To correctly specify a transformer with delta winding connections, you need to know how to do the related calculations.
The names for transformers configurations such as delta and wye derive from the way the windings are connected inside the transformer. These connections determine the way the transformer will behave, and so they also determine the methods of calculation required for properly applying a given transformer.
Delta-connected transformers have the windings of three 1Ø transformers connected in series with each other to form a closed circuit. The line conductors connected where two 1Ø transformers meet. This configuration is a “delta system,” because in an electrical drawing it looks like a triangle (Greek symbol Dfor the letter “delta”). Many call it a high-leg system because the voltage from Line 2 to ground is higher than that of the other legs. For example, a 120V delta transformer will have a 208V leg (Figure 12–1).
Graphics are not included in this newsletter.
Definitions
To avoid confusion, we need to clarify some key terms:
Line: The ungrounded (hot) conductor(s). See Figure 12–2 and Figure 12–3.
Line Current: The current on the ungrounded conductors (Figure 12–3, B1 and B2). In a delta system, the line current is greater than the phase current by the Square Root of 3 (which is approximately 1.732). In a wye system, the line current equals the phase current.
Line Voltage: The voltage between any two line (ungrounded) conductors (Figure 12–3, A1 and A2). In a delta system, the line voltage equals the phase voltage (Figure 12–11). But, the delta transformer also has a high-leg.
Phase Current: The current flowing through the transformer winding (Figure 12–3, D1 and D2). In a delta system, the phase current is less than the line current by the Square Root of 3. In a wye system, the phase current equals the line current.
Phase Load: The load on the transformer winding.
Phase Voltage: The internal transformer voltage generated across any one winding of a transformer. For a delta secondary, the phase voltage equals the line voltage. In a wye system, the phase voltage is less than the line voltage by the Square Root of 3 (Figure 12–3, A2 and C2).
Ratio: The number of primary winding turns divided by the number of secondary winding turns.
Unbalanced Load: (Neutral Current). The load on the secondary grounded (neutral) conductors.
With these terms defined, we are ready to tackle transformer calculations. We will look specifically at delta transformers.
Delta transformer current.
In a delta transformer, the line current does not equal the phase current (as it does in a wye transformer). Because each line from a delta configured transformer is connected to two transformer phases, the line current from a 3Ø load will be greater than the phase current by the Square Root of 3. Note these formulas:
ILine = IPhase x square Root of 3
ILine = VALine/(ELine x Square Root of 3)
IPhase = ILine/Square Root of 3
IPhase = VAPhase/EPhase
You can use the delta triangle (Figure 12–18) to calculate delta 3Ø line and phase currents. Place your finger over the desired item, and the remaining items show the formula to use.
If you plug some numbers in, you can more clearly see the effects of the delta configuration on currents. Let’s try this with a 240V, 36 kVA, 3Ø load (Figure 12–15).
Line: Total line power = 36 kVA
ILine = VALine/(ELine x Square Root of 3)
ILine = 36,000 VA/(240V x Square Root of 3)
ILine = 87A, or
ILine = IPhase x 3 x Square Root of 3
ILine = 50A x 1.732 = 87A
Phase: Phase power = 12 kVA (per winding)
IPhase = VAPhase/EPhase
IPhase = 12,000 VA/240V = 50A, or
IPhase = ILine/Square Root of 3
IPhase= 87A/1.732 = 50A
We can also use the formula: ILine = VALine / (ELine x Square Root 3). What is the secondary line current for a 480 to 240/120V, 150 kVA, 3Ø delta transformer (Figure 12–13)?
416A
360A
180A
144A
Answer: (b) 360A
ILine = VALine/(ELine x Square Root of 3)
ILine = 150,000 VA/(240V x 1.732) = 360A
You can calculate the phase current of a delta transformer winding by dividing the phase VA by the phase volts:
IPhase = VAPhase/EPhase
The phase load in VA of a 3Ø, 240V load = line load/3 (1/3 of load on each winding
The phase load in VA of a 1Ø, 240V load = line load (all on one winding)
The phase load in VA of a 1Ø, 120V load = line load (all on one winding)
Now, let’s work an example problem. What is the secondary phase current for a 480 to 240/120V, 150 kVA, 3Ø delta transformer (Figure 12–14)?
416A
360A
208A
104A
Answer: (c) 208A
Phase power = 150,000 VA/3 per phase = 50,000 VA per phase
IPhase = 50,000 VA/240V
IPhase = 208A
Try running these numbers with a 10A load and then with a 75A load. You’ll gain a greater appreciation for what happens in a delta system.
