Brain Filler Design Guide

Brainfiller®Guide for: Streamlined Electrical Design Based on the 2005 NEC® Jim Phillips, P.E. A Brainfiller® Publication www.brainfiller.com STREAMLINED ELECTRICAL DESIGN Edition 1.12 Jim Phillips, P.E. t2g technical training group Copyright © 2007 Technical Training Group All Rights Reserved ® brainfiller.com® ELECTRICAL DESIGN GUIDE DISCLAIMER This training notebook was developed by Technical Training Group / Jim Phillips, P.E., Canton, Ohio. Technical Training Group, has attempted to ensure that the information contained in this notebook is as accurate as possible. Information contained in this notebook is subject to change without notice and neither Technical Training Group nor Jim Phillips, P.E., assumes any responsibility for any errors, omissions or damages resulting from the use of the information contained within. This guide is only a compilation of data and information obtained from various sources and in no way constitutes an attempt by Technical Training Group or Jim Phillips, P.E. to render engineering or other professional services. If such services are required, a licensed professional engineer should be consulted. This information must be verified by the design professional in responsible charge of the design of the project before using it. The information contained in this training notebook is protected under United States Copyright laws. Much of the information is gathered from a variety of sources such as the National Electrical Code®. It must be noted that the tables apply to “typical” systems and may not account for issues such as but not limited to: elevated ambient temperatures, tap rules, special equipment requirements, and other exceptions or requirements that may be necessary. Permission is granted to reproduce this work provided it is not for sale and that the work is reproduced in its entirety with all pages including the cover pages, disclaimer, header and footers acknowledging Technical Training Group and Jim Phillips, P.E. Where trademarks or trade names are referenced, they are listed in the appendix in the back of this manual. The end user, by accepting this document or copy thereof, agrees to the conditions listed on this page. Did you receive a copy of this guide from someone else? Get your own registered copy free at: www.brainfller.com Jim Phillips, P.E. T 2G Technical Training Group® Page 3 of 37 3 Infinite Bus Short Circuit Calculations 4.5 Overcurrent Protection 2..0 6.6 Transformer Conductors 3.………………………………………………….4 Equipment Grounding Conductors 2.6 2.5 Calculation Worksheets Calculation of Conduit Fill.0 Introduction ……………………………………………………………….3(B) 3.7 Motor Schedules Coordination Analysis…………………….….3 Phase Conductors 2.11 3.1 Time Current Curves 7.….brainfiller. P.4 Magnetizing Inrush Current 3.com® ELECTRICAL DESIGN GUIDE Table of Contents 1.2 Motor Full Load Current Ratings 6.1 Motor Circuits 6.0 2.25 Motor Schedules………………………………………………………….…………………….2 Transformer Protection 3.…………..E.19 4.7 Conductor Schedules Transformer Circuits .……….2 Selective Coordination 3.2 Transformer Impedance 4.…….2 Load Current 2.5 Conductor Selection…………………………………………………….1 Transformer Size 3..1 NEC® Article 110.0 References and Trademarks Jim Phillips.1 Conductor Selection Criteria 2.9 Requirements 4.0 7..3 Motor Feeder Sizing 6.……………………………33 7.5 Motor Overload Protection 6.0 5.7 Transformer Schedules Infinite Bus Short Circuit Calculations…………………….0 4. T 2G Technical Training Group® Page 4 of 37 ..4 Motor Contribution 4.4 Motor Short Circuit Protection 6.6 Motor Time Current Curves 6.5 ANSI Thru Fault Protection 3.27 6..6 Additional Considerations 2.3 NEC® Table 450. T 2G Technical Training Group® Page 5 of 37 . voltage drop. There are many exceptions and other considerations to the information provided in this guide such as “tap rules” as allowed by the NEC®.L. harmonics. This guide provides many of the “typical” configurations of the building blocks in a series of tables that can be used in streamlining the electrical design process. Each of these “building blocks” generally has components such as cable and conduit selected based on a load current requirement. The experienced designer recognizes that a substantial portion of a design’s building blocks are always the same and can be standardized to some extent. and transformers. panels. The electrical power distribution system is actually made up of many “typical building blocks” such as feeders. In addition to using streamlined electrical design tables. P. As an example. Numerous codes and standards such as the National Electrical Code®.com® ELECTRICAL DESIGN GUIDE 1. U. Many of the programs also integrate a vast array of power system studies such as short circuit. This guide is in no way a substitute for having the design performed by a qualified electrical design professional.E. 40 HP motor feeder. dictate many of the design requirements. there are many commercially available computer programs available on the market that can assist in further accelerating the design process. The electrical design must always be performed by or under the direction of a licensed professional. ANSI Standards.brainfiller. motor circuits. conductor de-rating just to name a few. and others. Jim Phillips. coordination and arc flash into their packages allowing the use of the design data base for the study data base. The designer must be sure that all of the special considerations have been addressed and that the design follows all appropriate codes and standards.0 INTRODUCTION Designing an electrical power system can be quite a complex task. IEEE. there are only so many ways to size an 800 Amp circuit just like there are only so many ways to size a 460 Volt. conductors are only allowed to be continuously loaded up to 80% of their ampacity. The NEC® requires that circuits be designed to handle 125% of the continuous load and 100% of the noncontinuous load.E. The selection process usually begins with defining a load and then selecting a conductor with sufficient ampacity for that load. For devices with terminal ratings other than 75ºC other considerations need to be made. sizing a feeder for an 80 Amp continuous load would require the circuit be sized at least 125% of 80 Amps or 100 Amps. P.16 and are based on the 75ºC rating.0 CONDUCTOR SELECTION 2. A load that is expected to continue for three or more hours is defined as a continuous load. Conversely.2 Load Current The NEC® considers electrical loads to be either continuous or non-continuous.14(C) requires that the a conductor can not be loaded to a current that would produce a temperature greater than the device terminal rating which is typically 75ºC.brainfiller.3 Phase Conductors The phase conductors listed in the tables are based on ampacities listed in NEC® Table 310. However the actual selection process also involves: • • • • • • • Load requirements Continuous / non-continuous loading Insulation rating Ambient temperature Number of current carrying conductors in a raceway Temperature rating of device terminals Harmonics. Jim Phillips. T 2G Technical Training Group® Page 6 of 37 . As an example. Therefore the 100 Amp circuit can only be loaded to 80% X 100 Amps or 80 Amps continuously.com® ELECTRICAL DESIGN GUIDE 2. 2.1 Conductor Selection Criteria Selecting the correct conductor requires considering many variables. voltage drop and many other factors 2. a non-continuous load is expected to be operating less than three hours. Article 110. Looking at this in terms of the maximum circuit loading. This assumes the conductor is protected at it’s ampacity however the NEC® requires that the size of the equipment ground conductor be increased if the size of the phase conductor is increased such as in the case of solving a voltage drop problem. In this case. The NEC provides many exceptions to this simple concept. Many solid state circuit breakers have an adjustable long time pick up setting that allows a 1200 Amp breaker to be set at 1140 Amps. In this case.brainfiller. The basic concept of this protection article is that if you have a 50 Amp conductor. the NEC allows the use of the next higher standard overcurrent device rating which is 400 Amps. care must be given so the load does not exceed 80 percent of the conductors ampacity for continuous loads or 100 percent for non-continuous ratings. There are two options in this case. the maximum allowable continuous loading is 304 Amps (80% x 380 Amps) rather than 320 Amps (80% x 400 Amps). Option number two is to use an overcurrent device rated 1140 Amps or less. Where the next larger standard device is selected. The ampacity based on the 75 ºC rating of a 500 kcm conductor is 380 Amps but since a 380 Amp overcurrent device is not standard. Where the ampacity of the feeder conductor does not match a standard overcurrent device rating.122 and the overcurrent device rating listed in the table.com® 2. 