Thanks, you'll need to bear with me I'm slow lol (ask wifey)
load 1 i is 10/0
wouldn't it be the same for rtn to B with sign convention changed?
120/208 3-ph, 4W panelboard amps on A phase?
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Thanks, you'll need to bear with me I'm slow lol (ask wifey)
load 1 i is 10/0
wouldn't it be the same for rtn to B with sign convention changed?
Looking at it a bit more, I'm not sure my earlier assessment is correct. You are using a different angle convention than I'm used to. I know it's in there somewhere... :lol:
on the crux I agree with you
the kva method is suitable for this type of work
in fact if the currents are drawn up (always draw the picture echoes in my head from my old skool profs lol) and simply added are close enough for this application
for normal pf range
avg of vector/kva
a/b/c/n
29/17-19/10/10
simply summing
30/20/10/10
thanks for working with me
Perhaps the problem is your polar line 3. Currents should be at 30 (same as L-N load), 150, -90 respectively.
that changes A to 27.3 at 0 deg
I think the phase relationships are correct
at the top I offset all by +30 deg to normalize
so ph is 0/120/-120
neut is 30/150/-90
perplexed??
Verified...
try this
1 10/30 -10/0 ok
2 10/60 ok ok
3 10/30 ok ok
2 and 3 A must offset by 30
1 B only inverts sign not phase
1 and 3 A must be in phase
Feel like I'm having to learn a new language here..
hmy:Don't forget your reference voltage, e.g. VAB is VBA at other end, so same magnitude, opposite direction is technically more correct IMO. Its 10/0 at A end, 10/180 at B end... though -10/0 works just as well.
2 and 3 do not offset by 30°. L-N current is at same angle as 3Ø current... only L-L is shifted 30° (assuming all at same power factor).
Feel like I'm having to learn a new language here..
Don't forget your reference voltage, e.g. VAB is VBA at other end, so same magnitude, opposite direction is technically more correct IMO. Its 10/0 at A end, 10/180 at B end... though -10/0 works just as well.
2 and 3 do not offset by 30°. L-N current is at same angle as 3Ø current... only L-L is shifted 30° (assuming all at same power factor).
Ph does not change only reference direction when summing currents at each ph/node
1A and 1B are the same current, only the KCL sign has changed
2 is a neut voltage
3 is a ph voltage
as you pointed out in a previous post they are offset by 30 deg (not 60 as was asserted by another)
1 and 3 A must be in phase since they are both ph A-B neut V summed
I'm not going to argue about it. 10/180 is identical to -10/0. Anything beyond that is semantics.
The currents of L-N and 3Ø purely resistive loads are in phase. Those currents have a 30° phase shift from phase voltage. IOW, they match the phasing of line voltage.
Put a 10A L-N load on each leg. Add a 10A 3Ø load. Current on each leg is 20A. This would not be so if currents were not in phase with each other on each leg.
Yes
but it needs to be 120 deg offset from B 3
since it is derived from A and B 3 from ph B
your offset is 30
in addition you need to negate it when summing since is the same ref as ia
A 2 and A 3 can't be the same phase
A 2 = 120/30 / 12 = 10/30
A 3 = 208/0 / 20.8 = 10/0
the neut V is offset from the ph-ph by 30 as you noted
A 1 = 208/0 / 20.8 = 10/0
A 3 = the same or 208/0 / 20.8
even if your Vab basis is +30 (30/150/-90) they will be the same
Please humor me, no arguement
1 ok -10/0 ok
2 ok ok ok
3 10/0 10/120 10/-120
this puts 2A l-n offset 30 from ph a
makes 1B (derived from a) and 3B 120 offset
puts 1A and 3A in phase since both are ph-ph and derived from the same Vln
I think I'm failing Ingenieur Language 101... :lol:
I didn't publish my IN for this 10A circuits × 3 project. In fact, I got 10.0A@-180° using angle convention in diagram.
![image_5.jpeg](http://[URL="http://s39.photobucket.com/user/artm3/media/Mobile%20Uploads/image_5.jpeg.html"][IMG]http://i39.photobucket.com/albums/e156/artm3/Mobile%20Uploads/image_5.jpeg)
I think we may both be wrong lol, me more than you
I'm having a problem uploading
eg on my calcs I only took 1 current from the delta
ia = load 1 i + load 2 i + load 3 i
but load 3 i is the sum of 2 delta leg i's
I only accounted for one...knucklehead lol
your i neut should be the sum of your ia+ ib + ic
I summed them and that is what I got 10/30
see if I can upload
I don't care who is right/wrong
I want to know where I erred
oops
the initial conditions stated line current was 10 ie the sum of the delta legs
not the delta legs
goofball
move on, nothing to see here
my first calc stands
someone please review and punch a hole in it...please lol
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JasonCo
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Is this what I would have to learn as an engineer? Right now I'm in a 4 year electrical apprentice school, but you guys right now are talking a foreign language to me haha, it seams like fascinating stuff though. Are you guys engineers? How does one obtain the knowledge to have such an in debt conversation like this, just curious!
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Yes these are the basics
electricians and engineers work as a team if there is to be any chance of a successful project
not an engineer but I did stay at the holiday inn last night lol
j/k I'm an engineer much to my wifes consternation
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As I keep trying to tell you, line current of three phase load's A aligns with line current of line-to-neutral load. B and C line currents shift 120° to each side respectively
I3A,3B,3C = 10A@30°, @150°, @-90°
A=8.66+5j
B=-8.66+5.00j
C=0-10j
mivey
Senior Member
Pardon me if this is answered later but I started here.OK
I believe you
where did I screw up lol
excuse the scribble
IA does not equal I1 + I2 + I3A+I3C as you pictured. It is I1 + I2 + I3A-I3C. I quit looking after that.
Negative signs are to engineers as pimples are to teenagers.
Edit: typo
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mivey
Senior Member
Pardon me if this is answered later but I started here.
IA does not equal I1 + I2 + I3A+I3C as you pictured. It is I1 + I2 + I3A-I3C. I quit looking after that.
Negative signs are to engineers as pimples are to teenagers.
Edit: typo
It also appears you left out -ICA (IAC) in the IA sum.
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