bdms / psu1 mine drainage pa dep bureau of deep mine safety
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BDMS / PSU 1
Mine DrainageMine Drainage
PA DEP Bureau of Deep Mine PA DEP Bureau of Deep Mine SafetySafety
BDMS / PSU 2
• A sump is 300 feet long and 20 feet A sump is 300 feet long and 20 feet wide with a depth of 12 feet, with a flow wide with a depth of 12 feet, with a flow coming into the sump thru a 6 inch pipe coming into the sump thru a 6 inch pipe with a rate of 300 gallons per minute. with a rate of 300 gallons per minute. The pump has a efficiency rating of The pump has a efficiency rating of 60%. How long will it take you to pump 60%. How long will it take you to pump the sump dry? What size pump do you the sump dry? What size pump do you need?need?
BDMS / PSU 3
• Volume = rate x timeVolume = rate x time
• Rate = volume / timeRate = volume / time
• Time = volume / rateTime = volume / rate
BDMS / PSU 4
• Using a pipe that has a rate of 75 Using a pipe that has a rate of 75 gallons per minute, it took you 2.5 gallons per minute, it took you 2.5 hours to fill up a sump. What is the hours to fill up a sump. What is the volume of the sump?volume of the sump?
• Solution: Solution: Volume = Rate x TimeVolume = Rate x Time
Volume = 75 gpm x 2.5 hourVolume = 75 gpm x 2.5 hour
Volume = (75 gpm x 60 minutes x 2.5 hours)Volume = (75 gpm x 60 minutes x 2.5 hours)
Volume = 11,250 gallonVolume = 11,250 gallon
BDMS / PSU 5
12 “12 “
Large DiameterLarge DiameterSmall DiameterSmall Diameter( )55
N =N =
4 “4 “
( _12”__12”_4”4” ) 55N =N =
N =N = (
(
3”3”
)
) 55
N =N =
N =N = 243”243”
15.5 (4 INCH PIPES)15.5 (4 INCH PIPES)
Equivalent Flow: How Equivalent Flow: How many 4 inch pipes are many 4 inch pipes are needed to carry the flow needed to carry the flow from a 12” pipe?from a 12” pipe?
BDMS / PSU 6
Conversion Factors for Mine Drainage ProblemsConversion Factors for Mine Drainage Problems
Liquid Liquid Measure, in Measure, in
GallonsGallons
11
Volumetric MeasureVolumetric Measure
In Cubic In Cubic InchesInches
231231
In Cubic In Cubic FeetFeet
0.1340.134
Weight Weight Measure Measure in Poundsin Pounds
8.3428.342
11 1728 cu in1728 cu inGallons / Gallons / Cubic FootCubic Foot
7.4817.481
(use 7.5)(use 7.5)
Weight Measure, in Weight Measure, in PoundsPounds
62.562.5
1 Cubic Foot1 Cubic Foot1 Cubic Foot1 Cubic Foot
BDMS / PSU 7
• 1 gallon-water1 gallon-water == 8.345 pounds8.345 pounds• 1 gallon-water1 gallon-water == 231 cu inches231 cu inches• 1 gallon-water1 gallon-water == 0.134 cu feet0.134 cu feet• 1 cu. ft. of water1 cu. ft. of water == 1728 cu in 1728 cu in • 1 cu. ft. of water1 cu. ft. of water == 7.48 gallons (7.5 gal)7.48 gallons (7.5 gal)
• 1 cu. ft. of water 1 cu. ft. of water = = 62.425 lb.62.425 lb.
BDMS / PSU 8
• Volume.Volume.
• In this module we will expand on our In this module we will expand on our knowledge of calculating:knowledge of calculating:– Area.Area.– Volume of the various shaped containers Volume of the various shaped containers
that are components of a water handling that are components of a water handling system.system.
Mine DrainageMine Drainage
BDMS / PSU 9
Basic Three-dimensional ShapesBasic Three-dimensional Shapes
RectangleRectangle CylinderCylinder
Pyramid Pyramid or Prismor PrismSphereSphere
TrapezoidTrapezoid
BDMS / PSU 10
Calculating the Volume of a Calculating the Volume of a Rectangular SumpRectangular Sump
• The formula to calculate the volume of The formula to calculate the volume of a rectangle is:a rectangle is:
Volume = length Volume = length xx width width xx depth depth
Volume = (l) Volume = (l) xx (w) (w) xx (d) (d)
LengthLength WidthWidth
DepthDepth
BDMS / PSU
BDMS / PSU 11
1.1. Calculate the volume of a Calculate the volume of a rectangular sump with a rectangular sump with a lengthlength of of 2525 feet, a feet, a widthwidth of 10 feet and a of 10 feet and a depthdepth of 15 feet. of 15 feet.
