Water Pumps and Pumping Systems By Dr. Mohammed med Haggag Lecture 1
Course Content
Course Outlines 1. Introduction and objectives of the course.. Hydraulics of pumping systems: - Definitions - Estimation of the pumping head and efficiencies. - Pump characteristics and system curve - Pump selection and operating points. 3. Economics of pumping systems
Course Outlines (Cont.) 4. Cavitation and pump setting. 5. Applications of pumping systems: - Single pump with pipeline. - Equal pumps connected in parallel and in series - Unequal pumps in complex systems. 6. Unsteady flow in pumping systems
Course Objectives To understand the hydraulics of a pumping system and to be able to design/analyze a pumping system.
Course Outcomes - Student will be able to make a preliminary design of a pipeline (diameter, material, pressure rating ). - Student will be able to estimate the required head of a given system. - Student will be able to select the suitable pump from the economic point of view - Student will be able to carry out hydraulic analysis via spread sheet program (like Excel) - Student will be able to get the performance curves of a given pump in the laboratory. -Student will be able to carry out simple unsteady
Course Applications Covers
Toshka Pump Station 8
The beginning of Toshka canal 7-Stores Pump Station 6 Delivery Tunnels Underwater stories Lake Toshka Pump Station 9
Offices Delivery Tunnels Underwater floors Inlet X- Section in Toshka Pump Station 10
Single pump installation in Toshka Pump Station
Lab Works 1
Assignments 13
Exams 14
And Project with a Report 15
Finally Assignments Computer works Report Lab works 16
Course Web Site www.egypteducation.org Engineering Hydraulics Enrolment Key: 17
Lecture 1 Introduction
Hydraulic Pump (Definition) Water (oil) pump is a hydraulic machine. It is used to increase the total energy of the flowing water (oil) 19
Why do we need them? Why do we need to increase T.E.? - To left water from lower elevation to a higher one. 0
Why do we need them? Why do we need to increase T.E.? - To increase the flowing A discharge of a B given system. A Q B1
Question State three different applications in which Hydraulic pumps could be used?
3 cases for using a pump in the system: To raise water from a lower level to a higher one. To increase the discharge between two tanks, And to decrease the diameter of the pipe connected between two tanks. 3
Pump Classifications 4
Types of Pumps (w.r.t. Pumping Methodology) Positive displacement piston pump Diaphragm pump Rotary pumps gear pump two-lobe rotary pump screw pump Jet pumps Roto-dynamic axial-flow (propeller pump) radial-flow (centrifugal pump) mixed-flow (both axial and radial flow)
wrt Pumping Methodology Roto Dynamic Pumps Displacement Pump 6
Reciprocating action pumps Piston pump can produce very high pressures hydraulic fluid pump high pressure water washers diaphragm pump
Rotary Pumps Gear Pump fluid is trapped between gear teeth and the housing Two-lobe Rotary Pump (gear pump with two teeth on each gear) same principle as gear pump fewer chambers - more extreme pulsation trapped fluid
Rotary Pumps Disadvantages precise machining abrasives wear surfaces rapidly pulsating output Uses vacuum pumps air compressors hydraulic fluid pumps
Screw Pump Can handle debris Used to raise the level of wastewater Abrasive material will damage the seal between screw and the housing Grain augers use the same principle
Gear and Screw Pumps They are commonly used for - machinery lubrication - hydraulic elevators - - fuel oil transport and burner service -powering hydraulic machinery - for high temperature viscous products such as asphalt
Jet Pump eductor A high pressure, high velocity jet discharge is used to pump a larger volume of fluid. Advantages no moving parts self priming handles solids easily Disadvantage inefficient Uses deep well pumping pumping water mixed with solids
We have known that we have two main pump-types: - Displacement pumps and - Roto-dynamic Pumps 33
When Should I use Displacement Pumps? When Viscosity is high (over over 400 cstock) When Temperature is high When Having High Solids 34
Roto Dynamic Pumps 35
