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Motors--Losses, Loads and Operating Costs—SI Units - Transcript
Slide 1: Motors: Losses, Loads and Operating Costs—SI Units
Welcome to Motors: Losses, Loads and Operating Costs—SI Units
This course is the second course in the motors series
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This course was produced using content from the
US Department of Energy publication ―Improving Motor and Drive System Performance‖ and from the UK Carbon
Trust
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The screen controls allow you
to navigate through the eLearning experience
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Click on the attachments to download supplemental information for this course
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Slide 3: Objectives
At the completion of this course, you will be able to:
 Identify the ways that motors lose energy
 List the factors that influence energy efficiency of a motor
 Determine the loading of a motor, and you will be able to
 Calculate the operating costs of motor systems
Slide 4: Introduction
Electric motors, taken together, make up the single largest end use of electricity in many developed countries
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The cost of
running a motor can be as much as ten times to the purchasing cost of a motor
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Slide 5: What is a Motor?
An electric motor can be simply defined as a device for converting electrical energy into kinetic energy
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In our class on ―Efficient Motor
Control with Power Drive Systems‖, we introduce motors and look at them together with the gear and transmission
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AC motors are by far the most common in industry, and
we‘ll focus on them in this class
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In an induction motor the two
most important parts are the stator and the rotor
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Slide 7: Rotating Magnetic Field
The stator is equipped with three-phase windings, positioned at 120 degree intervals
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Motors—Losses, Loads and Operating Costs-SI Units
©2012 Schneider Electric
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All trademarks provided are the property of their respective owners
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The magnetic field in the rotor winding locks on to the rotating magnetic field in the stator windings and turns along
with it
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These are called synchronous motors because the rotor turns at the
same speed as the magnetic field
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The rotor current produces a second magnetic field, which tries to
oppose the stator magnetic field
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This is why these are called ―asynchronous‖
motors – the rotor is not synchronized with the rotating magnetic field
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Slide 10: Slip
Slip (S) is usually expressed as a percentage and the equation is shown here:

Where:
Ns = synchronous speed in RPM
Nb = base speed in RPM
(RPM = Revolutions Per Minute)
Slide 11: Calculating Synchronous Speed
Let‘s take some time and look at how to properly calculate synchronous speed
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This is called a ―Two Pole Motor‖ – since it has two poles per phase
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This would be a ―Four Pole Motor‖, since it has four poles
per phase and twelve poles altogether
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Motors—Losses, Loads and Operating Costs-SI Units
©2012 Schneider Electric
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All trademarks provided are the property of their respective owners
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So, be sure to check the definition carefully before following
the equation
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Slide 16: Effects of Slip
As we just saw, the rotor of a synchronous motor turns at the same speed as the magnetic field in the stator
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When the motor is unloaded, the slip will be very small
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Slide 17: Effects of Slip
However, as the motor load increases, slip becomes more significant – and that impacts energy consumption
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In this process, energy is lost in a
variety of ways
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Losses can vary from approximately 2 percent to 20 percent
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Slide 20: Power Losses
Power losses are the most significant
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These losses are caused when heat is generated from the current flowing in the windings and other electrical
fittings of the motor
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Slide 21: Power Losses
Higher efficiency motors include stators with more copper, to reduce electrical resistance and therefore minimize this
type of loss
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Higher efficiency motors have rotors with increased
Motors—Losses, Loads and Operating Costs-SI Units
©2012 Schneider Electric
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All trademarks provided are the property of their respective owners
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The fact that higher efficiency motors tend to have less slip, and therefore run faster, should be taken into account
when selecting the motor
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Slide 22: Magnetic Core Losses
Magnetic core losses are sometimes called ―iron losses‖
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Hysteresis is an effect where magnetic materials do not return to zero magnetization when the field is removed
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Eddy currents
are undesired electrical currents induced by the magnetic field
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Slide 23: Friction and Windage Losses
Friction losses occur in moving parts such as the motor bearings and cooling fan
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These are sometimes called ―mechanical losses‖
Slide 24: Stray Load Losses
Stray load losses can relate to a variety of effects, including harmonic energy generated when the motor is under
load, losses due to stray fluxes in the windings and conductor bars, as well as leakage in the laminate core of the
rotor
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Magnetic core losses as
well as friction and windage losses are present even when the motor is running under no load
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Loading of a motor is typically defined as the amount of work being done as compared to the maximum rated power
output at full load
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Motors—Losses, Loads and Operating Costs-SI Units
©2012 Schneider Electric
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All trademarks provided are the property of their respective owners
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But once the load
drops below 50% the efficiency decreases rapidly as shown in this figure
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High motor efficiencies and a power factor close to 1
are desirable for an efficient operation as well as for keeping costs down for the entire plant and not just the motor
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This method calculates the load as the ratio between the input power
(measured with a power analyser) and the rated power at 100 % loading
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The second is line current measurement
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This method is used when the power factor is not known
and only the amperage value is available
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For example, a 480 volt three-phase AC induction motor is rated at 35 kW or 47 HP power output
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6 amps at full load
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7 amps
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Motor loading = 39
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6 A = 75% load
The third is the slip method
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The accuracy of this method is limited but it can be used with the use
of a tachometer only—meaning no power analyser is needed
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There are three steps:
1
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Determine the input power at full load conditions (Pfull)
3
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Where:
Pinput = Three phase input power in kW, calculated from measurements taken at actual load conditions
V = RMS (root mean square) voltage in volts, mean line to line of 3 phases
I = RMS current in amps, mean of 3 phases, and
PF = the Power factor as a decimal
Slide 29: Determining Output Power
Here‘s how we would calculate the power used at full load
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horsepower by 0
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Then:
HP = The nameplate rated mechanical power output in horsepower
Slide 30: Calculating Load
From this the load of the motor is calculated like this:

Where:
Pinput = Input power in kW, measured at actual load conditions
Pfull = Input power in kW, used at full rated load
Logically, by combining these equations it is also true that load can be calculated as shown:

Where:
Pinput = Input power in kW, measured at actual load conditions
Prated = Nameplate rated mechanical power output in kW
η = Efficiency at full rated load
Or Prated can be replaced
Nameplate rated mechanical power output in HP multiplied by 0
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If the motor is fully loaded, then the LF term may be left off, since it is 100%
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Slide 31: Calculating Load From Input Power
(See this calculation in the addendum at the end of this document)
A 480V three-phase motor is drawing 36A
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What is the loading?
We start with the equation for input power:
Pinput kW = V * I * PF * √3 /1000
And we calculate as follows:

Which gives us the result:
25
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Then we use the equation for the input power at full rated load
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Finally we calculate the load
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Motors—Losses, Loads and Operating Costs-SI Units
©2012 Schneider Electric
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All trademarks provided are the property of their respective owners
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These loads are generally classified according to the relationship between speed,
torque and power
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Torque is
the rotational force, determined by the speed and the distance from the center of the shaft
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The classifications of loads are:
 Variable torque loads
 Constant torque loads, and
 Constant power loads
Slide 33: Variable Torque Loads
Variable torque loads include equipment such as fans and pumps
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For these loads, torque varies with speed squared and power varies with speed cubed
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If the speed was doubled--the motor will produce 4 times as much torque and eight times as much power
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Slide 34: Constant Torque Loads
A good example of a constant torque load is a conveyor
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power is directly proportional to speed
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Slide 35: Constant Power Load
Lastly, in a constant power load, torque varies with speed but power is constant
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Examples include machine tools and
centre winders
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Motors that are run below 50% of their rated load are dramatically less
efficient
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The decline will depend on
how the motor is utilised and how well the motor is maintained
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Capacity: As with most equipment, motor efficiency increases with the rated capacity
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Speed: Higher speed motors are usually more efficient
Type: For example, squirrel cage motors are normally more efficient than slip-ring motors
Temperature: Totally-enclosed fan-cooled (TEFC) motors are more efficient than screen-protected drip-proof (SPDP)
motors, and
Condition: Rewinding of motors can result in reduced efficiency
These factors should be remembered when multiple motors are being analysed
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to use an average motor efficiency for an analysis including motors that vary widely in their age, size, or speed
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Slide 39: Operating Costs Fully Loaded Motor-Metric
(After listening to the Introduction, click on each number below, to follow the problem step-by-step
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00 € / kW
 Energy - 0
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First, we calculate the annual operating hours, as shown
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Then we need to know how much power this motor draws, and how much energy it consumes
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The efficiency equation looks like this
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Now that we know the demand, we can calculate the energy used per year in kWh
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Motors—Losses, Loads and Operating Costs-SI Units
©2012 Schneider Electric
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All trademarks provided are the property of their respective owners
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All rights reserved
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Page 15

