Applications to Sizing Circuit Breakers (ANSI C37)

Fault Calculations

2.7 Applications to Sizing Circuit Breakers (ANSI C37) The application of fault calculations to power systems is considered, including the simplified method of short-circuit calculation (the E / X method), as utilized in the IEEE and ANSI standards for rating circuit breakers and fuses. However, the presentation is not a complete discussion of sizing and rating breakers and fuses. Short-circuit calculations for protective relaying are also briefly discussed.

2.7.1 Introduction There are several applications for short-circuit calculations in power systems, including: fuse and circuit breaker sizing, protective relay setting, and calculation of maximum values of electromagnetic forces applied to conductors. The classical theory of rotating machinery gives an accurate mathematical expression for the worst three-phase short circuit current for an initially unloaded synchronous generator with an external reactance of Xe.

isc(t) = 1.414 Iac cos( t + ) + Idc exp(-t / Ta) Idc = 1.414 E / (Xd" + Xe)

Iac = (I" - I') exp(-t / Td") + (I' - I) exp(-t / Td') + I

I = E / (Xd + Xe) = rms steady-state short-circuit current

I' = E / (Xd' + Xe) = rms transient short-circuit current

I" = E / (Xd" + Xe) = rms subtransient short-circuit current

E = prefault voltage (usually 1.00 per unit)

Xd = synchronous reactance

Xd' = transient reactance

Xd" = subtransient reactance

Td' = Tdo' (Xd' + Xe) / (Xd + Xe)

Td" = Tdo" (Xd" + Xe) / (Xd' + Xe)

Tdo' = open-circuit transient time constant

Tdo" = open-circuit subtransient time constant

Usually, the rms short-circuit current is the most interesting quantity to compute:

Irms(t) = (Iac2 + Idc2)1/2

Notice that the rms current decreases as time proceeds, due to decrements in both the AC and the DC components. This expression can be used for any short-circuit calculation, for example the momentary duty on a circuit breaker would be Irms at t = 1/2 cycle and the circuit breaker interrupting duty would be Irms at t = contact parting time.

This is a rigorous approach, but it is so laborious for systems with many generators in an extensive network that simpler methods are generally used in most applications.

-2.7.1-

C. W. Brice August 2002

2.7.2 Simplified methods The original ANSI (American National Standards Institute) standards for circuit breakers C37.4-1953, C37.5-1953, and C37.6-1953 used an interrupting rating that was based on the "... rms value including the DC component at the instant of contact separation as determined from the envelope of the current wave." This is known as the total current basis of rating a circuit breaker. A simplified method was included in the standards to allow the symmetrical short-circuit current calculation to be multiplied by a factor (depending on system characteristics and breaker speed) to obtain the worst-case total rms current at contact separation.

In the early 1950's the AIEE (one of the IEEE's predecessors) Switchgear Committee began work on revising the C37 standards and developing circuit breaker ratings based on symmetrical interrupting currents. This effort eventually led to the standards labeled ANSI C37.04, C37.06, C37.010, etc. These standards utilize the Symmetrical Current Basis of Rating contrasted to the previous Total Current Basis of Rating (which are labeled C37.4, C37.5, C37.6, etc.). At present both rating structures are still in use although the understanding is that new circuit breaker developments will be directed toward the symmetrical standards.

The simplified method of calculating short-circuit currents is to calculate the symmetrical short-circuit current produced by modeling each generator as a voltage source behind an appropriate reactance. Use of the subtransient reactance gives the initial symmetrical short-circuit current, and use of the transient reactance gives the short-circuit current a few cycles later. The DC component is ignored. The methods discussed previously may be used to perform this calculation using a digital computer program. Alternately, manual circuit analysis methods may be used for small systems.

The calculation of short-circuit currents by the simplified (or E / X) method is an acceptable approximation for most purposes, except for faults electrically close to large generating units. Most high-voltage lines and transformers have reactance values that are much larger than their resistance values. In cases where the resistance is significant (usually, but not always, low-voltage devices) the magnitude of the impedance Z may be substituted for the reactance X.

If the value of E / X does not exceed 80% (70% for a line-to-ground fault) of the symmetrical interrupting capability of a circuit breaker, the simplified method may always be used to size circuit breakers. This may result in excessively conservative design, though, and ANSI C37.010-1979 gives a set of factors, whose values depend on the system X / R ratio and the breaker speed, that may be used to adjust for AC and DC decrements. This will be discussed in more detail later.

It has been tacitly assumed that the simplified method of calculating the short-circuit current was to be used. This is consistent with usual practice, since multiplying factors can usually be used to estimate the total (or asymmetrical) current if needed.

