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Electrical Plan Review Point-To-Point Method Of Short ...

7 Electrical Plan ReviewPoint-To-Point Method Of Short -Circuit CalculationAdequate interrupting rating and protection of Electrical componentsare two essential aspects required by the NEC (B), , , , , , , and Table Note. The first step to ensure that system protective devices have the prop-er interrupting rating and provide component protection is to determinethe available Short -circuit currents. The application of the Point-To-Point Method can be used to determine the available Short -circuit cur-rents with a reasonable degree of accuracy at various points for either3for 1felectrical distribution systems. The example shown hereassumes unlimited primary Short -circuit current (infinite bus).Calculation Of Short -Circuit Currents Point-To-Point Of 3-Phase Short -Circuit CalculationBasic Short -Circuit Calculation 1 Determine amperesfrom either:a) Name plateb) Tables 3A & 3Bc) FormulaStep 2 Find transformermultiplierSee Note 3 Determine Short -circuit current(Formula or Table 5)See Note 1 and Note 4 Calculate f 5 Calculate M (multiplier) or takefrom Table 6 Compute theavailable Short -circuit current(RMS symmetrical)See Note 1, Note 2, and Note 53o transf.

Note: These values are equal to one over the impedance per foot and based upon resistance and reactance values found in IEEE Std 241-1990 (Gray Book), IEEE Recommended Practice for Electric Power Systems in Commerical Buildings & IEEE Std 242-1986 (Buff Book), IEEE Recommended Practice for Protection and Coordination of Industrial and ...

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Transcription of Electrical Plan Review Point-To-Point Method Of Short ...

1 7 Electrical Plan ReviewPoint-To-Point Method Of Short -Circuit CalculationAdequate interrupting rating and protection of Electrical componentsare two essential aspects required by the NEC (B), , , , , , , and Table Note. The first step to ensure that system protective devices have the prop-er interrupting rating and provide component protection is to determinethe available Short -circuit currents. The application of the Point-To-Point Method can be used to determine the available Short -circuit cur-rents with a reasonable degree of accuracy at various points for either3for 1felectrical distribution systems. The example shown hereassumes unlimited primary Short -circuit current (infinite bus).Calculation Of Short -Circuit Currents Point-To-Point Of 3-Phase Short -Circuit CalculationBasic Short -Circuit Calculation 1 Determine amperesfrom either:a) Name plateb) Tables 3A & 3Bc) FormulaStep 2 Find transformermultiplierSee Note 3 Determine Short -circuit current(Formula or Table 5)See Note 1 and Note 4 Calculate f 5 Calculate M (multiplier) or takefrom Table 6 Compute theavailable Short -circuit current(RMS symmetrical)See Note 1, Note 2, and Note 53o transf.

2 IFLA= ISCA= multiplier ISCA (L-L-L)= 46,273 x .852 = 39,425A(Use ISCA (L-L-L)at Fault #1 to calculate)** ISCA (L-L-L)= 833 X = 46,273 KVA x 1000EL-Lx x 1000EL-Lx 833 AKVA x 1000EL-Lx x 1000208 x faults f = x L x lL-L-LC x n x EL-Lf = x L x lL-L-LC x n x x 20 x 46,27322,185 x 2 x 208f = x 20 x 39,4255,907 x 1 x 2082 x L x lL-LC x n x EL-L100 Transf. % Z1o transf. IFLA=Multiplier =100 .9x Transf. % ZMultiplier =KVA x 1000EL-Lf =1 ISCA = ISCA x M1 + fM =11 + f11 + .174M =2 x L x lL-N*C x n x EL-Nf =1o line-to-line(L-L) faults See Note 51o line-to-neutral(L-N) faults See Note 2 andNote 5L= length (feet) of conduit to the conductor constant. See Tables 1, number of conductors per phase(Adjusts C value for parallel runs)I= available Short -circuit current inamperes at beginning of #1atfaultatbeginning of Short -circuit contribution, if significant, should be added at all faultlocations throughout the system.

3 A practical estimate of motor Short -circuit contri-bution is to multiply the total motor full-load current in amperes by 4. Values of 4 to6 are commonly acceptedNote single-phase center-tapped transformers, the L-N fault current is higherthan 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 terminalsIL-N= x IL-Lat Transformer TerminalsAt some distance from the terminals, depending upon wire size, the L-N fault current islower than the L-L fault current. The multiplier is an approximation and will theoret-ically vary from to These figures are based on change in turns ratio betweenprimary and secondary, infinite source available, zero feet from terminals of trans-former, and x %X and x %R for L-N vs.

