Transcript mine_drainage.ppt

Mine Drainage
PA DEP Bureau of Deep Mine
Safety
BDMS / PSU
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• 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?
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• Volume = rate x time
• Rate = volume / time
• Time = volume / rate
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• 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
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N=
N=
(
Large Diameter
Small Diameter
( )
_12”_
4”
)
5
5
Equivalent Flow: How
many 4 inch pipes are
needed to carry the flow
from a 12” pipe?
4“
N=
N=
( )
3”
5
12 “
( )
243”
N = 15.5 (4 INCH PIPES)
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Conversion Factors for Mine Drainage Problems
Volumetric Measure Weight
Liquid
Measure
Measure, in In Cubic
In Cubic
Inches
Feet
in Pounds
Gallons
1
231
0.134
1 Cubic Foot
1 Cubic Foot
1
1728 cu in
8.342
Gallons /
Cubic Foot
7.481
Weight Measure, in
Pounds
62.5
(use 7.5)
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•
•
•
•
•
1 gallon-water
1 gallon-water
1 gallon-water
1 cu. ft. of water
1 cu. ft. of water
=
=
=
=
=
8.345 pounds
231 cu inches
0.134 cu feet
1728 cu in
7.48 gallons (7.5 gal)
• 1 cu. ft. of water
=
62.425 lb.
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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.
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Basic Three-dimensional Shapes
Cylinder
Rectangle
Trapezoid
Pyramid
or Prism
Sphere
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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
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BDMS / PSU
Length
10
Example:
1. 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
Depth
15 ft
Length 25 ft
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BDMS / PSU
• The sump capacity in gallons will be:
7.5 gal/cu ft x 3,750 cu ft = 28,125 gallons
• What is the weight of the water in the
sump?
8.342 lbs/gal x 28,125 gal = 234,618.75 lb
– or 117.309 tons
• The weight of this water will be:
62.5 lbs/cu ft X 3,750 cu. Ft. = 234,375 lbs.
– Or 117.1875 tons
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Practice Exercise:
2. Calculate the volume of a rectangular
sump with a length of 50 feet, a width of
25 feet and a depth of 15 feet.
15 ft
50 ft
Answer: 18,750 cu ft
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Solution:
• Volume = length x width x depth
• Volume = 50 ft x 25 ft x 15 ft
• Volume = 18,750 ft3
15 ft
50 ft
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•
The sump capacity in gallons will be:
– 7.5 gal/cu ft x 18,750 cu ft = 140,625 gal
8.342 lbs/gal x 140,625 gal = 1,173,093.75lb
Or 586.5468 tons
•
The weight of this water will be:
62.5 lbs/cu ft X 18,750 cu. Ft. = 1,171,875 lbs
Or 585.9375 tons
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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 d
 = 3.1416
epth
Radius
Depth
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Example:
3. Calculate the volume of a cylinder with a
radius of 5 feet and a depth of 15 feet.
Volume =  x r2 x d
epth
Volume = 3.1416 x (5 feet)2 x 15 feet
Volume = 3.1416 x 25 ft x 15 ft
Volume = 1,178 cu ft
5 ft
15 ft
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• The sump capacity in gallons will be:
7.5 gal/cu ft x 1,178 cu ft = 8,835 gallons
• 8.342 lbs/gal x 8,835 gal =73,701.57 lb
Or 36.8507 tons
• The weight of this water will be:
62.5 lbs. X 1,178 cu. Ft. = 73,625 lbs.
– Or 36.8125 tons
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Practice Exercise:
4. Calculate the volume of a cylindrical
storage tank with a radius of 10 feet
and a depth of 30 feet.
10 ft
30 ft
Answer: 9,424.8 cu ft
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Solution:
10 ft
30 ft
•
•
•
•
Volume =  x r2 x d
Volume =  x (10 ft)2 x 30 ft
Volume = 3.1416 x 100 ft2 x 30 ft
Volume = 9,424.8 cu ft
epth
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Practice Exercise:
5. Calculate the volume of a
cylindrical storage tank with a
diameter of 10 feet and a depth of
30 feet.
