Electrical Plan Review
Objectives
1
By reviewing this brochure, the Electrical Inspector,Electrical Contractor, Plan Examiner, ConsultingEngineer and others will be able to . . .
■ Understand the importance of overcurrent protection.
■ Understand the meaning and importance of electricalterms commonly used relating to overcurrent protec-tion.
■ Understand and discuss the KEY National ElectricalCode® requirements regarding overcurrent protection.
■ Calculate short-circuit currents using the simplePOINT-TO-POINT method.
■ Check electrical plans to determine conformance toNational Electrical Code® sections that cover short-circuit currents, interrupting ratings, interruptingcapacities, short-circuit ratings, ground faults,grounding electrode conductors, equipmentgrounding conductors, etc.
■ Verify that circuit, feeder, service, grounding elec-trode conductors, equipment grounding conductors,and bonding conductors have adequate capacity toconduct safely ANY fault current likely to beimposed on them.
■ Determine let-through current values (peak & RMS)when current-limiting overcurrent devices are used.
■ Apply current-limiting data to protect downstreamelectrical components that have withstand ratingsLESS than the available fault-current at any givenpoint on the system.
■ Realize that whenever overcurrent protection isdiscussed, the two most important issues are:
— HOW MUCH CURRENT WILL FLOW?— HOW LONG WILL THE CURRENT FLOW?
■ Adopt a Form Letter and a Data Required Form thatcan be used to “log-in” the necessary data relatingto available fault currents, interrupting ratings, andlet-through currents.
■ Understand that the major sources of short-circuitcurrents are motors and generators.
■ Understand that transformers are NOT a source ofshort-circuit current.
■ Know how to ask the right questions.
Copyrighted April, 1999 by Cooper Bussmann, Inc., Printed in U.S.A.
250-96(a) General. Metal raceways, cable trays, cable armor, cable sheath, enclosures, frames, fittings, and other metal noncurrent-carrying parts that are to serve as grounding conductors, with or without the use of supplementary equipment grounding conductors, shall be effectively bonded where necessary to ensure electrical continuity and the capacity to conduct safely any fault current likely to be imposed on them. Any nonconductive paint, enamel, or similar coating shall be removed at threads, contact points, and contact surfaces or be connected by means of fittings designed so as to make such removal unnecessary.
Table 250-122 Minimum Size Equipment Grounding
Conductors for Grounding Raceway and Equipment.
NOTE: Where necessary to comply with Section 250-2(d), the equipmentgrounding conductor shall be sized larger than this table.
620-62 Selective Coordination. Where more than one driving machine disconnecting means is supplied by a single feeder, the overcurrent protective devices in each disconnecting means shall be selectively coordinated with any other supply side
overcurrent protective devices.
NOTE: Short-circuit currents may be calculated many ways. A simple methodis the Point-To-Point method as presented in Bussmann’s ElectricalProtection Handbook, Bulletin SPD. That data is included in this bulletin forease of use.
Electrical Plan Review
Critical National Electrical Code® Sections
2
15 14 12 20 12 10 30 10 840 10 860 10 8
100 8 6200 6 4300 4 2400 3 1500 2 1/0
600 1 2/0800 1/0 3/0
1000 2/0 4/01200 3/0 2501600 4/0 350
2000 250 4002500 350 6003000 400 6004000 500 8005000 700 12006000 800 1200
Size(AWG or kcmil)
Copper
Aluminum orCopper-Clad Aluminum
Rating or setting of Automatic Overcurrent Device in Circuit Ahead of Equipment, Conduit,
etc., Not Exceeding (Amperes)
110-3(b) Installation and Use. Listed or labeled equipment shall be installed and used in accordance with any instructions included in the listing or labeling.
110-9 Interrupting Rating. Equipment intended to interrupt current at fault levels shall have an interrupting rating sufficient for the nominal circuit voltage and the current that is available at the line terminals of the equipment.
Equipment intended to interrupt current at other than fault levels shall have an interrupting rating at nominal circuit voltage sufficient for the currentthat must be interrupted.
110-10 Circuit Impedance and Other Characteristics.The overcurrent protective devices, the totalimpedance, the component short-circuit current ratings, and other characteristics of the circuit to be protected shall be selected and coordinated to permit the circuit-protective devices used to cleara fault to do so without extensive damage to the electrical components of the circuit. This fault shallbe assumed to be either between two or more of the circuit conductors, or between any circuit conductor and the grounding conductor or enclosing metal raceway. Listed products applied in accordance with their listing shall beconsidered to meet the requirements of this section.
240-1 Scope (FPN). Overcurrent protection for con-ductors and equipment is provided to open the circuit if the current reaches a value that will cause an excessive or dangerous temperature in conductors or conductor insulation. See also Sections 110-9 and 110-10 for requirements forinterrupting ratings and protection against fault currents.
240-11 Definition of Current-Limiting Overcurrent Protection Device. A current-limiting overcurrent protective device is a device that, when interruptingcurrents in its current-limiting range, will reduce thecurrent flowing in the faulted circuit to a magnitude substantially less than that obtainable in the same circuit if the device were replaced with a solidconductor having comparable impedance.
250-2(d) Performance of Fault Current Path. The fault current path shall be permanent and electrically continuous, shall be capable of safely carrying the maximum fault likely to be imposed on it, and shall have sufficiently low impedance to facilitate the operation of overcurrent devices under fault conditions.
