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- 1. Civil Engineering Department Prof. Majed Abu-Zreig Hydraulics and Hydrology – CE 352 Chapter 5 Water Pumps10/30/2012 1
- 2. Definition• Water pumps are devices designed to convert mechanical energy to hydraulic energy.• They are used to move water from lower points to higher points with a required discharge and pressure head.• turbines turn fluid energy into electrical or mechanical
- 3. Pump Classification• Turbo-hydraulic (kinetic) pumpsCentrifugal pumps (radial-flow pumps)Propeller pumps (axial-flow pumps)Jet pumps (mixed-flow pumps)• Positive-displacement pumpsScrew pumpsReciprocating pumps
- 4. • This classification is based on the way by which the water leaves the rotating part of the pump.• In radial-flow pump the water leaves the impeller in radial direction,• while in the axial-flow pump the water leaves the propeller in the axial direction.• In the mixed-flow pump the water leaves the impeller in an inclined direction having both radial and axial components
- 5. In a centrifugal pump flow enters alongthe axis and is expelled radially. (Thereverse is true for a turbine.)An axial-flow pump is like a propeller;the direction of the flow is unchangedafter passing through the device.A mixed-flow device is a hybrid device,used for intermediate heads.
- 6. Schematic diagram of basic elements of centrifugal pump
- 7. Schematic diagram of axial-flowpump arranged in vertical operation
- 8. Jet pump
- 9. Screw pumps.• In the screw pump a revolving shaft fitted with blades rotates in an inclined trough and pushes the water up the trough.
- 10. Reciprocating pumps.• In the reciprocating pump a piston sucks the fluid into a cylinder then pushes it up causing the water to rise.
- 11. تبارك اهلل أحسن الخالقين
- 12. RECIPROCATING INERTIA (JOGGLE) PUMPS
- 13. RECIPROCATINGDISPLACEMENT PUMPSBasic principles of positive displacement pumps
- 14. Piston or Bucket Pumps
- 15. diaphragm pump
- 16. Different types of reciprocating displacement pumps
- 17. Selection of A PumpIt has been seen that the efficiency of a pump depends on the discharge,head, and power requirement of the pump. The approximate ranges ofapplication of each type of pump are indicated in the following Figure.
- 18. Centrifugal Pumps• Demour’s centrifugal pump - 1730• Theory – conservation of angular momentum – conversion of kinetic energy to potential energy• Pump components – rotating element - impeller – encloses the rotating element and seals the pressurized liquid inside – casing or housing
- 19. Centrifugal Pumps• Broad range of applicable flows and heads• Higher heads can be achieved by increasing the diameter or the rotational speed of the impeller Flow ExpansionDischarge CasingSuction Eye Impeller Impeller Vanes
- 20. Centrifugal Pump:• Centrifugal pumps (radial-flow pumps) are the most used pumps for hydraulic purposes. For this reason, their hydraulics will be studied in the following sections.
- 21. Main Parts of Centrifugal Pumps1. Impeller: • which is the rotating part of the centrifugal pump. • It consists of a series of backwards curved vanes (blades). • The impeller is driven by a shaft which is connected to the shaft of an electric motor.
- 22. Main Parts of Centrifugal Pumps2. Casing• Which is an air-tight passage surrounding the impeller• designed to direct the liquid to the impeller and lead it away• Volute casing. It is of spiral type in which the area of the flow increases gradually.
- 23. 3. Suction Pipe.4. Delivery Pipe.5. The Shaft: which is the bar by which the power is transmitted from the motor drive to the impeller.6. The driving motor: which is responsible for rotating the shaft. It can be mounted directly on the pump, above it, or adjacent to it.
- 24. Mechanics• Fluid enters at the eye of the impeller and flows outward. As it does so it picks up the tangential velocity of the impeller vanes (or blades) which increases linearly with radius (u = r ). At exit the fluid is expelled nearly tangentially at high velocity (with kinetic energy subsequently converted to pressure energy in the expanding volute). At the same time fluid is sucked in through the inlet to take its place and maintain continuous flow.
