EC&M’s Electrical Calculations Handbook - Chapter 5 potx

Short-Circuit CalculationsWhen the current flow path is directed correctly, the pres-sure of the source voltage forces normal current magnitudes to flow through the load impedances.. In

Short-Circuit Calculations

When the current flow path is directed correctly, the pres-sure of the source voltage forces normal current magnitudes

to flow through the load impedances During this time, the insulation surrounding the energized conductors prevents current from flowing through any path other than through the load impedance In this situation, the load impedance is large enough to limit the current flow to “normal” low val-ues in accordance with Ohm’s law:

E  I  Z

Problems arise, however, when the conductor insulation fails, permitting a shortened path for electron flow than

through the load impedance If the shortened path, or short circuit (also known as a fault), permits contact between a

phase conductor and an equipment grounding conductor,

this is known as a ground fault, or a phase-to-ground fault.

If, however, the shortened path instead permits contact between two or three phase conductors, then it is known as

a phase-to-phase fault.

If a solid connection is made between the faulted phase con-ductor and the other phase wire or the equipment grounding

conductor, then the short circuit is identified as a bolted fault.

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In bolted faults, little or no arcing exists, the voltage drop across the very low impedance of the almost-nonexistent arc

is very low, and the fault current is of high magnitude

If an arcing connection is made between the faulted phase conductor and the other phase wire or the equipment grounding conductor, then the short circuit is identified as an

arcing fault, with its associated lowered fault current flow In

certain systems, this arcing fault current can be so low that

it is not recognizable as a problem to the upstream overcur-rent device During such events, excessive heat buildup around the arc can occur, causing further damage to the oth-erwise sound electrical system or starting fires in nearby structures or processes Even though the fault may be small, heat energy flows from it over time to the surroundings, eventually causing high temperatures High temperatures applied to most wire insulation systems cause plastic defor-mation and failure, frequently resulting in an even more severe short circuit than the original one It is because of this consideration that everything possible must be done to inter-rupt current flow to a fault as soon as possible after the beginning of the fault or to limit fault current flow in some way, such as with transformers or grounding resistors Considering that fault current can reach several hundred thousand amperes, interrupting fault current flow can be a formidable task, so one rating of circuit breakers and fuses intended to interrupt fault currents is the ampere rating that each can interrupt These device ampere ratings fre-quently are given in terms of symmetrical current values (providing for no direct-current offset) or in terms of asym-metrical current values that include provision for direct-cur-rent (dc) offset The relative amount of dc offset and resulting asymmetrical current value multiplier used to multiply by the symmetrical current value are related to the

X/R value of the system at the point of the fault With a

highly resistive circuit having little reactance, the difference between the symmetrical and asymmetrical current values are minimal (there could be no difference at all), but with a highly reactive system containing only a small resistance value, the asymmetrical current value could be a multiple of three to five times the symmetrical value, or higher

Large values of current cause large magnetic forces When fault current flows through switchgear or a circuit breaker, the bus bars therein are attracted to one another and then are forced apart from one another at a frequency

of 50 or 60 times per second depending on the frequency of the power source This equipment must be braced to with-stand these forces, and this bracing is rated and is called the

withstand rating of the equipment Thus a circuit breaker

must be rated in full-load amperes for normal operation, it must have a withstand rating, and it must have an inter-rupting rating

The interrupting rating of a circuit breaker depends largely

on how fast its contacts can operate to begin to interrupt the current flow in the circuit A breaker that is extremely fast, with a contact parting time of, say, one cycle, must be able to

interrupt much more current in a system with a large X/R

value than would a slower breaker Fuses, on the other hand, are able to open and interrupt fault current in less than 14

cycle, so fuses must carry accordingly greater ratings Although the calculation of the asymmetrical current val-ues from the symmetrical current valval-ues is important, most of the short-circuit calculation work goes into determining the symmetrical short current availability at the prospective fault point in the electrical power system This symmetrical current calculation normally is done for a phase-to-phase fault, with the knowledge that this is normally the most demanding case Rarely (only where the fault occurs near very large electrical machines having solidly grounded wye connections) does the phase-to-ground fault exceed the phase-to-phase fault in cur-rent availability Therefore, concentration is on the phase-to-phase fault for the electrical power system with the largest expected quantity of rotating machines that will exist on the system in the foreseeable future

Short-circuit currents present a huge amount of destruc-tive energy that can be released through electrical systems under fault conditions These currents can cause serious damage to electrical systems and to equipment or nearby persons Protecting persons and electrical systems against damage during short-circuit conditions is required by the

National Electrical Code (Secs 110-9 and 110-10).