2.7.3 System X / R ratio calculation and significance To calculate the system X / R ratio for a fault at a given location, there are two different (but equivalent) methods:

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Fault Calculations

The first involves network reductions to find the Thevenin equivalent reactance and resistance. The proper procedure is to construct a network of reactances (neglecting all resistances regardless of value), then reduce the network to find the Thevenin reactance at the point of the fault. Next, construct a network of resistances (neglecting all reactances regardless of value), and then reduce this network to find the Thevenin resistance at the point of the fault. The X / R ratio for a fault in this location is the ratio of the Thevenin reactance to the Thevenin resistance

The second method is to use the Zbus matrix building algorithm that was discussed earlier. The program must be run twice, first for the reactances neglecting all resistance (call this result Xbus), then for the resistances neglecting all reactances (call this result Rbus). The X / R ratio for a fault at bus number i is then found by the ratio of the i-th diagonal element of Xbus over the i-th diagonal element of Rbus.

(X / R)i = Xbusii / Rbusii

The advantage of this approach is that X / R ratios for faults at every bus are determined from running the Zbus building algorithm only twice.

Note that the Zbus algorithm can handle the full R + jX impedance data, but the procedure outlined above should be used always for calculation of X / R. This will ensure that the X / R ratio is always over-estimated, thereby giving conservative results. If we were to use complex impedance data in the Zbus building algorithm there is a possibility of non-conservative results. Note also that the ANSI standards that we will use specify the separate network reductions for X's and then for R's, which is equivalent to running the Zbus building algorithms twice, first with only X's, then with only R's.

Practice Problem: For the system shown in Figure 2.7.1, calculate the X / R ratio for a three-phase short circuit at bus B1. Be sure to omit all resistances in the reactance calculation and to omit all reactances in the resistance calculation.

50 MILE LINE

0.60 + j 0.80OHMS PER MILE

G1: 100 MVA X = 15% R = 0.50% T1: 100 MVA X = 7.0% R = 0.35%

T2: 75 MVA X = 7.0% R = 0.35%G2: 75 MVA X = 10% R = 0.50%

T1

13.8 kV

G1

115 kV

B1 B2

115 kV 13.8 kV

T2

G2

Figure 2.7.1 Practice problem for calculation of X/R ratio.

Solution: Convert data to per unit on 100 MVA base:

G1: Xd" = .15, R = .005

G2: Xd" = .10 100/75 = .1333, R = .005 100/75 = .00667

-2.7.3-


T1: X = .070, R = .0035

T2: X = .070 100/75 = .0933, R = .0035 100/75 = .00467

115 kV line: X = .80 x 50 x 100/1152 = .302, R = .60 x 50 x 100/1152 = .227

For a fault at B1

Xthev = .220 x .529/(.220 + .529) = 0.1554

Rthev = .0085 x .238/(.0085 + .238) = 0.00821

X / R = 0.1554/.00821 = 18.9

For the sake of comparison, the calculation using full 60 Hz impedance data gives Zthev = 0.162/81.8o, and X / R = tan 81.8o = 6.94, which is smaller than the value computed above. Note that the first method is the correct one, in the sense that it is the more conservative approach and is consistent with the ANSI C37 standard. The differences between the two calculations are not always this great.

The reason for our interest in the X / R ratio is twofold: 1) we need it get appropriate multiplying factors in the ANSI C37 standards, and 2) it has fundamental physical significance. To illustrate the latter point, consider a simple series RL circuit excited by an ideal sinusoidal voltage source (an infinite bus). At t = 0, a switch closes connecting the source to the RL circuit.

e(t) = 1.414 E sin(t + ) = R i + L di/dt i(0) = 0

The complete solution (transient plus steady-state) is

i(t) = 1.414 I { sin(t + - ) - sin( - ) exp[-(R / X) t] } where = tan-1(X / R) and I = E / Z = (E / X) / [ (R / X)2 + 1 ]1/2 Note that in a real system with multiple generators and a network of lines, there is no single value of X / R. Our calculations are simply good engineering approximations.

2.7.4 First-cycle duties for fuses and low-voltage circuit breakers Usually, the first-cycle duty is the only value of short-circuit current needed for fuses and low-voltage breakers. The subtransient reactances are to be used for all generators and for all motors (assuming that data is available for the motors, which is sometimes not true for groups of low-voltage motors).

A group of low-voltage motors is often fed from a low-voltage substation. If the substation supplies only this group of motors, then the transformer kVA rating (self-cooled) is usually approximately equal to the total motor horsepower rating. If this is so, then the entire group of motors may be represented by a single reactance of 25% on a kVA base equal to the self-cooled rating of the transformer. If the total motor horsepower rating is not equal to the transformer kVA rating then the reactance value should be adjusted appropriately (as the motor HP rating

-2.7.4-

Fault Calculations

decreases, the contribution to fault current decreases, and the reactance should increase accordingly).

