1/ 9/ 13 I nduct ion M achines

1/ 1www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ i_m ain. ht m l

Fundamentals Design Aspects Starting Parameters Operation

Introduction Construction Principles of Operation Equiv alent Circuit

Power & Torque Torque-Speed Curv e Peak Torque & Power Examples

T hree-Phase Induction Machines

Like other electrical machines, induction machines can be operated as either generators or motors.

However, they are primarily used as induction motors. We will concentrate on motoring and later

consider applications where induction generation is attractive.

Induction machines are by far the most common type of motor used in industrial, commercial or

residential settings. (One notable exception may be in computer equipment, where there are significant

numbers of motors used for hard drives, DVD players etc.) Depending on location (which affects other

electrical energy uses, such as lighting, space heating) induction motors may consume up to 70% of all

electrical energy generated. There are two underlying reasons why induction motors are the general

purpose motor of choice:

Induction motors are cheap

Induction motors are robust

© Andy Knight Report an Error 06-Jan-201 3 1 0:59 PM

1/ 9/ 13 I nduct ion M achine Const r uct ion

1/ 3www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ basics/ const r uct ion. ht m l

Fundamentals Design Aspects Starting Parameters Operation

Introduction Construction Principles of Operation Equiv alent Circuit

Power & Torque Torque-Speed Curv e Peak Torque & Power Examples

Induction Machine Construction

Stator

The stator construction of a three-phase induction machine is similar to that of a three-phase

synchronous machine. A three-phase winding is placed in a number of slots in order to produce a

rotating sinusoidal mmf wave. As with other ac machines, the speed of rotation of the stator magnetic

field is described as the synchronous speed and is given by

Rotor

The rotor of an induction machine is different from other types of machine that we have considered so

far: there is no requirement for a power source on the rotor. The rotor of an induction machine can be

one of two types

1. Wound Rotor

2. Cage Rotor

Wound Rotor Machines

Wound-rotor induction machines have a three-phase winding, similar to the stator winding, on the

rotor. The rotor is usually wye-connected with the terminals of the three rotor phases connected to slip-

rings. In normal operation, the windings at the slip-rings are short-circuited to allow currents to flow.

An advantage of wound rotor machines is that external circuits can be connected to the rotor, allowing

external control of the machine. While all induction machines can be controlled to operate at different

torques and speeds, wound rotor control is particularly attractive in some application. Wound-rotor

induction machines are usually significantly more expensive than cage rotor machines. Possible

applications for wound-rotor machines include

1. speed control of very large machines (multi-MW)

2. reduced cost control of large machines

3. doubly-fed induction generation (used in some wind turbines)

Cage Rotor Machines

Cage rotor machines (also called squirrel cage machines) are the most common type of induction

motor. In a cage rotor design, there are solid conductors in slots on the rotor. The ends of the conductors

are short-circuited at each end of the rotor using an "end-ring". For small-medium sized machines (up

to a few hundred horsepower) the rotor conductors are cast using aluminum. This construction makes

the rotor relatively cheap to produce. In larger machines, rotors are usually made by manually

hammering solid copper bars into the rotor slots then manually brazing an end-ring in place. Fabricated

rotor cages are significantly more expensive that cast rotor cages.

Induction machine photos

The photographs below show different aspects of induction machine construction. Click the photo for a

1/ 9/ 13 I nduct ion M achine Const r uct ion

2/ 3www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ basics/ const r uct ion. ht m l

larger version

The above photos show (left to right): an old 15hp induction machine, with open stator housing; a 5kW

induction machine used in the lab benches; a modern high efficiency 2hp induction machines. Both of

the newer machines are of the "TEFC" type: Totally Enclosed Fan Cooled. There is no opening in the

stator housing, and there is no way for environmental material (water, dirt) to get into the motor. The

fan is outside the housing and blows air over the stator housing cooling fins. The fan can be seen in the

2hp photos below:

Stator

The stator of the motor is made up of many thin steel laminations stacked together and held in the rotor

housing. The conductors making up the coils in the stator windings are looped through slots in the stator

lamination. Coils in this machine insulated from the laminations using plastic sheets and held together

with string and paper to separate coil groups. The stator coils and laminations are then dipped in varnish

and baked to provide mechanical integrity.

Rotor

The rotor of the 2 hp motor is constructed using steel lamination and cast aluminum. If you look closely

at the rotor photos it is possible to see where the molten aluminum has leached out between the steel

laminations. In addition, the conductors in the rotor have been constructed with a "skew" of one

conductor pitch. The conductors are not arranged parallel to the axis of the rotor, but at an angle, this is

done to reduce torque vibrations and noise.

1/ 9/ 13 I nduct ion M achine Const r uct ion

3/ 3www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ basics/ const r uct ion. ht m l

The final two photos below highlight different rotor constructions. On the left are two rotors in the lab, a

cast cage rotor and a wound rotor, complete with slip rings. On the right is the cutaway motor from the

lab with a fabricated cage of copper bars, also with significant skew.

© Andy Knight Report an Error 06-Jan-201 3 1 0:59 PM

1/ 9/ 13 I nduct ion M achine Pr inciples

1/ 5www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ basics/ pr inciples. ht m l

Fundamentals Design Aspects Starting Parameters Operation

Introduction Construction Principles of Operation Equiv alent Circuit

Power & Torque Torque-Speed Curv e Peak Torque & Power Examples

Principles of Operation

The basic idea behind the operation of an induction machine is quite simple. Detailed mathematical

understanding of the interaction of magnetic fields and resultant torque is more complex. In many

cases, understanding the qualitative ideas and then applying a circuit model is sufficient. The qualitative

description is provided here, together with a more mathematical description for those who prefer that

type of approach. The mathematical description also includes some important definitions that are

required in order to develop a circuit mode.

Qualitative Description

1. The three-phase stator winding is connected to a three-phase supply

2. Currents flow in the stator winding, producing a rotating mmf and flux density

3. The stator flux density rotates at synchronous speed:

4. The magnetic field passes conductors on the rotor and induces a voltage in those conductors

5. Since the conductors are short circuited, current flows in the rotor conductors

6. The rotor currents produce a second rotor magnetic field, which acts to oppose the stator magnetic

field and also rotates at synchronous speed

7. With two magnetic fields rotating at constant speed, a torque is induced:

8. The rotor flux density will lag the stator flux density (flux density lags current by 90° electrically),

therefore the torque will be in the same direction as the rotation of the magnetic fields

9. The torque accelerates the rotor until synchronous speed is reached, at which time there is no

relative motion between the conductors and the stator flux density. Since the relative velocity is

zero, the induced voltage, rotor currents and flux density fall to zero and torque is also zero

Mathematical Principles

Definitions

supply frequency: fe in Hz or ωe in rad s-1

synchronous speed (radians per second): ωs

synchronous speed (revolutions per minute): ns

rotor mechanical speed (radians per second): ωm

rotor mechanical speed (revolutions per minute): nm

slip: difference between synchronous and mechanical speeds divided by synchronous speed:

slip speed:

in rpm: sns = ns-nm

in rad s-1 : sωs = ωs-ωm

1/ 9/ 13 I nduct ion M achine Pr inciples

2/ 5www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ basics/ pr inciples. ht m l

slip frequency: fsl = sfe in Hz or ωsl = sωe in rad s-1

Induced Rotor Voltage

As illustrated in the fundamental theory on rotating fields, passing balanced three-phase currents

through a balanced three-phase winding can produce a rotating mmf wave. Speed of rotation is set by

supply frequency and the number of poles in the machine. In an induction machine the air gap of the

machine is designed to be constant, therefore the rotating mmf will produce a rotating flux density. The

stator flux density can be defined in terms of either mechanical or electrical quantities:

In the above equation φm, φe are arbitrary phase angles in mechanical and electrical angles respectively.

We will set these to zero. θ is the location at which the flux density waveform is observed. (At a given

location, the flux density varies sinusoidally with time. At a given time, the flux density varies

sinusoidally with location.) Now, to understand how an induction machine works, we need to consider

the flux density seen by a conductor on the rotor.

