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
 Today, generating 1-hp
requires:
– 32 lb motor (30x less)
– 4 x 6 inches (12x less)
– costs $250 /year to
power (20x less)
 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
How Motors Work
 Three stator windings
located 120 apart in
cylinder around rotor.
 Rotor contains magnets,
called poles, and is
attached to shaft.
 Alternating current, also
offset by 120 , through
stator windings produces
electromagnetic field.
 EMF interacts with
magnets on rotor, which
rotates rotor and shaft.
o
o
Source: Douglas Wright, Department of Mechanical and Materials Engineering,
The University of Western Australia, www.mech.uwa.edu.au/DANotes/
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 31/2 / 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 31/2 / 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
h = 55% - 98%
 V-Belt drives
h = 92% smooth – 95% notched
 Synchronous belt drives
h = 98%
 Direct shaft couplings
h = 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 = 12.50 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 - 9.15 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 / $2,012 /yr x 12 mo/yr = 3 months
Motors: Energy Cost >> Purchase Cost
Purchase and Energy Costs
(20 hp motor at 8,000 hours/year over 20 years)
150,000
120,000
($)
90,000
60,000
30,000
0
Purchase
Energy
 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
Source: US DOE Motor Master+ 4.0 (2007)
Size (hp)
1
1.5
2
3
5
7.5
10
15
20
25
30
40
50
60
75
100
125
150
200
250
300
350
400
450
500
Efficiency
Rwd
74.0%
76.8%
79.3%
80.5%
83.0%
84.9%
85.7%
86.5%
88.3%
88.9%
89.2%
89.4%
91.1%
91.4%
91.5%
91.7%
91.1%
92.3%
92.8%
93.6%
93.7%
94.0%
93.1%
94.0%
93.7%
Cost
Rwd
$213
$223
$237
$249
$271
$319
$377
$463
$536
$614
$722
$858
$1,033
$1,171
$1,365
$1,716
$2,028
$2,420
$2,985
$3,571
$4,211
$4,734
$5,168
$5,722
$6,212
Efficiency
Premium Motor
85.6%
86.5%
87.0%
89.9%
90.4%
91.8%
92.0%
92.8%
93.5%
93.9%
94.1%
94.4%
94.9%
95.1%
95.5%
95.5%
95.3%
95.6%
96.2%
96.3%
96.1%
95.8%
95.8%
95.8%
95.8%
Cost
Premium
Motor
$333
$352
$406
$489
$495
$706
$805
$1,089
$1,286
$1,696
$2,104
$2,747
$3,183
$4,400
$5,118
$6,167
$8,075
$9,301
$11,281
$14,051
$19,846
$24,404
$26,750
$27,896
$29,436
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
– $1286 - $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 / $1,464 = 36 months
Replace or Repair?
Size
(hp)
1
5
10
15
20
30
50
60
75
100
150
200
250
300
500
Efficiency
Rewound
73
82
84.7
85.5
87.3
88.2
90.6
90.8
91
91.2
91.8
92.3
92.9
93.1
92.8
Cost
Rewound ($)
220
330
500
550
600
760
980
1,116
1,320
1,650
2,400
2,650
2,860
3,080
4,400
Efficiency
Engy Eff
84.6
89.8
91.7
92.6
93
93.8
94.4
94.8
95.3
95.4
95.5
95.7
95.8
96.1
96.6
Cost
Engy Eff ($)
275
432
686
911
1,071
1,553
2,482
3,280
4,476
5,645
8,624
10,680
13,043
14,084
25,725
Assuming 80% loaded, 6,000 hr/yr, $0.10 /kWh
Rew-Rep
($/yr)
68
191
324
484
505
731
800
1,004
1,339
1,738
2,279
2,771
2,933
3,621
7,630
Rep-Rew
($)
55
102
186
361
471
793
1,502
2,164
3,156
3,995
6,224
8,030
10,183
11,004
21,325
S. P.
(yr)
0.8
0.5
0.6
0.7
0.9
1.1
1.9
2.2
2.4
2.3
2.7
2.9
3.5
3.0
2.8
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.
 http://www1.eere.energy.gov/industry/bestpractices/software.html#mm
Payback for Replacing
Rather than Rewinding Motors
Operating Hours: 4,000 hrs/year
10
Simple Payback (Years)
9
$0.05 /kWh
8
$0.08 /kWh
7
$0.11 /kWh
6
5
4
3
2
1
0
0
50
100
150
200
Motor HP
Source: US DOE Motor Master+ 4.0
250
Payback for Replacing
Rather than Rewinding Motors
Operating Hours: 6,000 hrs/year
10
Simple Payback (Years)
9
$0.05 /kWh
8
$0.08 /kWh
7
$0.11 /kWh
6
5
4
3
2
1
0
0
50
150
100
Motor HP
Source: US DOE Motor Master+ 4.0
200
250
Payback for Replacing
Rather than Rewinding Motors
Operating Hours: 8,000 hrs/year
10
Simple Payback (Years)
9
$0.05 /kWh
8
$0.08 /kWh
7
$0.11 /kWh
6
5
4
3
2
1
0
0
50
100
150
200
Motor HP
Source: US DOE Motor Master+ 4.0
250