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LiftandEscalatorMotorSizingwithCalculationsandExamples

ARTICLE·FEBRUARY1999

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1AUTHOR:

LutfiAl-Sharif

UniversityofJordan

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Retrievedon:01January2016

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LIFT AND ESCALATOR MOTOR SIZING WITH

CALCULATIONS & EXAMPLES

Dr. Lutfi Al-Sharif Published in Lift Report, January/February 1999.

ABSTRACT This paper outlines the method of calculating the necessary size of motor needed to drive a lift or an escalator. To calculate the size of a motor two methods can be used: The steady state method and the dynamic method. The steady state method ensures that the motor can move the out of balance masses at the required steady state speed. The dynamic method ensures that the motor can accelerate (and possibly decelerate) the masses up to the required rated speed at the necessary rate (i.e., required acceleration). In lifts, both methods are usually used, due to the fact that one of the important performance criteria for a lift is that it is able to accelerate and decelerate in the required time. The steady state method is used to check the necessary power rating of the motor. As an additional check the dynamic method is used to ensure that the torque rating of the motor is sufficient to accelerate the motor in the necessary time. In escalators on the other hand, and due to the fact that escalators do not start and stop as frequently as lifts do, the only method usually used is the steady state method. The method of calculating the motor size needed for both lifts and escalators are highlighted. In addition, an outline of the energy efficient motor designs is given, showing the level of efficiency to be expected from modern designs. The reasons for derating motors driven by a solid state drive are given, and two methods are outlined.

1. INTRODUCTION When moving any load, two stages are encountered: Initial transient state and steady state. In the initial transient state, the speed of the load has to be increased up to the required speed. During the steady state, power has to be supplied to overcome the out of balance loads and keep masses moving at the rated speed. These two stages are shown diagrammatically in Figure 1.

2

Steady state

Transient stateSpeed

Time

Figure 1: Two states in moving an out of balance load.

During both of these stages, power needs to be supplied to overcome any friction and other related losses (e.g., air resistance). These two stages, can be related to the type of energy which needs to be supplied to the moving mass:

• Kinetic energy: During the initial acceleration phase, the motor needs to supply enough kinetic energy to the masses to accelerate them up to the required speed. In the case of lifts, these moving masses include the passenger load, the cabin mass, the counterweight, the ropes, the motor's rotor inertia, the handwheel (or the flywheel if one is fitted). In the case of an escalator or passenger conveyor, these masses would include in addition to the mass of the travelling passengers, the mass of the steps and the chain linking them.

• Potential energy: The motor has to be able to lift the out of balance load at the rated speed. In theory this energy is stored in these masses, and if the same masses come down again, then the energy should be returned to the supply.

• Frictional & other losses: In supplying the two elements of energy mentioned above, the motor needs to overcome the friction in the system (and the air resistance at high speeds) and the losses in the gearbox.

Based on the understanding of the three elements above, the motor selection criteria can be developed.

2. THE GENERAL APPROACH In developing a general approach to lift, escalator and passenger conveyor motor sizing, we can think of all three as special cases with different angles of inclination. the added difference is the counterweighting in lifts, which reduces the effective passenger load. Passenger conveyors can be dealt with as special cases of escalators. This difference in the angle of inclination is shown graphically in Figure 2. Lifts are shown to have an angle of inclination to the horizontal of 90°, although inclined lifts could have an angle as low as 15°. Escalators are shown to have either 30° or 35°, and passenger conveyors will have much smaller angles of inclination.

3

Lifts

30 deg.

35 deg.90 deg.

0 deg.

Escalators

Passengerconveyors

Figure 2: The general approach to motor selection using the angle of inclination.

3. LIFTS When selecting a motor, it is important to note that the rating of the motor (i.e., that on the nameplate) refers to net output power from the motor. This is equal to the product of the output net torque (i.e., the output mechanical torque less the friction and windage losses) and the output rotational speed in radians per second. This output mechanical power will obviously be less than the input electrical power to the motor, due to the following losses (taking an AC motor as an example):

• Stator copper losses. • Core losses (Eddy current and hysterisis losses). • Rotor copper losses. • Friction and windage losses.

