About Energy Efficiency of Induction Motor Applications: Two Reasons to Choose a Variable Speed Drive (VSD).

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Energy efficiency of an induction motor is a well established topic for academic as well as for industry. I will present in a series of posts on how VSD is On, to get the most of the energy.

Energy efficiency of induction motor applications: overview

The electrical instantaneous power (or active power) P_Elec is the derivative of the energy. Equivalently the accumulated electrical power over the time is the consumed energy E_Elec.

E_Elec = ∫ P_Elec∙dt

An electrical motor converts active power supplied by an AC source (e.g. by extension, a variable speed drive) to a mechanical power applied on a mechanical system, with some losses. It means that

P_Elec = P_Losses + P_Mechanical (1)

There are three origins of losses in this power conversion scheme:

• Mechanical losses
• Fundamental electrical machine losses (Joule effect)
• Variable Speed Drive-induced losses:
  • To the motor by fast commutations of power switches
  • To the mains supply power system by current harmonics
  • Itself to supply Automation Systems Smart Control as far as Power Conversion functions.

We will focus on the motor aspect of energy efficiency. This generic expression (1) may also be obtained from electrical induction motor model:

> dynamics
d/dt φ_s  =  u_s  –  R_s ∙ i_s  –  j ∙ ω_s ∙ φ_s
d/dt φ_r  = j ∙ np ∙ ω ∙ φ_r  –  R_r ∙ i_r  –  j ∙ ω_s ∙ φ_r

> magnetic coupling
φ_s = f_s ( i_s , i_r )
φ_r = f_r ( i_s , i_r )

> The electro-magnetic torque is given by the expression T_em = 3/2 ∙ np ∙ Imag( i_r* ∙ φ_r ).

with u: voltage, i: current, φ: flux, ω: the speed, R: resistance, f: coupling function that includes magnetic saturation effects, subscript s: stator, subscript r: rotor, np: the number of pairs of poles, *: the complex conjugate, and by calculating the active power, we get:

P_Elec = ( 3/2 ∙ R_s ∙ |i_s |^2 + 3/2 ∙ R_r ∙ | i_r |^2 ) + ( ω ∙ T_em ) (2)

These two terms (motor losses, and mechanical power) of the electric power in expression (2) explain how to optimize energy with an induction motor. At the nominal working point, losses represent between 3% and 15% of nominal power. The mechanical power is the major part of the power consumption.

It is summarized in the following control scheme:

We can see the two ways to optimize energy:

1. Optimize the mechanical power provided to the system to increase the system efficiency: the role of a VSD is key !
   It allows delivering all possible mechanical working points (speed, torque), and as a consequence the just needed mechanical power.

   For instance, instead of “regulating” a flow of a DOL pumping system by using a valve on a hydraulic circuit, a VSD allows to change the hydraulic working point by varying the speed on an efficient way.

2. Minimize the motor losses: the role of a VSD is key !
   A VSD equipped with a high efficiency control law will solve this point.

   For instance, instead of applying nominal magnetization to the motor at low level of load, an optimized-energy VSD allows to decrease magnetization in order to well balance the mechanical and the energy constraints.

In terms of analytics, we may define the instantaneous motor energy efficiency at a working point by

η_(P-MOT) = P_Mechanical / P_Active = P_Mechanical / ( P_Losses + P_Mechanical )

The motor losses power is depending on the mechanical working point (speed, torque). So, to evaluate the energy consumption efficiency of an electrical motor application, this instantaneous data does not make sense. Its interest is limited to the performance follow-up of the motor itself but does not provide consistent information about energy performance of the system.