2. Another example would be the use of 3 sets of 500 kcm conductor.4 which requires that conductors other than flexible cords.5 Overcurrent Protection The rating of the overcurrent protection listed in the table is based on NEC Article 240. P. Jim Phillips. An example would be the common use of 500 kcm copper conductors for a 400 Amp circuit. Option number one is to increase the circuit ampacity to meet or exceed 1200 Amps. you protect it with a 50 Amp overcurrent device.E. The ampacity of the total of 3 sets would be 3 X 380 Amps or 1140 Amps total. flexible cables.4 Equipment Grounding Conductors ELECTRICAL DESIGN GUIDE The equipment ground wire selection is based on NEC® table 250. since the ampacity is above the 800 Amp limit for using the next size standard overcurrent device a 1200 Amp device is not allowed. the next higher device rating is selected as long as it is no greater than 800 Amps in accordance with NEC® requirements. T 2G Technical Training Group® Page 7 of 37 . and fixture wires be protected at their ampacity unless one of the numerous exceptions apply such as the “tap rules”. over sizing the neutral due to harmonics. This requires that both the conductors and overcurrent protection (80% rated) would be subject to no more than 80% continuous load.19 as Adjusted for Ambient Temperature if Necessary 80 70 50 2. The ampacity is also based on the conductors being applied within the rated ambient temperature of 30ºC.1 Adjustment for More than 3 Current Carrying Conductors in a Raceway or Cable Excerpt from NEC® Table 310.16 through 310. The table in this section provides a list of phase and ground conductor sizes. P.6 Additional Considerations ELECTRICAL DESIGN GUIDE The conductor ampacity is also based on no more than three current carrying conductors per raceway. Jim Phillips. Table 2. When there are more than three current carrying conductors. Such exceptions might include derating due to harmonics. the ampacities must be derated based on Table 2. T 2G Technical Training Group® Page 8 of 37 . Additional derating must be applied to the ampacity for conductors used in higher ambient applications.15(B)(2)(a) Adjustment Factors for More Than Three Current-Carrying Conductors in a Raceway or Cable Number of Current-Carrying Conductors 4-6 7-9 10-20 Percent of Values in Tables 310.com® 2.7 Conductor Schedules The tables are based on “Typical” feeders and must be used with care.E. Both 3-wire circuits and 4-wire circuits with an individual ground conductor are included. There are numerous exceptions in the NEC that are not considered in the tables and should be addressed on a case by case basis. and overcurrent protection requirements for “typical” circuits.brainfiller. As an example. The phase conductor selections listed in the conductor schedules are based on serving continuous loads.1. a 100 Amp circuit would be loaded to no more than 80 Amps (80% x 100). tap rules as well as other design factors. 400 kcm 3 .brainfiller. Circuit ampacity is based on copper conductors.600 kcm 3 .500 kcm Conductor Ampacity 35 A 65 A 115 A 130 A 150 A 175 A 200 A 230 A 255 A 310 A 380 A 420 A 510 A 620 A 760 A 820 A 1005 A 1240 A 1680 A 1675 A 1900 A 2010 A 2280 A Equipment Ground 10 10 8 6 6 6 6 4 4 4 2 2 2 1 1/0 1/0 2/0 3/0 4/0 4/0 4/0 250 kcm 250 kcm ® Circuit ampacity is based on listed conductor ampacities from NEC Table 310. No.500 kcm 3 .#2 3 . 4.4/0 3 .250 kcm 3 . No more than 3 current carrying conductors in each conduit with minimal non-linear loads. P. 2. 6.2/0 3 .600 kcm 3 .600 kcm 3 . T 2G Technical Training Group® Page 9 of 37 .E.350 kcm 3 .3/0 3 . The table represents common circuit designs but does not represent all possible combinations.#10 3 .350 kcm 3 .500 kcm 3 . Sets 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 3 4 4 5 5 6 6 Phase Conductor 3 .400 kcm 3 .350 kcm 3 .1/0 3 .250 kcm 3 . 5.500 kcm 3 .2 Conductor Schedule . Conductors are assumed to be operating at rated ambient temperature with no correction applied.#6 3 . Device terminals must be suitable for 75ºC.16 based on 75ºC. Jim Phillips.400 kcm 3 . Other NEC® rules and exceptions still may apply.#1 3 .com® ELECTRICAL DESIGN GUIDE Table 2. 3.3 Phase 3 Wire Circuits Overcurrent Device Rating 30 60 100 125 150 175 200 225 250 300 400 400 500 600 800 800 1000 1200 1600 1600 1600 2000 2000 1. Device terminals must be suitable for 75ºC. Circuit ampacity is based on copper conductors. Sets 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 3 4 4 5 5 6 6 Phase and Neutral Conductor 4 . No more than 3 current carrying conductors in each conduit with minimal non-linear loads.#6 4 .3 Conductor Schedule . 2.3 Phase 4 Wire Circuits Overcurrent Device Rating 30 60 100 125 150 175 200 225 250 300 400 400 500 600 800 800 1000 1200 1600 1600 1600 2000 2000 1. T 2G Technical Training Group® Page 10 of 37 .4/0 4 .400 kcm 4 .500 kcm 4 .2/0 4 .350 kcm 4 .brainfiller.#10 4 .E. No.16 based on 75ºC.250 kcm 4 . 4.600 kcm 4 .#2 4 . Jim Phillips.500 kcm 4 . 3.400 kcm 4 .#1 4 .3/0 4 .350 kcm 4 . The table represents common circuit designs but does not represent all possible combinations. Other NEC® rules and exceptions still may apply. 6.600 kcm 4 .250 kcm 4 . Conductors are assumed to be operating at rated ambient temperature with no correction applied.500 kcm Conductor Ampacity 35 A 65 A 115 A 130 A 150 A 175 A 200 A 230 A 255 A 310 A 380 A 420 A 510 A 620 A 760 A 820 A 1005 A 1240 A 1680 A 1675 A 1900 A 2010 A 2280 A Equipment Ground 10 10 8 6 6 6 6 4 4 4 2 2 2 1 1/0 1/0 2/0 3/0 4/0 4/0 4/0 250 kcm 250 kcm ® Circuit ampacity is based on listed conductor ampacities from NEC Table 310.600 kcm 4 .400 kcm 4 .350 kcm 4 .com® ELECTRICAL DESIGN GUIDE Table 2.1/0 4 .500 kcm 4 . P. 5. Taking the total VA and dividing by 100 to convert it to kVA is one method for sizing a transformer.0 TRANSFORMER CIRCUIT TABLES 3. Most NEC® calculations for a facility’s load are based on total VoltAmperes or VA.e.brainfiller.732 As an example if the load is 1804 Amps per phase at 480 Volts.com® ELECTRICAL DESIGN GUIDE 3. Another method is if the total load is given in Amps. T 2G Technical Training Group® Page 11 of 37 .48 kVA = Current on each phase = Square root of three i. 480 Volts = 0. 