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those using air circuit breakers for short- 
circuit protection. Those using air circuit breakers for short-circuit 
protection are applied 04 exactly the same basis as low-voltage air circuit 
breakers in so far as short-circuit currents are concerned. 
Motor starters using fuses for short-circuit protection are applied on 
exactly the same basisas fuses in so far as short-circuit current is concerned. 
Low-voltage Motor Starters. 
AVAILABLE SHORT-CIRCUIT CURRENT 
In determining the short-circuit current, the impedance of the circuit 
protective device connected in the faulty feeder is neglected. The short- 
circuit current is determined by’ assuming that the protective device is 
shorted out by a bar of zero impedance (Fig. 1.27). The short-circuit 
/ 
SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 37 
current which flows in such a circuit is commonly called available short- 
circuit cumat. The procedure for determining the available short-circuit 
current is based on setting up impedance or reactance diagrams. The 
impedance of the short-circuit protective device that is nearest the short 
circuit (electrically) is omitted from the impedance diagram. 
Practically all protective devices are so rated and tested for short- 
circuit interrupting ability; hence this procedure may be followed in 
short-circuit calculations. This greatly simplifies the calculations and 
removes the effect of impedance variations between different types and 
makes of devices having the same interrupting rating. I t means that 
one set of short-circuit-current calculations for a given set of conditions 
is all that is needed for applying any type of protective device, regardless 
of the impedance of the devices themselves. 
GENERATOR 0 
TRANSFORMER 
MOTORS 
CABLE 
C A B L E 
SHORT 
CIRCUIT 
SHORT ClRCUlTED 81 
JUMPER OF ZERO 
IMPEDANCE 
FIG. 1.27 
circuit protective devices. 
Connections for determining available short-circuit current for testing rhort- 
38 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 
HOW TO MAKE A SHORT-CIRCUIT STUDY 
FOR DETERMINING SHORT-CIRCUIT CURRENT 
FORMULAS FOR SHORT-CIRCUIT STUDY' 
1. Changing ohms to per cent ohms, etc.: 
(ohms reactance) (kva.base) 
(kvt)*(lO) 
(ohms reactance)(kva base) 
(1.1) Per cent (%) ohms reactance = 
Per-unit (90 ohms reactance = (kv)*( 1000) (1.2) 
[see Eq. (1.34)] 
(1.3) 
(1.4) 
(% reactance)(kv)2(10) 
kva base Ohms reactance = 
Per-unit ohms reactance = per cent ohms reactance 100 
2. Changing per cent or per-unit ohms reactance from one kva base to 
another: 
% ohms reactance on kva base 2 
kva base 2 
kva base 1 - - X (% ohms reactance on base 1) (1.5) 
9f reactance on kva base 2 
kva base 2 
kva base 1 
- - X (% ohms reactance on kva base 1) (1.36) 
3. Converting utility-system reactance to per cent or per-unit ohms 
a. If given in per cent ohms reactance on a kva base different than that 
b. If given in short-circuit kva, convert to per-unit ohms thus: 
reactance on kva base being used in study: 
used in the study, convert according to Eq. (1.5). 
(1.6) 
kva base used in reactance diagram 
short-circuit kva of utility system 9 i reactance = 
c. If given in short-circuit amperes (rms symmetrical), convert t o per- 
unit ohms thus: 
Yi reactance = 
d. If only the kva interrupting rating of the incoming line breaker is 
* See pp. 54 to 57 for more prr-unit formulas 
1 kv = line-to-line kilovolts. 
(1.7) 
kva base used in reactance diagram 
(short-circuit current) (d$) (kv rating of system) 
known, 
SHORTT-CIRCUIT.CURRENT CALCULATING PROCEDURES 39 
9f ohms reactance 
(1.8) - 
kva base used in reactance diagram 
kva interrupting rating of incoming line breaker 
- 
4. Determining kva base of motors: 
The exact kva base of a motor = EI 43 (1.9) 
where E = name-plate voltage rating 
When motor full-load currents are not known, use the following kva bases: 
Induction motors: 
0.8-power factor synchronous motor: 
1.0-power factor synchronous motors: 
I = name-plate full-load current rating 
kva base = horsepower rating 
kva base = 1.0 (horsepower rating) 
kva base = 0.80 (horsepower rating) 
(1.10) 
(1.11) 
(1.12) 
5. Changing voltage base when ohms are used: 
Ohms on basis of voltage 1 
- - ')* X (ohms on basis of voltage 2) (1.13) 
In Eqs. (1.1) to (1.4), ohms impedance or ohms resistance may be sub- 
The final product is then per-unit or per 
(voltage 2)2 
stituted for ohms reactance. 
cent ohms impedance or resistance, respectively. 