4 L-L resistance and reactance L-N calculations at transformer secondary terminals, then proceed Point-To-Point . *300 KVA, 2%Z, 3f120/208V Secondary20', (2) 4-500 kcmil CUSteel conduit20', 4 - 2 AWG CUSteel conduitInfinitePrimaryAvailableFault #2 Fault #1 MAIN SERVICEPANELBRANCH CIRCUIT PANELStep 1 Step 2 Step 3 Step 4 Step 5 Step 6 Step 4 Step 5 Step 6==== .174= .852 (See Table 4)11 + f11 + === .474 (See Table 4)= Short -Circuit Current at Transformer Secondary3-Phase Short Circuit Current at Fault #1 ISCA (L-L-L)= 39,425 x .474 = 18,687A3-Phase Short -Circuit Current at Fault #2 FAULT #2**The motor contribution and voltage variance should be accounted for at thispoint. See Notes 1 and 4. Transformer %Z is multiplied by .9 to establish a worst case condition. See Note 3:The marked impedance values on transformers may vary 10% fromthe actual values determined by ANSI / ieee test.

5 See Standard , multiply transformer % Z by .9. Transformers constructed to ANSI standards have a impedance tolerance (two-winding construction).Note voltages may vary 10% for power , and for 120-volt light-ing services. Therefore, for worst case conditions, multiply values as calculatedin Step 3 by and/or 5:The calculated Short -circuit currents above represent the bolted faultvalues that approximate worst case conditions. Approximations of Bolted faultvalues as percentage of 3-Phase (L-L-L) bolted fault values are shown (L-L):87% Phase-Ground (L-G)25-125% (Use 100% near transformer, 50% otherwise)Phase-Neutral (L-N)25-125% (Use 100% near transformer, 50% otherwise)Note 6:Approximation of arcing fault values for sustained arcs as percentageof 3-Phase (L-L-L) bolted fault values are shown (L-L-L) Arching Fault Phase-Phase (L-L) Arcing Fault Phase-Ground (L-G) Arcing Fault 89% (maximum)74% (maximum)38% (minimum)8 Voltage(Line-to- Transformer KVA RatingLine)

6 15022530050075010001500 2000208125 208 3124166258331388 2082 27764164 5552220118 197 2953945907871312 1968 26243937 5249240108 180 2713615417221203 1804 24063609 4811440599814819729539465698413121968 2624460569414118828237762894112551883 2510480549013518027136160190212031804 240660043721081442172894817229621443 1925 Electrical Plan ReviewPoint-To-Point Method Of Short -Circuit CalculationAmpacityBuswayPlug-InFeederHi gh ImpedanceCopperAluminumCopperAluminumCop per225287002300018700 12000 40038900347002390021300 60041000383003650031300 80046100575004930044100 1000694008930062900562001560012009430097 1007690069900161001350119000104200901008 4000175001600129900120500101000909001920 0200014290013510013420012500020400250014 3800156300180500166700217003000144900175 400204100188700238004000 277800256400 Note:These values are equal to one over the impedance per foot forimpedance in a survey of KVA Rating255075100167250333500115/230109217 326435726108714482174120/240104208313417 696104213882083230/460541091632173635437 241087240/480521041562083485216941042 Calculation Of Short -Circuit CurrentsAt Second Transformer In the following procedure to calculate the levelof fault current at the secondary of a second,downstream transformer in a system when thelevel of fault current at the transformer primary ACalculate f (ISCA(P), known).

7 Step BCalculate M (multiplier) or takefrom Table CCalculate Short -circuitcurrent at secondaryof transformer.(See Note 1 under Basic Procedure )3o transformer (ISCA(P) andISCA(S) are 3o fault values).1o transformer (ISCA(P) andISCA(S) are 1o fault values;ISCA(S) is L-L.)ProcedureProcedure For Second Transformer in SystemFormulaISCA(P)= Available fault current at transformer (S)= Available fault current at transformer Primary voltage Secondary voltage = KVA rating of = Percent impedance of : To calculate fault level at the endof a conductor run, follow Steps 4, 5, and 6of Basic ,000 x KVAISCA(P)x VPx (%Z)ISCA(S)=VPx M x ISCA(P)f=100,000 x KVAISCA(P)x VPx (%Z)M=1 + f1 VSTable 3A. Three-Phase Transformer Full-LoadCurrent Rating (In Amperes)Table 1. C Values for BuswayTable 3B. Single-Phase Transformer Full-LoadCurrent Rating (In Amperes)MAIN UTILITYCONNECTIONKNOWNFAULTCURRENTKNOWNF AULTCURRENTT able 2.

