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

InputPower ElectricalOutputPower Mechanical

=EFF

• 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.

ExampleA motor delivers a shaft output power of 10 kW, and has an electrical power input of 12 kW. What is its efficiency?Solution

%3.83=kW12kW10

=Eff

• 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 8.2-2A 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

Do not confuse the power factor with the motor load factor defined next.

PF×80×380×3=000,40

wattsPF×I×V×3=P

%76=653,52000,40

=PF

• 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

Do not confuse the motor load factor with the power factor defined above.

%5.62=kW40kW25

=LF

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.

We used this electrical equation earlier to find the power factor of a motor.

WPF×I×V×3=P

kWPF×I×kV×3=P

Mechanical equation

• We need to know all of the mechanical operating parameters of the motor – kW, LF and the efficiency.

where NPKW is the nameplate kW rating of the motor.

kWEffLFNPKW

=INP

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

EffLFNPKW

=P

kW33.39=P89.070.0kW50

=P

ExampleCalculating 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

• Pressure (Head) and Speed

• Power and Speed

old

new

old

newRPMRPM

=LPSLPS

[ ][ ]

[ ][ ]2old

2new

2old

2new

old

new

LPS

LPS=

RPM

RPM=

PP

[ ][ ]3old

3new

old

new

LPS

LPS=

kWkW

Cube Law or Fan Law ExampleA 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.

[ ][ ][ ][ ]

kW25.1=newKW

81

=32

31=

10newkW

3oldLPS

3newLPS

=oldkWnewkW

Calculating annual savings for a given reduction in motor speed

( )3oldnewoldnew RPM÷RPM×kW=kW

kWh$

×YearDays

×Day

Hours×

EfficiencysavedkW

=Savings

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

( )3oldnewoldnew RPM÷RPM×kW=kW

kW2.7=RPM1450RPM1300

×kW10=kW3

new

Example Continued

kW8.2=kW)2.710(=SavedkW

yr/22.622$=kWh

10.0$×

yearshours2000

×90.0kW8.2

=

kWh$

×YearDays

×Day

Hours×

EfficiencysavedkW

=Savings

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 MotorsVariable 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 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

Variable Volume Selection

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

BeltsMotor and Location

Rated kW

EffRating # Type

(C/V)

UF Hrs/day

LF Age (yrs)

Rated RPM / RPM

VSD(Y/N)

Remarks

Motor ECMs

• Inefficient • Throttle valves• System requirements• Choose replacement

first• Motor optimization

Install efficient motorInstall variable speed driveInstall controlsUse MotorMaster

Use 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