- 25. Mechanics cont..• The analysis makes use of angular dynamics:• torque = rate of change of angular momentum• power = torque ´ angular velocity• The absolute velocity of the fluid is the vector sum of: impeller velocity (tangential)+ velocity relative to the impeller (parallel to the vanes)• Write:• u for the impeller velocity (u = wr )• n for the fluid velocity relative to the impeller• V = u + w for the absolute velocity
- 26. Pump Efficiency Pump output Power Po QH t p Power input to the pump Pi Tw Pi is the power input delivered from the motor to the impeller of the pump. Pi Qw (vto ro vti ri ) Q(vtouo vui )Motor efficiency: Pi m Pm is the power input delivered to the motor. PmOverall efficiency of the motor-pump system: o p m Po o Pm
- 27. Pump Efficiency• Example.• A pump lifts water from a large tank at a rate of 30 L/s. If the input power is 10 kW and the pump is operating at an efficiency of 40%, find:• (a) the head developed across the pump;• (b) the maximum height to which it can raise water if the delivery pipe is vertical, with D=100 mm and friction factor f = 0.015.• Answer: (a) 13.6 m; (b) 12.2 m
- 28. 5.4 Pump Characteristic Curves• Pump manufacturers provide information on the performance of their pumps in the form of curves, commonly called pump characteristic curves (or simply pump curves).• In pump curves the following information may be given: • the discharge on the x-axis, • the head on the left y-axis, • the pump power input on the right y-axis, • the pump efficiency as a percentage, • the speed of the pump (rpm = revolutions/min). • the NPSH of the pump.
- 29. NPSH - m 6 4 NPSH 2 0 70 Pump Curve 80% 60 70% 50 60%H (m) Efficiency cy cien 40 50% effi 40% % 30 20 10 0 100 200 300 400 Q (m3/hr)
- 30. A typical pump characteristics curve
- 31. Constant- and Variable-Speed Pumps• The speed of the pump is specified by the angular speed of the impeller which is measured in revolution per minutes (rpm).• Based on this speed, N , pumps can be divided into two types: • Constant-speed pumps • Variable-speed pumps
- 32. Constant-speed pumps• For this type, the angular speed , N , is constant.• There is only one pump curve which represents the performance of the pump NPSH - m 6 4 NPSH 2 0 70 Pump Curve 80% 60 70% Efficiency % 50 60% H (m) cy cien 40 50% effi 40% 30 20 10 0 100 200 300 400 Q (m3/hr)
- 33. Variable-speed pumps• For this type, the angular speed , N , is variable, i.e.; pump can operate at different speeds.• The pump performance is presented by several pump curves, one for each speed• Each curve is used to suit certain operating requirements of the system.
- 34. Similarity Laws: Affinity laws• The actual performance characteristics curves of pumps have to be determined by experimental testing.• Furthermore, pumps belonging to the same family, i.e.; being of the same design but manufactured in different sizes and, thus, constituting a series of geometrically similar machines, may also run at different speeds within practical limits.• Each size and speed combination will produce a unique characteristics curve, so that for one family of pumps the number of characteristics curves needed to be determined is impossibly large.
- 35. • The problem is solved by the application of dimensional analysis and by replacing the variables by dimensionless groups so obtained. These dimensionless groups provide the similarity (affinity) laws governing the relationships between the variables within one family of geometrically similar pumps.• Thus, the similarity laws enable us to obtain a set of characteristic curves for a pump from the known test data of a geometrically similar pump.
- 36. (a) Change in pump speed (constant size)• If a pump delivers a discharge Q1 at a head H1 when running at speed N1, the corresponding values when the same pump is running at speed N2 are given by the similarity (affinity) laws: 2 3 Q2 N 2 H2 N 2 Pi 2 N 2 Q1 N1 H1 N1 Pi1 N1 where Q = discharge (m3/s, or l/s). H = pump head (m). N = pump rotational speed (rpm). Pi = power input (HP, or kw).
- 37. • Therefore, if the pump curve for speed N1 is given, we can construct N1 the pump curve for the speed N2 using previous relationships. N2 Effect of speed change on pump characteristic curves.
