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Energy Efficient Motor

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Published in: Mechanical
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  1. Energy Efficient Motors
  2. Why High Efficiency Motors?é, • Electric motors responsible for 40% of global electricity usage Drive pumps, fans, compressors, and many other mechanical traction equipment International Energy Agency estimates 7% of global electricity demand could be saved by higher energy efficiency motors HEMs are cost effective solutions
  3. o or anagemen Why is electric motor management important? save energy reduce operating costs minimize downtime increase productivity
  4. Candidates for Audit Three-phase induction motors Non-specialty motor 10 hp to 600 hp > 3,500 hours per year of operation Constant load (not intermittent, not cyclic, or not fluctuating) Pumps Fans
  5. Candidates for Audit Underloaded motors For Replacement (retrofit, failed) Older or rewound standard efficiency motors
  6. Motor Efficiency
  7. Motor Losses No Load loss Magnetic core Friction and windage loss Load loss 10 stator copper loss rotor copper loss Stray Loss 1- iron 2- stator 3- rotor 4- friction & windage 5- stray O Rotor Loss • Core o Friction & Wondaqe LOOd
  8. Efficiency Efficiency — Poutput X 100% Pinput Efficiency - Efficiency - Efficiency = HP?? 746 x HP output x 100% Watts input Watts output x 1000/0 Watts output + Watts losses Watts input - Watts losses x 100% atts Input
  9. ffect of Loading on Efficiency Efficiency Power Factor 40 10 10 TYPICAL SQUIRREL CAGE MOTOR 20 30 50 GO 70 PER CENT LOAD - 60 to 80%, or 80 90 110 Optimum loading near 750/0
  10. Effect of Loading on Efficiency XE EFFICIENCY & POWER FACTOR VS LOAD 30 HR 1800 RPM 95 94 93 92 . 91 90 89 88 87 86 85 84 Source: http://www.reliance.com/mtr/b7087 FACTOR 1/2 95 90 85 75 70 65 60 55 314 LOAD 4/4 5/b7087 5 7.htm
  11. How Improving Efficiency Can Reduce Costs 1% improvement in efficiency will significantly affect kW savings For a 373 kW motor (500 hp) with 94% efficiency at 80% loading required input kW = 373kW x 0.80/0.94 = 317.45 kW For a 373 kW motor (500 hp) with 95% efficiency at 80% loading required input kW = 373kW x 0.80/0.95 = 314.11 kW kW Savings 3.34 Source: CDA: Understanding High Eff. Motors
  12. Determining Motor Loads
  13. Motor % Loading Input Power Line Current • Slip Method •I nput power method is preferred •Both Line Current and Slip Methods apply only at rated input voltage, hence measurements must be corrected. •Current at low load is not good load indicator. •Nameplate speed has some % tolerance.
  14. o or 0 oa Ing Input Power Motor Load = Pim x 1000/0 Measured kW Rated kW/ Efficiency Pim = Measured power input in kW x 100%
  15. Line Current Load 1m x Vm x 1000/0 measured input current, amperes Ir = Nameplate rated current, amperes Vm = measured line-to-line Vr = Nameplate rated voltage
  16. o or 0 oa The Slip Method Load = Slip x 100% Ing (ss — Sr) ( vr/V2) Slip = Synchronous speed - Measured speed in rpm SS = Synchronous speed in rpm Sr - Nameplate rated full-load speed Vr - Rated full load volts measured volts
  17. otor % Loading (Standard Motor) Example: 15hp or 11.25 kW standard motor with efficiency of 88% and a measured kW of 5.62 Input Power Method Input kW % Loading = Rated kW / Rated Efficiency 5.62 11.25/ 88.0 43.96 %
  18. otor Efficiency t Different Loading Conditions 100 S MOTORS PREMIUM REC ANCE S TAND.4RD S MOTORS STANÔARD 40 hp USMO R M RELIANCE XE US MOTORS- DARD 20 M RS - PREMIUM XE MAGNETCK - STANOARO us M?GNETEK si US MOTORS ST? US MOTORS - PREMIUM RELIANCE XE STANDARO US MOTORS STANDARD ource; ot aster ta ser Fun Load 95-e 95 4 93 n 02.4 04.5 90, 020 87.7 86.0 BO.S 35,5 8,40 Lead 96.1 957 03.8 94,9 94,1 88 0 92.7 93.0 39,6 92.2 89,5 66.5 84.0 50% Load 96,1 95,4 04.6 94-0 90.8 02.5 92.0 ea.3 88.5 80.5 65.5 82.0 94-3 93.2 62.0 89.5 04.0 82,5 84.3 75.0
  19. Efficiency at In ut kW Rated kW % Loadin Rated Efficienc Load Efficienc load level % Standard 5.62 43.3B aa.ü 84.53 Note: Efficiency at load level was computed by interpolation using the previous slide, table of efficiency at diff. loading conditions
  20. Motor Replacement
  21. Failure by Motor Component Unknown ( no root cause failure analysis performed), External (Environment, voltage, load, will likely occur again), Failure by Motor Component Rotor Bar, 5% Shaft Coupling, 2% Bearing 51% Stator Winding (may have been voltage, water, overload, Source: Effect of Repair/Rewinding on Motor Efficiency
  22. When Motor Fails? Replace with same standard type • Replace with proper size and type Replace with high efficiency type • Rewind