250-90 General. Bonding shall be provided where neces-sary to ensure electrical continuity and the capacity to conduct safely any fault current likelyto be imposed.
Electrical Plan Review
The Meaning of Short-Circuit Interrupting RatingFor a better understanding of interrupting rating, consider the following series of analogies.
3
Normal Current Operation
Short-Circuit Operation withInadequate Interrupting Rating
Short-Circuit Operation withAdequate Interrupting Rating
FLOOD GATESANALOGOUS TOOVERCURRENTPROTECTIVEDEVICE
RESERVOIR CAPACITYANALOGOUS TOAVAILABLE FAULT CURRENT
LOADCURRENT(100 GALLONSPER MINUTE)
OVERCURRENTPROTECTIVEDEVICE
AVAILABLE FAULTCURRENT (e.g., 50,000 AMPS)
FLOOD GATES AREDESTROYED BECAUSEOF INADEQUATEINTERRUPTING RATING
SHORT CIRCUITCURRENT(50,000 GALLONSPER MINUTE)
INADEQUATE INTERRUPTINGRATING. THEREFORE,OVERCURRENT PROTECTIVE DEVICE IS DESTROYED
AVAILABLE FAULTCURRENT (e.g., 50,000 AMPS)
Downstream components maynot be able to withstand theamount of let-through current.
Downstream componentscapable of withstandinglet-through current.
FLOOD GATES HAVEADEQUATE INTERRUPTINGRATING. FAULT CURRENTSAFELY INTERRUPTED
ADEQUATELY RATEDOVERCURRENT PROTECTIVEDEVICE IS UNDAMAGED
SHORT CIRCUITCURRENT SAFELYCLEARED
AVAILABLE FAULTCURRENT (e.g., 50,000 AMPS)
Electrical Plan Review
Ratings of Overcurrent Protective DevicesMost overcurrent protective devices are labeled with two current ratings.
4
The Fuse
The Circuit Breaker
(1) NORMAL CURRENT RATING
(2) INTERRUPTING RATINGS
(1) NORMALCURRENTRATING
(2) INTERRUPTING RATING
Electrical Plan Review
Data “Log In” Form Letter
5
CITY OF ANYWHERE, USA
DEPARTMENT OF ELECTRICAL INSPECTION
DATE
TO: ELECTRICAL CONTRACTORS, ENGINEERS, ARCHITECTS.
RE: ELECTRIC SERVICE PERMIT APPLICATION.
COMPLIANCE WITH THE NATIONAL ELECTRICAL CODE®, SECTIONS
110-3(b), 110-9, 110-10, 240-1, 240-11, 250-2(d), 250-90,
250-96(a), and TABLE 250-122 Note.
The City of
, Department of Electrical
Inspection is required to enforce the 1996 National Electrical Code.
To ensure compliance, attention will be given to the SHORT-CIRCUIT
RATINGS of the equipment and overcurrent devices to be installed.
To accomplish this with minimum effort and time, the attached
form(s) are required to be completed by the electrical contractor,
then submitted to the Electrical Inspection Department PRIOR to
actual installation. Include a one-line riser diagram showing conduc-
tor sizes,conduit sizes, distances, and fault currents at all panels,
motor control centers, and main service equipment.
This data will be reviewed for compliance and conformance to the
above Code sections. When approved, the data will be so marked,
and will be kept on file for future referen
ce.
Sincerely,
Chief Electrical Inspector
City
ofAnywhere
Electrical Plan Review
“Log In” Data Required Form
6
DEPARTMENT OF ELECTRICAL INSPECTIONCITY OF
Date
Permit
Electrical Contractor
Street Address
City State Zip
The following information is requested to determine that the electrical equipment to beinstalled at:
Name of occupant or owner
is in compliance with the National Electrical Code® as it relates to available short-circuitcurrents and interrupting ratings. See Sections 110-3(b), 110-9, 110-10, 240-1, 240-11,250-2(d), 250-90, 250-96(a), and Table 250-122 Note. This form is to be completedand returned to the Department of Electrical Inspection for approval prior to installa-tion. THE FOLLOWING INFORMATION IS TO BE SUPPLIED BY THE ELECTRICAL CON-TRACTOR OR OTHER RESPONSIBLE PARTY:
TRANSFORMER KVA IMPEDANCE % SECONDARY VOLTAGE
PHASE 3 OR 4 WIRE LENGTH OF SERVICE CONDUCTORS
SIZE & NUMBER OF SERVICE CONDUCTORS PER PHASE
TYPE OF CONDUCTORS: COPPER ■■ ALUMINUM ■■ CONDUIT SIZE STEEL ■■ NON-MAGNETIC ■■
TYPE, SIZE, AND INTERRUPTING RATING OF OVERCURRENT DEVICES IN SERVICE DISCONNECT
(MAIN DISTRIBUTION PANEL)
SIZE OF GROUNDING ELECTRODE CONDUCTOR BRACING OF SERVICE EQUIPMENT
(page 1 of 2)
Electrical Plan Review
“Log In” Data Required Form
7
1
2
3
4
5
6
7
8
9
10
11
12
LocationOfShort-Circuit Current
AT TRANSFORMERSECONDARY TERMINALS(INFINITE PRIMARY)
ON LINE SIDE OF MAINSERVICE EQUIPMENT
AT PANEL
AT PANEL
AT PANEL
AT PANEL
AT PANEL
AT PANEL
AT PANEL
AT PANEL
AT PANEL
AT PANEL
AmpereRating
InterruptingRating
Apparent RMSLet-ThroughCurrent
PeakLet-ThroughCurrent
Short-CircuitCurrent
Overcurrent Device
Use back of form or attach separate sheet for data on additional panels.