Turbomachines Demour s centrifugal pump - 1730 Theory conservation of angular momentum conversion of kinetic energy to potential energy in flow expansion Pump components impeller rotating element - encloses the rotating element and seals the pressurized liquid inside - casing or housing
Pressure Developed by Centrifugal Pumps Centrifugal pumps accelerate a liquid The maximum velocity reached is the velocity of the periphery of the impeller The kinetic energy is converted into potential energy as the fluid leaves the pump The potential energy developed is approximately equal to the velocity head at the periphery of the impeller A given pump with a given impeller diameter and speed will raise a fluid to a certain height regardless of the fluid density h p = V g
Radial Pumps also called centrifugal pumps broad range of applicable flows and heads higher heads can be achieved by increasing the diameteror the rotational speed of the impeller Discharge Suction Eye Flow Expansion Casing Impeller Impeller Vanes h p V = g
Axial Flow also known as propeller pumps low head (less than 1 m) high flows (above 0 L/s)
WRT Flow Direction (inside the Blades) Radial flow Axial flow Pump Pump (propeller) Mixed flow Pump Q H Examples Q H 40
wrt Pump Position Relative to Water Surface Non-Submersible Pump Submersible Pump 41
Electric energy Motor Mechanical energy Pump Water energy Motor efficiency x Pump efficiency = Overall efficiency η 1 x η = η t 4
TEL Concept Before Talking about How to Calculate The Pump Head, Let s First Review the TEL Concept
Energy Concept Nozzle losses
Energy Losses Darcy-Weisbach Formula Hazen-Williams Manning Equation
Comparison of Energy Losses Formula Aspect Darcy Weisbach Hazen William Manning Theoretical Basis Physically Based Empirically Based Semi- Empirically Based Range of Applications Place of application As Pipe gets rougher Laminar and Turbulent Turbulent Turbulent US Europe Australia f C n
Energy Losses Darcy-Weisbach Formula This is the most general formula for the pipe flow applications It was obtained experimentally (1845) Chezy`s equation derived from balancing the motivating and drag forces on the flow, can be reduced to the Darcy-Weisbach equation V S R C = = = = C h f L d 4 8 g f RS Reduces to fl V h = f d g h f : Loss of head due to friction (m) f: Friction factor (dimensionless) L: Length of pipe (m) d: Internal diameter of pipe (m) V: Mean velocity of flow in pipe (m/sec)
Friction Factor for Darcy-Weisbach Formula Friction factor depends on the state of the flow, which is classified according to the Reynolds number. R e = Vd ν d: Internal diameter of pipe (m) V: Mean velocity of flow in pipe (m/sec) ν: kinematic viscosity of fluid (m /sec) Laminar <000 Transition to turbulent 000-4000 Turbulent >4000
Smooth, Transition, Rough Turbulent Flow h f = f L V D g Hydraulically smooth pipe law (von Karman, 1930) Rough pipe law (von Karman, 1930) Transition function for both smooth and rough pipe laws (Colebrook) 1 R e f = lo g f.5 1 1 3.7 D = lo g f ε 1 ε D.5 1 = lo g + f 3.7 R e f (used to draw the Moody diagram)
Moody Diagram 0.1 f = C p friction factor D l laminar 0.05 0.04 0.03 0.0 0.015 0.01 0.008 0.006 0.004 0.00 ε D 0.001 0.0008 0.01 1E+03 1E+04 1E+05 1E+06 1E+07 1E+08 Re 0.0004 0.000 0.0001 0.00005 smooth
Pipe Roughness pipe material glass, drawn brass, copper 0.0015 commercial steel or wrought iron 0.045 asphalted cast iron 0.1 galvanized iron 0.15 cast iron 0.6 pipe roughness ε (mm) concrete 0.18-0.6 rivet steel 0.9-9.0 corrugated metal 45 PVC 0.1
Water Pumps and Pumping Systems By Dr. Mohammed med Haggag Lecture
Applications of the Darcy-Weisbach Equ. Type I: To Determine Head Loss Given (d, v or Q, material) Direct application Type II: To Determine Velocity or Flow Rate Given (d, h f, material) Since V or (Q) is not known, R e and f can not be determined directly. Iterative application Type III: To Determine Diameter Given (Q, limit of h f ) Since V or (Q) is not known, Re and f can not be determined directly. Iterative application
