Main Breaker Size

[jimmy_mcquade]

Hi everyone,

I am trying to learn industrial wiring and working with dad on a project he is
looking at trying to learn how he does things.
He has this project which is 208 three phase, with 12 motors.
The motors range from 1 1/2 hp to 10 hp.
The system when it is running will have all 12 motors on. To prevent a massive surge current, he plans to start each motor independent of each other on a time delay basis.

Table 430.250 tells me the amp draw of each motor.
Article 430.32 A(1) tells me how to size the overload protection.
I understand the timed delay starting to prevent a massive inrush current.

My question is this, how do you compute the main breaker size for these motors if the continuous run current of these motors is 136.6 amps and the total overcurrent protection is 170.75 amps?

Also, does this method of computing the breaker size also apply to other
voltages such as 220, 240, 440, 480 as well?

Thanks in advance,
Jimmy

 

[augie47]

You are embarking and a challenging and rewarding endeavor .
My opinion would be that you would size the "service" breaker as you would a feeder breaker.
Your main breaker would be sized by 430.62 (or 430.63 if other than motor loads are involved) which basically tells you to use use 430.52 to select the breaker for the largest motor and then add the sum of the FLA of the others.

 

[Hameedulla-Ekhlas]

Dear Friend,
please follow the below examples and hope to help u.


Example No. 1: Suppose you use a dual-element fuse for overload protection. What size fuse do you need for a 5-hp, 230V, single-phase motor with a service factor of 1.16 if the motor nameplate current rating is 28A?

(a) 25A
(c) 35A
(b) 30A
(d) 40A

The overload protection shall be sized according to the motor nameplate current rating [430.6(A), 430.32(A)(1), and 430.55].

You also have to consider another factor: nameplate temperature rise. For motors with a nameplate temperature rise rating not over 40?C, size the overload protection device no more than 125% of the motor nameplate current rating. Thus, 28A?1.25=35A [240.6(A)]

Example No. 2: Again, suppose you're using a dual-element fuse for the overload protection. What size fuse do you need for a 50-hp, 460V, 3-phase motor that has a temperature rise of 39?C and motor nameplate current rating of 60A (FLA)?

(a) 40A
(c) 60A
(b) 50A
(d) 70A

The overload protection is sized per the motor nameplate current rating, not the motor full load current (FLC) rating. Thus, 60A?1.25=75A. Overload protection shall not exceed 75A, so you need to use a 70A dual-element fuse [240.6(A) and 430.32(A)(1)].

Motors that don't have a service factor rating of 1.15 or higher or a temperature rise rating of 40?C and less must have an overload protection device sized at not more than 115% of the motor nameplate ampere rating (430.37).

Sizing branch-circuit conductors. Branch-circuit conductors that serve a single motor must have an ampacity of not less than 125% of the motor's FLC as listed in Tables 430.147 through 430.150 [430.6(A)]. You must select the conductor size from Table 310.16 according to the terminal temperature rating (60?C or 75?C) of the equipment [110.14(C)]. Let's reinforce this concept by working through a sample calculation


Example No. 3: What size THHN conductor do you need for a 2-hp, 230V, single-phase motor?

(a) 14 AWG
(c) 10 AWG
(b) 12 AWG
(d) 8 AWG

Let's walk through the solution:

Step 1: Conductor sized no less than 125% of motor FLC

Step 2: Table 430.148 shows the FLC of 2-hp, 230V, single-phase as 12A

Step 3: 12A ? 1.25 = 15A

Step 4: Per Table 310.16, you need to use 14 AWG THHN rated 20A at 60?C

The minimum size conductor the NEC permits for building wiring is 14 AWG [310.5]. However, local codes and many industrial facilities have requirements that 12 AWG be used as the smallest branch-circuit wire. So in this example you might need to use 12 AWG instead of 14 AWG


Branch-circuit protection for short-circuits and ground-faults. Branch-circuit short-circuit and ground-fault protection devices protect the motor, motor control apparatus, and conductors against short circuits or ground faults. They don't protect against an overload (430.51)

The short-circuit and ground-fault protection device required for motor circuits isn't the type required for personnel (210.8), feeders (215.9 and 240.13), services (230.95), or temporary wiring for receptacles (527.6).

