short circuit calculations

7Short-Circuit Calculations

7.1.0 Introduction7.1.1 Point-to-Point Method, Three-Phase Short-Circuit Calculations, Basic

Calculation Procedure and Formulas7.1.2 System A and System B Circuit Diagrams for Sample Calculations Using

Point-to-Point Method7.1.3 Point-to-Point Calculations for System A to Faults X1 and X27.1.4 Point-to-Point Calculations for System B to Faults X1 and X27.1.5 C Values for Conductors and Busway7.1.6 Shortcut Method 1: Adding Zs7.1.7 Average Characteristics of 600-V Conductors (Ohms per 100 ft): Two or

Three Single Conductors7.1.8 Average Characteristics of 600-V Conductors (Ohms per 100 ft): Three

Conductor Cables (and Interlocked Armored Cable)7.1.9 LV Busway, R, X, and Z (Ohms per 100 ft)7.1.10 Shortcut Method 2: Chart Approximate Method7.1.11 Conductor Conversion (Based on Using Copper Conductor)7.1.12 Charts 1 through 13 for Calculating Short-Circuit Currents Using Chart

Approximate Method7.1.13 Assumptions for Motor Contributions to Fault Currents7.1.14 Secondary Short-Circuit Capacity of Typical Power Transformers

7.1.0 Introduction

Of the four basic methods used to calculate short-circuit currents, the point-to-pointmethod offers a simple, effective, and quick way to determine available short-circuitlevels in simple to medium-complexity three-phase and single-phase electricaldistribution systems with a reasonable degree of accuracy. This method is bestillustrated by the figures and table that follow. Figure 7.1.1 shows the steps andequations needed in the point-to-point method. Figure 7.1.2 shows one-line diagramsof two systems (A and B) to be used as illustrative examples. Figures 7.1.3 and 7.1.4show the calculations for these two examples. And Table 7.1.5 provides the circuitconstants needed in the equations for the point-to-point method.

The point-to-point method is followed by two shortcut methods for determiningshort-circuit currents at ends of conductors, specifically, adding Zs and the chartapproximate method. These two methods make use of simplifications that are rea-sonable under most circumstances and almost certainly will yield answers thatare on the safe side.

Section

7.1

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7.1.1 Point-to-Point Method, Three-PhaseShort-Circuit Calculations, Basic CalculationProcedure and Formulas

7.2 Section Seven

7.1.1

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Short-Circuit Calculations 7.3

7.1.2 System A and System B Circuit Diagrams forSample Calculations Using Point-to-Point Method

7.1.2

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7.1.3 Point-to-Point Calculations forSystem A to Faults X1 and X2

7.4 Section Seven

7.1.3

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7.1.4 Point-to-Point Calculations forSystem B to Faults X1 and X2


7.1.4

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7.1.5 C Values for Conductors and Busway

7.6 Section Seven

TABLE 7.1.5

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7.1.6 Shortcut Method 1: Adding Zs

This method uses the approximation of adding Zs instead of the accurate methodof Rs and Xs (in complex form).

Example

1. For a 480/277-V system with 30,000 A symmetrical available at the line side of a conductorrun of 100 ft of two 500-kcmil per phase and neutral, the approximate fault current at theload side end of the conductors can be calculated as follows:

2. 277 V/30,000 A � 0.00923 � (source impedance).3. Conductor ohms for 500-kcmil conductor from Table 7.1.7 in magnetic conduit is 0.00546 �

per 100 ft. For 100 ft and two conductors per phase, we have4. 0.00546/2 � 0.00273 � (conductor impedance).5. Add source and conductor impedance, or 0.00923 � 0.00273 � 0.01196 � total.6. Next, 277 V/0.001196 � � 23,160 A rms at load side of conductors.

For impedance values, refer to Tables 7.1.7, 7.1.8, and 7.1.9.

7.1.7 Average Characteristics of 600-VConductors (Ohms per 100 ft): Two orThree Single Conductors


TABLE 7.1.7

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7.1.8 Average Characteristics of 600-V Conductors(Ohms per 100 ft): Three Conductor Cables (andInterlocked Armored Cable)

7.8 Section Seven

TABLE 7.1.8

TABLE 7.1.9

7.1.9 LV Busway, R, X, and Z (Ohms per 100 ft)

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7.1.10 Shortcut Method 2: ChartApproximate Method

The chart method is based on the following:

Motor Contribution Assumptions

120/208-V systems 50 percent motor load

4 times motor FLA contribution

240/480-V systems 100 percent motor load

4 times motor FLA contribution

Feeder Conductors. The conductor sizes most commonly used for feeders from mold-ed-case circuit breakers are shown. For conductor sizes not shown, Table 7.1.11 hasbeen included for conversion to equivalent arrangements. In some cases, it may benecessary to interpolate for unusual feeder ratings. Table 7.1.11 is based on using cop-per conductor.

