**3. Motor voltage asymmetry and its influence on inefficient energy usage**

Effects of Voltage Quality on Induction Motors' Efficient Energy Usage 135

**Figure 5.** Relation between derating factor and voltage asymmetry

**Figure 6.** Motor efficiency in dependence of motor load for different voltage asymmetries

voltage unbalance that could arise in the considered consumer network is presented.

**3.1. Equivalent circuit of induction motor for negative sequence** 

motor power losses be determined.

Electrical energy consumption is unnecessarily increased due to the lower motor efficiency, so maintenance of low voltage asymmetry (≤ 2%) is a measure of efficient energy usage. In that case, the influence of voltage asymmetry (negative sequence voltages) will be presented in detail in the paper as follows. At first, a procedure for calculation and analysis of negative sequence currents and corresponding power losses will be presented. Then, evaluation of

When the induction motor is supplied from the network with asymmetrical voltages, then the three-phase voltage system should be decomposed to positive, negative and zero sequences. Further, using equivalent motor circuits (Boldea & Nasar, 2002; Ivanov-Smolensky, 1982) separately for positive (Fig. 7a) and for negative sequences (Fig. 7b), calculations and analyses of motor energy and operation characteristics (currents, power losses, torques) are performed. At the end, with corresponding superposition of relevance values, their overall values are obtained. Only in that way could real (overall) values of

Analysis of the effect of unbalanced voltages on the three-phase induction motor is presented in the paper by Bonnett (2000). Since the unbalanced voltage of 2%, 3.5% and 5% causes an increase in losses, in the same order, for 8%, 25% and 50% of nominal power losses in the motors, it is a reasonable requirement to permit voltage asymmetry ≤2%, so this is the upper limit in most national and international standards. The truth is that with less load, the motor could safely work also at higher values of unbalanced voltage. The literature (Linders, 1972) states that the information given previously is determined from measurements and that they are higher than calculated values. However, it is explained here by the fact that the rotor inverse resistance is higher by 1.41 times compared to the rotor resistance in short-circuit mode (Kostic & Nikolic, 2010), since current frequency of the negative sequence in the rotor winding is twice as high (*fr,NS ≈ 2f1=2fr,SC*), i.e. it is higher by 1.41 times than corresponding values given in the literature. Thus, it is increasingly convincing that the requirements which are given in the most appropriate standards are justified. Performed analysis shows that there are some considerations that should be included in current standards. Motor operation is not generally allowed when voltage asymmetry (*U NS/UN*) is higher than 5%, because in the (rare) case that the direct and inverse component of the stator currents in one phase are collinear, increase of the current in that phase would be ≥ 1.38 times, and the increase of the losses in the windings of that phase would be ≥ 90% (Linders, 1972; Kostic & Nikolic, 2010).

The effect of voltage asymmetry on the three-phase induction motor is equivalent to the appearance of negative sequence voltage system that creates a rotating field which rotates contrary to the rotation of the positive sequence field and motor rotating direction. The consequence is that small values of negative sequence voltage produce relatively high values of negative sequence currents. By definition, the coefficient of asymmetry (*KNS*%) is the ratio of negative sequence voltage (*UNS*) and positive sequence voltage (*UDS*). For simplification, Standard NEMA use the following definition of voltage unbalance

### *Voltage unbalance=100· (maximum voltage deviation from average value)/average value*

For instance, for measured voltages of 396V, 399V and 405V, average value is 400V. Then, the highest variation from average voltage is determined (405V – 400V = 5V). At the end, the coefficient (percent) of voltage asymmetry is calculated as the quotient of highest variation and average value: *KNS*% = 100(5/400) = 1.25%. Since percentages of negative sequence currents are 6-10 times higher than corresponding voltage asymmetry, negative sequence currents could reach 10%. This causes additional motor heating and the appearance of inverse torque that reduces starting and maximum motor torque, and causes a small increase of motor slip. Because power losses and motor heating are increased, allowed motor loading is decreased. As the percent of asymmetry rises, the derating factor of nominal power decreases, according to NEMA MG1 (Bonnett, 2000), as shown in Fig. 5.

With an increase of voltage asymmetry coefficient, motor efficiency decreases under all load levels. Dependence of motor efficiency is given in Fig. 6 for the voltage unbalance of 0.00%, 2.50%, 5.00% and 7.50% (Bonnett, 2000).

**Figure 5.** Relation between derating factor and voltage asymmetry

would be ≥ 90% (Linders, 1972; Kostic & Nikolic, 2010).

2.50%, 5.00% and 7.50% (Bonnett, 2000).

**3. Motor voltage asymmetry and its influence on inefficient energy usage** 

Analysis of the effect of unbalanced voltages on the three-phase induction motor is presented in the paper by Bonnett (2000). Since the unbalanced voltage of 2%, 3.5% and 5% causes an increase in losses, in the same order, for 8%, 25% and 50% of nominal power losses in the motors, it is a reasonable requirement to permit voltage asymmetry ≤2%, so this is the upper limit in most national and international standards. The truth is that with less load, the motor could safely work also at higher values of unbalanced voltage. The literature (Linders, 1972) states that the information given previously is determined from measurements and that they are higher than calculated values. However, it is explained here by the fact that the rotor inverse resistance is higher by 1.41 times compared to the rotor resistance in short-circuit mode (Kostic & Nikolic, 2010), since current frequency of the negative sequence in the rotor winding is twice as high (*fr,NS ≈ 2f1=2fr,SC*), i.e. it is higher by 1.41 times than corresponding values given in the literature. Thus, it is increasingly convincing that the requirements which are given in the most appropriate standards are justified. Performed analysis shows that there are some considerations that should be included in current standards. Motor operation is not generally allowed when voltage asymmetry (*U NS/UN*) is higher than 5%, because in the (rare) case that the direct and inverse component of the stator currents in one phase are collinear, increase of the current in that phase would be ≥ 1.38 times, and the increase of the losses in the windings of that phase

The effect of voltage asymmetry on the three-phase induction motor is equivalent to the appearance of negative sequence voltage system that creates a rotating field which rotates contrary to the rotation of the positive sequence field and motor rotating direction. The consequence is that small values of negative sequence voltage produce relatively high values of negative sequence currents. By definition, the coefficient of asymmetry (*KNS*%) is the ratio of negative sequence voltage (*UNS*) and positive sequence voltage (*UDS*). For

*Voltage unbalance=100· (maximum voltage deviation from average value)/average value*  For instance, for measured voltages of 396V, 399V and 405V, average value is 400V. Then, the highest variation from average voltage is determined (405V – 400V = 5V). At the end, the coefficient (percent) of voltage asymmetry is calculated as the quotient of highest variation and average value: *KNS*% = 100(5/400) = 1.25%. Since percentages of negative sequence currents are 6-10 times higher than corresponding voltage asymmetry, negative sequence currents could reach 10%. This causes additional motor heating and the appearance of inverse torque that reduces starting and maximum motor torque, and causes a small increase of motor slip. Because power losses and motor heating are increased, allowed motor loading is decreased. As the percent of asymmetry rises, the derating factor of nominal power decreases, according to NEMA MG1 (Bonnett, 2000), as shown in Fig. 5.

With an increase of voltage asymmetry coefficient, motor efficiency decreases under all load levels. Dependence of motor efficiency is given in Fig. 6 for the voltage unbalance of 0.00%,

simplification, Standard NEMA use the following definition of voltage unbalance

**Figure 6.** Motor efficiency in dependence of motor load for different voltage asymmetries

Electrical energy consumption is unnecessarily increased due to the lower motor efficiency, so maintenance of low voltage asymmetry (≤ 2%) is a measure of efficient energy usage. In that case, the influence of voltage asymmetry (negative sequence voltages) will be presented in detail in the paper as follows. At first, a procedure for calculation and analysis of negative sequence currents and corresponding power losses will be presented. Then, evaluation of voltage unbalance that could arise in the considered consumer network is presented.

### **3.1. Equivalent circuit of induction motor for negative sequence**

When the induction motor is supplied from the network with asymmetrical voltages, then the three-phase voltage system should be decomposed to positive, negative and zero sequences. Further, using equivalent motor circuits (Boldea & Nasar, 2002; Ivanov-Smolensky, 1982) separately for positive (Fig. 7a) and for negative sequences (Fig. 7b), calculations and analyses of motor energy and operation characteristics (currents, power losses, torques) are performed. At the end, with corresponding superposition of relevance values, their overall values are obtained. Only in that way could real (overall) values of motor power losses be determined.