- 38. (b) Change in pump size (constant speed)• A change in pump size and therefore, impeller diameter (D), results in a new set of characteristic curves using the following similarity (affinity) laws: 3 2 5 Q2 D2 H2 D2 P 2 D2 i Q1 D1 H1 D1 P1 D1 i where D = impeller diameter (m, cm). Note : D indicated the size of the pump
- 39. 5.5 Hydraulic Analysis of Pumps and Piping Systems• Pump can be placed in two possible position in reference to the water levels in the reservoirs.• We begin our study by defining all the different terms used to describe the pump performance in the piping system.
- 40. Surface mounted centrifugal pump installation Below-surface (sump) centrifugal pump installation
- 41. Ht Dynamic Head•Ht (total dynamic head): it is the total headdelivered by the pump: Vd2 H t H stat hl 2g Vd2 H t H stat h f d hmd h f s hms 2g In the above equations; we define: hfs : is the friction losses in the suction pipe. hfd : is the friction losses in the discharge (delivery) pipe. hms : is the minor losses in the suction pipe. hmd: is the minor losses in the discharge (delivery) pipe. Vd : velcity of the dilevery pipe
- 42. Duty PointIn general the pump has to supply enough energy to:• lift water through a certain height the static lift Hs;• overcome losses dependent on the discharge, Q.Thus the system head is s losses H = Hs + hf. :Losses areproportional to Qm, so that the system characteristic is oftenquadratic:Ht =hs +kQm (m =2 or m=1.85)
- 43. System Curve 70 H p ( z2 z1 ) kQ m 60 50 hf kQ mH (m) 40 30 H t Z 2 Z1 hL 20 Static head (z2-z1)=Hs H t H stat hL 10 0 3 6 9 12 15 18 Q (m3 /hr)
- 44. Duty Point System Duty characteristic pointPumpcharacteristic
- 45. NPSH - m 6 4 2 0 70 Pump Curve 70% Efficiency % 60 ncy 60% 50 50% icieH (m) 40% eff 40 System Curve 30 20 10 0 3 6 9 12 15 18 Q (m3 /hr)
- 46. Selection of A Pump• In selecting a particular pump for a given system:• The design conditions are specified and a pump is selected for the range of applications.• A system characteristic curve (H-Q) is then prepared.• The H-Q curve is then matched to the pump characteristics chart which is provided by the manufacturer.• The matching point (operating point) indicates the actual working conditions.
- 47. • The pump characteristic curves are very important to help select the required pump for the specified conditions.• If the system curve is plotted on the pump curves in we may produce the following Figure: Matching the system and pump curves.• The point of intersection is called the operating point.• This matching point indicates the actual working conditions, and therefore the proper pump that satisfy all required performance characteristic is selected.
- 48. System Characteristic CurveThe total head, Ht , that the pump delivers includes theelevation head and the head losses incurred in the system. Thefriction loss and other minor losses in the pipeline depend onthe velocity of the water in the pipe, and hence the total headloss can be related to the discharge rateFor a given pipeline system (including a pump or a group ofpumps), a unique system head-capacity (H-Q) curve can beplotted. This curve is usually referred to as a systemcharacteristic curve or simply system curve. It is a graphicrepresentation of the system head and is developed by plottingthe total head, over a range of flow rates starting from zero tothe maximum expected value of Q.
