Point-To-Point Method Of Short-Circuit Calculation Calculation Of Short-Circuit Currents Point-To-Point Method. Adequate interrupting rating and protection of electrical components are two essential aspects required by the NEC 0.3(B), 0.9, 0.0, 240., 250.4, 250.90, 250.96, and Table 250.22 Note. The first step to ensure that system protective devices have the proper interrupting rating and provide component protection is to determine the available short-circuit currents. The application of the Point-To- Point method can be used to determine the available short-circuit currents with a reasonable degree of accuracy at various points for either 3f or f electrical distribution systems. The example shown here assumes unlimited primary short-circuit current (infinite bus). Basic Short-Circuit Calculation Procedure. Procedure Formula FAULT # Step Determine transf. KVA x 000 3o transf. I full-load amperes FLA = Step E L-L x.732 from either: a) Name plate KVA x 000 b) Tables 3A & 3B o transf. I FLA = Step 2 E L-L c) Formula Step 2 Step 3 Step 4 Step 5 Find transformer multiplier See Note 3. Determine transf. let-through shortcircuit current I SCA = Transf. FLA x multiplier (Formula or Table 5) See Note and Note 4. Calculate f factor. Calculate M (multiplier) or take from Table 4. Multiplier = 3o faults o line-to-line (L-L) faults See Note 5 o line-to-neutral (L-N) faults See Note 2 and Note 5 + f 00 Transf. % Z.732 x L x ll-l-l C x n x E L-L 2 x L x l L-L C x n x E L-L 2 x L x l L-N* C x n x E L-N L = length (feet) of conduit to the fault. C = conductor constant. See Tables, 2. n = number of conductors per phase (Adjusts C value for parallel runs) I = available short-circuit current in amperes at beginning of circuit. Step 6 Compute the available shortcircuit current fault beginning of circuit. I SCA = I SCA x M at at (RMS symmetrical) See Note, Note 2, and Note 5 Note. Motor short-circuit contribution, if significant, should be added at all fault locations throughout the system. A practical estimate of motor short-circuit contribution is to multiply the total motor full-load current in amperes by 4. Values of 4 to 6 are commonly accepted * Note 2. For single-phase center-tapped transformers, the L-N fault current is higher than the L-L fault current at the secondary terminals. The short-circuit current available (I) for this case in Step 4 should be adjusted at the transformer terminals as follows: At L-N center tapped transformer terminals I L-N =.5 x I L-L at Transformer Terminals At some distance from the terminals, depending upon wire size, the L-N fault current is lower than the L-L fault current. The.5 multiplier is an approximation and will theoretically vary from.33 to.67. These figures are based on change in turns ratio between primary and secondary, infinite source available, zero feet from terminals of transformer, and.2 x %X and.5 x %R for L-N vs. L-L resistance and reactance values. Begin L-N calculations at transformer secondary terminals, then proceed point-to-point. 7 Example Of 3-Phase Short-Circuit Calculation 300 KVA, 2%Z, 3f 20/208V Secondary Infinite Primary Available Step 3 Step 4 Step 5 Step 6 FAULT #2 Step 4 Step 5 Step 6 20', (2) 4-500 kcmil CU Steel conduit KVA x 000 300 x 000 I FLA = = = 833A E L-L x.732 208 x.732 Multiplier = 00.9x Transf. % Z ** I SCA (L-L-L) = 833 X 55.55 = 46,273 3-Phase Short-Circuit Current at Transformer Secondary I SCA (L-L-L) = 46,273 x.852 = 39,425A 3-Phase Short Circuit Current at Fault # (Use I SCA (L-L-L) at Fault # to calculate) 00.8.732 x L x ll-l-l.732 x 20 x 46,273 = =.74 C x n x E L-L 22,85 x 2 x 208 + f = Fault # MAIN SERVICE PANEL 20', 4-2 AWG CU Steel conduit +.74.732 x 20 x 39,425 5,907 x x 208 =. = Fault #2 BRANCH CIRCUIT PANEL = 55.55 =.852 (See Table 4) = =.474 (See Table 4) + f +. I SCA (L-L-L) = 39,425 x.474 = 8,687A 3-Phase Short-Circuit