Use back of form or attach separate sheet to show one-line diagram of service, feeders, and all related panels.
Attach let-through curves if current-limiting devices are used.
All current values in RMS line-to-line unless otherwise noted.
The undersigned accepts full responsibility for the values given herein.
SIGNED DATE
PHONE WHERE YOU CAN BE REACHED
Page 2 of 2
ITEM
Electrical Plan Review
Point-To-Point Method Of Short-Circuit Calculation
8
Adequate interrupting rating and protection of electrical componentsare two essential aspects required by the National Electrical Code inSections 110-9, 110-10, 240-1, 250-2(d), 250-90, 250-96(a) and Table250-122 Note. The first step to ensure that system protective deviceshave the proper interrupting rating and provide component protectionis to determine the available short-circuit currents. The application ofthe Point-To-Point method permits the determination of available short-circuit currents with a reasonable degree of accuracy at various pointsfor either 3o or 1o electrical distribution systems. This methodassumes unlimited primary short-circuit current (infinite bus).
Calculation Of Short-Circuit Currents —Point-To-Point Method.
Example Of Short-Circuit Calculation
Basic Short-Circuit Calculation Procedure.Procedure
Step 1 Determine transf.full-load amperesfrom either:a) Name plateb) Tables 3A & 3Bc) Formula
Step 2 Find transf.multiplier
Step 3 Determine transf.let-through short-circuit current(Table 5 orformula).
Step 4 Calculate“f” factor.
Step 5 Calculate “M”(multiplier) or takefrom Table 4.
Step 6 Compute theavailable short-circuit current(symmetrical) atthe fault.
3o transf. IFLA =
†ISCA = Transf.FLA x multiplier
ISCA = 46,273 x .742 = 34,343A
(Use ISCA @ Fault #1 to calculate)
** ISCA = 833 X 55.55 = 46,273
KVA x 1000EL-L x 1.73
KVA x 1000EL-L x 1.73
IFLA = = = 833AKVA x 1000 300 x 1000
100
3o faults f = 1.73 x L x lL-L-L
C x EL-L
f = 1.73 x L x lL-L-L
C x EL-L
1.73 x 20 x 46,27322,185 x 208
f = 1.73 x 20 x 34,3435,906 x 208
2 x L x lL-L
C x EL-L
100Transf. % Z
1o transf. IFLA =
— Multiplier =
100††.9x Transf. % Z
Multiplier =KVA x 1000EL-L
f =
1
ISCA = ISCA x M
1 + fM =
11 + f
11 + .347M =
2 x L x lL-N*C x EL-N
f =
1o line-to-line(L-L) faults on1o, center-tappedtransformers
1o line-to-neutral(L-N) faults on1o, center-tappedtransformers
L = length (feet) of conduit to the fault.C = constant from Tables 1, 2. For
parallel runs, multiply C values by the number of conductors per phase.
I = available short-circuit current in amperes at beginning of circuit.
Formula FAULT #1
atfault
atbeginning of circuit.
Note 1. Motor short-circuit contribution, if significant, may be added to the trans-former secondary short-circuit current value as determined in Step 3. Proceed withthis adjusted figure through Steps 4, 5, and 6. A practical estimate of motor short-cir-cuit contribution is to multiply the total load current in amperes by 4.Note 2. The L-N fault current is higher than the L-L fault current at the secondary ter-minals of a single-phase center-tapped transformer. 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 = 1.5 x IL-L at Transformer Terminals
At some distance from the terminals, depending upon wire size, the L-N fault current islower than the L-L fault current. The 1.5 multiplier is an approximation and will theoreti-cally vary from 1.33 to 1.67. These figures are based on change in turns ratio betweenprimary and secondary, infinite source available, zero feet from terminals of transformer,and 1.2 x %X and 1.5 x %R for L-N vs. L-L resistance and reactance values. BeginL-N calculations at transformer secondary terminals, then proceed point-to-point.
†
*
300 KVA, 2%Z
20', 500 MCM CUSteel conduit 20', #2 CU
Steel conduit
InfinitePrimaryAvailable
Fault #2
Fault #1
MAIN SERVICEPANEL
BRANCH CIRCUIT PANEL
Step 1
Step 2
Step 3
Step 4
Step 5
Step 6
Step 4
Step 5
Step 6
=
=
=
= .347
= .968
= .742 (See Table 4)
11 + f
11 + .968M = = = .508 (See Table 4)
= 55.55
At Transformer Secondary
Fault #1
ISCA = 34,343 x .508 = 17,447AFault # 2
FAULT #2
**For simplicity, the motor contribution and voltage variance was not included.See Notes 1 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 ±10% fromthe actual values determined by ANSI / IEEE test. See U.L. Standard 1561.Therefore, multiply transformer %Z by .9.
Note 4. Utility voltages may vary ±10% for power, and ±5.8% for 120-volt light-ing services. Therefore, for worst case conditions, multiply values as calculatedin Step 3 by 1.1 and/or 1.058 respectively.