Short-Circuit Current Readout. The readout obtained from the charts is the rms sym-metrical amperes available at the given distance from the transformer. The circuitbreaker should have an interrupting capacity at least as large as this value.

How to Use the Short-Circuit Charts

Step 1. Obtain the following data:

� System voltage� Transformer kVA rating� Transformer impedance� Primary source fault energy available in kVA

Step 2. Select the applicable chart from Figure 7.1.12 (Charts 1–13). The charts are grouped bysecondary system voltage, which is listed with each transformer. Within each group, the chart for thelowest-kVA transformer is shown first, followed in ascending order to the highest-rated transformer.

Step 3. Select the family of curves that is closest to the “available source kVA.” The upper-value-line family of curves is for a source of 500,000 kVA. The lower-value-line family of curves isfor a source of 50,000 kVA. You may interpolate between curves if necessary, but for values above100,000 kVA, it is appropriate to use the 500,000-kVA curves.

Step 4. Select the specific curve for the conductor size being used. If your conductor size is some-thing other than the sizes shown on the chart, refer to the conductor conversion table (Table 7.1.11).

Step 5. Enter the chart along the bottom horizontal scale with the distance (in feet) from thetransformer to the fault point. Draw a vertical line up the chart to the point where it intersectsthe selected curve. Then draw a horizontal line to the left from this point to the scale along theleft side of the chart.

Step 6. The value obtained from the left-hand vertical scale is the fault current (in thousandsof amperes) available at the fault point.


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7.1.11 Conductor Conversion (Basedon Using Copper Conductor)

7.10 Section Seven

TABLE 7.1.11

7.1.12 Charts 1 through 13 for CalculatingShort-Circuit Currents Using Chart ApproximateMethod

7.1.12

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7.1.12

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7.12 Section Seven

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7.1.13 Assumptions for MotorContributions to Fault Currents

To determine the motor contribution to the first half-cycle fault current when thesystem motor load is known, the following assumptions generally are made:

Induction motors: Use 4.0 times motor full-load current (impedance value of 25percent).

Synchronous motors: Use 5.0 times motor full-load current (impedance valueof 20 percent).

When the motor load is not known, the following assumptions generally are made:

208Y/120-V systems� Assume 50 percent lighting and 50 percent motor load.� Assume motor feedback contribution of 2.0 times full-load current of trans-

former.

240-480-600-V three-phase, three-wire systems� Assume 100 percent motor load.� Assume motors 25 percent synchronous and 75 percent induction.� Assume motor feedback contribution of 4.0 times full-load current of

transformer.

480Y/277-V systems in commercial buildings� Assume 50 percent induction motor load.� Assume motor feedback contribution of 2.0 times full-load current of

transformer or source.� For industrial plants, make same assumptions as for three-phase, three-wire

systems (above).

Medium-voltage motors� If known, use actual values. Otherwise, use the values indicated in the

preceding for the same type of motor.

Note on asymmetrical currents. The calculation of asymmetrical currents is a labo-rious procedure because the degree of asymmetry is not the same on all three phases.It is common practice to calculate the rms symmetrical fault current, with theassumption being made that the dc component has decayed to zero, and then apply amultiplying factor to obtain the first half-cycle rms asymmetrical current, which iscalled the momentary current. For medium-voltage systems (defined by the IEEE asgreater than 1000 V up to 69,000 V), the multiplying factor is established by NEMAand ANSI standards depending on the operating speed of the breaker; for low-voltagesystems, 600 V and below, the multiplying factor is usually 1.17 (based on generallyaccepted use of an X/R ratio of 6.6 representing a source short-circuit power factor of15 percent). These values take into account that medium-voltage breakers are ratedon maximum asymmetry and low-voltage breakers are rated on average asymmetry.


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7.1.14 Secondary Short-Circuit Capacityof Typical Power Transformers

7.14 Section Seven

TABLE 7.1.14

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