208/1 and 480/1 load calcs?

[rattus]

You guys are making this way too complicated. The simple fact is that a line to line load of 2228W or 2228VA will draw 10.7A at 208V. This current adds vectorially to any other phase currents, and I do not believe there is a fits-all formula for doing this. Why? Because such formulae are applicable only to balanced systems which this is not. To obtain an exact solution, we must express all currents as phasors and add them vectorially. Of course in practice, this may not be practical, so we approximate.

 

[ramsy]

But a _single_ phase load connected to a _three_ phase panel _cannot_ be a balanced load. The assumptions do not fit the example, and therefor the equations do not fit the example.

I found this in the 2002 NEC.

Minimum Size Feeder Required for Each Dwelling Unit (see 215.2)
For 120/208-V, 3-wire system (without ranges),
Net computed load of 3882 VA ? 2 legs ? 120 V/leg = 16.2 A

Net computed load of 2228 VA ? 2 legs ? 120 V/leg = 9.28 A (same formula as NEC)

In the NEC example above, the dwelling is served by two phase legs and a neutral 208v 3-phase svc.. It appears the the NEC prefers 208v 2-leg feeders sized by these same balanced assumptions and formula.

If it's accurate and a code compliant way to size each conductor, and the OCD's, why shouldn't we apply this same trick to motor loads --after adjusting for motor-power factors first-- (VA/PF)?

 

[steve66]

This scenario has stumped many an engineer, designer, and electrician. Some repsonses were getting close, and I thought Steve and Winnie were going to nail it, my hats off to their effort.

I think Winnie and Rattus and I have nailed it. (In particular, Winnie showed where I went wrong, and explained how 9.2 amps for a balanced panel can become 10.7 amps for an unbalanced panel).

I'll give the answer, then explain were it comes from - For a 2228W load connected across 2 phases of a 3 phase 208Y/120V panelboard (and no other load connected) the amount of current that will be seen on the Phase A, and Phase B feeder is: drum roll............................................ 6.2A.

Did you mean 9.2 amps? If so, you never answered my question about a single phase 208V source. It has to be the same 10.7 amps. If ETAP say's its 9.2 amps for a single phase load, it's wrong too. And if you really meant 6.2 amps, you are even farther off. How are you going to get 2228 watts with 208V and 6.2 amps?

Symetrical components aren't needed, and they don't provide any help for this problem. It's one voltage and one current in phase. There isn't even a neutral. Why make such an easy problem so hard?

Steve

 

[rattus]

I must agree with Steve. I don't remember studying symmetrical components although they are covered in my AC Circuits book. The comment is made however, that this method is more cumbersome than phasor analysis.

And, there is a single loop carrying a current of 10.7A everywhere in that loop. At least I have been led to believe that. No amount of analysis can change that fact.

 

[websparky]

The simple math method

The simple math method

I'll give the answer, then explain were it comes from - For a 2228W load connected across 2 phases of a 3 phase 208Y/120V panelboard (and no other load connected) the amount of current that will be seen on the Phase A, and Phase B feeder is: drum roll............................................ 6.2A.

I couldn't agree more!
2228 / (208 x 1.732) = 6.18A

Now, for the crux of the issue, if you were to put another 1114W load on Phase C, thereby blalancing the system, the current on each of the feeders goes to 9.3A. Again, I can confirm this using ETAP.

I couldn't agree more!
2228 + 1114 = 3342

3342 / (208 x 1.732) = 9.28A

 

[infinity]

I couldn't agree more!
2228 / (208 x 1.732) = 6.18A



I couldn't agree more!
2228 + 1114 = 3342

3342 / (208 x 1.732) = 9.28A

So if the supply came from a single phase 240/208 autotransformer fed from a 120/240 volt 3 wire system the current across the two conductors supplying 208 volts wouldn't be 10.7 amps? (2228/208=10.7)

 

[websparky]

Trevor,

When we have a single phase system, 120/240V, the calculations are straight forward.

2228va / 240 = 9.28A


However, I have never heard of a 240/208 transformer.

 

[websparky]

Trevor,

OK, now I get it.
If there were such a system, yes.

The other calculations are a result of the "magic" of 3 phase systems. :mrgreen:

 

[rattus]

I couldn't agree more!
2228 / (208 x 1.732) = 6.18A

I couldn't agree more!
2228 + 1114 = 3342

3342 / (208 x 1.732) = 9.28A

Sparky, you can't do that! There is no justification for the use of the 1.732 factor in this problem.

This is a simple problem with a simple solution:

Iload = P/V = 2228W/208V = 10.7A = Iline = Iphase

Eureka! I think I see the problem!!

Some of you are splitting the load and treating it as two line to neutral loads. This yields a current of 9.28A.