If motor contributions to low-voltage system faults cannot be calculated due to lack of data, the following "rule of thumb" may be used:

Imotor = 4 X (Sum of the Rated Current of Motors Connected to a Faulted Bus)

This is usually a satisfactory approximation, but actual calculation of motor currents is more accurate.

According to ANSI C37.13, short-circuit calculations for sizing low-voltage power circuit breakers usually do not need to consider the X / R ratio. The initial symmetrical short-circuit current I" can be used directly, since ample margins are built in to the breaker ratings for most purposes. There are some exceptions, however, which are listed below: 1) Local generation at circuit breaker voltage in sizes greater than 500 kVA, 2) Gas-filled or dry-type transformers in sizes greater than 1000 kVA, 3) Any transformer in sizes 2500 kVA and larger, 4) Network systems, 5) Transformers with impedances higher than those specified in ANSI C57 standard, 6) Current-limiting reactors at circuit breaker voltage on the source side, 7) Current-limiting busway at circuit breaker voltage on source side, 8) Any other application where available short-circuit current approaches 80% of the breaker short-circuit current rating (note: this is unlikely).

For any of these exceptions, use this table from the ANSI C37.13 standard:

Power Factor System X / R Multiplying Factor Unfused

Breaker Fused

Breaker 20% 4.90 1.00 1.00 15% 6.60 1.00 1.07 12% 8.27 1.04 1.11 10% 9.95 1.07 1.15 8.5% 11.72 1.09 1.18 7.0% 14.25 1.11 1.21 5.0% 20.00 1.15 1.26

Note that most low-voltage short-circuit duty calculations use the initial symmetrical short-circuit current.

Note also that for many low-voltage systems, the resistance should not be neglected. Instead of E/X, use E/Z, where

Z = (R2 + X2)1/2

-2.7.5-


Other relevant standards are NEMA AB 1 Molded-Case Circuit Breakers and NEMA SG 3 Low-Voltage Power Circuit Breakers, IEEE 141 Recommended Practice for Electric Power Distribution for Industrial Plants (The Red Book), IEEE 242 Recommended Practice for Protection and Coordination of Industrial and Commercial Power Systems (The Buff Book).

The C37 series contains several standards related to fuses, including ANSI C37.41, C37.42, C37.44, C37.45, and C37.46. Fuses may have either symmetrical or total current ratings. Use the subtransient reactances for rotating machines. For fuses with symmetrical current interrupting ratings, use the initial symmetrical short-circuit current I" to calculate the duty. If the fuse is rated on a total current basis, a multiplying factor is specified in ANSI C37.41, as follows:

Multiplying factor = 1.55 in most cases. Multiplying factor = 1.20 for special cases.

Special cases are (a) all distribution fuse cutouts and (b) power fuses (but not current-limiting fuses) that are remote from stations. The special case applies only if remote from generation, the system X / R ratio is less than 4, and the fuse is applied at 15 kV or less.

Example: X / R = 3 I" = 5000 A

Distribution fuse cutout with a total current rating. System is remote from all generation. Interrupting duty = 1.20 x 5000 A = 6000 A

Now replace the fuse cutout with one having a symmetrical current rating: Interrupting duty = 5000 A

Now replace with a current-limiting power fuse with a total current rating: Interrupting duty = 1.55 X 5000 A = 7750 A

If the current-limiting fuse has a symmetrical current rating: Interrupting duty = 5000 A

The point of this example is that the basis for the rating (symmetrical current or total current) must be known for a proper application.

Practice Problem: In the previous example, a current-limiting fuse is to be applied on the low-voltage side of the transformer, which serves an induction motor. The transformer has an X / R ratio of 8.00, and a reactance of 5.00% on a 100 kVA base. Calculate the three-phase short-circuit current through the fuse for a fault on the motor terminals. Calculate the X / R ratio and the interrupting duty (both symmetrical and total current duties).

Solution: 4000 V / 1.732 = 2309 V L-N Xsource = 2309 V / 5000 A = 0.462 ohms per phase Xtransf = .05 (4.00 kV)2 / 100 kVA = 8.00 ohms per phase Xtot = Xsource + Xtransf = 8.46 ohms per phase

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Fault Calculations

I" = 2309 V / 8.46 ohms = 273 A on high-voltage side I" = (4000 / 208) 273 A = 5250 A on low-voltage side R = 0.462 / 3 + 8.00 / 8 = 1.154 ohms per phase X / R = 8.46/1.154 = 7.33 Symmetrical: 5250 A. Asymmetrical: 1.55 x 5250 = 8140 A.

