What is the Formula for Sizing Circuit Breakers?

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Discussion Overview

The discussion revolves around the sizing of circuit breakers for a motor application, specifically addressing the protection against inrush currents in a 415VAC, 50Hz system. Participants explore various formulas and rules of thumb for determining appropriate circuit breaker ratings based on motor specifications and load conditions.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant inquires about a formula for sizing circuit breakers to protect against inrush currents, providing specific motor parameters.
  • Another participant suggests a circuit breaker rating of 40 Amps based on the motor's full load current, noting that inrush may increase the maximum amperage to 45 Amps.
  • Some participants propose that conductors should be sized at least 125% of the motor's full load current (FLC) for protection.
  • There is a suggestion that circuit breakers should be sized at 2.5 times the full load current to accommodate inrush currents, with a recommendation for a 100 Amp breaker in this context.
  • Concerns are raised about the adequacy of a 60 Amp main distribution panel for starting the motor, with suggestions that a higher rating may be necessary.
  • One participant emphasizes the importance of considering the inductive nature of motors and the complexities involved in sizing breakers compared to resistive loads.
  • Several participants engage in a mathematical exploration of power factor correction and its effects on motor current and KVA, with varying interpretations of the implications for circuit breaker sizing.

Areas of Agreement / Disagreement

Participants express differing views on the appropriate sizing of circuit breakers, with no consensus reached on a single formula or method. Some advocate for the 125% rule, while others argue for a higher multiplier based on inrush current considerations.

Contextual Notes

Participants reference various assumptions about motor characteristics, inrush currents, and the relationship between circuit breaker ratings and load conditions. There are unresolved mathematical steps and differing interpretations of how power factor correction affects current and KVA calculations.

Who May Find This Useful

This discussion may be useful for electrical engineers, technicians, and students interested in motor control, circuit protection, and electrical system design.

vsdguy
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Hi all,

I am curious to know if anyone can provide me with a formula or a rule
that sizes the circuit breaker so the circuit is protected during
inrush currents. The supply is 415VAC, 50Hz. The current coming from
the buildings main distribution panel is around 60 Amps. The
parameters on the motor nameplate are as followed:
415VAC
FLA=37.5 amps
Freq=50Hz
RPM=980


The VFD is a 40 HP AC drive.


Any help would be appreciated?


Kind Regards,


vsdguy
 
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Since I am a girl and as you can see the most brave one to be the first one to tackle your question I would say the breaker should be rated 40Amps bcs you say that nameplate specs are 37.5 Amps. If there is a high starting momentum then max Amperage is increased to 45 A. Your wire gauge= wire size should be #8. Hope it helps, or if someone knows more please help us.

*****One more thing; you say Distribution Panel and current coming is 60 Amps. What do you mean by that?
Let me analyse it; 60 Amps on the panel is disconnect rating of 60 Amps, or did you measure with the Amprobe and the reading is 60 A?
 
The cb protectes the conductors , i think the conductors should be sized at least 125% of motor flc.
 
wolram said:
The cb protectes the conductors , i think the conductors should be sized at least 125% of motor flc.

Yes CB protects conductors but the main protection is for the appliance and/or whatever is connected to those conductors and CBs.

Choosing CB we have to read manufacturers specs and if specs on the namplate are 37.5 Amps that would be the main starting factor in deciding CB size, IMO.

But, you right for 125%
 
Hi,

Thank you for your responses.
The current CB was already sized to 40amps. I think the 125% of FLC would suit for any CB rating.

****The current coming in the building is 60 amps which means that that 60 amps is split into 40 amps for the three phase and the remaining is 20 amps for lights, aircon, heat, recepticles, alarm system).

Sorry for not including details for the main supply.

Cherrio!
 
Hi again,

Just wanted to say thank you again Nikola-Tesla for your help along with Wolram.

Kind Regards,

vsdguy
 
Hmmmm...I'm not sure I agree with above answers.

Motors have a huge inrush of current essentially because the motor is just one long wire with very little resistance...until the magnetic field builds up.

Generally speaking...you need to size your circuit breaker 2.5 times the full load current.

That being said...a 100 amp circuit breaker with #2 wire would be most appropriate.

This is the general rule of thumb unless the nameplate or specs ...tell you the exact circuit breaker size.

The 125% rule is pretty much doesn't apply in this case since the breaker I am talking about is 250% of full load current.

More than likely the 40 amp circuit breaker that was mentioned above failed over and over on start up.
 
Also, you say that the main distribution panel is 60 amps...not enough.

You would need 100 amps just to start the motor alone.

If you are saying that 20 amps is for resistive elements...thats ok...use your 125% rule there...125 amp main breaker should work for whole system.

Ironically I just sized a 30 HP VFD with 38 full load amps for a AHU motor on a single line. 100 amp circuit breaker with #2 wire with a 80 amp fuse.

Psparky, P.E.
 
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and go to your breaker catalog for time-current curves. Motor will pull starting current until it accelerates the load to near synchronous speed.. you don't want nuisance trips.

On motor nameplate is a code letter that tells starting KVA, which is several times running .
That's why Sparky's number seems so high at first.

Google "motor kva code"
you can pay a little more and get motors that are easy to start.


old jim
 
  • #10
Ahhh...nice to see some sense in this thread. Although the younger engineers above are clearly stating what they learned in school and in early years of engineering.

Motors are a different flavor. The inductive nature of them makes them much trickier. The basic rules you know for lights and receptacles no longer apply.

KVA, power factor, reactive power, inrush current...and the use of starters ...are pieces to the puzzle.

Here's a question for the younger engineers.

Say a motor has a 10 KVA and full load current of 100 amps...with a power factor of .7. If you correct to power factor to .95 with a capacitor in parallel...

What is the new current and KVA of the motor?
 
  • #11
The above question should actually read...

Say you have a three phase 10 KVA motor with a full load current of 100 amps...with a power factor of .7. If you correct to power factor to .95 with a capacitor in parallel...

What is the new current and KVA of the motor?
 
  • #12
Really, no one?

Take a stab at it if you don't know it. You will learn something.
 
  • #13
FLA = kVA/[sqrt(3)*kV*pf]

sub in FLA = 100, kVA = 10, pf = 0.7

This yields your reference voltage, which is .0825kV, or 82.5V.

Now, change pf to 0.95. Your kVA will remain the same.

Solve for FLA. I got FLA = 74A
 
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  • #14
MG83 said:
FLA = kVA/[sqrt(3)*kV*pf]

sub in FLA = 100, kVA = 10, pf = 0.7

This yields your reference voltage, which is .0825kV, or 82.5V.

Now, change pf to 0.95. Your kVA will remain the same.

Solve for FLA. I got FLA = 74A

You are right about the KVA staying the same...however you missed on the full load amps. Don't feel bad...most people miss this question.

Nice try...try again or ask questions.

Draw a picture of a voltage source in parallel with a motor.
Now draw another picture of a voltage source in parallel with a capacitor...and a motor.
Does your amp calcluation still make sense?
 
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  • #15
Dear VSDguy,

the solution on your problem is this

40 amps x 125% of the FLA since it is a motor so you get 40x1.25 = 50 amp.
this is including the starting torque of the motor
 
  • #16
psparky said:
The above question should actually read...

Say you have a three phase 10 KVA motor with a full load current of 100 amps...with a power factor of .7. If you correct to power factor to .95 with a capacitor in parallel...

What is the new current and KVA of the motor?

The kVA will stay the same, while the current will change by the amount of amps required by the caps to provide the additional reactive power.

The initial reactive power,

Q_{i} = 10sin(arccos(0.7)) \approx 7kVAR

The final reactive power,

Q_{f} = 10sin(arccos(0.95)) \approx 3.1kVAR

Now,

Q_{cap} = Q_{f} - Q_{i} = \frac{V_{c}^{2}}{X_{c}}

Vc should be known since you should be able to solve for the voltage from the initial case and the voltage will remain the same whether or not the capacitors are present. Once you know Xc you know Ic.

The new PF will provide you with the new power (more than previous) given to the motor.

With this one can find the current flowing into the motor. (The capacitors aren't receiving any of this power)

The summation of these two currents will be the new full load current.

I am hesitant to run the actual numbers for everything as I am worried about forgetting those troublesome √(3)'s everywhere.

Hopefully my theory/understanding is in check! :rolleyes:
 
  • #17
The new PF will provide you with the new power (more than previous) given to the motor.

Almost...you are 99% correct...

This statement is not correct. The motor has no clue the capacitor is there and receives the same voltage, current and KVA as it did before the capacitor. But like you said...the power plant is now sending different current and less reactive power.
 
  • #18
psparky said:
Almost...you are 99% correct...

This statement is not correct. The motor has no clue the capacitor is there and receives the same voltage, current and KVA as it did before the capacitor. But like you said...the power plant is now sending different current and less reactive power.

The only reason I think the power has changed is because of the relationship shown within the power triangle.

Initially,

P_{i} = S cos(\theta_{i})

Finally,

P_{f} = S cos(\theta_{f})

If S is to remain the same, it must be such that,

P_{i} \neq P_{f}

How can this be true if you claim that the motor is receiving the same KVA, voltage and current?

Am I missing something?
 
  • #19
jegues said:
The only reason I think the power has changed is because of the relationship shown within the power triangle.

Initially,

P_{i} = S cos(\theta_{i})

Finally,

P_{f} = S cos(\theta_{f})

If S is to remain the same, it must be such that,

P_{i} \neq P_{f}

How can this be true if you claim that the motor is receiving the same KVA, voltage and current?

Am I missing something?

In simplest terms...we basically have a voltage source in parallel with a cap and motor.

What was the voltage across the motor before the cap?

What is the voltage across the motor after the cap?

The voltage across the motor before and after is obviously 480 volts. The load of the motor has not changed...

How can the current and KVA change in regards to the motor?

The current and reactive power coming from the source is now much different...but...the motor doesn't have a clue this is happening...think about it.
 
  • #20
psparky said:
In simplest terms...we basically have a voltage source in parallel with a cap and motor.

What was the voltage across the motor before the cap?

What is the voltage across the motor after the cap?

The voltage across the motor before and after is obviously 480 volts. The load of the motor has not changed...

How can the current and KVA change in regards to the motor?

The current and reactive power coming from the source is now much different...but...the motor doesn't have a clue this is happening...think about it.

I agree with what your saying and I believe the intuition is correct, but my problem is that I should see such agreement within my calculations.

Where is the discrepancy?
 
  • #21
jegues said:
I agree with what your saying and I believe the intuition is correct, but my problem is that I should see such agreement within my calculations.

Where is the discrepancy?

The current thru the motor has a lagging current vector...let's say -20 degrees. (not exact, but close enough)

Your capacitor has a current vector of +90 degrees pointing straight up.

The power company sees the combination of those two vectors turning the new single vector to let's say -5 degrees (not exact...but close enough) with a smaller magnitude.

All that being said...your motor still has a lagging current vector at -20 degrees.

The current and reactive power has now changed from the source and thru the capacitor...but not thru the motor.

I'm not sure how to fix your calcs...but if you understand what is happening...you should be able to adjust.

Here's another way to think of it...change the voltage across a motor and it doesn't like that. Change the current thru a motor and it doesn't like that. Change the KVA for a motor and it doesn't like that! Motors hate change!
 
  • #22
psparky said:
The current thru the motor has a lagging current vector...let's say -20 degrees. (not exact, but close enough)

Your capacitor has a current vector of +90 degrees pointing straight up.

The power company sees the combination of those two vectors turning the new single vector to let's say -5 degrees (not exact...but close enough) with a smaller magnitude.

All that being said...your motor still has a lagging current vector at -20 degrees.

The current and reactive power has now changed from the source and thru the capacitor...but not thru the motor.

I'm not sure how to fix your calcs...but if you understand what is happening...you should be able to adjust.

Here's another way to think of it...change the voltage across a motor and it doesn't like that. Change the current thru a motor and it doesn't like that. Change the KVA for a motor and it doesn't like that! Motors hate change!

As mentioned previously the intuition behind what's going on is clear, I just want to find the discrepancy within my calculations, or perhaps there is some sort of implied assumption within my work.
 
  • #23
I don't know how to help u with your calcs...sorry.

The power has definitely changed...just not thru the motor.
The power triangle of the motor has not changed. However, the power triangle of the entire system has changed.

Sleep on it...you'll get it.

Perhaps if you do actually do the calculations of the current thru the source before cap...and after cap. You will obviously have less current and lower power factor with the addition of the cap thru the source.

Forget about the square root of 3 if it scares out...just figure it at single phase.
Although three phase power is simple P = I*V*√3
 
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  • #24
psparky said:
I don't know how to help u with your calcs...sorry.

The power has definitely changed...just not thru the motor.
The power triangle of the motor has not changed. However, the power triangle of the entire system has changed.

Sleep on it...you'll get it.

If the power has definitely changed, can't we agree that the only element that can cause such a change is the motor?

Nothing else in our circuit has any relevance to power. (REAL POWER)
 
  • #25
jegues said:
If the power has definitely changed, can't we agree that the only element that can cause such a change is the motor?

Nothing else in our circuit has any relevance to power. (REAL POWER)

Motor cannot make this change. Motor hasn't changed.

I don't think real power has changed. Only reactive power has changed.
Picture the vector of the reactive power of the motor...small vector pointing straight up. Picture the vector of the reactive power of the capacitor...pointing straight down. Those two nearly cross each other out...All we are doing is making less reactive power...just not thru the motor.

Lower reactive power in this case makes for lower current and less full load amps from the power company.
 
  • #26
psparky said:
Motor cannot make this change. Motor hasn't changed.

I don't think real power has changed. Only reactive power has changed.

Lower reactive power in this case makes for lower current and less full load amps.

But not thru the motor...lol.

If you are telling me the reactive power has changed, then through the power triangle this implies either a change in complex power or real power does it not?
 
  • #27
jegues said:
If you are telling me the reactive power has changed, then through the power triangle this implies either a change in complex power or real power does it not?

Yes...complex power has changed. Real power has not.

Complex power and real power has stayed the same in motor...

See my post above...
 
  • #28
Let's say the system is running...source, cap and motor.

Let's say i remove the motor and leave the cap running.

The current and reactive power thru the cap is the same...however...the power company now sees a much higher reactive power compared to the entire system running.
 
  • #29
psparky said:
Yes...complex power has changed. Real power has not.

Complex power and real power has stayed the same in motor...

See my post above...

I see my mistake now(I hope), I was associating the complex power of the entire circuit to be the complex power of the motor.

These two are not the same.
 
  • #30
jegues said:
I see my mistake now(I hope), I was associating the complex power of the entire circuit to be the complex power of the motor.

These two are not the same.

Yatzee! That's the beauty of PF correction...it doesn't affect the motor in any way...just your utility bill!

It's tricky to get...took me a while too.

Like I said above...do all the calcs with current and voltage vectors...graph them out...
Graph out the current and voltage sin waves...
Calculate current and reactive power in motor before and after cap...
Calculate current and reacitve power thru source before and after cap...

Once you got it...you got it and will never forget.
 

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