1. demand factors.1 Transformer Size There are various methods that can be used for sizing transformers depending on factors such as capacity for future load growth. the three phase kVA would be: 1804 Amps X 0.48kV x √3 ] = 1804 Amps Jim Phillips.48kV X 1. the current rating for each phase would be: FLA = 1500kVA / [ 0. if a three phase transformer is rated 1500 kVA at 480 Volts. load diversity and others.E.e.732 = 1500 kVA To calculate the transformer’s ampacity given a transformers three phase kVA rating: FLA = kVA3phase / [ kVLine-Line x √3 ] As an example. the total load current can be converted from Amps to kVA as in the example below: To calculate the total 3 phase kVA given a three phase current on each phase we use the following formula: kVA3phase = Amps3phase X kVLine-Line X √3 Where: kVA3phase kVLine-Line Amps3phase √3 = total three phase kVA of load = Voltage line-line in kV i. P. Although this may be the most common selection. The tables in this section do not reflect every scenario or consider every variable and therefore. Many factors such as load. 167% can be used.3(B) defines the maximum ratings or settings of overcurrent protection for transformers rated 600 volts and less. Proper sizing of the transformer. a 75 kVA transformer is commonly used. consideration of the magnetizing inrush and short circuit protection all play an essential role in the design.3(B) NEC® Table 450. Jim Phillips.2 Transformer Protection Transformers are a very critical component in the electrical power distribution system since a failure often means losing a significant portion of the load creating costly downtime. The primary overcurrent device can be sized as large as 250% of the primary full load current as long as the secondary device is sized based on the 125% rule. a larger primary device may also require using larger primary conductors depending on how the tap rule exceptions are used. The transformer schedules in this section address secondary protection based on 125% of the full load current rating of the transformer. it is not the only selection and some may select a different size for a variety of reasons. if the transformer was sized to serve a 225 Amp panel rated 208 Volts. If the ampacity rating is 9 amps or less. ELECTRICAL DESIGN GUIDE Sizing the transformer is generally based on the load to be served. T 2G Technical Training Group® Page 12 of 37 . compliance with the NEC® Article 450.E. In addition. selection of its conductors and adequate protection are essential to a sound design. Sizing the primary device at it’s maximum of 250% can create special protection problems since it may not adequately protect the transformer against secondary short circuits. When a fault occurs between the transformer primary and secondary overcurrent device. For example.com® 3. the primary device must respond.brainfiller.3 NEC Table 450. P. This transformer’s secondary full load current rating known as FLA (as shown in Table 3. Where 125% of the transformer’s current rating does not correspond to a standard overcurrent device rating. the final design must be reviewed by an electrical design professional. it will typically be less sensitive creating insufficient protection of the transformer.3) is 208 Amps which is sufficient for the panel’s continuous current rating of 80% of it’s rated load ( 80% X 225 Amp = 180 Amps). 3. If it is too large. the next higher standard device rating may be used. 9 .7 .8 .5 .5 . the primary device only “sees” 58% of it’s normal current due to unbalances in the transformer windings.05 . and defines the upper limit for protection. The line to the right is labeled “ANSI Withstand” and represents the standard protection limit.2 .4 Magnetizing Inrush Current ELECTRICAL DESIGN GUIDE NEC® Table 450.1 .06 . dry type. P.1 Sec.05 .8 .0.2 1 . The withstand characteristic will vary depending on the type of transformer.02 TRANSFORMER 45 kVA INRUSH .08 . However. With the lower short circuit current.5 ANSI Thru Fault Protection CURRENT IN AMPERES X 10 AT 480 VOLTS When a short circuit occurs Maximum Minimum per NEC FLA downstream of a transformer. When a transformer is energized. When a secondary line to ground fault occurs on the secondary of a delta .wye grounded transformer.1 seconds so the overcurrent devices should be sized large enough and with enough time delay to not nuisance trip during transformer energization. the high currents associated Primary 70 A C/B with the fault will flow through the transformer possibly subjecting it to damaging magnetic forces and heat.5 .7 . i.6 .9 .01 5 6 7 8 9 10000 CURRENT IN AMPERES X 10 AT 480 VOLTS Jim Phillips.04 . For the 45 kVA dry type transformer in Figure 3.5 .08 .03 .6 .3 . the 1000 900 800 700 600 500 400 300 200 .1.04 . etc.03 . ANSI C57 defines a ANSI Adj.e.E.4 . Inrush 10 X 70 Amp C/B FLA at 0.brainfiller.48 . 3. the withstand characteristic is shown as two diagonal lines.6 . liquid filled.208 kV 5% 100 90 80 70 60 50 40 30 20 SECONDARY TIME IN SECONDS 4 3 2 4 3 2 1 . winding configuration and impedance. this does not necessarily mean that lower rated devices should always be used when protecting transformers.3 (B) provides the upper limits for sizing transformer protection.8 1 2 3 4 5 6 7 8 9 10 2 3 4 5 6 7 8 9 100 2 3 4 5 6 7 8 9 1000 2 . The withstand characteristic is defined as a time current 45 kVA 480 Volt curve and is based on the Delta Primary transformer’s size.1 .3 .wye grounded transformer.com® 3. Generally the smaller the device the better the protection since smaller devices tend to be more sensitive offering better protection.8 1 2 3 4 5 6 7 8 9 10 2 3 4 5 6 7 8 9 100 2 3 4 5 6 7 8 9 1000 2 3 4 TRANSFORMER FLA 5 6 7 8 9 10000 1000 900 800 700 600 500 400 300 200 PRIMARY 100 90 80 70 60 50 40 30 20 PRIMARY C/B SQD F Frame FA Frame = 100A (70AT) Trip = 70 PRIMARY C/B SQD FA 100/70 TRANSFORMER 45 kVA 0.6 .01 .07 .09 . ANSI transformers short circuit Withstand Withstand withstand characteristic that can be used in selecting the primary protective device. This inrush current can last approximately 6 cycles or 0. design.07 . it can draw a magnetizing inrush current through the primary device that can be as large as 8 to 12 times its primary full load current.09 .4 .02 PRIMARY C/B 17969A 3 4 .06 . T 2G Technical Training Group® Page 13 of 37 TIME IN SECONDS 10 9 8 7 6 5 TRANSFORMER 45 kVA 5% 10 9 8 7 6 5 . The parallel line just to the left is marked “Adjusted ANSI Withstand” and represents the withstand characteristic of a delta . a 70 Amp circuit breaker was selected which is large enough to carry the load current.brainfiller. T 2G Technical Training Group® Page 14 of 37 . The equipment grounding conductor(s) are sized based on NEC® Table 250. therefore the conductor selections listed in the table may not be the only acceptable choices. small enough to comply with the NEC® limits and slow enough to clear the magnetizing inrush current. Since each transformer can have a unique withstand characteristic and each overcurrent device can have a unique time current curve.brainfiller. it does not completely clear the adjusted ANSI characteristic. There are similar adjustments for deltadelta connections as well. In the example. Because of this situation. Adjusting the curve to the left by 58% means a smaller / faster primary device must be selected to satisfy the adjusted curve.com® ELECTRICAL DESIGN GUIDE primary device would operate more slowly than expected allowing damage to the secondary winding that is experiencing the short circuit current. P. For optimal device selection. Although the circuit breaker is fast enough to pass to the left of the ANSI withstand characteristic.com where you can order the video series on Protective Device Coordination Studies that contains an in depth example of ANSI curves and graphs. the individual transformer’s withstand characteristic and the selected overcurrent device’s time current curve based on the specific manufacturer should be used. The NEC allows for various exceptions to this sizing method based on numerous “tap rules”.122 and the overcurrent protection listed in the transformer schedule. however a smaller device may cause the inrush to become an issue.E. some may choose to select a 60 Amp device. the tables in this section represent general device selections. 3. visit: www. Jim Phillips.6 Transformer Conductors Transformer primary and secondary conductors listed in the table are sized according to the ampere rating of the primary and secondary overcurrent protection.16 copper conductors 75ºC ratings operating in an ambient temperature not exceeding 30ºC. Selecting overcurrent devices quite often requires compromising between competing objectives. In a few cases For a more detailed explanation of ANSI transformer protection. The conductor ampacity is based on NEC® Table 310. There are also exceptions where the size of the grounding conductor may need to be increased due to considerations such as very high short circuit current or an increase in the size of the phase conductor. typical overcurrent protection (more commonly used sizes) based on thermal magnetic circuit breakers and time delay fuses.Primary Time Delay Fuse Protection Table 3. maximum (not necessarily optimum) overcurrent protection based on NEC Table 450. The schedules are organized according to kVA rating.3 . The more detailed analysis usually requires developing time current curves which is beyond the scope of this guide.Primary Circuit Breaker Protection Table 3. Therefore. Other considerations such as a transformer’s maximum magnetic inrush current. The typical device sizes are based on a survey of breaker and fuse manufacturer’s data as well as various designs.Transformer Schedule .208Y/120 Volt transformers sizes are listed in the Transformer Schedules.3 (B).E.Transformer Schedule . Table 3.1 . 