**Figure 7.** Equivalent circuits of induction motor a) positive and b) negative voltage sequence, and c) completely circuit for negative sequence voltage

### **3.2. Parameters of equivalent circuit for negative sequence**

Stator winding resistance (*Rs*) and reactance (*Xs*) are almost the same for positive and negative voltage sequences. The parameters of equivalent circuits that are correlated to the rotor side and negative sequence voltage system (resistance *Rr,NS* and reactance *Xr,NS*) are substantially different than those for positive voltage sequences, because the frequencies of the rotor currents in negative sequences are higher for 50÷100 times:

$$f\_{\rm NS} = f\_1 \cdot (2 - s) \approx 2f\_1 \succ \sim s \cdot f\_1 \tag{13}$$

Effects of Voltage Quality on Induction Motors' Efficient Energy Usage 137

, , 2 *R R r NS r SC* (14)

, , 2 *X X r NS r SC* (15)

*r,SC*

,, , 2 *R kR kR r NS R NS r R SC r* (16)

,, , 2 *X kX kX r NS X NS r X SC r* (17)

**Figure 8.** Dependence of rotor resistance and inductance from ratio =Hb/

resistance and rotor reactance are higher by √2 times, Fig. 8.

The explanation is the following: for motors with powers higher than 5 kW (or with relative rotor conductor height *ξSC* = *Hb/∂SC* ≥ 1.2), already in short-circuit mode, the rotor induced currents not the entire cross section, or rotor conductor (bar) height *Hb*, (Kostic, 2010). From that it could be concluded that for the negative sequence currents (Ir,NS with frequency *fr,NS≈2f*) it is used for 2 times lower part of the section of the rotor conductor and rotor

Since usually values of rotor resistance (*Rr*) and reactance (*Xr*) are known in the nominal regime, then it is necessary to know values of the coefficient of rotor resistance increase (*kR*>1) and the coefficient of rotor inductance decrease (*kL* < 1), both in the short-circuit

where values of coefficients *kR,NS* and *kX,NS* are determined from Fig.8 for previously established ratio value *ζNS = Hb/∂NS*, where is *∂NS –*penetration depth of negative sequence field in rotor conductor. In such a manner approximate values of coefficients kR and kX are determined, for different rotor bar height *Hb* (mm)= 15, 20, 30, 40, 50 and given in Tab. 2.

From the quantitative review of corresponding values for *kR,SC* and *kR,NS*, and *kX,SC* and *kX,NS*, it could be seen that valid relations are: *kR,NS* ≈ 1.41*kR,SC* and *kX,NS* ≈ *kR,SC*/1.41, and it could be concluded that relations (14) and (15) are correct. In that way the author's statement that **"it** 

regime. From (14) and (15), the values for *Rr,NS* and *Xr,NS* could be calculated as:

(For example, for load slip s=0.01÷0.05: fr,NS=(1.90÷1.98)f1>>fr,PS=s·f1=(0.01÷0.05)f1).

Values of resistance *Rr,NS* and reactance *Xr,NS* are determined from the corresponding parameters in the short-circuit regime, *Rr,SC* and *Xr,SC*. In Boldea & Nasar (2002) and Ivanov-Smolensky (1982) it is noted that values of corresponding resistances and reactances are approximately equal to those in the short-circuit regime: *Rr,NS Rr,SC* and *Xr,NS Xr,SC*.

If we look carefully, we could find that this statement is not correct. Those parameter changes are dependent on the ratio of the rotor conductor height (*Hb*) and field penetration depth (*r,SC = (2/(2f1)*)1/2, i.e. the quotient  *= Hb/r,SC*; where: *f1* – frequency of supply voltage first order harmonic, - conductor specific resistance,  *= <sup>0</sup>* – magnetic permeability of conductor.

By Fig. 8 (Kostic, 2010), for *ζSC = Hb/∂SC* = 1.2÷3 are obtained: *Rr,SC* =1÷3*Rr* and *Lr,SC* =1÷0.50*Lr*, where *Rr* and *Lr* are resistance and inductance in a low slip regime, respectively, for instance in a nominal regime. Since current frequency of the negative sequence in the rotor conductor (bar) is twice as high (*fr,NS ≈ 2f1=2fr,SC*), the penetration depth of those currents is lower by √2 times then in a short-circuit regime. Corresponding values of resistance and reactance are higher by √2 times:

Effects of Voltage Quality on Induction Motors' Efficient Energy Usage 137

$$R\_{r,NS} = R\_{r,SC} \cdot \sqrt{2} \tag{14}$$

$$X\_{r,NS} = X\_{r,SC} \cdot \sqrt{2} \tag{15}$$

**Figure 8.** Dependence of rotor resistance and inductance from ratio =Hb/*r,SC*

136 Induction Motors – Modelling and Control

completely circuit for negative sequence voltage

**Figure 7.** Equivalent circuits of induction motor a) positive and b) negative voltage sequence, and c)

Stator winding resistance (*Rs*) and reactance (*Xs*) are almost the same for positive and negative voltage sequences. The parameters of equivalent circuits that are correlated to the rotor side and negative sequence voltage system (resistance *Rr,NS* and reactance *Xr,NS*) are substantially different than those for positive voltage sequences, because the frequencies of

Values of resistance *Rr,NS* and reactance *Xr,NS* are determined from the corresponding parameters in the short-circuit regime, *Rr,SC* and *Xr,SC*. In Boldea & Nasar (2002) and Ivanov-Smolensky (1982) it is noted that values of corresponding resistances and reactances are

If we look carefully, we could find that this statement is not correct. Those parameter changes are dependent on the ratio of the rotor conductor height (*Hb*) and field penetration


By Fig. 8 (Kostic, 2010), for *ζSC = Hb/∂SC* = 1.2÷3 are obtained: *Rr,SC* =1÷3*Rr* and *Lr,SC* =1÷0.50*Lr*, where *Rr* and *Lr* are resistance and inductance in a low slip regime, respectively, for instance in a nominal regime. Since current frequency of the negative sequence in the rotor conductor (bar) is twice as high (*fr,NS ≈ 2f1=2fr,SC*), the penetration depth of those currents is lower by √2 times then in a short-circuit regime. Corresponding values of resistance and

 *= Hb/*

1 11 (2 ) 2 *NS f f s f sf* (13)

 *=* 

 *Rr,SC* and *Xr,NS*

 *Xr,SC*.

*<sup>0</sup>* – magnetic permeability

*r,SC*; where: *f1* – frequency of supply

**3.2. Parameters of equivalent circuit for negative sequence** 

the rotor currents in negative sequences are higher for 50÷100 times:

approximately equal to those in the short-circuit regime: *Rr,NS*

*f1)*)1/2, i.e. the quotient

depth (*r,SC = (2*

of conductor.

*/(2*

reactance are higher by √2 times:

voltage first order harmonic,

(For example, for load slip s=0.01÷0.05: fr,NS=(1.90÷1.98)f1>>fr,PS=s·f1=(0.01÷0.05)f1).

The explanation is the following: for motors with powers higher than 5 kW (or with relative rotor conductor height *ξSC* = *Hb/∂SC* ≥ 1.2), already in short-circuit mode, the rotor induced currents not the entire cross section, or rotor conductor (bar) height *Hb*, (Kostic, 2010). From that it could be concluded that for the negative sequence currents (Ir,NS with frequency *fr,NS≈2f*) it is used for 2 times lower part of the section of the rotor conductor and rotor resistance and rotor reactance are higher by √2 times, Fig. 8.

Since usually values of rotor resistance (*Rr*) and reactance (*Xr*) are known in the nominal regime, then it is necessary to know values of the coefficient of rotor resistance increase (*kR*>1) and the coefficient of rotor inductance decrease (*kL* < 1), both in the short-circuit regime. From (14) and (15), the values for *Rr,NS* and *Xr,NS* could be calculated as:

$$R\_{r,NS} = k\_{R,NS} \cdot R\_r = \sqrt{2}k\_{R,SC} \cdot R\_r \tag{16}$$

$$X\_{r,NS} = k\_{X,NS} \cdot X\_r = \sqrt{2}k\_{X,SC} \cdot X\_r \tag{17}$$

where values of coefficients *kR,NS* and *kX,NS* are determined from Fig.8 for previously established ratio value *ζNS = Hb/∂NS*, where is *∂NS –*penetration depth of negative sequence field in rotor conductor. In such a manner approximate values of coefficients kR and kX are determined, for different rotor bar height *Hb* (mm)= 15, 20, 30, 40, 50 and given in Tab. 2.