10 ft
30 ft
Answer: 2,355 cu ft
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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
10 ft
30 ft
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• The sump capacity in gallons will be:
7.5 gal/cu ft x 2,356.2 cu ft = 17,671.5 gal
8.342 lbs/gal x 17,671.5 gal = 147,415.653 lb
Or 73.70 tons
• The weight of this water will be:
62.5 cu ft/lb X 2,356.2 cu.ft. = 147,262.5
lbs.
– Or 73.63 tons
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1000 ft
36 inches
•Volume =  x r2 x depth
•Volume = 3.1416 x (18 inches)2 x 1,000 ft
•Volume = 3.1416 x (1.5 ft)2 x 1,000 ft
•Volume = 7068.6 cu ft
What is the weight of this section of pipe, if
full of water?
7.5 gal / cu ft x 7068.6 cu ft = 53,014.5 gal
8.342 lb / gal x 53,014.5 = 442,246.95 lb
Or 221.12 ton
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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
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Length
25
Example:
6. Calculate the volume of a triangle
with a base of 8 feet, a height of 5
feet and a length of 8 feet.
Volume = Base x Height x Length
2
Volume = 8 ft x 5 ft x 8 ft
2
Volume = 160 cu ft
5 ft
8 ft
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8 ft
26
Practice Exercise:
Answer: 900 cu ft
10 ft
12 ft
15 ft
7. Calculate the volume of a triangle
with a base of 15 feet, a height of 10
feet and a length of 12 feet.
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Solution:
• Volume = Base x Height x Length
2
• Volume = 15 ft x 10 ft x 12 ft
2
• Volume = 900 ft3
10 ft
BDMS / PSU
15 ft
12 ft
28
• The sump capacity in gallons will be:
7.5 gal/cu ft x 900 ft3 = 6,750 gallons
• 8.342 lbs/gal x 6,750 gal = 56,308.5
Or 28.15425
• The weight of this water will be:
62.5 lbs/cu ft X 900 cu ft = 56,250 lbs.
– Or 28.125 tons
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Practice Exercise:
8. Calculate the volume of a triangle
with a base of 20 feet, a height of 15
feet and a length of 10 feet.
Answer: 1,500 cu ft
15 ft
10 ft
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20 ft
30
Solution:
15 ft
10 ft
20 ft
• Volume = base x height x length
2
• Volume = 20 ft x 15 ft x 10 ft
2
• Volume = 1,500 ft3
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• The sump capacity in gallons will be:
7.5 gal/cu ft x 1,500 cu ft = 11,250 gallons
• 8.342 lbs/gal x 11,250 gal = 93,847.5 lb
Or 46.92375 tons
• The weight of this water will be:
62.5 lbs. X 1,500 cu. Ft. = 93,750lbs.
– Or 46.875 tons
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Calculating the Volume of a Sphere
• The formula to calculate the volume of
a sphere is:
Volume =  x (diameter)3
6
Where  = 3.1416
BDMS / PSU
Diameter
33
Example:
• Calculate the volume of a sphere with a
diameter of 15 feet.
Volume = 3.1416 x (15 ft)3
6
15 ft
Volume = 1,767.15 cu ft
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• The sump capacity in gallons will be:
7.5 gal/cu ft x 1,767.15 cu ft = 13,253.62 gallons
• 8.342 lbs/gal x 13,253.62 gal = 110,561.73 lb
Or 55.2808 tons
• The weight of this water will be:
62.5 lbs. X 1,767.15cu. Ft. = 110,446.87lbs.
– Or 55.22 tons
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Practice Exercise:
9. Calculate the volume of sphere with a
diameter of 20 feet.
20 ft.
Answer: 4,187 cu ft
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Solution:
• Volume =  x (diameter)3
6
• Volume = 3.1416 x (20 ft)3
6
• Volume = 4,188.8 ft3
20 ft.
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• The sump capacity in gallons will be:
7.5 gal/cu ft x 4,188.8 cu ft = 31,416 gallons
• 8.342 lbs/gal x 31,416 gal = 262,072.27 lb
Or 131.03 tons
• The weight of this water will be:
62.5 lbs. X 4,188.8 cu ft = 261,800 lbs
– Or 130.9 tons
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Practice Exercise:
10. Calculate the volume of sphere with a
diameter of 12.5 feet.