Not only should short-circuit studies be performed when a facility electrical system is first designed, but they also should be updated when a major modification or renovation takes place and no less frequently than every 5 years Major changes would include a change by the electrical utility, a change in the primary or secondary system configuration within the facility, a change in transformer size or imped-ance, a change in conductor lengths or sizes, or a change in the motors that are energized by a system

When modifications to the electrical system increase the value of available short-circuit current, a review of overcur-rent protection device interrupting ratings and equipment withstand ratings should be made This may entail replacing overcurrent protection devices or installing current-limiting devices such as current-limiting fuses, current-limiting circuit breakers, or current-limiting reactors The key is to know, as accurately as possible, how much short-circuit current is available at every point within the electrical power system

Sources of Short-Circuit Current

Every electrical system confines electric current flow to

select-ed paths by surrounding the conductors with insulators of var-ious types Short-circuit current is the flow of electrical energy that results when the insulation barrier fails and allows cur-rent to flow in a shorter path than the intended circuit In nor-mal operation, the impedances of the electrical appliance loads limit the current flow to relatively small values, but a short-cir-cuit path bypasses the normal current-limiting load imped-ance The result is excessively high current values that are limited only by the limitations of the power source itself and by the small impedances of the conductive elements that still remain in the path between the power source and the short-circuit point Short-short-circuit calculations are used to determine how much current can flow at certain points in the electrical system so that the electrical equipment can be selected to with-stand and interrupt that magnitude of fault current In short-circuit calculations, the contribution of current sources is first determined, and then the current-limiting effects of

imped-ances in the system are considered in determining how much current can flow in a particular system part

There are three basic sources of short-circuit currents:

■ The electrical utility

■ Motors

■ On-site generators

There are two types of motors that contribute short-circuit current:

■ Induction motors

■ Synchronous motors

Between these sources of short-circuit current and the point

of the short circuit, various impedances act to limit (impede) the flow of current and thus reduce the actual amount of short-circuit current “available” to flow into a short circuit Naturally, the value of these impedances is different at every point within an electrical system; therefore, the magnitudes

of short-circuit currents available to flow into a short circuit

at different places within the electrical system vary as well Several calculation methods are used to determine short-circuit currents, and reasonably accurate results can be derived by system simplifications prior to actually perform-ing the calculations For example, it is common to ignore the impedance effect of cables except for locations where the cables are very long and represent a large part of the overall short-circuit current path impedance Accordingly,

in the most common form of short-circuit calculations, short-circuit current is considered to be produced by gener-ators and motors, and its flow is considered to be impeded only by transformers and reactors

The Ability of the Electrical Utility System to

Produce Short-Circuit Current

By definition, the source-fault capacity is the maximum out-put capability the utility can produce at system voltage

Generally, this value can be gotten from the electrical

utili-ty company by a simple request and is most often given in amperes or kilovoltamperes

Suppose that the utility company electrical system inter-face data are given as

MVASC 2500 at 138 kilovolts (kV) with an X/R

 7 at the interface point

For this system, the utility can deliver 2,500,000 kilo-voltamperes (kVA) ÷ [138 kV(兹3苶)], or a total of 10,459 sym-metrical amperes (A) of short-circuit current

The short-circuit value from the electrical utility company will be “added to” by virtue of contributions from the on-site generator and motor loads within the plant or building elec-trical power system That is, the short-circuit value at the interface point with the electrical utility will be greater than just the value of the utility contribution alone

Short-Circuit Contributions of On-Site

Generators

The nameplate of each on-site generator is marked with its

subtransient reactance X d ″ like this This subtransient

val-ue occurs immediately after a short circuit and only contin-ues for a few cycles For short-circuit current calculations, the subtransient reactance value is used because it produces the most short-circuit current

Determining how many kilovoltamperes an on-site genera-tor can contribute to the short-circuit current of an electrical power system is a simple one-step process, solved as follows:

Short-circuit kVA

For example, a typical synchronous generator connected to

a 5000 shaft horsepower (shp) gas turbine engine is rated at

7265 kVA, and its subtransient reactance X d″ is 0.17 The

generator kVA rating

d ″ rating

problem is to determine the short-circuit output capabilities

of this generator Thus

Short-circuit kVA  42,735 kVA

Note that when more than one source of short-circuit cur-rent is present in a system, the resulting amount of short-cir-cuit kilovoltamperes available to flow through a short cirshort-cir-cuit

is simply the arithmetic sum of the kilovoltamperes from the

various sources, and the total equivalent kilovoltamperes from

all these sources is simply the arithmetic sum of the individ-ual equivalent kilovoltampere values

