208/1 and 480/1 load calcs?

[Smart $]

I don't think you gave enough info.

Are we to assume circuits 9-11 and 13-15 also utilize a neutral?

No. Sorry for not specifying. I automatically assume there to be no neutral on 2-pole circuits UON.

If not I want to know how one side of a two pole breaker draws more than the other?

Through the breaker and branch circuit itself the current is 16A. When the current "arrives" at the bus bars it meets up with out-of-phase currents which have both an adding and cancelling effect. This is similar to how line-neutral currents balance out where they meet up with out-of-phase neutral currents. However, with phase conductors the angles of line-line loads aren't 120? out-of-phase with the source angle. They are ?30? out-of-phase... so only a partial "cancellation" occurs. Perhaps this is an oversimplification of the issue, but I find it suffices for practical application.

I must say I understand very little of how this occurs, but I do get a better understanding every time I run across reliable information. Haven't been pressed enough to actually buy a book that elaborates on the issue, though.

 

[Smart $]

Smart$'s table is IMHO confusing, because it appears to list the net increase in current flow caused by each load, skipping the intermediate step of showing the phase angles involved.

If one understands the concepts involved, the necessary info is in the table. The table itself was not offered as an explanation, rather only as the result.

For leg A, I get 62.4A with a phase angle of 7 degrees. (unity power factor would be 0 degrees)

For leg C, I get 46.55A with a phase angle of 230 degrees. (I'm rushing the math on this, and may have reversed a sign on my A and C phase angles.)

I don't understand how you are getting a reduction on the current for A and C feeder conductors. Is there a simple explanation? As I see it currently, the line-neutral loads do not provide any means for reduction and neither does the balanced three-phase circuit. I am of the impression all imbalances are carried on the feeder back to the transformer/generator...

 

[winnie]

Smart $,

As you note, the current flowing to the 208V single phase loads is 30 degrees out of phase with the 120V loads. On the 'B' phase in your example, you are adding 32A at 120 degrees, 16A at 90 degrees, and 16A at 150 degrees.

The current on the other side of the single phase load must be the same, except for polarity; so the 16A at 150 degree load on the B phase _must_ be matched by a 16A at 330 degree load on the 'A' phase.

On the 'A' phase you are adding 48A at 0 degrees and 16A at 330 degrees, thus a slight bit of cancelation.

-Jon

 

[kingpb]

Using symmetrical components:

View attachment 119

View attachment 120

View attachment 121

View attachment 122

View attachment 123

 

[rattus]

King, where do I place my Amprobe to measure 6.2A?

 

[iwire]

If one understands the concepts involved, the necessary info is in the table. The table itself was not offered as an explanation, rather only as the result.

I don't think the table is a good illustration.

IMO those two pole circuits should both be shown as 16 amps.

The result should only show up at the bottom of the table.

Where the breaker meets the bus it is still 16 amps per phase.

Once we get to the bus bar feed lugs that is where we should see the difference caused by the phase angles.

JMHO, Bob

BTW I readily admit I really do not understand this phase angle calculation stuff.

 

[Johnmcca]

"BTW I readily admit I really do not understand this phase angle calculation stuff" Iwire

Bob you're not alone on that point!

Why can't the powers that be reduce some of these concepts to a level that can be understood by the average electrician? (Or in Iwire's case much better than average.)

 

[steve66]

Kingpb:

You're still using three phase equations for what is really a single phase problem. Look at this sketch and please try to figure out where anything other than 10.7 amps or 0 amps would flow:

http://i49.photobucket.com/albums/f252/Sragan/1PHASEPANEL.jpg


Steve

 

[kingpb]

Kingpb:

You're still using three phase equations for what is really a single phase problem. Look at this sketch and please try to figure out where anything other than 10.7 amps or 0 amps would flow:

http://i49.photobucket.com/albums/f252/Sragan/1PHASEPANEL.jpg


Steve

Steve, I am using symmetrical components for unsymmetrical phasors. The calculations I presented can be verified by looking at Elements of Power System Analysis, William D Stevenson, Jr, 4th edition, and Handbook of Electric Power Calculations, H. Wayne Beaty, Third Edition. Other then by giving two industry standard publications to use as a go by, I don't know how to present the material any other way.

The confusion may be occuring because I believe a distinction has to be made between current across the load vs. the power delivered by the system. There are meters and protective relays that measure positive, negative, and zero sequence voltages and currents. I don't believe a standard amprobe would suffice, don't know if any of the fancy Flukes would have the capability or not.

 

[rattus]

King,

I can't see how this answers the original question which was, I think, "What are the secondary feeder currents in this unbalanced system?"

 

[rattus]

Don't Feel Bad:

Don't Feel Bad:

"BTW I readily admit I really do not understand this phase angle calculation stuff" Iwire

Bob you're not alone on that point!

Why can't the powers that be reduce some of these concepts to a level that can be understood by the average electrician? (Or in Iwire's case much better than average.)

Not to worry, the average engineer doesn't understand it either, and almost no one understands symmetrical components.

 

[kingpb]

King,

I can't see how this answers the original question which was, I think, "What are the secondary feeder currents in this unbalanced system?"