In the image shown above, there is a rotor conductor at position θm = α. If the rotor is stationary then

the rotor will observe the the stator flux density as

However, if the rotor is rotating at mechanical speed ωm the location of the conductor becomes

and the flux density seen by the conductor is given by

1/ 9/ 13 I nduct ion M achine Pr inciples

3/ 5www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ basics/ pr inciples. ht m l

Now, the voltage induced in a conductor of length l moving perpendicular to a magnetic field is given by

and the relative velocity of the conductor through the magnetic field is given by

Therefore the voltage induced in the conductor is given by

Two important results can be seen in the above equation:

The induced voltage is proportional to slip

The frequency of the induced voltage is proportional to slip

Rotor Currents and Field

Current Magnitude and Phase

Without knowing the full details of the rotor circuit, we can makes some assumptions about the circuit

to enable us to understand the behaviour of the induction machine. We will assume that the rotor

conductor is part of a circuit with constant resistance RR and inductance LR. (We will see later that

resistance can actually vary with slip, but will assume that it is constant for now). Now, if slip is low (s ≈

0) then the reactance associated with the inductance will be negligible:

In this case, though induced voltage is small, the induced currents may be significant since the

conductors are short circuited, so RR is low. Also the currents will be approximately in phase with the

induced voltage. If slip is high (s ≈ 1) then the rotor reactance will be significant. Due to the increase in

induced voltage rotor currents will be high, but will lag the induced voltage significantly due to the

inductance of the rotor circuit.

Rotor Flux Density

We know that the flux density produced by a set of ac currents rotates at a speed given by

In the case of the rotor currents, the above equation gives the speed of rotation relative to the

conductors. The actual speed of rotation of the flux density will be given by

1/ 9/ 13 I nduct ion M achine Pr inciples

4/ 5www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ basics/ pr inciples. ht m l

i.e. the rotor magnetic field rotates at synchronous speed. We can get an understanding of the relative

position of the rotor and stator fields by drawing phasor diagrams. The phasor diagram of the stator flux

density phasor can be drawn from either a stator reference frame, where it rotates at electrical speed ωe,

or from a rotor reference frame, where it rotates at electrical speed sωe

First consider the case where slip is low. Induced current lags induced voltage slightly, the rotor flux

density is almost 90° electrically behind the stator flux density.

From

at low slips, the angle between flux density phasors is close to 90° and the torque will be approximately

proportional to induced voltage and therefore proportional to slip.

Now consider the case where slip is close to 1.0, mechanical speed is close to zero. In this case, rotor

current lags induced voltage and the angle between rotor and stator flux densities is much greater.

From the torque equation, even though the magnitude of the induced currents is higher and the rotor

flux density phasor has a larger magnitude, torque will not necessarily be higher than it is a low slips.

Summary

1/ 9/ 13 I nduct ion M achine Pr inciples

5/ 5www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ basics/ pr inciples. ht m l

Induction machines produce torque at all speeds except synchronous speed

Induction machines can operate with only one electrical source, they do not require a source to be

connected to the rotor.

Rotor currents and torque are nonlinear functions of slip, a measure of the relative speed between

the stator magnetic field and the rotor mechanical speeds

© Andy Knight Report an Error 06-Jan-201 3 1 0:59 PM

1/ 9/ 13 St eady St at e Equivalent Cir cuit M odel f or I nduct ion M achines

1/ 2www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ basics/ cir cuit . ht m l

Fundamentals Design Aspects Starting Parameters Operation

Introduction Construction Principles of Operation Equiv alent Circuit

Power & Torque Torque-Speed Curv e Peak Torque & Power Examples

Induction Machine Equivalent Circuit Model

We have seen that induction machines (as you might guess from the name) operate on the principle of

induced currents. There are still two magnetic fields, one from each of the rotor and stator, but the rotor

field is induced by the stator field. Effectively, we can think of the induction machine as a rotating

transformer. The stator is like the primary of a transformer and creates the initial field, inducing

voltages and currents in the secondary rotor winding. The fundamental differences from a stationary

transformer are:

The secondary rotates

There is an airgap, therefore more mmf is needed for a given flux density

The secondary voltage and frequency depend on speed

As an aside, a wound rotor induction machine can actually be used as a variable frequency transformer.

For instance, a 60Hz system connected to the primary of an induction machine can transfer power to a

50hz system connected to the rotor if the machine is mechanically driven at a slip of 5/6.

The per-phase equivalent circuit model for an induction machine in steady state operation supplied by a

balances three-phase supply is based on the transformer model shown below

In the diagram above,

V1 = Phase RMS Voltage

I1 = Stator Phase Current

R1 = Stator Winding Resistance

X1 = Stator Winding Leakage Reactance

Xm = Magnetizing Reactance

Rc = Core Loss Resistance

E1 = Air Gap Voltage

I2 = Rotor Current Referred to Stator

ER = Rotor Induced Voltage (Actual)

IR = Rotor Current Voltage (Actual)

XR = Rotor Leakage Reactance (Actual)

RR = Rotor Resistance (Actual)

Rotor Circuit

We know from the operating principles that induced voltage ER and rotor leakage reactance XR both

depend on slip. To simplify the model we can define them both in terms of their values when the speed is

zero, slip s = 1.0

1/ 9/ 13 St eady St at e Equivalent Cir cuit M odel f or I nduct ion M achines

2/ 2www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ basics/ cir cuit . ht m l

where

ER0 = induced voltage at standstill

XR0 = rotor leakage reactance at standstill

With the above definitions we can write the equation for actual rotor current as

and the transformer model may be re-drawn as

In the above diagram, the effective turns ratio aeff is constant and equal to the effective turns ratio at

standstill. In a would rotor machine, aeff,XR0 and RR can be measured. In a cage machine these

parameters cannot be directly determined, there is no method to directly measure voltages or currents

on the rotor. To overcome this difficulty, the rotor (secondary) circuit can be referred to the stator

(primary) side.

Full Equivalent Circuit Model

In the above circuit

R2 = a2effRR, the rotor resistance referred to the stator

X2 = a2effXR0, the rotor leakage reactance referred to the stator

The symbols used in induction machine models vary depending on the text and the context in which the

circuit is being used. RI, Ro, Rfe, Rm can all be found as references to the iron loss resistance. In some texts

(especially from Europe), R2, X2 refer to actual rotor values with R'2, X'2 used for referred values. In

drives texts, it is common to find Rs, Rr for stator resistance and rotor resistance referred to the stator.

© Andy Knight Report an Error 06-Jan-201 3 1 0:59 PM

1/ 9/ 13 I nduct ion M achine Power and Tor que

1/ 2www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ basics/ pandt . ht m l

Fundamentals Design Aspects Starting Parameters Operation

Introduction Construction Principles of Operation Equiv alent Circuit

Power & Torque Torque-Speed Curv e Peak Torque & Power Examples

Power & T orque

Power Flow

The input power to a three-phase induction machine is given by

Output power can be found by subtracting the losses from the input power

Losses

1. Stator Copper Loss. The stator resistive losses

2. Rotor Joule Loss. The rotor resistive losses. This is often called rotor copper loss, but since the rotor

conductors are aluminum, rotor joule loss is the more correct terminology.

3. Core Loss, or Iron Loss. The losses due to eddy current and hysteresis losses in the laminations.

This can be calculated using the resistor Rc. Often, core losses are grouped with friction and

windage and stray loss as rotational losses.

Rotor Power

The power transferred to the rotor is called the "Airgap Power". Consider the equivalent circuit below

(the core loss resistance has been removed and core losses grouped into rotational loss).

From the above circuit, it can be seen that the total power transfer to the rotor is given by

To find the power converted to the mechanical system the rotor joule loss must be subtracted from the

total rotor power

From the above equations, it can be seen that power converters to the mechanical system is a function

of the airgap power and slip:

Final output power may be obtained by subtracting the rotational loss from Pconv.

1/ 9/ 13 I nduct ion M achine Power and Tor que

2/ 2www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ basics/ pandt . ht m l

Torque

As with all rotating mechanical systems in steady state, torque can be found from the power and

mechanical speed

In the case of an induction machine, the electromagnetic torque generated by the machine can be found

using

which gives

Writing the torque in terms of the rotor current:

Finally, to find the available shaft torque after rotational losses, the output power must be used.

