bdms / psu1 mine drainage pa dep bureau of deep mine safety

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


• 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


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


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



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.


BDMS / PSU 30



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


• 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.


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.


BDMS / PSU 36


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


BDMS / PSU 39


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 41

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 44

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 47

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 50

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 52

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



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).



PumpPump


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.



PumpPump


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


• 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


• 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


• 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


• 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

bdms / psu1 mine drainage pa dep bureau of deep mine safety

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