Energy Efficient Motor Drive Systems

Motor Electricity Use  Motors consume about 75% of all the electricity used by industry.  Their popularity is a testament to their reliability, versatility and efficiency.  Despite these attributes, the cost of powering motor driven systems in the US is over $90 billion per year.  Thus, increasing the efficiency of motor drive systems can lead to significant savings.

Motors: The Nature of Wealth  James Watt observed that a horse pulling 180 pounds of force made 144 trips around the circle in an hour, at an average speed of 181 feet per minute. Thus, the horse generated 33,000 ft. lbs. per minute, which Watt called one “horsepower”.  Generating 1 hp required: –1,000 lb horse –6 ft tall –costs $5,000 /yr to board  Today, generating 1-hp requires: –32 lb motor (30x less) –4 x 6 inches (12x less) –costs $250 /year to power (20x less)

How Motors Work  Three stator windings located 120 o apart in cylinder around rotor.  Rotor contains magnets, called poles, and is attached to shaft.  Alternating current, also offset by 120 o, through stator windings produces electromagnetic field.  EMF interacts with magnets on rotor, which rotates rotor and shaft. Source: Douglas Wright, Department of Mechanical and Materials Engineering, The University of Western Australia,

Determining Motor Input Power Estimating Input Power –Pin (kW) = Pout,rated (hp) x FracLoad x 0.75 (kW/hp) / Efficiency –Example: If a 100 hp motor is 80% loaded and 90% efficient, estimate input power: Input Power = 100 hp x 0.80 x 0.75 kW/hp / 0.90 = 67 kW Measuring Input Power –Pin (kW) = Current (A) x Voltage (V) x PF (kW/kVA) x 3 1/2 / 1,000 VA/kVA –Example: If a motor draws 100 A at 480 V with PF = 0.80, calculate input power: Input Power = 100 A x 480 V x 0.8 kW/kVA x 3 1/2 / 1,000 VA/kVA = 66 kW

Determining Motor Loading From Input Power Fraction Loaded = Actual Output Power / Rated Output Power Example: Calculate fraction loaded of a 100 hp, 95% efficient motor drawing 66 kW –Output Power = 66 kW x 0.95 / 0.75 kW/hp = 84 hp –Fraction Loaded = 84 hp / 100 hp = 84%

Inside Out Approach to Energy Efficient Motor Drive Systems  End Use –Turn off motors when not in use –Move motor use to off-peak shift  Distribution –Motor drives  Primary Energy Conversion –Right size motors –Purchase ‘Premium Efficiency’ motors

Plant Data  Plant operates 6,000 hours per year  Cost of electricity including demand is $0.10 /kWh  Cost of demand = $10 /kW-month

Turn Off Motors When Not In Use!  Stamping press motors –80% loaded while stamping –65% loaded during idle –81% of power dissipated as heat due to friction!  Example: Turn off idling 50-hp stamping press for 2,000 hr/yr. –Savings = ? $/yr

Turn Off Motors When Not In Use!  Stamping press motors –80% loaded while stamping –65% loaded during idle –81% of power dissipated as heat due to friction!  Example: Turn off idling, 90% efficient, 50-hp stamping press motor for 2,000 hr/yr. –50 hp x.65 / 90% x.75 kW/hp x 2,000 hr/yr = 54,167 kWh/yr –54,167 kWh/yr x $0.10 /kWh = $5,417 /yr

Turn Off Motors When Not In Use!  20-hp hydraulic system motor –8 kW while loaded –5 kW while unloaded –Draws 63% of loaded power when unloaded.  Example: Turn off idling hydraulic motor for 2,000 hr/yr. –Savings = ? $/yr

Turn Off Motors When Not In Use!  20-hp hydraulic system motor –8 kW while loaded –5 kW while unloaded –Draws 63% of loaded power when unloaded.  Example: Turn off idling hydraulic motor for 2,000 hr/yr. –5 kW x 2,000 hr/yr = 10,000 kWh/yr –10,000 kWh/yr x $0.10 /kWh = $1,000 /yr

Move Motor Use to Off-Peak Shift  Some motors used only during one shift  Move use to off-peak shift to reduce electrical demand charges  Example: Move 50-hp, 80% loaded, 90% efficient, grinder to off-peak shift –Savings = ? $/yr

Move Motor Use to Off-Peak Shift  Some motors used only during one shift  Move use to off-peak shift to reduce electrical demand charges  Example: Move 50-hp, 80% loaded, 90% efficient, grinder to off-peak shift –50-hp x 80% / 90% x 0.75 kW/hp = 33 kW –33 kW x $10 /kW-mo x 12 mo/yr = $4,000 /yr

Motor Startup and Demand Most motors –Operate at 900, 1,800 or 3,600 RPM. –Draw more current (power) as the motor comes up to speed –Come up to speed in a few seconds Most utilities –Calculate peak electrical demand as max 15 or 30 minute energy use –A few seconds of high power draw not enough time to significantly increase total use during demand period –Hence, negligible demand penalty from motor start up. Thus, turn off motors when not in use –Saves energy (kWh) with negligible affect on demand (kW)

Inside Out Approach to Energy Efficient Motor Drive Systems  End Use –Turn off motors when not in use –Move motor use to off-peak shift  Distribution –Motor drives  Primary Energy Conversion –Right size motors –Purchase ‘Premium Efficiency’ motors

Power Transmission  Gear drives  = 55% - 98%  V-Belt drives  = 92% smooth – 95% notched  Synchronous belt drives  = 98%  Direct shaft couplings  = 100%

Replace V-belts with Synchronous Belt Drive Example –Replace smooth V-belt with synchronous belt drive on 100 hp, 91% efficient motor if end use load is 75% of rated output Annual Savings –? $/yr Implementation Cost –$12 /hp x 100 hp = $1,200 Simple Payback –? $/yr

