1. PA DEP Bureau of Deep Mine Safety
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
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. Liquid Measure, in Gallons Weight Measure, in Pounds
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 = 231 cu inches 1 gallon-water = 0.134 cu feet
1 gallon-water = pounds
1 gallon-water = 231 cu inches
1 gallon-water = 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 = 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. Example:
Calculate the volume of a rectangular sump with a length of 25 feet, 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
Length 25 ft
Width
10 ft
Depth
15 ft
BDMS / PSU
BDMS / PSU

12. 7.5 gal/cu ft x 3,750 cu ft = 28,125 gallons
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, lb
or tons
The weight of this water will be:
62.5 lbs/cu ft X 3,750 cu. Ft. = 234,375 lbs.
Or tons
BDMS / PSU

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

14. Solution: Volume = length x width x depth
Volume = 50 ft x 25 ft x 15 ft
Volume = 18,750 ft3
50 ft
15 ft
25 ft
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 tons
The weight of this water will be:
62.5 lbs/cu ft X 18,750 cu. Ft. = 1,171,875 lbs
Or 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
 =
Radius
Depth
BDMS / PSU

17. 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 = x (5 feet)2 x 15 feet
Volume = x 25 ft x 15 ft
Volume = 1,178 cu ft
15 ft
5 ft
BDMS / PSU

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

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

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

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

22. Solution: Volume =  x r2 x depth Volume =  x (5 ft)2 x 30 ft
Volume = 2,356.2 cu ft
30 ft
10 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, lb
Or tons
The weight of this water will be:
62.5 cu ft/lb X 2,356.2 cu.ft. = 147,262.5 lbs.
Or tons
BDMS / PSU

24. What is the weight of this section of pipe, if full of water?
36 inches
Volume =  x r2 x depth
Volume = x (18 inches)2 x 1,000 ft
Volume = x (1.5 ft)2 x 1,000 ft
Volume = cu ft
What is the weight of this section of pipe, if full of water?
7.5 gal / cu ft x cu ft = 53,014.5 gal
8.342 lb / gal x 53,014.5 = 442, lb
Or 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. 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
Volume = 160 cu ft
8 ft
5 ft
BDMS / PSU

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

28. Solution: Volume = Base x Height x Length 2
Volume = 15 ft x 10 ft x 12 ft
Volume = 900 ft3
15 ft
10 ft
12 ft
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
The weight of this water will be:
62.5 lbs/cu ft X 900 cu ft = 56,250 lbs.
Or tons
BDMS / PSU

30. 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.
10 ft
15 ft
20 ft
Answer: 1,500 cu ft
BDMS / PSU

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

32. 7.5 gal/cu ft x 1,500 cu ft = 11,250 gallons
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 tons
The weight of this water will be:
62.5 lbs. X 1,500 cu. Ft. = 93,750lbs.
Or tons
BDMS / PSU

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

34. Example: Calculate the volume of a sphere with a diameter of 15 feet.
Volume = x (15 ft)3 6
Volume = 1, cu ft
15 ft
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35. 7.5 gal/cu ft x 1,767.15 cu ft = 13,253.62 gallons
The sump capacity in gallons will be:
7.5 gal/cu ft x 1, cu ft = 13, gallons
8.342 lbs/gal x 13, gal = 110, lb
Or tons
The weight of this water will be:
62.5 lbs. X 1,767.15cu. Ft. = 110,446.87lbs.
Or 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. Solution: Volume =  x (diameter)3 6 Volume = 3.1416 x (20 ft)3
Volume = 4,188.8 ft3
20 ft.
BDMS / PSU

38. 7.5 gal/cu ft x 4,188.8 cu ft = 31,416 gallons
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, lb
Or tons
The weight of this water will be:
62.5 lbs. X 4,188.8 cu ft = 261,800 lbs
Or tons
BDMS / PSU

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

40. Solution: Volume =  x (diameter)3 6 Volume = 3.1416 x (12.5 ft)3
Volume = 1, ft3
12.5 ft.
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 H QH
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
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
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

51. 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 H QH
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

52. BDMS / PSU

53. WHP = Flow(GPM) (X) Pressure(PSI) (/) 1714
Horsepower
The Horsepower required to operate a Positive Displacement Pump has two factors:
The Work Horsepower (WHP) - the actual work done
WHP = Flow(GPM) (X) Pressure(PSI) (/) 1714
The Viscous Horsepower(VHP) - the power required to turn the rotors, gears, etc. inside the viscous fluid. The Viscous Horsepower required is determined by the pump design and speed and is supplied by the pump manufacturer.
HP = WHP + VHP
BDMS / PSU