Note 5: Bolted fault approximations:L-L-L 100% of Step 6.L-L 87% of Step 6.L-G, L-N 25-125% of Step 6. (Use 50% as typical)
Note 6: Arcing fault approximation for sustained arcs (percentages of L-L-Lbolted fault values)
L-L-L
L-L
L-G
L-L-L with one primary open
480 Volts
89%
74%
38%
80%
208 Volts
12%
2%
—
—
Voltage(Line-to- Transformer KVA RatingLine) 150 167 225 300 500 750 1000 1500 2000208 417 464 625 834 1388 2080 2776 4164 5552220 394 439 592 788 1315 1970 2630 3940 5260240 362 402 542 722 1203 1804 2406 3609 4812440 197 219 296 394 657 985 1315 1970 2630460 189 209 284 378 630 945 1260 1890 2520480 181 201 271 361 601 902 1203 1804 2406600 144 161 216 289 481 722 962 1444 1924
Electrical Plan Review
Point-To-Point Method Of Short-Circuit Calculation
9
Ampacity BuswayPlug-In Feeder High ImpedanceCopper Aluminum Copper Aluminum Copper
225 28700 23000 18700 12000 — 400 38900 34700 23900 21300 — 600 41000 38300 36500 31300 — 800 46100 57500 49300 44100 — 1000 69400 89300 62900 56200 156001200 94300 97100 76900 69900 161001350 119000 104200 90100 84000 175001600 129900 120500 101000 90900 192002000 142900 135100 134200 125000 204002500 143800 156300 180500 166700 217003000 144900 175400 204100 188700 238004000 — — 277800 256400 —Note: These values are equal to one over the impedance per foot forimpedance in a survey of industry.
Voltage Transformer KVA Rating25 50 75 100 150 167 200 250 333 500
115/230 109 217 326 435 652 726 870 1087 1448 2174120/240 104 208 313 416 625 696 833 1042 1388 2083230/460 54 109 163 217 326 363 435 544 724 1087240/480 52 104 156 208 313 348 416 521 694 1042
Calculation Of Short-Circuit CurrentsAt Second Transformer In System.
Use the following procedure to calculate thelevel of fault current at the secondary of a sec-ond, downstream transformer in a systemwhen the level of fault current at the trans-former primary is known.
Step 1 Calculate “f”(IP(SCA), known).
Step 2 Calculate “M”(multiplier) or takefrom Table 4.
Step 3 Calculate short-circuitcurrent at secondaryof transformer.(See Note 1 under“Basic Procedure”)
3o transformer (IP(SCA) andI S(SCA) are 3o fault values).
1o transformer (IP(SCA) andIS(SCA) are 1o fault values;I S(SCA is L-L.)
—
—
Procedure
Procedure For Second Transformer in SystemFormula
IP(SCA) = Available fault current at transformer primary.IS(SCA) = Available fault current at transformer secondary.VP = Primary voltage L-L.VS = Secondary voltage L-L.
KVATRANS = KVA rating of transformer.%Z = Percent impedance of transformer.Note: To calculate fault level at the endof a conductor run, follow Steps 4, 5, and 6of Basic Procedure.
f=100,000 x KVATRANS.
IP(SCA) x VP x 1.73 (%Z)
IS(SCA) =VP x M x IP(SCA)
f=100,000 x KVATRANS.
IP(SCA) x VP x (%Z)
M =1 + f
1
VS
Table 3A. Three-Phase Transformer—Full-LoadCurrent Rating (In Amperes).
Table 1. “C” Values for Busway.
Table 3B. Single-Phase Transformer—Full-LoadCurrent Rating (In Amperes).
H.V. UTILITYCONNECTION
KNOWNFAULTCURRENT
KNOWNFAULTCURRENT
Table 2. “C” Values for Conductors (Note: These values are equal to one over the impedance per foot for impedances found in IEEE Std. 241-1990, IEEE Recommended Practice for Commercial Building Power Systems.)
