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pumps and thier characteristics study details
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- 2. CHAPTER 1 2- PumpsClassification 1- Function Of Pumps 3-Code and Standards
- 3. A wide varietyof pumps are used in petroleum industry. A pump is used to increase the total energy content of a liquid in the form of pressure increase. -Move liquids from low level to high level -Move liquids from low pressure location to high pressure location -Hydraulic Systems -To increase the flow rate of a liquid 1- FUNCTION OF PUMPS The pumps are used to perform one of the following jobs:
- 4. CENTRIFUGAL POSITIVE DISPLACEMENT DIAPHRAGM 2-Pumps Classification ROTARY GEAR VANE LOBE SCREW OTHERS RECIPROCATING PISTON PLUNGER OTHERS JET LIQUID RING ENERGY EXCHANGE MANY TYPES
- 5. Main Types Pumps HIGHLOW V. HIGH NO NO YES LOW HIGH V. HIGH YES NO NO Pressure P Flow Rate Q S .R .V Efficiency Maint. cost Pulsation Positive D.P. Centrifugal Axial Flow V. HIGH LOW LOW HIGH MEDIUM V. HIGH
- 6. 100 1000 10,000 20 50 200 500 2000 5000 2 5 2050 200 500 1000 2000 5000 10000 100 1000 10,000 20 50 200 500 2000 5000 100 10 Centrifugal multistage Centrifugal double suct. Axial flow Screw Gear Multi cylinder plunger GPM Ft of Liquid PSIg Centrifugal single stage
- 7. Centrifugal Pumps API 610 ASMEB73.1 & B73.2 Most common pumps API 685 Seal less Pumps Liquid Ring Vacuum Pumps API 681 Positive Displacement Pumps API 674 Reciprocating API 675 Controlled volume API 676 Rotary Firewater Pumps NFPA 20 3- Code and Standards
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- 11. Horizontal Split CaseFeed Pump 13000 gpm 7000 ft
- 12. Vertical Sump Pumps SingleStage Vertical Shaft 5000 gpm 300 ft
- 13. Vertical Cantilever Pump SlurryApplications 1000 GPM 110 ft Head 11 Ft Length
- 14. 1- Centrifugal pumpsTypes 2- Pumps arrangement
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- 19. Semi open impellerOpen impeller
- 20. Volute 1- WITHOUT DIFFUSER Volutefunction is to convert most of the Velocity energy to pressure P (v2/2g)
- 21. Volute Diffuser 2- WITH DIFFUSER Diffuserfunction is to decrease the turbulence losses and unify the direction of the outlet fluid
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- 23. Pumps – VerticalInline Centrifugal API 610, ASME B73.2
- 24. Very high Head VeryLow Flow Impellers Classification High Head High Flow Very high Flow Very Low Head
- 25. SOME TYPES OFCENTRIFUGAL PUMPS DOUBLE SUCTION IMPELLER SINGLE IMPELLER OPPOSITE IMPELLERS MULTI STAGE P0 P0 P4 P4 Balancing Drum MULTI STAGE
- 27. SINGLE IMPELLER PUMP MECHANICALSEAL HANGED BEAM IMPELLER
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- 29. SPLIT TYPE PUMPS Horizontally Split HighFlow Medium pressure
- 30. Vertically Split (DoubleBarrel) high pressure medium Flow
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- 32. Pd = P1+ P Centrifugal pumps in series Q Q Q P1 Pd P0 P1 P
- 33. Pumps arrangement Poor arrangementGood arrangement
- 34. Centrifugal pumps in parallel Q= 2 Q 1 Q1 P1 P1 Q 1 P Q P P
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- 36. FLUIDS FLOW KINAMATIC ENERGY v2<v1 P2 > P1 P1 P2 + 2 g V2 2 P2 + 2 g V1 2 P1 CONSTANT Thermal energy Plus
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- 39. DOUBLE SUCTION BALANCING HOLES P0 P0 P4 P4 BalancingDrum BALANCING DRUM OPPOSITE IMPELLERS
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- 45. P4 – P0 P4– P0 BALANCING DRUM P4 P0 P1 P2 P3 P0 P4 P0 P4 P1 – P0 P2 – P1 + P3 – P2 + P4 – P3 + P4 – P0 BALANCING LINE Balancing Drum Balancing Room P1 + –P0 – P1 + P2 + – P2 P3 –P3 P4