Current at Fault #2 ** The motor contribution and voltage variance should be accounted for at this point. See Notes and 4. Transformer %Z is multiplied by.9 to establish a worst case condition. See Note 3. Note 3: The marked impedance values on transformers may vary ±0% from the actual values determined by ANSI / IEEE test. See U.L. Standard 56. Therefore, multiply transformer %Z by.9. Transformers constructed to ANSI standards have a ± 7.5% impedance tolerance (two-winding construction). Note 4. Utility voltages may vary ±0% for power, and ±5.8% for 20-volt lighting services. Therefore, for worst case conditions, multiply values as calculated in Step 3 by. and/or.058 respectively. Note 5: The calculated short-circuit currents above represent the bolted fault values that approximate worst case conditions. Approximations of Bolted fault values as percentage of 3-Phase (L-L-L) bolted fault values are shown below. Phase-Phase (L-L): 87% Phase-Ground (L-G) 25-25% (Use 00% near transformer, 50% otherwise) Phase-Neutral (L-N) 25-25% (Use 00% near transformer, 50% otherwise) Note 6: Approximation of arcing fault values for sustained arcs as percentage of 3-Phase (L-L-L) bolted fault values are shown below. 3-Phase (L-L-L) Arching Fault Phase-Phase (L-L) Arcing Fault Phase-Ground (L-G) Arcing Fault 89% (maximum) 74% (maximum) 38% (minimum)
Point-To-Point Method Of Short-Circuit Calculation Calculation Of Short-Circuit Currents At Second Transformer In System. Use the following procedure to calculate the level of fault current at the secondary of a second, downstream transformer in a system when the level of fault current at the transformer primary is known. Procedure For Second Transformer in System Procedure Formula Step A Calculate f (ISCA(P), known). 3o transformer (ISCA(P) and ISCA(S) are 3o fault values). o transformer (ISCA(P) and ISCA(S) are o fault values; ISCA(S) is L-L.) ISCA(P) x VP x.732 (%Z) f= 00,000 x KVA ISCA(P) x VP x (%Z) f= 00,000 x KVA MAIN TRANSFORMER H.V. UTILITY CONNECTION ISCA(P) ISCA(P) KNOWN FAULT CURRENT KNOWN FAULT CURRENT ISCA(S) ISCA(S) Step B Step C Calculate M (multiplier) or take from Table 4. Calculate short-circuit current at secondary of transformer. (See Note under Basic Procedure ) + f I SCA(S) = VP x M x I SCA(P) I SCA(P) = Available fault current at transformer primary. ISCA(S) = Available fault current at transformer secondary. VP = Primary voltage L-L. VS = Secondary voltage L-L. VS KVA = KVA rating of transformer. %Z = Percent impedance of transformer. Note: To calculate fault level at the end of a conductor run, follow Steps 4, 5, and 6 of Basic Procedure. Table. C Values for Busway Ampacity Busway Plug-In Feeder High Impedance Copper Aluminum Copper Aluminum Copper 225 28700 23000 8700 2000 400 38900 34700 23900 2300 600 4000 38300 36500 3300 800 4600 57500 49300 4400 000 69400 89300 62900 56200 5600 200 94300 9700 76900 69900 600 350 9000 04200 9000 84000 7500 600 29900 20500 0000 90900 9200 2000 42900 3500 34200 25000 20400 2500 43800 56300 80500 66700 2700 3000 44900 75400 20400 88700 23800 4000 277800 256400 Note: These values are equal to one over the impedance per foot for impedance in a survey of industry. Table 2. C Values for Conductors 8 Table 3A. Three-Phase Transformer Full-Load Current Rating (In Amperes) Voltage (Line-to- Transformer KVA Rating Line) 45 75 2.5 50 225 300 500 750 000 500 2000 208 25 208 32 46 625 833 388 2082 2776 464 5552 220 8 97 295 394 590 787 32 968 2624 3937 5249 240 08 80 27 36 54 722 203 804 2406 3609 48 440 59 98 48 97 295 394 656 984 32 968 2624 460 56 94 4 88 282 377 628 94 255 883 250 480 54 90 35 80 27 36 60 902 203 804 2406 600 43 72 08 44 27 289 48 722 962 443 925 Table 3B. Single-Phase Transformer Full-Load Current Rating (In Amperes) Voltage Transformer KVA Rating 25 50 75 00 67 250 333 500 5/230 09 27 326 435 726 087 