Sym Opti Motor Energy By Tahir Saleem

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Optimization Of Sizing Of Motor For Energy Conservation by Tahir Saleem

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  • Sym Opti Motor Energy By Tahir Saleem

    1. 1. Optimization Of Sizing Of Motor For Energy Conservation
    2. 2. <ul><li>In the present energy crises the only solution is to save energy in major energy consumer </li></ul><ul><li>About 70% of world electrical load is consumed by electrical motors </li></ul><ul><li>High energy consumption is due to over sizing of motors. </li></ul><ul><li>Over 50% of drive power can be saved as stated by industrial research organization </li></ul>
    3. 3. <ul><li>Motors are generally run about 50% of its capacity </li></ul><ul><li>Losses accounts for 40% to 80% </li></ul><ul><li>Motors are largest abuser of energy consumption </li></ul>
    4. 4. Important Facts <ul><li>NASA in early 70 stated that motors could operate efficiently by effective control </li></ul><ul><li>Motors operate at its maximum efficiency at 80% of its full load </li></ul><ul><li>At lower load motor efficiency drops resulting in heat vibration noise etc </li></ul>
    5. 5. <ul><li>Motor consumed energy in less then three weeks of its cost </li></ul><ul><li>Motors can control its input power requirement according to its torque requirement </li></ul>
    6. 6. <ul><li>Generally motors are oversized to miscellaneous reasons. </li></ul><ul><li>Oversized motors on variable load with fixed speed , led to extremely energy inefficient </li></ul>
    7. 7. Motor Efficiencies Efficiency 0% 100% Motor Load RED - This is a standard motor efficiency. White - Motor which gets very efficient very quickly thus power factor increases dramatically for small load change. (High Efficiency Motor) Yellow - Motor runs very inefficiently and power factor has a gradual increase. (Poor Efficiency Motor) 50% 100%
    8. 8. Motor Losses <ul><li>Motor Losses at Fixed Voltage </li></ul>% Motor Loss 1 5 10 12 % Motor Load Copper Losses (I 2 R) 50 100 Copper Losses (I 2 R) Iron Losses (kvar) Friction Losses
    9. 9. Description of Losses Loss Type Description Iron Losses Also referred to as magnetizing or core losses. The magnetic field essential to the production of torque in the rotor causes hysteretic and eddy current losses. Copper Losses Also known as electrical losses. Current flowing through the stator and rotor winding produce resistive heating losses (1 2 R) which are appox. Proportional to the current time the winding resistance. Friction and Windage Losses Mechanical losses occurs in the Bearings and brushes of the motors. Stray Losses Include leakage Fluxes, space harmonics, and saturation effect in the stator and rotor.
    10. 10. Motor Design Improvement Technology Efficiency Improvement More Copper Improves cooling and reduces operating Larger Conductor Cross Section Temperatures and power losses Larger Rotor Bars and Rings Reduces resistance and lowers stator resistive heating losses Improved Core Design Lower flux density and increased cooling capacity reduces magnetic losses Lamination Steel Premium grade gives low core losses New Slot Design Improve both efficiency and power factor Optimized Air Gap Give lower current requirement and stray load looses Improved Fan Design Reduces windage losses and improve air flow
    11. 11. Summary Of Motor Losses:   Typical NEMA B Design Motor, 10 – 20 hp; 85% Efficiency Losses   Primary I2 R losses (stator) 5.6   Secondary I2 R losses (rotor) 2.7   Iron core losses 3.0   Friction and Wind age losses 1.4 Stray losses 2.3 Losses sub total 15.0 Useful Power 85.0 TOTAL 100%
    12. 12. <ul><li>Star Delta and DOL Current </li></ul><ul><li>Star Delta and DOL Torque </li></ul>Load Torque Curves X Nominal Motor Torque Speed 100% N 2.5 1 Delta Torque Star Torque Load Torque Excess Acceleration Torque 0.33 Speed 100% N Transition Peak up to 20 x In Excess Starting Current Motor Current 6-10 x In In
    13. 13. Electric Motor Systems Motor Sizing: Motor can be correctly sized to match the load they are expected to drive.
    14. 14. Motor Selection: High efficiency motors, with improved efficiency ratings of up to 10% can be substituted for standard construction motors. Motor Controls: Where the driven equipment does not have to meet a constant demand, controls can be used to reduce capacity and improve system efficiency.
    15. 15. Motor Drive Selection and Sizing Motors and driven machines must be looked at as a “system”. To achieve an energy efficient installation, when selecting a motor or a motor control system) it is essential to:
    16. 16. <ul><li>Determine the performance and load characteristics of the driven equipment. </li></ul><ul><li>Examine the characteristics of different types of motors in order to choose the one that will provide the most effective drive. </li></ul>
    17. 17. Motor Sizing: The efficiency can be expressed as Motor efficiency = Output Mechanical Power Input Electrical Power The efficiency of a standard motor is a function of many factors include.
