A QUANTITIES METHOD OF INDUCTION MOTOR UNDER UNBALANCED VOLTAGE CONDITIONS

The most complete definition for the voltage unbalance is using complex voltage unbalance factor (CVUF) that consists of its magnitude and angle. Unfortunately, the definition did not distinguish between undervoltage and overvoltage unbalance conditions. In this paper, t he analysis of the motor is performed using the method of symmetrical component and MATLAB software is used to investigate the performance of induction motor. The simulation results show that the International Electrotechnical Commission (IEC) definition of the voltage unbalance can be applied to evaluate total copper losses precisely. However, the phase angle of unbalance factor must be included for accurate predicting of peak currents and peak copper losses of the phase windings of the motor.The unbalanced conditions which distinguish between under and over voltage unbalance must also be taken into consideration for assessing all the quatities.


INTRODUCTION
Voltage unbalance is a power quality problem.It is a common phenomena found in a three-phase power system.Although three-phase voltage supply is balanced in both magnitude and phase-angle at generation and transmission levels, the voltages at distribution end and utilization side can become unbalance.It is practically impossible to be obviated due to the uneven distribution of single-phase loads in the three-phase supply system and asymmetry of transmission line and transformer winding impedances.
Such condition has severe impacts on the performance of an induction motor.This motor is designed and built to be able to tolerate only a low degree of voltage unbalance and has to be derated if the unbalance is excessive.
The negative impacts of voltage unbalance on an induction motor have been studied in depth [1]- [6].The voltage unbalance can increase stator and rotor losses of the induction motor, rise windings temperature and reduction the insulation life caused by overheating.The motor need to be derated, reducing its output horsepower so it can tolerate the extra heating imposed by the unbalanced voltage supply.According to this issue, Influence of magnitude of unbalance voltage factor on efficiency and motor losses has been investigated in [7].
In most previous studies, the method of evaluating the degree of unbalanced voltage is based on either National Electrical Manufactures Association (NEMA) standard or International Electrotechnical Commission (IEC) definition.Both the NEMA and the IEC definitions are only considering "magnitude" of voltage unbalance to describe the degree of unbalanced voltage.The three-phase voltage supply has not only magnitude but also phase-angle that giving contribution to the voltage unbalance.Hence, both definitions lead to a comparatively large error in predicting the performance of the induction motor [8].
A more precise approach in predicting the performance of an induction motor operation with unbalanced three-phase voltage is using complex quantity to specify the degree of voltage unbalance.This definition is also known as the complex unbalance voltage factor (CUVF) in some literatures.The CUVF is an extension of the IEC definition, first time evaluated by Wang [9], which consists of magnitude and angle of the voltage unbalance.However, Wang [9] did not distinguish between undervoltage and overvoltage unbalance conditions.
In this paper, the stator and rotor losses of an induction motor and its peak currents are studied precisely.The angle of the CUVF is taken account to estimate the peak currents and losses of the induction motor operating with both under-voltages unbalance and over-voltage unbalance conditions.The symmetrical component theory approach is used to analyze the operation of induction motor under such conditions.MATLAB program is also used for computer simulation.Finally, recommendations are suggested for accurate calculating the quantities of the induction motor operation under unbalanced voltage conditions.
Analysis of an induction motor operating with unbalanced voltage supply using symmetrical component approach [10] requires positive and negative sequence equivalent circuit of a three phase induction motor representing in Figure 1.Each circuit performs to both positive and negative sequence circuits.The only difference between the circuits is the load resistance defined by positive and negative sequence slips.Positive sequence slip is s 1 =s, negative sequence slip is s 2 =2-s and slip s is as shown equation ( 1), where n s is synchronous speed and n r is rotor speed.2), where i=1 for positive sequence and i=2 for negative sequence.
The sequence line voltages can be transformed directly to the sequence phase voltages by equation ( 4), where t= 3 1 * exp(-jπ/6), and its conjugate is t*= If the sequence phase voltages are applied to the sequence equivalent circuits (Figure 1), then the sequence currents can be computed.Positive sequence stator and rotor currents are shown in equation ( 5) and ( 6), and negative sequence stator and rotor currents are shown in equation ( 7) and (8). .
Calculating stator and rotor losses requires phase stator and rotor currents.The phase currents are determined by performing the transformation back.Transforming the stator currents using Fortescue matrix as shown in equation ( 9), and similarly the rotor currents of the induction motor can be calculated by transformation as shown in equation ( 10).In ( 10) and ( 11), there are no zero sequence currents (I s0 =I r0 =0) for the motor connecting in delta or ungrounded wye.Furthermore, stator copper losses of the motor can be calculated by equation ( 11) and its rotor copper losses is given by equation ( 12).Input active power of the motor is as shown equation ( 13) and and its input reactive power is defined by equation ( 14 As the motor has no core and mechanical losses, output power is As no core and mechanical losses, output power is The motor output torque is
The CVUF is defined as shown equation ( 20), where k v is the magnitude and θ v is the angle of the CVUF.
Let U sab , U sbc and U sca be the line voltages of stators.Their symmetrical components are given by (3) mentioned above.In contrary, the line voltages can be obtained from their positive and negative sequences, but minus the zero sequence component (U s0 =0).
By the CVUF definition, As described in [8], if the ratio of the line voltages is defined by Then the relations between (k u , θ u ) and (x,y) according to ( 19), ( 20) and ( 21 22) and ( 23) the relations between (k u ,θ u ) and (x,y) can be shown in x-y plane (Figure 2).The loci of constant k u looks like ellipses while the loci of constant θ u looks like lines.
Further more, the definition of the CVUF can be viewed more closely as a cylinder in three-dimensional.As example, the variations of terminal voltages (in per-unit value) for k u =5% with phase angle in 0-360° range is shown in Figure 3.
There are infinite combinations of terminal voltages of the induction motor for k u =5%.Every point of the outer surface of cylinder will give the same level of voltage unbalance.The large range of variation of the terminal voltages for a given value of voltage unbalance can be reduced by considering the phase angle θ u and the definition of under and over voltage conditions.For instant, the variations of terminal voltages under a 0% voltage unbalance is shown in Figure4.Considering to the definition above, there are two sections of the graphic.The first, k u =0% is for undervoltage if U ab , U bc , U ca < 1 and the other, k u =0% is for overvoltage if U ab , U bc , U ca > 1.
In this work, magnitude of the voltage unbalance factor of 0%, 1%, 2%, 3%, 4%, 5% and 6% are used for computation.Undervoltage condition is defined by 0.7≤ f < 1.0 and overvoltage condition is 1.0 < f ≤ 1.4.Torque of the machine is kept constant during computation process, T out =40.3578 N/m 2 for calculating peak current and peak losses to obtain rate current and T out =33.8046 N/m 2 for calculating total copper losses.Matlab programming is employed to calculate the currents and losses by using equations ( 1) -( 29).

