SIZING TRANSFORMERS WITH LARGE MOTOR LOADS (PART II)

LESSONS WE NEED TO LEARN

SIZING TRANSFORMERS

WITH LARGE MOTOR LOADS

(PART II)

By Doods A. Amora, PEE
(December, 2006)

(Last of Two Parts)

6.0: VOLTAGE SAGS IN LARGE MOTOR STARTING

The voltage dip at the transformer terminals is proportional to the motor load required in start-up. As discussed earlier, the voltage drop can be expressed as a percentage of the inrush motor load compared to the maximum capability of the transformer.

% Voltage Drop = (Motor Starting kVA) x 100 /(Motor Starting kVA + Short Circuit kVA)

EXAMPLE:

Now, supposing a lone 2,000 hp, 4.16 kV, Code Letter J motor (Starting kVA = 7.6 times its rated hp) is connected to the 7,500 kVA transformer described above, would it not be that the transformer is too large for the motor load? Probably yes, probably not! Let us see why.

A) IF THE MOTOR CONTROLLER IS ACROSS-THE-LINE (FULL-VOLTAGE CONTROLLER):

Starting kVA of 2,000 hp Motor (SkVA = 7.6):
Starting kVA of 2,000 hp motor = 2,000 kVA x 7.6 or 15,200 kVA

Three-Phase Short Circuit Capacity of the Transformer:
3Φ SC kVA of 7,500 kVA Trafo = 7,500 kVA/0.08 or 93,750 kVA

Note: In this case, the 7.5 MVA transformer has a maximum of 93.75 MVA
short circuit capability.

The voltage drop on motor inrush will be:

VD at Trafo Terminals = 15,200/(15,200 + 93,750) = 0.1395 or, 13.95%
VD at Trafo Terminals = 580 v
Trafo Terminal Volts During Motor Start-Up = 3,580 v

The transformer output voltage will drop from 4,160v to 3,580v (4,160 x 0.8605 = 3,580 v) as against a minimum requirement of 3,600v for a 4,000v motor. Thus, we can see that the transformer even at 7.5 MVA is small for a 2,000 hp motor!

Therefore, the transformer must be sized to a short circuit capability equal to or greater than 15,200 kVA times 10, or, 152,000 SC kVA in order to have a voltage drop of 10% or less.

The next higher standard size transformer at 10,000 kVA 8.0% impedance would have a short circuit output capability of 125,000 kVA which is still not sufficient. Or a 12.5 MVA transformer could be the choice!

In this particular application, the ratio of the selected standard size transformer kVA to motor kVA is 12,500 kVA/2,000 kVA = 6.25. Thus the transformer rating is 625% larger than the rating of the motor! Fantastic!

Note that the voltage sag in reality will even be bigger (actual voltage smaller) because the source can never be infinite in fault duty plus in addition, the voltage drop at the distribution cables from the transformer to the motor end. The total voltage sag is the sum of the sag in the secondary unit substation transformer and the secondary circuit. In the case of very large motors of several hundred to a few thousand horsepower, the impedance of the supply system (or the ‘available fault duty’) should be considered.


B) IF THE MOTOR CONTROLLER IS AN AUTOTRANSFORMER REDUCED VOLTAGE STARTER:

Now therefore, here comes the wisdom in using reduced voltage starting scheme and the choice of Code Letter specification of a motor. Normally in practice, large motors need to be equipped with Auto-Transformer RVS controllers while the motor Code Letter specification can be chosen.

Let us now see the effect with a 2,000 hp Code Letter E motor (Starting kVA = 4.8 times rating) is controlled by an Auto-transformer RVS. Establish further that the Auto-transformer RVS shall be at 65% tap.

From the table below, the line current at 65% tap draws 46% of the full voltage starting current.


Table 2: MOTOR STARTING METHODS FOR SQUIRREL CAGE INDUCTION MOTORS:
(Table re-configured for easy reference) in this blog:

 Full Voltage : 100%: Line Starting Current or kVA

 Reduced Voltage (RVS Autotransformer Type):

80% Tap: 68% of Line Starting Current or kVA of 100% of Full Voltage Starting
65% Tap: 46% of Line Starting Current or kVA of 100% of Full Voltage Starting

 Reduced In-Rush:

Wye-Delta: 33% of Line Starting Current or kVA of 100% of Full Voltage Starting

NOTE: In addition to methods listed above, users should consider solid-state soft-start motor controllers and/or adjustable speed drives.