To speak about energy efficiency of a system requires defining a mechanical cycle expected by machine or process systems: how speed, torque should evolve over the time to provide the expected work on the mechanical system.
If we define the mechanical energy E_mech as the accumulated mechanical power over the time, we may introduce two analytics dedicated to the motor performance:

1. the ‘capex’ system energy usage
   ν_(E-CA-SYST) = E_Mech / E_(Mech Nominal) = E_Mech / ( ∫ P_Nominal∙dt )
2. the ‘opex’ system energy efficiency
   η_(E-OP-SYST) = E_Mech/E_Elec = E_Mech/( ∫ P_Losses∙dt + E_Mech )

The first analytic will provide a view on the initial sizing efficiency of the motor: during operation, we expect that the required mechanical power to the motor is in average between 70% and 100% of the nominal motor power.
The second analytic will assess the operational efficiency of the motor: we expect that the required mechanical energy is provided by the motor with a minimum of electrical energy.

Induction motor efficiency: electrical motor losses optimization

We plot some curves at constant speed to support the discussion. We use the 4kW, 400V motor data in simulation, while we plot per unit data.
Figure 0 represents basically the motor efficiency at nominal flux in red, that has to be compared to the optimal efficiency in dashed blue. In more than these motor efficiencies, the orange curve is showing the ‘capex’ system energy efficiency.
A first comment is that an important gap exists between the optimal efficiency and the standard efficiency, for system efficiency lower than 40% … It could appear that a huge gain is possible in operation in this range of mechanical power, but we have to keep in mind that it corresponds to over sized systems.
For a general approach, it’s much more important to focus on the small gap that appears at higher than 40% of system efficiency. It is the working points that are the most consistently used on the field.

Let’s have a detailed look on electrical motor losses, figure 1. They depend on two parameters:

1. The resistances. By construction, the value may depend according to energy efficiency index. During operation, we have also to take into consideration that the resistances’ values can vary by +50% between cold state and hot state.

2. The consumed currents at the working point. As expressed in the motor equations, the torque is generated by the product of a flux (let assume a flux current) and a torque current.
For a given load torque, it exists an infinity of couples of currents that provides the electromagnetic torque. From those currents, we compute losses. In function of the magnetization current, the losses follow a convex curve, see figure below.

We can express many different things around losses and motor efficiency. We will develop three comments to conclude.

Comment 1/ Looking at the iso-torque curves on the figure 2, the losses have a minimum depending on magnetization level. By choosing nominal magnetization, we get less than 1-point more than the minimum losses value on a torque range from 70% to 180%.

We can see that these curves are flat, so that there is a natural bad conditioning for the automatic tracking of the optimal working point.

Comment 2/ With a 1% power measurement accuracy, we may not determine the optimal magnetization without a large uncertainty, on the figure 3.

• At 10% of nominal load, magnetization may vary from 23.5% to 80.6% for 1% of nominal power.
• At 100.% of nominal load, magnetization may vary from 85.3% to 104.5% for 1% of nominal power.

This comment has to be put in perspective of standard Variable Speed Drives technology, for which we can expect a reasonable 1%-to-5% power measurement accuracy.

Comment 3/ Let’s check the difference between the active power got at the default nominal flux and the active power got at the optimal flux.
We can observe on figure 4a that on the full torque range, the maximum difference is about 5.2% of nominal power. On the one hand, this maximum is reached at no load, where definitely the standard nominal motor energization is lost in a steady state approach. On the other hand, it does not make sense to compare on a static way the energy savings on the electrical motor, where there is no mechanical power provided on the rotor shaft. Definitely to switch off the system when not used is the best energy savings approach.
If we check the energy savings between 60% and 100% of nominal torque, we get an energy savings target at about less than 1.5% of nominal power.

We could also present the result in a relative way to say that an efficient optimization algorithm will allow saving up to 80%-90% of power, as shown on figure 4b. What you should have in mind is that if someone sells you 80% of energy savings in operation, the best energy savings is to resize your system as it definitely does not make sense to work continuously in a so small mechanical power range.

Energy efficiency of induction motor applications: conclusion

As I have introduced at the beginning of this post, there are two levels to optimize energy consumed by induction motor applications.

• define the speed/torque profile that provides the just enough mechanical power to the application system.
  Is the just enough mechanical power to provide to the application to get the expected result on the application system: flow control, flow deliver ?
• what is the right magnetization level of induction motor to optimize losses when providing mechanical power ?