Delta transformer balancing
To properly size a delta/delta transformer, the transformer phases (windings) must be balanced. You can do that with a two-step process:
Step 1. Determine the VA rating of all loads.
Step 2: Balance the loads on the transformer windings as follows
3Ø Loads: one-third of the load on each of the phases.
240V, 1Ø Loads: 100% of the load on Phase A or B. You can place some 240V, 1Ø load on Phase C when necessary for balance.
120V loads: 100% of the load on C1 or C2.
To size the panelboard and its conductors, you must balance the loads in amperes. Why balance the panel in amperes? Why not take the VA per phase and divide by phase voltage? Answer: Line current of a 3Ø load is calculated by the formula:
IL = VA/(ELine x Square Root of 3)
IL = 150,000 VA/(240V x 1.732) = 208A per line.
If you took the per-line power of 50,000 VA and divided by one line voltage of 120V, you would come up with an incorrect line current of 50,000 VA/120V = 417A.
Wednesday, September 8, 2010
Short Circuit Capacity: Basic Calculations and Transformer Sizing
Short circuit capacity calculation is used for many applications: sizing of transformers,
selecting the interrupting capacity ratings of circuit breakers and fuses, determining if a
line reactor is required for use with a variable frequency drive, etc.
The purpose of the presentation is to gain a basic understanding of short circuit capacity.
The application example utilizes transformer sizing for motor loads.
Conductor impedances and their associated voltage drop are ignored not only to present a
simplified illustration, but also to provide a method of approximation by which a plant
engineer, electrician or production manager will be able to either evaluate a new application
or review an existing application problem and resolve the matter quickly.
Literature containing a detailed discussion of short circuit capacity calculations are
available within the electrical power transmission industry. [1]
The following calculations will determine the extra kVA capacity required for a three
phase transformer that is used to feed a single three phase motor that is started with
full voltage applied to its terminals, or, "across-the-line."
Two transformers will be discussed, the first having an unlimited short circuit kVA
capacity available at its primary terminals, and the second having a much lower input
short circuit capacity available.
kVA of a single phase transformer = V x A
kVA of a three phase transformer = V x A x 1.732, where 1.732 = the square root of 3.
The square root of 3 is introduced for the reason that, in a three phase system,
the phases are 120 degrees apart and, therefore, can not be added arithmetically.
They will add algebraically.
Transformer Connected To Utility Power Line
The first transformer is rated 1000 kVA, 480 secondary volts, 5.75% impedance.
Rated full load amp output of the transformer is
1000 kVA / (480 x 1.732) = 1203 amps
The 5.75% impedance rating indicates that 1203 amps will flow in the secondary if
the secondary is short circuited line to line and the primary voltage is raised from
zero volts to a point at which 5.75% of 480 volts, or, 27.6 volts, appears at the
secondary terminals. Therefore, the impedance (Z) of the transformer secondary may
now be calculated:
Z = V / I = 27.6 volts / 1203 amps = .02294 ohms
The transformer is connected directly to the utility power lines which we will
assume are capable of supplying the transformer with an unlimited short circuit
kVA capacity. The utility company will always determine and advise of the short
circuit capacity available at any facility upon request.
With unlimited short circuit kVA available from the utility, the short circuit
amperage capacity which the transformer can deliver from its secondary is
480 volts / .02294 = 20,924 amps
An alternative method of calculating short circuit capacity for the above
transformer is:
1203 amps x 100 / 5.75% = 1203 / .0575 = 20,922 amps
Another alternative is to consult a reference manual. Cutler- Hammer Consulting
Application Catalog, 12th Edition, gives the specifications for the above mentioned
transformer and the value of the short circuit capacity in Table A25 on page A-59.
The short circuit capacity is given as 20,900 amps.
Now we are ready to apply a motor to the terminals of the transformer secondary.
We must determine the voltage drop which will be caused by the motor inrush on
start. If the voltage remains within the rated voltage of the motor, then no oversizing
of the transformer is required.
Motors rated for 460 volts are for use with distribution systems that are rated at
480 volts. The rating system allows a twenty volt drop in the distribution system
which may occur along the feeder cables which connect the power transformer to
the load. The NEMA specification for a standard motor is that it requires the motor 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 below 460 volts less 10%,
or, 414 volts.