    18. 18. <ul><ul><li>The design </li></ul></ul><ul><ul><li>The voltage level and voltage balance of the power supply </li></ul></ul><ul><ul><li>The load at which it operates </li></ul></ul><ul><ul><li>Its kilowatt rating, and Its synchronous speed </li></ul></ul>
    19. 19. Motor Over Sizing: <ul><li>In many applications motors will be oversized for the load they are serving due to a number of factors: </li></ul>
    20. 20. <ul><li>Designers add safety factors as insurance against failure in critical building systems, and calculated design load are usually conservative relative to actual operating loads. </li></ul><ul><li>Designers want the ability to increase the output of driven equipment such as fans, pumps or chillers, at the same time – in the future. </li></ul>
    21. 21. Motor Over Sizing: <ul><ul><li>The existing load is less than the initial design load due to energy management activities. </li></ul></ul><ul><ul><li>Large motors can override load fluctuations without dropping out. </li></ul></ul><ul><ul><li>Voltage imbalances in three phase power supplies can cause increases in motor losses and so a larger motor is required to meet the duty. </li></ul></ul>
    22. 22. <ul><li>The load on a motor can be estimated by comparing a measurement (s) of the input power with the motor’s rated input power. This would entail: </li></ul><ul><li>Obtaining the motor’s nameplate kilowatt rating. </li></ul>Diagnosing Over Sized Motors
    23. 23. <ul><li>Converting the nameplate rating, which refers to the rated mechanical power from the shaft and covering it to an input power, i.e. for a 7.5 kW motor with an efficiency of 85%, at rated (nameplate) output, the input is 7.5 kW / 0.85 – 8.8 kW. </li></ul>
    24. 24. <ul><ul><li>Using a portable power analyzer to determine the kilowatt motor loading (a clip-on ammeter does not give a good indication of motor powder consumption since the power factor drops sharply at low loads and amperage readings are greatly affected by power factor) if the motors load changes overtime a data logger should be used to record the load profile. </li></ul></ul>
    25. 25. <ul><ul><li>Determining the load factor from the ratio of the actual kilowatt input to the nameplate rating. Examples at the end will quantify the saving. </li></ul></ul>
    26. 26. Benefits of Correctly Sized Motors <ul><li>General guideline regarding motor down sizing can be summarized as: </li></ul><ul><li>A small change in motor efficiency can make a substantial change in total energy consumption due to the lengthy operating time. </li></ul>
    27. 27. <ul><li>Oversized motors used for a very low number of hours each year are unlikely to benefit markedly from a change in motor efficiency since the total amount of energy used is small and will not change dramatically. </li></ul>
    28. 28. <ul><li>Motors which are operating at over 50% of full load are generally not candidates because they are operating close to peak efficiency. </li></ul><ul><li>When determining the profitability of down size a motor, the annual energy cost savings and payback period can be calculated from: </li></ul>
    29. 29.   KW Saved =kW x %Load x 100 - 100_ Old eff. new eff. Annual Cost Savings = kW Saved x Operating Hours x Electricity Cost Simple Payback Period = Motor and Installation Cost Annual Cost Savings
    30. 30. Example: A 7.5 kW motor is rated at 1450 rpm at full load. The measured was 1480 rpm and the measured power 3.7 kW.   Fraction of full load = (1500rpm - 1480rpm) =40% (1500rpm - 1450rpm) Motor Efficiency = Output Mechanical Power Input Electrical Power   Motor Efficiency = 7.5 kW x 0.4 = 81% 3.7 kW
    31. 31. Example A 30 kW motor on a supply air fan operates at 30% of rated load (9 kW) which gives rise to an operating efficiency of 75%. The motor runs for 6000 hours per year. A 15 kW motor costing Rs. 30000/= to purchase and install, is proposed for this duty. The operating efficiency of the down sized motor will be 85%
    32. 32.   KW Saved = 30kW x 0.3 x ( 100 - 100 )= 1.413 kW           75 85 Annual cost savings =1.413kW x 6000hrs x Rs 6 =Rs 50868 Year kWh kWh   Simple payback period = six month
    33. 33. Conclusion <ul><li>Proper sizing of motors for drive system can save a lot of energy </li></ul><ul><li>In addition to above motor torque control with proper control can also save energy which is not discussed here </li></ul>
    34. 34. Conclusion <ul><li>Higher efficiency motors can also save energy unfortunately not made in Pakistan </li></ul><ul><li>Variable speed drives can also save power </li></ul><ul><li>Proper coupling mechanism will also save energy </li></ul><ul><li>Proper maintenance will also save energy </li></ul>
    35. 35. <ul><li>It is very much clear that there are many factors to save energy in motors each subject require through investigation before implementation. </li></ul><ul><li>Only aspect of over sizing is presented in this paper </li></ul>

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