Industrial skills
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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

Industrial Skills

Volume: Descriptions, Calculations & Industrial Applications


Description area a two dimensional quantity

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 volume a three dimensional quantity

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

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

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*


Industrial applications calculations

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

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


Other calculations1

Other Calculations:

Box B: Small Wood Box

Outside Dimensions: 16” x 16” x 20”

16” x 16” x 20” = 5,120 cu.in. divided by 1,728cu.in.

= 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

2.042 cu.ft. x 7.47 gallons = 15.254 gallons of water

15.254 gal. x 8.34 pounds = 127.218 pounds


Other calculations2

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


Other calculations3

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


Other calculations4

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


Industrial applications volume question sea salt dilution

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 dilution1

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

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”


Industrial applications container question space utilization1

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.)


  • Container question space utilization answer a

    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”)


    Industrial applications container question space utilization2

    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.)


  • Container question space utilization answer b

    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”)


    Industrial applications calculations volume of cylinders pipe

    Industrial Applications:Calculations: Volume of Cylinders (Pipe)

    Volume = πR²L or AL

    Volume is always expressed in cubic units.

    6.031”

    6.625”


    Industrial applications calculations volume of cylinders pipe1

    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


    Industrial applications calculations volume of cylinders pipe2

    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.


    Industrial applications calculations volume of pipe

    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.


    Industrial applications calculations volume of pipe1

    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.


    Industrial applications calculations volume of pipe2

    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.


    Industrial applications calculations volume of pipe3

    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.


    Industrial applications calculations volume of pipe4

    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.


    Industrial applications calculations volume of pipe5

    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.


    Industrial applications calculations volume of pipe6

    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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