© Andy Knight Report an Error 06-Jan-201 3 1 0:59 PM

1/ 9/ 13 I nduct ion M achine Tor que Speed Cur ves

1/ 3www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ basics/ t r q_speed. ht m l

Fundamentals Design Aspects Starting Parameters Operation

Introduction Construction Principles of Operation Equiv alent Circuit

Power & Torque Torque-Speed Curv e Peak Torque & Power Examples

T orque Speed Curve

General Curve

Using the induction machine power and torque equations it is possible to produce the torque speed curve

shown below.

The torque speed curve is approximately symmetric about synchronous speed. (The magnitude of the

peak negative torque is larger than the positive torque, the difference being a function of stator

resistance)

Operating Regions

The torque-speed curve brakes down into three operating regions:

1. Braking, nm < 0, s > 1

Torque is positive whilst speed is negative. Considering the power conversion equation

it can be seen that if the power converted is negative (from P = τ ω) then the airgap power is

positive. i.e. the power is flowing from the stator to the rotor and also into the rotor from the

mechanical system. This operations is also called plugging.

This mode of operation can be used to quickly stop a machine. If a motor is travelling forwards it

can be stopped by interchanging the connections to two of the three phases. Switching two phases

has the result of changing the direction of motion of the stator magnetic field, effectively putting

the machine into braking mode in the opposite direction.

2. Motoring, 0 < nm < ns, 1 > s > 0

Torque and motion are in the same direction. This is the most common mode of operation.

3. Generating, nm > ns, s < 0

In this mode, again torque is positive whilst speed is negative. However, unlike plugging,

indicates that if the power converted is negative, so is the air gap power. In this case, power flows

from the mechanical system, to the rotor circuit, then across the air gap to the stator circuit and

external electrical system.

T he torque equation

Using the equation

1/ 9/ 13 I nduct ion M achine Tor que Speed Cur ves

2/ 3www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ basics/ t r q_speed. ht m l

the torque speed curve can be solved by first solving the circuit model to find the rotor current. In the

absence of a computer, this is a tedious process at best. Multiple solutions of the above equation for

torque at different slips can be made simpler by simplifying the equivalent circuit model. Consider the

diagram below:

The stator part of the equivalent circuit (together with the magnetising branch) can be replaced by a

Thevenin equivalent circuit. In the Thevenin circuit, the stator phase voltage has been replaced by its

Thevenin equivalent,

and the impedances have been replaced by Thevenin equivalent impedances.

Incorporating the Thevenin model into the circuit model results in the Thevenin equivalent circuit

model of an induction machine.

In the above circuit, the calculation of rotor current is greatly simplified

The above expression for rotor current can be squared and substituted into the torque equation

Using the above equation, the variation of torque with slip can be plotted directly .

Note that if power or efficiency calculations are needed, the full equivalent circuit model should be used

1/ 9/ 13 I nduct ion M achine Tor que Speed Cur ves

3/ 3www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ basics/ t r q_speed. ht m l

(not the Thevenin version).

Motoring T orque Characteristic

The motoring region of the induction machine torque-speed curve is the region of greatest interest. The

plots below show a number of different torque speed curves, due to differences in the motor designs.

Common features of interest are noted and discussed below.

τrated . The rated torque of the machine. This is the design operating point.

τstart . The start torque of the machine, when the machine is at standstill.

τmax or τpo . The maximum torque or pull-out torque. Once a machine has reached rated operating

point, this is the maximum torque that can be applied without stopping the machine (pulling out).

In reality, since this torque is significantly more than the design rated torque, operation at

maximum torque is not possible due to thermal heating issues. (Currents will be above rated

values, I2R losses will be excessive.)

τpu . The pull-up torque of the machine. In some machines, the lowest point on the torque speed

curve between starting and pullout is not the start torque. In this case it is important to know the

pull-up torque. This is the minimum torque that the motor can accelerate up to the desired

operating speed.

In steady state operation, an induction motor will operate at the speed where the mechanical load

torque equals the torque developed by the motor. At low speeds, the difference between the motor torque

and the load torque accelerates the machine. Normal operation is to the right of the maximum torque.

In this region, an increase in the load torque will cause the motor to slow, increasing the motor torque

until an equilibrium is reached. To the left of the pullout torque, no such equilibrium can be reached.

Mechanically,

where J is the rotational inertia of the mechanical system.

© Andy Knight Report an Error 06-Jan-201 3 1 0:59 PM

1/ 9/ 13 I nduct ion M ot or Peak Tor que

1/ 2www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ basics/ peak_t r q. ht m l

Fundamentals Design Aspects Starting Parameters Operation

Introduction Construction Principles of Operation Equiv alent Circuit

Power & Torque Torque-Speed Curv e Peak Torque & Power Examples

Maxim um T orque and Output Power

Using the Thevenin torque equation:

it is possible to plot the torque-speed curve of an induction machine. However, it is useful to know the

magnitude of the maximum or "pullout" torque without plotting the full torque-speed curve. From the

equation above, the peak torque could be found be differentiating with respect to slip to find the slip

which gives maximum torque. However, a mathematically simpler and intuitively clearer answer can

be found be considering the power flow in the Thevenin equivalent circuit

Analysing the full equivalent circuit it was observed that

Therefore, since synchonous speed is constant, maximum torque occurs at the same slip as maximum

airgap power. Considering the Thevenin circuit, and applying maximum power transfer theory,

maximum airgap power and maximum torque will occur when

Re-arranging it is possible to obtain the slip for maxiumum torque, or pullout torque. Note that this

should not be called the "maxiumum slip".

Substituting the pullout slip into the Thevenin torque equation:

Discussion

From the two equations above it can be seen that

The slip at which maximum torque occurs is proportional to rotor resistance

The magnitude of the maxiumum torque is independent of rotor resistance

If all other parameters remain constant, increasing the rotor resistance will:

1/ 9/ 13 I nduct ion M ot or Peak Tor que

2/ 2www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ basics/ peak_t r q. ht m l

1. Reduce the speed at which maximum toruqe occurs

2. Increase the starting torque (until spo > 1, then it will reduce the start torque)

3. Increase slip for a given torque

4. Reduce the speed for a given torque

5. Increase the rotor losses at a given torque

The last point above can be shown by considering that the torque equation

is actually rotor copper loss divided by slip speed in radians per second. If slip increases, losses must

increase to maintaint the torque.

The diagram below plots torque speed curves for a motor with the following parameters, R2 is varied.

R1=0.5 Ω, X1=0.75 Ω, X2=0.5 Ω, Xm=100 Ω, f=60Hz, p=6, VLL=230V, Y-connection

© Andy Knight Report an Error 06-Jan-201 3 1 0:59 PM

1/ 9/ 13 I nduct ion M achine Exam ple Pr oblem s

1/ 5www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ basics/ basic_exam ples. ht m l

Fundamentals Design Aspects Starting Parameters Operation

Introduction Construction Principles of Operation Equiv alent Circuit

Power & Torque Torque-Speed Curv e Peak Torque & Power Examples

Equivalent Circuit Model Analy sis Exam ple

A 480V, 60 Hz, 6-pole, three-phase, delta-connected induction motor has the following parameters:

R1=0.461 Ω, R2=0.258 Ω, X1=0.507 Ω, X2=0.309 Ω, Xm=30.74 Ω

Rotational losses are 2450W. The motor drives a mechanical load at a speed of 1170 rpm. Calculate the

following information:

i. Synchronous speed in rpm

ii. slip

iii. Line Current

iv. Input Power

v. Airgap Power

vi. Torque Developed

vii. Output Power in Hp

viii. Efficiency

Comments

This machine has no iron loss resistance, so the equivalent circuit is as follows:

i. Synchronous speed is given by:

Therefore

ns = 1200 rpm

ii. Slip is given by

Using the rpm equation,

s = (1200-1170)/1200 = 0.025

iii. Now, phase current is given by

where phase impedance is given by

Using the above equation, Zin = 9.57 + j3.84 Ω

1/ 9/ 13 I nduct ion M achine Exam ple Pr oblem s

2/ 5www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ basics/ basic_exam ples. ht m l