The transformer will be asked to supply power to a motor which has a full load amp
rating of 1203 amps, which will fully load the transformer. Therefore, we will rate
the motor at 460 V x 1203 A x 1.732, or, 958.5 kVA. We will assume that our motor
will have an inrush of 600% of its full load rating which will cause an inrush of
460 V x 1203 A x 600% x 1.732 = 5751 kVA
The voltage drop at the transformer terminals will be proportional to the motor load.
The voltage drop will be expressed as a percentage of the inrush motor load compared
to the maximum capability of the transformer. [2] The transformer has a maximum kVA
capacity at its short circuit capability, which is
480 V x 20,924 A x 1.732 = 17,395 kVA
The voltage drop on motor inrush will be
5751 kVA / 17,395 kVA = .331, or, 33.1%
The transformer output voltage will drop to 480 x .669, or, 321 volts. Thus, we can
see that the transformer is much too small to use a motor that has a full load rating
equal to the full load capacity of the transformer.
The transformer must be sized so that its short circuit capabilty is equal to or
greater than 5751 kVA times 10, or, 57,510 kVA in order to have a voltage drop of 10%
or less. Therefore, the short circuit amperage capacity of the transformer to be used
must be a minimum of
57,510 kVA / (480 V x 1.732) = 69176 amps
A typical 2500 kVA, 5.75% impedance transformer will have a short circuit capacity of
52,300 amps. The next highest standard size transformer at 3750 kVA will have a 6.5%
impedance and would have a short circuit output capability of 69,395 amps which will be
sufficient.
In the particular application discussed, the ratio of the selected standard size
transformer kVA to motor kVA is 3750 kVA / 958.5 kVA = 3.91. Thus the transformer
rating is 391% larger, or, nearly four times, the rating of the motor. Note the non-linear
effect of the impedance rating of the transformers on their short circuit capacities.
Transformer Connected To An Upstream Transformer
The second transformer we will examine will have a finite short circuit capacity
available at its primary rather than an unlimited capacity. We will assume that a facility
derives its power from the same 1000 kVA transformer mentioned above and that the
second transformer is connected directly to the terminals of the 1000 kVA transformer.
Thus, feeder cables between the two transformers are eliminated and the impedance of
cables are not taken into account. However, the smaller the motor leads, the less will be
both the short circuit capacity and the voltage delivered to the motor terminals.
The second transformer, which will have a 480 volt primary and a 480 volt secondary, will
be used to power a 20 HP, 3 phase, 460 volt motor which will be started at full voltage. The
motor will be the only load on the transformer.
Sales catalogs by various manufacturers will invariably recommend a "minimum
transformer kVA" of 21.6 for use with a 20 HP motor. The minimum transformer kVA ratings
are for use with multiple motors on a single transformer. A multiple motor configuration
will be discussed in the next section of this article.
The 21.6 kVA is calculated as follows:
480 volts x 26 nominal amps x 1.732 = 21.6 kVA
The transformer manufacturers will give a 20 HP motor a nominal full load amp rating
of 27 amps, thus allowing no extra capacity:
460 volts x 27 nominal amps x 1.732 = 21.5 kVA
One motor manufacturer has rated a 20 HP motor at 26 Full Load Amps, 460 VAC,
205 Locked Rotor Amps, 81% Power Factor. The motor will present a load of
460 volts x 26 amps x 1.732 = 20.7 kVA
The starting motor kVA load with inrush current will be
460 V x 205 A x 1.732 = 163.3 kVA
We will consider using a 30 kVA general purpose transformer to supply the 20 HP motor.
The transformer will have a nominal impedance of 2.7% and an ouptut of 36.1 amps at
480 volts. The short circuit current capacity that can be delivered to the 21.6 kVA
transformer by the upstream 1000 kVA transformer is 20,924 amps, or, 17,395 kVA.
The short circuit amperage capacity of a transfomer with a limited system short circuit
capacity available at its primary is:
transformer full load amps / (transformer impedance + upstream system impedance as seen
by the transformer)
Where:
upstream system impedance as seen by the transformer =
transformer kVA / available primary short circuit capacity kVA
Therefore,
36.1 amps / [2.7% + (30 kVA / 17,395 kVA)] =
36.1 / (2.7% + .0017%) = 36.1 / .0287 = 1258 short circuit amps
The transformer output voltage drop upon motor inrush will be:
motor inrush kVA / short circuit kVA =
163.3 kVA / (480 V x 1258 A x 1.732) = 163.3 kVA / 1046 kVA =
.156 = 15.6 %
A 30 kVA transformer rating is too small as the motor voltage drop will exceed 10%.