CopperAWG Three Single Conductors Three-Conductor Cableor Conduit Conduitkcmil Steel Nonmagnetic Steel Nonmagnetic
600V 5KV 15KV 600V 5KV 15KV 600V 5KV 15KV 600V 5KV 15KV14 389 389 389 389 389 389 389 389 389 389 389 38912 617 617 617 617 617 617 617 617 617 617 617 61710 981 981 981 981 981 981 981 981 981 981 981 9818 1557 1551 1557 1558 1555 1558 1559 1557 1559 1559 1558 15596 2425 2406 2389 2430 2417 2406 2431 2424 2414 2433 2428 24204 3806 3750 3695 3825 3789 3752 3830 3811 3778 3837 3823 37983 4760 4760 4760 4802 4802 4802 4760 4790 4760 4802 4802 48022 5906 5736 5574 6044 5926 5809 5989 5929 5827 6087 6022 59571 7292 7029 6758 7493 7306 7108 7454 7364 7188 7579 7507 73641/0 8924 8543 7973 9317 9033 8590 9209 9086 8707 9472 9372 90522/0 10755 10061 9389 11423 10877 10318 11244 11045 10500 11703 11528 110523/0 12843 11804 11021 13923 13048 12360 13656 13333 12613 14410 14118 134614/0 15082 13605 12542 16673 15351 14347 16391 15890 14813 17482 17019 16012250 16483 14924 13643 18593 17120 15865 18310 17850 16465 19779 19352 18001300 18176 16292 14768 20867 18975 17408 20617 20051 18318 22524 21938 20163350 19703 17385 15678 22736 20526 18672 22646 21914 19821 24904 24126 21982400 20565 18235 16365 24296 21786 19731 24253 23371 21042 26915 26044 23517500 22185 19172 17492 26706 23277 21329 26980 25449 23125 30028 28712 25916600 22965 20567 17962 28033 25203 22097 28752 27974 24896 32236 31258 27766750 24136 21386 18888 28303 25430 22690 31050 30024 26932 32404 31338 283031000 25278 22539 19923 31490 28083 24887 33864 32688 29320 37197 35748 31959
Electrical Plan Review
Point-To-Point Method Of Short-Circuit Calculation
AluminumAWG Three Single Conductors Three-Conductor Cableor Conduit Conduitkcmil Steel Nonmagnetic Steel Nonmagnetic
600V 5KV 15KV 600V 5KV 15KV 600V 5KV 15KV 600V 5KV 15KV14 236 236 236 236 236 236 236 236 236 236 236 23612 375 375 375 375 375 375 375 375 375 375 375 37510 598 598 598 598 598 598 598 598 598 598 598 5988 951 950 951 951 950 951 951 951 951 951 951 9516 1480 1476 1472 1481 1478 1476 1481 1480 1478 1482 1481 14794 2345 2332 2319 2350 2341 2333 2351 2347 2339 2353 2349 23443 2948 2948 2948 2958 2958 2958 2948 2956 2948 2958 2958 29582 3713 3669 3626 3729 3701 3672 3733 3719 3693 3739 3724 37091 4645 4574 4497 4678 4631 4580 4686 4663 4617 4699 4681 46461/0 5777 5669 5493 5838 5766 5645 5852 5820 5717 5875 5851 57712/0 7186 6968 6733 7301 7152 6986 7327 7271 7109 7372 7328 72013/0 8826 8466 8163 9110 8851 8627 9077 8980 8750 9242 9164 89774/0 10740 10167 9700 11174 10749 10386 11184 11021 10642 11408 11277 10968250 12122 11460 10848 12862 12343 11847 12796 12636 12115 13236 13105 12661300 13909 13009 12192 14922 14182 13491 14916 14698 13973 15494 15299 14658350 15484 14280 13288 16812 15857 14954 15413 16490 15540 17635 17351 16500400 16670 15355 14188 18505 17321 16233 18461 18063 16921 19587 19243 18154500 18755 16827 15657 21390 19503 18314 21394 20606 19314 22987 22381 20978600 20093 18427 16484 23451 21718 19635 23633 23195 21348 25750 25243 23294750 21766 19685 17686 25976 23701 20934 26431 25789 23750 29036 28262 259761000 23477 21235 19005 28778 26109 23482 29864 29049 26608 32938 31919 29135
Table 5. Short-Circuit Currents Available from Various Size TransformersVoltage KVA Full % ShortAnd Load Impedance†† CircuitPhase Amps (Nameplate) Amps†
25 104 1.58 11,574371/2 156 1.56 17,351
120/240 50 209 1.54 23,1221 ph.* 75 313 1.6 32,637
100 417 1.6 42,478167 695 1.8 60,25525 69 1.6 4,79150 139 1.6 9,65275 208 1.11 20,821100 278 1.11 27,828150 416 1.07 43,198
120/208 225 625 1.12 62,0043 ph.** 300 833 1.11 83,383
500 1388 1.24 124,373750 2082 3.5 66,0951000 2776 3.5 88,1671500 4164 3.5 132,1902000 5552 5.0 123,3772500 6950 5.0 154,4441121/2 135 1.0 15,000150 181 1.2 16,759225 271 1.2 25,082300 361 1.2 33,426
277/480 500 601 1.3 51,3623 ph.** 750 902 3.5 28,410
1000 1203 3.5 38,1801500 1804 3.5 57,2612000 2406 5.0 53,4612500 3007 5.0 66,822
Single phase values are L-N values at transformer terminals. These figures arebased on change in turns ratio between primary and secondary, 100,000 KVAprimary, zero feet from terminals of transformer, 1.2 (%X) and 1.5 (%R)multipliers for L-N vs. L-L reactance and resistance values and transformerX/R ratio = 3.
Three-phase short-circuit currents based on “infinite” primary.
U.L. listed transformers 25 KVA or greater have a ±10% impedance toler-ance. Short-circuit amps reflect a “worst case” condition.
Fluctuations in system voltage will affect the available short-circuit current.For example, a 10% increase in system voltage will result in a 10% increasein the available short-circuit currents shown in the table.