54. Horsepower
Horsepower required to pump 400 GPM to an elevation of 300’ assuming the friction loss in the pipes amounted to 15% of the static head.
400 GPM x 345 x 8.5 = horsepower
33,000
BDMS / PSU

55. H = Hs + Hf + Hv + Hsh H = total head
Hs = is the vertical distance in feet from the suction liquid level to the discharge liquid level (total static head)
Hf = is the equivalent head, expressed as feet of liquid, required to overcome the friction caused by the flow through the pipe (friction head)
Hv = is the head, in feet required to create velocity of flow (velocity head)–
Note: in most cases, this value is negligible and is often ignored.
Hsh = is the head, in feet required to overcome the shock losses due to changes of water flow produced by fittings
BDMS / PSU

56. The vertical height difference from surface of water source to discharge point is termed as total static head
Static Discharge Head
Suction Line
A dropped rock or other object will gain speed rapidly as it falls. Measurements show that an object dropping 1 foot (ft) will reach a velocity of 8.02 feet per second (ft/s). An object dropping 4 ft will reach a velocity of ft/s. After an 8 ft drop, the velocity attained is ft/s. The force of gravity causes this gain in speed or acceleration, which is equal to 32.2 feet per second per second (ft/s2). This acceleration caused by gravity is referred to as g.
If water is stored in a tank and a small opening is made in the tank wall 1 ft below the water surface, the water will spout from the opening with a velocity of 8.02 ft/s. This velocity has the same magnitude that a freely falling rock attains after falling 1 ft. Similarly, at openings 4 ft and 8 ft below the water surface, the velocity of the spouting water will be and ft/s, respectively. Thus, the velocity of water leaving an opening under a given head, h, is the same as the velocity that would be attained by a body falling that same distance. The equation that shows how velocity changes with h and defines velocity head is:
This may also be written in velocity head form as:
Pump
Discharge Line
Static Suction Lift
Sump
BDMS / PSU

57. The vertical height difference from surface of water source to centerline of impeller is termed as static suction head or suction lift ('suction lift' can also mean total suction head).
Suction Line
A dropped rock or other object will gain speed rapidly as it falls. Measurements show that an object dropping 1 foot (ft) will reach a velocity of 8.02 feet per second (ft/s). An object dropping 4 ft will reach a velocity of ft/s. After an 8 ft drop, the velocity attained is ft/s. The force of gravity causes this gain in speed or acceleration, which is equal to 32.2 feet per second per second (ft/s2). This acceleration caused by gravity is referred to as g.
If water is stored in a tank and a small opening is made in the tank wall 1 ft below the water surface, the water will spout from the opening with a velocity of 8.02 ft/s. This velocity has the same magnitude that a freely falling rock attains after falling 1 ft. Similarly, at openings 4 ft and 8 ft below the water surface, the velocity of the spouting water will be and ft/s, respectively. Thus, the velocity of water leaving an opening under a given head, h, is the same as the velocity that would be attained by a body falling that same distance. The equation that shows how velocity changes with h and defines velocity head is:
This may also be written in velocity head form as:
Pump
Discharge Line
Static Suction Lift
Sump
BDMS / PSU

58. The vertical height difference from centerline of impeller to discharge point is termed as static discharge head.
Static Discharge Head
Suction Line
A dropped rock or other object will gain speed rapidly as it falls. Measurements show that an object dropping 1 foot (ft) will reach a velocity of 8.02 feet per second (ft/s). An object dropping 4 ft will reach a velocity of ft/s. After an 8 ft drop, the velocity attained is ft/s. The force of gravity causes this gain in speed or acceleration, which is equal to 32.2 feet per second per second (ft/s2). This acceleration caused by gravity is referred to as g.
If water is stored in a tank and a small opening is made in the tank wall 1 ft below the water surface, the water will spout from the opening with a velocity of 8.02 ft/s. This velocity has the same magnitude that a freely falling rock attains after falling 1 ft. Similarly, at openings 4 ft and 8 ft below the water surface, the velocity of the spouting water will be and ft/s, respectively. Thus, the velocity of water leaving an opening under a given head, h, is the same as the velocity that would be attained by a body falling that same distance. The equation that shows how velocity changes with h and defines velocity head is:
This may also be written in velocity head form as:
Pump
Discharge Line
Sump
BDMS / PSU

