Causes and Effects of Single-Phase Induction Motors

Causes and Effects of Single-Phasing Induction Motors

W. H. Kersting, Life Fellow, IEEE

Milsoft Utility Solutions

Abstract-- It is well known that a three-phase induction motor will continue to operate when a disturbance of some sort causes the voltages supplied to the motor to become single-phase. The single-phasing can occur as a result of a fuse blowing or protective device opening on one phase of the motor. Other possibilities include feeder or step-down transformer fuses blowing. Even though the motor will continue to operate in this condition, the motor will heat up very quickly and it is essential that the motor be removed from service by the opening of a motor circuit breaker or some other type of protective device.

This paper will describe three different ways in which an induction motor will operate in a single-phase condition. For purposes of this paper single-phase will include any condition in which the three line-to-line voltage phasors appear on the same line.

Index TermsInduction motors, single-phasing, losses, distribution systems, transformer connections

I. INTRODUCTION

When balanced or slighting unbalanced voltages are applied to a three-phase induction motor the motor will provide power to the shaft load. The analysis of an induction motor under a steady-state operation is well documented in the literature. [1], [2] The motor will operate within its rating when the voltages are balanced. When the voltages become unbalanced excessive heating will occur and the motor will have to be de-rated. [3]

A three-phase induction motor operating in the steady-state will continue to operate when a disturbance on the system causes the terminal voltages to become single-phase. This condition is referred to as single-phasing and will result in an operating condition that produces excess heating in the motor. Such a condition requires that the motor be provided with protection that will disconnect it from the system before the motor is permanently damaged.

The single-phasing can be the result of three possible disturbances on the system.

Open phase on the substation transformer primary

Open phase on the primary of the distribution step-down transformer bank

Open phase on the terminals of the motor

This paper will demonstrate how these three conditions lead to single-phasing of the motor. The paper will go into detail on the analyze of the most severe of these conditions, the open conductor at the terminals of the motor.

W. H. Kersting is Professor Emeritus of Electrical and Computer Engineering at New Mexico State University. He presently is a consultant to Milsoft Utility Services, Abilene, TX 79608.

(bjkersting@zianet.com)

II. THE SYSTEM

Figure 1 shows the system to be studied. The phasor diagrams of the voltages are sketched for a no-load condition in order to show the normal phase relationships on the three parts of the system.

Figure 1 The System

In Figure 1 the phasor diagrams demonstrate the standard 30 degree phase shift as the voltages are stepped down. Assume that the substation transformer voltage rating is 69,000 Delta 12,470 Wye and the distribution transformer bank voltage rating is 12,470 Wye 480 Delta. The resulting voltages under no-load will be:

Substation primary:

Assuming balanced conditions, the remaining voltages would have the normal 120 degree phase shifts. The analysis of the motor for all loading conditions can follow the method described in Reference [2].

The opening of a fuse on the primary side of the substation transformer is shown in Figure 2. Again the voltage phasors are displayed in this figure.

Figure 2 Open High Voltage Phase

With the open phase on the high voltage terminals, the voltages on the various windings, as shown in the phasor diagrams are:

Substation Primary:

Feeder Primary:

Motor Terminals:

Notice that the voltages applied to the motor have no phase shift between them. Therefore, the motor can not start under this condition but if it has already been running, then it will continue to run.

The opening of a fuse on the primary of the distribution step-down transformer is shown in Figure 3. The voltage phasors are displayed in this figure.

Figure 3 Open Phase on Step-Down Transformer

For this condition the line-to-line and line-to-ground voltages out of the substation are those of the original steady-state condition. Because of the open (shown here for phase-a), the voltage phasors for the step-down transformer are:

Step-down primary:

Motor terminals:

As in the previous case, the motor will not start with these applied voltages but will continue to run if the opening occurs during steady-state operating conditions.

Both the open phase on the substation transformer and the open phase on the step-down transformer can be analyzed using the method of Reference [2].

III. OPEN-PHASE MODEL

Figure 4 illustrates the open phase condition. In this figure and in the model development it is assumed that phase-a has been opened [1].

