Fluid Statics, Hydrodynamics and Hydraulic Machines Bobby Rauf 1
Fluid Facts 1) Liquids and gases can both be categorized as fluids. 2) Liquid fluids are assumed be incompressible. 3) Gaseous fluids are compressible 4) Fluid pressure, defined as force per unit area, is measured in pounds per square inch, psi, or pounds per square foot, psf, in US unit system. Pressure is measured in Pascals, Pa, or kilopascals, kpa, in the SI (metric) System. 5) Standard Atmospheric Pressure: 1.00 Atm; 14.7 psia; 407.1 inches of water gage; 29.92 in Hg; 760 mm Hg; 1.013 x 10 5 Pa; 101.3 kpa. 6) P abs = P gage + P atm 2
Fluid Facts 7) Fluid Density = = mass / volume = 1 / Specific Volume 8) Fluid Density,, in US units: lbm / ft 3 9) Fluid Density,, in SI units: Kg / m 3 10) Specific Gravity of Liquid = SG liquid = liquid / water 11) Specific Gravity of Gas = SG gas = gas / air 12) Specific Weight of Fluid = = g. ; in lbf / ft 3 or N / m 3 13) STP, Standard Temperature Pressure: 0 C ( 32 F); 101.3 kpa, or 1 atm, or 1 bar (14.7 psi). 14) r h = Hydraulic Radius = Area in Flow / Wetted Perimeter 3
FLUID STATICS - Manometer p 1 +.g.h = p 2 p 1 p2 h If p 1 represents atm. pressure, then: p 2 - p 1 = p gage, 2 =.g.h 4
Manometer & Multiple Fluid Densities p 0 + 1.g. h 1 + 2.g. h 2 = p p 0 p h 2 h 1 If p 0 represents atm. pressure, then: p - p 0 = p gage = 1.g. h 1 + 2.g. h 2 5
Problem 1 Inverted Fluid A glass filled with a fluid is inverted. The bottom is open. See figure 1. What is the pressure at the closed end A? The pressure, P A, at point A, plus the pressure (.g.h) exerted by the water equals the atmospheric pressure, P atm, outside the glass. 6
Problem 1 Figure 1 PA +.g.h = P atm PA = P atm -.g.h P A A.g.h = Wt. Of the Water h D P atm 7
HYDRODYNAMICS - Problem 2 Consider the holding tank shown in figure 2. Calculate the velocity of the water exiting to the atmosphere. 8
Problem 2; Figure 2 v 2m P 1 = P atm 1 P 2 = P atm 2 9
Problem 2, contd. Apply Bernoulli s equation between the free surface (point 1) and the exit (point 2) g.z 1 + v 12 /2 + p 1 / = g.z 2 + v 22 /2 + p 2 / {SI} p 1 = p 2 (both are, or are at, atmospheric pressure) v 1 = 0 (the free surface is stationary) g.z 1 + (0) = g.z 2 + v 22 /2 + (p 2 / - p 1 / ) g.z 1 = g.z 2 + v 22 /2 10
Problem 2, contd. g.z 1 = g.z 2 + v 22 /2 v 2 = 2.g. (z 1 - z 2 ) = (2).(9.81).(2) = 6.3 m/s 11
Problem 3 A pressure tank contains a fluid with weight density 81.5 lbf/ft 3 ( ). The pressure in the air space is 100 psia. Fluid exits to the atmosphere from the bottom of the tank. What is the exit velocity, v? 12
Problem 3 p v 1, z 1, p 10 ft v v, z 2, p atm 13
Problem 3 Apply Bernoulli s equation, for energy conservation, between the surface and the exit at the lower right end: p/ + g.z 1 +v 12 /2 = p atm / + g.z 2 + v 2 /2 {SI} v 1 =0 (at the free surface), z 2 =0 (at the exit) p/ + g.z 1 = p atm / + v 2 /2 14
Problem 3, contd. p/ + g.z 1 = p atm / + v 2 /2 v = 2.g.{(p - p atm ) /.g+z 1 )} Since p.g = = 81.5 lbf/ft 3, v = (2).(32.2ft/sec 2 )[(100 psi- 14.7psi).(144in 2 /ft 2 )/ 81.5lbf/ft 3 +10ft] v = 101.7 ft/sec 15
Problem 3 - i Buoyancy A hot air balloon is to used to transport a welder across a ravine. The balloon and the payload, including the welder and other accessories, have a mass of 200 kg. The ambient pressure and temperature are 102 kpa and 15.6 C. Assume wind to be calm, aloft and at ground elevation. (a) If density of ambient air is 1.2 kg/m 3 and if air in the balloon can be heated to a density of 1.0 kg/m 3, determine the volume and the radius of the balloon to achieve a lift off. (b) To what temperature must the air in the balloon be heated in order to achieve the density to 1.0 kg/m 3? 16
Problem 3 I Buoyancy
Problem 3 I, contd. Buoyancy Solution 3 i Hot Air = 1.0 kg/m 3 Amb. Air = 1.2 kg/m 3 P Amb = 102 kpa = 102,000 Pa Total Load, including the Payload & the Baloon = 200 Kg V HA = Volume of Hot Air in the Baloon = V AA = Volume of Ambient Air Displaced W payload + W balloon = 200 kg x g W hot air in balloon W amb. air displaced = 1.0 x g x V HA = 1.0 x g x V AA = 1.2 x g x V AA = 1.2 x g x V HA 18
Problem 3 I, contd. Buoyancy Solution 3 i (W payload + W balloon ) + W hot air in balloon = W amb. air displaced (200 kg x g) + 1.0 x g x V HA = 1.2 x g x V HA 200 kg + 1.0 kg/m 3 x V HA = 1.2 kg/m 3 x V HA V HA = V AA = 1000 m 3 Volume of a Spherical Baloon = V = 4/3 x 3.14 x r b 3 Radius of the Baloon = r b = {(V x 3/4) x (1/3.14)} 1/3 = 6.204 m 19
Problem 3 I, contd. Buoyancy Solution 3 i b) Calculate the temperature of the hot air in the balloon for a density of 1.0 kg / m 3 = T HA For ideal gases, p = x R x T HA T HA = p / ( R x ) = 355.8 K Or, T HA = 82.8 C Note: Max. Op. Temp for Nylon/Nymex Balloon: 120 C 20
Problem 3 I, contd. Buoyancy Solution 3 i Ancillary Note: Verification of the ambient air density assumption of 1.2 kg/m 3 For T Amb = 15.6 C = 288.6 K Ambient Air = p / (R x T Amb ) = 102,000 / (286 x 288.6) = 1.2 kg/m 3 21
Problem 4 Calculate the overturning moment, per unit width, acting on the dam shown below: h = 8 ft & h av = ½(0 + 8) h av = 8/2 ft = 4 ft 1/3h h r = 2/3h F P av = P =. g. h av =. h av Heel Toe = Specific weight of water = 62.4 lbf/ft 3 22
