3-Phase AC induction motors (part 2)

Electrical and mechanical performance

The angle between the two main stator components of voltage V and current IS is known as the power factor angle represented by the angle f and can be measured at the stator terminals. As shown, the stator current is the vector sum of the magnetizing current IM, which is in quadrature to the voltage, and the torque producing current IR, which is in phase with the voltage. These two currents are not readily available for measurement.

Consequently, the total apparent motor power S also comprises two components, which are in quadrature to one another,

kVA jQ + P = S

• Active power P can be calculated by…

• Reactive power Q, can be calculated by …

Where…

S = Total apparent power of the motor in kVA

P = Active power of the motor in kW

Q = Reactive power of the motor in kVAr

V = Phase-phase voltage of the power supply in kV

I_S = Stator current of the motor in amps

f = Phase angle between V and I_S (power factor = Cos f)

Not all the electrical input power PI emerges as mechanical output power PM. A small portion of this power is lost in the stator resistance (3.I^ 2.R_S) and the core losses (3.IM^ 2.RC) and the rest crosses the air gap to do work on the rotor. An additional small portion is lost in the rotor (3I^2R'R). The balance is the mechanical output power PM of the rotor.

Another issue to note is that the magnetizing path of the equivalent circuit is mainly inductive. At no-load, when the slip is small (slip s ? 0), the equivalent circuit shows that the effective rotor resistance R'R/s ? infinity. Therefore, the motor will draw only no-load magnetizing current. As the shaft becomes loaded and the slip increases, the magnitude of R'R/s decreases and the current rises sharply as the output torque and power increases.

This affects the phase relationship between the stator voltage and current and the power factor Cos f. At no-load, the power factor is low, which reflects the high component of magnetizing current. As mechanical load grows and slip increases, the effective rotor resistance falls, active current increases and power factor improves.

When matching motors to mechanical loads, the two most important considerations are the torque and speed. The torque-speed curve, which is the basis of illustrating how the torque changes over a speed range, can be derived from the equivalent circuit and the equations above. By reference to any standard guide on 3-phase AC induction motors, the output torque of the motor can be expressed in terms of the speed as follows:

This equation and the curve shows how the motor output torque TM varies when the motor runs from standstill to full speed under a constant supply voltage and frequency. The torque requirements of the mechanical load are shown as a dashed line.

====9: Torque-speed curve for a 3-phase AC induction motor.

A: is called the breakaway starting torque;

B: is called the pull-up torque;

C: is called the pull-out torque (or breakdown torque or maximum torque);

D: is the synchronous speed (zero torque)

At starting, the motor won’t pull away unless the starting torque exceeds the load breakaway torque. Thereafter, the motor accelerates if the motor torque always exceeds the load torque. As the speed increases, the motor torque will increase to a maximum T Max at point C.

On the torque-speed curve, the final drive speed (and slip) stabilizes at the point where the load torque exactly equals the motor output torque. If the load torque increases, the motor speed drops slightly, slip increases, stator current increases, and the motor torque increases to match the load requirements.

The range CD on the torque-speed curve is the stable operating range for the motor. If the load torque increased to a point beyond T Max, the motor would stall because, once the speed drops sufficiently back to the unstable portion ABC of the curve, any increase in load torque requirements TL and any further reduction in drive speed, results in a lower motor output torque.

The relationship between stator current IS and speed in an induction motor, at its rated voltage and frequency, is shown in the figure below. When an induction motor is started direct-on-line from its rated voltage supply, the stator current at starting can be as high as 6 to 8 times the rated current of the motor. As the motor approaches its rated speed, the current falls to a value determined by the mechanical load on the motor shaft.

====10: Current-speed characteristic of a 3-phase AC induction motor

Some interesting observations about the AC induction motor that can be deduced from the above equations are:

• Motor output torque is proportional to the square of the voltage.

Consequently, starting an induction motor with a reduced voltage starter, such as soft starters, star-delta starters, auto-transformer starters, etc, means that motor starting torque is reduced by the square of the reduced voltage.

• The efficiency of an induction motor is approximately proportional to (1 - s) i.e. as speed drops and slip increases, efficiency drops ...

The induction motor operates as a slipping clutch with the slip power being dissipated as heat from the rotor as 'copper losses'. On speed control systems that rely on slip, such as wound-rotor motors with variable resistors, slip-recovery systems, etc, speed variation is obtained at the cost of motor efficiency.

Efficient use of an induction motor means that slip should be kept as small as possible. This implies that, from an efficiency point of view, the ideal way to control the speed of an induction motor is the stepless control of frequency.

3-phase AC induction motors typically have slip values at full load of,

- 3% to 6% for small motors

- 2% to 4% for larger motors

This means that the speed droop from no-load to full load is small and therefore this type of motor has an almost constant speed characteristic.