From the quantitative review of corresponding values for *kR,SC* and *kR,NS*, and *kX,SC* and *kX,NS*, it could be seen that valid relations are: *kR,NS* ≈ 1.41*kR,SC* and *kX,NS* ≈ *kR,SC*/1.41, and it could be concluded that relations (14) and (15) are correct. In that way the author's statement that **"it**  **is not correct to believe that values of corresponding resistances and reactances for negative sequence currents are approximately equal to those for motor short-circuit regime"** is confirmed**,** as it is quoted in the literature (Boldea & Nasar, 2002; Ivanov-Smolensky, 1982). To the contrary, it is only correct that those values are in relation by (16) and (17). It is useful to specify common values for rotor conductor (bar) height *Hb* (mm) and frame sizes (axial height) for standard induction motors, as given in Tab. 3. From these facts more precise calculations and analyses of negative sequence voltage (and negative sequence currents) influencing the motor operation could be performed (Kostic & Nikolic, 2010).

Effects of Voltage Quality on Induction Motors' Efficient Energy Usage 139

*I U* (19)

(20)

(21)

voltage. In the following calculations and analyses the value *XNS* =0.13 (or *XSC* = 0.16, i.e.

1, 1, 1 1 7.7 *NS NS n N*

The value of increased losses in the phase windings of stator is proportional to the square of

, 1, 1,

The percentage of losses in rotor conductors could be higher by up to 3-6 times (2-5 times due to the higher rotor resistance for negative sequence currents and further up to 1.2 times due to the additional increase of rotor winding temperature under such a high power losses), i.e. *Rr,NS* ≈ *Rr,SC*=(2÷6) *Rr*. The equation for losses calculation in the rotor conductors

, 1, 1,

Assuming that negative sequence impedance is *ZSC,NS* ≈ *XSC,NS* = 0.13 (or *ISC*/*IN*=7.7) and negative sequence rotor resistance is *Rr,NS* = 5 *Rr*, the power losses values are calculated for voltage asymmetry of 2.5%, 3.5% and 5%. Such calculated values from (20) and (21), in percent of nominal losses, for particular motor parts (stator, rotor) and whole motor, are given in Tab. 4. Similar data are given in Linders (1972) where it was noted that measured values are 50% higher than calculated values. This difference was explained by an additional temperature increase of the rotor conductor, i.e. an increase of rotor resistance. Although an additional increase of rotor temperature by more than 100°C is not possible. The mentioned difference of measured and calculated values in Linders (1972) can be explained only by the fact that real rotor inverse resistance is 40% higher (based on (16) then its conventional value. Calculated values for permitted motor load are similar to those

Negative sequence voltage [%] 0.0 2.0 3.5 5.0 Negative sequence current [%] 0.0 15.0 27.0 38.0 Stator current (RMS) [%] 100.0 101.0 104.0 107.5 Increased stator winding losses [%] 0.0 2.4 7.4 15.0 Increased rotor winding losses [%] 0.0 12.0 37.0 75.0 Increased iron losses [%], Fig. 8 0.0 2.5 7.5 15.0 Increased total motor losses [%] 0.0 5.5 17.0 34.0 Permissible motor load P/Pn [%] 100 97 91 81

 

5 300 *Cur NS NS NS Cur n N N*

*PI U PI U*

60 *CuS NS NS NS CuS n N n*

*PI U PI U*

2 2

2 2

 

*I U*

negative sequence currents, and then from (19) it could be calculated as:

, 1

, 1

**Table 4.** Influence of negative sequence voltage on permissible motor load (*PN* ≥ 100 kW)

*ISC /IN* =6.25) is used, so:

for *Rr,NS* = 5*Rr*, is:

provided in NEMA standards (Fig. 5).


**Table 2.** Coefficients *kR* and *kX* in short-circuit regime and negative sequence currents


**Table 3.** Usual values for rotor bar height *Hb*, and frame sizes for standard induction motors

### **3.3. Negative sequence currents and power losses**

The negative effect on the motor operation due to the presence of negative sequence voltage is obvious for two reasons:


It is useful to express the value of negative sequence current (*I1,NS*) in the units of nominal positive sequence current *I1N,PS*:

$$\frac{I\_{1, \text{NS}}}{I\_{1n}} = \frac{\mathcal{U}\_{1, \text{NS}} / Z\_{M, \text{NS}}}{\mathcal{U}\_{1N} / Z\_{M,n}} \approx \frac{\mathcal{U}\_{1, \text{NS}} / X\_{M, \text{NS}}}{\mathcal{U}\_{1,n} / Z\_{M,N}} = \frac{\mathcal{U}\_{1, \text{NS}}}{\mathcal{U}\_{1N}} \cdot \frac{1}{X\_{M, \text{NS}}} \approx (6:8) \cdot \frac{\mathcal{U}\_{1, \text{NS}}}{\mathcal{U}\_{1N}} \tag{18}$$

since negative sequence motor impedance *ZM,NS* ≈ *XM,NS* ≈ (0.8-0.9) *XM,SC* and motor shortcircuit reactance *XM,SC* ≈ (0.15÷0.20) *ZM,N*, where *ZM,N* is motor impedance in the nominal regime. It could be seen from (18) that negative sequence currents in stator and rotor windings are 5÷8 times higher from the values of negative sequence voltage coefficients (*U1,NS /Un*). Since negative sequence currents are not dependent on motor load and slip, and then we suggest calculating the coefficient of asymmetry in the units of nominal motor voltage. In the following calculations and analyses the value *XNS* =0.13 (or *XSC* = 0.16, i.e. *ISC /IN* =6.25) is used, so:

138 Induction Motors – Modelling and Control

**is not correct to believe that values of corresponding resistances and reactances for negative sequence currents are approximately equal to those for motor short-circuit regime"** is confirmed**,** as it is quoted in the literature (Boldea & Nasar, 2002; Ivanov-Smolensky, 1982). To the contrary, it is only correct that those values are in relation by (16) and (17). It is useful to specify common values for rotor conductor (bar) height *Hb* (mm) and frame sizes (axial height) for standard induction motors, as given in Tab. 3. From these facts more precise calculations and analyses of negative sequence voltage (and negative sequence currents) influencing the motor operation could be performed (Kostic & Nikolic, 2010).

> *<sup>H</sup>* [mm] 15 20 30 40 50 *K R,SC* 1.30 2.00 3.00 4.00 5.00 *K R,NS* 2.00 2.80 4.20 5.60 7.00 *K X,SC* 0.90 0.75 0.50 0.38 0.30 *K X,NS* 0.75 0.54 0.36 0.27 0.22

**Table 2.** Coefficients *kR* and *kX* in short-circuit regime and negative sequence currents

*∂ Al,SC* = 10 mm; *∂ Al,NS* = 7 mm

**Table 3.** Usual values for rotor bar height *Hb*, and frame sizes for standard induction motors

windings, i.e. on the stator (*Rs*) and rotor (*Rr,NS*) resistance, and - inverse torque appears which is opposite to the motor operative torque.

The negative effect on the motor operation due to the presence of negative sequence voltage


It is useful to express the value of negative sequence current (*I1,NS*) in the units of nominal

1, 1, , 1, , 1, 1, 1 1 , 1, , 1 , 1 / / <sup>1</sup> (6 8) / / *NS NS M NS NS M NS NS NS n N M n n M N N M NS N*

since negative sequence motor impedance *ZM,NS* ≈ *XM,NS* ≈ (0.8-0.9) *XM,SC* and motor shortcircuit reactance *XM,SC* ≈ (0.15÷0.20) *ZM,N*, where *ZM,N* is motor impedance in the nominal regime. It could be seen from (18) that negative sequence currents in stator and rotor windings are 5÷8 times higher from the values of negative sequence voltage coefficients (*U1,NS /Un*). Since negative sequence currents are not dependent on motor load and slip, and then we suggest calculating the coefficient of asymmetry in the units of nominal motor

*I UZ UZ U X U* (18)

1.1 - 2.2 3 - 7.5 11 - 18.5 22 - 45 55 - 160 200 - 355 80 - 90 100 - 112 132 - 160 180 - 200 225 - 280 315 - 400 13 - 17 18 - 22 24 - 34 35 - 44 40 - 50 40 - 50

*I UZ UX U U*

**3.3. Negative sequence currents and power losses** 

Rotor slot depth

is obvious for two reasons:

Nominal power, *Pn* [kW]

Axial height [mm] Rotor slot depth, *H* [mm]

positive sequence current *I1N,PS*:

$$\frac{I\_{1, \text{NS}}}{I\_{1n}} = \nabla \mathcal{T} \cdot \frac{\mathcal{U}\_{1, \text{NS}}}{\mathcal{U}\_{1N}} \tag{19}$$

The value of increased losses in the phase windings of stator is proportional to the square of negative sequence currents, and then from (19) it could be calculated as:

$$\frac{P\_{CuS,NS}}{P\_{CuS,n}} = \left(\frac{I\_{1,NS}}{I\_{1N}}\right)^2 = 60 \cdot \left(\frac{\mathcal{U}\_{1,NS}}{\mathcal{U}\_n}\right)^2\tag{20}$$

