Electrical Motors
and
Variable Speed Drives
Electrical Motors
• Electrical motors are devices that convert
electrical energy into rotary mechanical
energy. Motors can be purchased to operate
on AC or DC power.
• Our interest is really only in AC motors, since
most facilities rarely use DC motors for their
chillers, fans, pumps and other large pieces
of equipment.
• Most AC motors are induction motors, which
are simpler, lighter and cheaper than the
alternative—an AC synchronous motor.
• There are many different sizes of motors,
rated by the amount of mechanical shaft
power they can provide.
• Induction motors range in size from 1/10
kW to over 500 kW.
• Almost all induction motors over 1 kW
use three phase power.
• There are several operating parameters
of induction motors that are of critical
importance to us in energy management.
Speeds of Induction Motors
• Synchronous Speed - The no-load
speed of an induction motor is called the
synchronous speed, since without any
load, the speed of the induction motor is
keyed to the power line frequency.
• Thus, the synchronous or no-load speeds
of induction motors are always multiples
of 50 in most of the world. In the US, line
frequency is 60 Hz, so motor speeds are
always multiples of 60.
Speeds of Induction Motors
(cont)
• Although an induction motor has a
synchronous speed, it is not a
synchronous motor. Synchronous motors
are physically very different from
induction motors.
• All induction motors have a rating which
is the no- load – or synchronous speed –
but this does not make them
synchronous motors.
• The no load or synchronous speed of an
induction motor is a function of its
mechanical construction, and how many
poles it was built with.
• A pole is a matching electromagnet on
the stator and the rotor of the motor.
• A 3000 rpm motor would have 2 poles,
and a 1500 rpm motor would have 4
poles.
• The equation for the speed of an induction
motor is related to the number of poles
and the frequency of the applied voltage in
Hz.
Speed = rpm = 60 x Hz / n
Where n is the number of pole pairs
• For a motor with 2 poles, or one pole pair,
the speed would be
Rpm = 60 x 50 / 1 = 3000 rpm
• For a motor with 4 poles, or two pole pairs,
the speed would be
Rpm = 60 x 50 / 2 = 1500 rpm
• Full Load Speed – The full load speed of
a motor is stamped on the nameplate, and
is 2 – 3% lower than the no load or
synchronous speed.
– For an 1500 rpm motor, the full load speed
might be 1470 rpm.
• Slip – Slip is the term for the difference
between the synchronous speed of a
motor and the actual speed of the motor at
any time.
– The full load slip is the largest value, and it is
the difference between the synchronous
speed and the full load speed.
– For a 1500 rpm motor, the full load slip is
around 30 to 50 rpm.
Efficiency, Power Factor and
Load Factor of Induction
Motors
• Three operating parameters of AC induction
motors that we need to understand and work
with are the motor’s efficiency, its power
factor and its load factor.
• Efficiency - Efficiency (EFF) of a motor is
defined as
Mechanical Power Output
EFF =
Electrical Power Input
• The more electrical power a motor can
convert to mechanical power, the higher its
efficiency.
• Larger motors usually have higher
efficiencies than smaller motors.
• Since electrical energy costs money, the
higher the efficiency of a motor, the lower
the energy operating cost of that motor.
•Many technological improvements have been
made to electric motors. From new materials
for inside windings to decreased friction from
well-designed ball bearings, the efficiency of
new motors increases almost every day.
Example
A motor delivers a shaft output power of 10
kW, and has an electrical power input of 12
kW. What is its efficiency?
Solution
10 kW
= 83 .3 %
Eff =
12 kW
• Power Factor – The power factor is an
electrical operating parameter of a motor,
and is found from the ratio of the real power
input in kW to the total power input in kVA.
• Larger motors usually have higher power
factors than smaller motors. Larger motors
will usually have power factors around 85%.
• Sometimes, electrical capacitors are installed
on motors to provide “power factor
correction.” This may be cost-effective for the
facility if their utility charges a substantial
penalty for low power factor.
Example 8.2-2
A three phase, 380 volt motor is drawing 80
amperes, and a real power of 40 kW. What is
the power factor of the motor?