- 46. 40 40 BALANCING DRUM 42 2 12 22 32 2 42 2 42 12–2 22– 12 + 32– 22 + 42– 32 + 42 – 2 BALANCING LINE Balancing Drum Balancing Room 12 + –2 – 12 + 22 + –22 32 – 32 42
- 47. BALANCING DISK P4 PS P1 P2P3 PS PS P4 Balancing disk P4 P1 – PS P2 – P1 P3 – P2 P4 – P3 + + + P4 – PS P4 – PS
- 48. P4 P4 P P Ps Ps Mechanical seal andbearings arrangement Balancing Pressure Room
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- 50. submergence Total lift Nozzle Gas main Risingmain Standing water level submergence Total lift = Ratio 1 300 1.3 200 1.6 100 Ratio Total lift ( ft) Gas lift Compressor Gas injected through the nozzle, mixed with the crude and forming a foam mixture with a very light density, that means long mixture column
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- 59. ROTARY PUMPS External Gear THE FLUIDIS TRAPPED BETWEEN ROTOR AND CASING
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- 68. TIMING GEAR FUNCTION 2-KEEPS NO CONTACT BETWEEN ROTORS 1- TRANSMIT MOTION TO OTHER ROTOR 3- PREVENT WEAR BETWEEN ROTORS
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- 72. Pulsation Dampener P Accumulator Q Relief Valve KEEPENOUGH QUANTITY IN SUCTION REDUCE PRESSURE FLUCTUATION
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- 75. PRESSURE Bar T time 1 23 4 5 6 7 8 9 10 Single Plunger Pump
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- 77. PRESSURE Bar Mean discharge pressure Ttime 1 2 3 4 5 6 7 8 9 10 Duplex Pump
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- 79. PRESSURE Bar T time 1 23 4 5 6 7 8 9 10 Triplex Pump Mean discharge pressure
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- 83. A A Side View PumpCover
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- 85. AFTER ROTATION This port is connected topump discharge Due to centrifugal force, a liquid ring will be formed This port is connected to pump suction
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- 88. Planned Downtime =Hours used for all planned jobs (TPM) Breakdown Time = Hours used for all unplanned jobs (TBD) Standby Time = Hours used for standby time (TSB) 4-Equipment performance Availability Reliability Utilization Total Period Hours = (TX)
- 89. Reliability = TX –( TPM + TBD ) TX (Re) Utilization = TX – ( TPM + TBD + TSB ) TX (U t) Av - TBD TX TX TX –TPM - TBD TX = = 1 - TPM TX Availability = TX – TPM TX (Av) = Re - TSB TX = TX – ( TPM + TBD) TX TSB TX - =
- 90. 1 - TPM TX Availability = (Av) Av- TBD TX Reliability = (Re) Utilization = (U t) Re - TSB TX
- 91. EXAMPLE MAINTENANCE STOPS FORA COMPRESSOR WAS AS FOLLOWS: TOTAL PERIOD OF 3 MONTHES PM = 216 HRS BD = 216 HRS SB = 532 HRS CALCULATE AV , Re , Ut = 0.8 Reliability = 216 2160 (Re) = 80 0 0 0.9 - Utilization = (U t) 0.8 - 532 2160 = 0.6 = 60 0 0 = 0.9 Availability = 216 2160 (Av) = 90 0 0 1 - Solution One year = 8640
- 92. EXAMPLE MAINTENANCE STOPS FORA COMPRESSOR WAS AS FOLLOWS: TOTAL PERIOD OF 3 MONTHES PM = 216 HRS BD = 0 HRS SB = 532 HRS CALCULATE AV , Re , Ut = 0.9 Reliability = 0 2160 (Re) = 90 0 0 0.9 - Utilization = (U t) 0.9 - 532 2160 = 0.7 = 70 0 0 = 0.9 Availability = 216 2160 (Av) = 90 0 0 1 - Solution One year = 8640
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- 96. Containment shell Outer magnetring Rotor chamber air gap Liquid gap inner magnet sheathing Secondary pressure casing inner magnet ring
- 97. Magnetic Drive Pump,Separately Coupled.