In this example:
Motor Starting kVA = 46% of 2,000 kVA x 4.8 or 4,416 kVA
Motor Starting kVA can therefore be placed at 4,416 kVA.

The voltage drop at transformer terminals on motor inrush will be:

VD at Trafo Terminals = 4,416/(4,416 + 93,750) = 0.045 or, 4.5%
VD at Trafo Terminals = 187.2 v
Trafo Terminal Volts During Motor Start-Up = 3,972.8 v

The transformer output voltage will drop from 4,160v to 3,972.8 (4,160 x 0.955 = 3,972 v) against a minimum requirement of 3,600v for a 4,000v motor.

Thus, we can see that the 7.5 MVA transformer now becomes sufficient for a 2,000 hp motor with a different code letter and with the use of an RVS. Note also that if the auto-transformer RVS is at 80% tap, the voltage drop as seen by the motor would still satisfy the 10% VD requirements.


7.0: MULTIPLE MOTORS IN A TRANSFORMER

What if there are smaller motors plus other loads among a few large ones in a single transformer source?

EXAMPLE:

Assume that in the 7.5 MVA Substation transformer described above, there are 2,150 kVA smaller 460 v motors in the system through a number of transformers downstream the 4,160 v bus of the substation. Other load, load growth included is 750 kVA. Assume also that there are 800 hp 4.16 kV Code Letter E motors directly connected at the 4.16 kv bus. How many 800 hp motors can be energized into the system complying with the 10% voltage sag requirements? Assume that RVS’s for the 800 hp motors are at 65% tap and the motors shall not be started at one time.

If there are several motors on one transformer:

TRAFO KVA = (Maximum Demand kVA of Small Group Motors) + (Maximum Demand kVA of Other Loads Including Load Growth) + (kVA Ratings of all Large Motors) + (Additional Trafo kVA capacity necessary to accommodate the inrush current of the largest motor)

Thus:

TRAFO KVA = 7,500 kVA
Max Demand of Small Group Motors = 2,150 kVA

Other Loads Including Load Growth = 750 kVA

kVA of Large Motors = KVA of LM (Unknown)
Additional Trafo kVA for 800 hp Start-Up = Unknown

Solving for the Starting kVA for 800 hp Code Letter E motor
with RVS at 65% tap:

800 hp Starting kVA = (0.46 x 800 x 4.8)
800 hp Starting kVA = 1766 kVA

Solving for the Additional Trafo kVA Needed for 800 hp Start-Up:

Trafo SC kVA Needed for one 800 hp Start-Up to maintain a 10% VD = 1,766 x10
Trafo SC kVA for one 800 hp Start-Up to maintain a 10% VD = 17,660 kVA

Additional Trafo kVA Needed for one 800 hp Start-Up = SC kVA x %IZ of Trafo
Additional Trafo kVA for one 800 hp Start-Up = 17,660 x 0.08 = 1,413 kVA

Solving for the Number of 800 hp Motors
that can be placed into the system:

Recalling: TRAFO KVA = (Max Demand kVA of Small Group Motors) + (Max Demand kVA of Other Loads Including Load Growth) + (kVA Ratings of all Large Motors) + (Additional Trafo kVA capacity necessary to accommodate the inrush current of the largest motor)

7,500 kVA = 2,150 kVA + 750 kVA + KVALM + 1,413 kVA

KVA Large Motors = 7,500 – 2,150 – 750 - 1,413
KVA Large Motors = 3,187 kVA

Number of 800 hp motors = 3,187/800
Number of 800 hp motors = 3.984, say 4

The transformer selected will be capable of running and starting all motors
provided that only one large motor is started at any one time. Additional capacity will be required for motors starting simultaneously.


8.0: BACK TO THE BOARD PROBLEM

A 250 hp motor driving a pump is served by 3 x 100 kVA 13.8kV- 480v transformer bank. Other data are as follows:

 Transformer Impedance: 3.7% IZ
 Short Circuit Capacity of Source: 200 MVA
 Motor Starting Current: 8 times of FLC
 Length of Cable from Transformer to Motor Controller: 50 ft

Solve for:

a) Size of Cable & Conduit
b) Branch Circuit Breaker & Controller Size if Auto-Transformer type RVS
c) Determine the voltage dip during motor start-up using full voltage starter
d) Determine the voltage dip during motor start-up using Auto-Transformer Reduced Voltage Starter tapped at 65%.

