Single phase load on 208/120Y calculation req'd by Oreg

Status
Not open for further replies.
Re: Single phase load on 208/120Y calculation req'd by Oreg

I am revising the sketch above showing the single phase load connected to the 208/120 volt transformer.

After thinking about this, I tend to agree with these points made by Bob, Ray, and others.

* The term "Va per phase" is irrelevant here because the load is not connected to the phase voltage. As far as this load is concerned, the phase voltage of 120 volts does not exist.

* The Power Factor of an AC system is determined by the type of connected load.

Ed
 
Re: Single phase load on 208/120Y calculation req'd by Oreg

Originally posted by bob:
[QB] rattus
I don't know where you are getting your information but you need to some other source. What you are putting out is just plain WRONG.
You have already admitted that you are confused by the terminology of Iline, Vline, Iphase and Iline. This is basic stuff.

Reply: Bob, we must be from different planets. I told Ed that it was confusing to use the same names for different variables. I know the difference.

The reason the apparent powers in VA are not equal is that the secondaries see a total of 240V, but the load sees only 208V.
Click to expand...
This is not correct. The voltage at the tranmsformer is 208 and not 240. You can not add the voltages at each coil and come out with 240. Its a vector addition.

Reply: The voltage across each xfrmr secondary is 120V. Apparent power delivered by each secondary is 2 x 120 x I. But the line to line voltage is 208 so the apparent power in the load is 208 x I.

True, Papp = Preal = 208 x I in the load, but not in the xfrmr.
Click to expand...
This is just wrong.
It is absurd to say that the power delivered at the load is not the same as at the transformer.

Reply: The real power is the same; the apparent power is not.

What is not obvious at first glance is that the PF in the secondaries is 87% because there are phase differences between voltage and current because the line to line voltage determines the phase angle of the current.
Click to expand...
This is also wrong. The power factor of the load is 1. Its not different at the transformer. Its just wrong to put out this kind of information.

Reply: There are three voltages and three phase angles and a single current. The current cannot be in phase with all three voltages. It is in phase with the 208V because the load is resistive.

There have been several posts that have tried to correct your mistakes but you have not seen fit to
take the hint. Suggest you get some one who is knowledgeable in that area of electrical power and see if they can help you.

Reply: There have been such claims, but I insist I know what I am talking about. Somehow I muddled through engine school with passing grades. I couldn't be that dumb.
Click to expand...
 
Re: Single phase load on 208/120Y calculation req'd by Oreg

Originally posted by bob:
Nice drawing Ed. Maybe you could have the drawing
showing the same information for the origional question regarding the single phase load being fed from 2 phases of the wye connected transformer.

Rattus
Eline, Ephase, Iline and Iphase are the standard way of showing the Line currents, Phase currents,
Line voltage and Phase voltage for Wye and Delta
connections. The numbers are not the same. No need to be confused.
Click to expand...
Bob,
I understand all that. It is my opinion though that Ephase and Iphase should be applied to the transformer only. If you use Iphase for example to describe the current from a wye and also use it to describe the current in a delta load, you have to expect trouble. I have seen it happen. Thousands of dollars were wasted with this careless mistake. This is not good engineering discipline. To say that the numbers are not the same is no justification for this practice.
 
Re: Single phase load on 208/120Y calculation req'd by Oreg

It seems that some of our members have missed the point of this discussion, and that point is that there is a basis for the infamous "Oregon factor". The crux of this matter is that the "exact" method for computing xfrmr loading is Vphase x Iphase. In practice I doubt that this makes any significant difference, but in this special case, the transformer would see some 15% more load than calculated from Vline x Iload.

Now if we convert the KVA rating of the transformer to a current rating, then there would be no argument.

Although this argument is technically correct, I see no point in tricking the testees with such an obscure point unless it is commonly taught in training programs, and I see no point in changing the way tranformer loading is computed.

OK Ed, I think this animal is ready for the glue factory.
 
Re: Single phase load on 208/120Y calculation req'd by Oreg

Ed, Rattus, Et: all

I feel I must set the record straight about my beliefs on the this thread. I HAVE been playing devils advocate on this entire question. The original question waaaay back on page one had to do with the way calculations are CONVERTED back and forth between LINEamps and PHASEva. It can, and does, get confusing if the terminology is not VERY precise.

Ed and Rattus, I fully agree with ALL you have said on this thread and on others. The single phase loading to the LINE conductors is just volts times amps. No factor, no magic factor, no muss, no fuss.

What the original question was REALLY looking foor was some validation that we were NOT nuts in the way we were all looking at this problem. It seems the difficulty here in Oregon will not go away quietly, but must be delt with in a larger forum. 50,000 electricians can't be wrong if there have been no problems with the installations using the OLD calculation methodology.