Thus the efficiency of the motor is the division of the output mechanical power by the input electrical power. By using the input rated current, voltage and power factor, and the output speed and torque, the efficiency of any motor can be calculated. Next, we examine the basic method of motor selection.

3.1 Steady state method (power only) The first and most commonly used method is to check that the motor supplies sufficient power to move the out of balance mass at the rated speed.

3.1.1 The formula The basic method can be used in most cases to select the size of the motor. The worst case is taken as that of lifting a fully loaded car in the up direction at the rated speed

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(assuming that the counterweight ratio is 50% or less). This calculated in the following stages:

• The fully loaded passenger mass is reduced by the amount of counterweight. This represents the out of balance mass.

• This is multiplied by the acceleration of gravity, to determine the force in Newtons needed to move this out of balance mass against gravity in the up direction.

• Multiplying this force by the rated speed gives the rate of flow of energy, or in other words the power in watts. This represents the net output power of the system.

• This calculated net output power has to flow through the system, and thus a larger value of power has to be supplied to account for all losses in the shaft and gearbox. Thus, the net output power is divided by the shaft efficiency and the gearbox efficiency to provide the required motor power output rating.

• The efficiency of the shaft and the gearbox (forward) can be taken as a combined figure of 85%, if no exact data is available. The forward efficiency of the gearbox is taken in this case because the motor is driving. The reverse efficiency is only taken if the braking calculation is being carried out. The forward and reverse efficiencies for worm gearboxes are different.

The formula for sizing the motor for a lift is as follows:

( )M

P s CF=

× × × × −75 9 81 1.η

Where: P is the rated passenger number in the car; 75 stands for 75 kg/passengers; 9.81 is the acceleration due to gravity; s is the rated top speed; CF is the counterweight factor (a factor less than 1); η is the total efficiency of the installation (taken around 85%). For a hydraulic lift, the same formula can be used by replacing CF by -1.

The counterweight ratio is important, and accounts for the fact that if the counterweight ratio is less than 50%, then the worst case scenario would be for a full car moving upwards. For example, if the counterweight ratio is only 40%, then when the car is full, only 40% weight of the passengers is compensated for by the weight of the counterweight, and the motor has to provide enough torque to lift the other 60%. Using a counterweight ratio of 40% is quite common. This is in recognition of the fact that the car rarely fills up to more than 80% of its rated load.

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3.1.2 Examples Three examples follow, to illustrate the above concepts. EXAMPLE 1 In a lift system which has an MG set supplying its DC hoist, calculate the size of the AC prime mover for a 49 passenger lift, running at 1.6 m/s, if the efficiency of the installation (including the MG set, the DC hoist motor and the shaft efficiency) is 70%, and the counterweight ratio is 40%. SOLUTION Applying the formula above gives:

( )MP s CF

=× × × × −

=× × × × −

=75 9 81 1 49 75 9 81 1 6 1 0 4

0749 44

. . . ( . ).

.η

kW

So, 50 kW (or 55 kW) motor can be selected.■ EXAMPLE 2 An 8 passenger hydraulic lift, runs at a speed of 1 m/s, and has an overall efficiency of 80%. If the mass of the car is equal to the rated load in the car, then calculate the required minimum size of the motor for the pump unit. SOLUTION As the mass of the car and associated equipment is equal to the rated load, then CF can be taken as -1. Thus, applying the formula gives a motor size of:

( ) ( )MP s CF

=× × × × −

=× × × × − −

=75 9 81 1 8 75 9 81 1 0 1 1

0 8147

. . . ( ).

