Electric Motors and Motor Management

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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



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)



Terminology

• NEMA – National Electrical Manufacturers Association

• EPACT – Energy Policy Act (Current standard is EPACT05)

• NLRPM – Synchronous speed

• FLRPM – running RPM at design load



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)



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.



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



$$$$$$

• 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.



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.



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



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



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 =?



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



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



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.



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



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 ______.



• 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.



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.


Date post:	12-Sep-2021
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Electric Motors and Motor Management

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