The percentage of losses in rotor conductors could be higher by up to 3-6 times (2-5 times due to the higher rotor resistance for negative sequence currents and further up to 1.2 times due to the additional increase of rotor winding temperature under such a high power losses), i.e. *Rr,NS* ≈ *Rr,SC*=(2÷6) *Rr*. The equation for losses calculation in the rotor conductors for *Rr,NS* = 5*Rr*, is:

$$\frac{P\_{\text{Cur},NS}}{P\_{\text{Cur},n}} = 5 \cdot \left(\frac{I\_{1,NS}}{I\_{1N}}\right)^2 = 300 \cdot \left(\frac{\mathcal{U}\_{1,NS}}{\mathcal{U}\_N}\right)^2\tag{21}$$

Assuming that negative sequence impedance is *ZSC,NS* ≈ *XSC,NS* = 0.13 (or *ISC*/*IN*=7.7) and negative sequence rotor resistance is *Rr,NS* = 5 *Rr*, the power losses values are calculated for voltage asymmetry of 2.5%, 3.5% and 5%. Such calculated values from (20) and (21), in percent of nominal losses, for particular motor parts (stator, rotor) and whole motor, are given in Tab. 4. Similar data are given in Linders (1972) where it was noted that measured values are 50% higher than calculated values. This difference was explained by an additional temperature increase of the rotor conductor, i.e. an increase of rotor resistance. Although an additional increase of rotor temperature by more than 100°C is not possible. The mentioned difference of measured and calculated values in Linders (1972) can be explained only by the fact that real rotor inverse resistance is 40% higher (based on (16) then its conventional value. Calculated values for permitted motor load are similar to those provided in NEMA standards (Fig. 5).


**Table 4.** Influence of negative sequence voltage on permissible motor load (*PN* ≥ 100 kW)

Given the pessimistic assumption, especially for **smaller motors' power (up to 10 kW)** when the rotor resistance is increased only by 2-3 times (up to 1.5-2.5 times due to the higher rotor resistance for negative sequence currents and even up to 1.2 times due to additional increase in temperature of the conductor rotor in such a large loss of power, i.e. *Rr,NS* ≈ 1.4 *Rr*,SC = (2 ÷ 3) *Rr*, calculation of losses in the rotor conductors, for *Rr,NS* = 3*Rr*, was conducted by the expression:

$$\frac{P\_{\text{Cur},NS}}{P\_{\text{Cur},n}} = 60 \cdot \left(\frac{I\_{1,NS}}{I\_{1n}}\right)^2 = 180 \cdot \left(\frac{\mathcal{U}\_{1,NS}}{\mathcal{U}\_n}\right)^2\tag{22}$$

Effects of Voltage Quality on Induction Motors' Efficient Energy Usage 141

windings of that phase could reach 90% = 100% (1.382-1). Otherwise, in practice it could rarely be the case when direct and inverse components of the current matching phase angle. For example, it is necessary to stress that the asymmetry is a consequence of only one phase voltage deviation of the values (not phased by the angle) and the phase angles of the direct and inverse impedance are a little different, which is rarely filled because tan(φNS)= 0.3 ÷ 0.4. But it is possible that the phase angle between these components is about 300, and a


respectively, for nominal load and 80% load, both for voltages' unbalance of 5%.

For these reasons the motor operation is not allowed when the values of the coefficient of

Experimental measurements (Boldea & Nasar, 2002) showed that the effect of unbalanced voltages on the iron losses increase more if the motor is powered with high voltage, Fig. 9:



**Figure 9.** Dependence of iron losses on voltage asymmetry for three values of supply voltage

factor), as well as reducing motor efficiency and increasing power consumption.

*NS*

**3.4. Causes and evaluation of inverse voltage values** 

This additionally has an influence on reducing the coefficient of nominal power (derating

By definition, the coefficient of asymmetry is the relationship between the inverse system voltage (*UNS*) and direct voltage systems (*UPS*). Thus, the percentage of unbalanced voltage is

% 100 *NS*

*<sup>U</sup> <sup>K</sup>*

*DS*

*<sup>U</sup>* (23)

corresponding increase in this phase will be lower:

negative sequence voltage is *U1NS* /*UN* ≥ 5%.

voltage, while

calculated using the formula:

since the inverse of impedance *Z1,NS* ≈ *XNS* ≈ 0.9 *XSC* = 0.13 (i.e. when the motor short-circuit reactance *XSC* = 0.143, or *ISC / In*= 7). Thus obtained data are given in Tab. 5 and they are more accurate for motors with power below 10kW. Based on these calculations, it was found that the effect of unbalanced voltage on power losses is smaller for motors with nominal power ≤ 10 kW. Thus, acceptable voltage asymmetry for these motors could be 3%.


**Table 5.** Influence of negative sequence voltage on permissible motor load (*PN* ≤ 10 kW)

Based on data given in Tab. 5, it is concluded that the effects of unbalance on increased power loss is less compared to the data specified in the relevant standards for motors of nominal power ≤ 10 kW. Thus, motor total losses increase is 9% (Tab. 5) for the voltage asymmetry of 3%. As it is less then permissible 10% for these motors, for the networks with motors below 10 kW may be accept a voltage asymmetry (*U1,NS* /*UN*) ≤ 3%.

Although the unbalanced voltage losses in the rotor conductors are much higher than the corresponding losses in the stator winding, the increase to the losses of one phase of the stator can be the greatest. Specifically, it is unfavorable in the case where the direct and inverse current component could be in phase in one phase. Current in this phase, at the voltages' unbalance of 5%, would be:



These corresponding power loss values, respectively, were higher than the nominal 100% (1.382-1) = 90% and 100% (1.182-1) = 39%. Then the current increase in that phase would be 1.38 times at the negative sequence voltage *U1,NS /UN*= 5% and increase of the losses in the windings of that phase could reach 90% = 100% (1.382-1). Otherwise, in practice it could rarely be the case when direct and inverse components of the current matching phase angle. For example, it is necessary to stress that the asymmetry is a consequence of only one phase voltage deviation of the values (not phased by the angle) and the phase angles of the direct and inverse impedance are a little different, which is rarely filled because tan(φNS)= 0.3 ÷ 0.4. But it is possible that the phase angle between these components is about 300, and a corresponding increase in this phase will be lower:


140 Induction Motors – Modelling and Control

conducted by the expression:

Given the pessimistic assumption, especially for **smaller motors' power (up to 10 kW)** when the rotor resistance is increased only by 2-3 times (up to 1.5-2.5 times due to the higher rotor resistance for negative sequence currents and even up to 1.2 times due to additional increase in temperature of the conductor rotor in such a large loss of power, i.e. *Rr,NS* ≈ 1.4 *Rr*,SC = (2 ÷ 3) *Rr*, calculation of losses in the rotor conductors, for *Rr,NS* = 3*Rr*, was

, 1, 1,

since the inverse of impedance *Z1,NS* ≈ *XNS* ≈ 0.9 *XSC* = 0.13 (i.e. when the motor short-circuit reactance *XSC* = 0.143, or *ISC / In*= 7). Thus obtained data are given in Tab. 5 and they are more accurate for motors with power below 10kW. Based on these calculations, it was found that the effect of unbalanced voltage on power losses is smaller for motors with nominal power ≤

Negative sequence voltage [%] 0.0 2.0 3.0 5.0 Negative sequence current [%] 0.0 15.0 22.0 38.0 Stator current (RMS) [%] 100.0 101.0 102.0 107.5 Increased stator winding losses [%] 0.0 2.4 5.4 15.0 Increased rotor winding losses [%] 0.0 7.0 16.0 45.0 Increased iron losses [%], Fig. 8 0.0 2.5 7.5 15.0 Increased total motor losses [%] 0.0 4.0 9.0 25.0 Permissible motor load P/Pn [%] 100 98 95 87

 

60 180 *Cur NS NS NS Cur n n n*

*PI U PI U*

, 1

10 kW. Thus, acceptable voltage asymmetry for these motors could be 3%.