3. it may not always be the most optimal selection for a given design. ANSI through fault protection and various other exceptions may require a different device rating.brainfiller. There are 3 Transformer Schedules listed in this section. Although the typical selection may be a more common size that is frequently used.7 Transformer Schedules The more common 3 phase 480 Volt Delta .Transformer Schedule . T 2G Technical Training Group® Page 15 of 37 .Secondary Overcurrent Protection Jim Phillips.2 .com® ELECTRICAL DESIGN GUIDE the NEC® requires a #3 equipment ground however this was changed to the more commonly used #2 conductor in the transformer schedules. P. the table is only a general guide and the final design must be reviewed by the electrical design professional in responsible charge of the project. 16 from the 75ºC column.480 Volts 3 Phase Delta Molded Case Circuit Breakers kVA 15 30 45 75 112.E. 5.500 kcm 4 sets . Maximum primary overcurrent protection is based on NEC Table 450. Transformer inrush current can be between 8 to 12 times the primary FLA and last approximately 6 cycles. Jim Phillips.5 150 225 300 500 750 1000 FLA 18 36 54 90 135 180 271 361 601 902 1203 Maximum OCP per NEC® 45 90 125 225 300 450 600 800 1200 2000 3000 Typical Circuit Breaker 30 50 70 125 175 225 350 450 800 1200 1600 Phase Wire #10 #6 #4 #1 2/0 4/0 350 kcm 2 sets . ANSI C57 short circuit protection and the use of the many exceptions found in the NEC such as the “tap rules”. T 2G Technical Training Group® Page 16 of 37 .com® ELECTRICAL DESIGN GUIDE Table 3.Primary Circuit Breaker Protection Transformer Primary .4/0 2 sets .1/0 4 .4/0 1.500 kcm Ground Wire #10 #10 #8 #6 #6 #4 #2 #2 2 . P. It is assumed no derating is required. The “typical “ circuit breaker size is based on a survey of many designers and manufacturers. 4. Many factors influence the final selection such as transformer magnetic inrush current. 3. The final selections must be reviewed and approved by the design professional in responsible charge of the design. 2.brainfiller. The phase and ground conductors are copper and sized based on the transformer primary full load current and the ampacities in table 310.1 Transformer Schedule .350 kcm 5 sets .3(B) which allows the primary overcurrent protection to be sized as large as 250% of the primary full load current as long as the secondary protection is sized no larger than 125% of the secondary full load current or next larger device rating.3/0 5 . Other conductor sizes may be selected in accordance with other NEC articles and exceptions. Many factors influence the final selection such as transformer magnetic inrush current. 3. ANSI C57 short circuit protection and the use of the many exceptions found in the NEC such as the “tap rules”.1/0 4 . 4. Maximum primary overcurrent protection is based on NEC Table 450.3/0 5 . The phase and ground conductors are copper and sized based on ampacities in table 310.2 Transformer Schedule . The “typical “ circuit breaker size is based on a survey of many designers and manufacturers.500 kcm Ground Wire #10 #10 #8 #6 #6 #4 #2 #2 2 . 2. Other conductor sizes may be selected in accordance with other NEC articles and exceptions. The final selections must be reviewed and approved by the design professional in responsible charge of the design.4/0 1.E. Jim Phillips.5 150 225 300 500 750 1000 FLA 18 36 54 90 135 180 271 361 601 902 1203 Maximum OCP per NEC® 45 90 125 225 300 450 600 800 1200 2000 3000 Typical Time Delay Fuse 25 45 70 125 175 225 350 400 800 1200 1600 Phase Wire #10 #6 #4 #1 2/0 4/0 350 kcm 500 kcm 2 sets .16 from the 75ºC column.500 kcm 4 sets .Primary Time Delay Fuse Protection Transformer Primary . Transformer inrush current can be between 8 to 12 times the primary FLA and last approximately 6 cycles.350 kcm 5 sets .3(B) which allows the primary overcurrent protection to be sized as large as 250% of the primary full load current as long as the secondary protection is sized no larger than 125% of the secondary full load current or next larger device rating. T 2G Technical Training Group® Page 17 of 37 . It is assume no derating is required. 5.480 Volts Time Delay Fuses kVA 15 30 45 75 112.com® ELECTRICAL DESIGN GUIDE Table 3.brainfiller. P. 3(B) which allows the secondary device to be sized up to 1.1/0 3 . The phase and ground conductors are copper and sized based on ampacities in table 310.500 kcm 5 sets . Other conductor sizes may be selected in accordance with other NEC articles and exceptions.2/0 5 .4/0 7 .500 kcm 8 sets . The maximum secondary overcurrent device rating is based on NEC 450. 3.16 from the 75ºC column. Transformer inrush current can be between 8 to 12 times the primary FLA and last approximately 6 cycles. the next larger device can be used. Other design considerations may affect the final size such as various NEC® exceptions.E.25 X secondary FLA and where this does not correspond to a standard overcurrent device rating.500 kcm 7 sets .350 kcm 2 sets . T 2G Technical Training Group® Page 18 of 37 .350 8 .500 kcm Ground Wire #8 #8 #8 #4 #3 2 .Secondary Overcurrent Protection Transformer Secondary Protection -208Y / 120 Volts Maximum OCP per NEC® 60 110 175 300 400 600 800 1200 2000 3000 4000 Typical Secondary Device Rating 60 100 150 225 400 600 800 1000 1600 2500 3000 kVA 15 30 45 75 112.3 Transformer Schedule . 4.#1 2 .com® ELECTRICAL DESIGN GUIDE Table 3.400 1. The “Typical” device rating corresponds to sizes typically used. The final selections must be reviewed and approved by the design professional in responsible charge of the design. P. Jim Phillips.500 kcm 3 sets .brainfiller. 2. It is assumed no derating is required.5 150 225 300 500 750 1000 FLA 42 83 125 208 312 416 625 833 1389 2083 2778 Phase Wire #6 #2 1/0 4/0 500 kcm 2 sets . 5% of the specified value. For the short circuit calculations to be accurate. The test begins with the transformer de-energized but then a very small percentage of rated primary voltage is applied.E. the actual transformer nameplate impedance must be used and not an assumed value. A very long time ago.com® ELECTRICAL DESIGN GUIDE 4.7.brainfiller. The percent of rated primary voltage that it took to produce rated full load current in the shorted secondary is the transformer percent impedance and is stamped on the nameplate.0 INFINITE BUS SHORT CIRCUIT CALCULATIONS 4. the transformer’s actual final tested impedance can vary by as much as +/.5%.75% therefore the Jim Phillips. the calculations can be based on using 0. The actual tested impedance can vary by +/. If the transformer has been specified but has not been delivered at the project site for inspection. P.5% of the specified 5.1 NEC Requirements NEC® Article 110. Multiplying Z X 100% = Z. T 2G Technical Training Group® Page 19 of 37 . These simplified calculations are often referred to as “infinite bus” calculations and are based on a minimal amount of data including the transformer’s three phase kVA rating (self cooled without fans). During the preliminary design stage of a project.2 Transformer Impedance The transformer percent impedance is derived from a factory test where a short circuit is applied on the secondary windings of the transformer. The primary voltage is slowly raised while monitoring the current circulating in the shorted secondary. As an example: A transformer is specified with a 5.9 requires that equipment intended to interrupt short circuit current must have a sufficient interrupting rating. When the current in the shorted secondary is equal to the transformer’s full load current