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

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

Hi All, I was finally able to perform the experiment on a 120/208 Y system today. We used a 29.725 ohm resistor from an old load bank as a load. according to the calculations this should develop a load of 1455va connected to 208 volts.

an actual voltage of 204 volts was available at the panel. On attaching the load and energizing the circuit, a current flow of 6.9 amps on a Fluke 33 true RMS meter was read.

204v* 6.9A =1407.6va OK so far.

calculate va from load resistor and 204v
204v /29.725ohms =6.86A (read 6.9 on meter)
204v*6.86A =1400va of load Looks pretty close to me.

no factor of ((2/3)*sqr(3))had to be used to calculate the amperage through the load. This was a pure resistive load so no voltage vs current phase shift more than the normal transformer inductive loading occured.

It is always nice to have the reality back up the theory and vice-versa. Now we know. Just use the normal ohms law calculations for ANY single phase load, whether its on a true single phase transformer or on a combination of 3 phase windings.

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

Maybe I am the only one who wants to know what question this factor is used for answering.

Correct me if I am wrong, but I think this started out as a way to point out the difference between the behavior of a 240V resistive load and a 240V inductive (motor) load when supplied with 208V.

When the 240V resistive "toaster" is supplied with only 208 volts the resistance is the same so current and power will be less by factor of 208/240 (or 1/1.154. so the 1.154 is in the denominator of the equation.

When the 240V rated inductive "motor" is supplied with only 208V the power is the same so current will increase by the factor of 240/208 (1.154/1). So the 1.154 is in the numerator of the equation

Which doesn't explain how it got on this test as described :roll:

Jim

PS I really enjoy the discussion and diagrams on the forum. You folks are great to spend the time it takes to hash all this out!
 
Re: Single phase load on 208/120Y calculation req'd by Oreg

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 adamate 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 Not only is this ridiculous, it is incorrect and many individuals have failed the State Exam because of the injustice being done in this State.
 
Re: Single phase load on 208/120Y calculation req'd by Oreg

Can these guys cite any references to back up this notion? No one on this panel agrees with them.
 
Re: Single phase load on 208/120Y calculation req'd by Oreg

Garrett,

Thanks for doing the experiment. This sure proves that no only is the State of Oregon wrong in their stand but the participants of this forum are more informed in regards to this issue than the State of Oregon.
 
Re: Single phase load on 208/120Y calculation req'd by Oreg

I agree with ccjersey that the only way this makes sense is that if the question stated that a 240 Volt motor load fed from a 208 V supply, how much current would it draw...maybe to show that if one is supplying a load that draws constant power from a supply that is providing a voltage that is less than the voltage rating of the load, then it will draw more current? In this case 208 V supplying a 240 volt motor, so it will draw more current by a factor of 1.154? :(
 
Re: Single phase load on 208/120Y calculation req'd by Oreg

David Z,

The motor thing does not work either since the 1.154 factor is in the denominator. It reduces the calculated current from 29 to 25A.

If the question is, "How much current is required to produce upteen KVA in any load" you simply divide by 208. No sense in injecting confusing fudge factors.

Someone needs to teach these guys Ohm's law!
 
Re: Single phase load on 208/120Y calculation req'd by Oreg

I would like to provide a comparison calculation that I would like all to look at and comment on.

There are five 1phase 208v 12kva loads (no 120vac) on a three phase 208/120 wye.

The loads are distributed as follows:
12kva/2=6kva on each leg. 6va/120vac=50amps
A B C A B C
6 6 50 50
6 6 50 50
6 6 50 50
6 6 50 50
6 6 50 50
-- -- -- -- -- --
18 24 18 150 200 150

12kva/120=500amps 150+200+150=500amps
18+24+18=60kva

12kva/(208*1.154)=50amps

This works out and makes sense.

The following does not work out.
12kva/208=58amps
A B C
58 58
58 58
58 58
58 58
58 58
-- -- --
174 232 174
174+232+174=580amps

Based on these examples, the 1.154 factor is required when using a single phase 208v load on a three phase 208/120wye.
 
Re: Single phase load on 208/120Y calculation req'd by Oreg

For some reason the last part of the previous post did not show the distributed loads on A, B, and C.
The following does not work out.
12kva/208=58amps
A B C
58 58
58 58
58 58
58 58
58 58
-- -- --
174 232 174
174+232+174=580amps

Based on these examples, the 1.154 factor is required when using a single phase 208v load on a three phase 208/120wye.
 
Re: Single phase load on 208/120Y calculation req'd by Oreg

Lets try one last time

A B C
58 58
58 58
58 58
58 58
58 58
-- -- --
174 232 174
 
Re: Single phase load on 208/120Y calculation req'd by Oreg

Specs, unless you can cite a source or provide a justification for the factor 1.154, you have proven nothing.
 
Re: Single phase load on 208/120Y calculation req'd by Oreg

Unfortunately the format I sent it in does not distribute the loads correctly under ABC.