**Table 5.** Influence of negative sequence voltage on permissible motor load (*PN* ≤ 10 kW)

motors below 10 kW may be accept a voltage asymmetry (*U1,NS* /*UN*) ≤ 3%.

voltages' unbalance of 5%, would be:


Based on data given in Tab. 5, it is concluded that the effects of unbalance on increased power loss is less compared to the data specified in the relevant standards for motors of nominal power ≤ 10 kW. Thus, motor total losses increase is 9% (Tab. 5) for the voltage asymmetry of 3%. As it is less then permissible 10% for these motors, for the networks with

Although the unbalanced voltage losses in the rotor conductors are much higher than the corresponding losses in the stator winding, the increase to the losses of one phase of the stator can be the greatest. Specifically, it is unfavorable in the case where the direct and inverse current component could be in phase in one phase. Current in this phase, at the

These corresponding power loss values, respectively, were higher than the nominal 100% (1.382-1) = 90% and 100% (1.182-1) = 39%. Then the current increase in that phase would be 1.38 times at the negative sequence voltage *U1,NS /UN*= 5% and increase of the losses in the

2 2

(22)


respectively, for nominal load and 80% load, both for voltages' unbalance of 5%.

For these reasons the motor operation is not allowed when the values of the coefficient of negative sequence voltage is *U1NS* /*UN* ≥ 5%.

Experimental measurements (Boldea & Nasar, 2002) showed that the effect of unbalanced voltages on the iron losses increase more if the motor is powered with high voltage, Fig. 9:


**Figure 9.** Dependence of iron losses on voltage asymmetry for three values of supply voltage

This additionally has an influence on reducing the coefficient of nominal power (derating factor), as well as reducing motor efficiency and increasing power consumption.

### **3.4. Causes and evaluation of inverse voltage values**

By definition, the coefficient of asymmetry is the relationship between the inverse system voltage (*UNS*) and direct voltage systems (*UPS*). Thus, the percentage of unbalanced voltage is calculated using the formula:

$$K\_{NS}\,\%\text{o} = 100 \frac{\text{U}\_{NS}}{\text{U}\_{DS}} \tag{23}$$

where the direct and inverse system voltage component are calculated using the formula

$$\mathcal{U}\mathcal{U}\_{PS} = \frac{\mathcal{U}\_{ab} + a\mathcal{U}\_{bc} + a^2\mathcal{U}\_{ca}}{\mathfrak{Z}} \tag{24}$$

Effects of Voltage Quality on Induction Motors' Efficient Energy Usage 143

*TUX* (28)

2

(29)

(30)

2

Where are values Rr, Rs, ZM,N, XM,SC, ZM,SC are defined in in 3.3 (page 12).

the relative values of the inverse torque is:

losses ΔPγ, NS= 50%.

rectifiers and inverters.

Assuming that RrNS=RrPS=Rr and Rr/(2-s) = Rr/2 and with less neglect, leads to a relationship:

, 1 2 , 1

*T U Z*

Sometimes it is convenient to express the inverse torque in units of the nominal torque i.e.:

, , 1 1 2 , 1 1 () 3 <sup>2</sup> *em NS NS M N NS em N N SC N*

This means that, for the coefficient of unbalanced voltage U1,NS / UN= 0.04, would be Tem,NS /Tem,N = 3·0.042 = 0.0048=0.48%. Slip and power losses will also be increased for 0.0048 times, or 0.5%. But if we assume that the inverse rotor resistance for 2-5 times higher (up to 2-4 times due to the higher rotor resistance for negative sequence currents and even up to 1.2 times due to additional increase in temperature of the conductor), then the expression for

> , 1 , 1 (6 15) *em NS NS em N N*

The lower value of the coefficient related to the motor power up to 10 kW, and the upper value for motor power above 100 kW. When coefficient of unbalanced voltage U1, NS/ UN = 0.05, inverse torque in units of the nominal torque could be at Tem,NS/Tem,N= 10 · 0.052 = 0.025=2.5%. Slip and power losses will be increased also for 0.025 times, or 2.5% PN. As this is a medium power motor, with ηN≈ 90% or with losses of PγN≈ 10% PN power losses are increased by ΔPγ= 25% PγN, which is less than half of the determined value of increasing

The explanation lies in the fact that in this way (i.e. through the power that is allocated to resistance Rr, NS/2, Fig.7c) covers only half the power losses in resistance Rr,,NS while the remaining half of the compensated part of mechanical power (Pm), which is obtained through the axis of direct voltage system, Fig. 10. Thus, another procedure confirmed

**4. Influence of motor non-sinusoidal voltage on efficient energy usage** 

Analysis of the effect of non-sinusoidal voltages on the three-phase induction motor is presented in this chapter. Power losses and reactive load are increased due to the presence of harmonics in supply voltage. Nowadays, the presence of harmonics in the supply voltage is even more frequent due to the growth of consumers which are supplied through the

adequate accuracy of quantitative estimates given in Table 5.

Two interesting cases have been analyzed (Kostic M. & Kostic B., 2011).

*T U T U*

*T Z U U T UX U*

( ) <sup>2</sup> *em NS NS PS em PS PS SC*

$$\mathcal{U}L\_{NS} = \frac{\mathcal{U}\_{ab} + a^2 \mathcal{U}\_{bc} + a \mathcal{U}\_{ca}}{\mathfrak{Z}} \tag{25}$$

Thus, in the case of symmetrical voltages at the motor *Uab* = *a2Ubc* = *aUca*, we get *UPS = Uab = Ubc = Uca* (24) and *UNS*=0 (25).

Asymmetry can arise for several reasons. One is the joining of large consumers to one or two phases. Thus, if a consumer who connected to one phase, e.g. phase "a", creates a voltage drop *ΔU* = 3%, then it causes the asymmetry of 1% (*UNS*= 1%), since the asymmetrical voltage system can be presented as the sum of the symmetric system voltage (*Uab = Ubc = Uca*) and the unbalanced system voltages (*Uab = ΔU* = 3%, *Ubc* = 0, *Uca* = 0). This second voltage system, according to (25), corresponds to unbalanced system voltage of *UNS* = 1%. If a purely inductive consumer is connected between two phases, so that the voltage levels on each phase of the impedance network is 3%, then a similar analysis leads to the conclusion that this causes asymmetry of 2% (*UNS* = 2%), at the motor connections. These cases are possible in practice and rarely exceed the specified quantitative values.

Asymmetry may be a consequence of fuse capacitor burning. The fall out of capacitors part between two phases, concerning of the voltage reduction, is equivalent to appear of inductive loads between the two the same phases. As a consequence of that is the appearance of the inverse voltage, which is value equal to 2/3 change of the phase voltages mentioned. The general assessment is that it is almost always the asymmetry coefficient *KNS* < 2%, except for the interruption of one phase in the network, or interruption in any of the motor phase windings.

### **3.5. Motor inverse torque**

The system of negative sequence voltages leads to the appearance of the inverse torque (Tem,NS), which is opposed to the torque that drives the motor (the torque that comes from the direct system voltages and currents, Tem,PS). Resultant driving torque is reduced, Tem= Tem,PS - Tem,NS and the direct torque must be increased to compensate that decrease. Therefore, the slip and direct current systems are increased, and also the corresponding power losses. Direct (Tem,PS) and inverse (Tem,NS) electromagnetic torques is:

$$T\_{em, PS} = \frac{3\mathcal{U}\_{1PS}^2 R\_r}{s\Omega \left[ (R\_s + R\_r/s)^2 + X\_{M,SC}^2 \right]} \tag{26}$$

$$T\_{em,NS} = \frac{\Im L\_{\rm INS}^2 R\_r / \left(2 - s\right)}{s\Omega \left[ \left(R\_s + R\_r / \left(2 - s\right)\right)^2 + X\_{M,SC}^2\right]}\tag{27}$$

Where are values Rr, Rs, ZM,N, XM,SC, ZM,SC are defined in in 3.3 (page 12).

142 Induction Motors – Modelling and Control

*UPS = Uab = Ubc = Uca* (24) and *UNS*=0 (25).

the motor phase windings.

**3.5. Motor inverse torque** 

where the direct and inverse system voltage component are calculated using the formula

*PS*

*NS*

in practice and rarely exceed the specified quantitative values.