One of the most fundamental and useful formulae for rotating machines is the one that relates the mechanical output power PM of the motor to torque and speed,

Where PM = Motor Output Power in kW TM = Motor Output torque in Nm N = Actual Rotational speed in rev/min.

Motor acceleration

An important aspect of correctly matching a motor to a load, is the calculation of the acceleration time of the motor from standstill to full running speed. Acceleration time is important to avoid over-heating the motor due to the high starting currents. So it’s often necessary to know how long the machine will take to reach full rated speed.

Manufacturers of electric motors usually specify a maximum starting time, during which acceleration can safely take place. This can be a problem during the acceleration of a high inertia load, such as a fan.

====9 shows the motor torque curve and the load torque curve plotted on the same graph for a speed range from standstill to full speed. Assuming DOL starting, the time taken to accelerate a mechanical load to full speed depends on:

• Acceleration torque (TA), which is the difference between the motor torque (TM) and the load torque (TL), TA = (TM - TL)

• Total moment of inertia (J Tot) of the rotating parts which is the sum of

- moment of inertia of the rotor

- referred value of the moment of inertia of the load

For acceleration to occur, the output torque of the motor must exceed the mechanical load torque. The bigger the acceleration torque, the shorter the acceleration time and vice versa. When the motor torque is less than the load torque, the motor will stall. ==== an example of the acceleration torque of a motor, started direct-on-line, driving a centrifugal pump load, whose torque requirement is low at starting and increases as the square of the speed.

The acceleration torque at starting is roughly equal to the rated motor torque, increases as the pump drive accelerates and then falls to zero as the motor reaches its rated speed. A steady state speed is reached when motor torque TM matches the load torque TL. The time taken from standstill to reach this stable speed is called the acceleration time.

====11: Acceleration torque during the starting of an AC induction motor

The rate of acceleration of the drive system also depends on the moment of inertia (J) of the rotating object. The higher the value of J, the longer it takes to increase speed.

Where…

J = Inertia of the drive system in kg-m^2

w = Rotational speed in radians/sec

If this formula is adjusted to convert the rotational speed from rad/sec to rev/min

Where n = Rotational speed in rev/min

Re-arranging ...

This is integrated with respect to speed, from starting speed (n1) to final speed (n2). The total acceleration time td is given by:

If the acceleration torque were constant over the acceleration period, this formula would simplify to:

Inertia can be calculated using the formula:

…

Where J = Moment of inertia of the rotating in kg-m

G = Mass in kilograms (kg)

D = Diameter of gyration in meters (m)

It’s not usually necessary to calculate the value of J from first principles because this may be obtained from the manufacturer of the motor as well as the driven machine.

AC induction generator performance

The performance of the 3-phase AC induction motor has been described for the speed range from zero up to its rated speed at 50 Hz, where it behaves as a motor. A motor converts electrical energy to mechanical energy. The induction motor will always run at a speed lower than synchronous speed because, even at no-load, a small slip is required to ensure that there is sufficient torque to overcome friction and windage losses.

If, by some external means, the rotor speed was increased to the point that there was no slip, the induced voltage and current in the rotor fall to zero and torque output is zero.

If the rotor speed is, by some external means increased above this, the rotor will run faster than the rotating stator field and the rotor conductors again start to cut the lines of magnetic flux. Induced voltage reappears in the rotor, but in the opposite direction. From Lenz's Law, this results in currents that oppose this change and the power flows in the opposite direction from the driven rotor to the stator. Power flows from the mechanical prime mover, through the induction machine into the electrical supply connected to the stator. When the speed of the machine exceeds the synchronous speed no, it then operates as an induction generator.

This situation can often occur in the case of cranes, hoists, inclined conveyors, etc, where the load 'over-runs' the motor. The torque-speed curve can be extended to cover the induction generator region as well. The shape of the curve in the generator region is identical to the motor region because exactly the same equivalent circuit applies. The only difference is that the slip is negative and active power is transferred back into the mains.

====12: Transition from induction motor to induction generator.

Efficiency of electric motors

The efficiency of a machine is a measure of how well it converts the input electrical energy into mechanical output energy. It’s directly related to the losses in the motor, which depend on the design of the machine. Referring to the equivalent circuit of an induction motor, the losses comprise the following:

• Load dependent losses: These are mainly the copper losses due to the load current flowing through the resistance of the stator and rotor windings and shown in the equivalent circuit as roughly I_S^2(R_S + R'_R). These losses are proportional to the square of the stator current.

• Constant losses: These losses are mainly due to the friction, windage and iron losses and are almost independent of load. They are represented in the equivalent circuit as IM^2 RC. Since the constant losses are essentially independent of load, while the stator and rotor losses depend on the square of the load current, the overall efficiency of an AC induction motor drops significantly at low load levels.