Problem 4 Typical Hydroelectric Dam Construction : 23
Problem 4, contd. F = P av. A =. g. h av. A =. h av A F = (62.4 lbf/ft 3 ). (4). ( 8 x 1) F = 1996.8 lbf F per foot = 1996.8 lbf /ft M = Moment (per foot) = F per foot x 1/3h M = Moment (per foot) = 1996.8 lbf/ft x (8/3 ft) M = 5325 ft lbf/ft Ancillary questions: What force is needed to prevent the dam from sliding and what constitutes it? 24
Problem 4 - i Calculate the horizontal and vertical components of the resultant force on a 2 meter wide inclined surface in the reservoir shown below. Note that this reservoir has just experienced a massive oil spill. This oil has a density of 900 kg/m 3 Oil Slick h o = 0.3 m h 2 = 2 m 30 Water 30 h 1 = 1 m h i = 1 m 25
Problem 4 i, contd. Given: Density of Oil, o = 900 kg/m 3 Density of Water, w = 1000 kg/m 3 Angle of the Inclined Plane, = 30 Height of the Inclined Surface, h i = 1 m Depth of Water Above Inclined Surface, h 1 = 1 m Total Height or Depth of Water, h 2 = h 1 + h i = 2.0 m Depth of the Oil Slick, h o = 0.3 m Avg. Depth of Water Above Incl. Surface = h w = 1/2 x (h 2 + h 1 ) = 1.5 m 26
Problem 4 i, contd. Length of the Incl. Surface = h i /Sin(30) = 2.0 m Width of the Inclined Surface = 2 m (Given) Area of the Inclined Surface Section = 2m x 2m = 4.0 m 2 Pressure Contributed by the Layer of Oil = o x g x h o = 2,649 N/m 2 or Pa Average Pressure Contributed by the Water = = 14,715 N/m 2 or Pa w x g x h w Total Average Pressure in the Water at Inclined Surface: 17,364 N/m 2 or Pa 27
Problem 4 I, contd. Resultant Force Component Analysis: R R y R x 28
Problem 4 i, contd. Perp. Resultant Force on any Point Along the Incl. Surface = Total Avg. Press x Area of Incl. Surface = 69,455 N Ry, the y - Comp. of the Resultant Force = R x Cos ( ) = 60,150 N Rx, the x - Comp. of the Resultant Force = R x Sin ( ) = 34,727 N 29
Problem 5 Figures 1 and 2, below, depict a pitot tube arrangement devised to measure the flow rate of water. Calculate the height of mercury column that would be sustained by the velocity pressure of the water. v = 10 m/s Water Mercury 30
Problem 5, contd. v 2 = v = 15 m/s 2 1 v 1 = 0 z 1 = z 2 and z 1 - z 2 = 0 z Hg = z water =? water = Density of water is 1000 kg / m 3 mercury = The density of mercury is 13,567 kg/m 3 31
Problem 5, contd. Apply Bernoulli s energy conservation equation at points 1 and 2, for fluid dynamics portion: p 1 / + g.z 1 +v 12 /2 = p 2 / + g.z 2 + v 22 /2 {SI} Since z 1 - z 2 = 0, v 1 = 0 and = water = 1000 kg/m 3 p 1 / = p 2 / + v 22 /2 p 1 - p 2 = (v 22 /2). =(225/2). 1000 = 112,500 pa 32
Problem 5, contd. Apply fluid statics principle to the pitot tube segment: p 2 + mercury. g. z Hg = p 1 + water. g. z water p 1 - p 2 = mercury. g. z Hg - water. g. z water z Hg = z water = (p 1 - p 2 ) / g / ( mercury - water ) z Hg = 112,500 / (9.81) / (13,567 1000) = 0.913 m 33
Problem 6 An irrigation water channel, or canal, is to be modified. The current and future specifications are listed below: Change Orig. Shape As % of Orig. New Shape Side Elevation or Vertical Component of the Side = 10 ft 10 ft Side Elevation Angle = 1.57 0.79 Horizontal Component of the Side Wall = 0 ft 10.00 ft Radians, Calct d. Length of Sides = 10.00 ft 14.14 ft, Calct d Length or Reach of the Water Channel, on Per Unit Basis: 1 30% 0.70 Fall or Drop on Per Unit Basis: 1.0 1.0 Slope, "S" (Given) = 0.05 0.0714 Calct d Basin Width = 40 ft 20 ft Roughness, "n" 0.025 0.025 34
Problem 6, contd. Original or Existing Design: 40 ft 10 ft Proposed or Final Design: 10 ft 20 ft 1 1 35
Problem 6, contd. Determine the following water channel performance parameters based on the data provided above: a) The slope of the modified water channel is: Slope S = Rise / Run Let Rise = R, and let Run = Length = L and L o = Orig. L Therefore, S o = Original Slope = R o / L o = 0.05 {Given}, & R o = 0.05 L o Since the total Rise or Fall stays constant, R o = R f = R = 0.05 L o Also, the length of the modified channel = L f = (0.7) L o New Slope or Slope of the Modified Channel = R / L f = R / (0.7) L o = 0.05 L o / (0.7) L o = 0.0714 36
Problem 6, contd. b) The volumetric flow rate in the original, or existing, design is : Manning s Equation : Volumetric Flow Rate = Q = (1.486/n).(A).(R 2/3 ).(S 1/2 ) Where, n = Orig. Roughness = Modified Channel Roughness = 0.025 {Given} A o = Original Area in Flow = 40 ft x 10 ft = 400 sq. ft Wetted Perimeter = 10 + 40 + 10 = 60 R o = Original Hydraulic Radius = Area in Flow / Wetted Perimeter = 400 / 60 = 6.7 S o = 0.05, {Given} Therefore, Volumetric Flow Rate o = Q o = (1.486/0.025).(400).(6.7 2/3 ).(0.05 1/2 ) = 18,832 cu-ft./sec 37
Problem 6, contd. c) The velocity of water in the original, or existing, design is : Volumetric Flow Rate = Q = V.A Where, V = Fluid Velocity, in ft/sec A = Area of cross-section or Area in Flow, in sq. ft. Therefore, Velocity of water in the original, or existing, design is = Q / A = 18,832 cu-ft./sec / 400 sq. ft. = 47.08 ft. / sec. 38