2

*U aU a U <sup>U</sup>* (24)

*U a U aU <sup>U</sup>* (25)

3 *ab bc ca*

2 3 *ab bc ca*

Thus, in the case of symmetrical voltages at the motor *Uab* = *a2Ubc* = *aUca*, we get

Asymmetry can arise for several reasons. One is the joining of large consumers to one or two phases. Thus, if a consumer who connected to one phase, e.g. phase "a", creates a voltage drop *ΔU* = 3%, then it causes the asymmetry of 1% (*UNS*= 1%), since the asymmetrical voltage system can be presented as the sum of the symmetric system voltage (*Uab = Ubc = Uca*) and the unbalanced system voltages (*Uab = ΔU* = 3%, *Ubc* = 0, *Uca* = 0). This second voltage system, according to (25), corresponds to unbalanced system voltage of *UNS* = 1%. If a purely inductive consumer is connected between two phases, so that the voltage levels on each phase of the impedance network is 3%, then a similar analysis leads to the conclusion that this causes asymmetry of 2% (*UNS* = 2%), at the motor connections. These cases are possible

Asymmetry may be a consequence of fuse capacitor burning. The fall out of capacitors part between two phases, concerning of the voltage reduction, is equivalent to appear of inductive loads between the two the same phases. As a consequence of that is the appearance of the inverse voltage, which is value equal to 2/3 change of the phase voltages mentioned. The general assessment is that it is almost always the asymmetry coefficient *KNS* < 2%, except for the interruption of one phase in the network, or interruption in any of

The system of negative sequence voltages leads to the appearance of the inverse torque (Tem,NS), which is opposed to the torque that drives the motor (the torque that comes from the direct system voltages and currents, Tem,PS). Resultant driving torque is reduced, Tem= Tem,PS - Tem,NS and the direct torque must be increased to compensate that decrease. Therefore, the slip and direct current systems are increased, and also the corresponding

> 2 1 , 2 2

*PS r*

3 / (2 )

*s r M SC*

*s r M SC*

3 ( /)

*s R Rs X*

2 1 , 2 2

( / (2 )) *NS r*

*s RR s X* 

,

,

(26)

(27)

power losses. Direct (Tem,PS) and inverse (Tem,NS) electromagnetic torques is:

*U R <sup>T</sup>*

*UR s <sup>T</sup>*

*em PS*

*em NS*

Assuming that RrNS=RrPS=Rr and Rr/(2-s) = Rr/2 and with less neglect, leads to a relationship:

$$\frac{T\_{em,NS}}{T\_{em,PS}} \approx (\frac{\mathcal{U}\_{1NS}}{\mathcal{U}\_{1PS}})^2 \cdot \frac{Z\_{PS}}{2X\_{SC}}\tag{28}$$

Sometimes it is convenient to express the inverse torque in units of the nominal torque i.e.:

$$\frac{T\_{em,NS}}{T\_{em,N}} \approx \left(\frac{\mathcal{U}\_{1NS}}{\mathcal{U}\_{1N}}\right)^2 \cdot \frac{Z\_{M,N}}{2X\_{SC}} \approx 3 \cdot \left(\frac{\mathcal{U}\_{1NS}}{\mathcal{U}\_{1N}}\right)^2\tag{29}$$

This means that, for the coefficient of unbalanced voltage U1,NS / UN= 0.04, would be Tem,NS /Tem,N = 3·0.042 = 0.0048=0.48%. Slip and power losses will also be increased for 0.0048 times, or 0.5%. But if we assume that the inverse rotor resistance for 2-5 times higher (up to 2-4 times due to the higher rotor resistance for negative sequence currents and even up to 1.2 times due to additional increase in temperature of the conductor), then the expression for the relative values of the inverse torque is:

$$\frac{T\_{em,NS}}{T\_{em,N}} \approx (6 \div 15) \cdot \left(\frac{\mathcal{U}\_{1NS}}{\mathcal{U}\_{1N}}\right)^2\tag{30}$$

The lower value of the coefficient related to the motor power up to 10 kW, and the upper value for motor power above 100 kW. When coefficient of unbalanced voltage U1, NS/ UN = 0.05, inverse torque in units of the nominal torque could be at Tem,NS/Tem,N= 10 · 0.052 = 0.025=2.5%. Slip and power losses will be increased also for 0.025 times, or 2.5% PN. As this is a medium power motor, with ηN≈ 90% or with losses of PγN≈ 10% PN power losses are increased by ΔPγ= 25% PγN, which is less than half of the determined value of increasing losses ΔPγ, NS= 50%.

The explanation lies in the fact that in this way (i.e. through the power that is allocated to resistance Rr, NS/2, Fig.7c) covers only half the power losses in resistance Rr,,NS while the remaining half of the compensated part of mechanical power (Pm), which is obtained through the axis of direct voltage system, Fig. 10. Thus, another procedure confirmed adequate accuracy of quantitative estimates given in Table 5.

### **4. Influence of motor non-sinusoidal voltage on efficient energy usage**

Analysis of the effect of non-sinusoidal voltages on the three-phase induction motor is presented in this chapter. Power losses and reactive load are increased due to the presence of harmonics in supply voltage. Nowadays, the presence of harmonics in the supply voltage is even more frequent due to the growth of consumers which are supplied through the rectifiers and inverters.

Two interesting cases have been analyzed (Kostic M. & Kostic B., 2011).

Effects of Voltage Quality on Induction Motors' Efficient Energy Usage 145

*RR R r h r sl h r e h* , , , (31)

*XX X r h r sl h r e h* , , , (32)

0.030 (p.u.) *R X r sl,sc r,sc* (33)

**Figure 11.** Motor equivalent circuit for harmonics: a) with separate rotor resistances and rotor

Increasing the order of harmonic causes increased frequency of induced currents in rotor conductors, compared to the one in the short–circuit regime. The skin effect is practically present only in the part of the conductor in the slot of the rotor, i.e. it leads only to an increase of rotor slot resistance (*Rr-sl,h*) and a reduction of rotor slot inductance (*Lr-sl,h*). As the depth of penetration is *∂Al,h*=5 = 4.5 mm, already for the fifth harmonic, *Rr-sl,h* is always equal to *Xr-sl,h,* Fig. 12. For this reason, similar to the corresponding scheme for short-circuit mode (Kostic, 2010), the rotor reactance (*Xr,h*) and resistance (*Rr,h*) are separated into two

where *Rr-e,h* is rotor winding end resistance and *Xr-e,h* is rotor winding end reactance ("e" in

Finally, resistance and reactance of stator windings, and resistance and reactance of rotor conductors outside of slots are grouped (Fig. 11b), for which the influence of skin effects can

In the paper by Caustic (2010) it is shown that values of rotor slot resistance (*Rr-sl,SC*) and rotor slot reactance (*Xr-sl,SC*) in the short-circuit mode are approximately the same for motors

Their values are within the narrow range of Xr-sl,SC = Rr-sl,SC ≈ 0.027÷0.033 p.u., respectively for

be neglected from the nominal regime to the short–circuit regime. These are:

The remaining resistance *Rr-sl,h* and reactance *Xr-sl,h*, are separated (Fig 11).

of all powers in a series and they are approximately equal to each other, i.e.:

the motors of large (> 100 kW), medium (11 - 50 kW) and low power (1 - 7.5 kW).

reactances and b) with grouped motor resistances and motor reactances

components in the equivalent circuit for the harmonics, Fig. 11a, i.e.:

the index comes from the abbreviation of the word "end").

**4.2. Parameters of equivalent circuit for harmonics** 


**Figure 10.** Electromagnetic torque characteristics (Tem) and the electromagnetic power (Pem) in the following regimes: generator (s < 0), motor (0 ≤ s ≤ 1) and braking (s> 1), as well as at point s = 2 (for the inverse voltage)


The reason for this (new analysis) lies in the fact that it is (wrongly) believed that the resistance of the smaller motor's rotor does not change for higher harmonic frequencies, i.e. that is identical for all harmonics (*Rr,h* = *Rr,*1 = *Rr* = *Const.*), which brings the difference mentioned above – and error. It is shown in Kostic M. & Kostic B. (2011) that values of the rotor resistance and the rotor reactance in a short-circuit regime could be estimated on the basis of the induction motor's catalogue date. Two important conclusions are established (probably for the first time) in Kostic (2010):


### **4.1. Equivalent circuit of induction motors for harmonics**

The equivalent circuit for the harmonics is identical to the corresponding equivalent circuit for a short-circuit regime for fundamental frequency *f*1. Only, instead of rotor short-circuit resistance (*Risk*) and rotor short-circuit reactance (*Risk*), the two rotor resistances, rotor end (*Rah*) and rotor slot resistance (*Rr-sl,h*), and two rotor reactances, rotor end (*Xr-e,h*) and rotor slot reactance (*Xr-sl,h*), are used for the harmonics order (h), Fig. 11a (Kostic M. & Kostic B., 2011).

**Figure 11.** Motor equivalent circuit for harmonics: a) with separate rotor resistances and rotor reactances and b) with grouped motor resistances and motor reactances

Increasing the order of harmonic causes increased frequency of induced currents in rotor conductors, compared to the one in the short–circuit regime. The skin effect is practically present only in the part of the conductor in the slot of the rotor, i.e. it leads only to an increase of rotor slot resistance (*Rr-sl,h*) and a reduction of rotor slot inductance (*Lr-sl,h*). As the depth of penetration is *∂Al,h*=5 = 4.5 mm, already for the fifth harmonic, *Rr-sl,h* is always equal to *Xr-sl,h,* Fig. 12. For this reason, similar to the corresponding scheme for short-circuit mode (Kostic, 2010), the rotor reactance (*Xr,h*) and resistance (*Rr,h*) are separated into two components in the equivalent circuit for the harmonics, Fig. 11a, i.e.:

$$R\_{r,h} = R\_{r-sl,h} + R\_{r-e,h} \tag{31}$$

$$X\_{r,h} = X\_{r-sl,h} + X\_{r-e,h} \tag{32}$$

where *Rr-e,h* is rotor winding end resistance and *Xr-e,h* is rotor winding end reactance ("e" in the index comes from the abbreviation of the word "end").