59. FRICTION LOSS The amount of pressure / head required to 'force' liquid through pipe and fittings.
Pressure Gauge
Suction Line
A dropped rock or other object will gain speed rapidly as it falls. Measurements show that an object dropping 1 foot (ft) will reach a velocity of 8.02 feet per second (ft/s). An object dropping 4 ft will reach a velocity of ft/s. After an 8 ft drop, the velocity attained is ft/s. The force of gravity causes this gain in speed or acceleration, which is equal to 32.2 feet per second per second (ft/s2). This acceleration caused by gravity is referred to as g.
If water is stored in a tank and a small opening is made in the tank wall 1 ft below the water surface, the water will spout from the opening with a velocity of 8.02 ft/s. This velocity has the same magnitude that a freely falling rock attains after falling 1 ft. Similarly, at openings 4 ft and 8 ft below the water surface, the velocity of the spouting water will be and ft/s, respectively. Thus, the velocity of water leaving an opening under a given head, h, is the same as the velocity that would be attained by a body falling that same distance. The equation that shows how velocity changes with h and defines velocity head is:
This may also be written in velocity head form as:
Pump
Discharge Line
Sump
BDMS / PSU

60. Friction Loss Hf = f L V2 D f is pipe coefficient of friction;
L is length of pipe;
V is velocity of water;
D is diameter of pipe
BDMS / PSU

61. Frictional Head
Is usually expressed by the following equation based upon upon the number of 100-ft lengths of pipe in the system:
Hf = (100/C)1.85[ (q1.85) ]
(d4.8655)
Where C is a constant, usually 100, accounting for surface roughness;
q is the flow in gallons per minute;
d is the inside diameter of the pipe in inches.
BDMS / PSU

62. Equivalent Number of Feet of Staight Pipe for Different Fittings
BDMS / PSU

63. Friction Loss in Feet for Old Pipe (C = 100)
BDMS / PSU

64. Velocity Head is the velocity head of liquid moving at a given velocity in the equivalent head through which it would have to fall to acquire the same velocity.
BDMS / PSU

65. A dropped rock or other object will gain speed rapidly as it falls.
Measurements show that an object dropping 1 foot (ft) will reach a velocity of 8.02 feet per second (ft/s).
An object dropping 4 ft will reach a velocity of ft/s.
After an 8 ft drop, the velocity attained is ft/s.
The force of gravity causes this gain in speed or acceleration, which is equal to 32.2 feet per second per second (ft/s2).
This acceleration caused by gravity is referred to as g.
BDMS / PSU

66. If water is stored in a tank and a small opening is made in the tank wall 1 ft below the water surface, the water will spout from the opening with a velocity of 8.02 ft/s.
This velocity has the same magnitude that a freely falling rock attains after falling 1 ft.
Similarly, at openings 4 ft and 8 ft below the water surface, the velocity of the spouting water will be and ft/s, respectively.
BDMS / PSU

67. Head Velocity
Hv is the velocity head of liquid moving at a given velocity in the equivalent head through which it would have to fall to acquire the same velocity.
Hv = V2 2g
Hv is velocity head in feet;
V is velocity of water in feet per second;
G is acceleration due to gravity, in feet per sec2.
BDMS / PSU

68. Thus, the velocity of water leaving an opening under a given head, H, is the same as the velocity that would be attained by a body falling that same distance. The equation that shows how velocity changes with H and defines velocity head is:
BDMS / PSU

69. S.G. Specific gravity. Weight of liquid in comparison to water at approx 20 deg c (SG = 1).
BDMS / PSU

70. Horsepower
Horsepower required to pump 400 GPM to an elevation of 300’ assuming the friction loss in the pipes amounted to 15% of the static head.
400 GPM x 345 x 8.5 = horsepower
33,000
BDMS / PSU

71. How long would it take 40 horsepower pump to pump 88,000 gallons to a total head of 360 feet?
33,000 x 40 = 439 GPM
360 x 8.35
88,000 = 200 minutes
439
BDMS / PSU

72. What is the weight of one cubic foot of Water?
Answer: Sixty-two and five tenths (62.5) pounds
What is the weight of one (1) gallon of water?
Answer: Eight and one third (8.342) pounds
BDMS / PSU

73. How many gallons are in one (1) cubic foot ?
Answer: Seven and five tenths (7.5) gallons
What is the pressure exerted by a column of water one (1) foot high and on one square inch of surface?
Answer: pounds pounds
144
BDMS / PSU

74. What is the volume of a body of dead water in a sump hole 25 foot deep by 500 feet by 1 foot?
Answer: Volume = Length x width x depth
V = 500 ft x 1 ft x 25 ft
V = 12,500 cu ft
BDMS / PSU

75. How long would it take a 40 H. P
How long would it take a 40 H.P. Pump to pump 88,000 gallons to a head of 360 feet?
GPM = 33,000 x HP
Head x 8.342
GPM = 33,000 x 40
360 x 8.342
GPM = 132,000
300,312
GPM =
Time = Volume__
GPM x 60
Time = 88,000____
x 60
Time = 3.33 hours
BDMS / PSU