**
††
†
*
*
f M f M0.01 0.99 1.50 0.400.02 0.98 1.75 0.360.03 0.97 2.00 0.330.04 0.96 2.50 0.290.05 0.95 3.00 0.250.06 0.94 3.50 0.220.07 0.93 4.00 0.200.08 0.93 5.00 0.170.09 0.92 6.00 0.140.10 0.91 7.00 0.130.15 0.87 8.00 0.110.20 0.83 9.00 0.100.25 0.80 10.00 0.090.30 0.77 15.00 0.060.35 0.74 20.00 0.050.40 0.71 30.00 0.030.50 0.67 40.00 0.020.60 0.63 50.00 0.020.70 0.59 60.00 0.020.80 0.55 70.00 0.010.90 0.53 80.00 0.011.00 0.50 90.00 0.011.20 0.45 100.00 0.01
Table 4. “M” (Multiplier).*
10
11 + f
M =
Electrical Plan Review
Peak Let-Through ChartsPeak let-through charts let you determine both the peak let-through current and the apparent prospective RMSsymmetrical let-through current. (These charts are commonly referred to as Current-Limitation Curves.)
11
400000
90008000
7000
70006000
6000
300000
200000
100000
5000
4000
4000
90000
3000
2000
1000
3000
7000
0
800007000060000
50000
6000
0
2000
00
2000
0
2000
4000
0
1000
00
5000
0
1000
0
5000
1000
3000
0
9000
080
000
40000
9000
8000
30000
20000
10000
3000
00
800000
500000
600000700000
9000001000000
B
A
INS
TAN
TAN
EO
US
PE
AK
LE
T T
HR
U C
UR
RE
NT
IN A
MP
ER
ES
PROSPECTIVE SHORT CIRCUIT CURRENT - SYMMETRICAL RMS AMPERES
601A
800A
1200A1600A2000A2500A
3000A4000A
6000A5000A
AM
PE
RE
R
ATI
NG
B
A
INS
TA
NTA
NE
OU
S P
EA
K L
ET-T
HR
OU
GH
CU
RR
EN
T I
N A
MP
ER
ES
15A
30A
60A
100A
200A
400A
600A
90008000
70
00
70006000
60
00
100000
5000
4000
40
00
90000
3000
2000
1000
30
00
70
00
0
800007000060000
50000
60
00
0
20
00
00
20
00
0
20
00
40
00
0
10
00
00
50
00
0
10
00
0
50
00
10
00
30
00
0
90
00
08
00
00
40000
90
00
80
00
30000
20000
100003
00
00
0
900800700600
500
400
300
200
100
10
0
40
0
20
0
30
0
70
0
50
0
60
0
90
08
00
PROSPECTIVE SHORT-CIRCUIT CURRENT - SYMMETRICAL RMS AMPERES
AM
PE
RE
R
AT
ING
LOW-PEAK® KRP-C Fuses LOW-PEAK® LPJ Fuses
400,000
100,000
1,000
INS
TAN
TAN
EO
US
PE
AK
LE
T-TH
RO
UG
H C
UR
RE
NT
IN A
MP
ER
ES
200,
000
RMS SYMMETRICAL CURRENTS IN AMPERES
100,
000
10,0
00
300,
000
10,000
1,00
0
A-B=ASYMMETRICAL AVAILABLE PEAK (2.3 x SYMM RMS AMPS)
B
A
600A
400A
200A
100A60A
30A
AM
PE
RE
RA
TIN
G
Electrical Plan Review
Peak Let-Through Charts
12
400000
90008000
7000
70006000
6000
300000
200000
100000
5000
4000
4000
90000
3000
2000
1000
3000
7000
0
800007000060000
B
A
50000
6000
0
PE
AK
CU
RR
EN
T IN
AM
PE
RE
S
2000
00
2000
0
2000
AVAILABLE RMS CURRENT IN AMPERES
4000
0
1000
00
5000
0
1000
0
5000
1000
3000
0
9000
080
000
40000
9000
8000
30000
20000
10000
600A
400A
200A
100A
30A
60A
AM
PE
RE
R
ATI
NG
400000
90008000
7000
70006000
6000
300000
200000
100000
5000
4000
4000
90000
3000
2000
1000
3000
7000
0
800007000060000
B
A
50000
6000
0
PE
AK
CU
RR
EN
T IN
AM
PE
RE
S
AVAILABLE RMS CURRENT IN AMPERES
2000
00
2000
0
2000
4000
0
1000
00
5000
0
1000
0
5000
1000
3000
0
9000
080
000
40000
9000
8000
30000
20000
10000
600A
400A
200A
100A
30A
60A
AM
PE
RE
R
ATI
NG
FUSETRON® FRS-R Fuses FUSETRON® FRN-R Fuses
LOW-PEAK® LPN-RK Fuses
60
30
A
B
600
400
200
AM
PE
RE
R
ATI
NG
400,000
100,000
10,000
1,000
1,00
0
10,0
00
100,
000
300,
000
100
RMS SYMMETRICAL CURRENTS IN AMPERES A-B=ASYMMETRICAL AVAILABLE PEAK (2.3 x SYMM RMS AMPS)
INS
TAN
TAN
EO
US
PE
AK
LE
T-TH
RO
UG
H C
UR
RE
NT
IN A
MP
ER
ES
LOW-PEAK® LPS-RK Fuses
Electrical Plan Review Work Sheet
Short-Circuit Calculations (Transformer)Short-Circuit at Secondary Terminals of Transformer (Infinite Primary)
13
A) FROM TABLES AmperesB) CALCULATED
Find Transformer Full-Load Amperes
(3 Phase)
Find Multiplier:
(1 Phase)
KVA x 1000 x 1000EL-L x 1.73 x 1.73
Amperes
STEP 1
==IFLA
STEP 2
Find Short-Circuit Current:STEP 3
100 100TRANS %Z
Amperes
Amperes
= =
=IFLAKVA x 1000 x 1000
EL-L
=
=
=
x= =
M
Short-Circuit Amperes
=
ISCA Transformer F.L.A. x M
A) FROM TABLES AmperesB) CALCULATED