And noting that the machine is delta connected, V1 = VLL = 480V

I1 = 43.1 - j17.4 A. |I1| =46.6 A, θ = -21.9°

Therefore IL = √3 × 46.6 = 80.6 A

iv. Input power is given by:

Therefore:

Pin = 62.2 kW

v. To find airgap power, There are two possible approaches:

a. Airgap power is the input power minus stator losses. In this case the core losses are grouped

with rotational loss. Therefore

Pgap = 62.2 kW - 3× 46.62 × 0.461

Pgap = 59.2 kW

b. Airgap Power is given by

This approach requires rotor current to be found. With no core loss resistance:

Giving I2 = 43.7 A. Substituting into the power equation

Pgap = 59.2kW

vi. Torque developed can be found from

where synchronous speed in radians per second is given by

giving

τ = 471 Nm

vii. Output power in horsepower is the output power in Watts divided by 746. (there are 746 W in one

Hp).

and

Therefore output power in Watts is:Pout = 55.3kW

1/ 9/ 13 I nduct ion M achine Exam ple Pr oblem s

3/ 5www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ basics/ basic_exam ples. ht m l

Pout = 74.1 Hp

viii. Efficiency is given by

Therefore

η = 55.3/62.2 = 88.9%

T hevenin Circuit Model Analy sis Exam ple

For the machine in the previous example, find:

i. Thevenin circuit parameters and Thevenin voltage

ii. Pullout slip

iii. Pullout Torque

iv. Start Torque

Using Matlab or Excel (or another computer program) plot the torque speed curve for slip in the range 0

to 1

i. Thevenin circuit parameters and Thevenin voltage:

Thevenin circuit parameters an voltage can be found using the equations provided on the formula

sheet, or from first principles. The Thevenin voltage is the voltage applied to the rotor assuming

that the rotor current is zero. Thevenin impedance is the impedance of the stator part of the

circuit, seen from the rotor, assuming that the stator supply is short circuited.

Substituting the equivalent circuit parameters in to the above equations gives:

VTH = 475.2 V, RTH = 0.452Ω, XTH = 0.313Ω

ii. Pullout slip

The slip at which maximum torque occurs can be found from maximum power transfer theory.

Maximum torque and maximum airgap power occur at the same slip, therefore maximum torque

occurs when

iii. Pullout Torque

Pullout torque can be found by substituting the above pullout slip into the Thevenin torque

equation

1/ 9/ 13 I nduct ion M achine Exam ple Pr oblem s

4/ 5www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ basics/ basic_exam ples. ht m l

or from the maximum torque equation directly

Substituting into the above equation:

iv. Start Torque

Start torque can be found by setting s=1 in the above equation for torque.

Either Matlab or Excel can be used to produce a torque speed plot with fairly minimal effort. The file

thevenin_torque_speed.m is a script which may be run to produce the Matlab plot below

The file excel_trq_speed_example.xls contains an example of how to produce an Excel plot of the curve

(shown below). Note that this file uses complex arithmetic, which is part of the Excel "Analysis ToolPak"

add-in. (Select Tools > Add-ins, Analysis ToolPak

1/ 9/ 13 I nduct ion M achine Exam ple Pr oblem s

5/ 5www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ basics/ basic_exam ples. ht m l

© Andy Knight Report an Error 06-Jan-201 3 1 0:59 PM

1/ 9/ 13 Ef f ect of I nduct ion M ot or Rot or Resist ance

1/ 3www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ design/ r ot or _inf luence. ht m l

Fundamentals Design Aspects Starting Parameters Operation

Rotor Resistance Motor Classes Design Trends & Efficiency Legislation & Standards

Rotor Resistance

Investigating the torque-speed curve it is apparent that the rotor circuit resistance has significant impact

on speed at which maximum torque occurs. The plots below illustrate two cases, with low rotor

resistance on the left and high rotor resistance on the right.

It can be seen from the plots that a high rotor resistance will provide a high starting torque, leading to

rapid acceleration of the mechanical system. This is desirable since during starting the stator current is

significantly above the rated current. Short acceleration times reduce the stress on the power system

caused by high currents.

While high starting torques are desirable, high rotor resistance results in a relatively high slip during

normal running operation. As torque is porportional to rotor joule losses divided by slip, high resistance

causes increased losses and reduced efficiency during normal operation.

The above points cause a problem. For most applications it is desirable to have:

high starting torque

high efficiency at rated speed

However, designs with high starting torque will have low efficiency at rated speed and designs with high

efficiency will have low starting torque. In order to resolve these confilicting requirements, two steps

must be considered:

1. Careful consideration of the application requirements

2. Designs of motors with variable rotor resistance.

Application Requirements

Many motor applications will not need both high start torque and high rated efficiency. Alternately, the

requirement for either high start torque or efficiency may be so significant that it over-rides other

requirements. Consider two examples:

Fan Load

Fans and rotary pumps typically have a torque requirement that varies as either the square or cube of

mechanical speed. When driving a fan, the motor must provide rated torque at rated speed, but at lower

speeds the torque demand is significantly lower. A fan application will therefore not usually require

significant starting torque, efficiency during steady operation at rated speed is the over-riding concern.

Variable-torque high-inertia loads

This type of load typically includes mechanical punches and reciprocating rod pumps used in oil

production ("nodding donkeys"). In the case of a reciprocating pump, the mechanical laod varies with

time, some of the time the motor is working against gravity to lift oil out of the ground, at other times, it

1/ 9/ 13 Ef f ect of I nduct ion M ot or Rot or Resist ance

2/ 3www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ design/ r ot or _inf luence. ht m l

is working with gravity as the rod falls. The speed range of this system is significant, requiring very hightorques at low speeds. When the motor is at high speed (as the rod falls) the torque (and efficiency)

requirement is minimal. In this case high torque at low speed is the over-riding requirement.

Varirable Rotor Resistance

Although it is common to think of low frequency conductors as having a constant resistance, the

resistance of all ac conductors is a function frequency. Induced currents in conductors act to oppose the

originating magnetic field. As a result, the depth of penetration of the magnetic field into the conductor

will vary with frequency. (The magnitude of the induced voltage is a function of rate of change of flux,

and therfore a function of frequency)

Skin Depth

The skin depth of a conductor is defined as the depth at which the magnitude of a magnetic field has

fallen to 1/e of the magnitude of the surface. We can approximate this as the depth at which currents are

actually flowing in the conductor. Skin depth is given by

where

δ = skin depth

μ = permeability of the conducting material, = μ0 in non-magnetic materials (copper, aluminum)

σ = conductivity

f = frequency of the magnetic field, relative to the conductor

Since the frequency of the stator magnetic field seen by the rotor conductors is a function of slip, the

actively conducting region of the rotor bars will be a function of slip. Hence, the effective roto resistance

will be a function of slip. The table below plots skin depth for aluminum (σ ≈ 2.9×107) in a machine with

a 60Hz supply.

Slip Slip Frequency (hz) δ (mm)

0.025 1 .5 7 6.3

0.05 3.0 54.0

0.083 5.0 42.8

0.1 67 1 0.0 29.6

0.333 20.0 20.6

0.50 30.0 1 7 .1

0.667 40.0 1 4.8

0.833 50.0 1 3.2

1 .0 60.0 1 2.0

The table above indicates a number of important points:

1. No matter how deep a rotor bar, only the top 12mm conducts at standstill

2. Medium-large machines with bars deeper than 12mm will have varying rotor resistance

3. Smaller machines with bars less than 12mm deep will have effectively constant rotor resistance

As an example, a machine with a rectangular bar 72 mm deep will have a resistance 6 times smaller at

low slips than it will at starting.