Volume = (l) Volume = (l) xx (w) (w) xx (d) (d)Volume = 25 ft Volume = 25 ft xx 10 ft 10 ft xx 15 ft 15 ftVolume = 3,750 cubic feetVolume = 3,750 cubic feet
Length 25 ftLength 25 ftWidthWidth10 ft10 ft
DepthDepth15 ft15 ft
Example:Example:
BDMS / PSU
BDMS / PSU 12
• The sump capacity in gallons will be:The sump capacity in gallons will be:7.5 gal/cu ft x 3,750 cu ft = 28,125 gallons7.5 gal/cu ft x 3,750 cu ft = 28,125 gallons
• What is the weight of the water in the What is the weight of the water in the sump?sump?8.342 lbs/gal x 28,125 gal = 234,618.75 lb8.342 lbs/gal x 28,125 gal = 234,618.75 lb
– or 117.309 tonsor 117.309 tons
• The weight of this water will be:The weight of this water will be:62.5 lbs/cu ft X 3,750 cu. Ft. = 234,375 lbs. 62.5 lbs/cu ft X 3,750 cu. Ft. = 234,375 lbs.
– Or 117.1875 tonsOr 117.1875 tons
BDMS / PSU 13
Practice Exercise:Practice Exercise:
2.2. Calculate the volume of a rectangular Calculate the volume of a rectangular sump with a length of 50 feet, a width of sump with a length of 50 feet, a width of 25 feet and a depth of 15 feet.25 feet and a depth of 15 feet.
50 ft15 ft
25 ft
Answer: 18,750 cu ft
BDMS / PSU
BDMS / PSU 14
Solution:Solution:
• Volume = length x width x depthVolume = length x width x depth
• Volume = 50 ft x 25 ft x 15 ftVolume = 50 ft x 25 ft x 15 ft
• Volume = 18,750 ftVolume = 18,750 ft33
50 ft15 ft
25 ft
BDMS / PSU 15
• The sump capacity in gallons will be:The sump capacity in gallons will be:– 7.5 gal/cu ft x 18,750 cu ft = 140,625 gal7.5 gal/cu ft x 18,750 cu ft = 140,625 gal
8.342 lbs/gal x 140,625 gal = 1,173,093.75lb8.342 lbs/gal x 140,625 gal = 1,173,093.75lb
Or 586.5468 tonsOr 586.5468 tons
• The weight of this water will be:The weight of this water will be:62.5 lbs/cu ft X 18,750 cu. Ft. = 1,171,875 lbs 62.5 lbs/cu ft X 18,750 cu. Ft. = 1,171,875 lbs
Or 585.9375 tonsOr 585.9375 tons
BDMS / PSU 16
• The formula to calculate the volume of The formula to calculate the volume of a cylinder is:a cylinder is:Volume = area of circle Volume = area of circle xx depth depth
OrOrVolume = Volume = xx r r2 2 xx d depthepth
= 3.1416= 3.1416
Calculating the Volume of a CylinderCalculating the Volume of a Cylinder
DepthDepth
RadiusRadius
BDMS / PSU 17
3.3. Calculate the volume of a cylinder with a Calculate the volume of a cylinder with a radiusradius of 5 feet and a of 5 feet and a depthdepth of 15 feet. of 15 feet.
Volume = Volume = x r x r2 2 xx d depthepth
Volume = 3.1416 Volume = 3.1416 xx (5 feet) (5 feet)22 xx 15 feet 15 feet
Volume = 3.1416 Volume = 3.1416 xx 25 ft 25 ft xx 15 ft 15 ft
Volume = 1,178 cu ftVolume = 1,178 cu ft
15 ft
5 ft
Example:Example:
BDMS / PSU 18
• The sump capacity in gallons will be:The sump capacity in gallons will be:7.5 gal/cu ft x 1,178 cu ft = 8,835 gallons7.5 gal/cu ft x 1,178 cu ft = 8,835 gallons
• 8.342 lbs/gal x 8,835 gal =73,701.57 lb8.342 lbs/gal x 8,835 gal =73,701.57 lbOr 36.8507 tonsOr 36.8507 tons
• The weight of this water will be:The weight of this water will be:62.5 lbs. X 1,178 cu. Ft. = 73,625 lbs. 62.5 lbs. X 1,178 cu. Ft. = 73,625 lbs.
– Or 36.8125 tonsOr 36.8125 tons
BDMS / PSU 19
Practice Exercise:Practice Exercise:
4.4. Calculate the volume of a cylindrical Calculate the volume of a cylindrical storage tank with a radius of 10 feet storage tank with a radius of 10 feet and a depth of 30 feet.and a depth of 30 feet.