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- 99. 2- Pumps Specificspeed 1- Centrifugal pumps Performance curve 3- Pumps Horse Power
- 100. CENTRIFUGAL PUMP PERFORMANCE CURVE(Q-H) H ft Q gpm. 100 200 300 400 500 ξ KW RPM = 3000 IMP.DIA = 10 Inch
- 101. CENTRIFUGAL PUMP PERFORMANCE CURVE(Q-H) a . b . Q gpm. 100 200 300 400 500 H ft KW ξ PUMP RPM = 3000 IMP. DIA. 10 .C 230 OPERATING POINT
- 102. Q gpm. 100 200300 400 500 H ft 60 50 N1 N3 N2 .70 40 ISO- EFFICIENCY CURVES
- 103. ISO- EFFICIENCY CURVES Qgpm. 100 200 300 400 500 H ft N1 N3 N2 60 50 .70 N1 N2 N3 Pump ( RPM ) N1 >N2 >N3
- 104. 1 (PSI) =2.31 (Ft) Pressure = Head (Ft ) x (SG) / 2.31 (PSI) Water 231 Ft x 1.0 / 2.31 = 100 PSI HCL 231 Ft x 1.2 / 2.31 = 120 PSI Gas oil 231 Ft x 0.80 / 2.31 = 80 PSI Gas oil For water
- 105. FRANCES RADIAL CAPLAN MIXEDFLOW PROPELLER NS = 500 800 1200 2000 3000 NS = Q 1/2 H 3 / 4 N Q= FLOW RATE (GALLONS. PER MIN). H= HEAD PER IMPELLER (FEET ) N = RPM PUMPS SPECIFIC SPEED NS
- 106. Radial flow 60Mixed flow 0 45 Mixed flow 0 45 0 60 0 90 0
- 108. 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 ξ 50 30 10 GPM 3000 500 300 200 100 1000 10000 GPM Q 5001000 1500 2000 2500 3000 NS N Q H 3/4 IF N = 2000 RPM Q = 1600 GPM H = 256 FT/ Impeller NS 2000 1600 256 3/4 NS 64 80000 1250 0.84 1250 IF N = 1500 RPM Q = 100 GPM H = 81 FT/ Impeller NS 1500 100 81 3/4 NS 27 15000 555 0.68 555
- 109. H P W = HQ ρ H P B = H Q ξ ρ H P W = WATER HORSEPOWER ρ = LIQUID DENSITY P = PUMP DIFF. PRESSURE Q = PUMP FLOW RATE ξ = PUMP EFFICINCY = BREAK HORSEPOWER BH P WHERE PUMPS POWER
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- 111. H P W = PQ 0.00058 P = p s i Q = GPM H P W = P Q 0.037 P = bar Q = M 3 hr HP B = P Q ξ 0.037 HP B = P Q ξ 0.00058 HOW TO ESTIMATE PUMP POWER 1 HP = 75 kg. m / sec 1 HP = 550 Ib. ft /sec WHP = Q.P 75 WHP = 75 Kg / cm2 M3 /hr WHP = 75 Kg m3 sec *3600 * 100*100 m2 WHP = 75 Kg 3600 * 100*100 m sec * WHP = Kg 0.037 m sec
- 112. FOR BOTH PUMPS WATER.HP. = 0.037 * 2*2000 HP. = 0.037 * 200*20 HP. = 148 HP EXAMPLE CALCULATE MOTOR HP. FOR 1-PUMP (A) HAS D.P = 2 bars Q = 2000 M3 / hr 2-PUMP (B) HAS D.P = 200 bars Q = 20 M3 / hr
- 113. 500 1000 15002000 2500 3000 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 ξ 50 30 10 GPM 3000 500 300 200 100 1000 10000 GPM Q 0.92 NS Q = 2000 * 4.4 = 8800 GPM H = 20 * 3.28 = 65.5 ft 4000 N S 1000 8800 65.5 3/4 N S 4000 PUMP (A)
- 114. BRAKE HP =148 /0.92 = 160 HP. Motor HP = 160*1.25 = 200 HP PUMP A NS = Q 1/2 H 3 / 4 N N = 1000 RPM D.P = 2O*3.28 = 65.5 ft Q = 2000*4.4 = 8800 GPM NS = 1000 * 8800 1/2 65.5 3/4 = 1000 * 93.8 23 4000 = ξ = 0.92