But this is wrong, because the midpoint of the load is not at zero! It is at 180V! Sketch it out for yourselves. The voltage across half the load is 104V.

I = 1114W/104V = 10.7A which should be no surprise.

Even my learned friend, kingpb, agrees that the current in the load is 10.7A. That being the case, the phase currents and line currents are also 10.7A.

Misapplying formulas can get you into trouble. The only formula we need here is,

I = P/V-----------Ohm's Law.

 

[ramsy]

Refering to an older thread.

Using methods and calculations that everyone in the industry knows and understands is the whole purpose of the Code. Isn't that why it is there? See; National Electric Code Examples D5(a) and D5(b). ..Why would you want me to use methods or calculations that were not standard to the industry? Arguing an issue as a point of theory is all fine and good but to implement your pet methodologies without warrant is a violation of the ideals of a uniform standard.

Those who arrived at 9.28A for this 208v 2228W load, have each cited different publications that guided their preference for the balanced results, using Pφ = Vφ* Iφ* cos(θ), or I? = P? / Pf / E?, rather than I = P/V.

1) ray94553 offers "Introductory Circuit Analysis 6th Edition by Robert L Boylstead ? 1990, Chapter 23"
2) kingpb offers the "Handbook of Electric Power Calculations, by H. Wayne Beaty, third edition, pg 13.9"
3) I also offered the NEC examples ray cited earlier.

Presumably these specific pages and chapters show some proof how values similar to 9.28 were derived, and thats why the citations were offered. If any proof or measurement exists that disputes the 9.28A, or NEC's use of Annex D ex.D(5), I believe this audience is sophisticated enough to recognize them.

When someone told science-fiction writer Theodore Sturgeon that 95% of science fiction was crap, he replied, 95% of everything is crap. This has gone down in science fiction circles as Sturgeon's law

95% of calculated results without proof is also crap. Although, calculation without measurement or proof can be disputed indefinately, the difference is those who rely on reproduceable measurements don't have to use the crap; while those looking exclusively at the numbers are getting people to eat it.

 

[infinity]

Trevor,

OK, now I get it.
If there were such a system, yes.

The other calculations are a result of the "magic" of 3 phase systems. :mrgreen:

I was trying to illustrate that if you have a simple two wire circuit, with a 208 volt supply, that simple ohms' law could be used to find the current.

 

[rattus]

I was trying to illustrate that if you have a simple two wire circuit, with a 208 volt supply, that simple ohms' law could be used to find the current.

Amen! And there are autotransformers which will drop the voltage from 240V to 208V. They are called Powerstats and Variacs.

The problem seems to be that some are using formulas which do not fit the problem.

 

[websparky]

rattus,

We are talking about feeders or services, right? No variacs needed! No phasers needed!

Please look at examples 5a and 5b in the NEC. You will see that the math and the methods are the same ones that some of us have been using all along.

 

[rattus]

rattus,

We are talking about feeders or services, right? No variacs needed! No phasers needed!

Please look at examples 5a and 5b in the NEC. You will see that the math and the methods are the same ones that some of us have been using all along.

Yes and no. The comment about Variacs simply supports infinity's rehetorical question about an autotransformer.

Phasors are needed to fully understand AC circuits especially three phase circuits, although this problem is so simple that Ohm's law will suffice--no need to refer to the Code.

This is really a single phase problem though. The voltage between any two feeders is 208V. The current needed to provide 2228W of power is 10.7A. That current flows in the two feeders and in the two secondary windings as well. Since this is a single phase problem, there is no reason to invoke the magic number "1.732" as you have done in a previous post.

We have three answers so far:

6.2A: obtained through the misapplication of sqrt(3)

9.2A: obtained by erroneously splitting the line to line load into line to neutral loads.

10.7A: obtained by using basic principles.

Steve66, help me out here!

 

[websparky]

3 Phase Load Calcs

3 Phase Load Calcs

Ok.....here is a simple 3 phase, 208Y/120V, service load scenario;

(A) Service panel with 12 - 20A breakers. All of the breakers are loaded to 16A.

What is the total amp draw and VA for this service?
Select the correct answer below.

(1.) 23040VA / 120V = 192A
(2.) 23040VA / 120V = 192A / 3 = 64A
(3.) 23040VA / (208 x 1.732) = 64A
(4.) None of the above.



(B) Service panel with 6 two-pole 20A breakers. All breakers are loaded to 16A

What is the total amp draw and VA for this service?
Select the correct answer below.

(1.) 19968VA / 208V = 96A
(2.) 19968VA / 208V = 96A / 3 = 36A
(3.) 23040VA / (208 x 1.732) = 64A
(4.) None of the above.



(C) Service panel with 4 three-pole 20A breakers. All breakers are loaded to 16A

What is the total amp draw and VA for this service?
Select the correct answer below.