1/ 9/ 13 Ef f ect of I nduct ion M ot or Rot or Resist ance

3/ 3www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ design/ r ot or _inf luence. ht m l

© Andy Knight Report an Error 06-Jan-201 3 1 0:59 PM

1/ 9/ 13 I nduct ion M ot or Design Classes

1/ 2www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ design/ m ot or _classes. ht m l

Fundamentals Design Aspects Starting Parameters Operation

Rotor Resistance Motor Classes Design Trends & Efficiency Legislation & Standards

Motor Design Classes

Since different applications need different characteristics from induction motors, the National Electical

Manufacturers Association (NEMA), a US organization, has specified different classes of induction

motor, with different characteristics. Typical torque speed-curves for classes A-D are sketched below.

Motor Classes A and D are obtained by designing the rotor resistance to be either low (class A) or high

(class D). Classes B and C are obtained by exploiting the effect of skin depth to obtain a variable

resistance rotor circuit. Example rotor conductor designs for classes A through D are shown below.

Comparing classes A and B, class B has a deeper bar to exploit skin depth effects. The bar width may

vary in order to enhance this effect. Class C rotors are typically fabricated with two seperate cages. Only

the outer cage will conduct at starting. There is air between the cages and at the top of the slot (to reduce

leakage flux). Note that cast rotor designs must have a closed slot design to prevent molten aluminum

from escaping. Class D has a small conductor, giving a high resistance at all slips. The class D example

shown has an open slot, for a fabricated rotor design.

A summary of the performance of the motor classes is provided in the table below

Class A Class B Class C Class D

Ty pe General Purpose General Purpose High Starting TorqueVery High Starting

Torque

StartTorque

1 00% rated for larger motors,200% rated, smaller motors

1 00% rated forlarger motors,200% rated,

smaller motors

Approx 250% rated > 27 5% rated

StartCurrent

˜800% rated500%-600%

rated

PulloutTorque

200%-300% rated ≥200% rated Slightly lower than class A

Pullout Slip <0.2 <0.2 High, can be as much as

1 .0

1/ 9/ 13 I nduct ion M ot or Design Classes

2/ 2www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ design/ m ot or _classes. ht m l

Rated Slip <0.05, lower than similar sizedclass B

must be <0.05,usually <0.03

<0.05, higher than class B High, ty pically 0.07 to0.1 1 , can be up to 0.1 7

ApplicationsFans, Blowers, Pumps,

Machine Tools As for Class ACompressors, pumps,

conv ey orsHigh inertia applications,e.g. mechanical punches

Notes

High starting inrush currentcauses power sy stem problems,it can cause the supply v oltage

to sag and requires specialstarting techniques. More

efficient than same sized classB

Replacemesntfor Class A dueto lower startcurrent. The

standard off-theshelf commodity

motor.

Applications that requirehigh start torques. Notethat the pullup and pull-

out torque can both belower than the start

torque. Less efficienct thanclass B

Very high inertiaapplications. e.g. in a

punch or reciprocal pumpwhere the slip may v ary

between 0 and 0.50 Muchless efficient than other

designs

© Andy Knight Report an Error 06-Jan-201 3 1 0:59 PM

1/ 9/ 13 Tr ends in I nduct ion M achine Design

1/ 3www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ design/ t r ends. ht m l

Fundamentals Design Aspects Starting Parameters Operation

Rotor Resistance Motor Classes Design Trends & Efficiency Legislation & Standards

Design T rends & Influences

History

The induction motor principle was patented by Tesla in 1888. (He also invented synchronous motors and

reluctance motors). By 1895 three-phase induction motors similar to today's designs were available.

Since then induction motors have become smaller and more efficient. These improvements are not

necessarily compatible with each other and have occurred for different reasons.

Smaller machines have been developed because it is in the manufacturers interest to do so. A

physically smaller induction machine with the same output capability will have lower material

costs and therefore can be made more profitably.

More efficient machines have been developed because of a combination of end user desire and

legislation. More efficient machines will have lower operating costs and require less generation=

which results in lower greenhouse gas emissions from fossil fuel power plants.

Energy Efficient Design

The techniques to make a more efficient motor are well understood:

1. Use thicker conductors. Increasing the cross-section of the conductors will result in lower I2R

losses.

2. Increase the length of the machine. A longer machine requires a lower torque density, which

means a lower flux density. Lower flux density will result in lower iron losses.

3. Increase the outer diameter of the stator. Increased outer diameter means an increased surface

area, allowing more effective cooling. This in turn means that a smaller, lower power fan can be

used

4. Use a low loss lamination steel. Lamination steels can be bought in different grades, with variable

hysteresis losses. Laminations can also be bought in various thicknesses, with thinner laminations

resulting in lower eddy current losses. A thin low-loss lamination will have significantly lower iron

loss than an thicker standard lamination

5. Ensure that the air gap length is constant. If the air gap surfaces are machined to give a constant

air gap, there will be smaller variations in flux density and therefore reduced likelihood of

concentrations of iron losses. (Eddy current losses are a function of flux density squared.)

Factors limiting the use of higher efficiency motors

The first limit on uptake of premium efficiency motors is cost. Reviewing the above options to improve

efficiency, 1-3 require more material to be put into the motor, increasing material costs. Option 4

requires more expensive laminations (thinner laminations and low loss steel are both more expensive).

In addition, if thinner laminations are used, more laminations are required for a given length of

machine, resulting in higher manufacturing costs (more lamination punches per machine, increased

wear on punches, more difficult handling). Option 5 adds an additional step to the manufacturing

process, again increasing manufacturing costs.

It can be seen that all the obvious options to improve motor efficiency result in more expensive

machines. To maintain profitability, manufacturers must therefore increase the purchase price of a

machine if it is more efficient.

From an end user point of view, the increased initial purchase price of a more efficient motor is more

1/ 9/ 13 Tr ends in I nduct ion M achine Design

2/ 3www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ design/ t r ends. ht m l

than offset by reduced lifetime operating costs. A premium efficiency motor will typically pay for itself inelectricity savings in about 4 years (assuming constant operation). Typical motor lifetimes are around

20 years, so from a lifetime perspective, premium efficiency motors are a sound financial choice.

If premium efficiency motors are such a sound choice, the question must be asked why people don't buy

premium efficiency motors. (They typically don't). A number of reasons can be found:

Most motor purchases are not made by the end user. They are made by "OEM"s: other equipment

manufacturers. OEM's don't pay the operating costs of the motor and must keep their costs down

to maintain their profitability.

New equipment is purchased from capital budgets. This is typically independent of any future

operating budget. Capital costs must usually be minimized during the development of a new

facility and are typically funded by one-time fixed budgets. Once a facility is built, the operating

costs are funded from another budget.

Large users generate their own electricity. In a regulated electricity market, large users do not

particularly care how much electricity they use, as long as it is lower than their generating

capacity. With the advent of de-regulated markets and the possibility of selling excess generating

capacity back to the grid, this is less of an issue.

A second limit on the development of premium efficiency motors has been standards. All motors in the

small-medium (up to several hundred Hp) size range must be made to meet industrial standards. This

limits the ability of manufacturers to be innovative with new motors. e.g. the maximum length and

outside diameter are specified in the standard, so cannot be increased beyond a certain point to improve

efficiency. In addition, standards specify minimum efficiency levels that must be met. Since all motors

that meet the standard are viewed as equivalent, manufacturer A has no incentive to exceed the

standard, as this would make its motors more expensive than those from manufacturer B who just

meets the standard. In order to remain in business, the manufacturer must aim to meet the standards at

minimum unit cost, which typically means just meeting the standard.

Increasing Market Penetration

There are number of sectors in society that are attempting to increase the use of premium efficiency

motors. The most influential in this area are governments, utilities and environmentally conscious

consumers.

Governments are concerned about reducing energy usage primarily because of commitments to reduce

greenhouse gases and other pollutants produced by fossil fuel power plants. In addition, a more

efficiency industry will, in the long term, be more globally competitive.

Utilities are concerned about energy usage due to power supply issues. e.g. in Ontario, the peak energy

usage is close to the peak generation capability. Reducing the end use of electricity will enhance power

system stability and reduce the probability of power outages.

An increasing number of consumers are demanding "greener" products. This significant market segment

has the purchasing power to influence the development of more efficient products. This type of

consumer is typically willing to pay more for "greener" products (e.g. hybrid cars, front load washing

machines)

Governments have the largest power to influence this area. In order to promote energy efficient

products to end consumers, the "Energy Star" approach has been used, highlighting the benefits of

energy efficient designs. This is particularly effective with consumer goods as the purchaser is usually

the end user.

