Delta amperage

[GoldDigger]

Going back for a moment to the simple resistive load delta case, you can either call it using vectors with very simple vectors or just using a formula, but if the current from A to B is x, and the current from B to C is y, then the line current in B is exactly (x+y)(sqrt3)/2.
I leave it as an exercise for the readers to extend this to other two line currents.

The more adept readers can then extend the formula to the case of equal displacement power factor loads and discover that the exact same formula applies when using the magnitude of the current, and amazingly the power factor of the line current is the same as the power factor of the line-to-line current.

Once you introduce unequal power factors for each load, you are finally in a situation where it is easier to just give up on formulas and start using vectors directly.

Regarding the spreadsheet, I suspect that the problem involves correctly entering the sign of the line-to-line currents. A proper formula in the spreadsheet should have taken care of that for you.

 

[david luchini]

Going back for a moment to the simple resistive load delta case, you can either call it using vectors with very simple vectors or just using a formula, but if the current from A to B is x, and the current from B to C is y, then the line current in B is exactly (x+y)(sqrt3)/2.

I do not think this is correct. For x=10 and y=5, the formula would yield 12.99

However, for x=10<0 and y=5<-120, Ib=y-x (using vector math) would yield 13.23<-160.89

 

[iceworm]

...if the current from A to B is x, and the current from B to C is y, then the line current in B is exactly (x+y)(sqrt3)/2. ...

I do not think this is correct. ..

Me neither. Took me a bit longer to see the line current is not linear with x and y. I'll pick an easier example than david.
for x = 1 and y = 0, (x+y)(sqrt3)/2 = .866. hummm, by inspection Line current in B must be 1.

I leave it as an exercise for the readers to extend this to other two line currents. ...

Okay, I'll pick:
|Ib| = sqrt(x^2 + y^2 - 2xycos(120)) = sqrt (x^2 + y^2 + xy)

mbrooke -
Assuming I didn't screw up the algebra (which I am capable of doing) - maybe this would work as a formula for the unbalanced load (resistive) phase currents.

ice

 

[david luchini]

|Ib| = sqrt(x^2 + y^2 - 2xycos(120)) = sqrt (x^2 + y^2 + xy)

This looks good to me. It appears that the 3phase calculator spreadsheet is using (x+y)(sqrt3)/2 to get incorrect line currents.

 

[Smart $]

First draft of a calculator for 480 and 208 wye... no 240/120.

Needs a thorough testing... :blink:

https://drive.google.com/file/d/0By0rzU3UuNh7dG5RMERnaWhWaVU/edit?usp=sharing

I don't know if it works with Google's spreadsheet app or Excel online. Recommend download and open with computer-based Excel.

 

[GoldDigger]

Me neither. Took me a bit longer to see the line current is not linear with x and y. I'll pick an easier example than david.
for x = 1 and y = 0, (x+y)(sqrt3)/2 = .866. hummm, by inspection Line current in B must be 1.



Okay, I'll pick:
|Ib| = sqrt(x^2 + y^2 - 2xycos(120)) = sqrt (x^2 + y^2 + xy)

mbrooke -
Assuming I didn't screw up the algebra (which I am capable of doing) - maybe this would work as a formula for the unbalanced load (resistive) phase currents.

ice

I concur.
I made the same foolish mistake that the author of the spreadsheet made, namely assuming that the current in the line would be at 120 degrees.
Instead it has to be at the angle of the vector sum/difference since the current has nowhere else to go.
Two major conclusions come from that:
1. Even with only resistive loads, some line currents will have a power factor less than 1 except when the loads are also balanced.
2. Although you can calculate the total power based on the line currents only, you need to know the PF on each line in order to do so.
Just the magnitudes are not enough unless you are able to use those magnitudes to calculate the corresponding vectors.
A major conceptual breakthrough, thank you!

Tapatalk!

 

[david luchini]

I concur.
I made the same foolish mistake that the author of the spreadsheet made, namely assuming that the current in the line would be at 120 degrees.
Instead it has to be at the angle of the vector sum/difference since the current has nowhere else to go.
Two major conclusions come from that:
1. Even with only resistive loads, some line currents will have a power factor less than 1 except when the loads are also balanced.
2. Although you can calculate the total power based on the line currents only, you need to know the PF on each line in order to do so.
Just the magnitudes are not enough unless you are able to use those magnitudes to calculate the corresponding vectors.
A major conceptual breakthrough, thank you!

Tapatalk!

I don't think its accurate to say that the "line currents" have a power factor less than 1. Power factor is the ratio of real power to the apparent power. I would think that the real power and apparent power are the same with a resistive load.

I know pf is the cos of the angle between the current and voltage, but for the line current A, what is the corresponding voltage? A-to-B, A-to-N? If A-to-N, what do you do in a delta?

 

[Smart $]

I don't think its accurate to say that the "line currents" have a power factor less than 1. Power factor is the ratio of real power to the apparent power. I would think that the real power and apparent power are the same with a resistive load.

I know pf is the cos of the angle between the current and voltage, but for the line current A, what is the corresponding voltage? A-to-B, A-to-N? If A-to-N, what do you do in a delta?

