# Analyzing premature winding failures (part 2)

Jan. 16, 2006
There is a shared responsibility when motors fail.

Thermo-mechanical stresses

Thermo-mechanical stresses in electric motors arise from thermal expansion of the metals used in its construction. The copper in the coils has a coefficient of thermal expansion quite a bit higher than that of the steel (see Table II).

 Material Coefficient of Expansion Copper 9.2 x 10-6/ °F Steel 6.5 x 10-6/ °F Aluminum 3.3 x 10-6/ °F Brass 1.0 x 10-6/ °F

The difference in thermal expansion is made much worse at motor start-up. The inrush current 5 or 6 times the full load current causes a rapid heating of the conductor during the first seconds of excitation. On the other hand, the stator core undergoes little heating during the initial period. The relative growth of the coils with respect to the stator structure causes mechanical stress in the end-turn region where the coil ends are constrained by surge rings and support brackets.

The growth of coils from thermal expansion and subsequent contraction during cooling tends to loosen coil end supports. The number of motor starts strongly influences such failures and  a motor that is started and stopped frequently will be more at risk of an early winding failure. Often plant personnel do not recognize the stresses created by operating procedures. In many cases it is possible to change the procedure to minimize the stress factor.

#### Dielectric stress

Dielectric stress (sometimes called voltage stress) is the stress placed upon a material when a voltage is placed across it. The voltage used in calculating voltage stress is the peak voltage or RMS x √2&RADIC>.

If the coil conductors shown in the figure are at 4,000 volts with 0.075 inches of insulation between coils and core and the core is at ground potential, the nominal stress is calculated as,
stress = 4,000/75 = 53 volts/mil.

This value of voltage stress is fairly typical for electric motors that are rated up to 4,000 V. The value becomes much greater at higher voltages. Motors at lower voltage ratings have proportionately more insulation and work at very low stress levels. Since the nominal working stresses are well under the values of dielectric strength shown for various materials in Table III, it might be assumed that dangerous electric stresses are unlikely to occur in motor coils. However, that is not the case.

The 53 volts/mil calculation assumes that the resulting stress is uniformly distributed across the 75 mils of insulation. This uniformity of distribution can develop only if all of the insulation space is occupied by material of the same dielectric constant. The fact is that a voltage placed across the space between the stator iron and coil conductor material divides in such a way that the stress on each insulating material is inversely proportional to the dielectric constant for the particular material. If the space is filled by mica insulation with air voids, the electric stress on the air can easily exceed its low break-down strength with resulting discharges.

#### Partial discharges in electric motors

Most people have observed an electrical discharge. Such a discharge occurs whenever switch contacts interrupt current into a circuit. It also happens when two exposed and energized conductors contact each other. In general, an arc causes severe damage on surfaces where it impinges. Switch contacts must be made from special metal alloys in order to withstand repeated arcing. Surfaces not designed to withstand an arc but nevertheless subjected to one incur heavy damage.

Discharges may also occur between surfaces if one or both are insulated. Such a discharge is called a partial discharge. It is different from an arc between two conductors because the insulation on at least one surface reduces current to a very low value. However, it does have a similar effect of damaging the surfaces upon which it impinges although at a much slower rate in proportion to the magnitude of current.

A partial discharge occurs when the dielectric strength of an air space between two charged surfaces is exceeded. The voltage that will cause a discharge to occur can be calculated by the combination of air-gap and insulation thickness. This value of voltage is called the inception voltage. It is calculated by the expression,

Vi = (1+ t⁄&FRASL>kg)Vp

where Vi is the inception voltage (the voltage that will cause a partial discharge), k is the dielectric constant of the insulating material, t  is the total thickness of the insulation material, g is an air-gap between the sides of two charged surfaces, Vp is the Paschen voltage corresponding to an air-gap g.

#### Slot discharge

A partial discharge can occur between the side of a core slot and a loose coil side if an air-gap exists there. The stress is greatest on those coils subject to the highest voltage. This means that the first coil position (the coil connected to the incoming line) in each coil group is most vulnerable to slot discharges.