The total cost is the annual demand and energy charges added together, plus tax
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Notice that we included the demand charges in this calculation
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However, if there is other equipment in the factory that has a higher demand,
the equipment with the highest demand and the time of day that the demand occurs will determine the actual demand
charges
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How would we calculate the cost if the motor was partially loaded?
Let‘s try:
A 20 kW fan motor is operated at 75% load and has a power factor of 85%
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075 €/kWh?
We can apply this equation to calculate the demand drawn by the partially loaded motor, and we see the answer is
16
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Motors—Losses, Loads and Operating Costs-SI Units
©2012 Schneider Electric
...
All trademarks provided are the property of their respective owners
...


Motors—Losses, Loads and Operating Costs-SI Units
©2012 Schneider Electric
...
All trademarks provided are the property of their respective owners
...

Changing a motor for a smaller one will only reduce demand charges if the peak demand of the plant is impacted
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This becomes important in tiered rate structures where successive blocks of power
or energy are less expensive
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If a measure saves 10,000 kWh, the savings should be based upon the cost of the last 10,000 kWh
consumed, and not an average cost
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A blended rate may be
calculated to combine various charges such as connection fee, demand, energy and power factor into one average
cost of energy
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Slide 43: Summary
Let‘s summarize some of the information that we have reviewed in this course
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Motors
typically run on direct current or alternating current
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In
particular we focused on the AC induction motor
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In each winding, a magnetic field forms
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The two most important parts of an induction motor are the stator and the rotor
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In this process, energy is lost
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Page 18

These intrinsic losses can be reduced only by changes in motor design and operating conditions
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The factors that influence the energy efficiency of a motor include age, capacity, speed, type, temperature, and
condition
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These loads are generally classified according to the relationship between speed,
torque and power
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Remember that the efficiency of large motors is different
than small motors, so it is not safe to make general assumptions about efficiency when making calculations
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Slide 44: Thank You!
Thank you for participating in this course
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Load from Input Power

Motor details
Voltage
Power output
Actual current
Power factor
Efficiency ηrated

480 V
35 kW
36
...
0

*
1000

85%

*

1
...
4 kW

=

Prated
ηrated

kW

OR

HP

*
ηrated

0
...
5

OR

38
...
4
38
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All rights reserved
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746

Fully Loaded Cost Metric
Motor details
Motor rated mechanical power
Operating hours
Load
Rated efficiency

25
...
00

€
kW

0
...
0
94%
26
...
6

kW

233,016

Demand

=

26
...
80

Demand

kW

kWh
year

kW
month

*

Cost

€
kW

*

12

months
year

kW
month

*

9
...
045

€

year

kWh

€
year

=

Total cost

10,485
...
80

€
year

+

10,485
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52

€
year

*

1
...
03

€
year

© 2012 Schneider Electric
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All trademarks provided are the property of their respective owners

Sales tax rate

6%

Partially Loaded Cost Metric
Motor details
Motor rated mechanical power
Operating hours
Load
Rated efficiency

20
kW
6200 hours per year
75%
90%

Electrical tariff
Average energy cost

0
...
67

Load

kW

=

Powerinput

kW

*

75%

Operating hours

hours
year

*

Cost

€
kWh

=

=

16
...
00

kW

*

6200

hours
year

$
year

© 2012 Schneider Electric
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All trademarks provided are the property of their respective owners

*

0

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