Per 430.52(C), you must size the short-circuit and ground-fault protection for the motor branch circuit ? except those that serve torque motors ? so they're no greater than the percentages listed in Table 430.52.

When the short-circuit and ground-fault protection device value that you find in Table 430.52 doesn't correspond to the standard rating or setting of overcurrent protection devices as listed in 240.6(A), use the next higher protection device size [430.52(C)(1) Ex. 1].

Did that statement stop you? Does it strike you as incorrect? That's a common response, but remember, motors are different than other system components. Motor overload protection devices, such as heaters and fuses, protect the motor and other items from overload. The short-circuit and ground-fault protection doesn't need to perform this function. Therefore, oversizing won't compromise protection. Undersizing will prevent the motor from starting.

Use the following two-step process to determine what percentage from Table 430.52 you should use to size the motor branch-circuit short-circuit ground-fault protection device.

Step 1: Locate the motor type on Table 430.52.

Step 2: Select the percentage from Table 430.52 according to the type of protection device, such as non-time delay (one-time), dual-element fuse, or inverse-time circuit breaker. Don't forget to use the next higher protection device size when necessary.

Let's see if you have this concept down with a short quiz. Of the following statements, which one is true? Use Table 430.52 to look up the numbers.


1.The branch-circuit short-circuit protection (non-time delay fuse) for a 3-hp, 115V, single-phase motor shall not exceed 110A.

2.The branch-circuit short-circuit protection (dual-element fuse) for a 5-hp, 230V, single-phase motor shall not exceed 50A.

3.The branch-circuit short-circuit protection (inverse-time breaker) for a 25-hp, 460V, 3-phase synchronous motor shall not exceed 70A.


Let's address each question individually. We'll be referring to 430.53(C)(1) Ex. 1 and Table 430.52.


1.Per Table 430.148, 34A?3.00=102A. The next size up is 110A. So this is true.

2.Per Table 430.148, 28A?1.75=49A. The next size up is 50A. So, this is also true.

3.Per Table 430.150, 26A?2.50=65A. The next size up is 70A. This is also true.


Remember the following important principles:


?You must size the conductors at 125% of the motor FLC [430.22(A)].

?You must size the overloads no more than 115% to 125% of the motor nameplate current rating, depending on the conditions [430.32(A)(1)].

?You must size the short-circuit ground-fault protection device from 150% to 300% of the motor FLC [Table 430.52].


If you put all three of these together, you can see the branch-circuit conductor ampacity (125%) and the short-circuit ground-fault protection device (150% to 300%) aren't related.

This final example should help you see if you've been paying attention


Example No. 4: Are any of the following statements true for a 1-hp, 120V motor, nameplate current rating of 14A? Refer to Fig. 6.

(a) The branch-circuit conductors can be 14 AWG THHN.

(b) Overload protection is from 16.1A.

(c) Short-circuit and ground-fault protection is permitted to be a 40A circuit breaker.

(d) All of these are true.

Walking through each of these, you can see:

(a) The conductors are sized per 430.22(A): 16A?1.25=20A; Table 310.16 requires 14 AWG at 60?C.

(b) Per 430.32(A)(1), overload protection is sized as follows: 14A (nameplate)?1.15=16.1A.

(c) Short-circuit and ground-fault protection is determined based on 430.52(C)(1): 16A?2.50=40A circuit breaker.

Therefore all three statements are true.

The 16A overload protection device protects the 14 AWG conductors from overcurrent, while the 40A short-circuit protection device protects them from short circuits. This example illustrates the sometimes confusing fact that when you're doing motor calculations, you're actually calculating overcurrent and short-circuit protection separately.