8 C Values for ConductorsCopperAWGT hree Single ConductorsThree-Conductor CableorConduitConduitkcmilSteelNonmagnet ic Steel Nonmagnetic600V 5kV15kV600V5kV15kV600V5kV15kV600V5kV15kV 14389--389--389--389--12617--617--617--6 17--10981--982--982--982--815571551-1559 1555-15591557-15601558-62425240623892430 2418240724312425241524332428242143806375 1369638263789375338303812377938383823379 8347744674457748114745467948204785472648 3348034762259075736557460445926580959895 9305828608760235958172937029675974937307 71097454736571897579750773641/0892585447 9739317903485909210908687089473937390532 /010755100629390114241087810319112451104 5105001170311529110533/01284411804110221 3923130481236013656133331261314410141191 34624/0150821360612543166731535114347163 9215890148131748317020160132501648314925 1364418594171211586618311178511646619779 1935218001300181771629314769208681897517 4092061720052183192252521938201633501970 4173851567822737205261867222646219141982 1249042412621982400205661823516366242972 1786197312425323372210422691626044235185 0022185191721749226706232772133026980254 4923126300962871225916600229652056717962 2803325204220972875227975248973215431258 2776675024137213871888929735264532340831 05130024269333460533315297351,000 2527822539199233149128083248873386432689 29320371973574931959 ISCA(P)ISCA(S)ISCA(P)ISCA(S)Note: These values are equal to one over the impedance per foot and based upon resistance and reactance values found in ieee Std 241-1990 (Gray Book), ieee recommended Practice for electric PowerSystems in Commerical Buildings & ieee Std 242-1986 (Buff Book), ieee recommended Practice for Protection and Coordination of Industrial and Commercial power Systems.

9 Where resistance and9 Electrical Plan ReviewPoint-To-Point Method Of Short -Circuit CalculationAluminumAWGT hree Single ConductorsThree-Conductor CableorConduitConduitkcmilSteelNonmagnet ic Steel Nonmagnetic600V 5kV15kV600V5kV15kV600V5kV15kV600V5kV15kV 14237--237--237--237--12376--376--376--3 76--10599--599--599--599--8951950-952951 -952951-952952-6148114761472148214791476 1482148014781482148114794234623332319235 0234223332351234723392353235023443295229 2829042961294529292963295529412966295929 4923713367036263730370236733734371936933 7403725370914645457544984678463245804686 466446184699468246461/057775670549358385 76656465852582057175876585257712/0718769 6867337301715369867327727171097373732972 023/088268467816391108851862790778981875 19243916489774/0107411016797001117410749 1038711185110221064211409112771096925012 1221146010849128621234311847127971263612 1151323613106126613001391013009121931492 3141831349214917146981397315495153001465 9350154841428013288168131585814955167951 6490155411763517352165014001667115355141 8818506173211623418462180641692119588192 4418154500187561682815657213911950318315 2139520607193142301822381209786002009318 4281648423451217181963523633231962134925 7082524423295750217661968517686259762370 2214372643225790237502903628262259761,00 0 2347821235190062877926109234822986529049 26608329383192029135 Table 5.

10 Short -Circuit Currents Available from Various Size Transformers(Based upon actual field nameplate data, published information, or from util-ity transformer worst case impedance)VoltageFull%ShortandLoadImpeda nce CircuitPhaseKVAAmps(nameplate)Amps ph.* ph.** ph.** ** **fM 4. M (Multiplier)*11 + fM =Table 5 Notes:Single phase values are L-N values at transformer terminals. These figuresare based on change in turns ratio between primary and secondary, 100,000 KVA primary, zero feet from terminals of transformer, (%X) and (%R)multipliers for L-N vs. L-L reactance and resistance values and transformerX/R ratio = Short -circuit currents based on infinite listed transformers 25 KVA or greater have a 10% impedance constructed to ANSI standards have a impedance tol-erance (two-winding construction). Short -circuit amps reflect a worst case condition (-10%).


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