Problem 6, contd. d) Which of the following factors influence the volumetric flow rate, Q, in Cu-ft/Sec? Intensity of storm, distance or total run of the water channel, size of the basin. Area, energy, type of terrain.. _ _ Area, type of terrain, slope, wetted perimeter.. Velocity, amount of rain, size of the basin, shape of water channel. 39
Problem 6, contd. e) The volumetric flow rate in the proposed or final design is : Manning s Equation : Volumetric Flow Rate = Q = (1.486/n).(A).(R 2/3 ).(S 1/2 ) Where, n = Orig. Roughness = Modified Channel Roughness = 0.025 {Given} A f = Modified or Prop. Area in Flow = (20 ft x 10 ft) + 2.(1/2 x 10 ft x 10 ft) = 300 sq. ft Wetted Perimeter, modified = 2.( 10 2 + 10 2 ) + 20 = 48.28 ft R f = Original Hydraulic Radius = Area in Flow / Wetted Perimeter = 300 / 48.28 = 6.2 S f = 0.0714, as calculated in part (a) Therefore, Final or proposed Volumetric Flow Rate = Q f = (1.486/0.025).(300).(6.2 2/3 ).(0.0714 1/2 ) = 16,117 cu-ft./sec 40
Problem 6, contd. f) The velocity of water in the modified or final design is : Volumetric Flow Rate = Q = V.A Where, V = Fluid Velocity, in ft/sec A = Area of cross-section or Area in Flow, in sq. ft. Therefore, Velocity of water in the final design is = Q / A = 16,117 cu-ft./sec / 300 sq. ft. = 53.72 ft. / sec. 41
Fluid Energy Flow Analysis Concepts; 1) Equivalent Diameter, D e, or Hydraulic Diameter, d h : The Equivalent Diameter, D e, or Hydraulic Diameter, d h are used to calculate the dimensionless Reynolds Number, R e, to determine if a flow is turbulent or laminar. D e = d h = 4. r h = 4 x Hydraulic Radius 42
Fluid Energy Flow Analysis Concepts; 2) Reynolds Number, Re: Reynolds Number, Re, is a dimensionless number representing the ratio between inertial forces and viscous forces in the fluid Re = D e. v / Where, v = Fluid Velocity = Kinematic Viscosity 43
Fluid Flow Energy Analysis : 3) Laminar and Turbulent Velocity Distributions in Circular Pipes: V=0 Parabolic Velocity Distribution V max Laminar Flow, where, Turbulent Flow, Re < 2300 (2100) where, Re > 4000 (10,000) Transient Flow for 2300 < Re < 4000 44
Fluid Flow Energy Analysis, Contd. : Energy Loss, or Head Loss due to friction, in turbulent flow situations is calculated using the Darcy Equation Method, based on the following formula: Where, H f = (f. L. V 2 ) / (2. D. g) f = Fluid Friction Factor read from Moody s Chart, using calculated Reynolds Number, Re. See the chart on the next page. L = Length of pipe, in ft. V = Fluid Velocity, in ft/sec D = Diameter of circular pipe or hydraulic diameter, of a non-circular vessel g = 32.2 ft/sec 2 45
Problem 7, contd. Moody s Graph 46
Fluid Flow Energy Analysis, Contd. : Notes : Darcy s Eq. can be applied for turbulent and laminar flow conditions. Challenge to accurately interpret the value of f from the Moody s Chart makes Darcy s method for calculation of frictional head loss less favorable. It is because of this reason that the Hazen-Williams equation and method are preferred choice under common situations, e.g. for Civil Engineering applications. 47
Fluid Flow Energy Analysis, Contd. : Energy Loss, or Head Loss calculation, in most Civil Engineering analysis is conducted using the Hazen-Williams Method, or equation. The Hazen-Williams equation is as follows: H L = (4.73. Q 1.85 L ) / (C 1.85. D 4.87 ) Where, Q = Volumetric Flow Rate in cu-ft./sec L = Length of pipe in ft. C = Hazen-Williams Roughness Coefficient, available for various pipes constructed from different types of material. See the chart in the next page. D = Dia. of circular pipe or hydraulic diameter, of a non-circular vessel, in ft. Alternate Form of H-W Equation, with Q in gpm and D in inches: H L = (10.44. Q 1.85 L ) / (C 1.85. D 4.87 ) 48
Fluid Flow Energy Analysis, Contd. : Material Hazen-Williams Coefficient - c - Asbestos Cement 140 Brass 130-140 Brick sewer 90-100 Cast-Iron - new unlined (CIP) 130 Cast-Iron 10 years old 107 113; Mean Value = 110 Cast-Iron 20 years old 89 100 Cast-Iron 30 years old 75-90 Cast-Iron 40 years old 64-83; Mean Value = 73.5 Cast-Iron, asphalt coated 100 Cast-Iron, cement lined 140 Cast-Iron, bituminous lined 140 Cast-Iron, wrought plain 100 Concrete 100-140 Copper or Brass 130-140 Corrugated Metal 60 Ductile Iron Pipe (DIP) 140 Fiber 140 Fiber Glass Pipe - FRP 150 Galvanized iron 120 Glass 130 Lead 130-140 Metal Pipes - Very to extremely smooth 130-140 49
Problem 7 A pump delivers water from a reservoir to a closed conduit system and an elevated water tank as shown in the figure below. The elevations are MSL, or with respect to the Mean Sea Level. The pump operates, continuously, at 1,000 gpm, gallons per minute. The pump efficiency is 65 % and the efficiency of the pump motor is 90%. The water level in the elevated water tank is varies between a 300 ft and 330 ft. The user demand in the closed conduit varies between 1,500 gpm and 500 gpm. The closed conduit system consists of 20 year old 10 cast iron pipe. 50
Problem 7, contd. Elevated Water Tank, EWT 2 330 MSL 300 MSL P 2 = P atm Z 2 =330 ft V 2 = 0 User Demand System C 1,200 P D Z D =330 ft V D = 2.04 ft/sec @ 500 gpm; Conversion factor: 0.13368 cu-ft/gal D 650 2,100 100 MSL 200 MSL Pump B A 1 P 1 = P atm Z 1 =110 ft V 1 = 0 110 MSL Reservoir Note: This slide is ref. to as 38 in the Audio. 51
