Mine Drainage PA DEP Bureau of Deep Mine Safety BDMS / PSU 1 • 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 2 • Volume = rate x time • Rate = volume / time • Time = volume / rate BDMS / PSU 3 • 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 4 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) BDMS / PSU 5 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) BDMS / PSU 6 • • • • • 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. BDMS / PSU 7 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 8 Basic Three-dimensional Shapes Cylinder Rectangle Trapezoid Pyramid or Prism Sphere BDMS / PSU 9 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 BDMS / PSU 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 BDMS / PSU 11 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 BDMS / PSU 12 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 BDMS / PSU 13 BDMS / PSU Solution: • Volume = length x width x depth • Volume = 50 ft x 25 ft x 15 ft • Volume = 18,750 ft3 15 ft 50 ft BDMS / PSU 14 • 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 15 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 BDMS / PSU 16 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 BDMS / PSU 17 • 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 18 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 BDMS / PSU 19 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 BDMS / PSU 20 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 BDMS / PSU 21 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 BDMS / PSU 22 • 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 23 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 BDMS / PSU 24 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 BDMS / PSU 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 BDMS / PSU 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. BDMS / PSU 27 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 BDMS / PSU 29 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 BDMS / PSU 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 BDMS / PSU 31 • 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 32 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 BDMS / PSU 34 • 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 35 Practice Exercise: 9. Calculate the volume of sphere with a diameter of 20 feet. 20 ft. Answer: 4,187 cu ft BDMS / PSU 36 Solution: • Volume = x (diameter)3 6 • Volume = 3.1416 x (20 ft)3 6 • Volume = 4,188.8 ft3 20 ft. BDMS / PSU 37 • 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 38 Practice Exercise: 10. Calculate the volume of sphere with a diameter of 12.5 feet. 12.5 ft. Answer: 1,022 cu ft BDMS / PSU 39 Solution: • Volume = x (diameter)3 6 • Volume = 3.1416 x (12.5 ft)3 6 • Volume = 1,022.65 ft3 12.5 ft. BDMS / PSU 40 BDMS / PSU 41 Pump Characteristic Curves E = ( 8.33 lb of water per gal) (33,000 ft-lb per min) (brake horsepower) BDMS / PSU 42 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 43 BDMS / PSU 44 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 45 • 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 46 BDMS / PSU 47 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 48 • 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 49 BDMS / PSU 50 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 51 BDMS / PSU 52 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 53 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 54 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 55 •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 56 •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 57 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 58 FRICTION LOSS The amount of pressure / head required to 'force' liquid through pipe and fittings. Pressure Gauge Suction Line Pump Discharge Line Sump BDMS / PSU 59 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 60 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 61 Equivalent Number of Feet of Staight Pipe for Different Fittings BDMS / PSU 62 Friction Loss in Feet for Old Pipe (C = 100) BDMS / PSU 63 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 64 • 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 65 • 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 66 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 67 • 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 68 • S.G. Specific gravity. Weight of liquid in comparison to water at approx 20 deg c (SG = 1). BDMS / PSU 69 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 70 • 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 71 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 BDMS / PSU 72 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 74 • 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 75 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 76 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 77 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 78 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 81 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 82 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 87 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