Replace V-belts with Synchronous Belt Drive Example –Replace smooth V-belt with synchronous belt drive on 100 hp, 91% efficient motor if end use load is 75% of rated output Annual Savings –100 hp x 75% / 0.91 x (1/.92-1/.98) x 0.75 kW/hp = 4.11 kW –4.11 kW x 6,000 hours/yr = 24,680 kWh/year –24,680 kWh/year x $0.10 /kWh = $2,468 /year Implementation Cost –$12 /hp x 100 hp = $1,200 Simple Payback –$1,200 / $2,468 = 6 months

Replace Smooth with Notched V-belts Example –Replace smooth with notched V-belts on 100 hp, 91% efficient motor if end use load is 75% of rated output Annual Savings –? $/yr Implementation Cost –Last 50% to 400% longer than smooth belts, but cost only 30% more than smooth belts, thus implementation cost is negligible Simple Payback –? months

Replace Smooth with Notched V-belts Example –Replace smooth with notched V-belts on 100 hp, 91% efficient motor if end use load is 75% of rated output Annual Savings –100 hp x 75% / 0.91 x (1/.92-1/.95) x 0.75 kW/hp = 2.12 kW –2.12 kW x 6,000 hours/yr x $0.10 /kWh = $1,273 /year Implementation Cost –Last 50% to 400% longer than smooth belts, but cost only 30% more than smooth belts, thus implementation cost is negligible Simple Payback –Immediate

Inside Out Approach to Energy Efficient Motor Drive Systems  End Use –Turn off motors when not in use –Move motor use to off-peak shift  Distribution –Motor drives  Primary Energy Conversion –Right size motors –Purchase ‘Premium Efficiency’ motors

Right-size Under-loaded Motors Example –Right-size 100-hp, 10% loaded motor. Annual Savings –? $/yr Implementation Cost –Right-sized motor costs $500 Simple Payback – ? months

Right-size Under-loaded Motors Example –Right-size 100-hp, 10% loaded motor. Annual Savings –Input power of 100-hp motor at 60% efficiency is: 100 hp x 10% / 60% x.75 kW/hp = kW –Output power is 100 hp x 10% = 10 hp. –Input power of 10 hp motor at 82% efficiency is: 10 hp / 82% x.75 kW/hp = 9.15 kW –Annual savings: (12.5 kW kW) x 6,000 hr/yr = 20,100 kWh/yr 20,100 kWh x $0.10 /kWh = $2,012 /yr Implementation Cost –10-hp motor costs $500 Simple Payback –$500 / $2,012 /yr x 12 mo/yr = 3 months

Motors: Energy Cost >> Purchase Cost  Consider 20-hp, 75% loaded, 93% efficient motor  Purchase cost: $1,161  Annual energy cost:  20 hp x 75% / 93% x.75 kW/hp x 6,000 hr/yr x $0.10 /kWh  = $7,258 /yr  After 1 year, energy cost is 6x purchase cost!  After 10 years, energy cost is 60x purchase cost!

1% Improvement in Efficiency Equals Purchase Cost Consider –20 hp motor, 93% efficient, 80% loaded, 6,000 hr/yr, 10 years Cost of electricity –If efficiency = 93%, then 10 year electricity cost = $77,500 –If efficiency = 94%, then 10 year electricity cost = $76,600 –Savings = $77,500 - $76,600 = $900 Cost of motor –Purchase cost = $1,000 Thus, –Purchase ‘premium efficiency’ motors!

Motor Efficiency and Costs

Replace Rather Repair Old Motors Example –Replace rather than rewind 75%-loaded 20-hp motor Annual Savings –? $/yr Implementation Cost –?$ Simple Payback –? months

Replace Rather Repair Old Motors Example –Replace rather than rewind 75%-loaded 20-hp motor Annual Savings –20 hp x 75% x (1/.883-1/.935) x 0.75 kW/hp = 0.71 kW –0.71 kW x 6,000 hours/yr = 4,251 kWh/year –4,251 kWh/year x $0.10 /kWh = $425 /year Implementation Cost –$ $536 = $750 Simple Payback –$750 / $425 = 21 months

Replace Rather Repair Old Motors Example –Replace rather than rewind 75%-loaded 100-hp motor Annual Savings –? $/yr Implementation Cost –?$ Simple Payback –? months

Replace Rather Repair Old Motors Example –Replace rather than rewind 75%-loaded 100-hp motor Annual Savings –100 hp x 75% x (1/.917-1/.955) x 0.75 kW/hp = 2.44 kW –2.44 kW x 6,000 hours/yr = 14,645 kWh/year –14,645 kWh/year x $0.10 /kWh = $1,464 /year Implementation Cost –$6,167 - $1,716 = $4451 Simple Payback –$4451 / $1,464 = 36 months

Replace or Repair? Assuming 80% loaded, 6,000 hr/yr, $0.10 /kWh

U.S. D.O.E. Motor Master Software  Over 25,000 motors from 18 manufacturers  Rapid data entry, sorting by condition, and rewind/replace recommendations.  Technical data to help optimize drive systems, such as: –Motor part-load efficiency, power factor –Full-load speed, locked-rotor, breakdown, and full-load torque.  Motor purchasing information, including list prices, warranty periods, etc.  Capability to calculate savings, payback, return-on-investment, etc. 

Payback for Replacing Rather than Rewinding Motors Source: US DOE Motor Master+ 4.0

Payback for Replacing Rather than Rewinding Motors Source: US DOE Motor Master+ 4.0

Payback for Replacing Rather than Rewinding Motors Source: US DOE Motor Master+ 4.0