30 ft
10 ft
Answer: Answer: 9,424.8 cu ft9,424.8 cu ft
BDMS / PSU 20
Solution:Solution:
• Volume = Volume = x r x r2 2 xx d depthepth
• Volume = Volume = x (10 ft) x (10 ft)22 x 30 ft x 30 ft• Volume = 3.1416 x 100 ftVolume = 3.1416 x 100 ft22 x 30 ft x 30 ft• Volume = 9,424.8 cu ftVolume = 9,424.8 cu ft
30 ft
10 ft
BDMS / PSU 21
Practice Exercise:Practice Exercise:
5.5. Calculate the volume of a Calculate the volume of a cylindrical storage tank with a cylindrical storage tank with a diameter of 10 feet and a depth of diameter of 10 feet and a depth of 30 feet.30 feet.
Answer: Answer: 2,355 cu ft2,355 cu ft30 ft30 ft
10 ft10 ft
BDMS / PSU 22
Solution: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
30 ft
10 ft
BDMS / PSU 23
• The sump capacity in gallons will be:The sump capacity in gallons will be:
7.5 gal/cu ft x 2,356.2 cu ft = 17,671.5 gal7.5 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 lb8.342 lbs/gal x 17,671.5 gal = 147,415.653 lb
Or 73.70 tonsOr 73.70 tons
• The weight of this water will be:The weight of this water will be:
62.5 cu ft/lb X 2,356.2 cu.ft. = 147,262.5 62.5 cu ft/lb X 2,356.2 cu.ft. = 147,262.5 lbs. lbs.
– Or 73.63 tonsOr 73.63 tons
BDMS / PSU 24
1000 ft36 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 Calculating the Volume of a Triangle:Triangle:• The formula to calculate the volume of a The formula to calculate the volume of a
triangular vessel or a trough is:triangular vessel or a trough is:• Volume = area of triangle x length of troughVolume = area of triangle x length of trough
OrOr
• Volume = Volume = base x height x lengthbase x height x length
22
HeightHeight
LengthLengthBaseBase
BDMS / PSU 26
6.6. Calculate the volume of a triangle Calculate the volume of a triangle with a base of 8 feet, a height of 5 with a base of 8 feet, a height of 5 feet and a length of 8 feet.feet and a length of 8 feet.
Volume = Volume = Base Base xx Height Height xx Length Length
22Volume = Volume = 8 ft 8 ft xx 5 ft 5 ft xx 8 ft 8 ft
22Volume = 160 cu ftVolume = 160 cu ft
8 ft8 ft
5 ft5 ft
8 ft8 ft
Example:Example:
BDMS / PSU 27
7.7. Calculate the volume of a triangle Calculate the volume of a triangle with a base of 15 feet, a height of 10 with a base of 15 feet, a height of 10 feet and a length of 12 feet.feet and a length of 12 feet.
Practice Exercise:Practice Exercise:
Answer: Answer: 900 cu ft900 cu ft
15 ft15 ft
10 ft10 ft
12 ft12 ft
BDMS / PSU 28
Solution:Solution:
• Volume = Volume = Base Base xx Height Height xx Length Length 22
• Volume = Volume = 15 ft x 10 ft x 12 ft15 ft x 10 ft x 12 ft 22
• Volume = 900 ftVolume = 900 ft33
15 ft15 ft
10 ft10 ft
12 ft12 ft
BDMS / PSU 29
• The sump capacity in gallons will be:The sump capacity in gallons will be:7.5 gal/cu ft x 900 ft7.5 gal/cu ft x 900 ft33 = 6,750 gallons = 6,750 gallons
• 8.342 lbs/gal x 6,750 gal = 56,308.58.342 lbs/gal x 6,750 gal = 56,308.5
Or 28.15425Or 28.15425
• The weight of this water will be:The weight of this water will be:62.5 lbs/cu ft X 900 cu ft = 56,250 lbs. 62.5 lbs/cu ft X 900 cu ft = 56,250 lbs.
– Or 28.125 tonsOr 28.125 tons
BDMS / PSU 30
Practice Exercise:Practice Exercise:
8.8. Calculate the volume of a triangle Calculate the volume of a triangle with a base of 20 feet, a height of 15 with a base of 20 feet, a height of 15 feet and a length of 10 feet.feet and a length of 10 feet.