.η

kW

Thus a motor sized at 16 kW could be used.■ EXAMPLE 3 A two speed lift has a rated speed of 1.2 m/s, and a car load of 13 passengers. If the overall system efficiency is 75%, and the counterweight ratio of 50% is used, calculate the size of the motor. SOLUTION Applying the above formula, gives:

( )MP s CF

=× × × × −

=× × × × −

=75 9 81 1 13 75 9 81 1 2 1 0 5

0757 65

. . . ( . ).

.η

kW

Thus, an 8.4 kW motor can be selected.■

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EXAMPLE 4 A VVAC driven lift has a rated speed of 1.0 m/s, and a car load of 78 passengers. If the overall system efficiency is 68%, and the counterweight ratio of 45%, calculate the size of the motor. SOLUTION Applying the above formula, gives:

( )M

P s CF=

× × × × −=

× × × × −=

75 9 81 1 78 75 9 81 10 1 0 450 68

46 4. . . ( . )

..

η kW

Thus, a 50 kW motor can be selected.■

3.1.3 The efficiency of the installation The efficiency of the system can also be determined based on the speed and whether it is geared or not. Figure 3 shows how the system efficiency can vary against speed for a geared system.

Relationship between lift system efficiency and speed (geared)

50%

55%

60%

65%

70%

75%

80%

0 0.5 1 1.5 2 2.5 3 3.5 4

Speed (m/s)

Effic

ienc

y

Figure 3: System efficiency against speed.

3.1.4 The duty cycle and the starts per hour The above examples have neglected the duty cycle and the number of starts per hour which the lift system is expected to run at. Manufacturers supply different tables depending on the number of starts per hour (e.g., Loher, Catalogue LN 15 e). Standard number of starts are usually 90, 120, 180, 240 starts per hour. After selecting the required power of the motor, a motor is picked from the relevant table.

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3.2 Dynamic method (Moment of inertia) The basic power method, only ensures that the motor can lift the out of balance masses at the rated speed. It thus only addresses the steady state situation (i.e., after all masses have started moving at the rated speed). However, the motor has to accelerate these masses up to the rated speed, and it has to do so at an acceptable value of acceleration. In this section we examine the moment of inertia method, as a checker for the basic method. It ensures that the motor is capable of supplying the required kinetic energy for translational and rotational masses at the required acceleration.

3.2.1 Translational masses For all translational masses, the motor has to supply them with the necessary kinetic energy to allow them to travel at the rated speed. The kinetic energy for a mass m moving at a velocity of v, is:

E m vkin = × ×12

2

mv

Figure 4: Translational mass m travelling at a speed v.

In addition, the motor when lifting out of balance masses, has to provide the potential energy. The potential energy is expressed as:

E m dpot = × ×9 81. where: m is the mass of the object in kg; 9.81 m/s2 is the acceleration due to gravity; d is the distance the object is lifted. The translational masses in a lift system to be taken into consideration are:

• Car mass. • Passenger mass. • Counterweight mass. • Rope mass.

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• Trailing cable mass (usually negligible in practice)

3.2.2 Rotational masses A mass m rotating around a certain point at a distance of r, and at a speed v, has a kinetic energy equal to:

( ) ( )E m v m r m rkin = × = × × × = × × ×12

12

12

2 2 2 2ω ω

Where: m is the mass of the object in kg; r is the radius of rotation in m; ω is the rotational speed in rad/s; By comparing this form to the translational form, it can be noticed that the linear speed corresponds to rotational speed in rad/s, the translational mass m corresponds to a rotational “mass” of mr2, which is called the moment of inertia. Figure 5 shows such a mass rotating at an angular speed, ω.

m

r

ω

Figure 5: Mass m rotating at an angular speed ω and at a radius r from the

centre point.

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The moment of inertia of the following items has to be included in the calculation: The rotor moment of inertia, the gearbox, the coupling, the brake drum, the handwheel or flywheel. The moment to of inertia of all masses (translational and rotational) has to be referred to the motor shaft.