Branch-circuit conductors and protection devices. Per 430.6(A), branch-circuit conductors to a single motor must have an ampacity of not less than 125% of the motor full load current (FLC) as listed in Tables 430.147 through 430.150. To illustrate this, let's size the branch-circuit conductors (THHN) and short-circuit ground-fault protection device for a 3-hp, 115V, single-phase motor. The motor FLA is 31A, and dual-element fuses for short-circuit and ground-fault protection are in use


?Per Table 430.148, the FLC current is 34A.

?34A?125%=43A.

?Per Table 310.16 (60?C terminals [110.14(C)(1)(a)]), the conductor must be a 6 AWG THHN rated 55A.

It continues please follow the part-2



best regards,
Hameedullah Ekhlas
Electrical Engineer
Kabul Afghanistan

 

[Hameedulla-Ekhlas]

Part-2


Fig. 1. Don?t make the mistake of using a motor?s FLA nameplate rating when using the short-circuit and ground-fault protection devices. You must use the FLC rating given in Table 430.148.
Per the motor FLC listed in Table 430.52, size the branch-circuit short-circuit and ground-fault protection devices by using multiplication factors based on the type of motor and protection device. When the protection device values determined from Table 430.52 don't correspond with the standard rating of overcurrent protection devices listed in 240.6(A), you must use the next higher overcurrent protection device. To illustrate this, let's use the same motor as in the previous example.


?Per 240.6(A), multiply 34A?175%
?You need a 60A dual-element fuse.

To explore this example further, see Example No. D8 in Annex D of the 2002 NEC. Once you've sized the motor overloads, branch-circuit conductors, and branch-circuit protective devices, you're ready to move on to the next step.

Motor feeder conductor calculations. From 430.24, you can see that conductors that supply several motors must have an ampacity not less than:


?125% of the highest-rated motor FLC [430.17], plus

?The sum of the FLCs of the other motors (on the same phase), as determined by 430.6(A), plus

?The ampacity required to supply the other loads on that feeder.



Motor feeder conductors shall be sized not less than 125% of the largest motor FLC plus the sum of the FLCs of the other motors on the same phase.
Use Fig. 2 and solve the following problem.

Example No. 1. For what ampacity must you size the feeder conductor if it supplies the following two motors? The terminals are rated for 75?C.


?One 7.5-hp, 230V (40A), single-phase motor

?One 5-hp, 230V (28A), single-phase motor


(a) 50A
(b) 60A
(c) 70A
(d) 80A

Let's walk through the solution.


?The largest motor is 40A.

?40A?1.25+28A=78A.

?80A is the closest selection that's at least 78A.


What size conductor would give us this ampacity?

(a) 2 AWG
(b) 4 AWG
(c) 6 AWG
(d) 8 AWG

Per Table 310.16, a 6 AWG conductor rated at 75?C provides 65A of ampacity, so it's too small. However, a 4 AWG conductor provides 85A of ampacity, which will accommodate the necessary 78A. Therefore, you need to size this feeder conductor at 4 AWG.

Next, we have to determine what size overcurrent protection device (OCPD) we must provide for a given feeder.


Fig. 3. To size overcurrent protection devices for each feeder, start by determining the ampacities required for each motor and move on from there.
Example No. 2. Using a slightly more complex example, try sizing the feeder conductor (THHN) and protection device (inverse-time breakers, 75?C terminal rating) for the following motors


?Three 1-hp, 120V, single-phase motors
?Three 5-hp, 208V, single-phase motors
?One wound-rotor, 15-hp, 208V, 3-phase motor

Refer to 240.6(A), 430.52(C)(1), Table 430.148, and Table 430.52. Start by determining the ampacities required for each size of motor, then walk through each step until you arrive at the correct OCPD size.


