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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.