Problem 7, contd. a) Identify the most appropriate value for the 20 year old, 10 diameter, cast iron pipe in the system. From the given Hazen-William Roughness Coefficient table: C, for 20 year old, 10 diameter, cast iron pipe is 100. Alternate, interpolation method for C calculation: C, for 40 year old, cast iron pipe is, approx. 80 C, for 10 year old, cast iron pipe is, approx. 110 C / Age = (110 80) / (40 10) = 1/1 = 1 C, for 20 year old, cast iron pipe is, approx. = 80 + (1/1). 20 = 100 52
Problem 7, contd. b) Using Hazen-Williams Equation for determination of Friction Head Loss, calculate the maximum pump horsepower developed. Assume a C, Hazen-William Roughness Coefficient, value of 105. The maximum Req. pump horsepower would be required under the following conditions: The pump is operating at a constant flow rate of 1,000 gpm. The flow to the user water distribution would be at a minimum flow rate of 500 gpm. The flow rate from point B to C is 500 gpm. The water tower is full or at maximum level, producing maximum head of 330 ft., for the pump to work against and sustain. 53
Problem 7, contd. Power Capacity of the Motor for Maximum Required Pump Head, in hp : hp = (. Q. H ) / ( 550. ) Where, = Sp. Weight of Water, in lbf/cu-ft. = 62.4 lbf/cu-ft Q = Flow Rate in Cu-ft. / sec.: 1000 gpm => 2.228 Cu-ft. / sec. = Hydraulic Pump Efficiency = 0.65 {Given} H = Pump Head in ft. =? - Calculate maximum H 54
Problem 7, contd. Apply Bernoulli s equation between the free surface (point 1) and the exit (point 2) z 1 + v 12 /2g c + p 1 / + H = z 2 + v 22 /2g c + p 2 / + H L {US, standard version divided by g } Note: p 2 / - p 1 / = 0, since p 1 & p 2, both, are at atmospheric pressure & v 1 = v 2 = 0; the free surface is stationary The Pump Head contrib. to cover the Vel. Hd. developed at Pt. D is negligible, at 0.06 ft, and therefore, left out of the Eq. The Pump Head contrib. to cover the Potential Hd. developed at Pt. D, is within the 330 ft. Elev. Tower Hd., and therefore out of the Eq. H = z 2 - z 1 + H L, or H = z 2 - z 1 + (H L-AB + H L-BC + H L-BD ) H = 330 110 + (H L-AB + H L-BC + H L-BD ) 55
Problem 7, contd. Head Loss Calculation: H L = (4.73. Q 1.85 L ) / (C 1.85. D 4.87 ) H L-AB = (4.73 x 2.228 1.85 x 2,100 ) / (105 1.85 x 0.833 4.87 ) = 19.37 ft. H L-BC = (4.73 x 1.114 1.85 x 1,200 ) / (105 1.85 x 0.833 4.87 ) = 3.07 ft H L-BD = (4.73 x 1.114 1.85 x 650 ) / (105 1.85 x 0.833 4.87 ) = 1.67 ft. Therefore, H = 330 110 + (HL AB + HL BC + HL BD ), or Max. Head Req. to be Delivered by Pump = H = 330 110 + (19.37+3.07+1.67) = 244.15 ft. And, Motor hp = (. Q. H ) / ( 550. ) = (62.4 lbf/ft 3 x 2.228ft 3 /s x 244.15ft.) / (550 lbf-ft/s/hp x 0.65) = 94.9 hp. Select a 95 hp motor. 56
Problem 7, contd. c) Calculate the maximum water pressure, in psig, available to the user. Maximum water pressure condition at user distribution point D purports the following: The pump is operating at a constant flow rate of 1,000 gpm. The flow to the user water distribution would be at a minimum flow rate of 500 gpm. The flow rate from point B to C is 500 gpm. The water tower is full or at maximum level, producing maximum head of 330 ft., for the pump to work against and sustain. Apply Bernoulli s Energy Equation at points D and (2). z 2 + v 22 /2g c + p 2 / = z D + v D2 /2g c + p D / + H L 57
Problem 7, contd. z 2 + v 22 /2g c + p 2 / = z D + v D2 /2g c + p D / + H L Note: Since v 2 = 0, v 22 /2g c = 0 Q D = 500 gpm => 1.114 ft 3 /s => v D = (1.114 ft 3 /s)/a = 2.04 ft/s Vel. Hd. at D = v D2 /2g c => 0.065 ft H L = H L-BC + H L-BD = 3.07 + 1.66 {As calc. in part (b) } p D / - p 2 / = z 2 - z D - v D2 /2g c H L = 330 200 0.065 3.07 1.66 = 125.2 ft p D-gage = (p D / - p 2 / ). = p D - p 2 = 125.2 ft x 62.4 lbf / ft 3 = 7813 lbf/ ft 2 => 54.3 psig 58
Problem 7, contd. d) Assume that the pump motor has just tripped. Calculate the maximum volumetric flow rate at user water distribution point D. Following conditions or parameters apply in this situation: The water tower is full and elevation head is 330 ft. Since the pump is de-energized, there is no water flow from A to B and Q BC = Q BD = Q =? To achieve maximum flow at D, the line must be open to atmospheric pressure; therefore, ( p D / - p 2 / ) = 0 v 22 /2g = 0; and calculated in previous parts, v D2 /2g = 0.06 0 Apply Bernoulli s Energy Equation at points D and 2. 0 0 z 2 + v 22 /2g + p 2 / = z D + v D2 /2g + p D / + H L Note: This slide is ref. to as 46 in the Audio. 59
Problem 7, contd. z 2 - z D = H BC + H BD H L = (H BC + H BD ) = ( 4.73. Q 1.85 L ) / (C 1.85. D 4.87 ) = { 4.73. Q 1.85. (1200 + 650) } / (105 1.85. 0.833 4.87 ) = 3.88 Q 1.85 Therefore, 3.88 Q 1.85 = z 2 - z D = 330 200 Q = {(330 200)/3.88} 1/1.85 Q = 6.68 cfs = 2997 gpm { Practice conversion in class} Note : This Q = 6.68 cfs => 12.25 ft/sec => v D2 /2g = 2.33 ft; which is still << Other heads. 60
Hydraulic Machines 61