rating. transformers used to have %IZ stamped on the nameplate and I X Z is a voltage. simplified short circuit calculations are often used to verify equipment’s adequacy prior to specification. the nameplate percent impedance and the secondary line-line voltage. Of course %Z does not equal %V according to Ohm’s Law. 4. To be even more conservative.925 X the specified impedance to account for the possibility of the transformer’s tested impedance being low by 7.75% impedance. Eventually the “I” was dropped since it represented 100% of the full load current from the test. the test is complete.7. Example: What is the worst case infinite bus short circuit current on the secondary of a transformer rated 1500 kVA with a nameplate impedance of 5.75 X 0. what would happen if 100% voltage was applied. % Zxfmr / FLAsecondary = 100 % / SCAinfinite Where: % Zxfmr FLAsecondary SCAinfinite = transformer nameplate impedance = transformer secondary full load current = maximum short circuit current assuming an infinite source.com® ELECTRICAL DESIGN GUIDE actual impedance could ultimately be as low as 92.925) of the specified value: 5. Jim Phillips. Although if you are located next to a 2000 MW power plant the short circuit current could become extremely high (although not infinite). The second step is to calculate the maximum three phase short circuit current based on the full load current and the nameplate percent impedance. use the base rating.5%) = 5.3 Infinite Bus Short Circuit Calculation Since we know that applying the rated percent impedance (voltage) on the primary of a transformer will result in rated full load current circulating in the shorted out secondary winding. Where a transformer has a base kVA rating and a fan cooled rating. This method leads to conservative results since utility systems do not have infinite short circuit current.7.5% (0. We can rearrange the two ratios and solve for SCAinfinite SCAinfinite = (FLAsec X 100) / % Zxfmr This method is commonly referred to as the “infinite bus” method because it is based only on the transformer impedance and ignores any other impedance on the primary such as the utility company. that means we assumed it was “0” and the current that would be available if limited by 0 ohms would be infinite.32% 4. This calculation ignores any other impedance such as the source of conductor.75 X (100% .E. P. The first step is to calculate the secondary full load current rating based on the transformer’s kVA rating.75% and a secondary voltage of 480 Volts? This problem can be solved in only two steps. You would have the maximum short circuit current.brainfiller. If the utility impedance is ignored.925 = 5. T 2G Technical Training Group® Page 20 of 37 . These two relationships are proportional. locations voltages less than 240 Volts.com 50% x 4 x transformer FLA. The magnitude of the motor contribution is based on the motor’s subtransient reactance Xd’’ and can be as high as each motors locked rotor (starting) current. all directly connected running motors can contribute short circuit current that flows towards the fault. T 2G Technical Training Group® Page 21 of 37 . www.e.732 ) FLAsecondary = 1804 Amps Step Two: SCAinfinite = (1804 Amps X 100 ) / 5.374 Amps ELECTRICAL DESIGN GUIDE 4.4 Motor Contribution During a short circuit. See serving motor load .E. Identifying all motors during the preliminary design stage can be a difficult task.brainfiller. A commonly used practice for a worst case approximation is to include a multiple of the transformer’s full load current in Attend Jim’s One Week Power System Engineering Class to the calculations as a “cushion” to account for learn how to conduct a motor contribution.48 kV X 1. i. P. Jim Phillips. This represents 50 % of the transformer capacity is serving motor load.brainfiller. 2 X transformer and dates at: FLA is often used. This course also includes approximated as 4 X transformer FLA.com® Step One: FLAsecondary = kVA / (kVline-line X √3) FLAsecondary = 1500 kVA ( 0.75% SCAinfinite = 31. This means that a typical motor can contribute short circuit current of approximately 4 to 6 times it’s full load current. This electrical design.not very likely! For the detailed agenda. protective very large and conservative approximation is device coordination studies based on the bulk of the transformer capacity and harmonic analysis. motor contribution can be detailed Short Circuit Study. For voltages of 240 Volts through 600 Volts. 5 Calculation Worksheets ELECTRICAL DESIGN GUIDE The short circuit calculation worksheets in this section are for approximations only. Therefore.brainfiller. They generally lead to very conservative results that can be much greater than the actual short circuit current based on more exact data. visit: www.com Jim Phillips.E. Although the worst case maximum short circuit calculations are conservative for selecting equipment interrupting ratings. T 2G Technical Training Group® Page 22 of 37 . it may also be desirable to perform a detailed short circuit study or detailed calculations for more exact results. For a free training video clip where Jim explains transformer percent impedance in detail. arc flash and harmonics studies. lower short circuit currents could possibly yield more significant problems.com® 4. it might not be conservative for use in other studies such as voltage drop. For these types of studies. P.brainfiller. com® ELECTRICAL DESIGN GUIDE Table 4.1 Transformer Infinite Bus Short Circuit Calculation Worksheet Steps INFINITE BUS SHORT CIRCIUT CALCULATION Transformer kVA3 Phase Rating (AA) Secondary VoltageLine-Line in kV Step 1 Enter Data = _______kVA Transformer % Impedance = _______%Z = _________kV Step 2 Determine FLAsecondary Step 3 Determine Short Circuit Amps (SCA) Step 4 Determine Approximate Motor Contribution SCAmotor Step 5 Determine Approximate Total Short Circuit Amps FLA = kVA3 phase ÷ (√3 x kVline-line) FLA = ___________kVA ÷ (√3 x __________kV) ________Amps SCAsecondary = (FLA x 100) ÷ %Z SCAsecondary = (___________Amps x 100) ÷ _____% _________SCA If Secondary Voltage < 240V If Secondary Voltage > 240V < 600V SCAmotor = FLAsecondary X 2 SCAmotor = ________ A X _____ Select either X “2” or “4” _______SCAmotor SCAmotor = FLAsecondary X 4 Total Short Circuit Amps = Step 3 + Step 4 Step 3 + Step 4 ________SCAtotal FLA SCAsecondary SCAmotor SCAtotal = Transformer secondary full load amps = Short Circuit Amps on the transformer secondary = Short circuit motor contribution = Total short circuit current at secondary bus Jim Phillips.E.brainfiller. T 2G Technical Training Group® Page 23 of 37 . P. 75% nameplate impedance and a secondary voltage of 480 Volts. Jim Phillips.48 kV Step 2 Determine FLAsecondary Step 3 Determine Short Circuit Amps (SCA) Step 4 Determine Approximate Motor Contribution SCAmotor Step 5 Determine Approximate Total Short Circuit Amps FLA = kVA3 phase ÷ (√3 x kVline-line) FLA = 1500 kVA ÷ (√3 x 0.374 SCA If Secondary Voltage < 240V If Secondary Voltage > 240V < 600V SCAmotor = FLAsecondary X 2 SCAmotor = 1804 AX 4 7.216 SCAmotor Select either X “2” or “4” SCAmotor = FLAsecondary X 4 Total Short Circuit Amps = Step 3 + Step 4 Step 3 + Step 4 38. P.2 Example Problem Using Transformer Infinite Bus Short Circuit Calculation Worksheet Steps INFINITE BUS SHORT CIRCIUT CALCULATION Transformer kVA3 Phase Rating (AA) Secondary VoltageLine-Line in kV Step 1 Enter Data = 1500 kVA Transformer % Impedance = 5.590 SCAtotal Example: Use this worksheet to determine the worst case 3 phase short circuit current at the 480 Volt secondary bus assuming an infinite source.E. The worksheet on this page is filled out to illustrate it’s use in performing the calculations for this problem.brainfiller. T 2G Technical Training Group® Page 24 of 37 .com® ELECTRICAL DESIGN GUIDE Table 4.75 % 31. The transformer is rated 1500 kVA with a 5.75 %Z = 0.48 kV) 1804 Amps SCAsecondary = (FLA x 100) ÷ %Z SCAsecondary = ( 1804 Amps x 100) ÷ 5. If the conductors are of different sizes.1158 x 1) = 0. NEC® Chapter 9 Table 1 states that the maximum percentage of the cross sectional area of conduit that can be