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Industrial Skills

Industrial Skills. Volume: Descriptions, Calculations & Industrial Applications. Description: Area A Two-Dimensional Quantity. Area measurements utilize linear measurements to calculate the number of unit squares within a given boundary.

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Industrial Skills

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

  2. 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.

  3. 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.

  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 = 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*

  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 = 8.3453 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” = 15,813.984cu.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

  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. = 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 2.308 cu.ft. x 7.47 gallons = 17.241 gallons of water 17.241 gal. x 8.34 pounds = 143.790 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. = 2.704 cubic feet Inside Dimensions: 14” x 14” x 16½” 14” x 14” x 16. 5” = 3,234cu.in. ÷ 1,728cu.in. = 1.872 cubic feet 1.872 cu.ft. x 7.47 gallons = 13.984 gallons of water 13.984 gal. x 8.34 pounds = 116.627 pounds

  9. 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 1.520 cu.ft. x 7.47 gallons = 11.354 gallons of water 11.354 gal. x 8.34 pounds = 94.692 pounds

  10. Other Calculations: Box D: Plain Cardboard Box Outside Dimensions: 15¼” x 15¼” x 15¼” 15.25” x 15.25” x 15.25” = 3546.578 cu.in. divided by 1,728cu.in. = 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 1.953 cu.ft. x 7.47 gallons = 14.589 gallons of water 14.589 gal. x 8.34 pounds = 121.672 pounds

  11. Other Calculations: • Stackable Fish Box: • Inside Dimensions (Rough): 11¼” x 16¾” x 29¾” • 11.25” x 16.75” x 29.75” = 5606.016 cu.in. divided by 1,728cu.in. • = 3.244 cubic feet • Subtract a total of .248 cu. ft. • True Inside Volume* = 2.996 cubic feet • 2.996 cu.ft. x 7.47 gallons = 22.380 gallons of water • 22.380 gal. x 8.34 pounds = 186.649 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*: 2.996 cu.ft. • Volume of Water: 22.380 gallons

  13. 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

  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. • 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.)

  16. Container Question – Space UtilizationAnswer 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 1.605 cu.ft. = 160.5 cu.ft.) • 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”)

  17. Industrial Applications:Container Question – Space Utilization • 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.)

  18. Container Question – Space UtilizationAnswer 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 1.605 cu.ft. = 168.525cu.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” = 53.25 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 = 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 inches 1199.8224”³ / 1728 ”³ = .6943 cubic feet

  21. Industrial Applications:Calculations: Volume of Cylinders (Pipe) What is the fluid volume of the pipe? 6.031” • 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* 42” Volume = 1199.8224”³ divided by 231”³ = 5.1940 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 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) 21.25” 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.

  23. Industrial Applications:Calculations: Volume of Pipe 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” 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) 202.3779”³ / 1728 ”³ = .11712 cubic feet (solid volume) 61.375” Fluid Volume: 202.3779”³ divided by 231”³ = .87610 gal. or .11712 cubic feet x 7.47 gal. = .87489 gal. Which can then be converted to fluid ounces by:.87489 x 128oz. = 111.99 fl. oz.

  24. Industrial Applications:Calculations: Volume of Pipe 1.033” 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? 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) 39.49452”³ / 1728 ”³ = .022856 cubic feet (solid volume) 47.125” Fluid Volume: 39.49452”³ divided by 231”³ = .17097 gal. or .022856 cubic feet x 7.47 gal. = .17073 gal. Which can then be converted to fluid ounces by:.17073 x 128oz. = 21.85 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 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) 24.2191”³ / 1728 ”³ = .01402 cubic feet (solid volume) 47” Fluid Volume: 24.2191”³ divided by 231”³ = .10485 gal. or .01402 cubic feet x 7.47 gal. = .10473 gal. Which can then be converted to fluid ounces by:.10473 x 128oz. = 13.41 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 0.164025” = 3.1416 x 0.164025” = 0.515300 sq.in. Volume = πR²L = .515300”² x 37.0” length = 19.0661 cubic inches (solid volume) 19.0661”³ / 1728 ”³ = .01103 cubic feet (solid volume) 37” Fluid Volume: 19.0661”³ divided by 231”³ = .08254 gal. or .01103 cubic feet x 7.47 gal. = .08239 gal. Which can then be converted to fluid ounces by:.08239 x 128oz. = 10.55 fl. oz.

  27. Industrial Applications:Calculations: Volume of Pipe 0.528” 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? Volume = πR²L or AL 0.528” Diameter = 0.264” Radius(.528”/2) Area = πR² = π x 0.264”² = π x 0.069696” = 3.1416 x 0.069696” = 0.21896 sq.in. Volume = πR²L = .21896”² x 46.375” length = 10.15427 cubic inches (solid volume) 10.15427”³ / 1728 ”³ = .00588 cubic feet (solid volume) 46.375” Fluid Volume: 10.15427”³ divided by 231”³ = .04396 gal. or .00588 cubic feet x 7.47 gal. = .04392 gal. Which can then be converted to fluid ounces by:.04392 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 51.375in. 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 0.069696” = 3.1416 x 0.069696” = 0.21896 sq.in. Volume = πR²L = .21896”² x 51.375” length = 11.24907 cubic inches (solid volume) 11.24907”³ / 1728 ”³ = .00651 cubic feet (solid volume) 51.375” Fluid Volume: 11.24907”³ divided by 231”³ = .048698 gal. or .00651 cubic feet x 7.47 gal. = .048630 gal. Which can then be converted to fluid ounces by:.048630 x 128oz. = 6.23 fl. oz.

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