Sizing Circuit Breakers?


by vsdguy
Tags: breakers, circuit, sizing
psparky
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#19
Jun14-12, 10:01 AM
P: 659
Quote Quote by jegues View Post
The only reason I think the power has changed is because of the relationship shown within the power triangle.

Initially,

[tex]P_{i} = S cos(\theta_{i})[/tex]

Finally,

[tex]P_{f} = S cos(\theta_{f})[/tex]

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

[tex]P_{i} \neq P_{f}[/tex]

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.
jegues
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#20
Jun14-12, 10:06 AM
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Quote Quote by psparky View Post
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?
psparky
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#21
Jun14-12, 10:29 AM
P: 659
Quote Quote by jegues View Post
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 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!
jegues
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#22
Jun14-12, 10:37 AM
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Quote Quote by psparky View Post
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!
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.
psparky
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#23
Jun14-12, 11:19 AM
P: 659
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
jegues
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#24
Jun14-12, 11:39 AM
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Quote Quote by psparky View Post
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)
psparky
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#25
Jun14-12, 11:45 AM
P: 659
Quote Quote by jegues View Post
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 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.
jegues
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#26
Jun14-12, 11:49 AM
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Quote Quote by psparky View Post
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?
psparky
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#27
Jun14-12, 11:50 AM
P: 659
Quote Quote by jegues View Post
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.........
psparky
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#28
Jun14-12, 11:55 AM
P: 659
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.
jegues
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#29
Jun14-12, 11:55 AM
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Quote Quote by psparky View Post
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.
psparky
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#30
Jun14-12, 11:59 AM
P: 659
Quote Quote by jegues View Post
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.
plumbersean
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#31
Jun4-13, 05:11 PM
P: 1
Although I agree that inductive loads have huge inrush, he states that this is running off of a drive that will severely limit inrush to the motor. Most panels I service with VFD use 1.25 X FLA and Part wind and across the line use 1.5 X FLA generally. I work in a lot of control panels for water/wastewater applications from fraction HP up to 250 HP motors. Just my 2 cents
Windadct
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#32
Jun5-13, 05:02 PM
P: 532
Hmmmm-- if you are using a 40HP VFD which no one seems to have mentioned / 30KW -- that will limit the starting current and provide the protection to the motor. The sizing of the feeder breaker should be > 40A and suited for the conductors. So I would use min of 50A and cable accordingly.

Depending on the electrical code in your locality - you may have an issue with the feeder being too small - for example you may not be allowed to have a 50A or 60 A breaker down stream from another 60 A Breaker.
hisham.i
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#33
Jun9-13, 08:19 AM
P: 177
Quote Quote by psparky View Post
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?
With PF = 0.7 -> Power = 7KW
& P= sqrt(3)*I*V , with I = 100 A, solve for V-> V = 7000/(sqrt(3)*I) -> V = 40V!!!!!!!
The given in not correct, if it is a 3 phase motor -> V = 380V then I will not 100A for 10KVA motor!

+ We go very far away from the question! it is regarding CB sizing.


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