Hydraulic Machines Two basic types of hydraulic machines, that convert energies between fluid and mechanical forms are: Pumps: Pumps convert mechanical energy into fluid energy; e.g. a hydraulic pump converts mechanical energy into pressure energy of the fluid which can then be used to perform work through hydraulic cylinders. Turbines: Turbines convert fluid energy into mechanical energy; e.g. turbines in hydroelectric power plants. 62
Hydraulic Machines 63
Hydraulic Machines Major Types of Pumps: 1) Positive Displacement (PD) Reciprocating Pumps. These pumps are suitable for viscous fluids and slurries; fluids sensitive to shear. These pumps provide fixed-displacement volume per cycle by entrapping a constant amount of fluid in each cycle. a) Power Pumps: It is a cylinder-operated pump. Power pumps can be Single Acting and Double Acting b) Direct Acting Pumps: Direct acting pumps, sometimes referred to as steam pumps, are double acting c) Diaphragm Pump: A diaphragm pump push s fluid mechanically through movement of a membrane diaphragm; check valves ensure flow of fluid in one direction. While diaphragm pumps are less efficient than other types of pumps they are suitable for pumping fluids that are sensitive to shearing. 64
Diaphragm Pump 65
Hydraulic Machines Major Types of Pumps, contd: 2) Rotary Pumps: Rotary pumps are positive displacement pumps that move fluid by means of screws, progressing cavities, gears, lobes, or vanes turning within a fixed casting. Rotary pumps produce smooth discharge. However, rotary pumps do experience slip, or leakage of some rotational volume back to the suction or intake side. This slip reduces the pump capacity. 66
Rotary Pump Types: Vane Type Rotary Pump Circumferential Piston Lobe Type 67
Hydraulic Machines; Major Types of Pumps contd: 3) Centrifugal Pumps: Centrifugal pumps are classified according to the way their impellers impart energy to the fluid. Speed range and specific application determine the type of centrifugal pump that should be selected. Centrifugal pumps are commonly used to move liquids through a piping system. The fluid enters the pump impeller along or near to the rotating axis and is accelerated by the impeller, flowing radially outward into a diffuser chamber, from where it exits into the downstream piping system. A centrifugal works on the principle of conversion of the kinetic energy of a flowing fluid (velocity pressure) into static pressure. Centrifugal pumps and reciprocating pumps are used extensively for pumping sludge. 68
Centrifugal Pump: 69
Hydraulic Machines, Major Types of Pumps, contd. Some of the centrifugal pump types are listed below: a) Radial Flow: Suitable for adding high pressure at low flow rates. Radial flow pumps can be single or double suction type. b) Axial Flow: Axial flow impellers impart energy to the fluid in form of pressure energy through compression. Suitable for adding low pressures at high fluid flow rates. 70
Hydraulic Machines Hydraulic Machine Formulae & Terminology : Circular Blade Pitch: Circular blade pitch is the impeller s circumference divided by the number of impeller vanes. Impeller Tip Speed, v tip : Impeller tip speed is the tangential velocity at the periphery of the impeller and is a function of impeller diameter and rotational speed. V tip =. D. n / 60 sec/min = D. ω / 2 Where, D = Diameter of the impeller n = Rotational speed in rpm ω = Angular speed in rads/sec 71
Hydraulic Machines Hydraulic Machine Formulae & Terminology, contd: Suction: Suction is the inlet point of a pump. Suction end parameters are subscripted with s. Discharge: Discharge is the outlet point of a pump. Discharge end parameters are subscripted with d. Friction Head, h f : Friction Head is the head required to overcome the resistance offered to flow of fluid in pipes, fittings, elbows, valves, entrances and exits. H f = (f. L. V 2 ) / (2. D. g) { Darcy Equation } 72
Hydraulic Machines Hydraulic Machine Formulae & Terminology, contd: Velocity Head, h v : The specific kinetic energy of the fluid. The velocity head is also referred to as dynamic head. h v = v 2 /2g Static Suction Head, h z(s) : Static Suction Head is defined as the vertical distance above the centerline of the pump inlet to the level of fluid in the reservoir. 73
Static Suction Head & Static Suction Lift Static Suction Head, h z(s) : Static Suction Lift, - h z(s 74
Static Discharge Head Static Discharge Head, h z(d) 75
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Hydraulic Machines - Formulae & Terminology, contd: Hydraulic machine parameters such as power factor, motor efficiency, motor input horsepower (EHp), brake horsepower (BHp), hydraulic horsepower (WHp), total efficiency, and pump efficiency are typically calculated or retrieved through reference tables. Additional Definitions: Head - a measure of the pressure or force exerted by the fluid. Power input - the electrical input to the motor expressed in kilowatts (kw). A measure of the rate at which work is done. Power factor - the ratio of the true power to the volt-amperes in an alternation current (ac) circuit. Motor efficiency - a measure of how effectively the motor turns electrical energy into mechanical energy. It is the ratio of power input to power output. 79
