3phase vs single phase, and proper useage of formulas

[hrc]

I've instigated some other discussions on these forums related to power factors and also sizing out of 3phase transformers related back to single phase loadings. But in some discussions, a few things are not quite clear.

The situation is this: I have a 208Y 3phase PDU, typical that is used all over the place in the states. This allows me to use either 208VAC (2 legs of the 3phase) or 120VAC (leg to neutral of course) to deliver to a bank of switching power supplies (this is a large network switch rack)so of course, the best design approach would be to try and put the same load one each pair of 208. Putting that aside, lets say we have 9 500W supplies, and each have a pf of .95 (just for reference). For a ballpark approximation, the simple relationship of P/V gives us 2.4A apparent draw for each supply. Adding in the pf value, gives the 'real' draw of 2.53A (correct?). IF we put a balanced load, each pair then would be seeing a load of approx 7.6A draw.

Now the end user is coming back and saying....you have a load of 4500W, and using the formula of I=W/(v*pf*1.73) is saying you only have a loading of 13A but I contend this is wrong because we don't have a pure 3phase load. If he wants to size out the transformer, he needs to use it based on the 7.6A * 3 = 22.7A total load, thus solving for watts he needs based on the 22.7A total load. This gives 7760 watts, and /3 would be 2.6KvA transformer.

Am I correct on this or is he correct?

 

[jghrist]

Your mistake is in using the calculated 7.6A load current as the line current. The load is Ø-Ø so the line current will be sqrt(3) times the load current or 13.1A, the same as what your end user got.

If you use your 7.6A·3 = 22.8A for calculating total load, you have to use Ø-Ø voltage. 22.8A·208V·0.95 = 4500W, which is exactly what you started out with (it better be or you violate conservation of energy). You size the transformer on VA, not on W anyway. Why not just take 9·500÷0.95 = 4737 VA and forget about calculating current?

 

[hrc]

That makes sense, but let me throw something else out.

I did some load tests on another unit (not 500W), where I measured the draw of the supply at 120V, and it measured at 1.55A, and did the same test using the two legs of 208V and it measured 0.938A. If the relationship of A=P/V, then its a .6X factor between 120 and 208 and it does hold true. If you take the 1.73 factor from the formula, and assuming the pf is constant between both, it *should* only show a draw of .54A, yet measured says the 1.73 didn't come into play. Thats the confusion I have on this.

 

[ScottyUK]

Hi hrc,

What did you measure the current with? Switching power supplies, especially of the size you are describing, tend to draw a load current that has a lot of harmonic content.

A measurement of true power requires an instrument which can measure the fundamental component of the current. The fundamental current component multiplied by the supply voltage, and the PF if they are not in phase, will give you the power in watts. You are almost certainly using a the value of a harmonic-laden current in your calculation, which is why your result does not appear to be correct.

Fluke, Yokogawa, and Voltech all make instruments capable of this type of measurement. You could also use a spectrum analyser with a current probe to measure the fundamental and harmonic components.



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[hrc]

The draw was only around a 180W, so not as much harmonic content. I was using a Fluke model 87 III, and measured the current inline. Given an apples to apples test, shouldnt the same amount of harmonics be present in both tests, or near the same given that the load was pretty light? The fact that measured of .94A vs the 'calculated' 0.54A is a pretty big difference, that can't be explained by just a measurement error due to harmonics?

 

[jghrist]

Line current = sqrt(3) times delta load current. sqrt(3)·0.938 = 1.62, pretty close to the measured line current of 1.55A.

 

[ScottyUK]

You Fluke 87 is a true-RMS instrument, but it will include all the harmonics in its calculation of current. It can't display just the fundamental.

If you need to convince yourself of the discrepancy, connect a constant load on the output of the power supply. The PSU will have a roughly constant efficiency over its rated input voltage range. From these two factors you can be pretty sure the true power is fairly constant, meaning that the product of voltage, current, and PF is also fairly constant. The 'PF' you use is actually the distortion factor, which is the phase angle between the voltage and the fundamental component of the current. PF for a sinusoidal voltage and complex current waveform is better expressed as the ratio of Power:VA.

If you follow this link it will allow you to download a lot of useful information from Voltech's website. You have to register, but it is free, and the information is worthwhile.

http://www.voltech.com/securearea/securehome/index.asp

Hope this helps.



-----------------------------------

Start each new day with a smile.

Get it over with.