3.2.3 Derivation of the formulae The formulae used to check that the motor is capable of accelerating all the masses are derived in this sub-section. They refer to Figure 6.

d

Motor

Flywheel

Gearbox

rω

v

CQ

C/W

Figure 6: Diagram of a traction lift identifying the masses and the relevant

parameters.

The moment of inertia of the translational masses referred to the motor shaft, can be expressed as follows:

( )J m dr

Q C C W drshaft Tot= ×

× = + + ×

×2

12

12 2

η η/

Where, Q is the rated load of the car; C is the mass of the car; C/W is the mass of the counterweight; η is the efficiency of the installation.

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Alternatively, if the diameter of the sheave is not fixed yet, or the gear ratio is not fixed, it might be easier to use the motor rotational speed and the linear lift speed, as follows:

( )Ratiov

nvn

vn

=×

= ×

=

602

602

9 55π π

.

Where, n is the motor speed in rev/min; v is the linear lift speed in m/s. Thus, the alternative form for the formula above becomes:

( )J m vn

Q C C W vntrans Tot= × ×

× = + + × ×

×9 55 1 9 55 12 2

. / .η η

The moment of inertia for rotational masses will include the motor moment of inertia, the brake drum and coupling moment of inertia, the gearbox moment of inertia, and the handwheel (or flywheel) moment of inertia.

J J J J Jrot mot gear coupling flywheel= + + + Where: Jmot is the moment of inertia of the motor; Jgear is the moment of inertia of the gearbox; Jcoupling is the moment of inertia of the coupling and the brake drum; Jflywheel is the moment of inertia of the flywheel or handwheel. The total moment of inertia referred to the motor shaft is the sum of the moment of inertia of translational masses, and the moment of inertia of rotational masses:

J J JTot trans rot= + Then the out of balance torque is calculated. The out of balance mass depends on the counterweight ratio and the rated passenger load in the car.

( )m P CFo b/ = × × −75 1

Where: mo/b is the out of balance masses; P is the rated number of passengers in the car; 75 is the mass per passengers in kg; CF is the counterweight ratio. From the out of balance m, the out of balance torque can be referred to the motor:

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T m dr

m vno b o b o b/ / /. . .= × ×

× = × × ×

×9 81

21 9 81 9 55 1η η

Where: To/b is the out of balance torque; mo/b is the out of balance mass; 9.81 m/s2 is the acceleration due to gravity; v is the rated speed of the lift in m/s; n is the motor rotational speed in rpm; d is the diameter of the sheave in m; r is the reduction ratio of the gearbox; η is the efficiency of the installation. The value of the rotational acceleration can then be calculated by dividing the net torque by the total moment of inertia referred to the motor shaft:

α =−T T

Jrad so b

Tot

max / / 2

Where: α is the rotational acceleration in rad/s2; JTot is the total moment of inertia referred to the motor shaft in kgm2; To/b is the out of balance torque in Nm; Tmax is the maximum permissible torque from the motor in Nm. The maximum permissible torque is usually taken as 2 to 2.4 times the rated torque of the motor. This leads to:

T Tratedmax = ×2

This rotational acceleration can be converted to linear acceleration by using the conversion ratio(s):

a dr

vn

m s= ×

= × ×

α α2

9 55 2. /

A value of acceleration of 0.8-1.0 m/s2 is acceptable. If the value is more, this is still acceptable if the drive is a variable speed drive (i.e., ACVV1, VVVF2, DC SCR3

If the acceleration is less than 0.6 m/s2, then the motor is not adequate, and a larger size motor with a higher torque needs to be selected, or the masses have to be reduced.

), because the drive the exact required voltage to achieve the required acceleration. If the drive is a two speed drive, then a flywheel might be needed to reduce the value of acceleration.

1 ACVV: Variable voltage AC drive. 2 VVVF: Variable voltage variable frequency drive. 3 DC SCR: Solid state DC drive, employing thyristors.