A 45 kVA transformer with a 2.4% impedance and an output of 54.1 amps at 480 volts
would have a short circuit capacity of 2034 amps. The voltage drop upon motor inrush
would be 9.66%.
For a single motor and transformer combination, one transformer manufacturer
recommends that the motor full load running current not exceed 65% of the transformer
full load amp rating. [3] Thus, for our 26 amp motor the transformer rating should be
a minimum of 40 amps, or, 33.3 kVA.
Multiple Motors On A Single Transformer
The minimum transformer kVA is given by transformer manufacturers so that a transformer
may be sized properly for multiple motors. If there are five motors on one transformer, add
the minimum kVA ratings and then add transformer capacity as necessary to accomodate the
inrush current of the largest motor.
The transformer thusly selected will be capable of running and starting all five motors
provided that only one motor is started at any one time. Additional capacity will be required
for motors starting simultaneously.
Also, if any motor is started more than once per hour, add 20% to that motor's minimum
kVA rating to compensate for heat losses within the transformer.
Motor Contribution to Short Circuit Capacity
When a fault condition occurs, power system voltage will drop dramatically. All motors
that are running at that time will not be able to sustain their running speed. As those motors
slow in speed, the stored energy within their fields will be discharged into the power line.
The nominal discharge of a motor will contribute to the fault a current equal to up to four
times its full load current.
With our 1000 kVA, 1203 amp transformer example given above, we will assume that all
1203 amps of load are from motors. The actual short circuit current will equal 20,924 amps
plus 400% of 1203 amps for a total of 25,736 short circuit amps.
When sizing the transformer for motor loads, the fault current contribution from the
motors will not be a consideration for sizing. However, the motor contribution must be
considered when sizing all branch circuit fuses and circuit breakers. The interrupting
capacity ratings of those devices must equal or exceed the total short circuit capacity
available at the point of application.
Motor contribution to short circuit capacity must be included when adding a variable
frequency drive to the system. See Variable Frequency Drives: Source Impedance
and Line Reactors
References:
[1] "Cutler-Hammer Consulting Application Catalog," published by Cutler-Hammer,
Division of Eaton Corporation, Pittsburgh, PA.
[2] "Short Circuit Capacity and Voltage Sag," IEEE (Institute of Electrical and Electronic
Engineers) Industry Application Society (IAS) Magazine, July/August 2000, page 38. You
might find this magazine in universities which have Electrical Enginneering programs.
Do not get it confused with IEEE Transactions on Industry Applications which is a
different publication.
[3] "Power Distribution Products" Catalog ATD-01, Acme Electric Corporation, 1995,
page 125.
Power Quality and Drives LLC
http://www.powerqualityanddrives.com
selecting the interrupting capacity ratings of circuit breakers and fuses, determining if a
line reactor is required for use with a variable frequency drive, etc.
The purpose of the presentation is to gain a basic understanding of short circuit capacity.
The application example utilizes transformer sizing for motor loads.
Conductor impedances and their associated voltage drop are ignored not only to present a
simplified illustration, but also to provide a method of approximation by which a plant
engineer, electrician or production manager will be able to either evaluate a new application
or review an existing application problem and resolve the matter quickly.
Literature containing a detailed discussion of short circuit capacity calculations are
available within the electrical power transmission industry. [1]
The following calculations will determine the extra kVA capacity required for a three
phase transformer that is used to feed a single three phase motor that is started with
full voltage applied to its terminals, or, "across-the-line."
Two transformers will be discussed, the first having an unlimited short circuit kVA
capacity available at its primary terminals, and the second having a much lower input
short circuit capacity available.
kVA of a single phase transformer = V x A
kVA of a three phase transformer = V x A x 1.732, where 1.732 = the square root of 3.
The square root of 3 is introduced for the reason that, in a three phase system,
the phases are 120 degrees apart and, therefore, can not be added arithmetically.
They will add algebraically.
Transformer Connected To Utility Power Line
The first transformer is rated 1000 kVA, 480 secondary volts, 5.75% impedance.