filled by conductors shall not exceed: From NEC Chapter 9 Table 1 Percent of Cross Section of Conduit and Tubing for Conductors Number of Conductors 1 2 Over 2 All Conductor Types 53% 31% 40% When sizing conduit where all conductors are the same size and have the same insulation. T 2G Technical Training Group® Page 25 of 37 . Multiply the conductor area by the number of conductors of that size. Jim Phillips. P.com® ELECTRICAL DESIGN GUIDE 5.brainfiller.E.1 on the next page. Select the conduit size using the appropriate conduit from Chapter 9.7073 x 4) = 2. Table 5 of the NEC®. Add these numbers to calculate the total area.0 CALCULATION OF CONDUIT FILL Pulling conductors in an overloaded conduit can result in an increased risk of insulation damage. The NEC® tables list the maximum number of conductors permitted in various types and trade sizes of conduit. Example: To calculate the conduit size for four 500 kcm and one #2 conductor with THHN insulation. Table 4 of the NEC® (500 kcm THHN area x 4) = (#2 THHN area x 1) = Total Area (0. Then look up the area of #2 THHN and multiply by one. select the conduit size from the Chapter 9. look up the conductor areas Chapter 9.8292 (0. Repeat for each size conductor and total all areas. Table 4 of the NEC®. select a 3 inch minimum size for rigid steel conduit.9450 From the 40% column in Table 5. Once the total area is calculated. use Table 5 from Chapter 9 of the NEC® and look up the area of 500 kcm THHN and multiply by four.1158 2. it is easiest to look up the conduit size in Annex C of the NEC®. 0 35.000 4.220 0.097 0.642 1.610 0.9 63.9 128.004 5.212 29.098 1.946 3. P.291 0.4 41.363 1.com® Table 5.470 0.473 0.153 8.809 1.7 102.039 > 2 – Wires 40% (IN2) 0.brainfiller.E.Article 344 .526 2. T 2G Technical Training Group® Page 26 of 37 .828 10.408 4.166 0.2 27..882 20.010 12.454 2 – Wires 31% (IN2) 0.974 5.579 3.314 0.8 Jim Phillips.056 1.2 78.663 ½” ¾” 1” 1 ¼” 1 ½” 2” 2 ½” 3” 3 ½” 4” 5” 6” 16.158 1 – Wire 53% (IN2) 0.9 154.508 2.887 1.325 3.2 52.499 10.125 0.549 0.866 7.071 3.Rigid Metal Conduit (RMC) Trade Size Inches Trade Size Metric Total Area 100% (IN2) 0.305 6.1 21.355 0.103 3.085 11.994 6.1 Rigid Metal Conduit Percent Area ELECTRICAL DESIGN GUIDE Excerpt from NEC Chapter 9 Table 4 .713 15.806 2.829 1.275 0.266 9.170 0.5 90. 3 Motor Feeder Sizing Motor feeder conductor sizes are based on a minimum of 125% of the motor full load amps per NEC® Article 430.250 and must be sized based on the motor’s actual nameplate current as well as the motor’s service factor. The conductor ampacity is based on NEC® Table 310.E. motor starting characteristics. The motor overload must also have sufficient time delay to allow the motor to start. Factors such as voltage drop. In addition.1 Motor Circuits NEC® Article 430. Motor overloads are not sized based on the currents of Table 430. The conductor size may need to be adjusted for variations from these selection factors.22. 6. Jim Phillips. the schedules contain the more common selections based on a survey of manufacturers and designs. These schedules provide data that can be used to determine the size of the phase conductors.0 MOTOR SCHEDULES 6. A common solution is to increase the size of the feeder. therefore the final design must be performed under the direction of a licensed design professional.22 defines the requirements for sizing motor feeders for single motors (although other requirements and exceptions exist in other articles as well).2 Motor Full Load Current Ratings NEC® Table 430.com® ELECTRICAL DESIGN GUIDE 6. This Section contains motor schedules to assist in motor circuit design.brainfiller. T 2G Technical Training Group® Page 27 of 37 .250 provides typical full load current values for three phase AC motors operating at speeds usual for belted motors with normal torque characteristics.16. 6. 75ºC copper conductors based on a 30ºC ambient temperature and no more than three current carrying conductors in a raceway. long motor feeders may result in an undesirable amount of voltage drop. These currents are listed in the Motor Schedules and are also the basis for selecting motor feeders and sizing the short circuit protection in accordance with the NEC®. temperature rating and thermal damage limits. P. short circuit protection based on circuit breakers and time delay fuses as well as NEMA starter sizes. ambient conditions and other design considerations could ultimately affect the final selection. Like other sections in this guide. 52 of the NEC Type of Motor Non Time Delay Fuse Dual Element Time Delay Fuse Instantaneous Trip Breaker Inverse Time Breaker Squirrel Cage Induction . Since different protective devices will have time current characteristics. Sizing the motor overload is based on the actual motor nameplate data rather than Table 430.com® 6. Motors can only survive this current for a specific amount of time (referred to as the safe stall time) and this current must be removed before the motor can sustain damage.250. the motor can draw a current equal to it’s locked rotor (starting) current that can typically range from 4 to 6 times the motor’s normal running current. P. the overload can be sized up to 125% of the motor nameplate full load current.brainfiller. During a stall condition. the NEC provides multipliers that are unique for each type of device. This can often be due to a problem with the process or load. such as a faulty bearing or other type of jam. Some of these multipliers are listed in the table below and are from NEC Table 430. The motor overload is designed with a time delay to allow the motor to start yet interrupt if the starting current lasts too long. Table 6.1 Motor Short Circuit Protection Maximum Rating or Setting of Motor Branch Circuit Short Circuit and Ground Fault Devices Excerpts from Table 430. Jim Phillips.4 Motor Short Circuit Protection ELECTRICAL DESIGN GUIDE Short circuit protection is based on a multiple of the motor full load current based on Table 430.250. For motors with a service factor 1.5 Motor Overload Protection Motor overload protection is designed to protect the motor against damaging currents that result from the motor stalling or significantly slowing down.52. T 2G Technical Training Group® Page 28 of 37 .15 or greater or a marked temperature rise of 40ºC or less.Other than Design B Energy Efficient Design B Energy Efficient 300 175 800 250 300 175 1100 250 6.E. The overload for other motors is 115% unless it falls under one of the exceptions in the NEC®. 2 Illustrate the protection of a 75 Hp motor.6 Motor Time Current Curves ELECTRICAL DESIGN GUIDE The operation of a motor and its protection can best be described with the use of time current curves. Voltage: 480 Current Scale x10^0 EQUIP1.1 and 6.brainfiller.2 are identical with only one difference.tcc Ref. P.10 current. The motor starting characteristic shows the locked rotor current (starting current) lasting for approximately 5 seconds and then lowering to the rated full load current.drw Figure 6. T 2G Technical Training Group® Page 29 of 37 . The typical motor can draw a locked rotor current often between 4 to 6 X (higher for premium 0.1 and 6. Figure 6.com® 6. 