Basic short circuit calculation procedure
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Transcript of Basic short circuit calculation procedure
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Basic Short-Circuit Calculation Procedure (Follow-Up)
June 26th, 2008
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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
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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.
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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.
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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.
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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.
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Calculation of Short-Circuit Currents-Point-To-Point Method
This method assumes unlimited primary short-circuit current (infinite bus).
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Step 1 (Ifla)
Determine Transformer full-load amperes fromName plateTransformer TableFormula= 3ph Transf Ifla= KVAx1000
E(L-L)
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Step 2 (Multiplier)
Find Transf. Mutilplier. Multiplier= __100__
Transf. %Z
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Step 3 (ISCA)
Determine transf. let-thru short-circuit current lSCA = IFLA x multiplier
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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
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Step 5 (“M”)
Calculate “M”M=__1__ 1+f
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Step 6 (ISCA @ fault)
Compute the available short-circuit current (symmetrical) at the fault ISCA = ISCA x M at
Fault
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Example of Short-Circuit Calculation
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Step 1 (Ifla)
Ifla = KVA x 100 EL-L x 1.73
Ifla = 300 x 1000 = 834 208 x 1.73
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Step 2 (Multiplier)
Multiplier=__100__Trans.%Z
Multiplier = 100 = 55.55 1.8
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Step 3 (ISCA)
ISCA = IFLA x MultiplierISCA = 834 x 55.55= 46,329A
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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
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Step 5 (“M”)
M= 1_ 1+f
M = ___1____ = .838 1 + .1927
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Step 6 (ISCA @ fault)
ISCA = ISCA X M Fault #1
ISCA = 46,329 x .838= 38,823 Fault #1
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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
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Step 5
M=_1_ 1+f
=_1__ = .437 1+1.29
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Step 6
ISCA = ISCA x M Fault #2 Fault #1
ISCA = 38,823 x .437 = 16,965A Fault #2
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Peak Let-Thru CurrentEquipment withstand ratings can be described as : How Much Fault Current can the equipment handle, and for How Long ?
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Peak Let-Thru CurrentBased 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.
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KRP-C Sp
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LPN RD SP
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LPS RK SP