12

To summarise the above derivation, all the formulae have been combined in one formula, shown here:

( )

( ) ( )a

T P CF vn

J J J J Q C C W vn

vn

m srated

mot gear coupling flywheel

=× − × × − × × ×

×

+ + + + + + × ×

×

× ×

2 75 1 9 81 9 55 1

9 55 19 55

22

. .

/ .. /

η

η

EXAMPLE 5 A lift system is designed to run at 1.75 m/s, with a car capacity of 28 passengers. The car mass is 1000 kg, and the counterweight ratio is 50%. Select a motor which will run at a speed of 920 rpm from the table below.

Motor Torque (Nm) Power (kW) Motor inertia (kgm2) A 138 13.5 0.55 B 167 16 0.62 C 200 19.5 0.77 D 248 24 1.1 E 286 27.5 1.3 F 325 31 1.52

SOLUTION Applying the power method, gives a required power of:

( )MP s CF

=× × × × −

=× × × × −

=75 9 81 1 28 75 9 81 175 1 0 5

07524

. . . ( . ).η

kW

The nearest motor is motor D. Next we need to check that this motor can provide the necessary acceleration.

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( )

( ) ( )

( )

( ) ( )

aT P CF v

n

J J J J Q C C W vn

vn

m s

a

rated

mot gear coupling flywheel

=× − × × − × × ×

×

+ + + + + + × ×

×

× ×

=× − × × − × × ×

×

+ + + + ×

2 75 1 9 81 9 55 1

9 55 19 55

2 248 28 75 1 0 5 9 81 9 55 175920

1075

11 0 15 2100 2000 3050 9 55

22

. .

/ .. /

. . . ..

. . .

η

η

×

×

× ×

=−

× =

175920

1075

9 55 175920

496 249 54 396

0 0182 1 02

2

2

..

. .

..

. . /a m s

Thus, the motor will also be capable of accelerating the system at the required acceleration.■ EXAMPLE 6 Taking the installation in EXAMPLE 4 again, find out the maximum value of linear acceleration it is capable of assuming the following additional parameters: Trated= 729 Nm Tmax = 2.34 Trated Jmot = 2.1 kg m2 Jcoupling = 0.25 kg m2 n = 610 rpm C = 8700 kg C/W = 11365 kg Applying these parameters in the formula for calculating the linear acceleration gives:

( )

( ) ( )

( )

( ) ( )

aT P CF v

n

J J J J Q C C W vn

vn

m s

a

rated

mot gear coupling flywheel

=× − × × − × × ×

×

+ + + + + + × ×

×

× ×

=× − × × − × × ×

×

+ + + +

2 34 75 1 9 81 9 55 1

9 55 19 55

2 34 729 78 75 1 0 45 9 81 9 55 10610

10 68

2 1 0 25 5900 8700 11365

22

. . .

/ .. /

. . . . ..

. .

η

η

× ×

×

× ×

=−

× =

9 55 10610

10 68

9 55 10610

1708 729117

0 0156 131

2

2

. ..

. .

.. . /a m s

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Thus, the motor will also be capable of accelerating the system at the required acceleration. In practice, the system will be set up to accelerate at a rate of 0.8 - 1.0 m/s2.■

4. ESCALATORS The sizing of the escalator motor depends on a number of factors

• The vertical rise of the escalator. • The efficiency of the escalator. • The efficiency of the gearbox. • The running speed of the escalator. • The angle of inclination of the escalator. • The number of passengers per step. • The rise of each step.

4.1 Developing the formula When sizing an escalator motor, it is only necessary to use the static method (i.e., to ensure that the motor can move all the masses at rated speed). The time it takes to accelerate up to full speed is not as critical as that in the case of a lift, because escalators only start a few times a day (or in some cases only once a day).

Following is the derivation of the formula for deciding the kW size of the motor. The weight of each passengers will be (assuming a 75 kg per passenger mass):

F m g= × Where:

m is the mass of one passenger in kg. g is the acceleration due to gravity.