Rated full load amp output of the transformer is
1000 kVA / (480 x 1.732) = 1203 amps
The 5.75% impedance rating indicates that 1203 amps will flow in the secondary if
the secondary is short circuited line to line and the primary voltage is raised from
zero volts to a point at which 5.75% of 480 volts, or, 27.6 volts, appears at the
secondary terminals. Therefore, the impedance (Z) of the transformer secondary may
now be calculated:
Z = V / I = 27.6 volts / 1203 amps = .02294 ohms
The transformer is connected directly to the utility power lines which we will
assume are capable of supplying the transformer with an unlimited short circuit
kVA capacity. The utility company will always determine and advise of the short
circuit capacity available at any facility upon request.
With unlimited short circuit kVA available from the utility, the short circuit
amperage capacity which the transformer can deliver from its secondary is
480 volts / .02294 = 20,924 amps
An alternative method of calculating short circuit capacity for the above
transformer is:
1203 amps x 100 / 5.75% = 1203 / .0575 = 20,922 amps
Another alternative is to consult a reference manual. Cutler- Hammer Consulting
Application Catalog, 12th Edition, gives the specifications for the above mentioned
transformer and the value of the short circuit capacity in Table A25 on page A-59.
The short circuit capacity is given as 20,900 amps.
Now we are ready to apply a motor to the terminals of the transformer secondary.
We must determine the voltage drop which will be caused by the motor inrush on
start. If the voltage remains within the rated voltage of the motor, then no oversizing
of the transformer is required.
Motors rated for 460 volts are for use with distribution systems that are rated at
480 volts. The rating system allows a twenty volt drop in the distribution system
which may occur along the feeder cables which connect the power transformer to
the load. The NEMA specification for a standard motor is that it requires the motor 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 below 460 volts less 10%,
or, 414 volts.
The transformer will be asked to supply power to a motor which has a full load amp
rating of 1203 amps, which will fully load the transformer. Therefore, we will rate
the motor at 460 V x 1203 A x 1.732, or, 958.5 kVA. We will assume that our motor
will have an inrush of 600% of its full load rating which will cause an inrush of
460 V x 1203 A x 600% x 1.732 = 5751 kVA
The voltage drop at the transformer terminals will be proportional to the motor load.
The voltage drop will be expressed as a percentage of the inrush motor load compared
to the maximum capability of the transformer. [2] The transformer has a maximum kVA
capacity at its short circuit capability, which is
480 V x 20,924 A x 1.732 = 17,395 kVA
The voltage drop on motor inrush will be
5751 kVA / 17,395 kVA = .331, or, 33.1%
The transformer output voltage will drop to 480 x .669, or, 321 volts. Thus, we can
see that the transformer is much too small to use a motor that has a full load rating
equal to the full load capacity of the transformer.
The transformer must be sized so that its short circuit capabilty is equal to or
greater than 5751 kVA times 10, or, 57,510 kVA in order to have a voltage drop of 10%
or less. Therefore, the short circuit amperage capacity of the transformer to be used
must be a minimum of
57,510 kVA / (480 V x 1.732) = 69176 amps
A typical 2500 kVA, 5.75% impedance transformer will have a short circuit capacity of
52,300 amps. The next highest standard size transformer at 3750 kVA will have a 6.5%
impedance and would have a short circuit output capability of 69,395 amps which will be
sufficient.
In the particular application discussed, the ratio of the selected standard size
transformer kVA to motor kVA is 3750 kVA / 958.5 kVA = 3.91. Thus the transformer
rating is 391% larger, or, nearly four times, the rating of the motor. Note the non-linear
effect of the impedance rating of the transformers on their short circuit capacities.
Transformer Connected To An Upstream Transformer
The second transformer we will examine will have a finite short circuit capacity
available at its primary rather than an unlimited capacity. We will assume that a facility
derives its power from the same 1000 kVA transformer mentioned above and that the
second transformer is connected directly to the terminals of the 1000 kVA transformer.
Thus, feeder cables between the two transformers are eliminated and the impedance of
cables are not taken into account. However, the smaller the motor leads, the less will be
both the short circuit capacity and the voltage delivered to the motor terminals.
The second transformer, which will have a 480 volt primary and a 480 volt secondary, will
be used to power a 20 HP, 3 phase, 460 volt motor which will be started at full voltage. The
motor will be the only load on the transformer.
Sales catalogs by various manufacturers will invariably recommend a "minimum
transformer kVA" of 21.6 for use with a 20 HP motor. The minimum transformer kVA ratings
are for use with multiple motors on a single transformer. A multiple motor configuration
will be discussed in the next section of this article.