?1-hp motor: FLC is 16A.
16A?250%=40A

?5-hp motor: FLC is 30.8A.
30.8A?250%=77A (Next size up is 80A.)

?15-hp motor: FLC is 46.2A.


Each motor?s FLC will come into play when sizing the conductor.
46.2A?150% (wound-rotor) 569A (Next size up is 70A.)

Now, let's look at the feeder conductor. Conductors that supply several motors must have an ampacity of not less than 125% of the highest-rated motor FLC (430.17), plus the sum of the other motor FLCs [430.6(A)] on the same phase (Fig. 4).

Continuing with this example, add up all the ampacities, multiplying the highest rated motor by 125%. Thus:


?(46.2A?1.25)+30.8A+30.8A+16A=136A.


Table 310.16 shows you need 1/0 AWG THHN because at 150A it's the smallest conductor that accommodates the 136A of ampacity we're working with. When sizing the feeder conductor, be sure to include only the motors that are on the same phase. For that reason, these calculations only involve four motors.

You must provide the feeder with a protective device with a rating or setting not greater than the largest rating or setting of the branch-circuit short-circuit and ground-fault protective device (plus the sum of the full-load currents of the other motors of the group) [430.62(A)]. Remember, motor feeder conductors must be protected against the overcurrent that results from short circuits and ground faults but not those that result from motor overload.

When sizing the feeder protection, be sure to include only the motors that are on the same phase.


In this example, the largest branch-circuit fuse or circuit breaker allowed for Motor 1 is 70A.
Refer to Fig. 5 for this sample motor feeder protection calculation.

Example No. 3. What size feeder protection (inverse-time breaker) do you need for the following two motors?


?5-hp, 230V, single-phase motor
?3-hp, 230V, single-phase motor

(a) 30A breaker
(b) 40A breaker
(c) 50A breaker
(d) 80A breaker

Let's walk through the solution.

Step 1: Get the motor FLC from Table 430.148.


?A 5-hp motor FLC is 28A.
?A 3-hp motor FLC is 17A.

Step 2: Size the branch-circuit protection per the requirements of 430.52(C)(1), Table 430.52, and 240.6(A)


?5-hp: 28A?2.5=70A

?3-hp: 17A?2.5=42.5A (Next size up is 45A.)


Step 3: Size the feeder conductor per 430.24(A).


?The largest motor is 28A.

?(28A?1.25)+17A=52A

?Table 310.16 shows 6 AWG rated 55A at 60?C as the smallest conductor with sufficient ampacity.


Step 4: Size the feeder protection per 430.62.


?It must not be greater than the 70A protection of the branch circuit plus the 17A of the other motor, which is the total of all loads on that feeder.

?70A+17A=87A


Choose the next size down, which is 80A.

How can you be safe if you're selecting the next size down instead of the next size up? Remember, you've already accounted for all the loads, and the NEC requires that you not exceed the protection of the branch circuit. Again, keep in mind that you aren't calculating for motor overload protection. Motor calculations are different from other calculations. With motor feeders, you're calculating for protection from short circuits and ground faults, only ? not overload.

Putting it all together. Motor calculations get confusing if you forget there's a division of responsibility in the protective devices. To get your calculations right, you must separately calculate the motor overload protection (typically near the motor), branch-circuit protection (from short circuits and ground faults), and feeder-circuit protection (from short circuits and ground faults). Remember that overload protection is only at the motor.

Any time you find yourself confused, just refer to NEC Figure 430.1. It shows the division of responsibility between different forms of protection in motor circuits. Example D8 in Annex D of the 2002 NEC illustrates this with actual numbers. Keeping this division of responsibility in mind will allow you to make correct motor calculations every time.


best regards,
Hameedullah Ekhlas
Electrical Engineer
Kabul Afghanistan

 

[jimmy_mcquade]

Augie47,
Thanks for your reply. But i must admit that i am still confused.

