electric motors and motor management - quia€¦ · electric motor management, why bother? •...
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Electric Motors and Motor Management
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Electric motor management, why bother?
• 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
• 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
• 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
• Design Slip = NLRPM – FLRPM
• True Slip = NLRPM – measured RPM
• % Load = True Slip / Design Slip
Example
• 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
• 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
• 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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$$$$$$
• Leave existing motors alone until they fail except: – 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
• 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: – 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:
Power Savings:
Energy Savings:
Brake HP (load on the shaft):
Efficiency
LFHPkW
746.
EEStd
e Efficiency
LFHP
Efficiency
LFHPkW
746.746.
.
TimegspowerSavinngsEnergySavi
LFHPBHP
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Manufacturer Specs.
Specifications: RM3003
SPEC. NUMBER: 34K15-895
CATALOG NUMBER: RM3003
FL AMPS: 1.3/.65
208V AMPS: 1.7
BEARING-DRIVE-END: 6203
BEARING-OPP-DRIVE-END: 6203
DESIGN CODE: B
DOE-CODE: --
FL EFFICIENCY: 64
ENCLOSURE: OPEN
FRAME: 48
HERTZ: 60
INSULATION-CLASS: B
KVA-CODE: L
SPEED [rpm]: 1725
OUTPUT [hp]: .25
PHASE: 3
POWER-FACTOR: 56
RATING: 40C AMB-CONT
SERIAL-NUMBER: --
SERVICE FACTOR: 1.35
VOLTAGE: 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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Drives
• 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
1212 /RPMRPMCFMCFM
21212 / RPMRPMSPSP
31212 / RPMRPMHPHP
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Variable Volume Options
0
20
40
60
80
100
120
140
20 40 60 80 100
Po
wer
inp
ut
rati
o (
%)
Load Fraction (%)
Chart Title
Constant Volume
Fan Law
VFD
Outled Damper
Variable Inlet Vane
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Fan Laws Example
• 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.
• New CFM = ?
• New HP requirement = ?
• These types of savings are why variable speed drives are so popular.
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Selection of best VAV option
• Outlet damper control • Simple and effective
• Not efficient, infrequently used except on pumping
• Great candidate for conversion to others
• Inlet vane control – Simple and effective
– More efficient than outlet damper, but significantly less than other options, fairly frequently used
– Great candidate for conversion to others
• VFD • 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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HVAC Performance measures • EER (Energy Efficiency Rating)
• EER = BTU of cooling output / Wh of electric input
• COP (Coefficient of Performance) • COP = EER / 3.412 Btu/Wh
• SEER 0-20 tons
• EER 21-99 tons
• COP 100+ tons
• kW/Ton = 12/EER = 3.517/COP
• 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 have 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 • 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
• A rooftop unit has an EER of 13.5. What is its kW/ton rating?
• 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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