2.7.5 First-cycle duties for medium and high voltage circuit breakers This section applies to circuit breakers for applications with nominal voltage greater than 1000 V. The first-cycle duty is for comparison with the breaker momentary rating (or closing and latching capability).

The rotating machine reactances to be used in this study are in the following table (from C37.5 and C37.010):

Turbo-generators, hydro-generators with damper

windings, and all condensers

1.00 Xd"

Hydro-generators with no damper windings

0.75 Xd'

Synchronous motors 1.00 Xd"

Induction motors

above 1000 hp at 1800 rpm or less

above 250 hp at 3600 rpm 1.00 Xd"

50 - 1000 hp at 1800 rpm or less

50 - 250 hp at 3600 rpm 1.20 Xd"

Below 50 hp Neglect

For induction motors Xd" is one divided by the per-unit locked-rotor current at rated voltage. Induction motors and small synchronous motors are often neglected in utility short-circuit calculations, but should be included on station service systems and on substations supplying large industrial customers. Industrial distribution systems must include motor contributions. These multiplying factors account for the fact that the motor current will decay with a subtransient time constant, and more accurate multipliers may be obtained from the manufacturer for specific large motors that contribute significantly to the fault current.

With these machine reactances, compute the initial symmetrical short-circuit current I". The first-cycle duty is 1.60 I". This quantity approximates the duty on the breaker during the first

-2.7.7-


half cycle of short-circuit current. The breaker must be able to close and latch, while withstanding the mechanical forces produced by this large current. This is particularly severe on systems with large motor contributions.

Breakers that are rated on a total current basis have a momentary rating, which must exceed the calculated first-cycle duty. Breakers that are rated on a symmetrical current basis have this same rating, but it is called the closing and latching capability.

Example: A circuit breaker on a 115 kV system has a rated maximum voltage of 121 kV, a rated continuous current of 1200 A, a rated interrupting time of 3 cycles, a maximum symmetrical interrupting capability of 20 kA, and a closing and latching capability of 32 kA. If I" = 15 kA then the first-cycle duty is 1.6 I" = 24 kA. Since the first-cycle duty is less than the closing and latching capability, the breaker is adequate in this regard. If this example were repeated for I" = 25 kA the first cycle duty would be 40 kA, and the breaker would not have a sufficient closing and latching capability. Note that the first-cycle duty is a total (or asymmetrical) current quantity, even though the breaker has a symmetrical interrupting rating.

Practice Problem: A circuit breaker on a 13.8 kV system has a maximum symmetrical interrupting capability of 48 kA, and a closing and latching capability of 77 kA. If the initial symmetrical short-circuit current is I" = 45 kA, calculate the first-cycle duty and evaluate the breaker's closing and latching capability.

Solution: The first-cycle duty is 1.60 x 45 kA = 72 kA. The closing and latching capability of 77 kA is adequate, but if future short-circuit currents are expected to increase significantly the breaker will not be adequate.

2.7.6 Interrupting duties for medium and high voltage circuit breakers For the interrupting duty calculation, we must know the contact parting time. Figure 2.7.2, which comes from ANSI C37.010, shows that the contact parting time is the sum of the tripping delay and the opening time. This is the total time between the initiation of the short circuit and the parting of the primary arcing contacts.

-2.7.8-

Fault Calculations

PRIMARYCONTACTS

PART

CONTACT PARTING TIME

INTERRUPTING TIME

INITIATIONOF SHORTCIRCUIT

TRIPCIRCUIT

ENERGIZED

TRIPPINGDELAY

OPENINGTIME

ARC ONPRIMARYCONTACTS

EXTINGUISHED

TIMEARCING

TIME

Figure 2.7.2 Events in the clearing of a fault by a circuit breaker.

The breaker contacts part, which initiates an arc that is subsequently extinguished by the breaker. The arcing time plus the opening time is the interrupting time. There is a relationship between the rated interrupting time for a given breaker and its minimum contact parting time. This is shown in the table below. Note that the actual contact parting time (but not the interrupting time) can always be increased by time delays in the protective relaying system.

