MCB SELECTION CHART

 

Calculations and Transformer Sizing and Short Circuit Capacity

Short circuit capacity calculation is used for many applications: sizing of transformers, selecting the interrupting capacity ratings of circuit breakers and fuses, determining if aline 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.

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 the full voltage applied to its terminals, or, "across-the-line."

Two transformers will be discussed, the first having an unlimited short circuit kV capacity available at its primary terminals, and the second has 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 shortcircuit 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

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 onstart. 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 at480 volts. The rating system allows a twenty-volt drop in the distribution system which may occur along with 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. 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 capability 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 of52,300 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 is 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, a 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 output of 36.1 amps at480 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 transformer 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 accommodate 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. The 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, the 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 all1203 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 Impedanceand Line Reactors

Calculating Motor Speed

 

Calculating Motor Speed:

A squirrel cage induction motor is a constant speed device. It cannot operate for any length of time at speeds below those shown on the nameplate without danger of burning out.

To Calculate the speed of a induction motor, apply this formula:

Srpm = 120 x F
            P

Srpm = synchronous revolutions per minute.
120   = constant
F       = supply frequency (in cycles/sec)
P       = number of motor winding poles

Example: What is the synchronous of a motor having 4 poles connected to a 60 hz power supply?

Srpm = 120 x F
            P
Srpm = 120 x 60
            4
Srpm = 7200
             4
Srpm = 1800 rpm

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Calculating Braking Torque:

Full-load motor torque is calculated to determine the required braking torque of a motor.
To Determine braking torque of a motor, apply this formula:

T = 5252 x HP
    rpm

T      = full-load motor torque (in lb-ft)
5252 = constant (33,000 divided by 3.14 x 2 = 5252)
HP    = motor horsepower
rpm = speed of motor shaft

Example: What is the braking torque of a 60 HP, 240V motor rotating at 1725 rpm?

T = 5252 x HP
    rpm
T = 5252 x 60
     1725
T = 315,120
     1725
T = 182.7 lb-ft

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Calculating Work:

Work is applying a force over a distance. Force is any cause that changes the position, motion, direction, or shape of an object. Work is done when a force overcomes a resistance. Resistance is any force that tends to hinder the movement of an object.If an applied force does not cause motion the no work is produced.

To calculate the amount of work produced, apply this formula:

W = F x D

W = work (in lb-ft)
F  = force (in lb)
D  = distance (in ft)

Example: How much work is required to carry a 25 lb bag of groceries vertically from street level to the 4th floor of a building 30' above street level?

W = F x D
W = 25 x 30
W = 750 -lb

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Calculating Torque:

Torque is the force that produces rotation. It causes an object to rotate. Torque consist of a force acting on distance. Torque, like work, is measured is pound-feet (lb-ft). However, torque, unlike work, may exist even though no movement occurs.

To calculate torque, apply this formula:

T = F x D

T = torque (in lb-ft)
F = force (in lb)
D = distance (in ft)

Example: What is the torque produced by a 60 lb force pushing on a 3' lever arm?

T = F x D
T = 60 x 3
T = 180 lb ft

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Calculating Full-load Torque:

Full-load torque is the torque to produce the rated power at full speed of the motor. The amount of torque a motor produces at rated power and full speed can be found by using a horsepower-to-torque conversion chart. When using the conversion chart, place a straight edge along the two known quantities and read the unknown quantity on the third line.

To calculate motor full-load torque, apply this formula:

T = HP x 5252
    rpm

T = torque (in lb-ft)
HP = horsepower
5252 = constant
rpm = revolutions per minute

Example: What is the FLT (Full-load torque) of a 30HP motor operating at 1725 rpm?

T = HP x 5252
    rpm
T = 30 x 5252
     1725
T = 157,560
     1725
T = 91.34 lb-ft

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Calculating Horsepower:

Electrical power is rated in horsepower or watts. A horsepower is a unit of power equal to 746 watts or 33,0000 lb-ft per minute (550 lb-ft per second). A watt is a unit of measure equal to the power produced by a current of 1 amp across the potential difference of 1 volt. It is 1/746 of 1 horsepower. The watt is the base unit of electrical power. Motor power is rated in horsepower and watts.
Horsepower is used to measure the energy produced by an electric motor while doing work.

To calculate the horsepower of a motor when current and efficiency, and voltage are known, apply this formula:

HP = V x I x Eff
        746

HP = horsepower
V    = voltage
I     = curent (amps)
Eff. = efficiency

Example: What is the horsepower of a 230v motor pulling 4 amps and having 82% efficiency?

HP = V x I x Eff
        746
HP = 230 x 4 x .82
        746
HP = 754.4
        746
HP = 1 Hp

Eff = efficiency / HP = horsepower / V = volts / A = amps / PF = power factor

Horsepower Formulas
To Find	Use Formula	Example
		Given	Find	Solution
HP	HP = I X E X Eff.        746	240V, 20A, 85% Eff.	HP	HP = 240V x 20A x 85%        746 HP=5.5
I	I = HP x 746      E X Eff x PF	10HP, 240V, 90% Eff., 88% PF	I	I = 10HP x 746       240V x 90% x 88% I = 39 A

To calculate the horsepower of a motor when the speed and torque are known, apply this formula:

HP = rpm x T(torque)
         5252(constant)

Example: What is the horsepower of a 1725 rpm motor with a FLT 3.1 lb-ft?

HP = rpm x T
         5252
HP = 1725 x 3.1
         5252
HP = 5347.5
         5252
HP = 1 hp

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Calculating Synchronous Speed:

AC motors are considered constant speed motors. This is because the synchronous speed of an induction motor is based on the supply frequency and the number of poles in the motor winding. Motor are designed for 60 hz use have synchronous speeds of 3600, 1800, 1200, 900, 720, 600, 514, and 450 rpm.

To calculate synchronous speed of an induction motor, apply this formula:

rpmsyn = 120 x f
              Np

rpmsyn = synchronous speed (in rpm)
f           = supply frequency in (cycles/sec)
Np       =  number of motor poles

Example: What is the synchronous speed of a four pole motor operating at 50 hz.?

rpmsyn = 120 x f
              Np
rpmsyn = 120 x 50
              4
rpmsyn = 6000
              4
rpmsyn = 1500 rpm