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Motor
Energy Saving Tips
Industrial
fans, pumps and air compressors use more than 50 percent
of the total motor related electricity used. According
to the U.S. Department of Energy (DOE), industry could
reduce energy used by motors by 11 to 18 percent if
it implemented all existing cost-effective technologies
and practices for improved efficiency.
Significant air emissions are released
when electricity is produced. In Minnesota, one-fourth
of the energy-related emissions of carbon dioxide, sulfur
dioxide, lead and mercury are from generating electric
power. Industry uses over 50 percent of this electricity.
Reducing electricity use by motors will help improve
Minnesota’s air quality.
As of October 1997, the Energy Policy
Act (EPACT) requires all motors sold in the U.S. to
meet efficiency standards. In 2001 a new class of premium
efficiency motors was designated, setting efficiency
standards for new motors beyond those of EPACT.
Although high efficiency motors have
been available for years, they make up less than 10
percent of all industrial motors in current use.
If your motors are not part of this
10 percent, they could be using excess electricity—increasing
your operating costs. This fact sheet will help you
calculate your motor operating costs, develop a policy
for motor repair and replacement and develop strategies
to reduce energy used by motors at your facility.
Motor Energy
Costs
Use the following equations to calculate demand and
energy costs. Together these equal your annual cost
to operate motor driven equipment. Once you have calculated
your annual cost, compare it to the annual cost to operate
EPACT and premium efficiency motors.
To calculate the energy cost per year, multiply:
- motor’s horsepower (hp)
- conversion factor 0.746 kW/hp
- number of operating hours per year (hr/yr)
- cost per kilowatt-hour ($/kWh)
- load factor (LF)
Divide the product by the motor’s efficiency. Load factor is the fraction of the motor’s horsepower actually used to drive a load. One way to determine load factor is to measure the amperage (amp) draw of the motor under load then divide by the motor’s full load rating.
| Energy cost per year = |
|
(hp)(0.746 kW/hp)(hr/yr)($/kWh)(LF)
|
|
|
|
motor efficiency |
To calculate the demand cost per year, multiply:
- motor’s horsepower (hp)
- conversion factor 0.746 kW/hp
- load factor (LF)
- cost per kilovolt amp per month ($/kVA/mo)
- 12 months per year (12 mo/yr)
Divide the product by:
- motor’s efficiency
- motor’s power factor
The cost per kVA is found on your electric bill. A power factor—ratio of true power used to the power drawn from the source—is found on a motor’s specification sheet.
|
Demand cost per year = |
| (hp)(0.746 kW/hp)(LF)($/kVA/mo)(12 mo/yr) |
 |
| (motor efficiency)(power factor) |
Example Calculations
For the example calculations
that follow use these factors:
- Motors are five horsepower
- Motors run 4,000 hours per year
- Motors run at 0.75 load factor
- Energy cost is $0.05 per kWh
- Unit demand cost is $3.00 per kVA
per month
| |
Standard |
EPACT |
Premium |
| Efficiency |
83% |
86% |
88% |
| Power Factor |
0.82 |
0.84 |
0.86 |
The following
is an example of how to calculate the annual energy,
demand and total cost of a standard efficiency motor.
|
Energy cost per year = |
|
(5 hp)(0.746 kW/hp)(4,000 hr/yr)($0.05/kWh)(0.75)
|
|
| 0.83 |
|
= $674 per year |
|
Demand cost per year = |
| (5 hp)(0.746 kW/hp)(0.75)($3.00/kVA/mo)(12 mo/yr) |
|
| (0.82)(0.83) |
|
= $148 per year |
The total annual cost to operate
the motor is $822.
After 10 years the motor’s operating
cost exceeds its purchase cost more than 20 times.
The table below summarizes the cost
for each motor. Payback periods for EPACT and premium
efficiency motors, compared to standard efficiency motors,
are based on purchase price and energy savings.
| |
Standard |
EPACT |
Premium |
| Purchase price |
$375 |
$445 |
$575 |
| Energy cost (yr) |
$674 |
$651 |
$636 |
| Demand cost (yr) |
$148 |
$140 |
$133 |
| Operating cost (yr) |
$822 |
$791 |
$769 |
| Payback years |
— |
2.3 |
3.8 |
Motor Repair and Replacement
Policy
Implement a motor replacement
policy to replace older, rewound motors. This can benefit
your operation in many ways, including:
- Increased efficiency, which lowers
operating costs
- Reduced downtime, which lowers
production costs
- Lower operating temperatures, which
lowers maintenance costs
Policy Example
The following policy was developed
by the Industrial Electrotechnology Laboratory. It covers
open drip-proof (ODP) and totally-enclosed fan-cooled
(TEFC) motor enclosures.
- When purchasing a new motor or
piece of equipment, specify energy efficient motors.
- When an existing motor fails:
- If it is energy efficient, send it for repair.
This applies to ODP and TEFC enclosures and one
to 200 hp motors.
- If it is not energy efficient and is an ODP
enclosure replace it with an energy efficient
model. This applies to one to 200 hp motors.
- If it is not energy efficient and is a TEFC
enclosure use the table below to select a
breakpoint horsepower—a motor size that
provides at least a two year payback—for
the
number of hours you operate motors.
|
Annual Operating
Hours |
Horsepower* |
| One shift (2,912 hr/yr) |
25 |
| Two shifts
(5,824 hr/yr) |
70 |
| Three shifts
(8,736 hr/yr) |
130 |
* For energy cost at $0.05 per kWh.
- When a motor is larger than the breakpoint
horsepower send it for repair. When a motor is
smaller than the breakpoint horsepower replace
it with an energy efficient motor.
- When repairing a motor will
cost more than 60 percent of the purchase cost of
a new energy-efficient motor, buy the new motor.
Conservation
Strategies
Energy efficient motors. When purchasing a new
motor choose the most energy efficient one you can afford.
Premium efficiency motors cost about 20 percent more,
but will payback in under four years with one-shift
operation and a cost of $0.05 per kWh. Payback will
be shorter for a 24-hour, seven-day-per-week operation.
Oversized motors. Motors are
oversized when they power end uses that require less
horsepower than the motor is capable of producing. For
example, when a 10 hp motor is used for an application
that calls for a five horsepower motor, the motor is
100 percent oversized, or operates at 50 percent full-load.
At smaller load factors motor efficiency is lower, leading
to increased operating costs. Select a lower power motor
and operate it at a higher load factor and near optimal
efficiency to help justify the motor replacement. Motors
operated at low load factors have lower power factors.
Motor replacement. Some motors
may warrant replacement before they fail. Evaluate motors
that are used for two or more shifts per day and are
oversized.
Synchronous belts. Optimize
transmission efficiency by using synchronous belts instead
of v-belts. V-belts can slip and deteriorate efficiency
at higher loads.
Variable speed drives. Consider
using a variable speed drive motor system instead of
traditional motors when loads vary significantly over
the course of daily use.
Voltage. The voltage at the
motor should be as close to the design limits, found
on the nameplate, as possible. Changes of more than
five percent can lead to two to four percent drops in
efficiency and increase temperatures, which decrease
the motor’s life. Voltage at the motor that is
not within the design limits leads to a decrease in
power factor. Low power factors may be penalized by
your power company.
For More Information
Additional resources are available online at the following
Web pages.
MnTAP has a variety of technical assistance services available to help Minnesota businesses implement industry-tailored solutions that maximize resource efficiency, prevent pollution, increase energy efficiency, and reduce costs.Our information
resources are available online. Or, call MnTAP at 612.624.1300
or 800.247.0015 from greater Minnesota
for personal assistance.
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