A discharge between the coil and slot does not constitute a ground fault. The level of current is too minute to trip ground fault protectors. However, there may be many discharges during each half cycle that the coil voltage exceeds the inception voltage. This produces a discharge pulse frequency of several thousand per second. The cumulative effect of millions of discharges is to eat a hole in the insulation wall.

The coil sides in the slots can be protected by coating them with a semi-conductive finish. The semi-conductive finish effectively puts the outside of the coil at ground potential as long as at least a part of the coil contacts the core. Discharges may occur if the protected coil loses contact with the side of the slot or if the protective surface is scraped off portions of it.

#### Internal discharge

Partial discharges can also occur in the interior of the coil if voids are present. Such discharges are caused by the same mechanism that causes slot discharge, that is, exceeding the dielectric strength of air. Internal partial discharge is at the same time more difficult to detect and more dangerous to the life of the motor. It is more difficult to detect because visual detection is impossible. Internal voids are not protected against even if the coil outer surface is well grounded. The voltage stress still occurs at the void with a resulting discharge across the void. The only protection is to eliminate voids. Internal partial discharge can greatly shorten the time to failure if the void is near the conductor There may be little insulation to erode until conductors are exposed. If the void is adjacent to two or more conductors and the conductor insulation is penetrated, a turn-to-turn short occurs.

The turn-to-turn short allows very high values of circulating current that, in turn, generates a lot of heat and causes a rapid decomposition of organic insulating materials. The result is a build-up of gas pressure and an explosive failure with a portions of the ground wall insulation blown away.

Formed coils that are pressed and cured before inserting into the slots have well consolidated insulation. The risk of having voids in the insulation wall in the straight part of the coil is, therefore, greatly minimized. However, the end-turn portion of the coils cannot be well consolidated prior to assembly into the stator. The bent part of the coil usually has voids. If the external surface becomes conductive, discharges result.

An external partial discharge usually takes years to develop into the final catastrophic failure. Internal voids, depending upon their location relative to the conductor, cause partial discharges that cause failure in a relatively short time (hundreds of hours).

#### Surge voltage effects

The normal voltage waveform for industrial power is the 60 Hertz commercial frequency. At 60 cycles per second, it takes a bit more than 0.004 second to go from zero voltage to peak voltage. Four milliseconds may seem to be a very short time. However, it is a very long time compared to the travel time of the electrical effect through a motor.

The speed with which electricity travels depends upon the substance of the space surrounding it. For example along a high voltage line in the air, electricity travels at the speed of light, a thousand feet in a millionth of a second. In a buried cable with a high value of capacitance, the advance of the electricity is slowed as it charges the cable. In this case, the speed is about half the speed of light. In an electric motor, the advancing electrification must also create a magnetic field in the core material. This process greatly slows the advance so that the speed through the motor winding, is about 25 feet in a millionth of a second.

An electric motor may have 1,000 feet of conductor in a phase winding. At 25 feet per microsecond, it takes 40 millionths of a second for the electricity to traverse the entire winding. Compare the travel time through the winding to the rise time of  0.004 second at 60 Hertz. The rise time of a 60 Hertz voltage is obviously very great compared to the travel time. What this means is that the voltage wave form is at essentially the same point through the winding. With such an arrangement, the voltage drop across each coil in a phase will be the same, i.e., if there are 24 coils connected in series in a 4,000 volt wye connected winding the RMS voltage measured across each coil will be,

If each coil consists of three turns, the turn-to-turn voltage is 96⁄&FRASL>3 = 32 volts. With 6 mils of turn insulation, the electric stress is E = 32⁄&FRASL>6 = 5.4 volts/mil.

Such a voltage stress does not come close to challenging the strength of the insulation. However, the internal voltage effects are altered greatly if the wave form is not a 60 Hertz sinusoidal waveform. For example, if the contacts on the motor controller happen to close when a voltage is at or near its peak value, then the peak voltage is transferred almost instantaneously to the motor winding. The winding does not see the gradual build-up of a sinusoidal voltage. The effect is the same as that obtained from a lightning strike. This is a surge voltage  a voltage that reaches the motor winding with a nearly zero rise time. There is actually some voltage surging every time contacts open or close.

