528cu.in. ÷ 1,728cu.in.

= 2.042 cubic feet



Other Calculations:

Box C: “Grainger” Cardboard Box

Outside Dimensions: 12½” x 12½” x 17¾”

12.5” x 12.5” x 17.75” = 2885.484 cu.in. divided by 1,728cu.in.

= 1.670 cubic feet

Inside Dimensions: 12¼” x 12¼” x 17½”

12.25” x 12.25” x 17.5” = 2,626.094 cu.in. ÷ 1,728cu.in.

= 1.5197 cubic feet



Other Calculations:

Box D: Plain Cardboard Box

Outside Dimensions: 15¼” x 15¼” x 15¼”


= 2.052 cubic feet

Inside Dimensions: 15” x 15” x 15”

15” x 15” x 15” = 3,375 cu.in. ÷ 1,728cu.in.

= 1.953 cubic feet



Other Calculations:

Stackable Fish Box:

Inside Dimensions (Rough): 11¼” x 16¾” x 29¾”


= 3.244 cubic feet

Subtract a total of .248 cu. ft.

True Inside Volume* = 2.996 cubic feet



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*: 2.996 cu.ft.

• Volume of Water: 22.380 gallons

Industrial Applications:Volume Question – Sea Salt Dilution

• 7 pounds of “Salt” to 25 gallons of water.• Fish Box: Volume*: 2.996 cu.ft.

• Volume of Water: 22.380 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 22.380 gal. x 4.48 oz. = 100.2624 oz. of salt 100.2624 divided by 16 oz. = 6.2664 pounds of salt

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”


• ANSWER A:• Trailer: Inside - 7ft.7in. by 4ft.9in. by 5ft.8in.

• 204.118 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” = 88.75 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 1.605 cu.ft. = 160.5 cu.ft.)

Container Question – Space UtilizationAnswer A:

57”

91”inside

inside

•20 Boxes on bottom. (5 boxes long x 4 boxes wide)•5 x 17.75” = 88.75 inches (possible 91”)•4 x 12.5” = 50 inches (possible 57”)

57”inside

68”Inside

height

•5 Layers of Boxes High. (5 layers x 20 boxes)•5 x 12.5” = 62.5 inches (possible 68”)

Answer: 100 Boxes (100 x 1.605 cu.ft. = 160.5 cu.ft.)


• ANSWER B:• Trailer: Inside - 7ft.7in. by 4ft.9in. by 5ft.8in.

• 204.118 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” = 53.25 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 1.605 cu.ft. = 168.525cu.ft.)

Container Question – Space UtilizationAnswer B:

57”

91”inside

inside

•21 Boxes on bottom. (7 boxes long x 3 boxes wide)•7 x 12.5” = 87.5 inches (possible 91”)•3 x 17.75” = 53.25 inches (possible 57”)

57”inside

68”Inside

height

•5 Layers of Boxes High. (5 layers x 21 boxes)•5 x 12.5” = 62.5 inches (possible 68”)

•Answer: 105 Boxes (105 x 1.605 cu.ft. = 168.525cu.ft.)

Industrial Applications:Calculations: Volume of Cylinders (Pipe)

Volume = πR²L or AL

Volume is always expressed in cubic units.

6.031”

6.625”


Volume = πR²L or AL *Always square radius first

6.031”

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” Diameter = 3.0155” Radius (6.031”/2)Area = πR² = π x 3.0155” ² = π x 9.0932” = 3.1416 x 9.0932” = 28.5672 sq.in.Volume = πR²L = 28.5672” ² x 42” length = 1199.8224 cubic inches1199.8224”³ / 1728 ”³ = .6943 cubic feet

42”


•Volume = 1199.8224 ”³

• = .6943 cubic feet

•1 cubic foot = 7.47 gallons of water*•1 gallon of water = 231cu.in. (1728 ”³/7.47gal.)

•1 gallon of water = 8.3453 pounds*

6.031”

What is the fluid volume of the pipe?

Volume = 1199.8224”³ divided by 231”³ = 5.1940 gal.

42”

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”

21.25”Volume = πR²L or AL

3.042” Diameter = 1.521” Radius (3.042”/2)Area = πR² = π x 1.521” ² = π x 2.3134” = 3.1416 x 2.3134” = 7.2679 sq.in.Volume = πR²L = 7.2679” ² x 21.25” length = 154.4427 cubic inches (solid volume)

154.4427”³ / 1728 ”³ = .08938 cubic feet (solid volume)

Fluid Volume: 154.4427”³ divided by 231”³ = .66858 gal. or .08938 cubic feet x 7.47 gal. = .66767 gal.

Which can then be converted to fluid ounces by: .66767 x 128oz. = 85.46 fl. oz.


A section of white 2 inch PVC pipe is 61.375 in. long. What is the solid and the fluid volume (water) of this pipe?

2.049”

61.375”

Volume = πR²L or AL 2.049” Diameter = 1.0245” Radius (2.049”/2)Area = πR² = π x 1.0245”² = π x 1.0496” = 3.1416 x 1.0496” = 3.2974 sq.in.Volume = πR²L = 3.2974” ² x 61.375” length = 202.3779 cubic inches (solid volume)





A section of white 1 inch PVC pipe is 47.125 in. long. What is the solid and the fluid volume (water) of this pipe?

1.033”

47.125”

Volume = πR²L or AL 1.033” Diameter = 0.5165” Radius (1.033”/2)Area = πR² = π x 0.5165”² = π x 0.26677” = 3.1416 x 0.26677” = 0.83808 sq.in.Volume = πR²L = .83808”² x 47.125” length = 39.49452 cubic inches (solid volume)





A section of white ¾ inch PVC pipe is 47in. long. What is the solid and the fluid volume (water) of this pipe?

0.810”

47”

Volume = πR²L or AL 0.810” Diameter = 0.405” Radius (.810”/2)Area = πR² = π x 0.405”² = π x 0.164025” = 3.1416 x 0.164025” = 0.515300 sq.in.Volume = πR²L = .515300”² x 47.0” length = 24.2191 cubic inches (solid volume)





A section of white ¾ inch PVC pipe is 37in. long. What is the solid and the fluid volume (water) of this pipe?

0.810”

37”






A section of gray, schedule 80 - ½ inch PVC pipe is 46.375in. long. What is the solid and the fluid volume (water) of this pipe?

0.528”

46.375”






A section of gray, schedule 80 - ½ inch PVC pipe is 51.375in. long. What is the solid and the fluid volume (water) of this pipe?

0.528”

51.375”





Documents

Industrial Skills Volume: Descriptions, Calculations & Industrial Applications