The effective number of steps on an escalator is equal to the vertical rise divided by the step rise:

Effective RR

E

s

number of steps = Vertical rise of escalatorRise of individual step

=

Where: RE is the escalator vertical rise. RS is the step vertical rise. Assuming that the number of passengers per step is n, then the total weight of all passengers on the escalator is:

15

F m g nRRTot

E

s

= × × ×

This force is then resolved into two components, one parallel to the direction of travel, and the other perpendicular to the direction of travel, as shown in Figure 7.

Passengers' weight

θ

Fθ

Tot

F cos( )θTot

F sin( )θTot

Figure 7: The resolution of the weight of the passengers.

Thus, the force in the direction of travel (denoted as Feff ) resulting from the weight of all passengers is:

F m g nRReff

E

s

= × × ×

× sin( )θ

Where: θ is the angle of inclination of the escalator. By multiplying this force by the linear speed of the escalator, the required power is found. However, the efficiency of the stepband and of the gearbox should be taken into consideration, as well as the power needed to keep the handrails moving. The final formula thus becomes:

Pm g n

RR

s PE

sH

S G

=

× × ×

× × +

× ×

sin( )θ

η η 1000

Where:

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P is the output power required from the motor in kW. m is the average mass per passenger in kg (usually 75 kg). g is the acceleration due to gravity (9.81 m/s2). n is the number of passengers per step (1, 1.5 or 2). RE is the vertical rise of the escalator in metres. RS is the step rise in metres (usually 0.2 m). θ is the angle of inclination of the escalator. s is the linear speed of the escalator in metres per second (0.5, 0.65 or 0.75 m/s). PH is the power in Watts needed to keep the handrails moving. ηS is the efficiency of the stepband. ηG is the efficiency of the gearbox.

Note that this is the output power provided by the motor, which is the rating of the motor. The actual electrical power into the motor will be higher, where the difference accounts to the losses inside the motor itself. EXAMPLE 7 An escalator has a rise of 20 m, is intended to run at a linear speed of 0.75 m/s, and will carry two passengers per step (i.e., n=2). Calculate the power rating of the necessary motor, assuming that the total efficiency of the gearbox and the stepband is 83%. Assume an angle of inclination of 30° to the horizontal. SOLUTION Let us assume that the power needed for the handrails is 4kw and that the step rise is 0.2 m. Applying the formula:

Pm g n

RR

s P

kW

E

sH

S G

=

× × ×

× × +

× ×

× × ×

× ° × +

×=

sin( )

..

sin( ) .

..

θ

η η 1000

75 9 81 2 200 2

30 0 75 4000

0 83 100071 3P =

The nearest size in this case will be 75kW.■

4.2 Non-standard running speed The last example has assumed that the motor is running near its synchronous speed. This assumption is important because the motor will only deliver its rated power, when running at the rated speed. Thus a motor with a 75 kW nameplate and 720 rpm, will deliver 75 kW only when running at 720 rpm. It will deliver 37.5 kW when running at 360 rpm...and so on. If a motor is to be run at less than its rated speed, the power its delivers is reduced by the ratio of the two speeds. Thus, the final formula becomes:

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Pm g n

RR

s P

nn

E

sH

S Gact

rat

=

× × ×

× × +

× × ×

sin( )θ

η η 1000

Where:

nact is the actual running speed in rpm. nrat is the rated running speed of the motor in rpm.

A motor may be run at lower than its rated speed by the use of an inverter. The reason why a motor might be run at a speed lower than its rated speed is to suit the ratio of the gearbox or the driving sprocket. The following example illustrates the above. EXAMPLE 8 A motor is to be used to drive an escalator with a rise of 12 m, running at 0.75 m/s, with an angle of inclination of 30° to the horizontal. The motor has to run at 630 rpm to achieve the required linear speed of the escalator. Select the size of the motor. SOLUTION The nearest synchronous speed to 630 rpm is 750 rpm, achieved by selecting a motor with 8 poles (4 pairs of poles). This motor is rated to run at 720 rpm, as this is the speed at which it will deliver its rated power. Again, assuming a step rise of 0.2 m and a 4 kW of power needed for the handrails, and applying the formula:

Pm g n

RR

s P

nn

P

E

sH

S Gact

rat

=

× × ×

× × +

× × ×

=× × ×

× ° × +

× × ×

=

sin( )

. ..

sin( ) .