Table showing the relationship between the interrupting time and the contact parting time:

Rated Interrupting Time Minimum Contact Parting Time 2 Cycles 1.5 Cycles 3 Cycles 2 Cycles 5 Cycles 3 Cycles 8 Cycles 4 Cycles

The rotating machine reactances to be used in interrupting duty calculations are in the following table (from C37.5 and C37.010):

-2.7.9-


Turbo-generators, hydro-generators with damper

windings, and all condensers

1.00 Xd"

Hydro-generators with no damper windings

0.75 Xd'

Synchronous motors 1.50 Xd"

Induction motors

above 1000 hp at 1800 rpm or less

above 250 hp at 3600 rpm 1.50 Xd"

50 - 1000 hp at 1800 rpm or less

50 - 250 hp at 3600 rpm 3.00 Xd"

Below 50 hp Neglect

Perform a short-circuit study using these machine reactances, and find the initial symmetrical short-circuit current I". Determine the X / R ratio. Remember to determine X, neglecting R's; then determine R, neglecting X's.

The procedure now depends on whether the breaker is rated on a total current basis or on a symmetrical current basis. We consider each case in turn.

A. Total current basis: If the fault is electrically close to a generator, both AC and DC decrements need to be considered. If the fault is remote from generation, then only DC decrements need to be considered. In this context, electrically close means that there is at most one transformer between the fault and the nearest generator, or that the external impedance between the fault and the generator is less than 1.50 times the Xd" value of the generator.

Using ANSI C37.5, look up the appropriate multiplying factor from one of three curves, which are reproduced in Figure 2.7.4. The total interrupting duty is the calculated value of the initial symmetrical short-circuit current I" times the multiplying factor. The first curve is for three-phase faults electrically close to generation. The second curve is for single line to ground faults electrically close to generation. The third curve is for either type of fault remote from generation.

Multiplying the appropriate factor times the initial symmetrical short-circuit current I" gives the total current interrupting duty. This is to be compared to the circuit breaker interrupting rating.

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Fault Calculations

Older breakers (1960's and before) were often rated in MVA. The example below is taken, in part, from Wagner [see bibliography at end of document].

Example: Breaker rating 5000 MVA at 69 kV. The breaker has a rated-voltage interrupting capability of

5000 MVA / (1.732 x 69 kV) = 42 kA

The maximum interrupting capability of the breaker is 44 kA. If the operating voltage is 67 kV, the breaker can interrupt a current proportionately larger:

42 kA x 69 kV / 67 kV = 43 kA

If the operating voltage is 64 kV, a proportionate increase in the interrupting capability would be 42 kA x 69 kV / 64 kV = 45 kA, but that would exceed the maximum capability of 44 kA, so the breaker interrupting capability is 44 kA at an operating voltage of 64 kV.

Now consider applying this breaker on a system with a Thevenin reactance of 0.05 per unit (on 100 MVA and 69 kV bases) and an X / R = 20, generation remote. Let the breaker have a rated interrupting time of 5 cycles, which corresponds to a contact parting time of 3 cycles. This gives a multiplying factor of 1.13 (from the third curve), and a breaker duty of

I" = 1 / 0.05 = 20 per unit

Ibase = 100 MVA / (1.732 x 69 kV) = 0.837 kA

I" = 20 x 0.837 kA = 16.7 kA

Breaker duty = 1.13 x 16.7 kA = 18.9 kA

Since this is well below the 69 kV rating of 42 kA, the application is satisfactory.

-2.7.11-


Figure 2.7.3 (a) Multiplying factors for circuit breakers rated on symmetrical basis. Three-phase short circuits including both AC and DC decrements, to be used when fault is electrically close to generators (no more than one transformation away).

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Fault Calculations

Figure 2.7.3 (b) Multiplying factors for circuit breakers rated on symmetrical basis. Single phase to ground faults near generators.

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Figure 2.7.3 (c) Multiplying factors for circuit breakers rated on symmetrical basis. All faults remote from generators.

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Fault Calculations

Figure 2.7.4 Multiplying factors for circuit breakers rated on asymmetrical (total) short-circuit current basis.

-2.7.15-


Practice Problem: A 13.8 kV, 8 cycle (4 cycle contact parting time) circuit breaker has an interrupting rating of 42 kA (total current) and a momentary rating of 80 kA. If applied on a 13.8 kV system with X / R = 30, and X = 11% (on 100 MVA, 13.8 kV base), evaluate the breaker interrupting rating.

Solution:

I" = 1 / 0.11 = 9.09 pu

Ibase = 100 MVA / (1.732 x 13.8 kV) = 4.18 kA

I" = 9.09 x 4.18 kA = 38 kA

If the breaker is fed predominantly from nearby generation, the multiplying factor is 1.04, but the factor is 1.16 if it is remote from generation.

1.04 x 38 kA = 39.5 kA

1.16 x 38 kA = 44 kA

The application is satisfactory only if it is fed predominantly from nearby generation. Note that some judgment may be required in interpreting the word "predominantly". The remote assumption is always the conservative one.