448 274 20/240 04 208 33 47 696 042 388 2083 230/460 54 09 63 27 363 543 724 087 240/480 52 04 56 208 348 52 694 042 Copper AWG Three Single Conductors Three-Conductor Cable or Conduit Conduit kcmil Steel Nonmagnetic Steel Nonmagnetic 600V 5kV 5kV 600V 5kV 5kV 600V 5kV 5kV 600V 5kV 5kV 4 389 - - 389 - - 389 - - 389 - - 2 67 - - 67 - - 67 - - 67 - - 0 98 - - 982 - - 982 - - 982 - - 8 557 55-559 555-559 557-560 558-6 2425 2406 2389 2430 248 2407 243 2425 245 2433 2428 242 4 3806 375 3696 3826 3789 3753 3830 382 3779 3838 3823 3798 3 4774 4674 4577 48 4745 4679 4820 4785 4726 4833 4803 4762 2 5907 5736 5574 6044 5926 5809 5989 5930 5828 6087 6023 5958 7293 7029 6759 7493 7307 709 7454 7365 789 7579 7507 7364 /0 8925 8544 7973 937 9034 8590 920 9086 8708 9473 9373 9053 2/0 0755 0062 9390 424 0878 039 245 045 0500 703 529 053 3/0 2844 804 022 3923 3048 2360 3656 3333 263 440 49 3462 4/0 5082 3606 2543 6673 535 4347 6392 5890 483 7483 7020 603 250 6483 4925 3644 8594 72 5866 83 785 6466 9779 9352 800 300 877 6293 4769 20868 8975 7409 2067 20052 839 22525 2938 2063 350 9704 7385 5678 22737 20526 8672 22646 294 982 24904 2426 2982 400 20566 8235 6366 24297 2786 973 24253 23372 2042 2696 26044 2358 500 2285 972 7492 26706 23277 2330 26980 25449 2326 30096 2872 2596 600 22965 20567 7962 28033 25204 22097 28752 27975 24897 3254 3258 27766 750 2437 2387 8889 29735 26453 23408 305 30024 26933 34605 3335 29735,000 25278 22539 9923 349 28083 24887 33864 32689 29320 3797 35749 3959 Note: These values are equal to one over the impedance per foot and based upon resistance and reactance values found in IEEE Std 24-990 (Gray Book), IEEE Recommended Practice for Electric Power Systems in Commerical Buildings & IEEE Std 242-986 (Buff Book), IEEE Recommended Practice for Protection and Coordination of Industrial and Commercial Power Systems. Where resistance and
Point-To-Point Method Of Short-Circuit Calculation Table 4. M (Multiplier)* f M f M 0.0 0.99.50 0.40 0.02 0.98.75 0.36 0.03 0.97 2.00 0.33 0.04 0.96 2.50 0.29 0.05 0.95 3.00 0.25 0.06 0.94 3.50 0.22 0.07 0.93 4.00 0.20 0.08 0.93 5.00 0.7 0.09 0.92 6.00 0.4 0.0 0.9 7.00 0.3 0.5 0.87 8.00 0. 0.20 0.83 9.00 0.0 0.25 0.80 0.00 0.09 0.30 0.77 5.00 0.06 0.35 0.74 20.00 0.05 0.40 0.7 30.00 0.03 0.50 0.67 40.00 0.02 0.60 0.63 50.00 0.02 0.70 0.59 60.00 0.02 0.80 0.55 70.00 0.0 0.90 0.53 80.00 0.0.00 0.50 90.00 0.0.20 0.45 00.00 0.0 * + f Table 5 Notes: * Single phase values are L-N values at transformer terminals. These figures are based on change in turns ratio between primary and secondary, 00,000 KVA primary, zero feet from terminals of transformer,.2 (%X) and.5 (%R) multipliers for L-N vs. L-L reactance and resistance values and transformer X/R ratio = 3. ** Three-phase short-circuit currents based on infinite primary. UL listed transformers 25 KVA or greater have a ±0% impedance tolerance. Transformers constructed to ANSI standards have a ± 7.5% impedance tolerance (two-winding construction). Short-circuit amps reflect a worst case condition (-0%). Fluctuations in system voltage will affect the available short-circuit current. For example, a 0% increase in system voltage will result in a 0% increase in the available short-circuit currents shown in the table. Table 5. Short-Circuit Currents Available from Various Size Transformers (Based upon actual field nameplate data, published information, or from utility transformer worst case impedance) Voltage Full % Short and Load Impedance Circuit Phase KVA Amps (nameplate) Amps 25 04.5 275 37.5 56.5 808 20/240 50 208.5 23706 ph.* 75 33.5 34639 00 47.6 42472 67 696.6 66644 45 25.0 3879 75 208.0 2332 2.5 32. 3259 50 46.07 43237 20/208 225 625.2 6960 3 ph.** 300 833. 83357 500 388.24 24364 750 2082 3.50 6609 000 2776 3.50 882 500 464 3.50 328 2000 5552 4.00 542 2500 6940 4.00 92764 75 90.0 0035 2.5 35.0 5053 50 8.20 6726 225 27.20 25088 300 36.20 3345 277/480 500 602.30 5463 3 ph.