Basic Short-Circuit Calculation Procedure (Follow-Up)

June 26th, 2008

Calculation of Short-Circuit Currents- Point-To-Point Method

Adequate interrupting capacity and protection of electrical components are two essential aspects required by the National Electric Code in Sections 110-9, 110-10, 230-65, 240-1, and 250-95

Interrupting Rating, InterruptingCapacity and Short-Circuit Currents

Interrupting capacity can be defined as “the actual short circuit current that a protective device has been tested to interrupt.”

The National Electrical Code Requires adequate interrupting ratings in section110-9.

NEC Section 110-9

Equipment intended to break current at fault levels shall have an interrupting rating sufficient for the system voltage and the current which is available at the line terminals of the equipment.

Calculation of Short-Circuit Currents-Point-To-Point Method

The first step to assure that system protective devices have the proper interrupting rating and provide component protection is to determine the available short-circuit currents.

Calculation of Short-Circuit Currents-Point-To-Point Method

The application of the point-to-point method permits the determination of available short-circuit currents with a reasonable degree of accuracy at various points for either 3ph or 1ph electrical distribution systems.

Calculation of Short-Circuit Currents-Point-To-Point Method

This method assumes unlimited primary short-circuit current (infinite bus).

Step 1 (Ifla)

Determine Transformer full-load amperes from

Name plateTransformer TableFormula= 3ph Transf Ifla= KVAx1000

E(L-L)

Step 2 (Multiplier)

Find Transf. Mutilplier. Multiplier= __100__

Transf. %Z

Step 3 (ISCA)

Determine transf. let-thru short-circuit current lSCA = IFLA x multiplier

Step 4 (“f” factor)

Calculate “f” factor 3ph faults f= 1.73 x L x ISCA

C x E(L-L)

L=length (feet) of circuit to the fault. C=Constant from Table 5-7-1. For parallel

runs, multiply C values by the number of conductors per phase.ISCA= available short-circuit current in amperes at beginning of circuit.

E= Voltage line to line

Step 5 (“M”)

Calculate “M”M=__1__ 1+f

Step 6 (ISCA @ fault)

Compute the available short-circuit current (symmetrical) at the fault ISCA = ISCA x M at

Fault

Example of Short-Circuit Calculation

Step 1 (Ifla)

Ifla = KVA x 100

EL-L x 1.73

Ifla = 300 x 1000 = 834

208 x 1.73

Step 2 (Multiplier)

Multiplier=__100__Trans.%Z

Multiplier = 100 = 55.55 1.8

Step 3 (ISCA)

ISCA = IFLA x Multiplier

ISCA = 834 x 55.55= 46,329A

Step 4 (“f” factor)

f=1.73 x L x I 2 x C x E(L-L)

f= 1.73 x 20 x 46,329 = .1927 2 x 20,000 x 208

Step 5 (“M”)

M= 1_ 1+f

M = ___1____ = .838 1 + .1927

Step 6 (ISCA @ fault)

ISCA = ISCA X M Fault #1

ISCA = 46,329 x .838= 38,823 Fault #1

Fault 2 Start at Step 4

Use ISCA @ Fault #1 to calculate F= 1.73 x L x ISCA

2 x C x E(L-L)

F= 1.73 x 20 x 38,823 = 1.29 5000 x 208

Step 5

M=_1_ 1+f

=_1__ = .437 1+1.29

Step 6

ISCA = ISCA x M Fault #2 Fault #1

ISCA = 38,823 x .437 = 16,965A Fault #2

Peak Let-Thru Current

Equipment withstand ratings can be described as : How Much Fault Current can the equipment handle, and for How Long ?

Peak Let-Thru Current

Based on present standards, data has been compiled through testing, resulting in Fuse Let-Thru Charts Peak let-thru current-mechanical forces Apparent prospective RMS symmetrical

let-thru current-heating effects.

KRP-C Sp

LPN RD SP

LPS RK SP