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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