Solution
P = 3 × V × I × PF watts
40,000 = 3 × 380 × 80 × PF
40,000
PF =
= 76 %
52,653
Do not confuse the power factor with the
motor load factor defined next.
• Motor Load Factor – The motor load
factor is a mechanical operating
parameter of a motor, and is found from
the ratio of the actual shaft power being
provided to the maximum shaft power
that could be provided by the motor.
– The maximum shaft power that can be
provided is the nameplate hp rating of the
motor.
– The actual shaft power being provided to the
load is determined by the load itself.
• A motor is what we call a “load driven
device.”
– This means that the motor only provides the
exact amount of power required by the load.
– If the load on a 20 kW motor is a fan requiring
only 10 kW to drive it, the load factor on the
motor is 10kW/20kW, or 50%.
• The input power to the motor will only be
what is needed to drive the actual load –
and most often it will not be the full rated
load power of the motor.
– Typical motor load factors on an annual basis
are in the range 40 – 60%.
Example
A 40 kW motor is connected to a 25 kW fan.
What is the load factor of the motor?
Solution
25 kW
LF =
= 62.5 %
40 kW
Do not confuse the motor load factor with
the power factor defined above.
Power Input to AC Induction Motors
• To understand how electrical energy is used
in a facility, we need to know how much
energy and power are used by the motors in
the facility.
• There are two basic equations we can use to
find this motor input power.
• One is the electrical equation which we have
already been using.
Electrical equation
We need to know all of the electrical
operating parameters – V, I, and PF.
P = 3 × V × I × PF W
P = 3 × kV × I × PF kW
We used this electrical equation earlier to find
the power factor of a motor.
Mechanical equation
• We need to know all of the mechanical
operating parameters of the motor – kW, LF
and the efficiency.
PIN =
NPKW
LF
Eff
kW
where NPKW is the nameplate kW rating of
the motor.
Example
A 50 kW motor with an efficiency of 89% is
operating at a load factor of 70%. What is the
input power in kW to the motor?
Solution
NPKW LF
P=
Eff
50 kW 0.70
P=
0.89
P = 39.33 kW
Example
Calculating Savings From Using a
High Efficiency Motor
Facility Information
• Facility has a 55 kW pump motor operating
at an annual load factor of 70%
• Existing motor is 87% efficient, and
operates 4000 hours each year
• Electric cost is $0.10 per kWh
• New 55 kW motor will be 94% efficient
• New high efficiency motor premium is
$1200
Economic Analysis of Motor EMO
• Savings is $1318.24
• Cost is $1200
• SPP = $1200/$1318.24 per year
= 0.9 years
Motor Nameplate Data
• Every motor has a nameplate which lists all of the
critical information about the operation of the motor.
• A typical motor nameplate is shown below.
• Parameters commonly found on motor
nameplates that we have not previously
defined include:
– FLA – Full load amps. The line current the
motor draws at full load.
– LRA – Locked rotor amps. This is a test value
for the motor, but it is useful for knowing the
maximum starting current surge into the motor.
(not shown on the example nameplate on the
previous page)
– Service Factor – A multiplier on the nameplate
horsepower of the motor. This tells how much
overload the motor can safely handle on a short
term basis.
Variable Speed Drives for
Motors
• AC induction motors are fixed speed
motors except for a small amount of slip
• Other speeds on the driven end have to be
engineered
• Because of the fan and pump laws, large
savings can be achieved using variable
speed drives for fans and pumps.
Centrifugal Fan and Pump
Laws
• Flow and Speed
LPS new RPMnew
=
LPS old
RPMold
• Pressure (Head) and Speed
Pnew [RPMnew ]2 [LPS new ]2
=
=
2
Pold
[RPMold ]
[LPS old ]2
• Power and Speed
kWnew [LPS new ]3
=
kWold
[LPS old ]3
Cube Law or Fan Law Example
A large office building presently has a 10
kW motor on a centrifugal fan in an air
handler unit. At full speed, the fan
supplies enough air flow to meet the
hottest day’s cooling needs. On a mild
day, half of that flow rate in L/s would
meet the cooling load. If a VFD is put on
that motor to run it at half speed, what is
the effect on the air flow rate and the
effect on the fan power needed?