Head Loss: Minor Losses In addition to the continuous head loss along the pipe length due to friction, local head losses occur at changes in pipe section, bends, valves and fittings. Head loss due to outlet, inlet, bends, elbows, valves, pipe size changes Can be disregarded in long pipes But significant for short lengths (< 50 m) Most minor losses can not be obtained analytically, so they must be measured
Head Loss: Minor Losses Equivalent Length Technique A fictions length of pipe is estimated that will cause the same pressure drop as any fitting or change in pipe cross section. This length is added to the actual pipe length Proportional to Kinetic energy technique The loss is considered proportional to kinetic energy head given by the following formula
Head Loss: Minor Losses
Head Loss: Minor Losses Flow expansions have high losses Kinetic energy decreases across expansion Kinetic energy potential and thermal energy Examples Hydraulic jump Vena contracta Losses can be minimized by gradual transitions
Head Loss due to Sudden Expansion: Conservation of Momentum M + 1 + M = W + F p + F 1 p M 1 x + M x = F p + F 1 x p x M x ρ V A 1 1 1 = 1 A1 M x = ρ V A ρ V ρ 1 A1 + V A = p 1 A p A x A1 V V 1 p 1 p A = γ g F ss A Apply in direction of flow Neglect surface shear Pressure is applied over all of section 1. Momentum is transferred over area corresponding to upstream pipe diameter. V 1 is velocity upstream. Divide by (A γ)
Head Loss due to Sudden Expansion: Conservation of Energy 1 l t p h g V z p g V z p + + + = + + 1 1 1 1 γ γ h l g V V p p + = 1 1 γ g V V p p h l 1 1 + = γ z 1 = z What is p 1 - p?
Energy Head Loss due to Sudden Expansion g V V p p h l 1 1 + = γ g A A V V p p 1 1 1 = γ 1 1 V V A A = g V V g V V V V h l 1 1 1 + = g V V V V h l 1 1 + = ( ) g V V h l 1 = 1 1 1 = A A g V h l 1 1 = A A K Momentum Mass
Contraction EGL HGL h c = K c V g Expansion!!! V 1 V vena contracta losses are reduced with a gradual contraction
Entrance Losses Losses can be reduced by accelerating the flow gradually and eliminating the vena contracta K e 1.0 K e 0.5 K e 0.04 reentrant h e = K e V g
Head Loss in Valves Function of valve type and valve position The complex flow path through valves often results in high head loss What is the maximum value that K v can have? h v =K v V How can K be greater than 1? g
Minor/Local Losses
Questions EGL HGL What is the head loss when a pipe enters a reservoir? V V g Draw the EGL and HGL K = A1 1 A
Questions Can the Darcy-Weisbach equation and Moody Diagram be used for fluids other than water? Yes ä Does a perfectly smooth pipe have head loss? Yes ä Is it possible to decrease the head loss in a pipe by installing a smooth liner? Yes
p γ 1 1 + z 1 + V 1 g = p γ + z + V g + h l cs 1 valve 100 m cs D=40 cm D=0 cm L=1000 m L=500 m Find the discharge, Q. What additional information do you need? V Apply energy equation 1 0 0 m g How could you get a quick estimate? Or spreadsheet solution: find head loss as function of Q. = + h l
Quiz In the rough pipe law region if the flow rate is doubled (be as specific as possible) What happens to the major head loss? What happens to the minor head loss? Why do contractions have energy loss?
Can you draw TEL and HGL? V /g h i h f h e H s H s =h i +h f +h e
Thus H s =h i +h f +h e Exit head loss Static head Inlet head loss Friction head loss Hs = ΣLosses
Can you get the required d to pass a given flow in the system? H s =h i +h f +h e Hs = Ki v g + f. L d v g + Ke v g Hs n = ( K + f. L ) d i= 1 v g 71
Can you get the required d to pass a given flow in the system? H s f. l Q ) d g( π. d n ( kmin or i= 1 + / 4) = n Hs = ( K + f. L ) d i= 1 v g ) d A V=Q/A Rn f Q(eq. ) Q Required d d 7
What if d > available pipe diameters in the market 73
In this case, One could think of using a pump 74
Can you draw TEL and HGL? H s H p H p =H s +h i +h fs +h fd +h e 75
76 Can you get now Hp? H p =H s +h i +h fs +h fd +h e g v d L f d L f K H H d d d s s s n i s p ).. ( 1 + + + = =
Can you draw TEL and HGL? H s H p 77
TEL Concept Note that: Note also that: For the shown Transmission 1-the TEL slope - Line TEL is Profile, constant Slope becomes Draw as long the steeper as TEL D, f and Q HGL are the as: same; What if we used a more powerful pump? Q gets Bigger; D gets Smaller; Pipe gets Rougher.
