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volt.s)* 
Per-unit ohms on old base volts (new base volts)* 
- 
and 
Per-unit ohms on new base volts = per-unit ohms on old base volts 
(1.38) 
Equations (1.37) and (1.38) may be used for per cent ohms as well as per- 
unit ohms. 
When using ohms 
instead of per cent or per-unit in the impedance diagram, it is important 
to convert the ohmic values to a common voltage base by Eq. (1.13). 
For example, if the short-circuit current is being calculated in a 480-volt 
system (supplied by transformers rated 480-volt secondary) fed through a 
cable and a transformer from a 2400-volt system, the ohms impedance of 
the cable in the 2400-volt circuit must be multiplied by 48O2/24OO2 to 
convert it to ohms on a 480-volt base. The transformer ratings, i.e., 480, 
240, etc., and not system ratings, if different from transformer rating, are 
used as the voltage base for short-circuit-current calculations. 
The utility system must be 
represented by a reactance in the impedance diagram. Sometimes this 
utility-system reactance is available in per cent on a certain base. If so, 
it is merely necessary to convert this value to the common base used in 
the impedance diagram. In some cases the 
utility engineers will give the short-circuit kva or current that the utility 
system will deliver a t the plant site. In otker cases, only the interrupting 
capacity of the incoming-line circuit breaker is known. In these cases to 
convert short-circuit kva, current, or incoming-line breaker interrupting 
rating to per cent reactance on the kva base used in the reactance diagram, 
proceed as follows: 
If given short-circuit kva, convert to per cent by using Eq. (1.6). 
If per-unit is desired, use also Eq. (1.4). 
If given short-circuit amperes (rms symmetrical), convert to per cent 
by Eq. (1.7) and to per-unit by Eqs. (1.7) and (1.4). 
If only the kva interrupting rating of the incoming line circuit breaker 
is known, convert to per cent by Eq. (1.8) and to per-unit by Eqs. (1.8) 
and (1.4). 
(old base volts)2 
(new base volts)2 
Converting Ohms to a Common Voltage Base. 
Representing the Utility Supply System. 
To do this, use Eq. (1.5). 
DETERMINING THE EQUIVALENT SYSTEM IMPEDANCE OR REACTANCE 
After completing the impedance diagram and inserting the values of 
reactance or impedance for each part of the diagram, it is necessary to 
reduce this network to one equivalent value. This can be done either by 
58 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 
longhand calculation or with the aid of a calculating board. Since so few 
engineers have access to calculating hoards and must use longhand meth- 
ods, this method will be covered in sufficient detail to enable solving the 
short-circuit problems commonly encountered. 
A d-c calculating board will permit 
accurate solution of all short-circuit problems where reactance only is 
considered. In most cases where resistance is a significant factor and 
must be considered, the d-c calculating board cannot be used readily. 
However, in some problems involving resistance, certain approximations 
can be made to obtain reasonably accurate answers on d-c calculating 
boards. For exact calculating-board solutions of problems factoring 
resistance and reactance, the a-c calculating board may he employed. 
A-c calculating boards have boxes to represent both the resistance and 
reactance of a circuit. The procedure for using calculating boards is 
beyond the scope of this book. 
Longhand methods of 
combining reactances vary in some respects. To illustrate the principles 
involved, refer to Figs. 1.37 and 1.39. 
Arbitrary values of reactance have been assigned to the various 
branches. Combining the various branches of the diagram is merely a 
question of reducing two or more series reactances to one value and 
reducing two or more parallel reactances to one value until one single 
equivalent value is obtained. 
Use of Calculating Boards. 
Longhand Method of Combining Reactances. 
The following shows how to combine reactances and resistances. 