For example, if two of the preceding 7625-kVA, 138-kV synchronous generators are cogenerating into an electrical power system that is also supplied from an electrical utility whose capabilities on the generator side of the utility trans-former are 300 million voltamperes (MVA), how many total fault kilovoltamperes are available at the terminals of the generator? The answer is

Total kVA 2 (42,735 kVA)  300,000 kVA  385,470 kVA

It is worthwhile to note that the kilowatt rating of the

gen-erator set is not used in this calculation Instead, the

kilo-voltampere rating of the electrical dynamo in the generator set is used because it (not the engine) determines how many short-circuit kilovoltamperes can be delivered momentarily into a fault This is so because most of the fault current is quadrature-component current having a very lagging power factor Another way of saying this is that the fault current does little work because of its poor power factor, and the size

of the engine determines the kilowatt value of real work the generator set can do

Short-Circuit Contributions of Motors

The nameplate of each motor is marked with its locked-rotor code letter as well as with its rated continuous full-load current

7265 kVA

and horsepower The locked-rotor value occurs immediately on motor energization and also immediately after a short circuit After a short circuit, this large amount of current flow through the motor only continues for a few cycles (even less time with induction motors) Unless specific information about the indi-vidual motor locked-rotor current characteristics is known, normally, a value of six times the motor full-load current is

assumed Dividing 100 percent current by 600 percent current produces the normally assumed X ″ motor value of 0.17 It is

this value that is used to determine the short-circuit power contribution of a motor or group of motors The characteristics

of small induction motors limit current flow more than those

of larger motors; therefore, for calculations involving motors smaller than 50 horsepower (hp), the 0.17 value must be increased to 0.20 In addition, the approximation that 1 hp equals 1 kVA normally can be made with negligible error,

par-ticularly with motors of less than 200 hp

Determining the short-circuit contribution from a motor is

a simple one-step process solved as follows:

Short-circuit power 

(Note: 1 hp  1 kVA for this calculation.)

Unless specific values of subtransient reactance or motor locked-rotor code letter are known, for motors of 50 hp and less, use 0.20 for the subtransient reactance For individual motors or groups of motors producing 50 hp and more, use 0.17 for the subtransient reactance Always assume that 1

hp  1 kVA.

For example, a typical 460-volt (V) motor is rated at 40 hp Determine the short-circuit contribution of this motor Thus

Short-circuit power   200 kVA

If there were 10 identical motors of this same rating con-nected to one bus, then their total short-circuit contribution

at the bus would be 10  200 kVA  2000 kVA.

40 kVA

0.20

motor horsepower rating

Note that for motors of 50 hp or larger, the X d″ rating of 0.17 should be used For example, a typical 4.16-kV motor is rated at 2000 hp Determine the short-circuit power contri-bution of this motor

Setting 1 hp equal to 1 kVA, we get

Short-circuit power   11,764 kVA

If all the short-circuit current contributors were at the same voltage, then short-circuit currents simply could be added to one another to determine the total amount avail-able to flow in a system However, because transformers greatly lessen the flow of fault current and also change volt-ages, the impact of transformers must be considered, and fault values transported through transformers must be cal-culated in terms of power instead of amperes

Let-Through Values of Transformers

For any transformer, the maximum amount of power (the fault capacity) that the transformer will permit to flow from one side of the transformer to the other side, in kilovoltam-peres, is calculated as

Let-through power 

For example, a 2000-kVA transformer has a nameplate impedance of 6.75 percent, and its voltage ratings are 13.8 kV-277/480 V, three phase Find the maximum amount of power that this transformer can let through if it is energized from an infinite power source The answer is

Let-through power 

 2000  29,630 kVA

0.0675

transformer kilovoltampere rating



%Z/100

transformer kilovoltampere rating

2000 kVA

If the power source upstream of the transformer is not infinite, then the amount of power that would be available

on the load side of the transformer would be less, and it is calculated using admittances as follows:

■ Utility short-circuit power is UP, and its admittance is 1/UP.

■ Maximum transformer let-through is T, and its admit-tance is 1/T.

■ Net power from the utility let through the transformer P

is calculated as

P

Where the utility can supply 385,000 kVA, the amount that the preceding 2000-kVA, 6.75 percent impedance trans-former can let through, or admit, is calculated as

Let-Through Values of Reactors

Similar to a transformer coil in its current-limiting charac-teristics, a reactor is a series impedance used to limit fault current In a three-phase circuit, there are normally three identical reactor coils, each connected in series with the phase conductors between the source and the electrical load The amount of power that a reactor will let through from an infinite power source to a short circuit on the output termi-nals of the reactor is calculated as

Let-through power

 1000 (phase-to-phase kilovolt circuit rating)reactor impedance in ohms per phase 2

1

(1/385,000)  (1/29,630)

1

(1/UP)  (1/T)