Because of price competition, AC motor manufacturers are under pressure to economize on the quality and quantity of materials used in the motor. Reducing the quantity of copper increases the load dependent losses. Reducing the quantity of iron increases iron losses. Consequently, high efficiency of motors usually cost more. On large motors, high efficiency represents a significant saving in energy costs, which can be offset against the higher initial cost of a more efficient motor.

For electric motors used in AC variable speed drive applications, additional harmonic currents result in additional losses in the motor making it even more desirable to use high efficiency motors.

====13: Efficiency of an AC induction motor vs load

Rating of AC induction motors

AC induction motors should be designed or selected to match the load requirements of any particular application. Some mechanical loads require the motor to run continuously at a particular load torque. Other loads may be cyclical or with numerous stops and starts.

The key consideration in matching a motor to a load is to ensure that the temperature inside the motor windings does not rise, as a result of the load cycle, to a level that exceeds the critical temperature. This critical temperature is that level which the stator and rotor winding insulation can withstand without permanent damage. Insulation damage can shorten the useful life of the motor and eventually results in electrical faults.

The temperature rise limits of insulation materials are classified by standards organizations, such as IEC 34.1 and AS 1359.3 2. These standards specify the maximum permissible temperatures that the various classes of insulation materials should be able to withstand. A safe temperature is the sum of the maximum specified ambient temperature and the permitted temperature rise due to the mechanical load.

For purposes of motor design, most motor specifications, such as IEC, AS/NZS, specify a maximum ambient temperature of 40 C. The temperature rise of the induction machine is the permissible increase in temperature, above this maximum ambient, to allow for the losses in the motor when running at full load. The maximum critical temperatures for each insulation class and the temperature rise figures, which are specified by IEC 34.1 and AS 1359.32 for rotating electrical machines, are as follows:

Insulation class E B F H Maximum temperature 120 dgr. Max temperature rise 70 dgr.

====14: Maximum temperature ratings for insulation materials.

From these tables, note that electrical rotating machines are designed for an overall temperature rise to a level that is below the maximum specified for the insulation materials.

For example, using class-F insulation, ...

Max ambient + Max temperature rise = 40 C …which gives a thermal reserve of 15 C. The larger the thermal reserve, the longer the life expectancy of the insulation material.

When operating continuously at the maximum rated temperature of its class, the life expectancy of the insulation is about 10 years. Most motors don’t operate at such extreme conditions because an additional safety margin is usually allowed between the calculated load torque requirements and the actual size of the motor chosen for the application. So life expectancy of a motor, which is correctly matched to its load and with suitable safety margins, can reasonably be taken as between 15 to 20 years.

If additional thermal reserve is required, the motor can be designed for an even lower temperature. It’s common practice for the better quality manufacturers to design their motors for class-B temperature rise but to actually use class-F insulating materials. This provides an extra 20 C thermal reserve that will extend the life expectancy to more than 20 years. This also means that the motor could be used at higher ambient temperatures of up to 50 C or more, theoretically up to 65 C. In manufacturer's catalogues, the rating of 3-phase AC induction motors are usually classified in terms of the following:

1. • Rated output power, in kW
2. • Rated speed, depends on the number of poles
3. • Rated for a continuous duty cycle S1, (see below)
4. • Rated at an ambient temperature not exceeding 40 C
5. • Rated at an altitude not exceeding 1000 m above sea level, which implies an atmospheric pressure of above 1050 mbar
6. • Rated for a relative humidity of less than 90%

====15: Summary of temperature rise for classes of insulation materials according to IEC 34.1

AC induction motors often need to operate in environmental conditions where the ambient temperature and/or the altitude exceed the basis for the standard IEC or AS rating.

Where the ambient temperature is excessively high, temperature de-rating tables are available from the motor manufacturers. An example of one manufacturer's de-rating table, for both temperature and altitude. As pointed out earlier, better quality AC induction motors have a built-in thermal reserve. In some cases, where the ambient temperature is only marginally higher than 40°C, this reserve may be used with no additional de-rating for temperature.

In motor mounting positions that are exposed to continuous direct sunshine, motors should be provided with a protective cover.

Ambient temperature --Permissible output % of rated output -- Altitude above Sea Level -- Permissible output % of rated output:

====16: Motor de-rating for temperature and altitude.

At high altitudes, where there is a reduced atmospheric pressure, the cooling of electrical equipment is degraded by the reduced ability of the air to remove the heat from the cooling surfaces of the motor. When the air pressure falls with increased altitude, the density of the air falls and, consequently, its thermal capacity is reduced. In accordance with the standards, AC induction motors are rated for altitudes up to 1000 meters above sea-level. Rated power and torque output should be de-rated for altitudes above that.

When a motor needs to be de-rated for both temperature and altitude, the de-rating factors given in the table above should be multiplied together. For example, for a motor operating at above 2500 m in an ambient temperature of 50°C, the overall de-rating factor should be (0.92 × 0.88) × 100%, or 81%.