Compare the calculations. I was reading some of Mike Holt's information and find that he calculates single phase loads on three phase 208/120 using the formula kva/2 to get the kva per leg and then he divides the kva per leg by 120. In this case the equations are all equal.(kva/2)/120 is equal to kva/(208*1.154) which is equal to kva/240. Perform the calculations and it works. No other way works.
 
Re: Single phase load on 208/120Y calculation req'd by Oreg

Specs1,
The solution to this is posted in the thread you started under the name "tck" on Feb 26. here

As for this -
Based on these examples, the 1.154 factor is required when using a single phase 208v load on a three phase 208/120wye.
Click to expand...
all I can say is -
:)
Ed
 
Re: Single phase load on 208/120Y calculation req'd by Oreg

The problem is dividing the kVA by 2. That is intuitive, but if you do that and then assume you have 120 volts applied to half of a 208 volt load, you are screwed! It may not look like it but this process has converted the calculation to a 240 volt supply and the 120/208 network just doesn't have it to give.

Then you have to use the infamous factor :mad:

specs Try your calculation dividing the kva by 2 and then use 104 instead of 120 to get amps

(edited to suggest halving the line-line voltage instead of using l-n)

Jim

[ February 28, 2005, 12:35 AM: Message edited by: ccjersey ]
 
Re: Single phase load on 208/120Y calculation req'd by Oreg

You guys are out of left field again. From what I can gather, we have a single, 6KVA load powered from 208V. Then,

Iline = Pa/V = 6KVA/208V = 29A

Ineut = 0

But, if we are talking of two 3KVA loads across two legs of the wye, then,

Iline = 3KVA/120V = 25A

Ineut = 25A

But, if this is the case, why not attack the problem directly instead of making up arcane formulas. It appears that someone thought two 120V loads returned to neutral are equivalent to the two loads in series from line to line. Not so!

I think we still have not seen the exact wording of the test question. If we have, then it is unclear.
 
Re: Single phase load on 208/120Y calculation req'd by Oreg

How do 2 - 120v loads on a transformer differ from 1-208v single phase load as far as the transformer loading is concerned?

Ex. Given 2-40A, 120v loads
VA = 120v * 40A * 2

Given 1-40A, 208V single phase load
We still have 40A on 2 of the phase windings as in the previous example. 120V * 40A * 2 = 208V * 40A * 1.15

Am I looking at this incorrectly?

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

Originally posted by lauraj:
How do 2 - 120v loads on a transformer differ from 1-208v single phase load as far as the transformer loading is concerned?

Ex. Given 2-40A, 120v loads
VA = 120v * 40A * 2

Given 1-40A, 208V single phase load
We still have 40A on 2 of the phase windings as in the previous example. 120V * 40A * 2 = 208V * 40A * 1.15

Am I looking at this incorrectly?

Laura
Click to expand...
Laura, for starters, the apparent power in one case is 9.6KVA and in the other it is 8320KVA.

Secondly, one case draws a neutral current, and the other does not because the phase angles of the load currents are different. Not the same animal at all.

It seems that all this hullabaloo is over the ratio 240/208, and I contend this is a clumsy formula. If you have a 120V problem, solve it as a 120V problem, don't complicate things with unneeded formulas. Stick with the basics.
 
Re: Single phase load on 208/120Y calculation req'd by Oreg

Laura,

You are changing it to a load defined by amperage. The original question was a single phase 208V load defined by the va of the load. Look at the posts above in this list. You are applying two different loads to the system in your example 8320 VA and 9600 VA and then you apply 1.15 to bring them to equality. Why would they need to be equal?. On 120/240 these would be the same loads, but not on 120/208.

Do you agree that 12kva = 12000VA/208V = 57.69Amps each line? This is line to line.

What about 2 line to neutral (120 Volt)loads of 6000 VA? 6000 VA/120 V = 50 amps for each load and for the sake of this discussion apply them to the same two lines you would apply the 208 V load to. So 50 amps each line.


When you take a 12kva @208V load and in your mind divide the kVA between the two lines for the ease of calculation, fine, but you haven't changed the load, it is still a 208 V line to line load. Now, say you want to know how many amps on each line- if you take these two imaginary 6 kVA loads and divide by the line to neutral voltage (120), you get only 50 amps per line. This cannot be right since the load hasn't changed.

The amperage calculation needs to be done for the l-n loads, the l-l single phase loads, and the 3 phase loads and then the amperages summed if you want to know the total line amps.

Jim

[ February 28, 2005, 01:58 PM: Message edited by: ccjersey ]
 
Re: Single phase load on 208/120Y calculation req'd by Oreg

Thanks CC, I have been properly schooled. You guys are completely right. I'm just trying to justify this 1.15 factor, as I also live in Oregon and will be taking that test at some point. Hmmm.

Laura
 
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