Finally, resistance and reactance of stator windings, and resistance and reactance of rotor conductors outside of slots are grouped (Fig. 11b), for which the influence of skin effects can be neglected from the nominal regime to the short–circuit regime. These are:


144 Induction Motors – Modelling and Control

inverse voltage)

whose amplitudes are equal *Uh* = 5%.

(probably for the first time) in Kostic (2010):

with high levels of harmonic voltage (*Uh,i* = 1/*hi*).

practically equally in a short-circuit regime, and

**4.1. Equivalent circuit of induction motors for harmonics** 

**Figure 10.** Electromagnetic torque characteristics (Tem) and the electromagnetic power (Pem) in the following regimes: generator (s < 0), motor (0 ≤ s ≤ 1) and braking (s> 1), as well as at point s = 2 (for the

1. The case that the voltage, containing harmonics of order *h* = 1, 5, 7, 11, 13, 17, ..., 37,

2. As the induction motors are supplied by a rectangular shape of the voltage inverter

The reason for this (new analysis) lies in the fact that it is (wrongly) believed that the resistance of the smaller motor's rotor does not change for higher harmonic frequencies, i.e. that is identical for all harmonics (*Rr,h* = *Rr,*1 = *Rr* = *Const.*), which brings the difference mentioned above – and error. It is shown in Kostic M. & Kostic B. (2011) that values of the rotor resistance and the rotor reactance in a short-circuit regime could be estimated on the basis of the induction motor's catalogue date. Two important conclusions are established

1. that rotor slot resistance and the rotor slot reactance values, *Rr-sl,SC* and *Xr-sl,SC*, are

2. that they have approximately equally values, per unit, for all motors of the same series.

The equivalent circuit for the harmonics is identical to the corresponding equivalent circuit for a short-circuit regime for fundamental frequency *f*1. Only, instead of rotor short-circuit resistance (*Risk*) and rotor short-circuit reactance (*Risk*), the two rotor resistances, rotor end (*Rah*) and rotor slot resistance (*Rr-sl,h*), and two rotor reactances, rotor end (*Xr-e,h*) and rotor slot reactance (*Xr-sl,h*), are used for the harmonics order (h), Fig. 11a (Kostic M. & Kostic B., 2011).

The remaining resistance *Rr-sl,h* and reactance *Xr-sl,h*, are separated (Fig 11).

### **4.2. Parameters of equivalent circuit for harmonics**

In the paper by Caustic (2010) it is shown that values of rotor slot resistance (*Rr-sl,SC*) and rotor slot reactance (*Xr-sl,SC*) in the short-circuit mode are approximately the same for motors of all powers in a series and they are approximately equal to each other, i.e.:

$$R\_{r-sl,sc} = X\_{r,sc} \approx 0.030 \text{ (p.u.)}\tag{33}$$

Their values are within the narrow range of Xr-sl,SC = Rr-sl,SC ≈ 0.027÷0.033 p.u., respectively for the motors of large (> 100 kW), medium (11 - 50 kW) and low power (1 - 7.5 kW).

The explanation is the following: for motors with powers higher than 5 kW (or with relative rotor conductor height ξSC = Hb/∂SC ≥ 1.5), already in short-circuit mode, the rotor induced currents not the entire cross section, or height of rotor bars H, (Kostic, 2010). Current frequencies of individual harmonics in the rotor winding are h times higher (fr,h ≈ h·f1 = h·fr,SC), so the actual depth of penetration of individual harmonic currents (∂h) is √h times lower. Therefore, the corresponding cross section of rotor conductor is √h times lower, so the values of rotor slot resistance are √h times higher and the values of rotor slot inductance are √h times lower as compared to those values for the fundamental harmonic in the shortcircuit regime. In general, rotor slot resistance (Rr-sl,h), rotor slot inductance (Lr-sl,h) and rotor slot reactance (Xr-sl,h), in the function of harmonic order h = f/f1, are shown in Fig. 12.

**Figure 12.** Dependencies of rotor slot resistance, inductance and reactance on harmonic order

If the ratio ξ = Hb/δr,SC ≥ 1.5, then equality Xr-sl,SC = Rr-sl,SC is already true for the fundamental harmonic. As for the harmonics of order h ≥ 5, the relative rotor conductor height is always equal to ξh = Hb/δ<sup>h</sup> ≥ 2, then for motors of all powers is shown in Fig. 12:

$$R\_{r-sl}(hf\_1) / R\_{r-sl,SC}(f\_1) = \sqrt{h} \tag{34}$$

Effects of Voltage Quality on Induction Motors' Efficient Energy Usage 147

, ,1 , 40 *R R R for h sh s s* (39)

*X X hX X sh r eh s r e* , , (41)

*R R RR sh r eh s r e* , , (42)

*R RR R h M h s r e r sl SC* , , (43)

*X X X hX h M h s r e r sl SC* , , (44)

2 2 *Z RX <sup>M</sup>*,*h Mh Mh* , , (45)

, ,, *I UZ I <sup>M</sup> <sup>h</sup>* 100 *Mh Mh N* % (46)

(47)

*r eh re* , , *L L* (40)

The values of penetration depth in the stator copper conductors are *δCu,* ≥ 1.6 mm for frequencies *f* ≤ 2000 Hz. As a rule, since the diameter of the stator winding conductors is *dCu* ≤ 2 mm, it can be assumed that the value of stator windings' resistance keeps almost the same value for all harmonics of order *h* ≤ 40 (or 2000 Hz/50 Hz), i.e. the following equation is

The same assumption approximately applies to the rotor end resistance (*Rr-e,h*) and to the

The total value of motor resistance (*RM,h*), motor reactance (*XM,h*) and motor impedance

The summary value of resistance (*Rs* + *Rr-e*) ≈ *Const* and the summary value of inductance (*Ls* + *Lr-e*) ≈ *Const*, retain approximately the same value in all modes: operating regime, the short-circuit regime and for regimes with harmonics. For calculation values of *RM,h* and *XM,h* by equations (43) and (44), values (*Rs+Rr-e*) and (*Xs+Xr-e*) ought to be determined from a

Harmonics currents (*IM,h*), due to the existence of the corresponding harmonics voltages

Harmonic losses in the motor, which are caused by harmonic currents through the windings of stator and rotor, are calculated using the formula (as a percentage of nominal motor

> 2 , , , 100 % cos *Mh Mh Cu h N R I P P*

 

(*UM,h*), are calculated using the formula (as a percentage of nominal current, %*IN*):

rotor end inductance (*Lr-e,h*). On this basis, the following equations can be written:

(*ZM,h*), for current harmonics, are given in the following expressions:

locked rotor test, and assumed value *Rr-sl, SC= Rr-sl,SC=0.030pu.* 

valid:

power, %*PN*):

$$\mathcal{L}\_{r-sl}(\mathcal{h}f\_1) / \mathcal{L}\_{r-sl,SC}(f\_1) = 1 / \sqrt{h} \tag{35}$$

According to this, it is concluded that the following equations can be written:

$$R\_{r-sl,h} = R\_{r-sl,SC} \cdot \sqrt{h} \tag{36}$$

$$L\_{r-sl,h} = L\_{r-sl,SC} \Big/ \sqrt{h} \tag{37}$$

$$X\_{r-sl,lt} = X\_{r-sl,SC} \cdot \sqrt{h} \tag{38}$$

This means that, based on given values of rotor slot resistance (*Rr-sl,SC*), rotor slot inductance (*Lr-sl,SC*) and rotor slot reactance (*Xr-sl,SC*) for fundamental frequency *f*1 in short-circuit mode, the corresponding parameter values for harmonics *h = fh/f*1 can be calculated, i.e. values: resistance (*Rr-sl,h*), inductance (*Lr-sl,h*) and reactance (*Xr-sl,h*).

