# Effective motor control for high-efficiency motors

**Power Parameters Pty Ltd**

By Tony Foster, Regional Sales Manager, Intelligent Energy Management Systems, Power Parameters Pty Ltd

Sunday, 08 May, 2016

Modern high-efficiency motors have the drawback of higher inrush and starting currents, which means protection equipment must be specified differently.

Other than for legacy installations, or special cases, it is generally mandatory to install high-efficiency induction motors in new installations. The current high-efficiency IE3 standard is being adopted globally but is not yet an enforceable standard in Australia. However, it will not indefinitely ‘stay away’ and the current-MEPS imposed top limit of 185 kW in Australia can also be expected to broaden. As for the idea that variable speed drives improve efficiency regardless of the efficiency of the motor, an argument can be made that there may well be an excessive use of this form of control, and that for many applications there is both an engineering case — as well as overall efficiency case — to be made for DOL or star-delta control. As will be explained in more detail, the challenge of providing effective control without spurious interruption because of initial high current (inrush and locked rotor current) must be met with appropriate control gear.

#### High-efficiency motors

There are many techniques employed in the physical construction of high-efficiency motors, but the overall aim is to reduce stator iron and copper losses, windage losses and rotor copper losses. In our CO_{2}-conscious world, the aim is worthy, although in terms of many practical applications it can be argued that notwithstanding improved efficiency of a motor, the effect of the attached drive chain may well bring to nought the efficiency dividend of the prime mover. One often sees this reasoning used as a counter argument.

On a practical note, manufacturers supplying international markets are increasingly locked into IE3 standards, and these motors are making their way into the Australian market and the even higher efficiency IE4 motors are, so to speak, just around the corner. As is the case in so many other fields of engineering, there are no silver bullet solutions — and high-efficiency motors are no exception. Thus engineering considerations have to take into account lowered starting torques and high inrush currents that can pose problems for adequate motor protection. The performance characteristics in terms of starting current (locked rotor) and inrush current relative to IE1 motors are shown in Figure 1.

*Figure 1: Starting current and locked rotor current relative to an IE1 motor. For a larger image click here.*

#### Back to school

A basic understanding of induction motor physics is helpful in understanding how the requirement of higher efficiency has its effect on motor operational parameters. In Figure 2, the basic equivalent circuit of an induction motor is presented. The circuit is shown on a per-phase basis. In essence an induction motor can be considered a rotary transformer, and under locked rotor conditions it is in fact a transformer with a shorted secondary (the squirrel cage). In the case of a wound-rotor induction motor, external resistance connected via slip rings approximates a transformer with an electrical rather than a mechanical load.

*Figure 2: Induction motor per-phase equivalent circuit. For a larger image click here.*

Let’s consider a motor under full load condition. Input power is given by:

where:

*P _{1}* is the watts per phase

*V*is the phase voltage

_{1}*I*is the phase current

_{1}*φ*is the phase angle.

The output mechanical power is given by:

where:

*T* is the torque (Nm)

*ω* is the angular speed (rad/s).

The efficiency, *η* is therefore given by:

where *m* is the number of phases.

If we assume that *V _{1}* and

*cos φ*(power factor) are constants, then for a higher efficiency motor,

*I*is reduced when compared with a lower efficiency motor. The

_{1}*cos φ*constancy assumption is a little brave, as it is essentially determined by the ratio of load current to magnetising current. However, bear in mind that in high-efficiency motors, iron losses are also reduced, and therefore this can easily account for any differential in

*cos φ*.

So far things look good for higher efficiency motors. However, the question of locked rotor torque arises. This is a very important parameter — particularly for high inertia loads — as it determines initial angular acceleration. At the locked rotor condition, slip is 100%, and therefore doesn’t figure in the torque equation. In Figure 2 the parameters to pay particular attention to are the leakage inductance and resistance per phase of the rotor.

The slip ratio is *σ* (equal to (*ω _{s}-ω_{l})/ω_{s}*)

where:

*ω _{s}* is the synchronous speed (rad/s) and determined by

*2πf*/

*p*, where

*p*is the number of pole pairs, and

*f*the supply frequency

*ω*is the load speed (rad/s).

_{l}The general torque equation is given by:

where:

*α * is a constant for a given motor, taking the number of phases into account

*V _{1}* is the phase voltage

*R*is the resistance per phase

_{2}*X*is the rotor leakage reactance per phase under locked rotor conditions.

_{2L}We can now examine the locked rotor torque, while assuming phase voltage to be constant. This yields a proportionality:

Note that the slip *σ* is 1 under the locked rotor condition.

In order to maximise torque:

or:

The above relationship is very important because by reducing the rotor resistance in order to limit *I ^{2}R* losses, we negatively affect the starting torque.

The rotor current, *I _{2}* is given by the following relationship:

The locked rotor current therefore is:

Reducing rotor resistance therefore increases locked rotor current, although in practice not greatly depending on the particular motor design.

**Inrush current**

The notion that the locked rotor current represents inrush current must be dispelled. It is in fact a quasi-steady state of operation. Inrush current, on the other hand is only partially determined by motor parameters — chiefly the effective leakage inductance and resistance. It should also be noted that under locked rotor conditions the mechanical load, represented by an equivalent resistor (*R _{L}*), becomes zero in value. Note: its value at a slip ratio of

*σ*is given by the formula:

Up till now we have concerned ourselves with the rotor, and locked rotor current. The rotor current is of course reflected into the stator circuit by virtue of the mutual inductance between them.