Solution:

a) Solving for the Cable Size & Conduit
Motor Full Load Current = 302 A (From NEC Table)
Size of Cable = 1.25 x 302 or 377 A
USE: 500 MCM THW in 4”Φ Conduit

b) Solving for the Branch Circuit Breaker & RVS Controller Size
Branch Circuit Breaker = 2.0 x 302 A or 604 A
USE: 600AT/600AF, 3P, 600V MCCB
USE: Auto-Transformer RVS NEMA Size 5

c) Solving for the Voltage Dip Using Full Voltage Starter:

SC kVA at Transformer Terminal

Let: kVA Base = 300 kVA
V Base = 480 v
I Base = 361 A

X @ 13.8 kV = kVA Base / SC KVA@ 13.8 kV Source
X @ 13.8 kV = 300 / 200,000
X @ 13.8 kV = 0.00150 pu

X T = %IZ/100 x kVA Base /Trafo kVA
X T = 3.7/100 x 300/300
X T = 0.037 pu

X EQ = 0.00150 + 0.037
X EQ = 0.0385 pu

I SC = IBASE / X EQ
I SC = 361 A / 0.0385
I SC = 9,377 A

Three-Phase Short Circuit Capacity of the Transformer:
KVA SC = √3 x 0.48 x 9,377
KVA SC = 7,796 kVA

Motor Starting kVA:

Motor Starting kVA = 250 hp x 8
Motor Starting kVA = 2,000 kVA

The voltage drop on motor inrush at the transformer terminals will be:

VD at Trafo Terminals = 2,000/(2,000 + 7,796) = 0.2042 or, 20.42%
VD at Trafo Terminals = 98 v
Trafo Terminal Volts During Motor Start-Up = 382 v

Voltage Drop at the Cables:

Impedance ZC of 500 MCM THW Cable in Metallic Conduit= 0.00575 Ω per 100 ft (From IEEE Table)

Impedance of C1: 50 ft, from transformer to Motor Controller:
ZC = 0.00575 Ω/100 ft x 50 ft
ZC = 0.002875 Ω

Voltage Drop along the Cable C:
VDC = √3 x (IMOTOR LINE START) x ZC1
VDC = √3 x (302 x 8) x 0.002875
VDC = 12.0 v

Total Voltage Drop at Motor Controller:
98.0 + 12.0 = 110 v

Voltage at Motor Controller Line Side Terminals during Start-Up = 370 v
Voltage Requirements during Start Up = 414 v for 460 v motors

The set-up can not satisfy voltage dip requirements. Not even 3 x 250 kVA (750 kVA). In fact a 3 x 333 kVA (1,000 kVA) set-up at 5.2%IZ can barely satisfy the requirement!

There could be two remedies for the situation. One is to change the motor controller from full voltage to auto-transformer RVS type. The other alternative is to change the Transformer from 300 kVA (3 x 100 kVA) to 1,000 kVA (3 x 333 kVA).

d) Solving for the Voltage Dip Using Auto-Transformer RVS:

If RVS Auto-Transformer Tap to 65%:
Motor Starting kVA = 0.46 x 250 hp x 8
Motor Starting kVA = 920 kVA

The voltage drop on motor inrush at the transformer terminals will be:

VD at Trafo Terminals = 920/(920 + 7,796) = 0.1055 or, 10.55%
VD at Trafo Terminals = 50.64 v
Trafo Terminal Volts During Motor Start-Up = 429.36 v

Voltage Drop along the Cable C1:
VDC1 = √3 x (IMOTOR LINE START) x ZC1
VDC1 = √3 x (302 x 8 x 0.46) x 0.002875
VDC1 = 5.534 v

Total Voltage Drop at Motor Controller:
50.64 + 5.534 = 56.174 v

Voltage at Motor Controller Line Side Terminals during Start-Up = 423.826 v
Voltage Requirements during Start Up = 414 v for 460 v motors

Therefore, the 3 x 100 kVA transformer need not be replaced if its Auto-Transformer RVS is tapped at 65%. Note that if the RVS is tapped at 80%, the transformer needed must be 750 kVA (3 x 250 kVA)!