Hydraulic Machines - Formulae & Terminology, contd: Motor input horsepower (EHp) - the power input to the motor expressed in horsepower. Brake horsepower (BHp) - the power delivered to the pump shaft expressed in horsepower. Hydraulic horsepower (WHp) - the pump output or the liquid horsepower delivered by the pump. Total efficiency - the ratio of the energy delivered by the pump to the energy supplied to the input side of the motor. Sometimes refered to as the wire to water efficiency. Pump efficiency - the ratio of the energy delivered by the pump to the energy supplied to the pump shaft. 80
Efficiency & Power Delivery Stages Electrical Power from Utility EHP > BHP > WHP Motor Eff. motor Pump Efficiency, pump HP, Delivered by the Utility to the Motor HP Del. By Utility = BHP / motor BHP, Delivered by the Motor to the Pump BHP = WHP / pump WHP, Delivered by the Pump to the Fluid Note: For a steam or hydro based electric power generating system, the power flow would be the reverse of the sequence depicted above. See Problem 9. 81
Machine Formulas & Terminology, contd: Pumping Power Formulas, contd: 2. Pumping Horsepower as a function of the head added, h A, volumetric flow rate, Q, and the Specific Gravity, SG : When h A is given in ft, Q, in gal/min, Pumping Power P p, in hp, can be calculated using the following equation: Hydraulic Horsepower = WHP, Water Horsepower : P p = h A. Q. SG / 3956. Eq. 2(a) 83
Hydraulic Machine Formulas & Terminology, contd: Pumping Power Formulas, contd: 3. Pumping Horsepower as a function of the Differential Pressure, P, and the volumetric flow rate, V, in ft 3 / sec : Hydraulic Horsepower = WHP, Water Horsepower : P p = P. V / 550, if P is in psf.. Eq. 3(a) P p = P. V / 3.819, if P is in psi.. Eq. 3(b) 84
Hydraulic Machine Formulae & Terminology, contd: Pumping Power Formulas, contd: 4. Pumping Horsepower as a function of W, Work or Energy (Specific) in ft-lbf /lbm and the volumetric flow rate: Hydraulic Horsepower = WHP, Water Horsepower : P p = W. Q. (SG) / 3956; where Q is in gal/min P p = W. ṁ. / 550; where ṁ is in lbm/sec. Eq. 4(a) Eq. 4(b) P p = W. V. (SG) / 8.814; where V is in ft 3 / sec Eq. 4(c) P p = h A. V. (SG) / 8.814; where V is in ft 3 / sec..eq. 4(d) 85
Hydraulic Machine Formulas & Terminology, contd: Specific Speed: The capacity and efficiency of a centrifugal pump are, partially, a function of the impeller design. Each impeller design will yield a certain flow rate and added head. The quantitative index used to optimize the impeller design is known as specific speed, n s. n s = n. V / (h A ) 0.75.. SI Units, Eq. SS-1 n s = n. Q / (h A ) 0.75.. US Units, Eq. SS-2 Note: n and n s are in rpm 86
Hydraulic Machine Formulas & Terminology, contd: Affinity Laws: If the impeller diameter is held constant and the speed is varied, the following ratios hold true: Q 2 / Q 1 = n 2 /n 1 h 2 /h 1 = (n 2 /n 1 ) 2 P 2 /P 1 = (n 2 /n 1 ) 3.Eq. AF(1) = (Q 2 /Q 1 ) 2...Eq. AF(2) = (Q 2 /Q 1 ) 3...Eq. AF(3) If the impeller speed is held constant and the impeller size is varied, the following ratios hold true: Q 2 / Q 1 = D 2 /D 1... Eq. AF(4) h 2 /h 1 = (D 2 /D 1 ) 2. Eq. AF(5) P 2 /P 1 = (D 2 /D 1 ) 3.. Eq. AF(6) 87
BLANK SLIDE 88
Problem 8 Case study of an elevated water tower, service pump and a fire pump. The service pump operates at constant 80 gpm. The inline fire pump can draw water, in case of fire, at the rate of 40 gpm. The local fire code requires that the fire pump must be able to sustain the required supply of water for a minimum period of 3 hours. The service demand per hour, for a 24 hour period, is given on the next slide. Calculate or determine the following based on the data provide: a) Maximum flow of water INTO the water tower. b) Average or uniform demand rate for the 24 hour period. c) Maximum flow OUT of the water tower in gpm. d) The hour when maximum water flows OUT of the water tower. e) The hour when the maximum amount of water flows INTO the water tower. f) Required, minimum, water tower volume or storage capacity. 89
Problem 8 Elevated Water Tower, EWT Fire Pump, 40 gpm Service Pump, 80 gpm 90
Problem 8, contd. a) Maximum flow of water into the water tower occurs when the service demand is the least and the no fire emergency conditions exist. The calculations required for this part and other are premised on conversion of service volumes demanded each hour to respective volumetric flow rates in gpm. This data and the demand to supply deficit for each hour are calculated and tabulated on Table 8.1. The net flow of water into the EWT, under these conditions would be = 80 gpm 23.33 gpm = 56.67 gpm { 1400 gph / 60m/h = 23.33 gpm} 91
Problem 8, contd. (b) Average or uniform demand rate for the 24 hour period is determined by summing the total demand over the 24 hour period and dividing by 24, see Table 8.1: = 86,700 gal / 24 hrs / 60 min / hr = 60.21 gpm (c) The conditions for maximum flow out of the EWT, in gpm, would be as follows, see Table 8.1: Maximum Service Demand, at 1700 hrs, of 8,000 gal, translating into = 8,000 gph / 60 min/hr = 133.33 gpm Full Demand by the Fire Pump = 40 gpm Therefore, the Max. Flow Out of EWT = 133.33 + 40 = 173.33 gpm {This is assuming that the supply pump is off line} 92