10 ft10 ft
15 ft15 ft
20 ft20 ft
Answer: Answer: 1,500 cu ft1,500 cu ft
BDMS / PSU 31
Solution:Solution:
• Volume = Volume = base base xx height height xx length length 22
• Volume = Volume = 20 ft x 15 ft x 10 ft20 ft x 15 ft x 10 ft 22
• Volume = 1,500 ftVolume = 1,500 ft33
10 ft10 ft
15 ft15 ft
20 ft20 ft
BDMS / PSU 32
• The sump capacity in gallons will be:The sump capacity in gallons will be:7.5 gal/cu ft x 1,500 cu ft = 11,250 gallons7.5 gal/cu ft x 1,500 cu ft = 11,250 gallons
• 8.342 lbs/gal x 11,250 gal = 93,847.5 lb8.342 lbs/gal x 11,250 gal = 93,847.5 lbOr 46.92375 tonsOr 46.92375 tons
• The weight of this water will be:The weight of this water will be:62.5 lbs. X 1,500 cu. Ft. = 93,750lbs. 62.5 lbs. X 1,500 cu. Ft. = 93,750lbs.
– Or 46.875 tonsOr 46.875 tons
BDMS / PSU 33
Calculating the Volume of a SphereCalculating the Volume of a Sphere
• The formula to calculate the volume of The formula to calculate the volume of a sphere is:a sphere is:
Volume = Volume = xx (diameter) (diameter)33
66
Where Where = 3.1416 = 3.1416
DiameterDiameter
BDMS / PSU 34
Example:Example:
• Calculate the volume of a sphere with a Calculate the volume of a sphere with a diameterdiameter of 15 feet. of 15 feet.
Volume = Volume = 3.1416 3.1416 xx (15 ft) (15 ft)33
66
Volume = 1,767.15 cu ftVolume = 1,767.15 cu ft15 ft15 ft
BDMS / PSU 35
• The sump capacity in gallons will be:The sump capacity in gallons will be:7.5 gal/cu ft x 1,767.15 cu ft = 13,253.62 gallons7.5 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 lb8.342 lbs/gal x 13,253.62 gal = 110,561.73 lbOr 55.2808 tonsOr 55.2808 tons
• The weight of this water will be:The weight of this water will be:62.5 lbs. X 1,767.15cu. Ft. = 110,446.87lbs. 62.5 lbs. X 1,767.15cu. Ft. = 110,446.87lbs.
– Or 55.22 tonsOr 55.22 tons
BDMS / PSU 36
Practice Exercise:Practice Exercise:
9.9. Calculate the volume of sphere with a Calculate the volume of sphere with a diameter of 20 feet.diameter of 20 feet.
20 ft.20 ft.
Answer: Answer: 44,187 cu ft,187 cu ft
BDMS / PSU 37
Solution:Solution:
• Volume = Volume = xx (diameter) (diameter)33
66
• Volume = Volume = 3.1416 x (20 ft)3.1416 x (20 ft)33
66
• Volume = 4,188.8 ftVolume = 4,188.8 ft33
20 ft.20 ft.
BDMS / PSU 38
• The sump capacity in gallons will be:The sump capacity in gallons will be:7.5 gal/cu ft x 4,188.8 cu ft = 31,416 gallons7.5 gal/cu ft x 4,188.8 cu ft = 31,416 gallons
• 8.342 lbs/gal x 31,416 gal = 262,072.27 lb8.342 lbs/gal x 31,416 gal = 262,072.27 lbOr 131.03 tonsOr 131.03 tons
• The weight of this water will be:The weight of this water will be:62.5 lbs. X 4,188.8 cu ft = 261,800 lbs 62.5 lbs. X 4,188.8 cu ft = 261,800 lbs
– Or 130.9 tonsOr 130.9 tons
BDMS / PSU 39
Practice Exercise:Practice Exercise:
10.10. Calculate the volume of sphere with a Calculate the volume of sphere with a diameter of 12.5 feet.diameter of 12.5 feet.
Answer: Answer: 1,022 cu ft1,022 cu ft
12.5 ft.12.5 ft.
BDMS / PSU 40
Solution:Solution:
• Volume = Volume = xx (diameter) (diameter)33
66
• Volume = Volume = 3.1416 x (12.5 ft)3.1416 x (12.5 ft)33
66
• Volume = 1,022.65 ftVolume = 1,022.65 ft33
12.5 ft.12.5 ft.
BDMS / PSU 42
Pump Characteristic CurvesPump Characteristic Curves
E = E = ( 8.33 lb of water per gal) ( 8.33 lb of water per gal)
(33,000 ft-lb per min) (brake horsepower)(33,000 ft-lb per min) (brake horsepower)
BDMS / PSU 43
Brake HorsepowerBrake Horsepower
• The product of the pressure head (H, ft) and the flow (Q, The product of the pressure head (H, ft) and the flow (Q, gpm) gives water horsepower or the theoretically minimum gpm) gives water horsepower or the theoretically minimum horsepower required to produce the desired results.horsepower required to produce the desired results.
• WHP = WHP = Q x 8.33 x HQ x 8.33 x H QHQH33,00033,000 or or 39603960
• Q is flow in gallons per minute and H is head in feet; 8.33 = 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 pounds per gallon of water and 33,000 = ft-lb/per min per horsepower.horsepower.
