1865 words - 7 pages

Short Circuit Capacity: Basic Calculations and Transformer SizingShort circuit capacity calculation is used for many applications: sizing of transformers,selecting the interrupting capacity ratings of circuit breakers and fuses, determining if aline reactor is required for use with a variable frequency drive, etc.The purpose of the presentation is to gain a basic understanding of short circuit capacity.The application example utilizes transformer sizing for motor loads.Conductor impedances and their associated voltage drop are ignored not only to present asimplified illustration, but also to provide a method of approximation by which a plantengineer, electrician or production manager will be able to either evaluate a new applicationor review an existing application problem and resolve the matter quickly.Literature containing a detailed discussion of short circuit capacity calculations areavailable within the electrical power transmission industry. [1]The following calculations will determine the extra kVA capacity required for a threephase transformer that is used to feed a single three phase motor that is started withfull voltage applied to its terminals, or, "across-the-line."Two transformers will be discussed, the first having an unlimited short circuit kVAcapacity available at its primary terminals, and the second having a much lower inputshort circuit capacity available.kVA of a single phase transformer = V x AkVA of a three phase transformer = V x A x 1.732, where 1.732 = the square root of 3.The square root of 3 is introduced for the reason that, in a three phase system,the phases are 120 degrees apart and, therefore, can not be added arithmetically.They will add algebraically.Transformer Connected To Utility Power LineThe first transformer is rated 1000 kVA, 480 secondary volts, 5.75% impedance.Rated full load amp output of the transformer is1000 kVA / (480 x 1.732) = 1203 ampsThe 5.75% impedance rating indicates that 1203 amps will flow in the secondary ifthe secondary is short circuited line to line and the primary voltage is raised fromzero volts to a point at which 5.75% of 480 volts, or, 27.6 volts, appears at thesecondary terminals. Therefore, the impedance (Z) of the transformer secondary maynow be calculated:Z = V / I = 27.6 volts / 1203 amps = .02294 ohmsThe transformer is connected directly to the utility power lines which we willassume are capable of supplying the transformer with an unlimited short circuitkVA capacity. The utility company will always determine and advise of the shortcircuit capacity available at any facility upon request.With unlimited short circuit kVA available from the utility, the short circuitamperage capacity which the transformer can deliver from its secondary is480 volts / .02294 = 20,924 ampsAn alternative method of calculating short circuit capacity for the abovetransformer is:1203 amps x 100 / 5.75% = 1203 / .0575 = 20,922 ampsAnother alternative is to consult a reference manual. Cutler- Hammer...

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