Find Transformer Full-Load Amperes
(3 Phase)
Find Multiplier:
(1 Phase)
KVA x 1000 x 1000EL-L x 1.73 x 1.73
Amperes
STEP 1
==IFLA
STEP 2
Find Short-Circuit Current:STEP 3
100 100TRANS %Z
Amperes
Amperes
= =
=IFLAKVA x 1000 x 1000
EL-L
=
=
=
x= =
M
Short-Circuit Amperes
=
ISCA Transformer F.L.A. x M
Short-Circuit at Secondary Terminals of Transformer (Infinite Primary)
=
=
Electrical Plan Review Work Sheet
Short-Circuit Calculations (Three o)
14
x = Amperes
A)
B)
C)
D)
1.73 x L x IL-L-L
C x EL-L
=f
11 + f
11 +
=Multiplier “M”
SCA = M x lL-L-L =
1.73 x xx
=
= =
=f
x = Amperes
A)
B)
C)
D)
Short-Circuit Current at
Short-Circuit Current at
1.73 x L x IL-L-L
C x EL-L
=f
11 + f
11 +
=Multiplier “M”
SCA = M x lL-L-L =
1.73 x xx
=
= =
=f
x = Amperes
A)
B)
C)
D)
Short-Circuit Current at
1.73 x L x IL-L-L
C x EL-L
=f
11 + f
11 +
=Multiplier “M”
SCA = M x lL-L-L =
1.73 x xx
=
= =
=f
15
Electrical Plan Review Work Sheet
Short-Circuit Calculations (Three o)
x = Amperes
A)
B)
C)
D)
1.73 x L x IL-L-L
C x EL-L
=f
11 + f
11 +
=Multiplier “M”
SCA = M x lL-L-L =
1.73 x xx
=
= =
=f
x = Amperes
A)
B)
C)
D)
Short-Circuit Current at
Short-Circuit Current at
1.73 x L x IL-L-L
C x EL-L
=f
11 + f
11 +
=Multiplier “M”
SCA = M x lL-L-L =
1.73 x xx
=
= =
=f
Short-Circuit Current at
x = Amperes
A)
B)
C)
D)
1.73 x L x IL-L-L
C x EL-L
=f
11 + f
11 +
=Multiplier “M”
SCA = M x lL-L-L =
1.73 x xx
=
= =
=f
Electrical Plan Review Work Sheet
Short-Circuit Calculations (Single o, Line-To-Line)
16
x = Amperes
A)
B)
C)
D)
2 x L x lL-L
C x EL-L
=f
11 + f
11 +
=Multiplier “M”
SCA = M x lL-L =
2 x xx
=
= =
=f
Short-Circuit Current at
x = Amperes
A)
B)
C)
D)
2 x L x lL-L
C x EL-L
=f
11 + f
11 +
=Multiplier “M”
SCA = M x lL-L =
2 x xx
=
= =
=f
Short-Circuit Current at
x = Amperes
A)
B)
C)
D)
2 x L x lL-L
C x EL-L
=f
11 + f
11 +
=Multiplier “M”
SCA = M x lL-L =
2 x xx
=
= =
=f
Short-Circuit Current at
17
Electrical Plan Review Work Sheet
Short-Circuit Calculations (Single o, Line-To-Line)
x = Amperes
A)
B)
C)
D)
2 x L x lL-L
C x EL-L
=f
11 + f
11 +
=Multiplier “M”
SCA = M x lL-L =
2 x xx
=
= =
=f
Short-Circuit Current at
x = Amperes
A)
B)
C)
D)
2 x L x lL-L
C x EL-L
=f
11 + f
11 +
=Multiplier “M”
SCA = M x lL-L =
2 x xx
=
= =
=f
Short-Circuit Current at
x = Amperes
A)
B)
C)
D)
2 x L x lL-L
C x EL-L
=f
11 + f
11 +
=Multiplier “M”
SCA = M x lL-L =
2 x xx
=
= =
=f
Short-Circuit Current at
18
Electrical Plan Review Work Sheet
Short-Circuit Calculations (Single o, Line-To-Neutral)
x = Amperes
A)
B)
C)
D)
C x EL-N
=f
11 + f
11 +
=Multiplier “M”
SCA = M x lL-N =
2 x xx
=
= =
=f
Short-Circuit Current at
x = Amperes
A)
B)
C)
D)
2 x L x lL-N*C x EL-N
=f
11 + f
11 +
=Multiplier “M”
SCA = M x lL-N =
2 x xx
=
= =
=f
Short-Circuit Current at
x = Amperes
A)
B)
C)
D)
2 x L x lL-N*C x EL-N
=f
11 + f
11 +
=Multiplier “M”
SCA = M x lL-N =
2 x xx
=
= =
=f
Short-Circuit Current at
2 x L x lL-N*
*See Note 2 on Page 8
19
x = Amperes
A)
B)
C)
D)
2 x L x lL-N*C x EL-N
=f
11 + f
11 +
=Multiplier “M”
SCA = M x lL-N =
2 x xx
=
= =
=f
Short-Circuit Current at
x = Amperes
A)
B)
C)
D)
2 x L x lL-N*C x EL-N
=f
11 + f
11 +
=Multiplier “M”
SCA = M x lL-N =
2 x xx
=
= =
=f
Short-Circuit Current at
x = Amperes
A)
B)
C)
D)
2 x L x lL-N*C x EL-N
=f
11 + f
11 +
=Multiplier “M”
SCA = M x lL-N =
2 x xx
=
= =
=f
Short-Circuit Current at
Electrical Plan Review Work Sheet
Short-Circuit Calculations (Single o, Line-To-Neutral)
*See Note 2 on Page 8
Electrical Plan Review
Grounding & Bonding of Service Equipment
20
EquipmentGroundingConductorMaterial: Section 250-118Install: Section 250-120
IMPORTANT:Effective Bonding andGrounding Required:
Section 250-2(d)Section 250-90Section 250-96(a)
Must have capacityto conduct safelyany fault currentlikely to beimposed on it.