Influencing the industrial use of energy efficient machines has proved more difficult:

1/ 9/ 13 Tr ends in I nduct ion M achine Design

3/ 3www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ design/ t r ends. ht m l

The EU is currently persevering with an "efficiency sticker" approach. Motors are grouped as low,

standard or high efficiency and sold badged into those groups. However, there is no requirement to

buy high efficiency designs. (It should be noted that the EU has lagged behind N. America in this

area)

Hydro-Quebec Energy Rebate Program. This was the initial N. American attempt to improve

energy efficiency. Hydro Quebec offered rebates to people who bought high efficiency motors,

offsetting the increase in initial costs. As an example, a $4000 rebate was available for a 600Hp

motor. This approach was eventually discontinued due to the high costs (Over $12 million)

Legislation. Specify the minimum efficiencies.

To understand the legislative approach, it is important to understand the roles of standards and

legislation

© Andy Knight Report an Error 06-Jan-201 3 1 0:59 PM

1/ 9/ 13 I nf luence of Legislat ion and St andar ds on I nduct ion M achines

1/ 2www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ design/ st andar ds. ht m l

Fundamentals Design Aspects Starting Parameters Operation

Rotor Resistance Motor Classes Design Trends & Efficiency Legislation & Standards

Legislation and Standards

Techniques to make motors more efficient are well known. However, it can be seen that the steps to

make a more efficient machine are not necessarily in the best interest of either the machine

manufacturer or purchaser. It can also be seen that in some instances, Standards could in the past have

been said to hinder the development of new, more efficient designs.

Standards

The standard that covers the design, construction and operating limits of induction machines is NEMA

MG1.

NEMA is the National Electrical Manufacturers Association, a US group which represents electrical

equipment manufacturers. MG-1 is the standard for motors and generators. Canadian motors also

follow MG-1.

NEMA MG-1 sets minimum performance values, (e.g. start, pullup, pullout torque, efficiency) and also

mechanical constraints. Motors are grouped by frame size and all motors of a certain frame must have

the same external dimensions, bolt hole locations etc. The standard also specifies the following

classifications

TEFC: Totally Enclosed Fan Cooled

TENV: Totally Enclosed Natural Ventilation

ODP: Open, Drip-Proof

Explosion-Proof

The benefit of the standard is that the consumer knows what they are buying, independent of the

manufacturer. This allows easy comparison between brands based on cost. e.g. all 5 Hp motors will be

the same physical size and shape and in the event of failure one motor can be replaced by another from

a different manufacturer without the need to redesign the mounting points, bolts etc.

Legislation

At times, Standards have limited the incentive to develop higher efficiency designs. As a result,

Governments have mandated legislated improvements over MG-1 (MG-1 was re-written to reflect the

legislation. Canada was one of the first jurisdictions to legislate for higher efficiency, introducing

standard CSA 390, which mandated a step change in motor efficiency. This was followed in the US by

NEPACT legislation, which duplicated the Canadian efficiency limits, resulting in a common North

American standard on efficiency.

Recent Developments

NEMA has recently (2004) introduced an extension to MG-1 to introduce efficiency standards for

"premium" efficiency motors. The premium efficiency motors program allows manufacturers to

develop, and consumers to purchase, motors at a higher efficiency than the minimum set by legislation.

Summary

Standards specify motor design and performance parameters, allowing consumers to make informed

choices about the motors they are buying. At times, the reluctance of industry to improve standards has

required legislators to force changes and impose minimum standards. Due to the relationship between

efficiency and motor cost, this is not usually in the best interest of motor manufacturers. Manufacturers

have recently pro-actively announced extensions to the standard to enable higher efficiency motors in

the marketplace.

1/ 9/ 13 I nf luence of Legislat ion and St andar ds on I nduct ion M achines

2/ 2www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ design/ st andar ds. ht m l

© Andy Knight Report an Error 06-Jan-201 3 1 0:59 PM

1/ 9/ 13 I nduct ion M ot or St ar t ing

1/ 1www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ st ar t ing/ st ar t . ht m l

Fundamentals Design Aspects Starting Parameters Operation

Starting Full Voltage Y - Δ Supply Circuitry

Starting Induction Motors

Induction Motors can usually be started by simply connecting them to the supply line voltage. This can,

however, cause power system problems. Start currents (or "in-rush" currents) are high, often about 6

times rated current. Starting a motor directly from the supply line can cause the supply voltage to dip, or

"sag", due to the higher than normal voltage drop across supply line and transformer impedances. In

severe cases, this voltage sag can cause supply instability or the failure of sensitive power electronics

connected to the system.

Wound Rotor Motors

Start behaviour of wound rotor machines can be controlled by adding resistance to the rotor circuit, as

shown below:

In the circuit model above, the actual resistance RE has been referred to the rotor as R′E. It can be seen

from the circuit model that the increased impedance will reduce the start current. In addition, the extra

rotor resistance will increase the motor torque during starting, reducing the start time and overall

impact of the in-rush current.

Cage Machines

Cage motors cannot have additional resistance added to the rotor circuit. A number of options are

available for cage motor starts:

Full Voltage, or direct-on-line start

Wye-Delta Start

Adding additional circuitry to the stator, including

Autotransformer

Additional resistance

Electronic soft-start or variable speed start

© Andy Knight Report an Error 06-Jan-201 3 1 0:59 PM

1/ 9/ 13 I nduct ion M ot or Full Volt age St ar t s

1/ 3www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ st ar t ing/ dol. ht m l

Fundamentals Design Aspects Starting Parameters Operation

Starting Full Voltage Y - Δ Supply Circuitry

Full Voltage Start

A full voltage start involves directly connecting the motor to the supply line. It can also be called a

"direct-on-line" start. A wiring schematic for a full voltage start is shown below.

In the above circuit, the motor is protected by 3 fuses, which will have a time-delay so that inrush

currents can flow for a short period. The motor is started by pressing the start button. This allows

current to flow through the relay M, in turn closing the normally open relay contacts M1 to M4. Contact

M1, M2 and M3 allow current to flow to the motor. M4 bypasses the start button and keeps current

flowing through the relay. The motor will stop when

a. The stop button is pressed, breaking the relay circuit and causing contacts M1 to M4 to open

b. The thermal overload resistors heat up enough to cause the normally closed O/L contact to open,

again breaking the relay circuit.

The primary advantages of the full voltage start are

the full torque is applied to the load, causing rapid acceleration and short transient

the start circuit is simple and low cost

The disadvantages of full voltage starting are

rapid acceleration may be undesirable for the mechanical system

High start currents during the transient

Start Currents

The graph below plots line current and torque against speed for the motor in the induction machine

examples. Rated torque for this machine is 477Nm, rated line current is 81.7A

1/ 9/ 13 I nduct ion M ot or Full Volt age St ar t s

2/ 3www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ st ar t ing/ dol. ht m l

It can be seen that the line current is significantly higher than the rated current for all speeds below

rated and doesn't begin to fall significantly until the slip is less than the pullout slip. It is typical for

currents to remain high until rated speed is reached. A small motor should be able to start a load of rated

torque in less than 10 seconds. For a larger machines, a rated load should be accelerated to rated speed

in less than 15 seconds

Start Codes

In order to quantify starting currents, which can vary with machine design, all induction motors are

assigned a start code letter (different from the letter used to designate the motor class). Start code letters

specify the start Voltamperes range in terms of the rated horsepower.

Code Letter Start kVA/rated Hp

A 0 - 3 .1 5

B 3.1 5 - 3 .55

C 3.55 - 4.0

D 4.0 - 4.5

E 4.5 - 5.0

F 5.0 - 5.6

G 5.6 - 6.3

H 6.3 - 7 .1

J 7 .1 - 8.0

K 8.0 - 9.0

L 9.0 - 1 0.0

M 1 0.0 - 1 1 .0

N 1 1 .0 - 1 2.5

P 1 2.5 - 1 4.0

R 1 4.0 - 1 6.0

S 1 6.0 - 1 8.0

T 1 8.0 - 20.0

U 20.0 - 22.4

V 22.4 +

To understand how to use start codes, consider the example below:

A 480V, 50Hp induction motor has start code H. Find the maximum current that may be expected at

starting.