Problem 8, contd. Table 8.1 I II Vol. Flow Rate in Pump Demand Minus Deficit to be Hour Peak Demand GPM Supply Rate Supply Offset by Tank in Gallons Gallons Ea. Hr. Stg. Capacity 0 1,400 23.33 4800 3,400 100 1,400 23.33 4800 3,400 200 1,400 23.33 4800 3,400 300 1,600 26.67 4800 3,200 400 2,100 35.00 4800 2,700 500 2,600 43.33 4800 2,200 600 2,900 48.33 4800 1,900 700 3,500 58.33 4800 1,300 800 3,900 65.00 4800 900 900 4,100 68.33 4800 700 1000 4,900 81.67 4800 (100) 100 1100 5,000 83.33 4800 (200) 200 1200 6,000 100.00 4800 (1,200) 1,200 1300 5,500 91.67 4800 (700) 700 1400 6,500 108.33 4800 (1,700) 1,700 1500 6,600 110.00 4800 (1,800) 1,800 1600 6,700 111.67 4800 (1,900) 1,900 1700 8,000 133.33 4800 (3,200) 3,200 1800 3,100 51.67 4800 1,700 1900 2,500 41.67 4800 2,300 2000 2,100 35.00 4800 2,700 2100 1,900 31.67 4800 2,900 2200 1,600 26.67 4800 3,200 2300 1,400 23.33 4800 3,400 86,700 10,800 Gallons 93
Problem 8, contd. (d) The hour when maximum water flows OUT of the water tower is 1700 hrs, see Table 8.1, at the rate of 133.33 gpm. (e) The hour when the maximum amount of water flows INTO the water tower are 0000, 0100, 0200 & 2300 hrs, see Table 8.1, at the rate of 23.33 gpm. (f) Required EWT storage capacity: The minimum required EWT storage capacity would have to comply with 3 hour fire pump supply code and must meet the service demand during the supply demand deficit periods; therefore, the EWT capacity components/constituents would be as follows: Volume of water required to operate the fire pump for 3 - hours: = 3 hrs x 40 gpm x 60 min/hr = 7200 gallons 94
Problem 8, part (f), contd. The EWT must have the capacity to meet the short fall during the periods when demand exceeds pump capacity. See Table 8.1. The deficit periods span from 1000 hrs through 1700 hours. The last column in the table lists the specific deficits associated with each of the eight (8) deficit hours. The total deficit from the last column in table 8.1 is = 10,800 gallons 95
Problem 8, part (f), contd. There fore the minimum required capacity of the EWT would be: = 10,800 gallons + 7,200 gallons = 18,000 gallons or = 2406.24 cu-ft. In other words, you would specify a water tank with the following radius: Since the EWT Capacity or volume is = 2,406.24 cu-ft And the volume of a sphere is = 4/3 x ᴨ x r 3 Radius of the EWT would be = r = 8.31 ft 96
Problem 9 Pressures on the intake and exit ends of a turbine are measured to be 50 psia and 10 psia, respectively. The volumetric flow rate for the fluid (water) is 120 ft 3 /sec. The turbine is driving a electric generator with nameplate efficiency of 90%. Calculate the following: a) Water Horsepower delivered by the water to the turbine. b) Brake Horsepower delivered by the turbine to the shaft driving the generator. The efficiency of the turbine is 65%. c) Maximum power generated, in KW s, by this hydroelectric power generating system. 97
Problem 9 Given: P in = 50 psia P out = 10 psia P = 40 psia P Conv. to psf = 40 lb/in 2 x 144 in 2 / ft 2 = 5760 psf V = Volumetric Flow Rate = 120 cu-ft / sec Turbine Efficiency = 65 % Generator Efficiency = 90 % a) WHP = Water Horsepower delivered by the water to the turbine: WHP = P p = P. V / 550, if P is in psf Eq. 3(a) P p = 5760 lbf/ft 2. 120 ft 3 /sec / 550 = 1256.73 hp 98
Problem 9, contd. b) Brake Horsepower delivered by the turbine to the shaft driving the generator; the turbine efficiency is 65%: BHP = WHP x Turbine Efficiency = 1256.73 hp x 0.65 = 816.87 hp c) Maximum power generated, in kw s, by this hydroelectric power generating system : Power Generated By The Hydro Elect. Power System, in kw's = BHP x Eff. x 0.746 kw / hp = 816.87 hp x 0.9 x 0.746 = 548.45 kw 99
Problem 10 Consider a hydroelectric reservoir where water is flowing through the turbine at the rate of 1100 ft3 / sec. The turbine exit point is 700 ft lower than the elevation of the reservoir water surface. The turbine efficiency is 90% and the total frictional head loss through the shaft (Draft Tube) and turbine system is 52 ft. a) Calculate the power output of the turbine in MW's. b) If the efficiency of the Electric Power Generator is 92%, what would the electric power output be for this hydroelectric power generating system? Given: specific weight = 62.4 lbf / ft 3 h f = 52 ft h z = 700 ft h A = h z - h f = 648 ft V = 1100 cu-ft / sec Turbine Efficiency 90 % Generator Efficiency 92 % 100
Problem 10, contd. a) Calculate the power output of the turbine in MW's. P p (in hp) = ( h A.. V ) / 550 Eq. 1(a) P p = WHP = Hydraulic Horsepower = (648 x 62.4 x 1100 ) / 550 = 80,870 hp P out = Power Output of the Turbine = (0.90 x 80,870) = 72,783 hp P out = Power Output of the Turbine in MW = ( 72,783 x 0.746 ) / 1000 kw per MW = 54 MW 101
Problem 10, contd. b) If the efficiency of the Electric Power Generator is 92%, what would the electric power output be for this hydroelectric power generating system? P Gen-out = Power Output of the Generator in MW = ( 54 MW x 0.92 ) = 50 MW 102