MAIN DISTRIBUTION PANEL
SupplementalGround(If Required)Section 250-50(a)
GroundingElectrodeSystemSection 250-50
BondingWhy? Section 250-90What? Section 250-92How? Section 250-94Material: Section 250-102(a)Size: Section 250-102(c)
METERBASE
NEUTRAL
800/800
Neutral Grounded ConductorSize: Sections 230-42, 220-22When Serving As GroundingConductor:What? Section 250-24(a)Size: Section 250-24(b)
Section 250-66
Neutral Disconnecting MeansSection 230-96
Main Bonding Jumper: Section 250-28, 250-96Material: Section 250-28(a)Size: Section 250-28(d)Connect: Section 250-28(c)
Grounding Electrode ConductorSize: Section 250-66Material: Section 250-62Install: Section 250-64Enclosure: Section 250-64(e)What: Section 250-24
Connection to ElectrodeSections 250-68, 8, 70
Bonding of Metal Water PipesSection 250-104
Grounded Neutral ServiceEntrance Conductors to PadMount Transformer
Electrical Plan Review
Typical Component—Short Circuit Current Ratings
21
Component Short-circuit current rating, kA
Clock-Operated Switch 5,000
HVAC Equipment
Single-Phase-Amps110-120V 200-208V 220-240V 254-277V9.8 or less 5.4 or less 4.9 or less — 2009.9-16.0 5.5-8.8 5.0-8.0 6.65 or less 1,00016.1-34.0 8.9-18.6 8.1-17.0 — 2,00034.1-80.0 18.7-44.0 17.1-40.0 — 3,500Over 80.0 Over 44.0 Over 40.0 Over 6.65 5,000
3-Phase-Amps200-208V 220-240V 440-480V 550-600V2.12 or less 2.0 or less — — 2002.13-3.7 2.1-3.5 1.8 or less 1.4 or less 1,0003.8-9.5 3.6-9.0 — — 2,0009.6-23.3 9.1-22.0 — — 3,500Over 23.3 Over 22.0 Over 1.8 Over 1.4 5,000
Meter Socket Base 10,000Motor Controller, Rated in Horsepower (kW)
a. 0-50 (0-37.3) 5,000b. 51-200 (38-149) 10,000c. 201-400 (150-298) 18,000d. 401-600 (299-447) 30,000e. 601-900 (448-671) 42,000f. 901-1600 (672-1193) 85,000
Photoelectric Switches 5,000Receptacle (GFCI Type) 10,000Receptacle (Other Than GFCI Type) 2,000Snap Switch 5,000Terminal Block 10,000Thermostat 5,000
Wire
Copper, 75° Thermoplastic Insulated Cable
Maximum Short-Circuit Withstand Current in Amperes
For For For For1/2 Cycle 1 Cycle 2 Cycles 3 Cycles
#14 2,400 1,700 1,200 1,000#12 3,800 2,700 1,900 1,550#10 6,020 4,300 3,000 2,450#8 9,600 6,800 4,800 3,900#6 15,200 10,800 7,600 6,200#4 24,200 17,100 12,100 9,900
CopperWire Size75° Thermoplastic
Electrical Plan Review
Work Sheet Problem—Main Distribution Panel
22
PRIMARY FUSE
Ground Buss
300 KVA Transformer by Utility120/208 Volt3 Phase, 4-Wire2% Impedance
(2) 3" C. each with 4-500 kcmils/XHHW - 20 Feet
4 - #8 THHN, 3/4"C. - 10 Feet
METER
MAIN SWITCH
800/800
200/200
200/150
100/100
100/90
100/70
100/
200/
4 - #3/0 THHN, 2" C. - 60 Feet
4 - #1/0 THHN, 1-1/2" C. - 15 Feet
4 - #3 THHN, 1-1/4" C. - 20 Feet
3 - #3 THHN, 1" C. - 35 Feet
3 - #4 THHN, 1" C. - 35 Feet
LPA
LPC
LPB
AC-1
AC-2
3
1
2
4
5
6
7
8
9
EMP
FLU
OR
.
FIXT
UR
E
3-#
12 T
HH
N
3-#
8 T
HH
N, 3
/4"
C.-
4 F
eet
Combination Motor
Controller
1/2"
C.-
30'
7-1/2
1
2
3
4
5
6
7
8
9
10