# Sizing Circuit Breakers?

1. Oct 19, 2006

### vsdguy

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 comming 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

2. Oct 20, 2006

### nikola-tesla

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 comming 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?

3. Oct 20, 2006

### wolram

The cb protectes the conductors , i think the conductors should be sized at least 125% of motor flc.

4. Oct 20, 2006

### nikola-tesla

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%

5. Oct 22, 2006

### vsdguy

Hi,

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

****The current comming 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!

6. Oct 23, 2006

### vsdguy

Hi again,

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

Kind Regards,

vsdguy

7. Oct 4, 2011

### psparky

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.

8. Oct 4, 2011

### psparky

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.

Last edited: Oct 4, 2011
9. Oct 4, 2011

### jim hardy

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 dont 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.

you can pay a little more and get motors that are easy to start.

old jim

10. Oct 4, 2011

### psparky

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. Oct 5, 2011

### psparky

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. Oct 10, 2011

### psparky

Really, no one?

Take a stab at it if you don't know it. You will learn something.

13. Mar 27, 2012

### MG83

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

Last edited: Mar 27, 2012
14. Mar 27, 2012

### psparky

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?

Last edited: Mar 27, 2012
15. Jun 14, 2012

### bong_101984

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. Jun 14, 2012

### jegues

The kVA will stay the same, while the current will change by the ammount 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! :uhh:

17. Jun 14, 2012

### psparky

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. Jun 14, 2012

### jegues

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. Jun 14, 2012

### psparky

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. Jun 14, 2012

### jegues

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. Jun 14, 2012

### psparky

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 lets 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. Jun 14, 2012

### jegues

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. Jun 14, 2012

### psparky

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

Last edited: Jun 14, 2012
24. Jun 14, 2012

### jegues

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. Jun 14, 2012

### psparky

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 eachother 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.