120/208 3-ph, 4W panelboard amps on A phase?

Ingenieur · Mar 14, 2016

Smart $ said:
This is where you are wrong. I(s) for load 3 (the 3Ø load) must be shifted 30°. Line current of a resistive 3Ø load is in phase with the line-to-neutral voltage. Perhaps you'll figure that out someday.

nope
l-l v is offset by 30 deg from l-g v
since Z is only R that holds true for i

Ingenieur · Mar 14, 2016

mivey said:
If you want 10 amp line currents then that is a different circuit. How far back to the post actually describing the circuit you are talking about?

Post 38 was done with line currents
our results match for
A
C
N

differ for B
the reason is obvious

Ingenieur · Mar 14, 2016

Smart $ said:
Don't need to draw the picture and I already did the hand part (via computer) back in post #45...

...using your angle convention. It agrees with my calculator, correcting for difference in angle convention.

That is incorrect
load 1 and 3 phase a must be in phase
just plug this in
load 1 ph a 10/30
load 2 ph a 10/60
load 1 ph b -10/30 or 10/210
take you 30 seconds

mivey · Mar 14, 2016

Ingenieur said:
That is incorrect

The final answer is correct. He was giving line currents. But if you work it the other way you get the same result:

Van = 120.000 at 30.000d
Vbn = 120.000 at 150.000d
Vcn = 120.000 at -90.000d

Vab = 207.846 at 0.000d
Vbc = 207.846 at 120.000d
Vca = 207.846 at -120.000d

I_L1ab = 10.000 at 0.000d
I_L2an = 10.000 at 30.000d
I_L3ab = 5.774 at 0.000d
I_L3bc = 5.774 at 120.000d
I_L3ca = 5.774 at -120.000d

Ia = 29.093 at 20.104d
Ib = 19.319 at 165.000d
Ic = 10.000 at -90.000d
In = 10.000 at 30.000d

mivey · Mar 14, 2016

mivey said:
Y'all must have been talking about two different things.

Ingenieur's circuit:

Van = 120.000 at 30.000d
Vbn = 120.000 at 150.000d
Vcn = 120.000 at -90.000d

Vab = 207.846 at 0.000d
Vbc = 207.846 at 120.000d
Vca = 207.846 at -120.000d

I_L1ab = 10.000 at 0.000d
I_L2an = 10.000 at 30.000d
I_L3ab = 10.000 at 0.000d
I_L3bc = 10.000 at 120.000d
I_L3ca = 10.000 at -120.000d

Ia = 36.327 at 22.089d
Ib = 26.458 at 160.893d
Ic = 17.321 at -90.000d
In = 10.000 at 30.000d

noting that:
I3a = 17.321 at 30.000d
I3b = 17.321 at 150.000d
I3c = 17.321 at -90.000d

Smart $'s circuit:

Van = 120.000 at 30.000d
Vbn = 120.000 at 150.000d
Vcn = 120.000 at -90.000d

Vab = 207.846 at 0.000d
Vbc = 207.846 at 120.000d
Vca = 207.846 at -120.000d

I_L1ab = 10.000 at 0.000d
I_L2an = 10.000 at 30.000d
I_L3ab = 5.774 at 0.000d
I_L3bc = 5.774 at 120.000d
I_L3ca = 5.774 at -120.000d

Ia = 29.093 at 20.104d
Ib = 19.319 at 165.000d
Ic = 10.000 at -90.000d
In = 10.000 at 30.000d

noting that:
I3a = 10.000 at 30.000d
I3b = 10.000 at 150.000d
I3c = 10.000 at -90.000d

Smart $ · Mar 15, 2016

Ingenieur said:
nope
l-l v is offset by 30 deg from l-g v
since Z is only R that holds true for i

True, but 3Ø i_PHASE currents combine at nodes: one at +30° and one at –30°. Net effect 1.732i@0°_Vln.

Ingenieur · Mar 15, 2016

I guess that clears that up lol

lauraj · Mar 16, 2016

I dreamed all night about how 10A + 10A + 10A = 29 or 36A depending on how you look at it. I have an EE degree but it's been at least 10 years since I've looked any phase shift problems, and I haven't missed them!

From a pure NEC service calculation perspective I believe the answer would be 30A. We don't take phase shift into account.

mivey · Mar 16, 2016

lauraj said:
I dreamed all night about how 10A + 10A + 10A = 29 or 36A depending on how you look at it. I have an EE degree but it's been at least 10 years since I've looked any phase shift problems, and I haven't missed them!