The values of penetration depth in the stator copper conductors are *δCu,* ≥ 1.6 mm for frequencies *f* ≤ 2000 Hz. As a rule, since the diameter of the stator winding conductors is *dCu* ≤ 2 mm, it can be assumed that the value of stator windings' resistance keeps almost the same value for all harmonics of order *h* ≤ 40 (or 2000 Hz/50 Hz), i.e. the following equation is valid:

146 Induction Motors – Modelling and Control

The explanation is the following: for motors with powers higher than 5 kW (or with relative rotor conductor height ξSC = Hb/∂SC ≥ 1.5), already in short-circuit mode, the rotor induced currents not the entire cross section, or height of rotor bars H, (Kostic, 2010). Current frequencies of individual harmonics in the rotor winding are h times higher (fr,h ≈ h·f1 = h·fr,SC), so the actual depth of penetration of individual harmonic currents (∂h) is √h times lower. Therefore, the corresponding cross section of rotor conductor is √h times lower, so the values of rotor slot resistance are √h times higher and the values of rotor slot inductance are √h times lower as compared to those values for the fundamental harmonic in the shortcircuit regime. In general, rotor slot resistance (Rr-sl,h), rotor slot inductance (Lr-sl,h) and rotor slot

reactance (Xr-sl,h), in the function of harmonic order h = f/f1, are shown in Fig. 12.

**Figure 12.** Dependencies of rotor slot resistance, inductance and reactance on harmonic order

equal to ξh = Hb/δ<sup>h</sup> ≥ 2, then for motors of all powers is shown in Fig. 12:

According to this, it is concluded that the following equations can be written:

resistance (*Rr-sl,h*), inductance (*Lr-sl,h*) and reactance (*Xr-sl,h*).

If the ratio ξ = Hb/δr,SC ≥ 1.5, then equality Xr-sl,SC = Rr-sl,SC is already true for the fundamental harmonic. As for the harmonics of order h ≥ 5, the relative rotor conductor height is always

This means that, based on given values of rotor slot resistance (*Rr-sl,SC*), rotor slot inductance (*Lr-sl,SC*) and rotor slot reactance (*Xr-sl,SC*) for fundamental frequency *f*1 in short-circuit mode, the corresponding parameter values for harmonics *h = fh/f*1 can be calculated, i.e. values:

1 ,1 ( )/ ( ) *R hf R f h r sl r sl SC* (34)

1 ,1 ( )/ ( ) 1/ *r sl r sl SC L hf L f h* (35)

*RR h r sl h r sl SC* , , (36)

*r sl h r sl SC* , , *LL h* (37)

*XX h r sl h r sl SC* , , (38)

$$R\_{s,h} \approx R\_{s,1} = R\_{s'} \quad \text{for} \ h \le 40 \tag{39}$$

The same assumption approximately applies to the rotor end resistance (*Rr-e,h*) and to the rotor end inductance (*Lr-e,h*). On this basis, the following equations can be written:

$$L\_{r-e,h} \approx L\_{r,e} \tag{40}$$

$$X\_{s,h} + X\_{r-e,h} = h \cdot \left(X\_s + X\_{r-e}\right) \tag{41}$$

$$R\_{s,h} + R\_{r-e,h} = R\_s + R\_{r-e} \tag{42}$$

The total value of motor resistance (*RM,h*), motor reactance (*XM,h*) and motor impedance (*ZM,h*), for current harmonics, are given in the following expressions:

$$R\_{M,h} = \left(R\_s + R\_{r-e}\right) + R\_{r-sl,SC} \cdot \sqrt{h} \tag{43}$$

$$X\_{M,h} = \left(X\_s + X\_{r-e}\right) \cdot h + X\_{r-sl,SC} \cdot \sqrt{h} \tag{44}$$

$$Z\_{M, \text{lr}} = \sqrt{\mathbf{R}\_{M, \text{lr}}^2 + \mathbf{X}\_{M, \text{lr}}^2} \tag{45}$$

The summary value of resistance (*Rs* + *Rr-e*) ≈ *Const* and the summary value of inductance (*Ls* + *Lr-e*) ≈ *Const*, retain approximately the same value in all modes: operating regime, the short-circuit regime and for regimes with harmonics. For calculation values of *RM,h* and *XM,h* by equations (43) and (44), values (*Rs+Rr-e*) and (*Xs+Xr-e*) ought to be determined from a locked rotor test, and assumed value *Rr-sl, SC= Rr-sl,SC=0.030pu.* 

Harmonics currents (*IM,h*), due to the existence of the corresponding harmonics voltages (*UM,h*), are calculated using the formula (as a percentage of nominal current, %*IN*):

$$I\_{M,h} = 100 \cdot \mathcal{U}\_{M,h} \Big/ Z\_{M,h} \text{ (\%}I\_N\text{)}\tag{46}$$

Harmonic losses in the motor, which are caused by harmonic currents through the windings of stator and rotor, are calculated using the formula (as a percentage of nominal motor power, %*PN*):

$$P\_{Cu,h} = 100 \cdot \frac{R\_{M,h} \cdot I\_{M,h}^2}{\eta \cdot \cos \phi} \text{ (\%P}\_N\text{)}\tag{47}$$

Commonly, these power losses are calculated as a percentage of nominal power losses in motor windings (%*PCuN*). Thus, assuming that losses *PCuN* make up one half of the total power losses in the motor, their value can be determined from the formula:

$$P\_{\rm Cu,h} = 100 \cdot \frac{R\_{\rm M,h} \cdot I\_{\rm M,h}^2}{\eta \cdot \cos \phi} \cdot \frac{2\eta}{1-\eta} \left(\%P\_{\rm CuN}\right) \tag{48}$$

Effects of Voltage Quality on Induction Motors' Efficient Energy Usage 149

[%*I <sup>n</sup>* ] [%*P <sup>n</sup>* ] [%*P Cun* ]

*ΣP M,h ΣP M,h* 0.112 - 0.186 4.256 - 2.109

For motors with power 5 kW - 400 kW, in the same order, the ranges of parameter values are




Application of the suggested method is illustrated by Tab. 6. The results show the amounts of increase in power losses due to the presence of harmonics in a given amount (*Ui* = 5%,

*h=f/f <sup>1</sup> U h,i R <sup>s</sup> R r,h R M,h X M,h Z M,h I M,h P M,h P M,h*

5 0.05 0.015 - 0.05 0.067 0.072 - 0.117 0.735 0.739 - 0.744 6.748 0.039 - 0.074 1.482 - 0.839 7 0.05 0.015 - 0.05 0.079 0.094 - 0.129 1.018 1.022 - 1.053 4.822 0.026 - 0.042 0.988 - 0.476 11 0.05 0.015 - 0.05 0.099 0.114 - 0.144 1.579 1.583 - 1.586 3.157 0.014 - 0.020 0.532 - 0.227 13 0.05 0.015 - 0.05 0.108 0.123 - 0.158 2.416 2.419 - 2.421 2.066 0.006 - 0.010 0.228 - 0.113 17 0.05 0.015 - 0.05 0.124 0.129 - 0.174 2.694 2.643 - 2.646 1.891 0.006 - 0.009 0.228 - 0.102 19 0.05 0.015 - 0.05 0.131 0.146 - 0.181 3.249 3.252 - 3.254 1.537 0.004 - 0.006 0.152 - 0.068 23 0.05 0.015 - 0.05 0.144 0.159 - 0.194 3.526 3.534 - 3.536 1.414 0.004 - 0.005 0.152 - 0.057 25 0.05 0.015 - 0.05 0.150 0.165 - 0.200 3.833 3.836 - 3.838 1.303 0.003 - 0.005 0.114 - 0.057 29 0.05 0.015 - 0.05 0.212 0.177 - 0.212 4.080 4.084 - 4.086 1.224 0.003 - 0.005 0.114 - 0.057 31 0.05 0.015 - 0.05 0.167 0.182 - 0.217 4.357 4.361 - 4.362 1.147 0.003 - 0.004 0.114 - 0.045 35 0.05 0.015 - 0.05 0.177 0.192 - 0.227 4.910 4.919 - 4.920 1.016 0.002 - 0.003 0.076 - 0.034 37 0.05 0.015 - 0.05 0.182 0.197 - 0.232 5.187 5.191 - 5.192 0.963 0.002 - 0.003 0.076 - 0.034

**Table 6.** Values of harmonic resistances (*RM,h*), reactances (*XM,h*) and impedances (*ZM,h*) and

Total *THD u =* 7.3% *THD <sup>i</sup>* =9.87%

corresponding currents and harmonic losses for motors > 100 kW (left) and < 5 kW (right), for the given




1 1.00 0.015 - 0.05 0.030 0.045 - 0.080 0.161 0.167 - 0.180

**Example 1** 

0.875,

and the

given (Kravcik, 1982) for:

*Rs,h* =*Rs* =0.015*ZN*÷0.050*ZN* = *Const*.,


*PM,h* [%*PCu,N*] according to (48).

*i* = 5, 7, .., 37) in the supply voltage.

values of voltage harmonics *Uh,i* (p.u.) = 5%