• The efficiency which is output over input or E = WHP/bhp The efficiency which is output over input or E = WHP/bhp can be expressed:can be expressed:
• E = E = Q (GPM) x H (ft)Q (GPM) x H (ft)3960 x bhp3960 x bhp
BDMS / PSU 45
HPHPoutout = = Head(feet) x Capacity(gpm) x 8.33(lbs./gallon) x SGHead(feet) x Capacity(gpm) x 8.33(lbs./gallon) x SG
33,000 (foot pounds / minute)33,000 (foot pounds / minute)
Head = 160 feet Head = 160 feet
Capacity = 300 gallons per minuteCapacity = 300 gallons per minute
8.33 = the weight of one US gallon 8.33 = the weight of one US gallon
SG = specific gravity of water at 68 degrees FSG = specific gravity of water at 68 degrees F
33,000 = the conversion of foot 33,000 = the conversion of foot pounds / minute to HPpounds / minute to HP
BDMS / PSU 46
• HP = HP = 160 x 300 x 8.33160 x 300 x 8.33 = = 399,840399,840 = 12.1 HP = 12.1 HP
33,00033,000 33,000 33,000
• If we had the pump curve supplied by the pump If we had the pump curve supplied by the pump manufacturer we would learn that he had manufacturer we would learn that he had calculated that it will take 20 horsepower to do calculated that it will take 20 horsepower to do this, so our efficiency would be:this, so our efficiency would be:
• 12.1 HP12.1 HPoutout = .60 or 60% efficient = .60 or 60% efficient
20 (Hp20 (Hpinin ) )
BDMS / PSU 48
HPHPoutout = = Head(feet) x Capacity(gpm) x 8.33(lbs./gallon) x SGHead(feet) x Capacity(gpm) x 8.33(lbs./gallon) x SG
33,000 (foot pounds / minute)33,000 (foot pounds / minute)
Head = 160 feet Head = 160 feet
Capacity = 300 gallons per minuteCapacity = 300 gallons per minute
8.33 = the weight of one US gallon 8.33 = the weight of one US gallon
SG = specific gravity of water at 68 degrees FSG = specific gravity of water at 68 degrees F
33,000 = the conversion of foot 33,000 = the conversion of foot pounds / minute to HPpounds / minute to HP
BDMS / PSU 49
• HP = HP = 160 x 300 x 8.33160 x 300 x 8.33 = = 399,840399,840 = 12.1 HP = 12.1 HP
33,00033,000 33,000 33,000
• If we had the pump curve supplied by the pump If we had the pump curve supplied by the pump manufacturer we would learn that he had manufacturer we would learn that he had calculated that it will take 20 horsepower to do calculated that it will take 20 horsepower to do this, so our efficiency would be:this, so our efficiency would be:
• 12.1 HP12.1 HPoutout = .60 or 60% efficient = .60 or 60% efficient
20 (Hp20 (Hpinin ) )
BDMS / PSU 51
Brake HorsepowerBrake Horsepower
• The product of the pressure head (H, ft) and the flow (Q, The product of the pressure head (H, ft) and the flow (Q, gpm) gives water horsepower or the theoretically minimum gpm) gives water horsepower or the theoretically minimum horsepower required to produce the desired results.horsepower required to produce the desired results.
• WHP = WHP = Q x 8.33 x HQ x 8.33 x H QHQH33,00033,000 or or 39603960
• Q is flow in gallons per minute and H is head in feet; 8.33 = 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 pounds per gallon of water and 33,000 = ft-lb/per min per horsepower.horsepower.
• The efficiency which is output over input or E = WHP/bhp The efficiency which is output over input or E = WHP/bhp can be expressed:can be expressed:
• E = E = Q (GPM) x H (ft)Q (GPM) x H (ft)3960 x bhp3960 x bhp
BDMS / PSU 53
Horsepower
• The Horsepower required to operate a Positive The Horsepower required to operate a Positive Displacement Pump has two factors:Displacement Pump has two factors:
• The Work Horsepower (WHP) - the actual work doneThe Work Horsepower (WHP) - the actual work done
• WHP = Flow(GPM) (X) Pressure(PSI) (/) 1714WHP = Flow(GPM) (X) Pressure(PSI) (/) 1714
• The The Viscous Horsepower(VHP)Viscous Horsepower(VHP) - the power required - the power required to turn the rotors, gears, etc. inside the viscous fluid. to turn the rotors, gears, etc. inside the viscous fluid. The Viscous Horsepower required is determined by The Viscous Horsepower required is determined by the pump design and speed and is supplied by the the pump design and speed and is supplied by the pump manufacturer.pump manufacturer.