B. Symmetrical current method: The newer ANSI standards have the multiplying factors built in to the circuit breaker ratings, so that the breaker has a symmetrical current interrupting rating (ANSI C37.010). The new standard is based on X / R = 15. If X / R exceeds 15, or if it is unknown, then the Simplified Method (E / X Method) can not be used. The exception is that if the symmetrical short-circuit current is < 80% of the breaker three-phase fault interrupting capability (70% for line-to-ground fault) then the application is satisfactory regardless of the X / R ratio.

If X / R < 15, the symmetrical short-circuit current can reach 100% of the breaker interrupting capability. If X / R > 15, the standard C37.010 contains curves of multiplying factors, which are to be multiplied by the symmetrical short-circuit current I" to obtain the breaker interrupting duty. These curves are duplicated in Figure 2.7.3. The use of these curves is called the E / X Method with Adjustments for Decrements.

The new breaker ratings are contained in ANSI C37.06, which defines 1) rated maximum voltage, 2) rated voltage range factor K, 3) rated short-circuit current, among others. The rated maximum voltage is the highest voltage for which the breaker is designed. For example, a system with nominal voltage of 115 kV would require a breaker with a maximum voltage of 121 kV, while a nominal voltage of 138 kV would require 145 kV maximum voltage.

Circuit breakers with maximum voltages above 72.5 kV have voltage range factors of K = 1.0, which means that their interrupting rating is independent of the operating voltage. In other words, the interrupting rating is a current rating. Breakers with maximum voltages of 72.5 kV and below, have voltage range factors 1.0 < K < 3.75. For these, the interrupting capability is inversely proportional to the operating voltage, up to a limit of K times the rated short-circuit current.

-2.7.16-

Fault Calculations

Example: A circuit breaker has a rated maximum voltage of 15 kV, a voltage range factor of K = 2.27, and a rated short-circuit current of 19.0 kA. Apply on a system with a maximum operating voltage of 13 kV. The interrupting capability is 19 kA X 15 kV / 13 kV = 21.9 kA. But K times the rated short-circuit current = the maximum interrupting capability = 2.27 X 19.0 kA = 43.1 kA. The actual interrupting capability is 21.9 kA.

Apply the same breaker on a 6 kV system. 19.0 kA X 15 / 6 = 47.5 kA. The maximum interrupting capability is 43 kA. The actual capability is 43 kA.

Note also that the close and latch capability is 1.6 K times the rated short-circuit current.

Practice Problem: A 13.8 kV system has a circuit breaker rated (on a symmetrical basis) as follows: rated maximum voltage = 15 kV, voltage range factor K = 1.30, rated interrupting time = 5 cycles (contact parting time = 3 cycles), maximum symmetrical interrupting capability = 48 kA, rated short-circuit current (at rated maximum voltage) = 37 kA, closing and latching capability = 77 kA. If the system X / R is unknown, and X = 14% (on 100 MVA, 13.8 kV base), evaluate the breaker interrupting capability.

Solution:

I" = 1 / 0.14 = 7.14 pu

Ibase = 4.18 kA

I" = 7.14 x 4.18 kA = 30 kA

The interrupting capability at 13.8 kV is 37 kA (15/13.8) = 40 kA. 80% of 40 kA is 32 kA, which is greater than the 30 kA interrupting duty. Thus, the breaker is satisfactory, regardless of the X / R ratio.

Example: A 5 cycle breaker has ratings as follows: maximum voltage = 38 kV, K = 1.65, rated short-circuit current = 22 kA. It is operated at 33 kV. The three-phase symmetrical current capability is 22 kA X 38 / 33 = 25.3 kA. Max. capability = 36.3 kA. Actual capability = 25.3 kA.

Note that the single line-to-ground capability is 1.15 times the three-phase capability, but not to exceed the maximum symmetrical capability.

Going back to the previous example, the line-to-ground fault capability is 1.15 X 25.3 kA = 29.1 kA, and the maximum capability is 36 kA. The actual capability is 29.1 kA.

If there is a need to calculate the asymmetrical (total) current capability, ANSI C37.010 gives S-factors to accomplish this. For example, a 5 cycle breaker (3 cycle contact parting time) has S = 1.10, which would give in the last example the following capabilities:

three phase fault: 1.1 X 25.3 = 27.8 kA

line to ground fault: 1.1 X 29.1 = 32 kA.