100 TIME IN SECONDS Jim Phillips. The NEC® allows this device to be set as high as 8 X the motor full load current for 1 standard squirrel cage induction motors. The 8 X factor 75HP Start.1 allows the motor sufficient margin for starting without operating the short circuit protective device. The starting current is shown as 6 times the Full Load Current. commonly MS referred to as a motor circuit protector for short 10 circuit protection. Figures 6.01 0.5 1 10 100 1K 10K efficiency and some other designs).E.1 uses an instantaneous only circuit breaker. excluding Design B energy efficient motors which can have a much larger starting 0. The black vertical band in Figure 1 represents the time current characteristic of the motor circuit protector. Figure 6. The motor stall time is shown as a “+” on each graph and the overload (green crosshatched band) must have enough time delay for the motor to start yet operate quickly CURRENT IN AMPERES enough to protect against 1000 stall outs. The motor circuit protector can be set as high as 8 X the Full Load Current to allow the motor to start. Jim Phillips.5 1 10 100 1K 10K is used. 6. The selections are typical and can vary depending on many other variables and design conditions. Care must be given so the fuse will allow the motor to start and preferably allow the motor’s overloads to trip first in the event of a stalled motor or overload condition.10 right is the same as Figure 6.7 Motor Schedules Motor schedules are provided for 460 Volt and 230 Volt standard induction motors. P. The final selection should be reviewed by a licensed design professional responsible for the project. Voltage: Figure 6. For motors with long starting times.2 shows a time delay fuse that has a slope that is inversely proportional to the current. and include reviewing time current curves to verify correct protection and operation for the specific application.drw 480 Current the motor’s full load current.brainfiller.e. 175% X 40 Amps would allow a maximum fuse size of 70 Amps.1 on the previous page but instead of a motor circuit protector. The NEC® allows the next higher standard rating. a time delay fuse 0. For circuits with significant voltage drop. ELECTRICAL DESIGN GUIDE CURRENT IN AMPERES 1000 100 MS 10 TIME IN SECONDS As an example.com® Figure 6. i. The curve on the 0.E. larger devices may be required. less current .more time. larger conductors might be used. more current . T 2G Technical Training Group® Page 30 of 37 .less time.2 Scale x10^0 EQUIP1.01 0. The fuse in Figure 2 can be sized up to 175% X 75HPFUSE. a 40 HP motor rated 460 Volts has a full load current rating of 40 1 Amps according to NEC Table 430. Correct overload protection must also be provided in accordance with the NEC®.250. The NEC allows this type of device to be sized as large as 175% of the motor full load current.tcc Ref. Long motor starting times may require alternate device selections. low speed. Actual motor nameplate data must be used for the selection of overload protection. 6. 7. 8.25 X FLA per NEC 430.based on 250% of FLA or next standard rating based on NEC 430.5 10 15 20 25 30 40 50 60 75 100 125 150 200 1. 12. Conductor ampacity is based on no more than three current carrying copper conductors in a raceway. Maximum Dual Element Fuse . Other conductor sizes may be necessary for special conditions such as voltage drop or conductor derating.based on 175% of FLA or next standard rating based on NEC 430. 4.22 for single motors. 75ºC insulation and 75ºC device terminal rating.based on survey of several manufacturer’s recommended RK-5 time delay fuse.com® Table 6.52. Phase conductor size is based on 1. Although No. it is common to use No. Jim Phillips. 10. 14 conductors can be used with some smaller motors. P. high starting currents and other special conditions may require additional consideration in the selection of protection. Motor full load current rating (FLA) is based on NEC table 430.based on a survey of several manufacture’s recommended circuit breakers. 11. FLA 7. Typical Inverse Circuit Breaker . continuous duty applications.6 11 14 21 27 34 40 52 65 77 96 124 156 180 240 Maximum Inverse Time Breaker 20 30 35 60 70 90 100 150 175 200 250 350 400 450 600 Phase Conductor Size 12 10 10 10 10 8 8 6 4 3 1 2/0 3/0 4/0 350 kcm Minimum NEMA Starter Size 0 1 1 2 2 2 3 3 3 4 4 4 5 5 5 The selections in the table are “typical” for motors operating at normal speed and torque.52.brainfiller. It is possible that smaller devices than those shown can be used and still allow motor to start.1 460 Volt Motor Circuit Schedules Short Circuit Protection Typical Dual Element Fuse 10 15 20 30 40 50 60 70 100 125 150 175 200 225 300 Maximum Dual Element Fuse 25 35 45 70 90 110 125 175 200 250 300 400 500 600 800 Typical Inverse Time Breaker 15 25 30 45 60 70 80 100 110 150 175 200 225 250 350 ELECTRICAL DESIGN GUIDE Hp 5 7.E. 5. 3.. Maximum Inverse Breaker . T 2G Technical Training Group® Page 31 of 37 . 2. Typical Dual Element Fuse . Motors that have special starting characteristics such as high torque. 9.250. 12 instead. P.25 X FLA per NEC 430.2 230 Volt Motor Circuit Schedules Short Circuit Protection Typical Dual Element Fuse 10 15 20 35 40 60 80 100 125 150 175 225 250 350 450 500 600 Maximum Dual Element Fuse 15 20 30 40 50 80 100 125 150 200 250 300 350 450 600 700 1000 Typical Inverse Circuit Breaker 15 20 30 45 60 80 90 100 125 150 200 225 300 400 500 600 800 ELECTRICAL DESIGN GUIDE Hp 2 3 5 7. 2.based on 250% of FLA or next standard rating based on NEC 430. 4. high starting currents and other special conditions may require additional consideration in the selection of protection. 9. T 2G Technical Training Group® Page 32 of 37 . Conductor ampacity is based on no more than three current carrying copper conductors in a raceway. 7.6 15. Typical Inverse Circuit Breaker . Long motor starting times may require alternate device selections. 14 conductors can be used with some smaller motors. 10. Jim Phillips.based on 175% of FLA or next standard rating based on NEC 430.com® Table 6.250. Motors that have special starting characteristics such as high torque. It is possible that smaller devices than those shown can be used and still allow motor to start.8 9. 8.5 10 15 20 25 30 40 50 60 75 100 125 150 200 1. Maximum Inverse Breaker . Phase conductor size is based on 1. 5. low speed. Typical Dual Element Fuse .. 11.brainfiller. FLA 6. 75ºC insulation and 75ºC device terminal rating.based on a survey of several manufacture’s recommended circuit breakers. continuous duty applications. 6. 12 instead.based on survey of several manufacturer’s recommended RK-5 time delay fuse. it is common to use No.52. 3. Other conductor sizes may be necessary for special conditions such as voltage drop or conductor derating.2 22 28 42 54 68 80 104 130 154 192 248 312 360 480 Maximum Inverse Circuit Breaker 20 25 40 60 70 110 150 175 200 300 350 400 500 700 800 1000 1200 Phase Conductor Size Minimum NEMA Starter Size 0 0 1 1 2 2 3 3 3 4 4 5 5 5 6 6 6 12 12 12 10 10 6 4 4 3 1 2/0 3/0 250 kcm 350 kcm 2 sets . Maximum Dual Element Fuse .22 for single motors. Motor full load current rating (FLA) is based on NEC table 430.E.3/0 2 sets . Although No. Actual motor nameplate data must be used for the selection of overload protection. 12.4/0 2 sets -350 kcm The selections in the table are “typical” for motors operating at normal speed and torque.52. 09 .4 . only the device closest to the problem trips and the other devices remain closed.5 .com® ELECTRICAL DESIGN GUIDE 7. Graph 1 at the left represents a 20 Amp circuit breaker. time current curves are used. 2007 ® Graph 1 20 Amp Circuit Breaker Jim Phillips.1 .E.5 and 7 Seconds Instantaneous / Magnetic Part of the Curve Larger currents cause the breaker’s instantaneous to trip .01 5 6 7 8 9 10000 CURRENT IN AMPERES X 10 AT 480 VOLTS T2G Technical Training Group EasyPower ® TIME-CURRENT CURVES Brainfiller.6 . 1000 900 800 700 600 500 400 300 200 .9 .02 .05 .5 .6 .5 . Look at the bottom horizontal scale for the number “1”.6 . This scale is frequently multiplied by 10 or 100 to allow graphing higher magnitudes of current. The graph’s horizontal scale is defined as current in Amps.1 Time Current Curves A coordination study is an “attempt” to select devices and determine optimum setting adjustments so that during an overload or short circuit.7 .2 1 .7 .3 .3 .09 .1 .07 .06 . T 2G Technical Training Group® TIME IN SECONDS 10 9 8 7 6 5 10 Amp line never intersects the breaker’s curve 10 9 8 7 6 5 Every overcurrent device has a time current curve that defines how it will respond to various levels of current.06 .6 .8 .8 .9 . is 20 Amps. The curves are logarithmic in nature which represents orders of magnitude so the graph scale increases by factors of 10 rather than just single digits such as 1. Following the 10 Amp line up to the top of the graph. The current scale CURRENT IN AMPERES X 10 AT 480 VOLTS for the graphs in this section are all 2 X 10 = 20 Amps multiplied by 10.brainfiller.2 .04 . P.4 .04 .8 1 2 3 4 5 6 7 8 9 10 2 3 4 5 6 7 8 9 100 2 3 4 5 6 7 8 9 1000 2 3 4 5 6 7 8 9 10000 1000 900 800 700 600 500 400 300 200 100 90 80 70 60 50 40 30 20 Overload / Thermal Part of the Curve 100 90 80 70 60 50 40 30 20 TIME IN SECONDS 4 3 2 4 3 2 1 . To evaluate coordination.03 . it never intersects the circuit breaker Page 33 of 37 .07 . Since this is also multiplied by 10. the “1” represents 10 Amps.8 1 2 3 4 5 6 7 8 9 10 2 3 4 5 6 7 8 9 100 2 3 4 5 6 7 8 9 1000 2 3 4 .05 .03 . 