I agree, this topic went to left field.

My opinion, is that I would size the "Feeders and Panel" based on 3-phase balanced system.

As far as branch loads, the panel schedule sounds like it is based on balanced, which is not necessarily accurate, but it would provide an additional margin.

Use an "adjusted load KW" upward by using a 0.867 pf to get an adjusted KVA. this of course is based on splitting the load equally, but uses the 10.7A instead 9.2A. Again, additional margin on the panel.

 

[LarryFine]

Ifailed an inspection on a residential job because the in inspector (in the city of san jose CA) claims that #14 wire and #12 wire are not supposed to be in the same box; #12 romex on 20a breaker with #12 going in and out of switch boxes I used #14 romex as switchlegs Idon't see anything wrong with installation I've been doing that for years I need help!!!!!!!!!!!!!!!!!!!1

You DO need help. The inspector is correct, but he explained it poorly. There's no problem with different ratings' conductors in the same box. What you're doing wrong is using #14 wire on 20a breakers, even for switch legs for low-current loads.

That you've been doing it for a long time is a bad thing, not a good thing, because it means you've done it many times.

 

[LarryFine]

Ifailed an inspection on a residential job because the in inspector (in the city of san jose CA) claims that #14 wire and #12 wire are not supposed to be in the same box; #12 romex on 20a breaker with #12 going in and out of switch boxes I used #14 romex as switchlegs Idon't see anything wrong with installation I've been doing that for years I need help!!!!!!!!!!!!!!!!!!!1

You DO need help. The inspector is correct, but he explained it poorly. There's no problem with different ratings' conductors in the same box. What you're doing wrong is using #14 wire on 20a breakers, even for switch legs for low-current loads.

That you've been doing it for a long time is a bad thing, not a good thing, because it means you've done it many times.

 

[infinity]

Calibay,

Welcome to the forum. First off what does your question have to do with the topic of this thread? You could have started your own thread so this one doesn't go off on a tangent. If you're having trouble starting a thread just ask and we will help you. But since you're a neophyte here we can answer your question. The inspector is correct about not having #14 switch legs on the 20 amp circuit, but incorrect about them being within the same box. The #14 switch leg on the 20 amp circuit does not comply with the tap rules in Article 240.

 

[Smart $]

...

In my mind if I had to locate a 'place' for this change I would say it happens in the bus bars between the breakers, not at the point where the breaker hits the bus.

...

I believe you are partially correct... the same amount that I am partially incorrect Portions of out-of-phase currents opposite polarity. During these portions—highlighted in yellow in image below—the change occurs between breaker poles. At least that is my current (pun intended) assessment ...until shown otherwise

View attachment 129

PS: Is there some reason why in trying to respond to post #80, my response keeps getting inserted as post #76...???

 

[Smart $]

Smart $,

As you note, the current flowing to the 208V single phase loads is 30 degrees out of phase with the 120V loads. On the 'B' phase in your example, you are adding 32A at 120 degrees, 16A at 90 degrees, and 16A at 150 degrees.

The current on the other side of the single phase load must be the same, except for polarity; so the 16A at 150 degree load on the B phase _must_ be matched by a 16A at 330 degree load on the 'A' phase.

On the 'A' phase you are adding 48A at 0 degrees and 16A at 330 degrees, thus a slight bit of cancelation.

-Jon

Got it!

Had a hard time getting it in my head. It took plotting out three currents (image below) and adding the waveforms to realize how this happens...

View attachment 125

 

[Smart $]

I don't think the table is a good illustration.

Perhaps it is not a "good" illustration... but I wasn't implying that it be a new industry standard

Where the breaker meets the bus it is still 16 amps per phase.

Where breaker poles meet the bus is where the change occurs.

Once we get to the bus bar feed lugs that is where we should see the difference caused by the phase angles.

Perhaps so, but I believe there would need to be some indication as to how the adjustment was made. Without a means by which to check the math?easily? how can one determine the validity of the value stated as an adjusted current?

 

[sceepe]

Thank you guys for your reply. My next question would be to confirm an error in the excel file I'm using. This panel schedule has me insert the amperage for phase instead of watts per phase into each pole. For the total connected VA, it's adding the total Amps per phase and mulitplying it by the Line to Neutral voltage. This method will not produce the wattage of 2228W when using 10.7Amps per phase of the two pole breaker. I'm trying to think of a formula to use that will produce the correct VA, but can't think of a simple formula to use. Is there a simple formula that I'm over looking?

It would be so much easier if the panel used watts instead of amps for load values.

Thanks

Throw away your panel schedule program its garbage. You need to work in watts when doing panel schedules. Send me a private message with your email and I will send you a spreadsheet that works.

 

[iwire]

Where breaker poles meet the bus is where the change occurs.

Well we are arguing about something very strange here.

We just look at this differently.

In my mind if I had to locate a 'place' for this change I would say it happens in the bus bars between the breakers, not at the point where the breaker hits the bus.

I imagine a real fun time for you engineering types would be to figure the current in each section of bus between each breaker pole.