- 49. System Characteristic Curve Ht H stat h L
- 50. Selected Pump
- 51. Elevated Tank
- 52. Selected Pump
- 53. System Curve & Pump Curve cases Pump selection Pump Curve System Curve Pump Curve System Curve Pump Curve System Curve
- 54. Example 2For the following pump, determine the required pipes diameter topump 60 L/s and also calculate the needed power.Minor losses 10 v2/2gPipe length 10 kmroughness = 0.15 mmhs = 20 m Q 70 60 50 40 30 20 10 0 L/s Ht 31 35 38 40.6 42.5 43.7 44.7 45 P 40 53 60 60 57 50 35 -
- 55. To get 60 L/s from the pump hs + hL must be < 35 m Assume the diameter = 300mm Then:A 0.070m 2 , V 0.85m / sRe 2.25 10 5 , K S / D 0.0005, f 0.019 0.019 10000 0.85 2hf 23.32m 0.3 19.62 10 V 2 10 0.85 2hm 0.37m 2g 2ghs h f hm 43.69m 35m
- 56. Assume the diameter = 350mm Then: A 0.0962m 2 ,V 0.624m / s Re 1.93 10 5 , K S / D 0.00043, f 0.0185 h f 10.48m, 10 V 2 10 0.624 2 hm 0.2m 2g 2g hs h f hm 30.68m 35m QH t 1000 9.81 1000 35 60Pi 38869.8W 38.87kW p 0.53
- 57. Example 3A pump was designed to satisfy the following system Q (m3/hr) 3 6 9 hf (m) 12 20 38Check whether the pump is suitable or not atm. pressure head 10.3 m Vapour pressure head 0.25m Pipe diameter is 50mm hd 13m 24 V 2 suction Part hL 2g
- 58. 1- Draw the system curve and check the operation pointHSTAT h d hS 13 7 20m
- 59. There are an operation point at: Q = 9 m3/hr H =58m NPSHR =4.1 Then Check NPSHA Q 9 / 3600V 1.27m/s π A 0.05 2 4 24 1.27 2hL 2.0m 2g Patm PVapor(NP SH)A h S h f S h mS γ air γ Vapor(NP SH)A 7 2 10.3 0.25(NP SH)A 1.05 4.1pump is not suitable, the cavitation will occur
- 60. Multiple-Pump Operation• To install a pumping station that can be effectively operated over a large range of fluctuations in both discharge and pressure head, it may be advantageous to install several identical pumps at the station. Pumps in Parallel Pumps in Series
- 61. (a) Parallel Operation• Pumping stations frequently contain several (two or more) pumps in a parallel arrangement. Manifold Qtotal Qtotal =Q1+Q2+Q3 Pump Pump Pump Q1 Q2 Q3
- 62. • In this configuration any number of the pumps can be operated simultaneously.• The objective being to deliver a range of discharges, i.e.; the discharge is increased but the pressure head remains the same as with a single pump.• This is a common feature of sewage pumping stations where the inflow rate varies during the day.• By automatic switching according to the level in the suction reservoir any number of the pumps can be brought into operation.
- 63. How to draw the pump curve for pumps in parallel???• The manufacturer gives the pump curve for a single pump operation only.• If two or pumps are in operation, the pumps curve should be calculated and drawn using the single pump curve.• For pumps in parallel, the curve of two pumps, for example, is produced by adding the discharges of the two pumps at the same head (assuming identical pumps).
- 64. Pumps in series & Parallel j nPumps in Parallel: Q Q1 Q 2 Q3 Q n Q j1 H m H m1 H m2 H m3 H mn
- 65. (b) Series Operation• The series configuration which is used whenever we need to increase the pressure head and keep the discharge approximately the same as that of a single pump• This configuration is the basis of multistage pumps; the discharge from the first pump (or stage) is delivered to the inlet of the second pump, and so on.• The same discharge passes through each pump receiving a pressure boost in doing so
- 66. Pump Pump Pump QQ Htotal =H1+H2+H3
- 67. How to draw the pump curve for pumps in series???• the manufacturer gives the pump curve for a single pump operation only.• For pumps in series, the curve of two pumps, for example, is produced by adding the heads of the two pumps at the same discharge.• Note that, of course, all pumps in a series system must be operating simultaneously
- 68. H3H1 Three pumps in series H12H1 Two pumps in series H1H1 Single pump H1 Q Q1
- 69. Pumps in series and parallel Dischrge Q (l/s)
- 70. Cavitation of Pumps and NPSH• In general, cavitation occurs when the liquid pressure at a given location is reduced to the vapor pressure of the liquid.• For a piping system that includes a pump, cavitation occurs when the absolute pressure at the inlet falls below the vapor pressure of the water.• This phenomenon may occur at the inlet to a pump and on the impeller blades, particularly if the pump is mounted above the level in the suction reservoir.