Figure 4- Open Phase-a Serving an Induction Motor

In Figure 4 the following definition of terms are:

V12, V23, V31 = system line-to-line voltages

Vsab, Vsbc, Vsca = operating line-to-line voltages at the motor terminals

Vsan, Vsbn, Vscn = stator line-to-neutral voltages

Isa, Isb, Isc = line currents

For this operating condition the line currents must satisfy:

(1)

The symmetrical component sequence currents therefore are:

(2)

It is important to recognize in Equation 2 that:

(3)

In order to satisfy Equation 3, the positive and negative sequence networks for the induction motor must be connected as shown in Figure 5.

Figure 5 Sequence Network Connection

In Figure 5 note that the input voltage is given by:

(4)

Substitute the expansion of the positive and negative sequence voltages in terms of the phase line-to-neutral voltages:

(5)

The result of Equation 5 is very important. Recognize that the voltage input to the connection of the two sequence networks is a function of the actual line-to-line terminal voltage . Since the motor is connected to a three-wire delta line, it is good that the needed voltage is a known line-to-line voltage.

IV. CIRCUIT ANALYSIS

The analysis of the circuit of Figure 5 is best accomplished by first computing the equivalent positive and negative sequence impedances at the terminals of the networks. In order to do this, a value of slip must be assumed.

Assume the positive sequence slip:

Then the negative sequence slip is given by:

(6)

The load resistances of the two networks are:

(7)

Note in Equation 7 that the negative sequence load resistance will be negative since the positive sequence slip is always a value of one or less. Typical positive sequence slip values for a motor operating near its rating will range from 0.02 to 0.04.

The general equation for the input equivalent impedance at the terminals of the networks is given by:

(8)

where: i = 1 for positive sequence

i = 2 for negative sequence

The stator sequence currents are then:

(9)

Because the motor is connected to a three-wire delta line, the zero sequence stator current must be zero.

The stator input line currents are computed by:

(10)

In general form Equation 11 is written as:

(11)

Referring to Figure 2, the sequence rotor currents are determined by:

(12)

Motor Terminal VoltagesThe sequence line-to-neutral stator voltages are:

(13)

The equivalent line-to-neutral stator terminal voltages are:

Or:

(14)

The line-to-line stator voltages are:

(15)

It needs to be pointed out that the stator line-to-line voltages will not be the same as the secondary line-to-line voltages. Because of this a voltage will appear across the switch (fuse) and is given by:

(16)

Converted Power

The total converted sequence powers are given by:

(17)

Recall that the negative sequence load resistance is a negative number so that the converted negative sequence power will be negative which adds to the effective rotor power loss.

The total converted power is:

(18)

Stator and Rotor Power Losses

The total stator and rotor power losses are:

(19)

Total complex stator power input:

(20)

V. EXAMPLE

To illustrate the effects of the various single-phasing conditions, the following induction motor will be analyzed:

150 kVA, 4-pole, Wye connected, 480 volt

watts

per-unit

per-unit

per-unit

The motor will be analyzed using a slip of 0.0365 for the following operating conditions:

1. Balanced three-phase terminal voltages

2. Open phase-C on the substation transformer

3. Open phase-a on the step-down transformer

4. Open phase-a on the motor terminals

The motor impedances are converted to ohms using the kVA and voltage ratings of the motor as the bases. For a the specified slip, the sequence input motor impedances are:

Table 1 Summary of Analyses

ParameterCase-1Case-2Case-3Case-4

Motor LL Voltages

V12480/0240/00359.4/-1.0

V23480/-120480/-180415.7/-150480/-120

V231480/120240/0415.7/30438.5/81.8

Motor LN Voltages

V1n277.1/-300138.6/-150214.1/-41.8

V2n277.1/-150240/180138.6/-170.7241.9/-145.7

V3n277/90240/0277.1/30282.1/82.8

Stator Currents

Ia179.5/-63.5477.1/11.7462.5/82.50

Ib179.5/176.5628.6/135.3564.2/-170.7274.1/-158.7

Ic179.5/56.5539.2/-92.1617.6/-36.5274.1/21.3

Rotor Currents

Ia154.9/-37.2495.3/10.8447.2/78.266.4/48.2

Ib154.9/-157.2601.8/139.4571.3/-167.6276.3/-144.3

ic154.9/82.8485.5/-93.5562.9/-34.0212.0/31.8

Input Complex Power

kVA149.2256.7256.7131.6

Power Factor0.8340.3990.3990.781

Losses-kW

Stator4.9947.1547.157.75

Rotor4.3651.0351.037.6

Total9.3598.1898.1815.35

Converted Power-kW115.024.254.2587.35

Shaft HP149.81.341.34117.1

Table 1 summarizes the results of the analyses for the four operating conditions. Case 1 is the motor operating in the steady-state with balanced three-phase voltages applied. Note that the input kVA and shaft horsepower are approximately equal. A usual approximation is to assume the two are equal. The total winding losses for this condition are 9.35 kW.

Case 2 is the analyses of the operating conditions when phase-C on the substation transformer is opened. As seen from the Table, this is a very severe condition on the motor. The shaft output power is greatly reduced (1.34 HP) and the losses are 10 times greater than Case 1. The increased losses will result in a great heating of the stator and rotor windings leading to burn-out if the motor is not disconnected quickly from the source.

Case 3 occurs when the phase-a fuse on the step-down ungrounded wye-delta transformer bank blows. It is interesting to note that there is little if any difference in the operating conditions compared to Case 2. It is not clear if this will always be the case or just because of the particular parameters used in the example.

Case 4 occurs when phase-a at the motor terminals is open. This is the true single-phasing case. Note that this operating condition isnt as severe as Cases 1 and 2. While the additional losses are not significantly greater than Case 1, they are large enough to cause excessive heating of the

windings. Again, the motor must be disconnected from the system.

It is somewhat surprising that the most severe operating conditions are the opening of phases either at the substation transformer or the step-down transformer. In the case of the open phase on the substation transformer the protective scheme for the transformer should take the transformer out of service very quickly. If that protective scheme fails, many induction motors on the feeder are going to be subjected to a very severe heating problem that may result in many motor burn-outs. In the case of the open phase on the step-down transformer there will normally not be an overall protective scheme that will disconnect the transformer bank from service. For this case only the motors connected to the secondary of the bank will experience over heating.

The three single-phasing cases demonstrate the critical need for a protective scheme that will disconnect the motor during any of these cases. The need for the protective scheme is also needed when the supply voltages become unbalanced. [3].

VI. CONCLUSIONS

Three different ways in which an induction motor can operate in the single-phase condition have been described and analyzed. For purposes of this paper single-phasing is assumed whenever the three terminal line-to-line voltages at the motor terminals lie on a straight line. These phasors are displayed in Figures 1,2 and 3.

For the cases of the open phases on the substation transformer or the step-down transformer, the method of analysis follows that of Reference [2]. The open phase at the terminals of the motor requires a different method of analysis and is developed in the paper.

The results of the example problem are summarized in Table 1. These results illustrate the great increase in winding losses and the need for a protective scheme that will disconnect the motor during a single-phase condition.

References

1. Wagner, C.F. and Evans, R. D. , Symmetrical Components, McGraw-Hill, New York, 1933.

2. Kersting, W. H., Phillips, W. H. Phase frame analysis of the effects of voltage unbalance on induction machines, IEEE Transactions on Industrial Applications, March/April 1997.

3. Kersting, W. H. , Causes and effects of unbalanced voltages serving an induction motor, IEEE Transactions on Industrial Applications, January/February 2001.

W. H. Kersting (SM64, F89) was born in Santa Fe, NM. He received the BSEE degree from New Mexico State University, Las Cruces, and the MSEE degree from Illinois Institute of Technology. He joined the faculty at New Mexico State University in 1962 and served as Professor of Electrical Engineering and Director of the Electric Utility Management Program until his retirement in 2002. He is currently a consultant for Milsoft Utility Solutions. He is also a partner in WH Power Consultants, Las Cruces, NM.

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