6 . Determining the symmetrical short-circuit kva: 
Symmetrical short-circuit kva = ~ (kva base) (1.14) 
% X * 
-~ - y? '& (kva base) (1.15) 
(1.16) 
(1.16a) 
(line-to-neutral voltage)2 
ohms reactance X 1000 
kv2 X lo00 
ohms reactance 
= 3 
- - . 
7. Determining the symmetrical short-circuit current: 
(1.17) (100) (kva base) 
(% X*)(v%(kvt) 
(% X*)(&)(kvt) 
Symmetrical short-circuit current = 
(1.18) 
(1.19) 
- kva base 
- - kvt X lo00 
- 
( d ) ( o h m s reactance) 
* X = reactance or impedanoe. 
t kv = line-&line kilovolts. 
L 0 TABLE 1.3 Factor ( K ) to Convert Ohms to Per Cent or Per-unit Ohms for Three-phase Circuits* 
216Y/125 
240 
480 
600 
2,400 
4.1 60 
4,800 
6.900 
7,200 
l1,OOO 
11.500 
12,000 
12,500 
13.200 
13,800 
23,000 
37.4M) 
46,000 
69,OCU 
loot 
P.r c*nt 
'14 
73 
43.4 
27.7 
1.73 
0.56 
0.435 
0.210 
0.193 
0.0825 
0.0755 
0.0695 
0.064 
0.0574 
0.0525 
0.0187 
0.00711 
0.00471 
0.0021 2 - 
Per-""it 
2.14 
1.73 
0.434 
0.277 
0.0173 
0.00576 
0.00435 
0.0021 
0.001 93 
0.000825 
0.000755 
0.000695 
0.00064 
0.000574 
0.000525 
0,0001 87 
0.000071 I 
0.0000471 
0.0000212 
Per cent 
321.5 
260.4 
65.21 
4.166 
2.604 
0.808 
0.651 
0.315 
0.289 
0.123 
0.113 
0.104 
0.096 
0.086 
0.0787 
0.0283 
0.0107 
0.00708 
0.0031 5 
- 
1 50 
Per-""it 
3.215 
2.604 
0.6521 
0.4166 
0.02604 
0.00808 
0.00651 
0.0031 5 
0.00289 
0.00123 
0.00113 
0.00104 
0.00096 
0.00086 
0.000787 
0.000283 
0.000106 
0.0000708 
0.000031 5 
Base kvo 
200 
Per cant 
128 
147 
86.8 
55.5 
3.47 
1.15 
0.868 
0.42 
0.386 
0.165 
0.151 
0.138 
0.127 
0.114 
0.105 
0.0378 
0.0142 
0.00945 
0.0042 
Por-un1t 
4.28 
3.47 
0.868 
0.555 
0.0347 
0.0115 
0.00868 
0.0042 
0.00386 
0.00165 
0.00151 
0.00138 
0.00127 
0.00114 
0.00105 
0.000378 
0.000142 
0.0000945 
0.000042 
300 
Per cent 
t43 
a1 
30.2 
83.3 
5.21 
1.72 
1.302 
0.63 
0.579 
0.247 
0.226 
0.208 
0.192 
0.172 
0.157 
0.0567 
0.0213 
0.0141 
0.0063 
kva base , kva base 
kv' X 10 
* For per-unit, K = For per cent, K = kv = line-to-line kilovolts 
kv' X 1wO 
- 
Per-""it 
6.43 
5.21 
1.302 
0.833 
0.0511 
0.0172 
0.01302 
0.0063 
0.00579 
0.00247 
0.00226 
0.00208 
0.001 92 
0,00172 
0.001 57 
0.000547 
0.00021 3 
0.0001 41 
0.000063 
500 
__ 
Per cent 
07 1 
868 
217 
I38 
8.68 
2.88 
2.17 
1.05 
0.965 
0.413 
0.377 
0.347 
0.32 
0.286 
0.262 
0.045 
0.0355 
0.0236 
0.0105 
Per-""it v) 
I 
- 2 
2 
2.17 K 
0.71 
8.68 
E 
1.38 2 
0.0868 
0.0288 
0.0105 n 
0.00965 
0.00413 
0.00377 5 
0.00347 f 
0 
0.0032 
0.00286 
0.00262 
0.00045 c 
0.000355 R 
0.000236 v, 
0.000105 
0.0217 B -I 
2 
6 
new base in kva, 
100 
t To determine multiplying factors far any other base use figures under 100-kvs base columns multiplied by 
SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 
8. Determining the asymmetrical short-circuit current: 
INFINITE 
H BUSES