Thank You ALL for the interesting thread and posts. I think we can FINALLY let this horse Rest In Peace. Long Live Ohms Law!
 
Re: Single phase load on 208/120Y calculation req'd by Oreg

Looking through the thread, fsimmons states the problem a little differently than others prior to that.

QUOTE-- "I teach the Supervisor Prep Class that prepares individuals for taking the State of Oregon Test. I personally have talked with Chief Inspector John W. Powell and Acting Assistant Chief Stan Penrose and they are adamant that in order to convert single phase amperes, 208 volt to volt amperes you have to do the following when the load if fed from a three phase 208/120 system.

Example of a motor load - 8.8 amperes

8.8 X (208 X 1.154) = 2,112vA "

This seems to fit what rattus has maintained in his analysis of the question in that it increases the calculated transformer/feeder loading.

It is an algebraic transformation of the way others presented the question...

Amps = kVA/(208 * 1.154)

Rattus, I understand what you are saying that by using this you will be calculating a reduced permissible amperage loading of the transformer, but...

If this is used to calculate the additional, incremental transformer/feeder current imposed by the specified 208 V single phase L-L load, the amperage will be reduced, not increased as you maintain happens in reality.

Jim
 
Re: Single phase load on 208/120Y calculation req'd by Oreg

From ccjersey:
If this is used to calculate the additional, incremental transformer/feeder current imposed by the specified 208 V single phase L-L load, the amperage will be reduced, not increased as you maintain happens in reality.

Reply: No, ccjersey, I do not claim that it increases any current anywhere. The method merely computes the allowable apparent power (read current) from the transformer, and it does so correctly. Again, I do not claim that this should be standard practice. I only claim that the OFF has a valid basis.


[ March 19, 2005, 03:37 PM: Message edited by: rattus ]
 
Re: Single phase load on 208/120Y calculation req'd by Oreg

Ok,
I'm not as smart as most of the rest of you, but I've been trying to get a handle on this discussion. I have come up with another observation I would like to have picked apart.

If you connect 3 120-volt resistive loads (of 10 amps each) to a 120/208 wye transformer, each winding will have 10 amps and we will have 3600 VA working through the resisters.

If you connect 3 208-volt resistive loads (of 5 amps each, one load each on phases AB,AC,BC) to the same transformer (without the previous load), each winding will have 10 amps and we will have 3120 VA working through the resisters(208*5*3). This is a balanced load on the transformer, so using the formula of 10(amps)* 208(volts)*1.732(sqrt 3) you have a calculated load of 3603.

It appears we have a loss in the transformer of .07% 3603/3600 when we connect three 120-volt loads, and a 13.39% loss when we connect three 208-volt loads.

So.... how am I looking at this wrong, or why is my math wrong?

Thanks.
 
Re: Single phase load on 208/120Y calculation req'd by Oreg

I just noticed.

3 208-volt loads @ 5 amps each are = 3120 VA.

10 amps through each of three windings in a 3-phase transformer (wye) is 3602.56 VA.

3602.56 / 3120 = 1.154........ the Oregon factor?
 
Re: Single phase load on 208/120Y calculation req'd by Oreg

If you connect 3 208-volt resistive loads (of 5 amps each, one load each on phases AB,AC,BC) to the same transformer (without the previous load), each winding will have 10 amps and we will have 3120 VA working through the resisters(208*5*3)..
Click to expand...
The Iline is 5 amps per phase not 10.
5 amps x 208 x 1.73 = 1799 va
This is a balanced load on the transformer, so using the formula of 10(amps)* 208(volts)*1.732(sqrt 3) you have a calculated load of 3603.
Click to expand...
I am not sure where the 10 amps comes from but even if it were correct 10 x 208 x 1.73 = 3598 va.
The load amps is 5 and the va = 1799 va.


3 208-volt loads @ 50 amps each are = 31200 VA.
Click to expand...
50 amps x 208 x 1.73 = 17992 va
100 amps through each of three windings in a 3-phase transformer (wye) is 36020.56 VA.
Click to expand...
100amps x 120 v x 3 = 36000.
Hardworker
This is just a number someone came up with to try
to make a caculation give a correct answer.
I have not seen anyone post information that this
number has any scientific meaning.

[ March 19, 2005, 09:18 PM: Message edited by: bob ]
 
Re: Single phase load on 208/120Y calculation req'd by Oreg

stiff, you added the 5A currents arithmetically. You must add them vectorially, then the phase currents and line currents will be 8.66A. Then the apparent power per phase is 1039VA which is approximately 208V x 5A.

Let the line to line load currents be 5.77A, then the phase currents are 10A, and the apparent power is 1200VA computed at the load or computed at the xfrmr. This is to be expected with balanced loads. Try it.

Don't fret the small differences. We are presenting irrational numbers with 3 significant figures, so you can expect some roundoff errors.