From a pure NEC service calculation perspective I believe the answer would be 30A. We don't take phase shift into account.

Well young Jedi, consider not the phase shift and know not do you if it is 10 amps or 17 amps. The difference in the two circuits it was.

Smart $ · Mar 16, 2016

mivey said:
Well young Jedi, consider not the phase shift and know not do you if it is 10 amps or 17 amps. The difference in the two circuits it was.

Did what say you?

Ingenieur · Mar 16, 2016

a/b/c/n
vector
29/19/10/10

just summation w/o regard to phase angles (or power factor)
30/20/10/10

depending on line i being 10 on the 3 ph rather than on each delta leg

Ingenieur · Mar 16, 2016

Smart $ said:
Did what say you?

line vs delta at being 10 amps?

Smart $ · Mar 16, 2016

lauraj said:
I dreamed all night about how 10A + 10A + 10A = 29 or 36A depending on how you look at it. I have an EE degree but it's been at least 10 years since I've looked any phase shift problems, and I haven't missed them!

From a pure NEC service calculation perspective I believe the answer would be 30A. We don't take phase shift into account.

Jedi speak not...

Typical service calculation would use the kVA method mentioned earlier. On a 208/120 system, Line A would have the greatest draw, and that would be
(10A×208÷2)+(10A×120)+(10A×208×sqrt(3)÷3)=
1040VA+1200VA+1200VA=3440VA
3440VA÷120V=28.7A

Sahib · Mar 17, 2016

Smart $ said:
(10A×208÷2)+(10A×120)+(10A×208×sqrt(3)÷3)=
1040VA+1200VA+1200VA=3440VA
3440VA÷120V=28.7A

What if you take individual KVA, find current and sum up arithmetically?

Sahib · Mar 17, 2016

Yes, the result will be the same. But it would be equal or greater than the vectorial sum of the same currents. Isn't it?

Smart $ · Mar 17, 2016

Sahib said:
What if you take individual KVA, find current and sum up arithmetically?

That is individual kVA... divided per line... how it's done on a panel schedule calculation for unbalance.

Sahib said:
Yes, the result will be the same. But it would be equal or greater than the vectorial sum of the same currents. Isn't it?

Actually it's less than the vectorial sum.

Sahib · Mar 17, 2016

Smart $ said:
Actually it's less than the vectorial sum.

I do not see how it can be. Will you please demonstrate it?

Smart $ · Mar 17, 2016

Sahib said:
I do not see how it can be. Will you please demonstrate it?

I gave both the kVA method and vectorial sum magnitudes back in post #35:
http://forums.mikeholt.com/showthread.php?t=175384&p=1722247#post1722247

Because you doubt, I leave the proof or disproof as an exercise for you. FWIW, most is demonstrated here already.

Sahib · Mar 17, 2016

Smart $ said:
I gave both the kVA method and vectorial sum magnitudes back in post #35:
http://forums.mikeholt.com/showthread.php?t=175384&p=1722247#post1722247

Because you doubt, I leave the proof or disproof as an exercise for you. FWIW, most is demonstrated here already.

No. You have not demonstrated it because

GoldDigger said:
the magnitude of the vector sum will always be less than or equal to the sum of the magnitudes.

Smart $ · Mar 17, 2016

Sahib said:
No. You have not demonstrated it because

GoldDigger said:

the magnitude of the vector sum will always be less than or equal to the sum of the magnitudes.

Click to expand...

That's not what I said in post #96 (http://forums.mikeholt.com/showthread.php?t=175384&p=1723539#post1723539). I said the kVA method results in a value less than the vectorial sum. Here's the resultant magnitudes of the three methods discussed in this thread. If you disagree with any of the values, please demonstrate your proof.

A=30A, B=20A, C=10A Arithmetic sum of individual load currents

A=29.1A, B=19.3A, C=10.0A Vectorial sum of load currents

A=28.7A, B=18.7A, C=10.0A Individual load kVA summing method

120/208 3-ph, 4W panelboard amps on A phase?

Senior Member

Senior Member

Senior Member

Senior Member

Senior Member

Esteemed Member

Senior Member

Senior Member

Senior Member

Esteemed Member

Senior Member

Senior Member

Esteemed Member

Senior Member

Senior Member

Esteemed Member

Senior Member

Esteemed Member

Senior Member

Esteemed Member