Depending upon the connection between the switchgear and the motor, overhead line, buried cable, or duct bar, the wave front of the voltage surge has some rise time but it might be less than 1 microsecond. Going back to the speed at which the voltage penetrates the winding, it takes about 40 microseconds to travel through 24 coils, or about 1.67 microseconds per coil. At this rate of penetration, with a rise time of 1 microsecond the voltage into the line lead coil is already at its peak value while the output lead is still at zero voltage. The full incoming voltage falls across the first coil. If there are three turns per coil, the voltage is dissipated across two turn insulation interfaces. If the conductor of a 4,000 volt motor has 6 mils of insulation, the input voltage is divided by 6 x 2 mils of insulation. The resulting stress is,

The surge voltage stress on the turn-to-turn insulation in this case is about 50 times the continuous stress value. Frequently the voltage surge is several times the normal magnitude and the voltage stress may exceed the strength of the material. In that case, a turn-to-turn failure occurs. The turn insulation failure goes unnoticed but ultimately develops into a catastrophic ground fault with possible severe damage to the structural parts of the motor.

#### Conclusions

It is not always easy to determine who is at fault when premature failures occur. The motor repair/rebuilder has a responsibility to ensure the quality of construction with voidless coils, coil tightness in slots, and so forth. The vendor should be accountable for failures traceable to shortcomings in these areas even if the failure occurs after a warranty expires.

On the other hand, the motor user also must accept responsibility for those factors under his control. These include installation (surge protection), operation (limiting motor starts), and maintenance (cleaning stator coils periodically).

#### Electrical properties of insulation

When considering the role of electrical stresses in causing early winding failures, understanding the properties of the insulating materials is essential. Following are definitions of the important properties.
Dielectric strength is specified in units of volts per mil. It is the voltage stress the material can withstand before breaking down.

Permittivity of free space is 2.25 x 10-13  farads per inch. This parameter is a measure of the electric charge that can occupy free space (air).

Relative permittivity, often called dielectric constant, is specified as a simple number in the range of 1 to 7. The number is a measure of the amount of electric charge that can be stored in a particular material as compared to air. For example, a mica capacitor with a dielectric constant of 6 can store six times the charge of an air capacitor of the same size.

Volumetric resistivity of a material is specified in terms of ohm-inches. Resistivities of insulation materials are characterized by very large numbers. For example, the resistivity of paper, which is low among the insulation materials is 108 ohm-inches, a quadrillion times greater than that of copper.

Table III show a number of commonly encountered insulation materials with values for the properties defined in the preceding paragraphs.

 Dialectric Constant Dielectric Strength Material Resistivity Air 1 75 Infinite Insulating Paper 2 500 108 Polyethylene Film 2.3 1,000 1017 Polycarbonate Film 3 600 1015 Epoxy Resin 3.3 500 1014 Molded Phenolic 4 200 1011 Porcelain 6 900 1015 Mica 6.5 1,000 1012

The insulation resistance can be approximated for any motor with clean dry windings using the formula

R = (rho x t⁄&FRASL>A) x 10-6

where R is the insulation resistance in megohms, rho is the resistivity for the particular insulation material, t is the insulation thickness between conductor and ground, A is the total slot area between conductors and the core (do not include duct area in core)

#### Sample calculation:

Calculate the insulation resistance of a 9,000 hp. The core has 72 slots and is 40 inches long with 16 air ducts 0.375 inches wide. The slot is configured as in the figure. Use 1014 for resistivity.
The total slot area A= [(4 x 1.1) + 0.68] x  72 x [ 40 - (16 x 0.375)]
= 12,435 square inches
R = (1014 x 0.165⁄&FRASL> 12435) x 10-6 = 1,327 megohms

As a winding becomes dirty with surface contamination, the measured insulation resistance may reduce to values of a few meg-ohms.

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