- 71. • Under this condition, vapor bubbles form (water starts to boil) at the impeller inlet and when these bubbles are carried into a zone of higher pressure, they collapse abruptly and hit the vanes of the impeller (near the tips of the impeller vanes). causing: • Damage to the pump (pump impeller) • Violet vibrations (and noise). • Reduce pump capacity. • Reduce pump efficiency
- 72. How we avoid Cavitation ??• To avoid cavitation, the pressure head at the inlet should not fall below a certain minimum which is influenced by the further reduction in pressure within the pump impeller.• To accomplish this, we use the difference between the total head at the inlet P V , and the water vapor pressure head 2 s s Pvapor 2g
- 73. Where we take the datum through the centerline of the pump impeller inlet (eye). This difference is called the Net Positive Suction Head (NPSH), so that 2 Pvapor Ps Vs NPSH 2g There are two values of NPSH of interest. The first is the required NPSH,denoted (NPSH)R , that must be maintained or exceeded so that cavitationwill not occur and usually determined experimentally and provided by themanufacturer.The second value for NPSH of concern is the available NPSH, denoted(NPSH)A , which represents the head that actually occurs for the particularpiping system. This value can be determined experimentally, or calculated ifthe system parameters are known.
- 74. How we avoid Cavitation ??• For proper pump operation (no cavitation) : (NPSH)A > (NPSH)R
- 75. Determination of datum(NPSH)A hsapplying the energy equation betweenpoint (1) and (2), datum at pumpcenter linePatm PS VS2 hS hL air 2gPS VS2 Patm hS hL 2 g airPS VS2 PVapor Patm P hS hL Vapor 2 g Vapor air Vapor Patm PVapor( NPSH ) A hS hL air Vapor
- 76. P Pvapor ( NPSH ) A hs h f s hm s atm Note that (+) is used if hs is above the pump centerline (datum).at T 20 o Patm 10.14 kN / m 2 Vapor 2.335 kN / m 2 P
- 77. Thoma’s cavitation constantThe cavitation constant: is the ratio of (NPSH)R tothe total dynamic head (Ht) is known as the Thoma’scavitation constant ( ) ( NPSH )R HtNote: If the cavitation constant is given, we can find themaximum allowable elevation of the pump inlet (eye)above the surface of the supply (suction) reservoir.
- 78. Head analysis in the suction side of the pump hp Hs hc hfs V2/2g H’s
- 79. Example 1A Pump has a cavitation constant = 0.12, this pump was instructedon well using UPVC pipe of 10m length and 200mm diameter, thereare elbow (ke=1) and valve (ke=4.5) in the system. the flow is 35m3and The total Dynamic Head Ht = 25m (from pump curve)f=0.0167Calculate the maximum suction headatm. pressure head 9.69 mVapour pressure head 0.2m
- 80. σ 0.12 NPSH R σ H t 0.12 25 3 Patm P (NPSH)A hS h f S hmS Vapor γair γVapor Q 0.035VS 1.11 m/s A π 0.22 4 VS2 1.112 VS2 1.112he 0.063 hV 4.5 4.5 0.283m 2g 2g 2g 2g L V2 10 1.112h fS f 0.0167 0.053m D 2g 0.2 2 g Patm P(NPSH)A hS h f S hmS Vapor γair γVapor3 hS 0.053 0.283 0.063 9.69 0.2hS 6.088m
- 81. Specific Speed• Pump types may be more explicitly defined by the parameter called specific speed (Ns) expressed by: N Q Ns 3 H 4Where: Q = discharge (m3/s, or l/s). H = pump total head (m). N = rotational speed (rpm).