Another example: A 46 kV system, a 5 cycle breaker with maximum voltage of 48.3 kV, K = 1.21, rated short-circuit current = 17 kA. At 46 kV the three-phase capability is 17.9 kA, the line

-2.7.17-


to-ground capability is 20.5 kA, and the maximum capability is 20.6 kA. If the three-phase fault current is

I" = E / X = 16.5 kA

and the line-to-ground fault current is

I"LG = 3E / (2X1 + X0) = 17.0 kA

The three-phase capability is (16.5/17.9) 100% = 92% > 80%

The line-to-ground capability is (17/20.5) 100% = 83% > 70%

The multiplying factors must be used. Assume the breaker is remote from generation and the system X / R ratio is 25 for three-phase faults, and X / R = 30 for line-to-ground faults. Using minimum contact parting time of 3 cycles, the three-phase factor is 1.10 and the line-ground factor is 1.14 (from the third set of curves).

The three-phase duty is 1.10 x 16.5 kA = 18.2 kA.

The line-ground duty is 1.14 x 17.0 kA = 19.4 kA.

Since the three-phase capability is only 17.9 kA, this application is unsatisfactory. The line-ground capability of 20.5 kA is marginally satisfactory.

Practice Problem: Same as the previous practice problem, but X / R = 30, and X1 = 11%, X0 = 10%, a) electrically close to generation, b) remote from generation. Evaluate interrupting capability for three-phase and for line-to-ground short circuits.

Solution:

Three-phase short circuits:

I" = (1 / 0.11) x 4.18 kA = 38 kA

a) close to generation: multiplying factor is 1.03

Interrupting duty = 38 kA x 1.03 = 39 kA. Interrupting capability at 13.8 kV = 40 kA; application is satisfactory.

b) remote from generation: multiplying factor is 1.13

Interrupting duty = 38 kA x 1.13 = 43 kA.

Application is not satisfactory.

For a single line-to-ground short circuit:

I" = 3 x 1.00 / (.22 + .10) = 9.375 per unit

9.375 x 4.18 kA = 39.2 kA

Interrupting capability = 1.15 x 40.2 kA = 46.2 kA

a) close to generation:

Interrupting duty = 1.10 x 39.2 kA = 43.1 kA. Application is satisfactory.

-2.7.18-

Fault Calculations

b) remote from generation:

Interrupting duty = 1.13 x 39.2 kA = 44.3 kA. Application is satisfactory.

Although the application is satisfactory with regard to line-to-ground faults, it is satisfactory with regard to three-phase faults only if it is fed predominantly from generation.

A serious problem in sizing circuit breakers is unanticipated growth in the available short-circuit MVA. Any new construction of generating plants and lines tends to decrease the reactance of the system, which will result in an increase in the available short-circuit MVA. The only defenses against this problem are to size the breaker somewhat larger than the present system would require, or to install current limiting apparatus such as reactors.

Automatic reclosing of circuit breakers is a useful method of decreasing the severity of system faults. This is true since the majority of faults on many systems are temporary. Some breakers (especially oil circuit breakers) with more than two operations, or less than 15 second delay before reclosing, will require derating of the breaker. SF6 breakers usually do not require derating. This method of derating the breaker is given in ANSI C37.06.

Note that this is not a complete discussion of circuit breaker application. In particular, we have not mentioned transient recovery voltage (the voltage across the open breaker contacts after operation), or switching capacitors. See ANSI standards and Wagner for more information.

The last practice problem is an industrial power system with utility feed and in-plant generation. This problem is very similar to an illustrative example in IEEE 141 (the Red Book) but is not as extensive.

Practice Problem: For the system of Figure 2.7.5, perform a short-circuit study for a three-phase fault on the 13.8 kV bus. Calculate the first-cycle duty and the interrupting duty on the 13.8 kV breakers. Note that different reactances are used for the first-cycle duty than for the interrupting duty. This requires two short-circuit studies. Note also that the duty on all breakers connected to a bus is the total fault current into a bus fault. This is because the actual fault may be on either side of a given breaker, so that each one must be able to handle the total fault current at the bus. Select one of the circuit breakers listed below. All have a nominal voltage class of 13.8 kV rms, a rated maximum voltage of 15 kV rms, and a rated interrupting time of 5 cycles.

-2.7.19-


3 IND.MOTORS

69 kV SYSTEM: 1000 MVA SHORT CIRCUIT AVAILABLE

G1: 20 MVA GENERATOR Xd" = 0.09 X/R = 40

T1: 20 MVA X = 0.07 X/R = 21

EACH T2: 1.5 MVA X = 0.055 X/R = 10

EACH LV SUBSTATION HAS MOTOR LOAD:

0.5 MVA TOTAL MOTORS EACH 50 TO 150 HP

1.0 MVA TOTAL MOTORS EACH < 50 HP

Xd" = 0.25 X/R = 9

1750 HP IND. MOTORS, EACH:

1800 RPM Xd" = 0.17 X/R = 30

500 HP IND. MOTORS, EACH:

1800 RPM Xd" = 0.18 X/R = 20

EACH 1750 HP

SUBSTATIONS

69 kV SYSTEM

LOAD

STATIC

T2

T1

T2

UNIT

2 LV

T2 T3

G1

4 IND.EACH 500 HP

4.16 kV BUS

13.8 kV BUS

MOTORS

Figure 2.7.5 Example system for short-circuit calculations.

