where Emax is the no-load line-to-neutral maximum voltage of the generator. Alternatively, using the root mean square (rms) values,
′′ = ′′
I E X
g d
(4.12)
′ = ′ I E
X
g d
(4.13)
I E X
g d
= (4.14)
where
Eg = no-load line-to-neutral rms voltage
Iʺ = subtransient current,* rms value without dc offset Iʹ = transient current, rms value without dc offset
I = steady-state current, rms value
Note that, the importance of the reactances given by Equations 4.9 through 4.11 depends on what percentage they represent of the short-circuit impedance. For example, if the fault occurs right at the terminals of the generator, they are very important; however, if the fault is remote from the genera- tor, their importance is smaller.
The fault current will be lagging in power in a system where X ≫ R. Table 4.1 gives the typical values of the reactances for synchronous machines.
It is interesting to observe in Figure 4.6a that the total alternating component of armature current consists of the steady-state value and the two components that decay with time constants Td′ and Td′′. It can be expressed as
I I I t
T I I t
d T
ac= ′′ − ′ exp − exp
′′
+ ′ − −
′
( ) ( ) ++I (4.15)
where all quantities are in rms values and are equal but displaced 120 electrical degrees in the three phases.
component of the fault current, and the initial symmetrical rms current value is increased by a specific factor depending on the speed of the circuit breaker.* For example, if the circuit breaker opening time is eight, three, or two cycles, then the corresponding multiplying factor is 1.0, 1.2, or 1.4, respectively.
Therefore, the interrupting capacity (or rating) of a circuit breaker is expressed as
Sinterrupting= 3(VprefauIt)( )ζI′′ ×10−6MVA (4.16) where
Vprefault = prefault voltage at point of fault in volts Iʺ = initial symmetrical rms current in amperes ζ = multiplying factor
Note that, Equation 4.16 includes only the ac component. The multiplying factors and the reac- tance types are given in Table 4.1 [7]. As discussed before, the asymmetrical current wave decays gradually to a symmetrical current, the rate of decay of the dc component being determined by the L/R of the system supplying the current. The time constant for dc component decay can be found from
Tdc = circuit L/R s (4.17a)
or
T L R
dc
circuit /
cycles
= 2π (4.17b)
The momentary duty (or rating) of a circuit breaker is expressed as
Smomentary= 3(Vprefault)( )( . )I′′ 1 6 ×10−6MVA (4.18) This equation includes the dc component. Thus, the rms momentary current can be expressed as
Imomentary = 1.6 × Iʺ A (4.19)
Here, the Imomentary current is the total rms current that includes ac and dc components, and it is used for oil circuit breakers of 115 kV and above. The circuit breaker must be able to withstand this rms current during the first half-cycle after the fault occurs. Note that, if the Iʺ is measured in peak amperes, then the peak momentary current is expressed as
Imomentary = 2.7 × Iʺ A (4.20)
In the United States, the ratings of circuit breakers are given in the American National Standards Institute (ANSI) standards [8] based on symmetrical current,† in terms of nominal voltage, rated maximum voltage, rated voltage range factor K, rated continuous current, and rated short-circuit current. The required symmetrical current-interrupting capability is defined as
Required symmetrical
current-interrupting capabilitty Rated short-
circuit current Rated maxim
= uum voltage
Operating voltage
*Note that, the fault megavolt-ampere is often referred to as the fault level.
† The ratings of the circuit breakers can also be based on total current, which includes the dc component.
The standard dictates that for operating voltages below 1/K times the rated maximum voltage, the required symmetrical current-interrupting capability of the circuit breaker is equal to K times the rated short-circuit current. Table 4.2 gives outdoor circuit breaker ratings based on symmetrical current. Note that, the rated voltage factor K is defined as the ratio of the rated maximum voltage to the lower limit of the range of operating voltage in which the required symmetrical and asym- metrical interrupting capabilities vary in inverse proportion to the operating voltage [9]. Therefore, TABLE 4.1
Reactance Quantities and Multiplying Factors for Application of Circuit Breakers
Reactance Quantity for Use in X1
Multiplying Factor
Synchronous Generators and
Condensers
Synchronous Motors
Induction Machines A. Circuit Breaker Interrupting Duty
1. General case:
Eight-cycle or slower circuit breakersa 1.0 Subtransientb Transient Neglect
Five-cycle circuit breakers 1.1
Three-cycle circuit breakers 1.2
Two-cycle circuit breakers 1.4
2. Special case for circuit breakers at generator voltage only. For short-circuit calculations of more than 500,000 kVA (before the application of any multiplying factor) fed predominantly direct from generators, or through current-limiting reactors only:
Subtransientb Transient Neglect
Eight-cycle or slower circuit breakersa 1.1
Five-cycle circuit breakers 1.2
Three-cycle circuit breakers 1.3
Two-cycle circuit breakers 1.5
3. Air circuit breakers rated 600 V and less 1.25 Subtransient Subtransient Subtransient B. Mechanical Stress and Momentary Duty of Circuits Breakers
1. General case 1.6 Subtransient Subtransient Subtransient
2. At 500 V and below, unless current is fed predominantly by directly connected synchronous machines or through reactors
1.5 Subtransient Subtransient Subtransient
Source: Westinghouse Electric Corporation, Electrical Transmission and Distribution Reference Book. WEC, Pittsburgh, 1964.
a As old circuit breakers are slower that modern ones, it might be expected that a low multiplier could be used with old circuit breakers. However, modern circuit breakers are likely to be more effective than their slower predecessors, and therefore, the application procedure with the older circuit breakers should be more conservative than with modern circuit breakers. Also, there is no assurance that a short circuit will not change its character and initiate a higher current flow through a circuit breaker while it is opening. Consequently, the factors to be used with older and slower circuit breakers well may be the same as for modern eight-cycle circuit breakers.
b This is based on the condition that any hydroelectric generators involved have amortisseur (damper) windings. For hydro- electric generators without amortisseur windings, a value of 75% of the transient reactance should be used for this calcula- tion rather than the subtransient value.
general expressions (which take into account the rated voltage range factor K) for the rms and peak momentary currents, respectively, are
Imomentary = 1.6 × K × Iʺ A (4.21)
and
Imomentary = 2.7 × K × Iʺ A (4.22)
Notice that, in Table 4.2, the factor K is 1 for the nominal voltages of 115 kV and above. Therefore, Equations 4.21 and 4.22 become the same as Equations 4.19 and 4.20, respectively.
TABLE 4.2
Outdoor Circuit Breaker Ratings Based on Symmetrical Current
Nominal rms Voltage Class (kV)
Rated Maximum
rms Voltage
(kV)
Rated Voltage
Range Factor,
K
Rated Continuous
rms Current
(kA)
Rated Short- Circuit rms Current (at
Rated Maximum
kV) (kA)
Rated Interrupting
Time (Cycles)
Rated Maximum rms Voltage
Divided by K (kV)
Maximum rms Symmetrical Interrupting Capabilitya
(kA)
14.4 15.5 2.67 0.6 8.9 5 5.8 24
14.4 25.5 1.29 1.2 18 5 12 23
23 25.8 2.15 1.2 11 5 12 24
34.5 38 1.65 1.2 22 5 23 36
46 48.3 1.21 1.2 17 5 40 21
69 72.5 1.21 1.2 19 5 60 23
115 121 1 1.2 20 3 121 20
115 121 1 1.6 40 3 121 40
⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮
115 121 1 3 63 3 121 63
138 145 1 1.2 20 3 145 20
138 145 1 1.6 40 3 145 40
138 145 1 2 40 3 145 40
⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮
138 145 1 3 80 3 145 80
161 169 1 1.2 16 3 169 16
161 169 1 1.6 31.5 3 169 31.5
⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮
161 169 1 2 50 3 169 50
230 242 1 1.6 31.5 3 242 31.5
230 242 1 2 31.5 3 242 31.5
⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮
230 242 1 3 63 3 242 63
345 362 1 2 40 3 362 40
345 362 1 3 40 3 362 40
500 550 1 2 40 2 550 40
500 550 1 3 40 2 550 40
700 765 1 2 40 2 765 40
700 765 1 3 40 2 765 40
a Equal to K times the rated short-circuit rms current.
EXAMPLE 4.1
A circuit breaker has a rated maximum rms voltage of 38 kV and is being operated at 34.5 kV.
Determine the following:
(a) The highest symmetrical current-interrupting capability
(b) The operating voltage at the highest symmetrical current capability (c) The associated rms momentary current rating
(d) The associated peak momentary current rating Solution
(a) From Table 4.2, the rated voltage range factor K is 1.65, the rated continuous rms current is 1200 A, and the rated short-circuit rms symmetrical current at the rated maximum rms voltage of 38 kV is 22,000 A. However, since the circuit breaker is used at 34.5 kV, its sym- metrical current-interrupting capability is
( , )
. ,
22 A 38 kV
34 5kV 24 232 A 000
=
The highest symmetrical current-interrupting capability is (22,000 A) K = 22,000 × 1.65 ≅ 36,000 A (b) Which is possible when the operating voltage is
38 38
1 65 23
kV kV
K = ≅
.
(c) Note that, at lower operating voltages, the highest symmetrical current-interrupting capabil- ity of 36,000 A cannot be exceeded. Hence, the associated rms momentary current is
Imomentary K I
1 36 A
57 6 A rms
= × × ′′
=
= 1 6
6 000 00 .
. ( , ) ,
(d) The associated peak momentary current rating is Imomentary 2 7 K I
2 7(36 A) 97 2 A peak
= × × ′′
=
= .
. ,
, 000 00
A simplified procedure for determining the symmetrical fault current is known as the
“E/X method” and is described in Section 5.3.1 of ANSI C37.010. This method [6,9] gives results approximating those obtained by more rigorous methods. In using this method, it is necessary first to make an E/X calculation. The method then corrects this calculation to take into account both the dc and ac decay of the current, depending on circuit parameters X/R.
The approximation basically results owing to the use of curves.
EXAMPLE 4.2
Consider the system shown in Figure 4.7 and assume that the generator is unloaded and running at the rated voltage with the circuit breaker open at bus 3. Assume that the reactance values of the generator are given as Xd′′ =X1=X2=0 14. pu and X0 = 0.08 pu based on its ratings. The trans- former impedances are Z1 = Z2 = Z0 = j0.05 pu based on its ratings. The transmission line TL23 has Z1 = Z2 = j0.04 pu and Z0 = j0.10 pu. Assume that the fault point is located on bus 1, and determine the subtransient fault current for a three-phase fault in per units and amperes. Select 25 MVA as the megavolt-ampere base and 8.5 and 138 kV as the low-voltage and high-voltage bases.
Solution
′′= ′′
= ∠ °
= I E
X
j j
f g
d
1 0 0 0 14 7 143
. . . pu The current base for the low-voltage side is
I S
B VB
B ( )
( )
, ( . )
.
LV
LV
kVA kV A
=
=
= 3 25 000
3 8 5 1698 1 Therefore,
′′= ′′ =
=
If If ( . )( . ) , . 7 143 1698 1 12 129 52 A
Example 4.3
Use the results of Example 4.2 and determine the following:
(a) The possible value of the maximum of the dc current component (b) Total maximum instantaneous current
G
1 2 3
TL23 138 kV 8.5 kV
25 MVA
8.5/138 kV 25 MVA
Load
FIGURE 4.7 Transmission system for Example 4.2.
(c) Momentary current
(d) Interrupting capacity of the three-cycle circuit breaker if located at bus 1
(e) Momentary interrupting capacity of the three-cycle circuit breaker if located at bus 1 Solution
(a) Let
Idc,max = peak-to-peak amplitude then
I I
I
f dc
rms
pu pu
max
( )
( . )
.
=
( )
′′=
=
= 2 2 2 7 143 10 1 (b) From Figure 4.5a,
′′ =
=
( )
′′=
I I
If
max ,
. 2
2 2pu
dc max
2 2 0
(c) Imomentary current represents the summation of Iac and Idc, where Idc is about 50% of Iac. Hence,
Imomentary If 16(7 143pu) 1143pu
= ×
=
= 1 6.
. . . (d)
Sinterrupting= V If f′′ ×
=
3 10−
3 8500 12 129 52 ζ 6
( )( , . )(11 2 10. ) 6 .
×
=
−
214 3 MVA (e)
Smomentary= V If f′′ ×
=
3 1 6 10−
3 8500 12 129 52 ( . ) 6
( )( , . )(( . ) .
1 6 10× 6
=
−
285 7 MVA