Answer the Following Identify the flow direction for the given pipe: P A =48 psi P A =6 psi Z B = 11 m Z A = 100 m
Answer the Following Can you help me to identify the leakage location (if any)? Note that d,f are const. For the given TEL of the shown transmission line: Q
Determination of Pumping Head Calculation of Hp, A numerical Example 81
Calculation of H req & Electric Power Example 60 o, r/d= Q=100 l/s d= 300 mm L = 4.8 km G.Steel pipe (0.50) a) Calculate the required pumping head (11.40) b) Electric motor power (η t =0.78) c) Monthly and annual cost of working electrical power if: - Daily working hours = 1 hrs - Cost of 1kw-hour=0.18 L.E. 8
H p n = H s + ( i= 1 K + f. L d ) v g H s =0.5-11.4 =9.1 m.(0.3) A = π = 0.07065m 4 Q 0.1 V = = = 1.415m / s A A V = 0.1011197m g Pipe material Galvanized S. ε = 0.15 mm ε/d = 0.0005 83
V. d Rn = = 1.41548x0.3 4.x10 1.011x10 6 = υ From Moody Diagram get f f 0.0165 5 84
Determination of Local Losses Type of local losses Inlet Exit Bend Valves Others ΣKi Ki 1.0 1.0 0.1x3 0.5+3 5.88 85
H p n = H s + ( i= 1 K + f. L d ) v g H H p p = 9.1+ = 9.1+ 0.0165 *4800 (5.88 + )*0.101119 0.3 (7.448 + 0.6001) = 37.6m 86
Efficiencies t p Electric energy W. H. P η p = M. H. P M. H. P ηm = E. H. P W. H. P ηt = E. H. P η = η xη m Motor Mechanical energy Pump Water energy 87
Calculation of EHP Cost WHP = WHP = 49.48 γ. Q. H γ=1000 kg/m 3, p Q = 0.1 m 3 /s, 75 EHP= 49.48/0.78 = 63.44 EMP = EHPx0.746=48 KWatt Monthly cost of E.P.=E.M.PxTxCost H p =37.6m Monthly cost of E.P.=48x1(hrs)x30(day)x0.18 = 3100 L.E. (η t =0.78) Annual Cost of E.P.= 3100x1(month)=37000 L.E. 88
Any Questions? 89
How to get f? Moody Diagram (Chart) Friction Coefficient Roughness Ratio 0.05 (k/d) f / 4 0.0 0.018 0.016 0.014 0.01 0.010 0.008 0.006 0.06 0.04 0.0 0.01 0.006 0.004 0.00 ε d 0.005 0.004 0.003 0.001 0.0006 0.000 0.0001 0.005 0.00004 0.00 V.d υ 10 3 10 4 10 5 10 6 10 7 10 8 Reynolds Number 0.00001 90
PVC Pipe Material Roughness Height ε(mm) 0.003 Glass fiber Reinforced Plastic (GRP) Galvanized Steel Cast Iron Reinforced Concrete 0.03 0.15 0.5 1.00 91
Inlet losses K i = 0.5 K i = 1.0 9
Exit losses K e = 1.0 93
Bend/Elbow losses r θ o, r/d Bend θ r/d.5 45 60 90 135 180 1 0.11 0.19 0.5 0.33 0.41 0.48 1.5 0.1 0.17 0. 0.9 0.36 0.43 0.09 0.16 0.1 0.7 0.35 0.4 3 0.08 0.15 0. 0.6 0.35 0.4 4 0.08 0.15 0.19 0.6 0.35 0.4 θ o θ K elbow.5 0.17 30 0. 45 0.4 60 0.7 75 1 90 1.5 Elbow 94
Valve Losses Fully Open Partially Open Gate Valve Check valve (Non Return Valve) K gate = 0.5 K check = 3 Check valve Gate valve 95
Radial Flow Centrifugal Pump Settings Delivery level Delivery pipe, ld Suction pipe, ls hs Suction level 96
Axial Flow Propeller Pump Settings Delivery level Delivery pipe, ld Motor 97
Water Pumps and Pumping Systems By Dr. Mohammed med Haggag Lecture3: : Pumping Cost Two Types of Problems Existing Systems To Analyze the existing system New Systems So what do you mean by design?! To Design the system from scratch