(1.) 13312VA / 208V = 64A
(2.) 13312VA / 208V = 64A / 3 = 21.3A
(3.) 23040VA / (208 x 1.732) = 64A
(4.) None of the above.

 

[cpal]

Ok.....here is a simple 3 phase, 208Y/120V, service load scenario;

(A) Service panel with 12 - 20A breakers. All of the breakers are loaded to 16A.


are they single pole breakers

Charlie

 

[Smart $]

Ok.....here is a simple 3 phase, 208Y/120V, service load scenario;

(A) Service panel with 12 - 20A breakers. All of the breakers are loaded to 16A.

What is the total amp draw and VA for this service?
Select the correct answer below.

(1.) 23040VA / 120V = 192A
(2.) 23040VA / 120V = 192A / 3 = 64A
(3.) 23040VA / (208 x 1.732) = 64A
(4.) None of the above.

The correct answer is (1.) Note this response assumes single pole breakers.
Choices (2.) and (3.) would be correct if you would have stated amp draw per phase with loads balanced. (2.) is not your typical expression, but it is equivalent to (3.):

120 x 3 = 120 x sqrt(3) x sqrt(3) = 208 x sqrt(3) = 208 x 1.732​

(B) Service panel with 6 two-pole 20A breakers. All breakers are loaded to 16A

What is the total amp draw and VA for this service?
Select the correct answer below.

(1.) 19968VA / 208V = 96A
(2.) 19968VA / 208V = 96A / 3 = 36A
(3.) 23040VA / (208 x 1.732) = 64A
(4.) None of the above.

The correct answer here is (4.). You have 6 single phase circuits at 208V, 16A each, for a total VA of 19968. On an assumed balanced system...

VA = 1.732 x E x I​

...where E is L-L voltage and I is phase current. This is the formula you used in choices (2.) and (3.) of question (A). Therefore:

I = 19968VA / (208V x 1.732) = 19968VA / (120V x 3) = 55.5A/phase.​

(C) Service panel with 4 three-pole 20A breakers. All breakers are loaded to 16A

What is the total amp draw and VA for this service?
Select the correct answer below.

(1.) 13312VA / 208V = 64A
(2.) 13312VA / 208V = 64A / 3 = 21.3A
(3.) 23040VA / (208 x 1.732) = 64A
(4.) None of the above.

Again the correct answer is (4.) Here you have four 3-phase circuits drawing 16A on each phase. Therefore:

4 circuits x 16A/circuit-phase = 64A/phase

VA = 1.732 x 208V x 64A = 3 x 120V x 64A = 23040​

Choice (3.) would be correct if you had asked for amp draw per phase...

.....

Now here's a real world question:

Panel is 208Y/120. On the left side, top down, all circuits are drawing 16A, (4) 1P, (2) 2P, and (1) 3P. The right side has yet to be loaded. What is the amp draw on each feeder conductor?

 

[Smart $]

...
Now here's a real world question:

Panel is 208Y/120. On the left side, top down, all circuits are drawing 16A, (4) 1P, (2) 2P, and (1) 3P. The right side has yet to be loaded. What is the amp draw on each feeder conductor?

No takers?

My answer:

View attachment 108

 

[iwire]

I don't think you gave enough info.

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

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

My answer off the cuff based on the limited info is

A - 64

B - 64

C - 48

 

[winnie]

iwire,

As I understand Smart$'s example, the loads on the two pole breakers do not involve a neutral. The current through the pole in slot 9 must therefore be equal to the current through the pole in slot 11, similarly for 13 and 15.

But the load added to the _feeder_ by this current flow is _not_ equal to the current flowing through the breaker. This is simply because we are adding a load with a phase angle that is not necessarily the same as the phase angle of the current already flowing on the feeder.

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.

Circuit 3 puts a 16A, 120 degree load on leg B
Circuit 9,11 puts a 16A, 90 degree load on leg B
Circuit 13,15 puts a 16A, 150 degree load on leg B
Circuit 17,19,21 puts a 16A, 120 degree load on leg B

The total current ends up being 59.7A, 120 degrees on the feeder for leg B, in agreement with Smart$.

Circuit 13,15 is also placing a 16A, 330 degree load on leg A.
Circuit 9,11 is putting a 16A, 270 degree load on leg C.
Thus I disagree with Smart$'s totals for legs A and C.

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.)

The same current flows through both poles of the breakers in a given circuit. But the effect on the total phase load is _different_ because of the different phase angles already on the legs.

In all honesty, this is an intellectually interesting analysis, but IMHO it exhibits an apparent precision that would not really be justified by real world loads. IMHO it is accurate enough to simply assign VA to each leg and divide by the leg to neutral voltage.

-Jon