12.5 ft.
Answer: 1,022 cu ft
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Solution:
• Volume =  x (diameter)3
6
• Volume = 3.1416 x (12.5 ft)3
6
• Volume = 1,022.65 ft3
12.5 ft.
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Pump Characteristic Curves
E = ( 8.33 lb of water per gal)
(33,000 ft-lb per min) (brake horsepower)
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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
33,000
or
QH
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
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BDMS / PSU
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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
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• HP = 160 x 300 x 8.33 = 399,840 = 12.1 HP
33,000
33,000
• If we had the pump curve supplied by the pump
manufacturer we would learn that he had
calculated that it will take 20 horsepower to do
this, so our efficiency would be:
• 12.1 HPout = .60 or 60% efficient
20 (Hpin )
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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
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• HP = 160 x 300 x 8.33 = 399,840 = 12.1 HP
33,000
33,000
• If we had the pump curve supplied by the pump
manufacturer we would learn that he had
calculated that it will take 20 horsepower to do
this, so our efficiency would be:
• 12.1 HPout = .60 or 60% efficient
20 (Hpin )
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BDMS / PSU
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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
33,000
or
QH
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
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BDMS / PSU
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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
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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
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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
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•The vertical height
difference from
surface of water
source to discharge
point is termed as
total static head
Static Discharge Head
Suction Line
Pump
Discharge Line
Static Suction Lift
Sump
BDMS / PSU
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•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
Pump
Static Suction Lift
Discharge Line
Sump
BDMS / PSU
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The vertical height
difference from centerline of
impeller to discharge point
is termed as static
discharge head.
Static Discharge Head
Suction Line
Pump
Discharge Line
Sump
BDMS / PSU
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FRICTION LOSS
The amount of
pressure / head
required to 'force'
liquid through pipe and
fittings.
Pressure
Gauge
Suction Line
Pump
Discharge Line
Sump
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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
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Frictional Head
• Is usually expressed by the following equation
based upon upon the number of 100-ft lengths of
pipe in the system:
Hf = 0.2083 (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
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Equivalent Number of Feet of Staight
Pipe for Different Fittings
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Friction Loss in Feet for Old Pipe (C = 100)
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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.
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• 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
16.04 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 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
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• 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 16.04 and 22.68 ft/s,
respectively.
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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.
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• 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:
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• S.G.
Specific gravity. Weight of liquid in
comparison to water at approx 20 deg c
(SG = 1).
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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
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• 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
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1. What is the weight of one cubic foot
of Water?
Answer: Sixty-two and five tenths (62.5)
pounds
2. What is the weight of one (1) gallon
of water?
Answer: Eight and one third (8.342) pounds
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3. How many gallons are in one (1)
cubic foot ?
Answer: Seven and five tenths (7.5) gallons
4. What is the pressure exerted by a
column of water one (1) foot high
and on one square inch of
surface?
Answer: 0.4340 pounds
BDMS / PSU
62.5 pounds
144
73
5. 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
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• 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 = 439.54
Time = Volume__
GPM x 60
Time = 88,000____
439.54 x 60
Time = 3.33 hours
BDMS / PSU
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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?
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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 1in. column of water to a height of:
•
Pressure = weight of water column
• 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
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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
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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
79
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
80
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
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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)
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Head = Pressure x 2.31
Specific Gravity
Head = 14.7 psi x 2.31 = 33.96 Feet
1.0
BDMS / PSU
83
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
84
Solution Problem 2:
• HPB = QH (8.33)
33,000 E
• HPB = (150gpm)(370 ft)(8.33)
(33,000)(.7)
• HPB = 462315
23100
• HPB = 20.01 hp
BDMS / PSU
85
Cost to Pump Water – Electric
$ per hour = gpm x head in feet x 0.746 x rate per KWH
3960 x Pump Efficiency x Electric Motor Efficiency
BDMS / PSU
86
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
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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
88