### Motors - University of Dayton Academic Webserver

```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
Motors: The Nature of Wealth
 James Watt observed that a horse pulling
180 pounds of force walked at 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 (20x less)
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
Turn Off Motors When Not In Use!
 Stamping press motors
– 65% of power dissipated as
heat due to friction!
 Example: Turn off 50-hp
stamping press for 2,000
hr/yr.
– 50 hp x .65 x .75 kW/hp x
2,000 hr/yr x \$0.10 /kWh =
\$4,875 /yr
Turn Off Motors When Not In Use!
 Hydraulic system motors
– Draws 63% of loaded power
 Example: Turn off 20-hp
hydraulic motor for 2,000
hr/yr.
– 5 kW x 2,000 hr/yr
x \$0.10 /kWh = \$1,000 /yr
Move Motor Operation to Off-Peak Shift
 Motor used only during first shift
 Move motor use from 1st to 2nd
shift to reduce electrical demand
 Example: Move use of 50-hp,
grinder to off-peak shift
– 50-hp x 0.75 kW/hp x 80% / 90% =
33 kW
– 33 kW x \$14 /kW-mo x 12 mo/yr =
\$5,544 /yr
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
Replace Smooth with Notched V-belts
Notched V-belts
– 3% more efficient than smooth belts
– Last 50% to 400% longer than smooth belts
– Cost only 30% more than smooth belts
Example
– 25-hp motor, 91% efficient, 75% loaded
– Savings = 25 hp x 0.75 kW/hp x 75% / .91 x
(1/.92 - 1/.95 ) = 0.5 kW
– Savings = 0.5 kW x 6,000 hours/yr = 3,000
kWh/year
– Savings = 3,000 kWh/year x \$0.10 /kWh =
\$300 /year
h = 92%
h = 95%
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
– Replace rather than repair older failed motors
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
• 20-hp, 93% eff, 75% loaded, 8,000 hrs/year, \$0.10 /kWh, cost = \$1,161
• Annual energy cost = 20 hp x 75% x .75 kW/hp / 93% x 8,000 hr/yr
x \$0.10 /kWh = \$9,677 /yr
• Over 1 yr, energy cost is 8x greater than purchase cost
• Over 12-yr life, energy cost is 100x greater than purchase cost!
Consider
– 15 hp motor, 80% loaded, 6,000 hr/yr, \$0.10 /kWh
– Standard Eff = 0.91 = \$889
Premium Eff = 0.93 = \$1,010
Cost of electricity
– Savings = 15 hp x .8 x .75 kW/hp x 6,000 hr/yr x \$0.10 /kWh x
(1/.91 – 1/.93)
– Savings = \$127 /yr
Incremental Cost of Premium Efficiency Motor
– \$1,010 - \$889 = \$121
Simple Payback
– \$127 / \$121 /yr = 1 year
Replace or Repair Older Failed Motor?
S ize
(h p )
1
5
10
15
20
30
50
60
75
100
150
200
250
300
500
E ffic ie n c y
R ewound
73
82
8 4 .7
8 5 .5
8 7 .3
8 8 .2
9 0 .6
9 0 .8
91
9 1 .2
9 1 .8
9 2 .3
9 2 .9
9 3 .1
9 2 .8
C ost
R e w o u n d (\$ )
220
330
500
550
600
760
980
1 ,1 1 6
1 ,3 2 0
1 ,6 5 0
2 ,4 0 0
2 ,6 5 0
2 ,8 6 0
3 ,0 8 0
4 ,4 0 0
E ffic ie n c y
E n g y E ff
8 4 .6
8 9 .8
9 1 .7
9 2 .6
93
9 3 .8
9 4 .4
9 4 .8
9 5 .3
9 5 .4
9 5 .5
9 5 .7
9 5 .8
9 6 .1
9 6 .6
C ost
E n g y E ff (\$ )
275
432
686
911
1 ,0 7 1
1 ,5 5 3
2 ,4 8 2
3 ,2 8 0
4 ,4 7 6
5 ,6 4 5
8 ,6 2 4
1 0 ,6 8 0
1 3 ,0 4 3
1 4 ,0 8 4
2 5 ,7 2 5
Assuming 80% loaded, 6,000 hr/yr, \$0.10 /kWh
R e w -R e p
(\$ /yr)
68
191
324
484
505
731
800
1 ,0 0 4
1 ,3 3 9
1 ,7 3 8
2 ,2 7 9
2 ,7 7 1
2 ,9 3 3
3 ,6 2 1
7 ,6 3 0
R e p -R e w
(\$ )
55
102
186
361
471
793
1 ,5 0 2
2 ,1 6 4
3 ,1 5 6
3 ,9 9 5
6 ,2 2 4
8 ,0 3 0
1 0 ,1 8 3
1 1 ,0 0 4
2 1 ,3 2 5
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
Replace Rather than Rewind 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
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
 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
Employ Energy Efficient Flow Control
Inefficient Flow Control
Fan w/ Inlet Vanes
By-pass loop
(No savings)
By-pass damper
(No savings)
Throttling valve
(Small savings)
Inlet vanes
(Moderate savings)
Efficient Flow Control
Close Bypass Valve
dP
VFD
Trim impellor for
constant-flow
pumps
Slow fan for
constant-flow
fans
VFD for
variable-flow pumps
or fans
Power and Flow Control
100%
Power (%)
80%
60%
40%
20%
0%
0%
20%
40%
60%
80%
Volume Flow Rate (%)
By-pass
Outlet Damper
Variable Inlet Vane
Variable Frequency Drive
100%
Case study: Large Cooling Towers
Large Cooling Loop Pumps
Worlds Largest Bypass Pipe
For Constant Flow Pumping:
Trim Pump Impellor and Open Throttling Valve
For Constant Flow Fan:
Slow Fan Speed by Increasing Pulley Diameter
For Variable Flow:
Install VFD & Control with Difference Pressure
b yp ass /
p ressu re
relief
valve
co o lin g
to w er
dP
co o lin g
w ater to
p ro cess
7 .5 h p
pump
city w ater
m ake-u p
25 hp
pump
reservo ir
w arm
w ater
co o l
w ater
p ro cess w ater retu rn
VSD
• W2 = W1 (V2/V1)3
• Reducing flow by
50% reduces
pumping costs by
87%
```