1. Combining reactance and resistance to determine impedance, 
z = r + j z (1.39) 
2. Adding series reactance of circuits where resistance is neglected add 
z = m 
wherej = 47 
reactances arithmetically, i.e., 
x, + x2 + xa = x. = equivalent reactance 
z,, z2, and x3 = reactances of circuit components 
zs = equivalent reactance 
3. Combining parallel reactances, 
zo = equivalent reactance 
For two reactances only x, and z2 
(1.40) ( d ( z 2 ) 
21 + 2 2 XI = - 
For combining several parallel reactances 
(1.41) 1 1 1 1 1 1 
2. 2, 2 2 X I 2, 
- = _ + - + - + - + E 
SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 59 
ONE-LINE DIAGRAW 
INFINITE 
c. 
REACTANCE DIAGRAM OF CIRCUIT 
SHOWN IN ONE LINE DIAGRAM TO 
THE LEFT. 
P ~ T & T( $*, 
mP.T* EQUIVALENT Y 
CONVERT PITI , PITI e c , TO 
c. EQUIVALENT Y. 
S T E P # I 
COMBINE SERIES REACTANCES 
P I ~ T I , R B T ~ , E T C . 
STEP x z 
--&- I 
&Pa. C t a"' 3 + c * 
L COMBINE 2 + Ct , 3+ C+ AND 
UNTIL ONE EOUIVALENT 
cs c4 
THEN REPEAT STEPS 2.3 e 4 
DRAW NEW DIAGRAM REACTANCE IS OBTAINED. 
STEP-* 3 STEP tt4 
FIG. 1.39 
into a single resultant value. 
Example of the method of combining remtmces of a network-type system 
MI SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 
Some systems are such that they cannot he reduced by merely com- 
bining series and parallel rgactances. For example, take the one-line 
diagram of a circuit as show in upper left-hand corner of Fig. 1.39. The 
reactance diagram is shown the ypper righehand corner of Fig. 1.39. 
In addition to combining serieszind parallel reactances, it is necessary to 
convert a triangle of reactances such as PI, TI, Pzr T , and C1 to an equiva- 
lent Y of reactances by the formulas of Fig. 1.40. By these conversions, 
\ 
I 
ob + ac + be 
b 
a b + a c + bc 
B = 
c = 
A = o b + o c + b c 
a 
0c 
A + B + C 
a=- 
b : " 
A + B + C 
A 8 
A + B + C C : 
FIG. 1.40 Formula for converting a triangle or delta of three impedances to a Y of three 
equivalent impedances, and vice verso. 
any commonly encountered system reactance diagram can be reduced to 
one equivalent reactance. 
Sometimes i t is desirable to consider the 
resistance and reactance of a circuit. This involves combining imped- 
ances. The procedure for combining impedances is outlined here. The 
combining of parallel impedances necessitates multiplication and division 
of impedances (complex quantities) and is outlined here. 
When two or more impedances are in 
series, the resistance and reactance components are added separately to 
combine the series into one equivalent value. 
Combining Impedances. 
Adding Series Impedances. 
Refer to Fig. 1.41. The three series impedances are 
z1 = TI + jzl 
za = 72 i- jxa 
zz = Tp + ja 
SHORT-ClRCUIT+CURRENT CALCULATING PROCEDURES 61 
3 SERIES EQUIVALENT 
IMPEDANCES IMPEDANCE 
FIG. 1.41 Example illustrating the combining of series impedances. 
The equivalent impedance 
2 % = rl + V Z + 73 + j(z1 + zz + 4 (1.42) 
Using the numerical values of Fig. 1.41, 
2 , = 1 + j 2 
22 = 2 + j 3 
= 0.5 +jl 
21 = (1 + 2 + 0.5) + j ( 2 + 3 + 1) = 3.5 + j G 
The above is applicable when impedances are expressed in ohms, per- 
Combining Parallel Impedances. Parallel impedances may be 
unit or per cent. 
reduced to one equivalent impedance as follows (see Fig. 1.42): 
TWO PARALLEL IMPEDANCES EQUIVALENT IMPEDANCE 
FIG. 1.42 Example illustrating Le combining of parallel impedances. 
61 WORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 
(1) Reduce the per cent values of resistance and reactance in each of 
the given parallel circuits to a per-unit basis by dividing per cent figures 
by 100 or convert