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

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

Volume = rate x time Rate = volume / time Time = volume / rate BDMS / PSU

Using a pipe that has a rate of 75 gallons per minute, it took you 2 Using a pipe that has a rate of 75 gallons per minute, it took you 2.5 hours to fill up a sump. What is the volume of the sump? Solution: Volume = Rate x Time Volume = 75 gpm x 2.5 hour Volume = (75 gpm x 60 minutes x 2.5 hours) Volume = 11,250 gallon BDMS / PSU

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Solution: Volume =  x r2 x depth Volume =  x (5 ft)2 x 30 ft Volume = 2,356.2 cu ft 30 ft 10 ft BDMS / PSU

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

BDMS / PSU

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

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

BDMS / PSU

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

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

BDMS / PSU

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

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

BDMS / PSU

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

BDMS / PSU

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

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

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

The vertical height difference from surface of water source to discharge point is termed as total static head Static Discharge Head Suction Line A dropped rock or other object will gain speed rapidly as it falls. Measurements show that an object dropping 1 foot (ft) will reach a velocity of 8.02 feet per second (ft/s). An object dropping 4 ft will reach a velocity of 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. 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. Thus, the velocity of water leaving an opening under a given head, h, is the same as the velocity that would be attained by a body falling that same distance. The equation that shows how velocity changes with h and defines velocity head is: This may also be written in velocity head form as: Pump Discharge Line Static Suction Lift Sump BDMS / PSU

The vertical height difference from surface of water source to centerline of impeller is termed as static suction head or suction lift ('suction lift' can also mean total suction head). Suction Line A dropped rock or other object will gain speed rapidly as it falls. Measurements show that an object dropping 1 foot (ft) will reach a velocity of 8.02 feet per second (ft/s). An object dropping 4 ft will reach a velocity of 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. 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. Thus, the velocity of water leaving an opening under a given head, h, is the same as the velocity that would be attained by a body falling that same distance. The equation that shows how velocity changes with h and defines velocity head is: This may also be written in velocity head form as: Pump Discharge Line Static Suction Lift Sump BDMS / PSU

The vertical height difference from centerline of impeller to discharge point is termed as static discharge head. Static Discharge Head Suction Line A dropped rock or other object will gain speed rapidly as it falls. Measurements show that an object dropping 1 foot (ft) will reach a velocity of 8.02 feet per second (ft/s). An object dropping 4 ft will reach a velocity of 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. 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. Thus, the velocity of water leaving an opening under a given head, h, is the same as the velocity that would be attained by a body falling that same distance. The equation that shows how velocity changes with h and defines velocity head is: This may also be written in velocity head form as: Pump Discharge Line Sump BDMS / PSU

FRICTION LOSS The amount of pressure / head required to 'force' liquid through pipe and fittings. Pressure Gauge Suction Line A dropped rock or other object will gain speed rapidly as it falls. Measurements show that an object dropping 1 foot (ft) will reach a velocity of 8.02 feet per second (ft/s). An object dropping 4 ft will reach a velocity of 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. 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. Thus, the velocity of water leaving an opening under a given head, h, is the same as the velocity that would be attained by a body falling that same distance. The equation that shows how velocity changes with h and defines velocity head is: This may also be written in velocity head form as: Pump Discharge Line Sump BDMS / PSU

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

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

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

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

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

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

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. BDMS / PSU

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

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

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

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

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

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

How many gallons are in one (1) cubic foot ? Answer: Seven and five tenths (7.5) gallons What is the pressure exerted by a column of water one (1) foot high and on one square inch of surface? Answer: 0.4340 pounds 62.5 pounds 144 BDMS / PSU

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

How long would it take a 40 H. P How long would it take a 40 H.P. Pump to pump 88,000 gallons to a head of 360 feet? GPM = 33,000 x HP Head x 8.342 GPM = 33,000 x 40 360 x 8.342 GPM = 132,000 300,312 GPM = 439.54 Time = Volume__ GPM x 60 Time = 88,000____ 439.54 x 60 Time = 3.33 hours BDMS / PSU

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

Solution Problem 1: At sea level, atmospheric pressure is equal to 14.7 psi. If a perfect vacuum were to be created in a suction line, atmospheric pressure could push a 1- in. column of water to a height of: Pressure = weight of water column Divide atmospheric pressure at sea level by 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

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

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

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

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

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) This shows that the maximum total suction lift, when pumping water at sea level, is approximately 34 ft. This figure is theoretical and can never be obtained in practice. That is why good mining practice dictates that the total dynamic suction lift should never exceed a value of approximately 20 ft. for trouble-free operations BDMS / PSU

Head = Pressure x 2.31 Specific Gravity Theoretically we could come out of the water tank, go straight up into the air with the hose to a height of almost 34 feet and then down into the container and the water would flow with no trouble. I say theoretically because there is friction in the hose that offers resistance to the flow and that friction has to be considered any time you are trying to calculate the possibility of using a siphon to move liquids. Head = 14.7 psi x 2.31 = 33.96 Feet 1.0 BDMS / PSU

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

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

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

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

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