So if my 12 motors consisted of the following: (2) 7 1/2 hp motors, (3) 1 hp motors, (3) 1 1/2 hp motors, (2) 3 hp motors, and (2) 5 hp motors i would do this.

The largest motor is 7 1/2 hp which is 24.2 amps * 1.25 = 30.25 amps.
the sum of the other motors is 112.4 amps. adding them up gives us 142.5
amps.

The next bresker size is 150 amps (i think) and that is the size i would pick.

But aren't all breakers rated at 80%?
For example a 20 amp breaker is rated at 16 amps, 16* 125% = 20 amps.

So wouldn't i select a breaker that would be rated at 142.5 * 1.25 = 178 amps. The next available size is 200 amps.

Thanks,
Jimmy

 

[Hameedulla-Ekhlas]

no,no what you are pointing it is regarding breaker size for continues loads but calculation for motors are completely different.
Please follow the examples step by step and then you will find all your questions in this.
the largest motor*2.5 + the other motors current.

Hopes this will help you.

best regards,
Hameedullah Ekhlas
Electrical Engineer
Kabul Afghanistan

 

[augie47]

as noted, you are sizing Ground-Fault/Short Circuit Protection (assuming you are not using the breaker as overload protection), so your multilpier is based on 430.52 (in most cases 250%) + the sum of the FLA of the other motors.
This is covered in 430.62 & 430.63

 

[ceb58]

Hameedulla, you need to come here to teach

 

[dana1028]

Excellent

Excellent

Hameedullah - great examples and effort to explain an often confusing subject.

Thank you for taking the time and effort to provide all of us some great information.

 

[DARUSA]

I will say use a 175 Amps Inverse Time Breaker then size your OL protection for each motor based on the Nameplate FLA

 

[augie47]

Unless I made a math mistake, I came up with 172.9. Unless the load was continuous, you don't need to consider your 80%.
Next standard size is 175, next "normal" is 200.
Keep in mind 175 would be a minimum, but that does not consider any load diversity. I would think it would doubtful that all would run at the same time.

 

[RETRAINDAILY]

Hi everyone,

I am trying to learn industrial wiring and working with dad on a project he is
looking at trying to learn how he does things.
He has this project which is 208 three phase, with 12 motors.
The motors range from 1 1/2 hp to 10 hp.
The system when it is running will have all 12 motors on. To prevent a massive surge current, he plans to start each motor independent of each other on a time delay basis.

Table 430.250 tells me the amp draw of each motor.
Article 430.32 A(1) tells me how to size the overload protection.
I understand the timed delay starting to prevent a massive inrush current.

My question is this, how do you compute the main breaker size for these motors if the continuous run current of these motors is 136.6 amps and the total over current protection is 170.75 amps?

Also, does this method of computing the breaker size also apply to other
voltages such as 220, 240, 440, 480 as well?

Thanks in advance,
Jimmy

WHAT SHE SAID .... NAME PLATE
THEN ITS 125% OF THE LARGEST MOTOR PLUS THE SUM OF ALL OTHER MOTOR AND LOADS. ME I NEVER GO WITH MIN because THEN I ALWAYS SEEM TO BE LOOKING FOR MORE. if I am tring to figure site unseen I always go better.

 

[jimmy_mcquade]

Thanks for the replies.

Augie47,
I got with dad and he said that all motors will be running at the same time
once the system is started. This is an assembly line with various pieces of
equipment from start to finish.

So, My question still is, Will you have to consider the 80% rule for the circuit breaker. if my calculations are correct, we will be running 136.6 amps and a 175 amp breaker at 80% is 140 amps. This is very close. I just don't know.

But if the amps were 150, would the 80% rule apply?

regards,
jimmy

 

[dana1028]

Thanks for the replies.

So, My question still is, Will you have to consider the 80% rule for the circuit breaker. if my calculations are correct, we will be running 136.6 amps and a 175 amp breaker at 80% is 140 amps. This is very close. I just don't know.