1/ 9/ 13 I nduct ion M ot or Full Volt age St ar t s

3/ 3www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ st ar t ing/ dol. ht m l

From the start code table, class H has a maximum start kVA of 7.1 × the rated horsepower. Therefore

Sstart = 7.1 × 50 = 355kVA

Therefore the start current is

IL = 355 × 103/( √3 × 480)

IL = 427A

© Andy Knight Report an Error 06-Jan-201 3 1 0:59 PM

1/ 9/ 13 I nduct ion M ot or Wye- Delt a St ar t

1/ 1www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ st ar t ing/ wye_delt a. ht m l

Fundamentals Design Aspects Starting Parameters Operation

Starting Full Voltage Y - Δ Supply Circuitry

Wy e - Delta Start

A Y-Δ start is only applicable to motors that normally operate in delta connection and that can be re-

connected to operate in wye configuration. If the motor operates in delta:

Now, if the motor is connected in wye-configuration:

Comparing the above equations, it can be seen that the wye phase current is √3 times smaller than the

delta phase current and that the wye line current is 3 times smaller than the delta phase current.

In a wye-delta start, the motor is started in wye configuration, with line currents one third of those that

would have occurred with a delta start. Once the machine has accelerated to low slip, the wye supply is

removed and the motor is re-connected in delta configuration.

There are disadvantages to a wye-delta start. If the phase currents are reduced by 1/√3, the torque will

also be reduced by one third. This may cause the start transient to become unacceptably long. In

addition, there can be a significant short term transient when the delta connection is applied.

In the graph below, line currents and torque are plotted for the example motor connected in both wye

and delta.

© Andy Knight Report an Error 06-Jan-201 3 1 0:59 PM

1/ 9/ 13 Addit ional Cir uit r y f or I nduct ion M ot or St ar t ing

1/ 1www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ st ar t ing/ add_cct . ht m l

Fundamentals Design Aspects Starting Parameters Operation

Starting Full Voltage Y - Δ Supply Circuitry

Additional Supply Circuitry

Inrush current can also be controlled by controlling the supply to the motor.

Autotransformer Start

Using an autotransformer with a number of secondary "taps" it is possible to control the supply voltage

to the motor. At low speeds, the supply voltage is kept low to maintain a low current. As slip reduces and

the motor phase impedance increases, the supply voltage can be increased by changing the transformer

taps. Once close to rated speed, the autotransformer can be by-passed altogether.

The autotransformer approach has the advantage that it is not necessary to access the terminals of the

motor windings, the control can be applied remotely to the motor. (unlike Y-Δ). This is a significant

advantage in many operating environments. However, like Y-Δ, reducing the phase supply voltage

reduces the currents, torque and therefore increases the length of time of the start transient. In addition,

autotransformers are expensive additional equipment that is only used in starting.

Additional Resistance

Adding additional resistance in line with the stator increases the overall impedance of the motor circuit

during starting, again limiting current at the expense of torque. Again, one of the significant

disadvantages of this method is the cost of the additional equipment.

Electronic Soft Start or Variable Speed Drives

An electronic soft start works by sing power electronics to gradually increase the voltage applied to the

motor. This effect is somewhat like using an autotransformer, but carried out with electronics rather

than magnetic circuitry. The power electronics work similarly to a light dimmer, reducing the rms

voltage, but at fixed frequency.

Variable speed drives adjust both the voltage and frequency of the supply to the motor. As a result, it is

possible to accelerate a load at rated torque without exceeding rated current. Drives are generally beyond

the scope of this course, but will be covered briefly.

© Andy Knight Report an Error 06-Jan-201 3 1 0:59 PM

1/ 9/ 13 Det er m ining t he Equivalent Cir cuit Par am et er s of I nduct ion M achines

1/ 2www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ par am et er s/ par am s. ht m l

Fundamentals Design Aspects Starting Parameters Operation

Determ ining Induction Machine Param eters

Three tests are needed to determine the parameters in an induction machine model. Detailed testing is

quite involved and is specified in IEEE Standard 112.

DC Resistance Test

If a DC voltage is applied to the stator terminals, there will be no potential difference across any

inductance and no induced voltage on the rotor. As a result, the per-phase circuit is reduced to the stator

winding resistance. Applying a DC voltage across two motor terminals and measuring the current, the

stator resistance can be found. Considering the figures below, it can be seen that the apparent impedance

will depend on whether the machine is wye or delta connected.

Locked Rotor Test

In the locked rotor test, the rotor of the machine is prevented from rotating and the supply voltage

gradually increased until rated current is reached. When the rotor is stationary , the slip, s = 1 and the

equivalent circuit can be drawn as

Analysing the circuit:

Therfore, if the stator power, current and reactive power are measured, the resistance and leakage

reactances can be found.

Ideally, the locked rotor test should be performed at a reduced frequency to account for skin depth and

more accurately predict the rotor resistance under load conditions. Typically, this test is done at 1/4

rated frequency. With the value of stator resistance from the DC test, the rotor resistance referred to the

stator can be found.

If the locked rotor test is done at reduced frequency, then the reactances (which are proportional to

frequency) must be scaled to find the correct value at rated frequency:

1/ 9/ 13 Det er m ining t he Equivalent Cir cuit Par am et er s of I nduct ion M achines

2/ 2www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ par am et er s/ par am s. ht m l

At this point, only the total leakage reactance is known. The relative values of stator and rotor leakage

must be found from experience and are summarised in the table below

Percentages of leakage reactance due to stator or rotor

Motor Type X1 X2

Wound Rotor 50% 50%

Class A 40% 60%

Class B 40% 60%

Class C 30% 7 0%

Class D 50% 50%

No Load Test

During the no-load test, the machine is allowed to accelerate up to synchronous speed with no load

applied. The machine will run close to synchronous speed, with s → 0 If slip approaches zero, the rotor

current will fall to zero and the equivalent circuit can be drawn as shown below.

Remembering that this is a per-phase circuit we can write

where rotational losses are defined as

PF&W are friction and windage losses, Pcore are iron core losses and Pstray are stray load losses.

Again, measuring the stator current, power and reactive power, the remaining paramters can be found.

Using the power, if stator resistance is known (from the DC test), the total rotational losses can be found

Using the value of X1 from the locked rotor test, the magnetising reactance, Xm, may be found from the

no load reactive power.

© Andy Knight Report an Error 06-Jan-201 3 1 0:59 PM

1/ 9/ 13 I nduct ion M achine O per at ing M odes

1/ 1www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ oper at ion/ m odes. ht m l

Fundamentals Design Aspects Starting Parameters Operation

Modes of Operation Pole Changing Variable Frequency Generating

Modes of operation

As mentioned when considering the induction machine torque-speed curve, the three main modes of

operation are braking, motoring and generating. We have concentrated on analysis of a machine being

used as a motor, but the analysis is general for other modes of operation.

Motoring: Small Slips

In usual operation of an induction machine, as a motor attached to a fixed frequency supply, there is

little control over the operating point: the motor will operate at the speed where the load torque is equal

and opposite to the motor torque.

Considering again the torque speed equation:

If the slip is small then

and the torque equation can be re-written as

It can be seen that at small slips, torque is proportional to slip, doubling the load will approximately

double the slip.

It is clear from the above analysis that there is very little speed variation in an induction machine if the

synchronous speed is constant. The rotor will rotate at a small slip, slightly below synchronous speed. If

operation at more than one fixed speed is required, it is necessary to change the synchronous speed.