76. Problem 1:
If atmospheric pressure pushes mine water up a suction line due to the vacuum created by a pump, is there a limitation as to the maximum length of suction line? If so, what is the value?
BDMS / PSU

77. Solution Problem 1:
At sea level, atmospheric pressure is equal to 14.7 psi. If a perfect vacuum were to be created in a suction line, atmospheric pressure could push a 1- in. column of water to a height of:
Pressure = weight of water column
Divide atmospheric pressure at sea level by lb/in3 (the weight of one cubic inch of water) to obtain the theoretical suction lift.
14.7 (lb/in2) / (lb/in3) = (inches) (inches) / 12 (inches per foot) = 33.9 (ft)
BDMS / PSU

78. Theoretical Suction Lift
At sea level the atmosphere exerts a force of 14.7 lb/in2 (PSI) on the earth's surface.
The weight of the atmosphere on a body of water will prevent lift from occurring unless an area of low pressure is created.
BDMS / PSU

79. Theoretical Suction Lift
In tube (A) atmospheric pressure is the same inside the tube as it is outside: 14.7 PSI. Since the weight of the atmosphere is being exerted equally across the surface, no change occurs in the water level inside the tube.
BDMS / PSU

80. Theoretical Suction Lift
In tube (B) a perfect vacuum is created making atmospheric pressure greater on the water outside the tube. The resulting differential causes water, flowing naturally to the area of lowest pressure to begin filling the tube until it reaches a height of 33.9 feet.
BDMS / PSU

81. Theoretical Suction Lift
Why is 33.9 feet the highest water can be lifted in this example? Because at this point the weight of the water inside the tube exerts a pressure equal to the weight of the atmosphere pushing down on the ocean's surface. This height represents the maximum theoretical suction lift and can be verified using the following calculation.
BDMS / PSU

82. Theoretical Suction Lift
Divide atmospheric pressure at sea level by lb/in3 (the weight of one cubic inch of water) to obtain the theoretical suction lift.   14.7 (lb/in2) / (lb/in3) = (inches) (inches) / 12 (inches per foot) = 33.9 (ft)
This shows that the maximum total suction lift, when pumping water at sea level, is approximately 34 ft. This figure is theoretical and can never be obtained in practice. That is why good mining practice dictates that the total dynamic suction lift should never exceed a value of approximately 20 ft. for trouble-free operations
BDMS / PSU

83. Head = Pressure x 2.31 Specific Gravity
Theoretically we could come out of the water tank, go straight up into the air with the hose to a height of almost 34 feet and then down into the container and the water would flow with no trouble. I say theoretically because there is friction in the hose that offers resistance to the flow and that friction has to be considered any time you are trying to calculate the possibility of using a siphon to move liquids.
Head = 14.7 psi x 2.31 = Feet
1.0
BDMS / PSU

84. Problem 2:
What is the required brake horsepower to pump 150 gpm (gallons per minute) against a total dynamic head of 370 ft if the pump operates at 70 % efficiency?
BDMS / PSU

85. Solution Problem 2: HPB = QH (8.33) 33,000 E
HPB = (150gpm)(370 ft)(8.33)
(33,000)(.7)
HPB =
23100
HPB = hp
BDMS / PSU

86. Cost to Pump Water – Electric
$ per hour = gpm x head in feet x x rate per KWH
3960 x Pump Efficiency x Electric Motor Efficiency
BDMS / PSU

87. WHP = Flow(GPM) (X) Pressure(PSI) (/) 1714
Horsepower
The Horsepower required to operate a Positive Displacement Pump has two factors:
The Work Horsepower (WHP) - the actual work done
WHP = Flow(GPM) (X) Pressure(PSI) (/) 1714
The Viscous Horsepower(VHP) - the power required to turn the rotors, gears, etc. inside the viscous fluid. The Viscous Horsepower required is determined by the pump design and speed and is supplied by the pump manufacturer.
HP = WHP + VHP
BDMS / PSU

88. H = Hs + Hf + Hv + Hsh H = total head
Hs = is the vertical distance in feet from the suction liquid level to the discharge liquid level (total static head)
Hf = is the equivalent head, expressed as feet of liquid, required to overcome the friction caused by the flow through the pipe (friction head)
Hv = is the head, in feet required to create velocity of flow (velocity head)–
Note: in most cases, this value is negligible and is often ignored.
Hsh = is the head, in feet required to overcome the shock losses due to changes of water flow produced by fittings
BDMS / PSU