** 750 903 3.50 28672 000 204 3.50 38230 500 806 3.50 57345 2000 2408 4.00 66902 2500 30 4.00 83628 Aluminum AWG Three Single Conductors Three-Conductor Cable or Conduit Conduit kcmil Steel Nonmagnetic Steel Nonmagnetic 600V 5kV 5kV 600V 5kV 5kV 600V 5kV 5kV 600V 5kV 5kV 4 237 - - 237 - - 237 - - 237 - - 2 376 - - 376 - - 376 - - 376 - - 0 599 - - 599 - - 599 - - 599 - - 8 95 950-952 95-952 95-952 952-6 48 476 472 482 479 476 482 480 478 482 48 479 4 2346 2333 239 2350 2342 2333 235 2347 2339 2353 2350 2344 3 2952 2928 2904 296 2945 2929 2963 2955 294 2966 2959 2949 2 373 3670 3626 3730 3702 3673 3734 379 3693 3740 3725 3709 4645 4575 4498 4678 4632 4580 4686 4664 468 4699 4682 4646 /0 5777 5670 5493 5838 5766 5646 5852 5820 577 5876 5852 577 2/0 787 6968 6733 730 753 6986 7327 727 709 7373 7329 7202 3/0 8826 8467 863 90 885 8627 9077 898 875 9243 964 8977 4/0 074 067 9700 74 0749 0387 85 022 0642 409 277 0969 250 222 460 0849 2862 2343 847 2797 2636 25 3236 306 266 300 390 3009 293 4923 483 3492 497 4698 3973 5495 5300 4659 350 5484 4280 3288 683 5858 4955 6795 6490 554 7635 7352 650 400 667 5355 488 8506 732 6234 8462 8064 692 9588 9244 854 500 8756 6828 5657 239 9503 835 2395 20607 934 2308 2238 20978 600 20093 8428 6484 2345 278 9635 23633 2396 2349 25708 25244 23295 750 2766 9685 7686 25976 23702 2437 26432 25790 23750 29036 28262 25976,000 23478 2235 9006 28779 2609 23482 29865 29049 26608 32938 3920 2935 reactance values differ or are not available, the Buff Book values have been used. The values for reactance in determining the C Value at 5 KV & 5 KV are from the Gray Book only (Values for 4-0 AWG at 5 kv and 4-8 AWG at 5 kv are not available and values for 3 AWG have been approximated). 9
0
Work Sheet Problem Main Distribution Panel PRIMARY FUSE 300 KVA Transformer by Utility 20/208 Volt, 3 Phase, 4 Wire 2% Impedance METER (2) 3" C. Each with 4-500 kcmils/xhhn - 20 Feet 4-8 AWG THHN, 3/4"C. - 0 Feet 8 EMP 2 MAIN SWITCH Ground Bus -2/0 AWG 2 3 800/800 200/200 4-3/0 AWG THHN, 2" C. - 60 Feet FLUOR. 20V FIXTURE 3-2 AWG THHN /2" C. - 30' LPA 9 4 200/50 4 - /0 AWG THHN, -/2" C. - 5 Feet 3 LPC 4 5 6 7 8 00/00 00/90 00/70 00/ 4-3 AWG THHN, -/4" C. - 20 Feet 3-3 AWG THHN, " C. - 35 Feet 3-4 AWG THHN, " C. - 35 Feet 5 6 7 LPB AC- AC-2 3-8 AWG THHN, 3/4" C.- 4 Feet 0 9 200/ 7-/2 Combination Motor Controller Note: Assume steel conduit.
Short-Circuit Calculations - Worksheet () Transformer (Secondary Terminals Assuming Infinite Primary) Find: Transformer Full-Load Amperes - I FLA (3 Phase): I FLA = Find: Multiplier M SCA = Motor Contribution* = * Note: Calculate additional motor short-circuit contribution. Assume 50% (400A) of the total load is from all motors. Multiply total motor FLA by 4 (400 x 4 =,600A). In theory, the additional motor short-circuit contribution should be calculated at all points in the system, and may vary depending upon the location. (2) MDP Short-Circuit Current at beginning of run (Transformer Secondary Terminals with voltage variance) = (3) LPA Short-Circuit Current at beginning of run (MDP with voltage variance) = (4) LPC Short-Circuit Current at beginning of run (MDP with voltage variance) = 2
Short-Circuit Calculations - Worksheet (5) LPB Short-Circuit Current at beginning of run (MDP with voltage variance) = (6) AC- Short-Circuit Current at beginning of run (MDP with voltage variance) = (7) AC-2 Short-Circuit Current at beginning of run (MDP with voltage variance) = 3
Short-Circuit Calculations - Worksheet (8) EMP Short-Circuit Current at beginning of run (MDP with voltage variance) = (9) Fluorescent Fixture Short-Circuit Current at beginning of run (LPA with voltage variance) = *Ignore motor contribution for this step (0) Combination Motor Controller Short-Circuit Current at beginning of run (LPC with voltage variance) = 4