- 82. • This expression is derived from dynamical similarity considerations and may be interpreted as the speed in rev/min at which a geometrically scaled model would have to operate to deliver unit discharge (1 l/s) when generating unit head (1 m).• The given table shows the range of Ns values for the turbo- hydraulic pumps: Pump type Ns range (Q - l/s, H-m) centrifugal up to 2600 mixed flow 2600 to 5000 axial flow 5000 to 10 000
- 83. Specific speed variation
- 84. Specific Speed vs Efficiency
- 85. Example 5• A centrifugal pump running at 1000 rpm gave the following relation between head and discharge: Discharge (m3/min) 0 4.5 9.0 13.5 18.0 22.5 Head (m) 22.5 22.2 21.6 19.5 14.1 0• The pump is connected to a 300 mm suction and delivery pipe the total length of which is 69 m and the discharge to atmosphere is 15 m above sump level. The entrance loss is equivalent to an additional 6m of pipe and f is assumed as 0.024.1. Calculate the discharge in m3 per minute.2. If it is required to adjust the flow by regulating the pump speed, estimate the speed to reduce the flow to one-half
- 86. 1) System curve:• The head required from pump = static + friction + velocity head Vd2 H t H stat h f d hmd h f s hm s 2g• Hstat = 15 m• Friction losses (including equivalent entrance losses) = 8 f LQ 2 h fs hms h fd hmd 2 g D5 8 0.024 (69 6) 2 Q g (0.3) 2 5 61.21Q 2 where Q in m3/s
- 87. 2 V 1 Q 2• Velocity head in delivery pipe = d 10.2Q 2 2g 2g A 3where Q in m /sThus:• H t 15 71.41Q 2 where Q in m3/sor• H t 15 19.83 10 3 Q 2 where Q in m3/min• From this equation and the figures given in the problem the following table is compiled: Discharge (m3/min) 0 4.5 9.0 13.5 18.0 22.5 Head available (m) 22.5 22.2 21.6 19.5 14.1 0 Head required (m) 15.0 15.4 16.6 18.6 21.4 25.0
- 88. Pump and Sytem Curves 28 26 24 22 20 18Head,Ht (m) 16 14 12 10 8 6 4 Pump Curve 2 System Curve 0 0 2 4 6 8 10 12 14 16 18 20 22 24 Discharge, Q (m3 /min)
- 89. From the previous Figure, The operating point is: • QA = 14 m3/min • HA = 19 m• At reduced speed: For half flow (Q = 7 m3/min) there will be a new operating point B at which: • QB = 7 m3/min • HB = 16 m• HomeWork How to estimate the new speed ?????
- 90. Pump and Sytem Curves 28 26 24 22 20 A 18Head,Ht (m) 16 14 B 12 10 8 Pump Curve 6 System Curve 4 A 2 B 0 0 2 4 6 8 10 12 14 16 18 20 22 24 Discharge, Q (m3 /min)
- 91. 2Q2 N 2 H2 N 2 Q1 N1 H1 N1 2H Q Q HB B 16 2H 2 Q 0.327Q 2 7This curve intersects the original curve for N1 = 1000 rpmat C where Qc= 8.2 m3/ hr and Hc= 21.9 m, thenQB N 2 7 N2 N2 = 855rpmQC N1 8.2 1000
- 92. Pump and Sytem Curves 28 26 24 22 C 20 A 18Head,Ht (m) 16 14 B 12 Pump Curve 10 System Curve 8 A 6 B 4 C 2 0 0 2 4 6 8 10 12 14 16 18 20 22 24 Discharge, Q (m3 /min)
- 93. Pumps Group
- 94. Home Work
- 95. Pumps in branching pipes
- 96. Concept of Pumps in branching pipeHp(net) = Hp - hfs hf1HSH1 = Hs1 + hf1 hf2HSH2= Hs2 + hf2 Hp Hp(net) Hs1Hp(net) =HSH1=HSH2 Hs2 hfs
- 97. Pumps in branching pipes Example 5.5 Example 5.6Q Hp Hsh1 Hsh2 H Q1+Q2 Net Hp0 80 10 30 30 20 8010 78.5 15.12 37.68 40 35 77.4720 74 30.48 60.72 50 43 69.8830 66.5 56.08 99.12 60 51 57.2340 56 91.92 152.88 70 56 39.5250 42.5 138 222 80 62 16.7560 20 194.32 306.48
- 98. Example 5.5 and 5.6 Hp Hsh1 Hsh2 H1+H2 Net Hp 100 90 80 70 60H (ft) 50 40 30 20 10 0 0 10 20 30 40 50 60 70 Q (cfs)

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