• HP = WHP + VHPHP = 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 = 39.92 horsepower
33,000
BDMS / PSU 55
H = HH = Hss + H + Hff + H + Hvv + H + Hshsh
• H = total headH = total head
• HHss = is the vertical distance in feet from the suction liquid level = is the vertical distance in feet from the suction liquid level
to the discharge liquid level (total static head)to the discharge liquid level (total static head)
• HHf f = is the equivalent head, expressed as feet of liquid, = is the equivalent head, expressed as feet of liquid,
required to overcome the friction caused by the flow through required to overcome the friction caused by the flow through the pipe (friction head)the pipe (friction head)
• HHvv = is the head, in feet required to create velocity of flow = is the head, in feet required to create velocity of flow
(velocity head)– (velocity head)– Note: in most cases, this value is negligible and is often ignored.Note: in most cases, this value is negligible and is often ignored.
• HHshsh = is the head, in feet required to overcome the shock losses = is the head, in feet required to overcome the shock losses
due to changes of water flow produced by fittingsdue to changes of water flow produced by fittings
BDMS / PSU 56SumpSump
Suction LineSuction Line
PumpPumpDischarge LineDischarge Line
Static Discharge HeadStatic Discharge HeadStatic Discharge HeadStatic Discharge Head
Static Suction LiftStatic Suction LiftStatic Suction LiftStatic Suction Lift
•The vertical height The vertical height difference from difference from surface of water surface of water source to discharge source to discharge point is termed as point is termed as total static headtotal static head
BDMS / PSU 57SumpSump
Suction LineSuction Line
PumpPump
Discharge LineDischarge Line
Static Suction LiftStatic Suction LiftStatic Suction LiftStatic Suction Lift
•The vertical height difference The vertical height difference from surface of water source from surface of water source to centerline of impeller is to centerline of impeller is termed as static suction head termed as static suction head or suction lift ('suction lift' can or suction lift ('suction lift' can
also mean total suction head).also mean total suction head).
BDMS / PSU 58SumpSump
Suction LineSuction Line
PumpPump
Discharge LineDischarge Line
Static Discharge HeadStatic Discharge HeadStatic Discharge HeadStatic Discharge Head
The vertical height The vertical height difference from centerline of difference from centerline of impeller to discharge point impeller to discharge point is termed as static is termed as static discharge head.discharge head.
BDMS / PSU 59SumpSump
Suction LineSuction Line
PumpPump
Discharge LineDischarge Line
FRICTION LOSSFRICTION LOSSThe amount of The amount of pressure / head pressure / head required to 'force' required to 'force' liquid through pipe and liquid through pipe and fittings.fittings.
Pressure Pressure GaugeGauge
BDMS / PSU 60
Friction LossFriction Loss
• HHff = = f L Vf L V22
DD
• f is pipe coefficient of friction;f is pipe coefficient of friction;
• L is length of pipe;L is length of pipe;
• V is velocity of water;V is velocity of water;
• D is diameter of pipeD is diameter of pipe
BDMS / PSU 61
Frictional HeadFrictional Head
• Is usually expressed by the following equation Is usually expressed by the following equation based upon upon the number of 100-ft lengths of based upon upon the number of 100-ft lengths of pipe in the system:pipe in the system:
Hf = 0.2083 (100/C)Hf = 0.2083 (100/C)1.851.85[[ (q (q1.851.85) ]) ]
(d(d4.86554.8655))– Where C is a constant, usually 100, accounting for Where C is a constant, usually 100, accounting for
surface roughness; surface roughness; – q is the flow in gallons per minute; q is the flow in gallons per minute; – d is the inside diameter of the pipe in inches.d is the inside diameter of the pipe in inches.
BDMS / PSU 62
Equivalent Number of Feet of Staight Equivalent Number of Feet of Staight Pipe for Different FittingsPipe for Different Fittings
BDMS / PSU 63
Friction Loss in Feet for Old Pipe (C = 100)Friction Loss in Feet for Old Pipe (C = 100)
BDMS / PSU 64
Velocity Head is the Velocity Head is the velocity head of liquid velocity head of liquid moving at a given moving at a given velocity in the equivalent velocity in the equivalent head through which it head through which it would have to fall to would have to fall to acquire the same acquire the same velocity.velocity.
BDMS / PSU 65
• A dropped rock or other object will gain speed A dropped rock or other object will gain speed rapidly as it falls. rapidly as it falls.
• Measurements show that an object dropping 1 foot Measurements show that an object dropping 1 foot (ft) will reach a velocity of 8.02 feet per second (ft) will reach a velocity of 8.02 feet per second (ft/s). (ft/s).
• An object dropping 4 ft will reach a velocity of An object dropping 4 ft will reach a velocity of 16.04 ft/s. 16.04 ft/s.