- 115. N S 1000 88 655 3/4 5001000 1500 2000 2500 3000 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 ξ 50 30 10 GPM 3000 500 300 200 100 1000 10000 GPM Q 0.3 NS Q = 88 GPM H = 6550/10 = 655 ft N S 73 73 PUMP (B)
- 116. BRAKE HP =148 /0.3 = 500 HP. Motor HP = 500 * 1.25 = 625 HP PUMP B NS = Q 1/2 H 3 / 4 N N = 1000 RPM D.P = 2O00*3.28 = 6550 ft D.P/impeller (10 imp) = 655 ft Q = 20*4.4 = 88 GPM NS = 1000 * 88 1/2 655 3/4 = 1000 * 9.38 130 73 = ξ = 0.3
- 117. Q 5000 gpmBHP 1818 P/stage 169 psi MHP 2182 N 1800 rpm H/stage 391 ft Ns 1448 ξ 0.81 END PRESSURE 500 psi D.P. 507.8 psi n 3 stages ELEVATION/2.31 5 psi EFFICIENCY AS A FUNCTION OF SPECIFIC SPEED AND CAPACITY 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 500 1000 1500 2000 2500 3000 50 30 10 GPM 5 10000 3000 500 1000 300 200 100 N S GPM 0.81
- 118. 2- Pumps AffinityLaws 1- Theory Of Pressure Increase In Impeller
- 119. FLUIDS FLOW KINAMATIC ENERGY v2<v1 P2 > P1 P1 P2 + 2 g V2 2 P2 + 2 g V1 2 P1 CONSTANT Thermal energy Plus
- 120. DIMENTIONS 2 V 2g ft 2 sec 2 ft 2 sec = (ft ) = ft 2 sec ft 2 sec = P density ft 3 2 ft = ( ft ) = Lb Lb 2 ft 3 ft =
- 122. velocity volute suction Impeller shroud + 2 g V2P FLOW KINAMATIC ENERGY
- 123. THE FLOW RATEWILL BE Q 2 Q 1 N 2 1 N IF THE PUMP SPEED CHANGES FROM N1 TO N2 PUMPS AFFINITY LAWS THE DISCH PRESS. WILL BE 2 P2 P1 N2 1 N THE HORSEPWER WILL BE 3 N2 1 N 2 1 P H P H
- 124. Exercise s m 9 , 0 1 5 , 0 45 , 0 Q D D Q n n D D c B D c B D Q Q 3 1 1 2 2 2 1 2 1 2 m 2 2 1 m 1 1 2 1 = = = = = = m 81 100 5 , 0 45 , 0 H D D H 2 1 2 1 2 2 = = = kW 90 123 5 , 0 45 , 0 P D D P 3 1 3 1 2 2 = = = •Find the flow rate, head and power for a centrifugal pump impeller that has reduced its diameter • Given data: hh = 80 % P1 = 123 kW D1 = 0,5 m H1 = 100 m D2 = 0,45 m Q1 = 1 m3/s
- 125. Exercise s m 1 , 1 1 1000 1100 Q n n Q 3 1 1 2 2 = = = m 121 100 1000 1100 H n n H 2 1 2 1 2 2= = = kW 164 123 1000 1100 P n n P 3 1 3 1 2 2 = = = • Find the flow rate, head and power for a centrifugal pump that has increased its speed • Given data: hh = 80 % P1 = 123 kW n1 = 1000 rpm H1 = 100 m n2 = 1100 rpm Q1 = 1 m3/s
- 126. Initial N1 ORD1 1000 New N2 OR D2 1500 Initial Q1 Flow rate 120 Initial P1 Pressure 10 Initial HP1 Horse power 100 New Q2 Flow rate 180 New P2 Pressure 23 New HP2 Horse power 338 N = PUMP RPM D = PUMP IMPELLER DIAMETER PUMPS AFFINITY LAWS C:Documents and Settings11030D esktophydraulics .xls - 'Flow affinity law'!A1