3 etc. 2.02 100 Amp line intersects the breaker’s curve between 2.5 . This allows clearing the problem circuit while maintaining service continuity to the remainder of the system.01 .com TCC-1 FAU LT: D ATE: BY : R EVISION: 1 Jan 31.08 .08 . The upper left portion of the curve aligns at “2” which when multiplied by 10.0 Coordination Analysis 7. 01 5 6 7 8 9 10000 ® FAULT: D ATE: BY : Jan 31.6 .08 .02 .com® ELECTRICAL DESIGN GUIDE curve.09 .07 . Now look at 10 on the horizontal current scale.8 .1 .5 .04 .3 .5 . would not 225 A Breaker Set “Low” 1000 900 800 700 600 500 400 300 200 . 2007 R EVISION : 1 Jim Phillips. it will trip between 2.02 . A short CURRENT IN AMPERES X 10 AT 480 VOLTS circuit or overload ® T2G Technical EasyPower Brainfiller.4 .05 . it will trip instantaneously in about 1 cycle. Graph CURRENT IN AMPERES X 10 AT 480 VOLTS 2 illustrates the time current curve of an adjustable 225 Amp main breaker and a 20 Amp branch breaker.brainfiller. The instantaneous region of the breaker’s curve is represented by the long flat horizontal portion of the curve on the right side of the graph.09 . This indicates that if a current above 900 Amps flows through Instantaneous both the 20 Amp set “Low” and 225 Amp devices.7 . It actually represents 10 Amps X 10 or 100 Amps.7 .04 .E. the time current curves of all of the devices in a particular circuit must be drawn together.8 1 2 3 4 5 6 7 8 9 10 2 3 4 5 6 7 8 9 100 2 3 4 5 6 7 8 9 1000 2 3 4 100 90 80 70 60 50 40 30 20 40 30 20 5 6 7 8 9 10000 1000 900 800 700 600 500 400 300 200 100 90 80 70 60 50 TIME IN SECONDS 4 3 2 4 3 2 1 . If the breaker sees a larger current such as several thousand Amps.6 .5 .5 and 7 seconds.9 .05 . both might possibly trip.6 .8 1 2 3 4 5 6 7 8 9 10 2 3 4 5 6 7 8 9 100 2 3 4 5 6 7 8 9 1000 2 3 4 . T 2G Technical Training Group® Page 34 of 37 TIME IN SECONDS 10 9 8 7 6 5 10 9 8 7 6 5 . In this example.4 .03 .07 .6 .8 .03 . P.1 .06 .2 . This means that if the 20 Amp breaker carries 100 Amps.2 Selective Coordination To evaluate selective coordination between two or more devices. 7. the circuit breaker curve is intersected between 2. both curves begin to overlap at 900 Amps.9 . This indicates that Both curves completely both devices will overlap at 1500 Both curves very likely trip Amps begin to overlap at together for all 900 Amps currents above 1500 Amps.01 .06 .08 .3 .5 and 7 seconds.com TCC-1 Training Group above 1500 Amps TIME-CURRENT CURVES Graph 2 on the 20 Amp 20 A and 225 A Breakers circuit. The curves for both devices completely overlap for currents above 1500 Amps. Moving up the vertical scale on this line.2 1 . This means the breaker does not trip which is what you would expect when a 20 Amp breaker only carries 10 Amps.5 . 7 .6 . Graph 3 illustrates the preferred coordination between the 20 Amp circuit breaker and the main 225 Amp breaker.3 .03 . Although not perfect. faster settings are typically more sensitive and may provide better overall protection.9 .06 .01 5 6 7 8 9 10000 CURRENT IN AMPERES X 10 AT 480 VOLTS T2G Technical Training Group EasyPower ® TIME-CURRENT CURVES Brainfiller.com TCC-1 FAU LT: D ATE: BY : R EVISION: 1 Jan 31.8 1 2 3 4 5 6 7 8 9 10 2 3 4 5 6 7 8 9 100 2 3 4 5 6 7 8 9 1000 2 3 4 .2 1 .1 .5 . Changing the setting to “high” as illustrated in Graph 3. However. Graph 2 illustrates the instantaneous setting set “low” which means the overlap with the 20 Amp breaker occurs at a 900 Amps.8 .02 With the higher instantaneous.com® ELECTRICAL DESIGN GUIDE only trip the 20 Amp circuit.09 . moves the beginning of the overlap from 900 Amps to 1800 Amps. This setting change would increase coordination so the devices now coordinate with each other for currents up to 1800 Amps. P.8 1 2 3 4 5 6 7 8 9 10 2 3 4 5 6 7 8 9 100 2 3 4 5 6 7 8 9 1000 2 3 4 100 90 80 70 60 50 40 5 6 7 8 9 10000 1000 900 800 700 600 500 400 300 200 100 90 80 70 60 50 40 30 20 A comprehensive coordination study requires taking these basic concepts and applying them to other devices in the power distribution system. Coordination is improved somewhat with the CURRENT IN AMPERES X 10 AT 480 VOLTS higher setting since there is less overlap between the two curves.01 . 1000 900 800 700 600 500 400 300 200 . There will always be trade offs in determining the optimum settings.03 . Since the 225 Amp main breaker is adjustable.5 . Overlapping of the curves is an indication that both devices could trip together.05 .08 . it is possible to “move” it’s instantaneous trip setting which is defined by the vertical part of the curve.02 .2 .4 . 30 20 TIME IN SECONDS 4 3 2 4 3 2 1 . T 2G Technical Training Group® Page 35 of 37 TIME IN SECONDS 10 9 8 7 6 5 Instantaneous set “High” 10 9 8 7 6 5 .6 .5 .08 . 2007 ® Graph 3 20 A and 225 A Breaker 225 Amp Breaker Set “High” Jim Phillips.E. this increases the likelihood that only the 20 Amp breaker would respond to the fault. but the 225 Amp main would also trip removing the entire panel from service instead of just the individual 20 Amp circuit.1 .07 .3 .07 .05 .5 .4 .6 .brainfiller.04 . Slower settings for upstream devices often improve coordination with downstream devices.9 .7 .04 .09 .6 . now the overlap does not begin until 1800 Amps .8 .06 . E.brainfiller. T 2G Technical Training Group® Page 36 of 37 .. Quincy. P. Canton. Inc. Ohio ® National Electrical Code® and NEC® are registered trademarks of the National Fire Protection Association.com® REFERENCES AND TRADEMARKS ELECTRICAL DESIGN GUIDE T2G Technical Training Group and Brainfiller® are registered trademarks of Technical Training Group. MA Jim Phillips. !EEE 1584 Electrical Safety / Arc Flash . Jim consistently receives excellent reviews for his "unique" teaching style and ability to answer the tough questions in an easy to understand manner. He is nationally known for his power system analysis background and ability to analyze complex problems. P. He has also written for Consulting-Specifying Engineer Magazine and holds a BSEE from The Ohio State University.Brainfiller. Illumination Engineering Society. Inc.E. A Brainfiller® Publication www. • • • • • • • Technical Articles Continuing Education Courses Distance Learning On Site Training Training Seminars Scheduled Nationwide Work Sheets Training Videos Tap into Jim’s experience with any of these programs that he has developed: • • • • • • • • • • • • • How to Perform an Arc Flash Study . Jim has conducted more than 1600 live seminars for tens of thousands of people from the United States and around the globe. NFPA.com . He is a member of the IEEE 1584 working group “IEEE Guide for Performing Arc Fl h Hazard Calculations” and is a regular contributor for NEC Digest® the NFPA’s Official NEC® Magazine. He is also the founder of Phillips Engineers + Consultants. He is a licensed Professional Engineer in many states. Early in his career he worked for Ohio Edison Company and was a Project Engineer for Square D Company's Power System Analysis Group. He has taught classes at the college level and is a Senior Member of IEEE where he has been a distinguished lecturer.NFPA 70E Power Distribution Equipment Design of Electric Power Systems National Electrical Code ® Short Circuit Studies Protective Device Coordination Analysis Emergency and Standby Power Systems Protective Relaying Medium Voltage Power Systems Symmetrical Components and Per Unit Analysis Grounding and Power Quality Harmonic Analysis and Power Factor For over 25 years. and many other organizations. past member of the Energy Policy Committee. P.E. His vast experience in the electrical industry makes him a highly sought after speaker on every facet of electric power systems.brainfiller. JIM PHILLIPS.com Your source for training by Jim Phillips.