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Joined: Dec 2004
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The "no pain, no gain" statement has come back to haunt me.......
I'm almost out of smokes, I'm getting snowed in, went through alot of scrap paper, and my fingers hurt from erasing so much, but I'm still thinking...
I hope to have these phase angles figured out by the end of this weekend, then I'll throw another example at you, if you guys don't mind, so I can double check myself...
Redsy, I may take you up on the book offer...
Dnk..
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Joined: Sep 2003
Posts: 650
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JBD,
I agree with you that there is only one phase angle for the current flowing in the wire.
To calculate the total current flow, you have to account for the phase angle differences in the summation.
I do agree with you that the result that you get by simply adding the single phase and three phase currents without considering phase angle will probably get you as close as calculating in the angles without knowing the power factor of the loads and thus the actual phase current angles.
-Jon
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Joined: Feb 2002
Posts: 182
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posted by JBD "You can not have a phase angle difference on a single conductor. Yes the current flowing in the C-B winding will be at a different phase angle from the current in the C-A winding, but there is still only 1 current in the C phase conductor."
That is correct. All load on winding AC are in phase. They are not in phase with AB and BC.
"Go ahead and work out the math for the summation of these two winding currents at the C conductor and then tell me if it is substantially different from my method. I don't think it is worth the effort for sizing a conductor and overcurrent protective device." If you add the phase load in AC to the single phase load you get 625 amps. Just adding phase AC to AB you get 625 + 333 = 958 amps. Same for BC. If load is contineous 958 x 1.25 = 1197. Doing it with vectors you get Ia = 835A Ib = 577A Ic = 890A 890 x 1.25 = 1112. The higher the single phase load the greater the error.
[This message has been edited by Bob (edited 02-12-2006).]
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Joined: Sep 2003
Posts: 650
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Please, lets not get caught up too much in this calculation; remember that all these calculations are _always_ approximations. We assume sinusoidal loads. We assume nominal voltage. We have been assuming unity power factor. We assume lots of things.
Lets try things by 'ignoring' phase angles. First we calculate the current for the three phase load, using VA/ V / root(3) A 'pure' 240kVa 240V three phase load is 577A/phase (approximately!) Of course, this calculation automatically includes the phase angle differences between the various three phase legs.
Then we calculate the single phase loads using VA/V A 34 KVA 120V single phase load is 283A A 41 KVA 120V single phase load is 342A
Then we just add the numbers up on the various legs. This, of course, neglects the phase angle difference between the single and three phase load, giving. A: 860A B 577A C 919A
This as compared to the calculation which includes the phase angle differences: A:835A B577A C 890A
(Bob and I appear to be getting exactly the same numbers using different calculation methods, which IMHO is a rather good check)
In other words, by ignoring the phase difference one gets about 3% difference in the calculated currents versus including the phase angles but assuming that the three phase load has a unity power factor.
Now, let us change our assumptions, and use a 0.8 lagging power factor for the 3 phase load. This is a lagging phase angle of about 37 degrees. If I put the new phase angles into the calculation, I get: A:919A B 577A C 771A
That is to say that our initial assumption of unity power factor for the three phase load leads to a much larger potential error than just adding currents up and ignoring the phase difference between the single and the three phase loads.
-Jon
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Joined: Jul 2001
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Jon, That's all I was trying to say. We are splitting hairs over how precise our inprecise guess is. Without acknowledging the assumptions that have been made.
I have no problem with people trying to learn the correct methods, but there is a time and a place for everything. I just felt this was the time to learn how to do " engineering based estimation" which is also a valuable tool.
[This message has been edited by JBD (edited 02-12-2006).]
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Joined: Oct 2000
Posts: 2,723 Likes: 1
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So, whaddaheck is the correct answer??? Ironically, I have been performing A LOT of Load Calcs + Panel Schedules -vs- Actual In-Field Amperage Tests (before and after the Calcs + Schedules were done), and so far things have been very accurate. A couple systems (4, to be exact) were 4 Wire "Closed" Delta (3 Transformers, middle one with center tap), and an additional two were 4 Wire "Open Delta Vee Connections" (2 Transformers, largest has center tap). Had several 3 Wire Deltas (2 Separately Derived Systems, 3 from the Utility guys), ranging from 240V 3Ø 3 Wire Corner Grounded Deltas, to 600V 3Ø 3 Wire Ungrounded Deltas. Most systems are 4 Wire Wye setups - either 208Y/120V or 480Y/277V. Anyhow, what's the verdict? Scott35
Scott " 35 " Thompson Just Say NO To Green Eggs And Ham!
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Joined: Sep 2003
Posts: 650
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Verdict? We don't need no steenkin' verdict. I think that the best that we can do is cover the possible different assumptions, and play with different calculation techniques. I was describing using vector addition of the various currents, but that requires making assumptions about the power factors of the loads. As JBD notes, a rough and ready estimate simply using addition and ignoring phase angles will give us a number that falls right into the range of possible values that you get with the "more accurate" vector calculations as you change the power factor assumptions. IMHO a very useful reminder that sometimes you don't really care if 86 angels or 94 angels can dance on a particular pin, and "about 90" is the correct answer to use -Jon
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Posts: 404
Joined: March 2007
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