We mean by designing a pumping system, To do the following... Design pipeline Pipe size, number Pipe material Pressure rating and Pipe thickness Pipe layout and profile Locations of valves Design pumping units Q, H EHP Pump type and model Number of pumps, arrangement and setting Working hours and control Water hammer analysis if necessary Cost of the whole system Cavitation check (if any) Pump protection 3 We mean by designing a pumping system, To do the following... Different Alternatives Should be proposed and select the optimum alternative Based on Design pipeline Pipe size, number Pipe material Pressure rating and Pipe thickness Pipe layout and profile Locations of valves Water hammer analysis if necessary Cost of the whole system Design pumping units Q, H EHP Pump type and model Number of pumps, arrangement and setting Working hours and control The Minimum Total Annual Cost Cavitation check (if any) Pump protection 4
Pumping Cost 5 Pumps and Pipeline Economics The annual cost includes: Pipeline cost C 1 Pump and motor cost. C Cost of power. C 3 Cost of Fittings. Operation & Maintenance Cost. 6
Pumps and Pipeline Economics Capital Cost of Fitting =15% Capital Cost of Pipeline O&M of Pipeline = 1% of Capital cost of pipeline O&M of Pump and Motor = 5% of Capital cost of pump & motor 7 Pumps and Pipeline Economics (Cont.) We need to: Set The Annual Cost (C 1 +C +C 3 ) To a minimum value 8
Significance of changing pipe size What do you expect if we select bigger pipe sizes? Cost of pipeline (C 1 ) Hp required Cost of pump (C ) Cost of power (C 3 ) d1 C 1 d1 Hp1 C d1 C 3 d1 < < > > > d C 1 d Hp C d C 3 d 9 Significance of changing pipe size (Cont.) Cost (C 1 +C +C 3 ) C 1 C C 3 d 10
Annual Cost of pipeline (C 1 ) Capital Cost of pipeline =L x Cost of 1m Annual Cost of pipeline (C 1 ) =Capital cost x Recovery factor 11 Annual Cost of pipeline (C1) (Cont.) n I(1 + I) Recovery Factor = n (1 + I) 1 Where: I = inflation rate or interest rate, n = Life time (yrs) 1
Annual Cost of pipeline (C1) (Cont.) C 1 = n I(1 + I) C. *. n (1 + I) 1 Capital cost Recovery factor 13 Annual Cost of pump and motor (C ) Calculate Electric MHP Capital Cost of Motor and Pump Electric MHP 14
Annual Cost of pump and motor (C) (Cont.) C = n I(1 + I) C. *. n (1 + I) 1 Capital cost Of Pump and Motor Recovery factor 15 Annual Cost of Power (C3) C 3 = γ. Q. H η t p / 75 x(0.746) x( daily. working. hrs) x365 xcost. of 1kw _ hr 16
Example (1.5) (.5) Plan Q=00 l/s L=8. km G.S. I = 8% n= 30, 15 Yrs Compute: V for (d=300:600 mm) Hp, C 1,C,C 3 Plot the total annual cost versus d. 17 Example (Cont.) pipe size (mm) 300 400 500 600 A(m) V=Q/A (m/s).894 1.59155 1.01859 0.70736 ks/d Rn=V.d/ν f (moody) Hf=fl/d.v/g Hp required WHS=γ.Q.Hp/75 Motor power Cost of pipe/m Capital Cost of pipe Capital Cost of pump C1 (n=30yrs, I=.08) C (n=15yrs, I=.08) C3 Total (C1+C+C3)(LE) 1139810 501810 43140 46005018
Example (Cont.) Cost(LE) 100000 1000000 800000 600000 400000 00000 3.5 1.5 1 0.5 V(m/s) 0 0 300 350 400 450 500 550 600 d(mm) 19 OK To get the pipe size, Take V=1.5m/s get d min Choose d (from the market) d min Calculate V act =Q/(π.d /4) Make sure that: 1.0 V act 1.5 V=1.5m/s Total Cost V=1 m/s d min d 0
Mark the following statements ( or X): The main objective of a water pump is to increase the kinetic energy of the pumped water. Priming is necessary for all pump types. Local losses can be neglected in all the pumping system. Friction losses are always higher than local losses. For a given motor power and constant efficiency, it is expected that Q will decrease when the pipeline gets older. 1 Any Questions?
Radial Flow Centrifugal Pump Settings Delivery pipe, ld Delivery level Suction pipe, ls hs Suction level 3 Axial Flow Propeller Pump Settings Delivery pipe, ld Delivery level Motor 4
Pressure Rating and Pipe Thickness
Water Pumps and Pumping Systems By Dr. Mohammed med Haggag Lectures 4: Operating Point
Operating Point It is the point of intersection for the system head curve and the pump s characteristic curve. Why we are interested in getting the Operating Point?! To get Q act. for the existing systems. To Select the relevant pump for the new systems
How to Get the Operating Point? System head Curve Representing the behavior of the piping system Pump characteristic Curve Representing the behavior of the pump system 3
How to get the System Curve H p n = H s + ( i= 1 K + fs. L d s s + fd. L d d d ) v g H. L. L n s s d d p = H s + ( K + + ) i= 1 ds dd f f Q ga Q V R n f H f H p H p H p 0 H s H p1 Q 1 H p1 H s Q H p Q 4 Q 1 Q
Special System Curves H s s d d s H p H s + ( K + + ) n = f. L. L i= 1 ds dd f Q ga a) Short pipe or smooth pipe (ideal flow) b) No Static Head (real flow) H s 5
System Curve (Cont.) Dynamic Head Static Head Dynamic Head H s >> L p >> 6
Factors Affecting System Curve H. L. L n s s d d p = H s + ( K + + ) i= 1 ds dd f f Q ga f H K d p Q 7
Pump (Characteristic) Performance Curves What are the pump performance curves? Hp Power η Q Q Q Pumping head versus discharge Brake horsepower versus discharge Efficiency versus discharge These relations are derived from actual tests on a given pump 8
How to Measure Pump Performance Curves? Hp h md Manometric head: H m = h md -h ms Q h ms Sluice valve H p =H m +V d /g-v s /g Orifice meter 9
Photo Gallery of Actual Pump Performance Test Delivery pipe 10
Photo Gallery of Actual Pump Performance Test 11
Photo Gallery of Actual Pump Performance Test H 30 5 0 15 10 5 0 H(m) = -0.3089Q - 1.415Q + 6.57 R = 0.9979 0 4 6 8 Q(l/s) 1
Three important points on the pump characteristic curve
Types of Curves
Example of Pumps with Dropping Curves
Three important points on the pump characteristic curve The shut-off head, this is the maximum head that the pump can achieve and occurs at zero flow. The pump will be noisy and vibrate excessively at this point. The pump will consume the least amount of power at this point. The best efficiency point B.E.P. this is the point at which the pump is the most efficient and operates with the least vibration and noise. This is often the point for which pumps are rated and which is indicated on the nameplate. The pump will consume the power corresponding to its B.E.P. rating at this point. The maximum flow point, the pump may not operate past this point. The pump will be noisy and vibrate excessively at this point. The pump will consume the maximum amount of power at this point.
Factors Affecting Pump Curve Motor Speed N 17
Factors Affecting Pump Curve Impeller Size 18
Factors Affecting Pump Curve Pump Type From the consistency point of view, which pump type is more preferred? 19
Factors Affecting Pump Curve Pump Type (Cont.) Best Efficiency Point (BEP) BEZ = -0%Q bep to +10%Q bep 0
Effect of Operation Outside BEZ Radial Forces Significantly Increases
Effect of Operation Outside BEZ Vibration Significantly Increases
How to Select the Relevant Pump Type? We can easily answer this question if we know the differences between pump types Pump Specific Speed (Ns) N s can be used to differentiate between pumps N s = N.Q 0.5 /H 3/4 Where: N in rpm, Q in m3/s, H in m Kindly Note that: N s is calculated only at the BEP
Major Differences in Three Classes of Centrifugal Pumps
Example How to Identify Relevant Pump Type A designer needs to select a pump type for pumping of raw wastewater for a local head of 30m and the maximum design capacity is of 0.5 m3/s at an operating speed of 100 rpm. Identify the relevant pump type. Type: Radial (N s <80), Francis Vane Type
Example: How to Identify Relevant Pump Type
Constant Versus Variable Speed Pumps Item Constant Speed Variable Speed Capability N is constant N could be variable Cost Less expensive More expensive Reliability Most reliable Less reliable Applications Suits mainly the Suits mainly the case of non case of continuous continuous flow flow No. of Required Pumping Units Water Hammer Generally Higher Generally Lower Less harmful in some cases Pumping stations having large variations in flow rates might require more pumping units under constant speed system than those under variable speed system
Any Questions?
Water Pumps and Pumping Systems By Dr. Mohammed med Haggag Lecture 5: : Pumps Configuration
Pump Configuration Pump configuration means how many pumping units are required to satisfy the design requirements (Discharge and Head). The pumping units can be arranged in any of the following configuration: 1- Single Pump. Pumps in Parallel 3. Pumps in Series 4. Pumps in Parallel-Series
Single Pump 3
How to Select the Relevant Pump? Step#1: Calculate H req and identify pump type via Ns Step#: Get the relevant pump from the pump group chart H H req P I P II P III P V P IV P VI Given: Hs = 0m, Q req = 700 m 3 /hr, L p = d p = Q req Q 4
How to Select the Relevant Pump? (Cont.) PII- Curves If PII Type is a Variable Speed Pump Alternative #1 000 1500 1100 Step#3: Draw system curve Step#4: Set the point (Q req,h req ) on the system curve. Step#5: Check the possible alternatives. Alternative # 5
How to Select the Relevant Pump? (Cont.) If PII Type is a Constant Speed Pump 6
Recirculation Line Most centrifugal pumps should not be used at a flow rate less than 50% of the B.E.P. (best efficiency point) flow rate without a recirculation line. (What is the B.E.P.?) If your system requires a flow rate of 50% or less, then use a recirculation line to increase the flow through the pump keeping the flow low in the system, or install a variable speed drive.
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9
10
Pumps in Parallel 11
Set of Pumps in Parallel Why? P III H H req P II P VI P I P IV P V Q Q req 1
Set of Pumps in Parallel Why? Q t Demand is not constant 13
Situations Require Using Pumps in Parallel (summary) 1- If Q des is significantly larger than the available pumps in the market. - If the demand is significantly variable with time. 3- For emergency cases (fire fighting). 4- For pump station upgrade to fulfill future increased demand. 14
Operating Point for Pumps in Parallel The main idea is as follow: -convert the set of parallel pumps into one single (virtual) pump. - generate the pump curve of the virtual pump knowing that: H const & Q is added up. - get the operating point (virtual pump curve with the system curve) p 1& p // Q p1 Q tot =Q p1+ Q p Q p p 1 p 15
Operating Point for Pumps in Parallel P III H H req P II P VI P I P IV P V Q req / Q Q req 16
Operating Point for Pumps in Parallel PIV- Curves Q p 1pw,H,H p 1pw Q tot Q1pw (Q p, H p ) (Q tot, H tot ) (Q 1pw,H 1pw ) (Q req, H req ) (Q req /,H req ) 17
General Comments (Q 1p, H 1p ) (Q p, H p ) (Q 3p, H 3p ) 18
the capacity of three pumps running will not be much greater than that of two pumps.running 19
Important Comment Can I use this pump? H Area prone to cavitations Q 0
Important Comment Assume we have two pumps operating in parallel H pumps in // single pump If the pump runs individually, it will experience cavitations Q 1
Variation of Operating Point with Time H pumps in // Pump# Pump#1
Variation of Operating Point with Time 70 60 50 40 30 0 10 0 0 50 100 150 00 50 300 350 400 3
Variation of Operating Point with Time 70 60 50 40 30 0 10 0 0 50 100 150 00 50 300 350 400 4
Variation of Operating Point with Time 70 60 50 40 30 0 10 Pumps with such characteristic Curves should be avoided 0 0 50 100 150 00 50 300 350 400 5
Necessary Fittings of parallel pumps 6
Pumps in Series 7
Set of Pumps in Series Why? H H req P I P II P V P III P IV P VI Q req Q 8
Operating Point for Pumps in Series The main idea is as follow: -convert the set of series pumps into one single (virtual) pump. - generate the pump curve of the virtual pump knowing that: Q const & H is added up. - get the operating point (virtual pump curve with the system curve) Q tot =Q p1 =Q p Q p Q p H tot =H p1 +H p p 1& p H p1 H p p p 1 9
Operating Point Adjustment H Suppose that the required discharge is different from the existing actual operating point what can we do? Operating point Required Design Point Q 30
Operating Point Adjustment Four solutions are available: H New Operating point after throttling Operating point 1- Use shorter pumping time with Q op instead of Q des. - Use throttle valve/not gate valve to decrease Qop new to Q des 3- Adjust pump speed if possible based on Affinity Laws. 4- Use smaller impeller d (if possible). Required Design Point Q des Q op Q 31
Affinity Laws H Equal Efficiency Curve Operating point Required Design Point Q des Q op Q 3
Any Questions?