• After an 8 ft drop, the velocity attained is 22.70 ft/s.After an 8 ft drop, the velocity attained is 22.70 ft/s.• The force of gravity causes this gain in speed or The force of gravity causes this gain in speed or
acceleration, which is equal to 32.2 feet per second acceleration, which is equal to 32.2 feet per second per second (ft/sper second (ft/s22). ).
• This acceleration caused by gravity is referred to This acceleration caused by gravity is referred to as as gg..
BDMS / PSU 66
• If water is stored in a tank and a small If water is stored in a tank and a small opening is made in the tank wall 1 ft below opening is made in the tank wall 1 ft below the water surface, the water will spout the water surface, the water will spout from the opening with a velocity of 8.02 from the opening with a velocity of 8.02 ft/s. ft/s.
• This velocity has the same magnitude that This velocity has the same magnitude that a freely falling rock attains after falling 1 ft.a freely falling rock attains after falling 1 ft.
• Similarly, at openings 4 ft and 8 ft below Similarly, at openings 4 ft and 8 ft below the water surface, the velocity of the the water surface, the velocity of the spouting water will be 16.04 and 22.68 ft/s, spouting water will be 16.04 and 22.68 ft/s, respectively. respectively.
BDMS / PSU 67
Head VelocityHead Velocity
• HHvv is the velocity head of liquid moving at a is the velocity head of liquid moving at a given velocity in the equivalent head through given velocity in the equivalent head through which it would have to fall to acquire the which it would have to fall to acquire the same velocity.same velocity.
• HHvv = = VV22
22gg
• HHvv is velocity head in feet; is velocity head in feet;• V is velocity of water in feet per second;V is velocity of water in feet per second;• G is acceleration due to gravity, in feet per G is acceleration due to gravity, in feet per
secsec22..
BDMS / PSU 68
• Thus, the velocity of water leaving an Thus, the velocity of water leaving an opening under a given head, opening under a given head, HH, is the , is the same as the velocity that would be same as the velocity that would be attained by a body falling that same attained by a body falling that same distance. The equation that shows distance. The equation that shows how velocity changes with how velocity changes with HH and and defines velocity head is:defines velocity head is:
BDMS / PSU 69
• S.G.S.G.Specific gravity. Weight of liquid in Specific gravity. Weight of liquid in comparison to water at approx 20 deg c comparison to water at approx 20 deg c (SG = 1).(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 = 39.92 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
1.1. What is the weight of one cubic foot What is the weight of one cubic foot of Water?of Water?
Answer:Answer: Sixty-two and five tenths (62.5) Sixty-two and five tenths (62.5) poundspounds
2.2. What is the weight of one (1) gallon What is the weight of one (1) gallon of water?of water?
Answer:Answer: Eight and one third (8.342) pounds Eight and one third (8.342) pounds
BDMS / PSU 73
4.4. What is the pressure exerted by a What is the pressure exerted by a column of water one (1) foot high column of water one (1) foot high and on one square inch of and on one square inch of surface?surface?
3.3. How many gallons are in one (1) How many gallons are in one (1) cubic foot ?cubic foot ?
Answer:Answer: Seven and five tenths (7.5) gallons Seven and five tenths (7.5) gallons
Answer:Answer: 0.4340 pounds 0.4340 pounds 62.5 pounds62.5 pounds 144 144
BDMS / PSU 74
5.5. What is the volume of a body of dead What is the volume of a body of dead water in a sump hole 25 foot deep by 500 water in a sump hole 25 foot deep by 500 feet by 1 foot?feet by 1 foot?
Answer: Volume = Length x width x depthAnswer: Volume = Length x width x depth
V = 500 ft x 1 ft x 25 ftV = 500 ft x 1 ft x 25 ft
V = 12,500 cu ftV = 12,500 cu ft
BDMS / PSU 75
• How long would it take a 40 How long would it take a 40 H.P. Pump to pump 88,000 H.P. Pump to pump 88,000 gallons to a head of 360 gallons to a head of 360 feet?feet?
• GPM = GPM = 33,000 x HP33,000 x HP
Head x 8.342Head x 8.342
GPM = GPM = 33,000 x 4033,000 x 40
360 x 8.342360 x 8.342
GPM = GPM = 132,000132,000
300,312300,312
GPM = 439.54GPM = 439.54
• Time = Time = Volume__Volume__
GPM x 60GPM x 60
Time = Time = 88,000____88,000____
439.54 x 60439.54 x 60
Time = 3.33 hoursTime = 3.33 hours
BDMS / PSU 76
Problem 1:Problem 1:
• If atmospheric pressure pushes mine If atmospheric pressure pushes mine water up a suction line due to the water up a suction line due to the vacuum created by a pump, is there a vacuum created by a pump, is there a limitation as to the maximum length of limitation as to the maximum length of suction line? If so, what is the value? suction line? If so, what is the value?