Solution
• The first law says that if the fan motor runs
at half speed, it produces half the air flow
rate in L/s.
• The third law – the cube fan law says that
the fan power needed is (1/2) cubed, or 1/8
of the 10 kW. This is only 1.25 kW.
kWnew [LPSnew ]3
=
kWold
[LPSold ]3
kWnew [1]3 1
=
=
3
10
8
[2]
KWnew = 1.25 kW
Calculating annual savings for a
given reduction in motor speed
3
(
)
kWnew = kWold × RPMnew ÷ RPMold
kW saved Hours Days
$
Savings =
×
×
×
Efficiency
Day
Year kWh
Speed Reduction Savings
Example
A 10 kW motor is presently driving a
centrifugal fan at 1450 RPM. What would
the savings in kW and USD be if the fan
pulley was changed so that the new
speed was 1300 RPM? The motor
efficiency is 90%.
Solution
3
(
)
kWnew = kWold × RPMnew ÷ RPMold
kWnew
1300 RPM
= 10 kW ×
1450 RPM
= 7.2 kW
3
Example Continued
kW Saved = (10
7 .2 ) kW = 2 .8 kW
kW saved Hours Days
$
Savings =
×
×
×
Efficiency
Day
Year kWh
2.8 kW 2000 hours $0.10
=
×
×
0.90
years
kWh
= $622.22 / yr
Variable Volume Options for Air
Flow
• Outlet damper control (see next page, location
1)
• Inlet vane control (see next page, location 2)
• Magnetic clutching (see next page, location 3)
– Eddy current clutch
– Permanent magnetic clutch
• Variable Frequency Drives (see next page,
location 4)
• Hydraulic drives, variable sheaves, etc.
Electric Motors
Variable
Volume
Options
Sketch
1
2
3
4
Variable Volume Selection
• Outlet damper control
– Simple and effective
– Not efficient. Infrequently used
– Great candidate for conversion to other methods
of flow control
• Inlet vane control
– Simple and effective
– More efficient than outlet damper but significantly
less efficient than other options. Fairly frequently
used
– Great candidate for conversion to variable speed
drives on fan
Variable Volume Selection
• Variable Frequency Drive (VFD)
– Probably most efficient
– Competitive cost
– Harmonic concerns (input and output)
– Remote (clean area) installation
• Magnetic clutches (permanent magnet or eddy
current)
– Bulky and heavy on motor shaft
– No harmonics
– Close to same savings as VFDs
Typical Power Consumption of Various
Control Systems
Source: www.drives.com
Variable Volume Selection
• Choose the technology that your staff
understands and likes to use
– Ensure they are onboard
• You may not want to mix technologies in a
given facility
• Most efficient selection is a VFD followed
closely by magnetic clutching followed (way
back) by inlet and outlet vane controls
Variable Speed Drive Applications
• Any large centrifugal blower or pump that
runs a lot with a variable load!
– Put VSD on existing constant volume system
– Put VSD on variable volume system that
presently has inlet or outlet control
• Chilled water pumps in large facilities
• Cooling water pumps
• VAVs using inlet vane
• Forced draft (blower) cooling towers
Centrifugal Pump
Centrifugal Pump
Motor Auditing Checklists and
Forms
Motor and
Location
Rate
d kW
Eff
Ratin
g
Belts
#
Type
(C/V)
UF
Hrs
/da
y
LF
Age
(yrs
)
Rated
RPM /
RPM
VSD
(Y/N)
Remarks
Motor ECMs
ÄInstall efficient motor
Inefficient
ÄInstall variable speed drive
Throttle valves
System requirements ÄInstall controls
Choose replacement ÄUse MotorMaster
first
• Motor optimization
ÄUse latest engineering
tools
•
•
•
•
Guidance for Implementing an
Effective Motor Management
Program
• Establish preventive maintenance program
– Research Best Practices
– Temperature recordings/thermography scans
– Motor vents clean
– Redirect ventilation
– Greasing – correct amount
• Establish repair/replacement policy
– Have a lot of Input from Facility Staff in development
– Know what you’ll purchase if it fails – speeds facility
recovery