9.0: THE REAL ENGINEERING ECONOMICS

While first costs are very important to clients & plant owners; correct engineering practice should not be sidestepped in selecting best systems. More often than not, too much cheap engineering is in fact, costly. By making the first cost not objectionable, it does not mean that the engineer shall use cheap, undersized and substandard materials or equipment. Applying correct Value Engineering, the design must fundamentally conform to good engineering practice, codes and standards. This fundamental or minimum requisite must not be compromised under the cloak of costs.

It is important for the Industrial Power Systems Designer to understand that First Cost is not enough in determining the economics of the project. Total Life Cycle Cost of the power system depends on equipment purchase price & quality, construction & installation costs, operating costs including losses, outage costs, repair costs, useful lifetime of the equipment and administrative costs.

In the USA & other advance countries, the insurance companies play lead roles in the safety and soundness in any installations. That’s why they are strongly behind the National Electrical Code and the National Fire Protection Code. In the Philippines, if it’s “faulty electrical wirings”, the insurance companies pay without much fuss, and nobody seems to ask how standards were complied.

But alas, more often than not, what design engineers consider as “best solutions” may not correspond to what owners perceive. In the name of cost, the ‘economical solutions’ that clients want usually end up into sickly systems that are vulnerable to failure or worse, violations to the Code and Correct Engineering Practice for which we engineers are supposed to profess. Note that there is a common misguided thought in electrical engineering design economics. For instance, why use an expensive RVS when Across-the-Line Starter will do? Or why a 1,000 kVA transformer for a 250 hp motor? In this case, we say that firstly, the engineer must know his codes & standards. The standards must first be satisfied before any value engineering can be focused. Sensitive to safety & system operability in the ultimate analysis, it is the engineer's responsibility to find solutions that satisfy both parties.

Doods A. Amora
December, 2006

SIZING TRANSFORMERS WITH LARGE MOTOR LOADS (Part 1)

LESSONS WE NEED TO LEARN

SIZING TRANSFORMERS

WITH LARGE MOTOR LOADS

By Doods A. Amora, PEE
December 1, 2006)

(First of Two Parts)

1.0: INTRODUCTION

The purpose of this article is to help expand the engineer’s basic understanding on ‘short circuit capacity’ – how it affects voltage sags during starts-up of motors of significant sizes.

But why are Short Circuit & Motor Starting Calculations involved in transformer sizing?

One clue is:

"Starting up one thousand-1.0 hp motors is world apart from starting up one-1,000 hp motor"

SIDEBAR 1:

When I took my PEE board exams 22 years ago, there was one problem under the subject “ELECTRICAL DESIGN & CONSTRUCTION” which ran similar as follows:

A 250 hp motor driving a pump is served by 3 x 100 kVA 13.8kV- 480v transformer bank. Other data are as follows:

 Transformer Impedance: 3.7% IZ
 Short Circuit Capacity of Source Utility: 200 MVA
 Motor Starting Current: 8 times of FLC
 Length of Cable from Transformer to Motor Controller: 50 ft

Solve for:

a) Size of Cable & Conduit
b) Branch Circuit Breaker & Controller Size if Auto-Transformer type RVS
c) The voltage dip during motor start-up using full voltage starter
d) The voltage dip during motor start-up using Auto-Transformer Reduced Voltage Starter tapped at 65%.

To be honest, I wasn’t ready to solve the (c) and (d) parts of the problem. To my defense mechanism, I told myself, “the problem was irrelevant. Nandiyan na ‘yan, naka-install na kunyari; bakit kwentahin pa ang voltage dip?“

To make the story short, lumipas ang mahabang panahon… Then here comes now a real situation. What if one is asked to design a set-up for a 500 hp motor driving a pump somewhere? Would a 750 kVA transformer be sufficient? Maybe yes, maybe not! Try solving the problem and one will discover things that are assumed easy but are actually not quite simple!

SIDEBAR 2:

In one brewery I happened to work with, we had two – 450 hp 4 kV motors driving ammonia compressors controlled by Auto-Transformer RVS. Later, three – 700 hp motors were added but interestingly this time with across-the-line starters. The system was directly connected to the Power Plant’s 4,160 v bus through feeders. No problem was imagined because the Power Plant as a standard operating procedure, used to operate with significant spinning reserve capacity at any given time – and there were nine (9) generating units to support the loads.

However, when we started-up for the first time one of the three 700 hp compressors, a huge voltage dip occurred and the neighboring Glass Plant tripped off all its operating air compressors thus shutting down the entire plant. Pandemonium followed as the Power Plant struggled to recover the voltage by shedding off other major loads. Power Plant finally recovered but with all major manufacturing plants served by it at shutdown, including the brewery.