The voltage drop of the line conductor?

 

[GoldDigger]

The line current is the current in that line period. And the phase angle is the angle of that line current to the line to ground/neutral/center point voltage. Agreed it is clearer in the case of a wye than a delta.
But if you pick any arbitrary voltage reference point, the total power will be the sum of the signed power values calculated based on the vector dot product of the line current and that line's voltage from the common reference point.
That is why/how power meters work.

Tapatalk!

 

[mbrooke]

Yeah - that happens to me sometimes in power analysis. (color me minorly embarassed)



I think you picked a good one. The shortcomings don't look important. The sheet is protected so I can't see the algolrithm. I tried three different delta currents. The line currents came up fine. The sheet assumes the loads are resistive - no phase angle information.

The calculated power not so good. For the 480D selection, the power is not correct. I didn't look at the other selections - there is no 240D.

Since your main concern is the line currents, it should work for what you want. I think you got it.

ice

Well I tried it out and to be honest either I don't understand delta currents or it has to many shortcomings. For example, when I do an open delta or only one load across a leg Im seeing phase currents less than what they should be. For example if I add 100 amps only across A-B and not B-C or C-A Im getting 86.6 amps. Where do the other 15 amps go? Is this correct? Similar readings are confusing the heck out of me. (See pics)


And oh, it ok. Im not the best with terminology:ashamed1:.

It seems line some of the line current calculations come up wrong as well...Try entering 10, 0, 0 for L-L load currents...You get 8.66, 8.66 and 0 for line currents...On "split phase," two 10a l-n loads give a 20A neutral current.

It's a real nice interface though. Be great if the errors were corrected. I'd be nice to see the hidden part of the sheet.

I saw that, and I don't think its correct either but an expert would know better. I only say so because Ive closed 208 volt loads into a panel board one at a time while amp clamping getting the same readings on the breaker as well as the incoming 2 legs only when the load was connected across the 2 phases with no other loads on other legs.


Just tried the neutral current and its definitely wrong, its adding rather than subtracting.

 

[Smart $]

...its definitely wrong...

Hint... http://forums.mikeholt.com/showthread.php?t=161205&p=1562971#post1562971

You probably want 240/120... easier to modify the draft as a separate calculator. Prefer the draft tested before doing that. If there are any calculation errors, I'd only have to fix one rather than two.

 

[mbrooke]

Hint... http://forums.mikeholt.com/showthread.php?t=161205&p=1562971#post1562971

You probably want 240/120... easier to modify the draft as a separate calculator. Prefer the draft tested before doing that. If there are any calculation errors, I'd only have to fix one rather than two.

Your calculator appears ok thus far. Just has no straight delta calcs But I like where your going with it, its really really helpful!


The one that I think is wrong is this one: http://www.google.com/url?sa=t&rct=j...-DSeeNlpdhvH1A

 

[Smart $]

Just has no straight delta calcs

Just don't enter any line-to-neutral loads.

 

[david luchini]

For example if I add 100 amps only across A-B and not B-C or C-A Im getting 86.6 amps. Where do the other 15 amps go? Is this correct? Similar readings are confusing the heck out of me. (See pics)

It's not correct. With 100A only across A-B, you should see 100A on Line A and Line B

 

[iceworm]

The line current is the current in that line period. And the phase angle is the angle of that line current to the line to ground/neutral/center point voltage. Agreed it is clearer in the case of a wye than a delta.
But if you pick any arbitrary voltage reference point, the total power will be the sum of the signed power values calculated based on the vector dot product of the line current and that line's voltage from the common reference point.
That is why/how power meters work. ...

This is about how I see it. I sketched out a delta, balanced, 1.0pf diagram as I would expect if I connected a power analyzer, Vab, Vbc, Vca, Ia, Ib, Ic. I would expect the reference vector to be as shown on the delta, Van, where "n" is a calculated center point.

ice

 

[mbrooke]

It's not correct. With 100A only across A-B, you should see 100A on Line A and Line B

Good to know. That really was throwing me off. If the spread sheet had the correct computation it would be gold. :happyyes:

I guess I shouldn't be looking through Google then :happyno:

 

[mbrooke]

Ok digging through the internet I came across this. Ironically on a previous thread.


This is just what I was looking for! Great for frequent use rather than crunching out the numbers every time. However, can anyone confirm its accuracy?


Its an excel, however you can enter any phase amperage under the magnitude cells (grey) getting the leg amperage at the top:

https://docs.google.com/file/d/0BxN...tMDllODgyNWRmNjAy/edit?authkey=CNrTzt0F&hl=en#

 

[mbrooke]

Correct in function?

https://docs.google.com/file/d/0BxN...tMDllODgyNWRmNjAy/edit?authkey=CNrTzt0F&hl=en#

 

[iceworm]

Correct in function?

Virus protection says it's a malware site - so I can't tell you. What do you get when you plug in some of the test cases listed in this thread?

ice

 

[mbrooke]

Correct in function?

Virus protection says it's a malware site - so I can't tell you. What do you get when you plug in some of the test cases listed in this thread?

ice

Here is one screen shot: