ESTADO DE FLUJOS DE EFECTIVO CONSOLIDADO (En miles de euros)
TOTAL consolidado
values.
Figure shows a three-winding transformer, and Fig. shows its equivalent circuit. The following equations are easily derived and are the proper ones to use in short-circuit studies:
Equivalent Circuit of Three-winding Transformer.
Designers give reactance values between pairs of windings.
-
= 2
(1.46)
+ - +
2-
2
=
=
All must be on
N
OTE:
The equivalent circuit and equations for a four-winding trans- former are more complicated and will not he evident by simple analogy from Eq. (1.46).,
AFIG. 1.44 transformer.
diagram and equivalent circuit of three-winding
EXAMPLES OF SHORT-CIRCUIT-CURRENT CALCULATIONS'
The following examples are indicative of methods of applying the short- circuit-current calculating procedures outlined in the foregoing.
Systems 600 Volts and Below. The system shown in Fig. 1.45 involves one source of supply through a transformer from a primary sys- tem. The kva for the short-circuit calculations is taken as the
Numbers in parentheses in Figs. 1.45 and 1.47 to 1.50 refer to numbers of formulas used.
SHORT-CIRCUIT-CURRENT PROCEDURES
INCOMING LINE
A
67
0.25 MOTORS SOURCE
TRANSFORMER
750 KVA 5.5 x
REACTANCE DIAGRAM USE 750 KVA BASE
FOR CALCULATIONS VOLTS
SOURCE REACTANCE ON 750 KVA BASE - 0.0075%
0.0625
I 750
X AMPERES SYMMETRICAL 1.18)
18,000 AMPERES ASYMMETRICAL
FIG. 1.45 Illustration of procedure for calculation of short-circuit in radial
rating of the transformer. The kva of the connected motors is assumed to be 750 with an equivalent reactance of 25 per cent. Only reactances are used in these calculations. This problem is the type on which
1.5 is based.
Large Systems. Problems, particularly those involving secondary-network systems in the downtown area of the large cities or in large buildings, require the determination of the short-circuit current on a basis. In these systems it is particularly important that the reactance of all elements, however small, be taken into account, as they have a much more significant effect in reducing the short-circuit current a t volts than a t 480 or 600 volts.
P L A N
FEEDERS BREAKERS
CHANNEL A
I
NETWORK PROTECTOR NETWORK TRANSFORMER
VOLTS
I
ALINE
YVA SC
L-
nus o'
CIRCUIT BREAKER
.
E L E V A T I O N
FIG. 1.46 of equipment for network
SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES
The one-line diagram is shown in Fig. which the
data on t h e circuit The diagram is i n Fig.
Figure the condensed diagram to illustrate rela- tive distribution of reactance in t h e system. It be noted the overhead bus has 70 per cent as much impedance as of all the transformers ahead of it,. this item cause a serious error in t h magnitude of short-circuit,
T h e intermediate steps etween Figs. and can be
out by oing text.
T h e short circuit is just ahead of the 4000-amp circuit breaker as this the available short-circuit, which this circuit breaker must interrupt. As pointed out previously, air circuit breakers are applied the basis of and therefore
the short,-rirruit duty them, of breaker is not included.
T h e in Fig. 1.48
is typical of might, be a steel mill. The base chosen is 100,000 Precise data are available large motors are
used the Since the large only
part of the motor load, the remaining motor load is estimated. For short circuits on the system motor load is assumed to be equal to t h e capacity supplying each 22-kv bus, or 62,500 k r a 20,000 kva.
Should more precise data be available regarding load, these data should be used for simulating motor for faults the 22-kv system. In example, the connected horsepower t h e 6.0-kv mas known be as shown in diagram.
To check t h e momentary at the the primary sys-
tem should be represented by its nf
12.2 per cent. For interrupting d u t y on the bus, primary should be represented by a reartanre t o the
duty on 22-kv system, or 17.5 per
These large complicated should he set up a calculating board to accurate he obtained easily.
The equipment for this example is as Fig. 1.46.
Large High-voltage Power System.
SHORT CIRCUITS IN SINGLE-PHASE LIGHTING A N D WELDING POWER SYSTEMS (600 VOLTS A N D LESS)
A common is t o use single-phase to
three-phase primary systems supply single-phase for welders and for rircuits in some of the older
When determining the short-circuit current a t of these transformers, is necessary t o use t h e proper impedance t o the primary system. I n three-phase short-circuit calculations, the reactance
70
FIG. 1.47 One-line reactance diagram,
SHORT-CIRCUIT-CURREM CALCULATING PROCEDURES 71
and short-circuit-current calculation procedure for network system in Fig.
of a conductor is the reactance from the center of the condurtor t o the theoretical neutral. Assume that for phase the rurrent leaves on the phase conductor and through the neutral. a three-phase short circuit, the three balance; so there is no rurrent in the neutral. With single-phase line-to-line short the leaves on one phase conductor and returns the other. Therefore this rurrent
sees the reactance of two as series. for
phase conuected line-to-hie on the primary, the primary system impedance must be used t o represent i t in a true
t o the rest of the circuit. The remaining are essentially the same as for three-phase circuits using the transformer and
circuit reactances.
Single-phase tramformers used for supplying single-phase lighting circuits usually have the available for t o
and ground the user and are usually relatively low kva.
These small transformers have a relatively high
ratio compared with three-phase of a higher voltage rating of larger kva rating.
KVA 3 PHASE SHORT CIRCUIT OUTY
BASE 500 KVA
PRIMARY SYSTEM REACTANCE ON 3-PHASE BASIS.
PRIMARY SYSTEM REACTANCE ON SINGLE
BASIS 2 0.01
PRIMARY SYSTEM
TRANSFORMER X
TOTAL
26000 AMP SYMMETRICAL 0.0192
MODIFIED)
1.25 26000 3 2 5 0 0 A M P
FIG. 1.49 Short-circuit-current calculating procedure for single-phase two-wire 480-volt
SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 7.5
The most severe short-circuit condition in this case is a line-to-neutral short circuit because it involves a much higher primary-to-secondary turn ratio than does a line-to-line short circuit. Hence, this is the basis on which protective equipment should be selected.
Since the reactance and resistance of the transformers are given on the basis of a full winding, it is necessary to convert to the proper values only one-half the secondary winding is involved as is the case when a line-to-line neutral short circuit occurs. The reactance is increased by a factor of 1.2 and the resistance by a factor of 1.41. Therefore, the pub- lished reactances and resistances of these transformers are multiplied by those figures.
Figure 1.49 a typical example where reactance only is used, as would be the case for a relatively large 480-volt transformer supplying a welder circuit. I n these calculations it is necessary to use the to-neutral reactance of the primary system. In the example of Fig. 1.50 use twice the line-to-neutral reactance of the primary. Use the proper
000 KV4 3
3 % FULL
5 0 KV4
SYSTEM ON 3
0.00198
61
0 0 3 6 % X
SYSTEM ON 4
OF R
00172%
1.25 X 10300 I2900 11.201
FIG. .SO Short-circuit-current procedure for single-phase three-wire
76 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES
reactance and resistance for line-to-neutral short circuit a t the secondary of the transformer. I n cases there is assumed to be no motor feedback.
TABLES AND CURVES FOR ESTIMATING