The event’s aftermath saw signs posted on all compressor control panels that reads: “NOTIFY POWER PLANT BEFORE STARTING”. Why? Because Power Plant had to add more generators on stream before any starting can be done. This was to provide the on-line generating capacity large enough to support the starting kVA needed, thus averting objectionable voltage dips. Interestingly, after the start-up the additional generators had to be shutdown. In the end, the full voltage starters were replaced with RVS units. In the succeeding breweries being built thereafter, MV Soft Starters were employed.

Again, there’s more to it than meets the eye. Recalling back my own PEE board exams, now I understand that after all, the problem on voltage dip was not irrelevant but a test of engineering competence why I should be given a license as PEE.


2.0: SUBSTATION TRANSFORMERS VS. LOADS

Generators and transformers react to large motor starting as manifested in voltage dips. Let us first discuss transformers.

In general terms, transformers in an electrical system are usually larger than the maximum demands they serve, in some instances even larger than the connected loads. In the industrial plant scenario, the obvious reason at first glance for this apparent oversizing is the anticipation for future load growth. Fine…

But more often than not, sizing the transformer with extra kVA capacity unwittingly addresses voltage sag problems, not for load growth for which it is intended originally. That’s why for newly constructed plants where load growth is not yet there, the problem of starting significantly large motors may not surface out. Why? Because the extra kVA capacity intended for load growth is taking care of it.

3.0: SYSTEM BEHAVIOR IN EVENTS OF FAULTS

Short circuit capacity calculation is used in many applications, some of which are the following:

a) selecting the interrupting capacity ratings of circuit breakers and fuses,
b) specifying the short-time withstand of switchgears,
c) determining if a line reactor is required,
d) and a host of many others.

But unknown to many, Short Circuit Calculations together with
Motor Starting Calculations are also used in Sizing Transformers.

Short Circuit condition brings down the voltage very dramatically. The amount of voltage during a full three phase fault at the terminals of the transformer is determined by the %IZ of the transformer, i.e., if the transformer is connected to an infinite source of short circuit currents.

Take for instance a 7,500 kVA, 8.0% IZ, 69-4.16 kV transformer. The 8.0% impedance rating would mean that full load currents of 1,041 amps will flow in the secondary if the secondary is short circuited and the primary voltage is raised from zero volts to a point at which 8.0% of 4,160 volts, or 332.8 volts, appears at the secondary terminals. Therefore, the impedance (Z) of the transformer secondary may now be calculated:

Z = V / I = 332.8 volts/1,041 amps = 0.3197 ohms

The exercise described above is what is known in textbooks as the “Short Circuit Test” which is actually performed in laboratories or electrical shops (drying up transformers). What if the short circuit at the secondary is real – meaning, the short circuit happened at full 4160v at the inception of the fault? If the transformer is connected directly to a source capable of supplying the transformer with an unlimited short circuit kVA capacity, the short circuit amperage capacity which the transformer can deliver from its secondary is:

4,160 volts /0.3197 = 13,012.5 amps

Another method of calculating short circuit capacity for the above transformer is:

SC Capacity = Trafo FLC/%IZ
= 1,041 A / 0.08
= 13,012.5 amps,

or terms of kVA:

SC Capacity = 7,500 kVA/0.08 = 93,750 kVA

Such three-phase fault will bring down the voltage at the secondary terminals from 4,160v to approaching zero, while at the same time delivering 93.75 MVA of short circuit power to the point of fault. This condition however must of course be short-lived because some sort of protection upstream must operate to interrupt the circuit quickly, otherwise the transformer will fail. Transformer damage (known as the transformer short-time withstand) is dependent on the amount of fault current and the impedance of the transformer. Usually this transformer short-time withstand could only last for a few seconds.

4.0: SYSTEM BEHAVIOR DURING LARGE MOTOR START-UP

Why is motor starting related to short circuit condition? Partly similar to short circuit condition, motors have a high initial inrush current when energized and draw heavy load at a low power factor (0.30 lagging for 100 hp & about 0.16 for 1,000 hp motor) for a very short time. This sudden increase in the current flowing to the load causes a momentary increase in the voltage drop at the supply transformer terminals, the voltage drop along the distribution system, and a corresponding reduction in the voltage at the utilization equipment.