Considering the equation for synchronous speed

the options available are to change the number of poles or change the supply frequency

© Andy Knight Report an Error 06-Jan-201 3 1 0:59 PM

1/ 9/ 13 I nduct ion M ot or Pole Changing

1/ 1www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ oper at ion/ pole_change. ht m l

Fundamentals Design Aspects Starting Parameters Operation

Modes of Operation Pole Changing Variable Frequency Generating

Pole Changing

Changing the number of poles in a machine gives a set of discrete operating speeds. e.g. if a machine can

have 2 or 6 poles, it can operate at approximately 1200 rpm or 3600 rpm (with a 60Hz supply)

Consequent Poles

The basic idea of consequent poles is to split a single phase winding into two groups of coils, and to have

the capability of reversing the connections of one of the coil groups. It is best explained with the aid of a

diagram as shown below:

In the diagram on the left, a single 2-pole winding is shown. This same flux pattern can be obtained

using two windings connected to operate in parallel, as shown in the centre diagram. If the number of

poles in the machine must be increased from 2 to 4, the supply to one of the windings can be reversed,

resulting in a 4 pole field.

Multiple windings

The idea of consequent poles helps if the speed must change by 2, but if other speeds are needed, one

option is two wind more than one set of stator windings onto a machine, but only operate one at a time.

As an example, it is possible to wind both a 6 pole winding and a 2 pole winding onto the same stator. If

the 6-pole winding is excited, the speed will be 3 times lower than when the 2-pole winding is excited.

This idea is used in washing machines to switch between wash and spin cycles. This approach is

expensive as only half of the copper in the machine is active at any one time and the stator slots must be

made large enough to fit two sets of windings, rather than just one.

© Andy Knight Report an Error 06-Jan-201 3 1 0:59 PM

1/ 9/ 13 Var iable Fr equency O per at ion of I nduct ion M achines

1/ 2www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ oper at ion/ var iable_f r eq. ht m l

Fundamentals Design Aspects Starting Parameters Operation

Modes of Operation Pole Changing Variable Frequency Generating

Variable Frequency Operation

If a variable frequency supply is available, the synchronous speed of an induction machine may be

chosen at any desired value. However, caution must be taken when operating an induction machine at

frequencies below the rated value. Consider the induction machine equivalent circuit in terms of

inductances, rather than reactances:

If the machine is operated at reduced frequency the impedance of the circuit will be reduced. If the

voltage is held at the rated value then the currents will be above the rated value and the machine will

overheat. In order to maintain constant current through an inductor with variable frequency operation,

the voltage and frequency must be adjusted together. This is called constant Volts per Hertz operation.

If the machine is operated at low frequencies (e.g. less than 1/4 rated) the voltage drop across the stator

resistance will be significant. (As reactances get smaller, R1 gets proportionally bigger.) Unless the stator

supply voltage is compensated to allow for additional stator resistance voltage drop, the available torque

will fall.

If the machine is to be operated above rated frequency, the voltage cannot be increased above rated

voltage and the magnetizing current in the machine will be reduced. This mode of operation is called

field weakening operation.

Summary

Below rated frequency, keep the ratio V/f constant at rated values.

At low frequencies, the V/f ratio must be boosted to maintain rated torque.

Above rated frequency, maintain rated voltage.

Constant Volts-per Hertz operation

As an approximation, assume that a constant supply voltage:frequency ratio results in a constant

Thevenin Voltage:frequency ratio. (Due to stator resistance, this assumption doesn't hold at low

frequencies). If slip is small, then it has been shown that the torque equation

can be approximated as

which, for constant V:f may be written as

i.e. if V/f is constant, torque is approximately proportional to slip speed.

1/ 9/ 13 Var iable Fr equency O per at ion of I nduct ion M achines

2/ 2www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ oper at ion/ var iable_f r eq. ht m l

The above statement can be used as a basis for relatively simple speed control.

Example

A 480V, 60Hz, 4-pole motor has rated speed of 1750 rpm and rated torque of 10Nm. If a torque of

10Nm is needed at a mechanical speed of 1500 rpm, find the synchronous speed, supply frequency and

line-line supply voltage.

At rated torque, the slip speed will be the rated value. For a 4-pole 60hz machine, synchronous speed is

1800 rpm, therefore rated slip speed =1800-1750=50rpm. When operating at 1500rpm, 10Nm, slip

speed will still be 50rpm and the synchronous speed is given by

With the synchronous speed, the supply frequency can be found

Finally, if V/f is constant, the supply voltage must be 480 × 51.67/60 = 413.3V

Torque-Speed Curves

The figure below plots torque speed curves at different supply frequencies for the example motor with

constant V/f ratio.

Variable Voltage Variable Frequency Supply

Most modern drives use Pulse Width Modulation, (PWM) supplies to approximate a variable voltage and

frequency sinusoidal supply. In a PWM supply, a DC voltage is switched rapidly to approximate the

shape of the desired waveform. An example is shown below.

© Andy Knight Report an Error 06-Jan-201 3 1 0:59 PM

1/ 9/ 13 I nduct ion G ener at or s

1/ 3www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ oper at ion/ gener at ing. ht m l

Fundamentals Design Aspects Starting Parameters Operation

Modes of Operation Pole Changing Variable Frequency Generating

Generating

If the rotor of an induction machine rotates above synchronous speed, slip is negative, as are torque,

mechanical output power and air gap power. i.e. the machine is operating as a generator. When

considering synchronous machines it is normal to re-define the direction of positive current when

switching between generators and motors (keeping a positive power flow for both cases). With induction

machines we will not do this. If slip is negative, the "input power" to the electrical terminals will be

negative, implying that power is flowing out of the electrical terminals.

Consider the machine used in the basic examples:

A 480V, 60 Hz, 6-pole, three-phase, delta-connected induction motor with the following parameters:

R1=0.461 Ω, R2=0.258 Ω, X1=0.507 Ω, X2=0.309 Ω, Xm=30.74 Ω

Rotational losses are 2450W.

Now, suppose that the machine is being driven by a mechanical system such that it is rotating at 1224

rpm

Calculate the following information:

i. slip

ii. Line Current

iii. Power and Reactive Power at the terminals

iv. Airgap Power

v. Torque Developed

vi. Mechanical Power

vii. Efficiency

Much of the theory for this problem is identical to a motor problem:

i. Slip is given by

Using the rpm equation,

s = (1200-1224)/1200 = -0.02

ii. Now, phase current is given by

where phase impedance is given by

Using the above equation, Zin = -10.3 + j5.29 Ω

And noting that the machine is delta connected, V1 = VLL = 480V

I1 = -36.7 - j18.8 A. |I1| =41.4 A,

Therefore IL = √3 × 41.4 = 71.7 A

1/ 9/ 13 I nduct ion G ener at or s

2/ 3www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ oper at ion/ gener at ing. ht m l

Note that the real part of phase current is negative, indicating that real power flows out of the

terminals. Also, the imaginary part of the phase current, which indicates that reactive power must

flow into the terminals. An induction generator cannot operate without a reactive power supply,

either from a power grid or capacitor bank

iii. Using complex notation,

S=3VI*

or

Pelec =3 &times V×Re{I} = -53.0kW

Qelec =-3 &times V×Im{I} = +27.2kVAR

iv. Airgap Power is given by

This approach requires rotor current to be found. With no core loss resistance:

Giving I2 = 37.8 A. Substituting into the power equation

Pgap = -55.4kW

v. Torque developed can be found from

where synchronous speed in radians per second is given by

giving

τ = -441 Nm

vi. Mechanical power can be found using

and

Therefore mechanical power in kW is:Pout = -55.4 × (1.0+0.02) - 2.45

Pmech = -58.9 kW

vii. Efficiency is given by

Therefore

η = 53.0/58.9 = 89.9 %

1/ 9/ 13 I nduct ion G ener at or s

3/ 3www. ece. ualber t a. ca/ ~knight / elect r ical_m achines/ induct ion/ oper at ion/ gener at ing. ht m l

Summary

It can be seen from the above analysis that the equations for an induction motor can all be applied to an

induction generator. (As long as output and input power are correctly described as either electrical or

mechanical).. Induction generation used to be relatively rare. However, it is becoming increasingly

common as induction generators are the generator of choice for large wind turbines. The requirement of

an induction generator to have an independent reactive power supply has caused significant research

into the impact of large wind farms of power system stability.

© Andy Knight Report an Error 06-Jan-201 3 1 0:59 PM