Let’s take the example of centrifugal pump working under a flow requirement, described on figure 5a and 5b.

A traditional way to control the flow will be to use a control valve. Let’s define to use cases at 85% of nominal flow and at 50% of nominal flow.

85%, WP1/WP2:
Staying at nominal speed, the motor head will increase (on pump curve) up to working point WP1, defined by mechanical data (1500 rpm, 3.76 kW) and hydraulic data (19.3 m^3/h, 35.3 MWC).
Using a VSD the flow is controlled through the motor speed. To reach the same flow value, the motor head will decrease (on system curve) up to working point WP2, defined by mechanical data (1275 rpm, 2.46 kW) and hydraulic data (19.3 m^3/h, 22.7 MWC).
By using a VSD, we need to provide 34.6 % less mechanical energy to the hydraulic system compared to a valve control at 85% nominal flow.

50%, WP1’/WP2′:
Staying at nominal speed, the motor head will increase (on pump curve) up to working point WP1′, defined by mechanical data (1500 rpm, 3.03 kW) and hydraulic data (11.4 m^3/h, 41.1 MWC).
Using a VSD the flow is controlled through the motor speed. To reach the same flow value, the motor head will decrease (on system curve) up to working point WP2′, defined by mechanical data (750 rpm, 0.50 kW) and hydraulic data (11.4 m^3/h, 7.9 MWC).
By using a VSD, we need to provide 83.4% less mechanical energy to the hydraulic system compared to a valve control at 50% nominal flow.

Let’s now focus on induction motor efficiency, on mechanical working point as described on figure 5c.

85%, WP1/WP2:
To deliver a mechanical working point (1275 rpm, 2.46 kW), a standard constant flux control law will consume 2.89 kW to provide mechanical torque. An optimized energy control VSD will be able to decrease magnetization, and then will consume 2.85 kW.
By using a VSD that embeds energy control law, we get 34.5% energy savings on electrical consumption instead of only 33.5% energy savings without motor control optimization, at 85% nominal flow.

50%, WP1’/WP2′:
To deliver a mechanical working point (750 rpm, 0.50 kW), a standard constant flux control law will consume 0.74 kW to provide mechanical torque. An optimized energy control VSD will be able to decrease magnetization, and then will consume 0.60 kW.
By using a VSD that embeds energy control law, we maintain 82.9% energy savings on electrical consumption instead of only 78.9% energy savings without motor control optimization, at 50% nominal flow.

Summary for the flow need equals to 85%

	id.	Mechanical power	Electrical power
WP1	I	3.759 kW	4.353 kW
WP2, standard VSD	II	2.458 kW	2.894 kW
WP2, optimal VSD	III	2.458 kW	2.853 kW

Comments:

• Relatively to I, III allows 34.5% energy savings. In absolute, III allows 37.5% nominal power savings
• Relatively to II, III allows 1.4% energy savings. In absolute, III allows 1.0% nominal power savings

Summary for the flow need equals to 50%

	id.	Mechanical power	Electrical power
WP1′	I	3.030 kW	3.489 kW
WP2′, standard VSD	II	0.500 kW	0.737 kW
WP2′, optimal VSD	III	0.500 kW	0.597 kW

Comments:

• Relatively to I, III allows 82.9% energy savings. In absolute, III allows 72.3% nominal power savings
• Relatively to II, III allows 19.0% energy savings. In absolute, III allows 3.5% nominal power savings

As a conclusion of the example, the usage of VSD to adjust mechanical power is a way for several tens percents of energy savings.
With an optimized control law, VSD allows adding a few percents of energy savings for induction motor.

In a next post, I will focus on the motor control law to achieve induction motor energy savings. I will apply the energy efficiency principle on different use-cases, in particular for machine or process applications. I will give performance results:

1. Mechanical dynamic response
2. Energy savings

on realistic (speed, torque) profiles, demonstrating the key role of VSD in energy efficiency of induction motor applications.