Thus with locked rotor conditions applying we* *have an equivalent impedance of:

The effective motor impedance at the start, *Z _{M}*, should be considered as the per phase value. Unsurprisingly the inrush current depends on the phase angle of the supply voltage when connection is being made — and there are three phases. To keep things mathematically consistent, we will write for

*V*

_{1}_{:}

where:

*v _{1}* is the instantaneous voltage, at

*ωt*

*δ*is the phase angle of the voltage when connected.

The angle *δ* is also related to time, ie, *ωt _{0}* where

*t*is an initial time. However,

_{0}*δ*is an initial condition, whereas

*ωt*is the variable.

The general expression for the current after connection of voltage to an inductive circuit is of the form:

where:

*v* is instantaneous voltage, *R* is lumped resistance and *L*, lumped inductance.

This all looks unfamiliar to the motor equivalent circuit and its equations. Solving this equation, yields the expression below.

where:

*V _{1p}* is the peak voltage

*Z*is the motor impedance under locked rotor conditions (startup)

_{M}*δ*is the phase lag of the voltage at the moment of connection

*θ*is the arctangent of

_{1}*R/X.*

The second term in the above equation is an exponentially decaying baseline on which the oscillating (at power frequency) current rests. The bigger in value *R* is, and the smaller in value *X* is, the sooner the decay takes place. This is shown graphically in Figure 3.

*Figure 3: Instantaneous per-phase startup current waveform for an induction motor.*

#### Back to the practical world

The above theory might explain why motors with a low resistance are subject to higher inrush currents than those with higher effective resistance, but the challenge really is to have effective control and protection gear. The ratio of resistance to reactance (*R*/*X*) is of importance but in practice what we are dealing with is the short-circuit impedance of the supply circuit as well, since this adds to the* *motor impedance,

*Z*. For mission-critical applications, using a high sampling rate waveform recorder to measure inrush current is strongly suggested. The advantage is that it is a site test and therefore takes into account supply circuit impedance. Not only can the maximum amplitude of starting current be gauged, but also the duration. This is equally important as we now know that a combination of low supply and motor impedance, and a low

_{M}*R*/

*X*ratio, will result in a high inrush current. In addition the exponential decay component will take longer to return to zero (the steady state).

Motor control gear for high-efficiency motors needs to meet appropriate electromechanical parameters. Electrodynamic attraction on contact sets can cause welding, particularly on a frequent start regime subject to starting currents in excess of 20 times full load current. The overload protection curve for the motor and motor circuit should be carefully selected so that the extreme inverse characteristic of the protection caters for the very high inrush current possible, present for up to two cycles. The current-time envelope usually controlled by a magnetic trip should be such as to accommodate the run-up time for the mechanical load.

In view of reduced staring torques, the run-up time is important, because if too long, the thermal inverse protection region may be reached causing early tripping. A first estimation can be made on the basis of full load power and starting torque. Full load power equals:

where:

*T _{fl}* is full load torque in Nm required by the load

*ω*

_{fl}is full load speed in rad/s.

Initial acceleration is determined by:

where:

*dω*/*dt* is angular acceleration in rads^{-2}

*T _{s}* is starting torque in Nm

*J*is the moment of inertia of the load in kgm

^{2}.

A first approximation of the run-up time is to divide the full load speed by the acceleration. More accurate estimates can be made by consulting the torque-speed curve of the motor, and dividing this up into smaller linear portions, calculating time increments for each in order to arrive at a total run-up time.

The very important points being made here can be summarised as follows:

Irrespective of a contractor’s or designer’s point of view for the necessity or otherwise for high-efficiency motors — or for that matter whether IE3 is to be preferred over the current MEPS classifications — the reality is that many motors now commercially available are subject to IE2 and IE3 classifications (with IE4 around the corner). Consequently high-efficiency motor control is no longer a seldom-occurring situation.

The higher rating DOL-start motors will often be subject to the analytical approach described above. For lower rated motors the task of picking the correct protection current can be simplified given a suitable suite of protective devices to choose from — in short, by allowing the setting of the rated motor current to be placed toward the lower limit of the current rating range. This is best achieved by the availability of overlapping motor-current rating ranges. The designer is therefore able to choose a lower to upper* *current range whereby the short circuit multiplier used on the upper level current value (yielding the maximum permitted protection current before tripping) falls above the estimated inrush current. A side benefit of the choice of a range where the rated motor current is closer to the lower level current is reduced power loss in the motor starter.

#### Example

Take the following example of two motor starter protectors for a motor rated at 15 A.

- Motor starter protector A: Setting scale 10–16 A, multiplier of 13.
- Motor starter protector B: Setting scale 14–20 A, multiplier of 13.

Motor starter protector B (14–20 A) is recommended since it has a 5 A clearance to the top protection setting as opposed to protector A with a 1 A clearance. The power loss of motor starter protector B is approximately 35% lower than that of motor starter protector A.

In the case of motor starter protector A, the response value of the short-circuit release is 208 A (13 x 16 A). With a rated motor current of 15 A, the short-circuit release is 13.86 times the setting current (208/15 = 13.86).

In the case of motor starter protector B, the response value of the short-circuit release is 260 A (13 x 20 A). With a rated motor current of 15 A, the short-circuit release is 17.33 times the setting current (260/15 = 17.33). Because the inrush current to rated current ratio is significantly increased for the B case, it is the better choice for an IE3 motor.

The objective of highlighting the technical aspects of IE3 motors is to alert designers of motor control centres to the necessity of considering anew the field of motor protection and starting. The ‘suck it and see’ method is likely to bring grief. Proper analysis of motor control parameters is the only way of specifying the technical requirements for the motor control and protection equipment.

*Image credit: ©stock.adobe.com/au/imantsu*

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