  23. Rewind a Failed Motor?
  24. Rewound Motors 'Study indicates that losses of rewound motors increased by 18% or approx. 1.5 to 2.5% decrease in efficiency 'Motors 15 years old (or prev. rewound) have significantly lower efficiencies, hence it is best to replace them •If the rewind cost exceeds 50% to 65% of a new HEM price, buy a new HEM. Increased reliability and efficiency should quickly recover price premium Source: Buying an Energy-Efficient Electric Motor, Motor Challenge
  25. igh Efficiency •Improved steel properties •Thinner laminations ncrease conductors volume •Modified slot design 'More efficient fan design Motors e/7hSerM Courtesy of Copper Dev. Centre
  26. Comparison of HEM Standard Motors Figure 2: Motor Construction Standard Efficiency vs EPACT Efficiency 87.5% Premium Efficiency 902%
  27. Energy Losses Windage + Friction 13 % Air resistance and bearing friction are essentially independent of motor load. Can be reduced by improving bearing and seal selection, air flow and fan design. Energy efficient motors produce less heat and use a smaller fan Stray Loss 9 % The result of leakage fluxes induced by load currents. Can be improved by improving slot geometry Rotor losses 20% Heating in the rotor winding can be reduce by increasing the size of the conductive bars and end rings to produce lower resistance Iron loss 23% Energy required to overcome opposition to changing magnetic fields in the core material. Can be decreased by using better quality steel and by lengthening the core to reduce magnetic flux density
  28. Motor Efficiencies To summarise we can state that higher efficiency is reached by: Smaller joule losses in stator and rotor by using more copper; Smaller iron losses because of better iron core; Smaller mechanical losses because of better ventilation fans and bearings. Source: Leonardo Energy — Power quality and Utilisation Guide
  29. Motor Efficiencies Apart from that the efficiency depends on the following factors: The capacity: larger motors have a higher efficiency; The number of pair of poles: the more pair of poles, the lower the efficiency; The load: the lower the load, the lower the efficiency. Source: Leonardo Energy — Power quality and Utilisation Guide
  30. Comparison of HEM vs Standard Motors Sample Comparison of Standard Vs HEM Typical Losses (Watts) Loss Components 1) Iron 2) Stator 12 r 3) Rotor 12 r 4) Friction & Windage 5) Stray Loss Load Total Std. 220 530 218 71 131 1,170 High Eff. 104 298 192 70 101 765 AC Motor Components of Motor Loss Typical Design B Motor 10 HP, 1750 RPM, TEFC
  31. igh Efficiency vs Standard Moto Sample Estimated Savings Re lace•nent Schelne Standard Pro Size HEM Pro Size In Hit kW Rated kW % Loadin( Rated Efficien Efficien @ load level kW Savin( s kWh Savin( s Peso Savin s Ori( inal 5.62 11.25 43.96 88.00 84.53 5.62 7.5 64.44 86.00 87.16 0.20 1,156 8,089 5.62 7.5 68.71 91.7 90.25 0.42 2,427 16,990 Note: Efficiency at load level was computed by interpolation using the previous slide, table of efficiency at diff. loading conditions
  32. Estimated Annual Peso Savings kW Saved = 1 EffSTD 1 EffHE - 5.62 kW (1/0.8453 - 1/0.9025) - kWh Saved = k W saved x hours = 0.4214 x 5,760 = 2,427 kWh Peso Savings = 2,427 x *P7.O/ kWh = P16,990 0.4214 kW @P9/kWh: Savings = P21 ,843 Peso/ kWh, may vary depending on applicable rate operating hours = 20 hours/ day, 24 days/ month, 12 months/ year
  33. sing High Efficiency Motor Simple Payback Period Payback Period = Cost of High Efficiency Motor Peso Savings Php 16,990Php 2.65 years Only 2 years if electricity rate is P9/kWh
  34. Sample - HVAC Applications A Library and Archive premises uses a number of chillers and air conditioning units to maintain a constant cool environment for stored films and video tapes. A detailed study was undertaken on four motors so that the performance of high-efficiency motors could be compared directly. Table 2- 3 gives details of the motors selected & table 2-4 shows the comparison between Standard and HEM in these 4 applications . The overall payback period on the replacements was 1.1 years Table 2-3 Motor Details Motor Application Fan No 1 Fan No 2 Fan 3 Pump No 1 Motor Applicatio Fan No 1 Fan No 2 Fan No 3 Pump No 1 Overall Oliginal Motor Details 4.0 kW 2.2 kW 2.2 kW 1.5 kW 4 pole 4 pole 4 pole 4 pole Annual Operatmg Hours 8,760 8, 760 8,760 8, 760 Operatmg Load 26.8 48.6 56.8 39.3 lable 2-4 Comparative Motor Perfonnance Standard Motor Operation High-efficiency Motor Operation Operating Load (%) 26.8 48.6 56.8 39.3 42.9 Efficiency 54.6 79.3 79.3 73.0 71.6 Power Factor 0.42 0.56 0.71 0.45 0.54 Efficiency 83.6 82.9 83.3 86.0 Power Factor 0456 056 0477 0453 0T61 Cost Saving Oppoltunity Use 2.7 kW HEM Use 1.1 kWHEM Use 2.2 kW HEM Use 1.1 kWHEM Power Saving (%) 34.7 4.40 5.10 14.8 14.8
