Electric Motors and
Motor Management
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Electric motor management,
why bother?
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Electric motors use over ½ all
U.S. electricity
Motor driven systems use over
70% electric energy for many
plants
Motor driven systems cost about
$90 billion to operate per year
A heavily used motor can cost
10 times its first cost to run one
year
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Electric Motor Management,
why so difficult
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Load on most driven systems is
unknown, at least on retrofits
Very difficult to determine load
accurately through
measurements
Electric motor management is
FULL of surprises
Yet, savings can be large
(small percentage of a big
number is a big number)
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Terminology
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NEMA – National Electrical
Manufacturers Association
EPACT – Energy Policy Act
(Current standard is EPACT05)
NLRPM – Synchronous speed
FLRPM – running RPM at design
load
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Electric Motor Basics,
Slip
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Design Slip = NLRPM – FLRPM
True Slip = NLRPM – measured RPM
% Load = True Slip / Design Slip
Example
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FLRPM = 1760 (off name plate)
Design HP = 50 (off name plate)
Measured RPM = 1776
NLRPM = ?
Design Slip = ?
True Slip = ?
% Load = ? (Load factor)
True Load = ? (HP times % load)
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Voltage Imbalance
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Problems can occur because of voltage
imbalance between three phases. This can be a
serious problem in motors.
Percent voltage imbalance is found as the ratio
of the largest phase voltage difference from
average, divided by the average voltage.
For example, if we have 220, 215, and 210
volts, the voltage imbalance is 5/215 = .023 or
2.3% (greatest difference between voltages
from average voltage e.g.. 215 = average, 220
= +5 and 210 = -5 so 5V is the difference.
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Single Phasing
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Loss of one phase in a three phase system
Worst case of voltage imbalance
Causes:
In plant
Pole hits
Tree limbs
Animals
Lightning
In other words, this does happen
• Each 10° C rise in temperature reduces motor
life 50%
• Phase current increases exponentially
• NEC 430 is a good article for this
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Page 7
$$$$$$
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Leave existing motors alone until they fail except:
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Exceptionally oversized motors (25% loading or so)
Sizes that are needed elsewhere (requires inventory)
When they fail, maybe buy new energy efficient
motors (EPACT or Premium) instead of paying for
rewind.
Rewind motors over 20HP at a typical cost of
60% of a new motor.
If financial incentives are available, much more
may be done. Premium efficiency motors usually
need economic help.
Rewinds must follow NEMA specifications and
require periodic tests.
Failed non energy efficient ODP motor – scrap for
copper and buy new efficient motor.
Failed non energy efficient TEFC motor – scrap for
copper and buy efficient motor unless >75HP.
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Energy Efficient Motors
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More copper – less resistance losses or heat
because of larger wire
Better fans and bearings, more carefully
lubricated
Longer and heavier
Save energy and reduce demand
Reduce load on cables, transformers, etc.
Speed is slightly higher
Significant larger inrush current
Watch retrofits for possible issues:
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Faster speed = more volume (work)
Watch circuit breakers and LRA
Energy efficient motors (used 2000 hrs. or more
per year) are almost always cost effective for
new purchases, as well as alternative to rewinds
except as discussed earlier.
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Motor Calculations
Power Input:
HP  .746  LF
kW 
Efficiency
Power Savings:
 HP  .746  LF 
 HP  .746  LF 
kW e   Efficiency    Efficiency 
Std .
EE
Energy Savings:
EnergySavings  powerSavings  Time
Brake HP (load on the shaft):
BHP  HP  LF
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Page 10
Manufacturer Specs.
Specifications: RM3003
SPEC. NUMBER:
CATALOG NUMBER:
FL AMPS:
208V AMPS:
BEARING-DRIVE-END:
BEARING-OPP-DRIVE-END:
DESIGN CODE:
DOE-CODE:
FL EFFICIENCY:
ENCLOSURE:
FRAME:
HERTZ:
INSULATION-CLASS:
KVA-CODE:
SPEED [rpm]:
OUTPUT [hp]:
PHASE:
POWER-FACTOR:
RATING:
SERIAL-NUMBER:
SERVICE FACTOR:
VOLTAGE:
34K15-895
RM3003
1.3/.65
1.7
6203
6203
B
-64
OPEN
48
60
B
L
1725
.25
3
56
40C AMB-CONT
-1.35
230/460
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Motor Sample “B”
A recent advertisement said a premium efficiency
50HP motor is available at 94.5%. It would replace
a motor that presently runs at 90.7%. Given the
parameters below, calculate the cost of operating
both motors and the savings for conversion:
-Motor runs 8,760 hours/year
-Demand cost is $10 per kW month
-Energy cost is $0.06 per kWh
-Motor runs at 80% load at all time
Demand savings = ?
Energy Savings = ?
Energy cost = ?
Power savings =?
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Page 12
Drives
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Motors are fixed speed devices likely running
between NLRPM and FLRPM
Other speeds on the driven end have to be
engineered (which will affect the load on the
motor)
Because of the “Fan” laws (pumping or blowing)
centrifugal devices are desired applications for
varying CFM or GPM
CFM 2  CFM 1 RPM 2 /RPM 1 
SP2  SP1 RPM 2 / RPM 1 
2
HP2  HP1 RPM 2 / RPM 1 
3
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Variable Volume Options
Chart Title
140
120
Power input ratio (%)
100
80
Constant Volume
Fan Law
VFD
60
Outled Damper
Variable Inlet Vane
40
20
0
20
40
60
Load Fraction (%)
80
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Fan Laws Example
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A 40 HP centrifugal blower is on a forced draft
cooling tower. It is basin temperature controlled
but conversion to a variable speed drive is being
considered. When the blower is running at ½
speed, what is the HP requirement.
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New CFM = ?
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New HP requirement = ?
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These types of savings are why variable speed
drives are so popular.
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Selection of best VAV option
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Outlet damper control
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Inlet vane control
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Simple and effective
Not efficient, infrequently used except on pumping
Great candidate for conversion to others
Simple and effective
More efficient than outlet damper, but significantly less
than other options, fairly frequently used
Great candidate for conversion to others
VFD
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Probably most efficient
Competitive cost
Remote (clean area) installation
Multiple motors may be connected to one drive providing
higher savings, but sizing is critical
Motors and loads must be agreeable to VFD’s
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Page 16
HVAC Performance measures
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EER (Energy Efficiency Rating)
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EER = BTU of cooling output / Wh of electric input
COP (Coefficient of Performance)
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COP = EER / 3.412 Btu/Wh
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SEER 0-20 tons
EER 21-99 tons
COP 100+ tons
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kW/Ton = 12/EER = 3.517/COP
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1) 10 ton rooftop A/C with an EER of 12 will
have a COP of ______.
2) A 10 ton rooftop A/C with an EER of 12 will
have a kW at full load of ______.
3) A 10 ton rooftop A/C with an EER of 8.5 will
have a COP of ______.
4) A 10 ton rooftop A/C with an EER of 8.5 will
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haveFree
a kW
at full load
of ______.
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• One ton of A/C = 12,000Btu/h
• A ton is a measure of A/C power, and is used
when sizing systems, or when determining
electrical demand.
• One ton-hour of A/C = 12,000Btu
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A ton-hour is a measure of A/C energy, and is
used when sizing storage tanks for thermal
energy storage (TES) systems, or when
determining electrical energy consumption.
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Examples
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A rooftop unit has an EER of 13.5. What is its
kW/ton rating?
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How many kWh is used to provide 120 million
ton-hours of air conditioning with a system
having a COP of 3.0? Use kWh/ton-h =
3.517/COP.
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