RESULTS AND ANALYSIS
Faiz et.al in [7] reported that copper losses of the induction motor increases related to magnitude of voltage unbalance factor.They did not investigate the influence of phase angle of the unbalance factor.Here, Figs. 5 and 6 describe the effects of phase angle of the voltage unbalance factor on stator and rotor copper losses.The phases a, b and c of the stator losses (P isa , P isb , P isc ) and the rotor losses (P ira , P irb , P irc ) vary from θ v =0° to 360° for a constant k v =6% and output torque=40.3578N/m 2 at slip s=0.0224.The losses in the three phases are extremely non-uniform under the unbalance voltage condition and vary significantly with θ v , but total of stator and rotor losses remain constant and are not dependent on the phase angle of unbalanced factor.The peak losses of each phase refer to peak currents.
The peak current of stator occurs when the phase angle of unbalanced factor is The phase angle of the unbalanced factor for the rotor peak current is where CD 1 and CD 2 are positive and negative sequence current dividers and defined by If at an angle of θ v the current in phase-a has a peak value, then in the next 120° interval (= θ v + 120°), phase-b would have a peak and phase-c would have its peak current in the subsequent 120° interval (= θv + 240°).Once a phase has the value of peak current above the rate current, it will lead to excessive heating in the motor's windings.The motor is not allowed to operate at above the rated current or it will harm to the stator and rotor windings.
Table 1 shows variations of peak losses and peak currents occurred in the rotor and stator phases of the induction motor operation under k v =6% and output torque=40.3578N/m 2 with considering over or under voltage unbalance conditions.Comparing Tables 1 and 2, it can be found that peak losses and peak current in one of stator and rotor phase windings become very high if the induction motor is operated with either over voltage unbalance or under-voltage unbalance conditions.Under the same value of k v , the under-voltage unbalance gives higher peak losses and peak currents than the three-phase over voltage unbalance does.A large value of unbalanced may create the current flowing in one of the phase windings exceeds the rated value.This condition will damage the insulation of the motor's windings and may shorten the motor's life for long-term operation.Precise derating for induction motors with unbalanced voltage investigated by Faiz et.al. in [8].They found that both the NEMA and IEC definitions lead to large error in predicting the performance of the machine when operation with unbalanced voltage conditions.It is true for calculating the motor derating factor.Moreover, they [8],[9] did not define unbalanced condition to distinguish between over and under voltage unbalance to obtain more accurate results.Figure 6 shows that the IEC definition of voltage unbalance combined with the unbalanced condition can be applied to evaluate total copper losses precisely.The total copper losses are summation of the stator and rotor copper losses and do not depend on the phase angle of voltage unbalance.At the same degree of voltage unbalance, the total losses depend upon unbalance conditions.As seen in Figure 6, the motor operates in undervoltage unbalance condition gives greater losses compared to it is in over voltage unbalance condition at k v =0%, 1%, 2%, 3%, 4%, 5% and 6%.

CONCLUSIONS
The degree of voltage unbalance is a quantity required in studying the performance of a three phase induction motor operating with an unbalanced voltage supply.It has been asserted that the prevailing operation conditions of the motor are not accurately assessed without knowing the phase angle of unbalanced factor and the voltage unbalance conditions (under or over voltage unbalance).
The IEC definition of voltage unbalance combined with the unbalanced condition can be applied to evaluate total copper losses precisely.However, the phase angle of unbalanced factor must be included for accurate predicting of peak current and peak copper losses of the phase windings.

FigureFigure 2 .
Figure 1.Sequence equivalent circuit for the positive and negative sequence circuit are shown in equation (

Figure 4 .
Figure 3. Terminal Voltage of a Three-Phase Induction Motor with k u =5%

Figure 6 .
Figure 5. Variations of phase a, b and cRotor Losses with θ v

Table 1 .
Comparison of peak losses and peak current under k v =6% and T out =40.3578 N/m 2

Table 2 .
Peak losses and peak current under balanced voltage and T out =40.3578 N/m 2