[ March 19, 2005, 09:42 PM: Message edited by: rattus ]
 
Re: Single phase load on 208/120Y calculation req'd by Oreg

Well I don't live in Oregon anyway and don't know anyone in the hierarchy of the state government.

Jim
 
Re: Single phase load on 208/120Y calculation req'd by Oreg

Originally posted by rattus:
stiff, you added the 5A currents arithmetically. You must add them vectorially,
Click to expand...
I must admit I don't know what this exactly means.

When A phase is 120 volts positve (RMS) in relation to neutral, then B and C phases are both -88 volts in relation to neutral. If I am correct in this understanding, then I conclude that loads connected between AB and AC would cause A to draw the sum of the two loads, but you are saying that is incorrect?
 
Re: Single phase load on 208/120Y calculation req'd by Oreg

Stiff, yes, that is incorrect. You must use trig to add these currents, and since we are dealing with RMS values, we do not consider instantaneous values.

To simplify things, the vector sum of two currents of the same magnitude and 120 degrees apart is,

1.732 x I.

The vector DIFFERENCE of two equal voltages 120 degrees apart (as in the wye) is,

1.732 x Vphase = 120V x 1.732 = 208V.

Then in this case,

Iphase = Iline = Iload x 1.732

rattus
 
Re: Single phase load on 208/120Y calculation req'd by Oreg

stiff,
No, Oregon is recognizing the fact that, in this case, the apparent power computed at the load is some 15% less than the apparent power computed at the two transformers. All of these calculations are done with RMS voltages and currents. Nothing instantaneous about it.

It boils down to the ratio 240/208. They are asking for the ALLOWABLE current in terms of VA rating, line voltage, and the OFF. It is a clumsy way to do it in my opinion, but it is technically sound.

Power, real or apparent, computed from RMS values is an average power.
 
Re: Single phase load on 208/120Y calculation req'd by Oreg

Ok, I'm really getting in over my head now.

I drew out a 3-phase sine wave and measured the voltage from A-phase at its apex to B&C phases at that moment in time (which B&C phases were crossing each other and had 0 volts between them) and the voltage measurement was not 208 like I would have expected it to be, but it was more like 199 (actually, in my drawing it was 198.7).

Will you more educated people tell me if this voltage is correct? If not, what should it be?

Thanks,
HWS
 
Re: Single phase load on 208/120Y calculation req'd by Oreg

Stiff, you need to be sure that you have the peak voltages correct. For 208Vrms that would be +/- 294V. For 120V it would be +/- 170V.

Remember too that you have to take the DIFFERENCE of the two phase voltages. The peak would occur where they are equal and opposite. Zero would occur where they cross.

I helped ronaldrc on this same problem. You might contact him privately for assistance.

rattus
 
Re: Single phase load on 208/120Y calculation req'd by Oreg

Originally posted by rattus:
Stiff, you need to be sure that you have the peak voltages correct. For 208Vrms that would be +/- 294V. For 120V it would be +/- 170V.

rattus
Click to expand...
I drew it two ways, one with the top and bottom of the sine waves being 120 volts from 0, and the other with the sine waves being 170 volts from 0.

When A was +120 volts (apex), B & C were both -78.7 volts in relation to 0 (which would make it 198.7 from A, instead of 208).

When A was +170 volts (apex), B & C were both -111.3 volts in relation to 0. If you multiply 281.3 (111.3+170) by .707 you come up with 198.9, not 208 like you would expect.

Can anyone else draw this out and tell me if they come up with the same observations?

[ March 20, 2005, 03:01 PM: Message edited by: hardworkingstiff ]
 
Re: Single phase load on 208/120Y calculation req'd by Oreg

Originally posted by hardworkingstiff:

I drew it two ways, one with the top and bottom of the sine waves being 120 volts from 0, and the other with the sine waves being 170 volts from 0.

When A was =120 volts (apex), B & C were both -78.7 volts in relation to 0 (which would make it 198.7 from A, instead of 208).

When A was +170 volts (apex), B & C were both -281.3 volts in relation to 0. If you multiply 281.3 by .707 you come up with 198.9, not 208 like you would expect.
Click to expand...
Reply: Stiff, the peaks of the 208V waves do not conincide with the peaks of the 120V waves. Let's try a little trig.

Let
Van(t) = 170cos(wt) and
Vbn(t) = 170cos(wt + 120)
wt = 30 degrees, then
Van(t) = 170cos(30)= 170 x 0.866 = 147V
Vbn(t) = 170cos(150) = 170 x -0.866 = -147V

The difference between these two voltages is 294V, the peak. Multiply that by 0.707 for the RMS value of 208V.

Please note that 2cos(30) = 1.732

Sketch that out if you need to.

rattus



[ March 20, 2005, 03:14 PM: Message edited by: rattus ]
 
Status
Not open for further replies.
Top