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Mine Drainage

Learn how to calculate water volume in mine sumps using depth, width, and length parameters. Discover the efficiency of pumps in draining mine sumps and solve conversion problems related to liquid measures and weights.

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Mine Drainage

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  1. Mine Drainage PA DEP Bureau of Deep Mine Safety BDMS / PSU

  2. A sump is 300 feet long and 20 feet wide with a depth of 12 feet, with a flow coming into the sump thru a 6 inch pipe with a rate of 300 gallons per minute. The pump has a efficiency rating of 60%. How long will it take you to pump the sump dry? What size pump do you need? BDMS / PSU

  3. Volume = rate x time • Rate = volume / time • Time = volume / rate BDMS / PSU

  4. Using a pipe that has a rate of 75 gallons per minute, it took you 2.5 hours to fill up a sump. What is the volume of the sump? • Solution: Volume = Rate x Time Volume = 75 gpm x 2.5 hour Volume = (75 gpm x 60 minutes x 2.5 hours) Volume = 11,250 gallon BDMS / PSU

  5. ( ) Equivalent Flow: How many 4 inch pipes are needed to carry the flow from a 12” pipe? 5 N = Large Diameter Small Diameter ( ) 5 N = _12”_ 4” 4 “ N = ( ) 5 3” ( ) 12 “ N = 243” N = 15.5 (4 INCH PIPES) BDMS / PSU

  6. Conversion Factors for Mine Drainage Problems Liquid Measure, in Gallons Volumetric Measure Weight Measure in Pounds In Cubic Inches In Cubic Feet 1 231 0.134 8.342 1 Cubic Foot 1 Cubic Foot Gallons / Cubic Foot Weight Measure, in Pounds 1 1728 cu in 7.481 (use 7.5) 62.5 BDMS / PSU

  7. 1 gallon-water = 8.345 pounds • 1 gallon-water = 231 cu inches • 1 gallon-water = 0.134 cu feet • 1 cu. ft. of water = 1728 cu in • 1 cu. ft. of water = 7.48 gallons (7.5 gal) • 1 cu. ft. of water = 62.425 lb. BDMS / PSU

  8. Mine Drainage • Volume. • In this module we will expand on our knowledge of calculating: • Area. • Volume of the various shaped containers that are components of a water handling system. BDMS / PSU

  9. Basic Three-dimensional Shapes Cylinder Rectangle Trapezoid Pyramid or Prism Sphere BDMS / PSU

  10. Calculating the Volume of a Rectangular Sump • The formula to calculate the volume of a rectangle is: Volume = length x width x depth Volume = (l) x (w) x (d) Depth Length Width BDMS / PSU BDMS / PSU

  11. Depth 15 ft Width 10 ft Length 25 ft Example: • Calculate the volume of a rectangular sump with a lengthof 25feet, a width of 10 feet and a depth of 15 feet. Volume = (l) x (w) x (d) Volume = 25 ft x 10 ft x 15 ft Volume = 3,750 cubic feet BDMS / PSU BDMS / PSU

  12. The sump capacity in gallons will be: 7.5 gal/cu ft x 3,750 cu ft = 28,125 gallons • What is the weight of the water in the sump? 8.342 lbs/gal x 28,125 gal = 234,618.75 lb • or 117.309 tons • The weight of this water will be: 62.5 lbs/cu ft X 3,750 cu. Ft. = 234,375 lbs. • Or 117.1875 tons BDMS / PSU

  13. 15 ft 50 ft 25 ft Practice Exercise: • Calculate the volume of a rectangular sump with a length of 50 feet, a width of 25 feet and a depth of 15 feet. Answer:18,750 cu ft BDMS / PSU BDMS / PSU

  14. 15 ft 50 ft 25 ft Solution: • Volume = length x width x depth • Volume = 50 ft x 25 ft x 15 ft • Volume = 18,750 ft3 BDMS / PSU

  15. The sump capacity in gallons will be: • 7.5 gal/cu ft x 18,750 cu ft = 140,625 gal 8.342 lbs/gal x 140,625 gal = 1,173,093.75lb Or 586.5468 tons • The weight of this water will be: 62.5 lbs/cu ft X 18,750 cu. Ft. = 1,171,875 lbs Or 585.9375 tons BDMS / PSU

  16. Calculating the Volume of a Cylinder • The formula to calculate the volume of a cylinder is: Volume = area of circle x depth Or Volume =  x r2 x depth  = 3.1416 Radius Depth BDMS / PSU

  17. 5 ft 15 ft Example: • Calculate the volume of a cylinder with a radius of 5 feet and a depth of 15 feet. Volume =  x r2 x depth Volume = 3.1416 x (5 feet)2x 15 feet Volume = 3.1416 x 25 ft x 15 ft Volume = 1,178 cu ft BDMS / PSU

  18. The sump capacity in gallons will be: 7.5 gal/cu ft x 1,178 cu ft = 8,835 gallons • 8.342 lbs/gal x 8,835 gal =73,701.57 lb Or 36.8507 tons • The weight of this water will be: 62.5 lbs. X 1,178 cu. Ft. = 73,625 lbs. • Or 36.8125 tons BDMS / PSU

  19. 10 ft 30 ft Practice Exercise: • Calculate the volume of a cylindrical storage tank with a radius of 10 feet and a depth of 30 feet. Answer: 9,424.8 cu ft BDMS / PSU

  20. 10 ft 30 ft Solution: • Volume =  x r2 x depth • Volume =  x (10 ft)2 x 30 ft • Volume = 3.1416 x 100 ft2 x 30 ft • Volume = 9,424.8 cu ft BDMS / PSU

  21. 10 ft 30 ft Practice Exercise: • Calculate the volume of a cylindrical storage tank with a diameter of 10 feet and a depth of 30 feet. Answer: 2,355 cu ft BDMS / PSU

  22. 10 ft 30 ft Solution: • Volume =  x r2 x depth • Volume =  x (5 ft)2 x 30 ft • Volume = 3.1416 x 25 ft2 x 30 ft • Volume = 2,356.2 cu ft BDMS / PSU

  23. The sump capacity in gallons will be: 7.5 gal/cu ft x 2,356.2 cu ft = 17,671.5 gal 8.342 lbs/gal x 17,671.5 gal = 147,415.653 lb Or 73.70 tons • The weight of this water will be: 62.5 cu ft/lb X 2,356.2 cu.ft. = 147,262.5 lbs. • Or 73.63 tons BDMS / PSU

  24. 1000 ft 36 inches • Volume =  x r2 x depth • Volume = 3.1416 x (18 inches)2 x 1,000 ft • Volume = 3.1416 x (1.5 ft)2 x 1,000 ft • Volume = 7068.6 cu ft What is the weight of this section of pipe, if full of water? 7.5 gal / cu ft x 7068.6 cu ft = 53,014.5 gal 8.342 lb / gal x 53,014.5 = 442,246.95 lb Or 221.12 ton BDMS / PSU

  25. Calculating the Volume of a Triangle: • The formula to calculate the volume of a triangular vessel or a trough is: • Volume = area of triangle x length of trough Or • Volume = base x height x length 2 Height Base Length BDMS / PSU

  26. 5 ft 8 ft 8 ft Example: • Calculate the volume of a triangle with a base of 8 feet, a height of 5 feet and a length of 8 feet. Volume = Base x Height x Length 2 Volume = 8 ft x 5 ft x 8 ft 2 Volume = 160 cu ft BDMS / PSU

  27. 10 ft 12 ft 15 ft Practice Exercise: • Calculate the volume of a triangle with a base of 15 feet, a height of 10 feet and a length of 12 feet. Answer: 900 cu ft BDMS / PSU

  28. 10 ft 12 ft 15 ft Solution: • Volume = Base x Height x Length 2 • Volume = 15 ft x 10 ft x 12 ft 2 • Volume = 900 ft3 BDMS / PSU

  29. The sump capacity in gallons will be: 7.5 gal/cu ft x 900 ft3 = 6,750 gallons • 8.342 lbs/gal x 6,750 gal = 56,308.5 Or 28.15425 • The weight of this water will be: 62.5 lbs/cu ft X 900 cu ft = 56,250 lbs. • Or 28.125 tons BDMS / PSU

  30. 15 ft 10 ft 20 ft Practice Exercise: • Calculate the volume of a triangle with a base of 20 feet, a height of 15 feet and a length of 10 feet. Answer: 1,500 cu ft BDMS / PSU

  31. 15 ft 20 ft 10 ft Solution: • Volume = base x height x length 2 • Volume = 20 ft x 15 ft x 10 ft 2 • Volume = 1,500 ft3 BDMS / PSU

  32. The sump capacity in gallons will be: 7.5 gal/cu ft x 1,500 cu ft = 11,250 gallons • 8.342 lbs/gal x 11,250 gal = 93,847.5 lb Or 46.92375 tons • The weight of this water will be: 62.5 lbs. X 1,500 cu. Ft. = 93,750lbs. • Or 46.875 tons BDMS / PSU

  33. Diameter Calculating the Volume of a Sphere • The formula to calculate the volume of a sphere is: Volume =  x (diameter)3 6 Where  = 3.1416 BDMS / PSU

  34. 15 ft Example: • Calculate the volume of a sphere with a diameter of 15 feet. Volume = 3.1416 x (15 ft)3 6 Volume = 1,767.15 cu ft BDMS / PSU

  35. The sump capacity in gallons will be: 7.5 gal/cu ft x 1,767.15 cu ft = 13,253.62 gallons • 8.342 lbs/gal x 13,253.62 gal = 110,561.73 lb Or 55.2808 tons • The weight of this water will be: 62.5 lbs. X 1,767.15cu. Ft. = 110,446.87lbs. • Or 55.22 tons BDMS / PSU

  36. Practice Exercise: • Calculate the volume of sphere with a diameter of 20 feet. 20 ft. Answer: 4,187 cu ft BDMS / PSU

  37. 20 ft. Solution: • Volume =  x (diameter)3 6 • Volume = 3.1416 x (20 ft)3 6 • Volume = 4,188.8 ft3 BDMS / PSU

  38. The sump capacity in gallons will be: 7.5 gal/cu ft x 4,188.8 cu ft = 31,416 gallons • 8.342 lbs/gal x 31,416 gal = 262,072.27 lb Or 131.03 tons • The weight of this water will be: 62.5 lbs. X 4,188.8 cu ft = 261,800 lbs • Or 130.9 tons BDMS / PSU

  39. 12.5 ft. Practice Exercise: • Calculate the volume of sphere with a diameter of 12.5 feet. Answer: 1,022 cu ft BDMS / PSU

  40. 12.5 ft. Solution: • Volume =  x (diameter)3 6 • Volume = 3.1416 x (12.5 ft)3 6 • Volume = 1,022.65 ft3 BDMS / PSU

  41. BDMS / PSU

  42. Pump Characteristic Curves E = ( 8.33 lb of water per gal) (33,000 ft-lb per min) (brake horsepower) BDMS / PSU

  43. Brake Horsepower • The product of the pressure head (H, ft) and the flow (Q, gpm) gives water horsepower or the theoretically minimum horsepower required to produce the desired results. • WHP = Q x 8.33 x HQH 33,000 or 3960 • Q is flow in gallons per minute and H is head in feet; 8.33 = pounds per gallon of water and 33,000 = ft-lb/per min per horsepower. • The efficiency which is output over input or E = WHP/bhp can be expressed: • E = Q (GPM) x H (ft) 3960 x bhp BDMS / PSU

  44. BDMS / PSU

  45. HPout = Head(feet) x Capacity(gpm) x 8.33(lbs./gallon) x SG 33,000 (foot pounds / minute) Head = 160 feet Capacity = 300 gallons per minute 8.33 = the weight of one US gallon SG = specific gravity of water at 68 degrees F 33,000 = the conversion of foot pounds / minute to HP BDMS / PSU

  46. HP = 160 x 300 x 8.33 = 399,840 = 12.1 HP 33,000 33,000 • If we had the pump curve supplied by the pump manufacturer we would learn that he had calculated that it will take 20 horsepower to do this, so our efficiency would be: • 12.1 HPout = .60 or 60% efficient 20 (Hpin ) BDMS / PSU

  47. BDMS / PSU

  48. HPout = Head(feet) x Capacity(gpm) x 8.33(lbs./gallon) x SG 33,000 (foot pounds / minute) Head = 160 feet Capacity = 300 gallons per minute 8.33 = the weight of one US gallon SG = specific gravity of water at 68 degrees F 33,000 = the conversion of foot pounds / minute to HP BDMS / PSU

  49. HP = 160 x 300 x 8.33 = 399,840 = 12.1 HP 33,000 33,000 • If we had the pump curve supplied by the pump manufacturer we would learn that he had calculated that it will take 20 horsepower to do this, so our efficiency would be: • 12.1 HPout = .60 or 60% efficient 20 (Hpin ) BDMS / PSU

  50. BDMS / PSU

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