. ..

θ

η η 1000

75 9 81 1 5 120 2

30 0 75 4000

0 95 0 87 1000 630720

39 86 kW

The nearest size standard motor would be a 40 kW motor.■

5. DERATING MOTORS IN ACVV AND VVVF APPLICATIONS When a motor is driven from a voltage phase controller (i.e., an ACVV drive), or an inverter (VVVF), the waveform fed to the motor contains harmonics. These cause extra heating and loss of torque, based on the following argument. The first main reason is the harmonic content of the waveform. The 3rd, 6th, 9th... harmonics do not induce any torque in the motor and just lead to heating. This

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is because the third harmonics of all three phases are in phase, and no phase shift exists between them. The 5th, 11th, 17th...etc., cause generate what is called a negative phase sequence. The usual positive phase sequence is when the sequence of the phases is RST: Phase R leads phase S by 120 degrees, phase S leads phase T by 120 degrees. A negative phase sequence is said to exist if the phase sequence is RTS: Phase R leads phase T, and phase T leads phase S. Notice that a positive phase sequence can be written as RSTRSTRST.., so that any of the following phase sequences are positive: RST, STR, TRS and equivalent. In the same way a negative phase sequence can written as: RTSRTSRTS.., so that any of the following phase sequences are negative and equivalent: RTS, TSR, SRT. By plotting the 5th harmonics, it can be seen that a negative phase sequence results (see Figure 8). This induces a negative (i.e., a torque that opposes the direction of rotation). The effect of this opposing torque would be to reduce the effective torque produced by the motor. Thus, the motor will need to draw more current to compensate for this reduction. This will result in further heating.

Negative phase sequence for 5th harmonic

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 30 60 90 120

150

180

210

240

270

300

330

360

Angle (degrees)

R1 S1 T1

T5R5 S5 S5 T5

Negative Phase sequence

Figure 8: Negative phase sequence generated by 5th harmonic.

The 7th, 13th, 19th....harmonics generate a positive phase sequence. This effectively boosts the torque of the motor, and thus does not have any detrimental effect on the performance. This is shown in Figure 9.

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Negative phase sequence for 7th harmonic

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

10 30 60 90 120

150

180

210

240

270

300

330

360

Angle (degrees)

R1 S1 T1

T7R7 S7 S7 T7

Positive Phase sequence

Figure 9: Positive phase sequence generated by 7th harmonic.

Moreover, the high frequency content of the waveforms, causes extra Eddy current and hysterisis losses in the core. The second main reason for heating when feeding a motor from an ACVV or VVVF drive occurs if the motor is cooled by a fan mounted on the rotor shaft. When the motor runs at speed lower than its rated speed, the fan does not circulate the air as efficiently as it should do. This problem does not occur when the motor is forced cooled by a dedicated motorised fan. For these two reasons, motors needs to be derated when used with these drives to allow for the extra heating effects. This is usually done by selecting one frame size up, or reducing the value of the torque from the motor used in the calculation by 5%, based on an assumption that the above effects will reduce the rated torque by that amount. Another method for derating a motor is to use a derating curve which depends on the speed of operation (Brook Crompton, 1996).

6. ENERGY EFFICIENT MOTORS With modern designs of induction motor, higher levels of efficiency are possible, which lead to lower losses in the motor, and thus less heat is generated and smaller frame sizes are possible. The main sources of inefficiency in an induction motor are as follows (Purdue University, 1998):

• Stator copper losses: These are proportional to the square of the current and are around 33% of the total losses in the motor.