 

[hrc]

>Line current = sqrt(3) times delta load current. sqrt(3)>·0.938 = 1.62, pretty close to the measured line current >of 1.55A.

But the 1.55A was the load running at 120V, while the .938 was running at 208V. By saying delta, I am assuming you mean the difference, not a delta configuration.

 

[rbulsara]

hrc:

Your end user is correct, (9) 500W units amounts to 4500W of power no matter how you feed it. The formula he gave you is also correct for a 3 phase balanced load, assuming you connect (3) units between each pair of lines.

Also the your statement that ".....Adding in the pf value, gives the 'real' draw of 2.53A (correct?). If we put a balanced load, each pair then would be seeing a load of approx 7.6A draw. " is NOT correct. The addition of current in 3 phase lines is not an arithmetic sum. Look up some basic text books on the 3 phase AC circuits.

Also see thread237-89634 and other thread embedded in it.

 

[rbulsara]

hrc:

In regard to your other post, when you compare a load conected between 120V or 208V only (between two wires, or ends of the load) it is still a single phase load and the formula for 3 phase load does not apply, even if those two wires are part of a 3 phase, 4 wire system.

3 phase formula, VA=1.732*I*V is true for a balanced 3 phase system only.

If you have unbalanced load, then you need to figure your wire size ans source (transformer)based on the maximum unbalanced load. For example, if you connect (5) 500W units between a pair of phases, say A and B, and you turn on only those five units,which will give maximum unbalanced current using a single phase formula I=W/(V*pf)and the voltage is the voltage between the two wires A and B in this case. Now if you use 3 phase transformer rated for this current. If you look closely you will find that you will need a larger 3 phase transformer to feed unbalanced 3phase loads, compared to one required for a balanced 3 phase load.

 

[jghrist]

hrc,

I'm afraid I don't understand your second example. I assumed a 120/208 volt source with the load connected delta; three loads connected phase-to-phase at 208 volts. If the current through the three 208 volt connected loads are are 0.938A each, then the three line currents coming from the transformer will be 1.62A each. I don't understand your A=P/V comment or your 0.6X factor.

 

[hrc]

rbulsara stated exactly what I was thinking, but this seems to contradict almost everything else that has been posted.

"3 phase formula, VA=1.732*I*V is true for a balanced 3 phase system only."


So in this case, where I have no control over what loads will be active at anytime, so it is an unbalanced system, the formula of I=W/v*pf is the only one that does hold true. If we take the worse case example, that all loads are on at the same time, and each 'grouping' of loads draw (for sake of simplicty) 100A each, then how do you calculate the total kva requirements for the transformer? This is confusing because we go from single phase unbalanced loads to 3phase. jghrist has indicated that a phase to phase current draw, would relate back to I*1.73 for the transformer requirements. Did I interpret that correctly?

Perhaps this has gotten a bit convoluted, but if what rbulsara stated is true, then this confirms what my measurments indicated as well.


In response to jghrist that example was to show that a measured current at 208V vs. 120V was .6x that of 120V. In other words, 120/208 = .6, and the currents were 1.55A/0.938A also = 0.6, as expected since wattage was constant.

 

[jghrist]

If you are starting with a load of known power (watts), then calculating the current is unnecessary and confusing. Just size the transformer for the load watts divided by the power factor.

With regard to the .6X factor, 208/120=sqrt(3). 1/sqrt(3)=0.577 appr. = 0.6.

 

[DanDel]

hrc, for an unbalanced system the individual phase currents must be multiplied by the individual phase voltages(L-N, or L-L/1.73). This will give you the individual phase VA. Then, simply add the three individual pahse values together to get the total VA.

 

[rbulsara]

hrc:

Each of the 3 phases of the a 3-phase transformer is of the same rating and equals 1/3 of the total 3 phase rating.

So if you have one (or two) phases loaded more than others, you need to size the tranformer 3 times the greatest load on any one of the phases. You can not exceed the rated current on any of the phases at any given time. Therefore it is imoprtant that you balance the load as much as you can.

Say your loading comes out to be phase A= 100A, B=150A and C=120A the rated current of a 3 phase transformer will have to be 150*3=450A or 1.732*450A*230V/1000 =180kVA. Use the next higher standard size and allow for some cushion.

 

[rbulsara]

A CORRECTON:

I am sorry for a big mix up in my past post.

The rating will be 1.732*150*230/1000=60kVA.

I mistakenly multipled 150A with 3. I was thinking of multiplying kVA rating of each phase by 3.