  35. Other Benefits of Good Motor Loadin and High Efficiency •Improved Power factor •Less kW demand ,Reliability
  36. Energy Sa vings Through Variable Speed Drives
  37. Variable Speed Drives In applications that require flow or pressure, particularly in systems with high friction loss, the most 'energy effective' technique is Often variable speed control. This is because the consumed power is proportional with n3, n being the number of revolutions.
  38. Suitable Prime Movers for Variable Speed include: Electronic Inverter Drive Slip ring and commutator type AC electric motors • Slip coupling DC electric motors • Variable V-belt drives Steam turbine and reciprocating motors Multi-speed dual wound or pole changing electriC motors
  39. Electronic I nverter Drives • Major advantages of electronic variable frequency drives of speed control over other techniques: Reliable soft start-up and shut-down procedures; Independent torque and speed control; Lots of motor safety controls.
  40. Motor Power Input using Variab Speed Drives Input Power2 Input Powerl Input Power2 1 00 watts 3 Fan Speed2 Fan Speedl 3 650 rpm 500 rpm --------lnput-æowerz---------=.-----æo-watts--------------------------------------------------—
  41. Sample - Variable Speed Drive Application INPUT Paw,'ER VARIABLE SPEED ONIVE vso INPUT POWER 43 CONVENTIONAL PUMPING SYSTEM SYSTEM EFFICIENCY = STAN ORO MOTOR Effie STD PUMP ENERGY-EFFICIENT PUMP ING SYSTEM SYSTEM EFFICIENCY = HIGH EFFICIENCY MOTOR HEM MORE EFFICIENT PUMP COUPLING THROTTLE Efficiency—a % PIPE OUTPUT POWER 31 OF OUTPUT R-ATEO FLOW COUPLING LOW-FRICTION PIPE aUTPUT POWER 31 OF OUTPUT RATED FLOW Figure I - a) Conventional pumping system (total effioeney 31 b) Energy-efiicjent pumping system cornbjnjng em:jent technologies (total efiicjeney
  42. Power Quality
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  44. Additional Notes on HEM
  45. Why Choose High Efficieny Motors Three factors to keep in mind whether you're replacing an old worn out motor or specifying for a piece: Energy efficient motors only provide savings when they're running, and the more the motors run, the more energy and money they save. Maximum savings ( and the fastest returns on investment) are attained in regions where utility rates are highest. Even so, energy-efficient motors are highly recommended even in low energy-cost areas because they provide savings that justify their initial cost over time. Select motor for its intended application . Every new installation should only be made after conducting a thorough analysis of the economic and technical factors involved.
  46. Energy Efficient Motors "An electric motor can consume electricity to the equivalent of its capita/ cost within the first 500 hours of operation - a mere three weeks of continuous use, or three months of single shift working. Every year, the running cost of the motor will be from four to sixteen times its capital cost. Over its working life, an average of thirteen years, it may consume over 200 times its capita/ cost in energy. Clearly, the lowest overall cost will not be achieved unless both capita/ and running costs are considered together. " CDA Publication 116
  47. Barriers to HEM Usage Lack of awareness Energy expenses are invisible to management — hidden in general overhead Low priority among capital investment and operating objectives Subsidies on electricity price First cost vs. long term operating costs Reluctance to change a working process due to lack Of experience
  48. Good Day!! Thank You. Rolando C Constantino, PEE ENPAP - AEMAS

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