But if the amps were 150, would the 80% rule apply?

regards,
jimmy

The 80% rule always applies to continuous loads. [Now that I've said that, someone will prove me wrong about the 'always' comment].

 

[augie47]

Interesting, when you look at conductors for a single motor [430.22(A)]
the Code speaks of 125% of the FLA per [430.6(A)], for a group of motors, 430.24 only requires the 125% of the largest motor.
So it appears, as far as conductor sizing, the 125% of the largest motor takes care of the "continuous" rating and no additional multipliers be applied.
Likewise, using the procedures set forth in 430.62 for feeder SCGF protection, there appears to be no need for any multiplier.

Applying 430.24 and 430.62, my figures would be a minimum conductor rated at 142.5, and a minimum breaker of 175 amps (next largest above 172.9). I don't see where any other multipliers need be applied.

additional comments are more than welcome

 

[bob]

Augie47,
Thanks for your reply. But i must admit that i am still confused.

So if my 12 motors consisted of the following: (2) 7 1/2 hp motors, (3) 1 hp motors, (3) 1 1/2 hp motors, (2) 3 hp motors, and (2) 5 hp motors i would do this.

The largest motor is 7 1/2 hp which is 24.2 amps * 1.25 = 30.25 amps.
the sum of the other motors is 112.4 amps. adding them up gives us 142.5
amps.

The next bresker size is 150 amps (i think) and that is the size i would pick.

But aren't all breakers rated at 80%?
For example a 20 amp breaker is rated at 16 amps, 16* 125% = 20 amps.

So wouldn't i select a breaker that would be rated at 142.5 * 1.25 = 178 amps. The next available size is 200 amps.

Thanks,
Jimmy

From Table 430.250
7.5 hp FLA = 24.2, 5 hp FLA = 16.7, 3 hp FLA = 10.6, 1 1/2 hp FLA = 6.6
1 hp FLA = 4.6

From table 430.52 Max Inverse Breaker = 250% of FLA
From 420.62 The Max feeder Breaker =
2.5 x 24.2 + 24.2 + 2 x 16.7 + 2 x 10.6 + 3 x 6.6 + 3 x 4.6 = 173 amps
or 175 amps breaker.

 

[wawireguy]

Our friend from Afghanistan has it correct. A good place for you to start would be Mike Holts NEC Exam prep(inexpensive one : ). I don't believe it goes over the exact scenario you are trying to build but it does go over the basics of sizing OCP and conductors for multiple motors. Get down the basics then scrutinize the NEC to see if it matters that you won't be starting all the motors at the same time.

 

[skeshesh]

Interesting, when you look at conductors for a single motor [430.22(A)]
the Code speaks of 125% of the FLA per [430.6(A)], for a group of motors, 430.24 only requires the 125% of the largest motor.
So it appears, as far as conductor sizing, the 125% of the largest motor takes care of the "continuous" rating and no additional multipliers be applied.
Likewise, using the procedures set forth in 430.62 for feeder SCGF protection, there appears to be no need for any multiplier.

Applying 430.24 and 430.62, my figures would be a minimum conductor rated at 142.5, and a minimum breaker of 175 amps (next largest above 172.9). I don't see where any other multipliers need be applied.

additional comments are more than welcome

The conductor sizing rule for several motors (430.24) assumes that two or more motors will not be starterd simultaneously. Obviously under this assumption a 125% of the largest motor FLA is the biggest load the conductors can experience. I would say if two or more machines are to start simultaneously, which is not the case in this situatino as described by OP, the sizing of the conductor would have to account for the in-rush created by the start-up of more than one motor. Furthermore, the protective devices setting may also need to be adjusted to avoid nuisance tripping. I've been involved in a few projects for larger scale facilities where a coordinated/selective protection study was done using SKM, ETAP, or similar software to determine the most effective protective scheme.

BTW, I'm convinced Hameedullah is using us as a practice audience for a machine theory book he's working on