The magnitude of transient current involved in motor starting is however very much lower than the short circuit condition. But in effect, switching “on” to energize a large motor can be likened to a “soft short circuit”. Like short circuits, the effect of starting large motors results to voltage dips. The voltage drop at the transformer secondary terminals is proportional to motor starting kVA over the short-circuit capacity of the transformer. When motor starting kVA is drawn from a system, the voltage drop in percent of the initial voltage is approximately equal to the Motor Starting kVA divided by the Sum of this kVA and the Short Circuit kVA.

% Voltage Drop = (Motor Starting kVA) x 100 /(Motor Starting kVA + Short Circuit kVA)

5.0: MOTOR STARTING KVA

Thus, we need to know the starting kVA of the subject motor. In general, IEEE Red Book says, “the starting current or starting kVA of a standard motor draws several times its full-load running ratings. Without any specific information on a subject motor, a motor is always assumed to require about 1 kVA for each motor horsepower in normal operation, so the starting current of the average motor will be about 5 kVA for each motor horsepower. When the motor rating in horsepower approaches 5% of the secondary unit substation transformer capacity in kilovolt-amperes, the motor starting apparent power approaches 25% of the transformer capacity which, with a transformer impedance voltage of 6.07%, will result in a noticeable voltage sag on the order of 1%. This sudden increase in the current drawn from the power system may result in excessive drop in voltage unless it is considered in the design of the system. The motor-starting load in kilovolt-amperes, imposed on the power supply system, and the available motor torque are greatly affected by the method of starting used”.

The specific starting values for ac motors over hp are indicated in terms of Code Letters on the nameplates of North American-made motors. NEMA Code Letters indicate the motor starting characteristics as presented in the following table.

TABLE 1: NEMA CODE LETTER DESIGNATIONS (STARTING KVA’s)

CODE LETTER KVA per HP

A: 3.15
B: 3.16 – 3.55
C: 3.56 – 4.0
D: 4.1 – 4.5
E: 4.6 – 5.0
F: 5.1 – 5.6
G: 5.7 – 6.3
H: 6.4 – 7.1
J: 7.2 – 8.0
K: 8.1 – 9.0
L: 9.1 - 10.0
M: 10.1 - 11.2
N: 11.3 - 12.5
P: 12.6 - 14.0
R: 14.1 - 16.0
S: 16.1 - 18.0
T: 18.1 - 20.0
U: 20.1 - 22.4
V: 22.5 - and up

The above table means that a 200 hp motor with a NEMA Code Letter ‘G’ requires a starting kVA of 1,140 to 1,260 kVA (1,200 kVA on the average) when using Full Voltage (Across-the-Line) motor controller. This starting kVA is also known as “Locked Rotor kVA” or sometimes “Locked Rotor Amperes”.

With the starting kVA of the motor such as described above, we must then determine the voltage dip caused by the motor inrush on start-up. The ensuing voltage during start-up must stay within the allowable operating voltage of the system. If such condition happens, then no oversizing of the transformer is required. But when the voltage dip exceeds the operating requirement of the system, then the transformer must provide extra kVA.

North American motors are rated for 230, 460, 2,300, 4,000, 6,600 or 13,200 volts for use with distribution sub-systems that are rated at 240, 480, 2400, 4,160, 6,900 or 13,800 volts respectively. Note the difference in motor nameplate voltages viz-a-viz the transformer terminal voltages. The apparent lower motor voltages than distribution nominal voltages are deliberately established by manufacturers to deal with the inherent voltage drops in the system such as: internal voltage drop of the transformer as dictated by its voltage regulation capability, voltage drop along the distribution cables and the impedance of the system. This could mean that the 4,160 v at the transformer at no-load condition may only be 3,800 v at the motor terminals when the system is already heavily loaded. In this condition, when a large motor is started-up somewhere in the system, the lower will the voltage be as felt by other loads in the system.

That’s why NEMA standard motors are designed to be capable of operating at plus or minus 10% of nameplate voltage. Therefore, the voltage drop on inrush should not be allowed to drop more than -10% of the rated voltage. This means 208v for 230v or 414v for 460 volt motors. Likewise, 2.07 kV or 3.6 kV for 2.3 or 4.0 kV motors, respectively. It means that a 4 kV motor can still operate satisfactorily at 3,600 v but any disturbance in the system that brings the system voltage lower, the affected motors may trip off as provided for by its protection – or if not, the motor burns.

(To be continued…)

Doods A. Amora
December, 2006