The 21.6 kVA is calculated as follows:
480 volts x 26 nominal amps x 1.732 = 21.6 kVA
The transformer manufacturers will give a 20 HP motor a nominal full load amp rating
of 27 amps, thus allowing no extra capacity:
460 volts x 27 nominal amps x 1.732 = 21.5 kVA
One motor manufacturer has rated a 20 HP motor at 26 Full Load Amps, 460 VAC,
205 Locked Rotor Amps, 81% Power Factor. The motor will present a load of
460 volts x 26 amps x 1.732 = 20.7 kVA
The starting motor kVA load with inrush current will be
460 V x 205 A x 1.732 = 163.3 kVA
We will consider using a 30 kVA general purpose transformer to supply the 20 HP motor.
The transformer will have a nominal impedance of 2.7% and an ouptut of 36.1 amps at
480 volts. The short circuit current capacity that can be delivered to the 21.6 kVA
transformer by the upstream 1000 kVA transformer is 20,924 amps, or, 17,395 kVA.
The short circuit amperage capacity of a transfomer with a limited system short circuit
capacity available at its primary is:
transformer full load amps / (transformer impedance + upstream system impedance as seen
by the transformer)
Where:
upstream system impedance as seen by the transformer =
transformer kVA / available primary short circuit capacity kVA
Therefore,
36.1 amps / [2.7% + (30 kVA / 17,395 kVA)] =
36.1 / (2.7% + .0017%) = 36.1 / .0287 = 1258 short circuit amps
The transformer output voltage drop upon motor inrush will be:
motor inrush kVA / short circuit kVA =
163.3 kVA / (480 V x 1258 A x 1.732) = 163.3 kVA / 1046 kVA =
.156 = 15.6 %
A 30 kVA transformer rating is too small as the motor voltage drop will exceed 10%.
A 45 kVA transformer with a 2.4% impedance and an output of 54.1 amps at 480 volts
would have a short circuit capacity of 2034 amps. The voltage drop upon motor inrush
would be 9.66%.
For a single motor and transformer combination, one transformer manufacturer
recommends that the motor full load running current not exceed 65% of the transformer
full load amp rating. [3] Thus, for our 26 amp motor the transformer rating should be
a minimum of 40 amps, or, 33.3 kVA.
Multiple Motors On A Single Transformer
The minimum transformer kVA is given by transformer manufacturers so that a transformer
may be sized properly for multiple motors. If there are five motors on one transformer, add
the minimum kVA ratings and then add transformer capacity as necessary to accomodate the
inrush current of the largest motor.
The transformer thusly selected will be capable of running and starting all five motors
provided that only one motor is started at any one time. Additional capacity will be required
for motors starting simultaneously.
Also, if any motor is started more than once per hour, add 20% to that motor's minimum
kVA rating to compensate for heat losses within the transformer.
Motor Contribution to Short Circuit Capacity
When a fault condition occurs, power system voltage will drop dramatically. All motors
that are running at that time will not be able to sustain their running speed. As those motors
slow in speed, the stored energy within their fields will be discharged into the power line.
The nominal discharge of a motor will contribute to the fault a current equal to up to four
times its full load current.
With our 1000 kVA, 1203 amp transformer example given above, we will assume that all
1203 amps of load are from motors. The actual short circuit current will equal 20,924 amps
plus 400% of 1203 amps for a total of 25,736 short circuit amps.
When sizing the transformer for motor loads, the fault current contribution from the
motors will not be a consideration for sizing. However, the motor contribution must be
considered when sizing all branch circuit fuses and circuit breakers. The interrupting
capacity ratings of those devices must equal or exceed the total short circuit capacity
available at the point of application.
Motor contribution to short circuit capacity must be included when adding a variable
frequency drive to the system. See Variable Frequency Drives: Source Impedance
and Line Reactors
References:
[1] "Cutler-Hammer Consulting Application Catalog," published by Cutler-Hammer,
Division of Eaton Corporation, Pittsburgh, PA.
[2] "Short Circuit Capacity and Voltage Sag," IEEE (Institute of Electrical and Electronic
Engineers) Industry Application Society (IAS) Magazine, July/August 2000, page 38. You
might find this magazine in universities which have Electrical Enginneering programs.
Do not get it confused with IEEE Transactions on Industry Applications which is a
different publication.
[3] "Power Distribution Products" Catalog ATD-01, Acme Electric Corporation, 1995,
page 125.
Power Quality and Drives LLC
http://www.powerqualityanddrives.com
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