-2.7.20-

Fault Calculations

Nominal 3-phase MVA

class

Rated voltage range factor

K

Rated symmetrical short-circuit current

(at rated max. voltage) kA, rms

Maximum symmetrical short-circuit interrupting

capability kA, rms

Closing & latching

capability kA, rms

250 2.27 9.3 21 34

500 1.30 18 23 37

750 1.30 28 36 58

1000 1.30 37 48 77

Solution:

Transformer reactances in per unit on 10 MVA base:

T1: X = .07 (10/20) = .035

T2: X = .055 (10/1.5) = .367

T3: X = .055 (10/7.5) = .0733

Rotating machine subtransient reactances in per unit on 10 MVA base:

69 kV system: I" = 100 per unit, X = 1/100 = .01

Generator G1: X" = .09 (10/20) = .045

1750 HP Induction Motor (kVA rating = HP rating): X" = .17 (10/1.75) = .971

500 HP Induction Motor (kVA rating = HP rating): X" = .18 (10/0.50) = 3.60

LV motor group of 1.0 MVA, less than 50 HP each: X" = .25 (10/1) = 2.50

LV motor group of 0.5 MVA, 50 to 150 HP each: X" = .25 (10/0.50) = 5.00

Reactances for AC high-voltage circuit breaker first-cycle duties. Use subtransient reactances, except for:

500 HP motor, use 3.60 x 1.20 = 4.32 LV motor group, 50 to 150 HP, use 5.00 x 1.2 LV motor group, less than 50 HP, omit.

Reactances for AC high-voltage circuit breaker interrupting duties. Use subtransient reactances, except for:

1750 HP motor, use .971 x 1.50 = 1.457 500 HP motor, use 3.60 x 3.00 = 10.8 LV motor group, 50 to 150 HP, use 5.00 x 3.00 = 15.0 LV motor group, less than 50 HP, omit.

-2.7.21-


X / R ratio and resistances for interrupting duties. T1: X / R = 21, X = .0350, R = .001667 T2: X / R = 10, X = .367, R = .0367 T3: X / R = 14, X = .0733, R = .00524 69 kV system: X / R = 22, X = .010, R = .000455 G1: X / R = 40, X = .045, R = .001125 1750 HP IM: X / R = 30, X = 1.457, R = .0486 500 HP IM: X / R = 20, X = 10.8, R = .540 LV motor group: X / R = 9, X = 15.0, R = 1.667 1 / .045 = 22.2 1 / .322 = 3.11 1 / 3.18 = .314

For a three-phase short circuit on the 13.8 kV bus, I" = 22.2 + 22.2 + 3.11 + .314 = 47.8 per unit

First-cycle duty = 47.8 x 1.6 x 10 MVA / (1.732 x 13.8 kV) = 32 kA 1 / .045 = 22.2 1 / .485 = 2.06 1 / 7.68 = .13

For a three-phase short circuit on the 13.8 kV bus, I" = 22.2 + 22.2 + 2.06 + 0.13 = 46.6 per unit

Symmetrical interrupting duty = 46.6 x 10 MVA / (1.732 x 13.8 kV) = 19.5 kA

The 750 MVA, 15 kV breaker has an interrupting capability of 28 kA (15/13.8) = 30 kA at 13.8 kV. 80% of this value is 24 kA, which is sufficient for this application. The closing and latching capability is 58 kA, which is sufficient (compare to only 32 kA first-cycle duty).

To see if this is excessively conservative, calculate the X / R ratio:

Resistance network for interrupting duty. 1/R = 1/.00212 + 1/.001125 + 1/.0197 + 1/.852 R = .000708 per unit X / R = .02144/.000708 = 30.3

If remote from generation, interrupting duty = 1.13 x 19.5 = 22 kA. If close to generation, 1.03 x 19.5 = 20 kA.

The 500 MVA breaker has a symmetrical interrupting capability of 18 (15/13.8) = 19.6 kA, which is not adequate. Select the 750 MVA breaker. The 1000 MVA breaker might be justified if short-circuit current levels are expected to rise in the near future.

-2.7.22-

Documents

Applications to Sizing Circuit Breakers (ANSI C37)