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• Rotor copper losses (these are referred to as copper losses as an expression, although the rotor bars would probably be Aluminium): These are also proportional to the square of the rotor current, and are around 15% of the total losses.

• Core or magnetisation losses: These are a function of the stator voltage and the frequency, and are around 16% of the total losses.

• Rotational losses: These are caused by windage and friction losses, and are a function of speed. They amount to 14% of the total losses.

• Stray losses: The other 22% are losses which are not accounted for. In order to reduce these losses, energy efficient motors are designed with the following features:

• In order to reduce the core losses, thinner laminations are used which reduce the losses caused by Eddy currents. A more precise air gap is also used. The smaller the air gap, the lower the magnetisation current needed and hence the lower the losses.

• In order to reduce the copper losses, larger cross sectional areas of copper conductors are used in the stator. Also larger bars in the rotor are used, or even replaced with copper bars.

• Rotational losses are reduced by selecting better bearings and improving the fan design.

The purchase cost of these motor is usually 20-25% higher than standard motors, but they can achieve around a 6% improvement in efficiency. As an example, for a 7.5 kW motor, an increase in efficiency from 84% to 91% can be achieved.

It is important to emphasise the use of an energy efficient motor, does not reduce the power requirements of the load, but would reduce the energy consumption, and might allow a smaller frame size of motor to be selected.

7. CONCLUSIONS Two methods for sizing lift motors are used. One ensures that the motor can move the required load at the rated speed. This is referred to as the steady state method. It is usually used as a first pass to select a certain motor. The second method is used as a check to ensure that the motor can accelerate all the masses at the required acceleration. This is usually referred to as the dynamic method, and is usually carried out as a second pass, once the motor has been selected using the first method. The size of motor for a lift is basically a function of the load in the car, the speed and the efficiency in the system and the coun As for escalators, only the static method is usually used. This is because the acceleration of a starting escalator is not very critical, and because an escalator is not strated as often as a lift is. The size of the motor for an escalator mainly depends on the veritcal rise, the angle of inclination, the linear speed of the escalator, the number of passengers per step and the efficiency of the system.

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When the motor is being driven by a solid state drive, there might be a need to derate the motor or use a larger size frame, to counteract the extra heat generated. A derating curve can be used for this purpose, which is a function of the speed of the motor. Energy efficient motors are now more widely used, and employ design features which reduce the internal heat losses. They have an efficiency around 6% higher than standard motors.

AUTOBIOGRAPHICAL NOTES

The author graduated in Electrical Engineering in 1987, and worked for two years as an electrical and electronic lift systems design engineer. He received his M.Sc. in Remote Lift Monitoring in 1990, and his Ph.D. in Artificial Intelligence Applications in 1992 from UMIST (Manchester, U.K.). He was then appointed as Senior Electrical Engineer for Lifts & Escalator at London Underground, and is still working for London Underground, currently as Team Leader in the Lift & Escalator Department. As a second role within London Underground, he is the developer of the model structures for the Whole Life Asset Plans for the Underground’s assets, and the chief modeller for the station based assets. He is also a Chartered Electrical Engineer, and a part time lecturer in electronics and electronic systems at the South Bank University.

REFERENCES & BIBLIOGRAPHY BSI, 1987, “BS 4999, 1987: General requirements of rotating electrical machines:

Part 111: Specification for build in thermal protection for electric motors rated at 660 volts a.c. and below”.

Costello, P., 1994, “Communication advances raise motor protection”, Drives & Controls, October 1994.

Loher, 1996, “Three Phase elevator motors”, Catalogue number LN 15 e. Ziehl Abegg, 1987, “Thermistor motor protection relay: TUS”, Technical Information

6.03.058 e. Ziehl Abegg, 1996, “Spreadsheet for motor selection”. Purdue University, 1998, “EET231 - Lecture 21: Induction motor equivalent circuit &

efficiency”, http://www.tech.purdue.edu/eet/courses, 20/08/1998.