Basic short circuit calculation procedure
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This document provides information and steps for calculating short-circuit currents using the point-to-point method. It defines key terms like interrupting capacity and discusses National Electric Code requirements. The point-to-point method calculates available short-circuit currents at various points on electrical distribution systems assuming unlimited primary current. It provides a 6 step process for determining the available short-circuit current at the fault, including calculating the transformer full load amps, multiplier, short-circuit current, "f" factor, "M" value, and final short-circuit current at the fault. An example calculation is shown. The document also discusses equipment withstand ratings and peak let-through currents.
Basic short circuit calculation procedure
- 1. Basic Short-Circuit Calculation Procedure (Follow-Up) June 26th , 2008
- 2. 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
- 3. Interrupting Rating, Interrupting Capacity 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 section 110-9.
- 4. 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.
- 5. 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.
- 6. 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.
- 7. Calculation of Short-Circuit Currents-Point-To-Point Method This method assumes unlimited primary short-circuit current (infinite bus).
- 8. Step 1 (Ifla) Determine Transformer full-load amperes from Name plate Transformer Table Formula= 3ph Transf Ifla= KVAx1000 E(L-L)
- 9. Step 2 (Multiplier) Find Transf. Mutilplier. Multiplier= __100__ Transf. %Z
- 10. Step 3 (ISCA) Determine transf. let-thru short-circuit current lSCA = IFLA x multiplier
- 11. 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
- 12. Step 5 (“M”) Calculate “M” M=__1__ 1+f
- 13. Step 6 (ISCA @ fault) Compute the available short-circuit current (symmetrical) at the fault ISCA = ISCA x M at Fault
- 15. Step 1 (Ifla) Ifla = KVA x 100 EL-L x 1.73 Ifla = 300 x 1000 = 834 208 x 1.73
- 16. Step 2 (Multiplier) Multiplier=__100__ Trans.%Z Multiplier = 100 = 55.55 1.8
- 17. Step 3 (ISCA) ISCA = IFLA x Multiplier ISCA = 834 x 55.55= 46,329A
- 18. 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
- 19. Step 5 (“M”) M= 1_ 1+f M = ___1____ = .838 1 + .1927
- 20. Step 6 (ISCA @ fault) ISCA = ISCA X M Fault #1 ISCA = 46,329 x .838= 38,823 Fault #1
- 21. 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
- 22. Step 5 M=_1_ 1+f =_1__ = .437 1+1.29
- 23. Step 6 ISCA = ISCA x M Fault #2 Fault #1 ISCA = 38,823 x .437 = 16,965A Fault #2
- 24. Peak Let-Thru Current Equipment withstand ratings can be described as : How Much Fault Current can the equipment handle, and for How Long ?
- 25. 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.
- 26. KRP-C Sp
- 27. LPN RD SP
- 28. LPS RK SP
Editor's Notes
- #14: Motor short-circuit contribution, if significant, may be added to the transformer secondary short-circuit current value as determined in Step 3. Proceed with this adjusted figure through stepts 4,5,and 6. A practical estimate of motor short-circuit contribution is to multiply the total load current in amperes by 4.