BDMS / PSU 77
Solution Problem 1:Solution Problem 1:
• At sea level, atmospheric pressure is equal to 14.7 At sea level, atmospheric pressure is equal to 14.7 psi. If a perfect vacuum were to be created in a psi. If a perfect vacuum were to be created in a suction line, atmospheric pressure could push a 1- suction line, atmospheric pressure could push a 1- in. column of water to a height of:in. column of water to a height of:
• Pressure = weight of water columnPressure = weight of water column• Divide atmospheric pressure at sea level by 0.0361 Divide atmospheric pressure at sea level by 0.0361
lb/inlb/in33 (the weight of one cubic inch of water) to (the weight of one cubic inch of water) to obtain the theoretical suction lift.obtain the theoretical suction lift.
• 14.7 (lb/in14.7 (lb/in22) / 0.0361 (lb/in) / 0.0361 (lb/in33) = 407.20 (inches)) = 407.20 (inches)407.20 (inches) / 12 (inches per foot) = 33.9 (ft)407.20 (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 0.0361 lb/in3 (the weight of one cubic inch of water) to obtain the theoretical suction lift. 14.7 (lb/in2) / 0.0361 (lb/in3) = 407.20 (inches)407.20 (inches) / 12 (inches per foot) = 33.9 (ft)
BDMS / PSU 83
Head =Head = Pressure x 2.31Pressure x 2.31
Specific GravitySpecific Gravity
Head = Head = 14.7 psi x 2.3114.7 psi x 2.31 = 33.96 Feet = 33.96 Feet
1.01.0
BDMS / PSU 84
Problem 2:Problem 2:
• What is the required brake What is the required brake horsepower to pump 150 gpm horsepower to pump 150 gpm (gallons per minute) against a total (gallons per minute) against a total dynamic head of 370 ft if the pump dynamic head of 370 ft if the pump operates at 70 % efficiency?operates at 70 % efficiency?
BDMS / PSU 85
Solution Problem 2Solution Problem 2::
• HPHPB B = = QH (8.33)QH (8.33)
33,000 E33,000 E
• HPHPB B = = (150gpm)(370 ft)(8.33)(150gpm)(370 ft)(8.33)
(33,000)(.7)(33,000)(.7)
• HPHPB B = = 462315462315
2310023100
• HPHPB B = 20.01 hp= 20.01 hp
BDMS / PSU 86
Cost to Pump Water – ElectricCost to Pump Water – Electric
$ per hour =$ per hour = gpm x head in feet x 0.746 x rate per KWHgpm x head in feet x 0.746 x rate per KWH
3960 x Pump Efficiency x Electric Motor Efficiency3960 x Pump Efficiency x Electric Motor Efficiency
BDMS / PSU 87
Horsepower
• The Horsepower required to operate a Positive The Horsepower required to operate a Positive Displacement Pump has two factors:Displacement Pump has two factors:
• The Work Horsepower (WHP) - the actual work doneThe Work Horsepower (WHP) - the actual work done
• WHP = Flow(GPM) (X) Pressure(PSI) (/) 1714WHP = Flow(GPM) (X) Pressure(PSI) (/) 1714
• The The Viscous Horsepower(VHP)Viscous Horsepower(VHP) - the power required - the power required to turn the rotors, gears, etc. inside the viscous fluid. to turn the rotors, gears, etc. inside the viscous fluid. The Viscous Horsepower required is determined by The Viscous Horsepower required is determined by the pump design and speed and is supplied by the the pump design and speed and is supplied by the pump manufacturer.pump manufacturer.
• HP = WHP + VHPHP = WHP + VHP
BDMS / PSU 88
H = HH = Hss + H + Hff + H + Hvv + H + Hshsh
• H = total headH = total head
• HHss = is the vertical distance in feet from the suction liquid level = is the vertical distance in feet from the suction liquid level
to the discharge liquid level (total static head)to the discharge liquid level (total static head)
• HHf f = is the equivalent head, expressed as feet of liquid, = is the equivalent head, expressed as feet of liquid,
required to overcome the friction caused by the flow through required to overcome the friction caused by the flow through the pipe (friction head)the pipe (friction head)
• HHvv = is the head, in feet required to create velocity of flow = is the head, in feet required to create velocity of flow
(velocity head)– (velocity head)– Note: in most cases, this value is negligible and is often ignored.Note: in most cases, this value is negligible and is often ignored.
• HHshsh = is the head, in feet required to overcome the shock losses = is the head, in feet required to overcome the shock losses
due to changes of water flow produced by fittingsdue to changes of water flow produced by fittings
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