Water Pumps and Pumping Systems By Dr. Mohammed med Haggag Lecture 6: Pump Cavitation
What is Cavitation? The Heart Attack of a Pump Cavitation comes from the Latin word, Cavus which means cavity. Cavitations take place when negative pressure decreases below a certain value called vapor pressure.
Vapor Pressure Two ways to boil water: One way: Increase temperature (while P=constant). Another way: Decrease pressure (while T=constant). 3
Cavitation in Water Pumps water vapor bubbles form when the pressure is less than the vapor pressure of water very high pressures (800 MPa or 115,000 psi) develop when the vapor bubbles collapse Vapor pressure e (Pa) 8000 7000 6000 5000 4000 3000 000 1000 0 0 10 0 30 40 Temperature (C)
What Happens when Cavitation Takes place? Formation of vapor bubbles at the eye of the impeller (the lowest pressure point). Vapor bubbles are rapidly collapsed. The rapid collapse will cause small shock waves that impact the impeller surface leading to impeller pitting and erosion. 5
Locations of Low Pressure Areas inside the Impeller of a Centrifugal Pump So What can we get from that profile?! 6
We Can get the following The pressure might be considerably less inside the impeller. i.e. local pressure drop takes place inside the impeller Not at the suction pipe. 7
Impeller Cavitations Region 8
Steps of Cavitation 9
Collapse of Bubbles and Pitting Formation 10
Symptoms of Cavitations Abnormal sound, noise and vibration; General reduction in pump efficiency; Physical damage of impeller, volute casing. Reduction in the pumped flow, might reach to zero. 11
Reduction of Pump Efficiency 1
Photographic Evidence of Cavitation 13
Photographic Evidence of Cavitation 14
Net Positive Suction Head NPSH R - absolute pressure in excess of vapor pressure required at pump inlet to prevent cavitation given by pump manufacturer determined by the water velocity at the entrance to the pump impeller NPSH A - pressure in excess of vapor pressure available at pump inlet determined by pump installation (elevation above reservoir, frictional losses, water temperature) If NPSH A is less than NPSH R cavitation will occur
Cavitation Check h s h ms H atm NPSHA h vap 16
Criteria for Checking Cavitations NPSHA = H atm -h ms -h vap NPSHA NPSHR 17
NPSH Problem Determine the maximum h s (or the minimum reservoir level) Knowing that T =18 o C, Q=15 l/s, ds = 150 mm, for the given pump type. NPSHA = H atm -h ms -h vap NPSHA = NPSHR NPSHR = H atm -h ms -h vap h ms_max = H atm -h vap - NPSHR 18 C hs=? h ms_max = 10.33-000/9810-.5=7.6 m h s_max = h ms_max -Sum Losses h s_max < 7.6 m Pvap =000 pa
Important Comments Can I use this pump? H Area prone to cavitations Q 19
Pump Diagnostic Investigation Problem Statement: A Pump station has two identical pumping units setting in parallel. During the high flow event, no problems have been identified. During the low flow event, significant reduction in the efficiency and increase in the power consumption with some vibrations and noise have been noted. Can you justify with neat sketches what might cause this problem? 0
Important Comment Assume we have two pumps operating in parallel H pumps in // single pump If the pump runs individually, it will experience cavitations Q 1
Mitigations for No Cavitations Do not use valves Do not use Concentric reducer Use Eccentric one In suction side Avoid high static suction (h s ) Long pipe should be avoided Do not use undersized diameter
Concentric Versus Eccentric Reducer 3
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