Problem 11 Consider a hydraulic machine application scenario where the system performance requirement is governed by the mathematical expression: H = 50 + Q 2. Where H represents the required system head in ft. and Q is the volumetric flow rate in cu-ft / sec. The scenario considers and compares two (2) pumps of different size. The pump performance curves for these two pumps are shown in figure 11(a). These performance curves represent operation at 1400 rpm. a) Determine the flow rate Q for Pump - 1 when it is operating at 1400 rpm, to meet the system head requirement. b) What would the system head be at the performance point in part (a)? c) Determine the flow rate Q for Pump - 2 when it is operating at 1400 rpm. 103
Problem 11 d) What would the system head be, with Pump - 2, at the performance point in part (c)? e) Calculate the Brake Horse Power, in hp, required to drive Pump - 1, under part (a) conditions. f) If Pump -1 is powered by a VFD and the shaft speed is reduced to 1200 rpm, what would the new flow rate be in the system? g) Which pump would you select for your application if a head of approx. 220 ft is required at a volumetric flow rate of 10 cu-ft/sec.? 104
Problem 11 a) Determine the flow rate Q for Pump - 1 when it is operating at 1400 rpm, to meet the system head requirement. Given: Conversion Factor: 448.8 gpm per cu-ft/sec System Head Requirement Eq.: H (Q) = 50 + Q 2 Fig. 11(a), Q vs. h A Q: Flow Rate System Head, H s 0 50 2 54 5 75 7 99 10 150 15 275 20 450 25 675 30 950 35 1275 Head, H, in ft. 250 (g) 200 150 100 50 The Pump Performance Curves are Based on 1400 rpm Operation System & Pump Curve Pump #1 Perf. Curve Pump #2 Perf. Curve 0 13 cuft/sec 0 11 cuft/sec Q, Flow Rate in 10 20 30 40 cu-ft./sec 105
Problem 11, contd. a) Determine the flow rate for Pump - 1, at the required system head, when it is operating at 1400 rpm Identify the point on the graph at which Pump-1's performance curve intersects with the system head requirement curve. Extrapolate a line (dashed line on the graph) to the horizontal axis representing Q, in Cu-ft./sec. The projected line intersects the Q- axis at 11 cu-ft/sec. Therefore, the required flow rate Q: = Q, in gals/min, or, gpm = 11 cu-ft/sec 106
Problem 11, contd. b) System Head at the performance point in part (a), at Q of 11 cuft./sec = H (Q) = 50 + Q 2 = 171 ft c) Determine the flow rate Q for Pump - 2 when it is operating at 1400 rpm. Identify the point on the graph at which Pump-2's performance curve intersects with the system head requirement curve. Extrapolate a line (dashed line on the graph) to the horizontal axis representing Q, in cu-ft./sec. The projected line intersects the Q- axis at 13 cu-ft/sec. Therefore, the required flow rate, Q = 13 cu-ft/sec Q, in gals/min, or, gpm = 13 cu-ft/sec x 448.8 gpm/(cu-ft/sec) = 5,834 gpm 107
Problem 11, contd. d) System Head at the performance point in part (a), at Q of 13 cuft/sec = H (Q) = 50 + Q 2 = 219 ft e) Calculate the Brake Horse Power, in hp, required to drive Pump - 1, under conditions stated in part (a). Approach: Calculate the WHP, Water Horse Power, first, using Eq. 4 (d); then derive the BHP using the pump efficiency. The pump efficiency is not given and must be retrieved from the given pump curve. Since, WHP = P p = h A. V. (SG) / 8.814; the power that needs to be delivered to the fluid. h A, as calculated in part (b) = 171 ft. V, as determined in part (a) = 11 cu-ft/sec,or = 4937 gpm SG for water = 1 108
Problem 11, contd. Part (e) contd. n = 1400 rpm Therefore, WHP = P p = h A. V. (SG) / 8.814 WHP = P p = 213 hp Now, BHP = WHP / Eff. p The pump efficiency Eff. p can be obtained through the Specific Speed Pump Efficiency curve "E." From equation SS-2, the Specific Speed is: n s = n. Q / (h A ) 0.75 Where, n and n s are in "rpm", Q is in "gpm" and h A is in "ft Therefore, n s = 1400. 4,937/ (171) 0.75 or, n s = 2080 rpm Then, from Pump Efficiency curve "E," the pump efficiency = 87.5% Hence, BHP = WHP / Eff. p = 213 / 0.875 = 244 hp 109
Problem 11, contd. Specific Speed vs. Pump Efficiency Curve 110
Problem 11, contd. f) If Pump -1 is powered by a VFD and the shaft speed is reduced to 1200 rpm, what would the new flow rate be in the system? From Affinity Laws; Equation AF (1): Q 2 /Q 1 = n 2 /n 1 and Q 2 = (n 2 /n 1 ). Q 1 Given and as calculated earlier: Q 1 = 4937 n 1 = 1400 n 2 = 1200 Therefore, Q 2 = (n 2 /n 1 ). Q 1 = (1200/1400) x 4937 = 4232 gpm 111
Problem 11, contd. g) Which pump would you select for your application if a head of approx. 220 ft is required at a volumetric flow rate of 10 cu-ft/sec.? The question is premised on the assumption that the system head curve has changed such that it passes through a point on the graph where Q = 10 cu-ft/sec and the head h A is approximately 220 ft. By inspection of the Q vs. h A graph, figure 11 (a), we can see that Pump #1 is capable of providing only 185 ft at Q of 10 cu-ft/sec. Therefore, Pump #2 would be more suitable for meeting the specified requirement. 112