Revisit: 3 phase 208, tapping for single phase loads, current on retur

[hrc]

I've touched on this discussion in the past, and need to revisit it now that I have some additional information.

3 phase 208, type a power supply across phases A-B and C-B so in both cases the 'return' is on phase B. Current flow thru phase A to the load reads 310mA without the other load active on C-B. The load on C-B is then turned on without the other and measures 229mA (of course on both feed and return).

Now, both loads are on and I measure 445mA on phase B, the return phase. I assume that some vector cancelation is occuring here, so how does one calculate what the current will be in phase B?

 

[dpc]

The A-B and B-C currents are separated by approximately 120 degrees. The measured current in B will be the phasor (vector)sum of the two currents. So if you draw two vectors 120 degrees apart of appropriate magnitude and add them together, you get the resultant current.

But the angle will be exactly 120 degrees only if the power factor of the two loads are equal. If the power factors are not equal, the angle between the two currents will be something other than 120 degrees.

 

[hrc]

guess my post never made it...let me try again.

Lets say you have three loads, each of which are 11A each, connected across AB, BC and AC. First off, the power for each would be 208*11A = 2288W. So if we look at the fact that each phase will be the combined vectors of two loads, so something less than 22A, measured as single phase. Since power is power and no getting around that (good ol laws of physics) you would think that each phase is then 20A*208=4160W. Wait, how can that be? But if you take 4160/1.72=2418W (pretty close to the 11A worth of power). So, if we look at the current draw for each phase, this has to be related back to 3 phase current calculations. Thus the approx 22A we measure under a single phase condition is actually only 22/1.72 or 12.7A and given that it is probably closer to 20A that makes it even closer to the 11A single phase.

OK, quite a bit to digest, but am I correct in these assumptions?

 

[Mstrvb19]

hrc-As dpc mentioned, the phase currents are (assuming unity power factor) 120 degrees apart. Therefore, from your situation above, phase B current B= 310 sin (377-0)-229 sin(377-2pi/3). However, you will see when you work this out that your loads are not operating at the same powerfactor. The 445mA on phase B seems reasonable. In response to your second post, you are making a couple of faulty assumptions, but you seem to be on the right track. You started off with currents that are "inside the delta" meaning that these are the equivalent currents that would flow to one leg of the load if it were operating as a single phase load. However, you calculated the power in your second equation using the 20A and the line to line voltage. In a delta configuration, each line is both a supply and a return, and the 20A of current is not all going to one load, but to two of the delta loads. You appear to be somewhat on the right track, though. In a balanced delta system, the line current will be sqrt(3) times greater than the current "within the delta" on a cyclic steady state basis. This approximation can be applied to an open delta system such as the one in your first post if you remember that the common phase behaves exactly as if you had three phase power (phase B) while the other, unconnected phases behave as if you are using only single phase power. It gets tricky with unbalance loads. To approximate it, you can take the average of the two (310+229)/2=269.5mA *1.72=463.54mA. However, as I mentioned, this does not exactly work becuase your two loads are not operating at unity power factor. Therefore, I suggest that you use the trig equations I stated above, subtracting off the additional angle due to power factor in the argument of each sin().

 

[hrc]

Starting to make sense to a digital/firmware guy. Not to sound too stupid but in the equation

current B= 310 sin (377-0)-229 sin(377-2pi/3)

where did the 310 & 377 and 229 values come from?

The problem I have also is i need to explain this in real simple terms or at least have the backup calculations to show how it all relates. The explaination of the open loop was good, least proves theory and practice.

However, the final configuration will be a closed loop system (does it matter if its wye or delta? I thought all 208 phase to phase was wye) will be the totals for each phase/leg then divided bt sqrt(3).

Thus in the most simplistic terms, not acccounting for power factors, the 20A component will require 11.6A from one phase of the 3phase feed circuit.

 

[Mstrvb19]

hrc- The numbers come from the following: the 310 and 229 are in mA (I was assuming you wanted your answer in mA also) and are the magnitudes of each sinusoidal current. Which reminds me, these should have been peak values and I bet you gave rms values, so they really should have been 438 and 323, respectively. The 377 is the number of radians per second (I assumed you were operating at 60hz). This comes from the relationship that radian frequency (w) is equivalent to 2*pi*f= 2*pi*60=377. Remember that a 208 system is really only 208 if you are connected in delta. If you are operating a wye connected load, it is 120. Also, the current rules for a wye load are different- A wye load pulls the exact same magnitude of current as the current that appears on the incoming lines. So a 20A load will show 20A on the line. Make sense? The Handbook of Electric Power Calculations (McGraw-Hill) has a fairly good discusssion of this and I suggest you give it a look. Unfortunately, I just changed jobs and I don't have access to a link to provide you. I hope this is helping.

 

[hrc]

Thanks for all the great info. However I just found out that I made a bad assumption on the fact the source power was delta, and it turns out to be WYE.

So, now if I connect (for simplicity sake) a balanced load of 10A on between each of the nodes, am I going to see a 20A draw for each leg (sumation of the two loads on each node)?

Phase A B C
10A(ab 10A(ab)
10A(bc) 10A(bc)
10A(ac) 10A(ac)

Such that the transformer has to be capable of 20A per winding, or in this case the UPS has to have 20A per phase avalable.

 

[rbulsara]

See the followinig link..

http://www.eece.ksu.edu/~starret/581/3phase.html#6

This site is not a place to learn basic engineering from scratch...

refer to some basis text books in AC Circuits and read up polyphase circuits.

 

[Mstrvb19]

hrc- In theory it doesn't matter whether your source is delta or wye. If your load is balanced and your line to line voltage is 208, the problem is the same. As I mentioned, the currents in each leg of the delta are not in phase, so they do no add arithmetically (i.e. 10A per phase does not make 20A per line). That is why I suggested the sinusoidal equation. To get a better understanding of what I am saying, try taking a graphing calculator and plotting 10sin(377), 10sin(377-2pi/3), and [10sin(377)-10sin(377-2pi/3)] all over two periods or so. I suggest picking up a textbook as rbulsara stated. It is not easy to explain this without the use of figures. Good luck!

 

[rbulsara]

hrc:

Refer to figure 1 in the link I posted earlier.
When you measure current in any of the lines, you are measureing the 'rms' or a kind of average value of the ac current which in fact is sine wave.

Although the 'rms' values in each phase may be identical (in balanced system), the 'instantaneous' values differ becasue of the phase angle difference (120 degrees in our case).
In other words, the polarity (positive or negative OR supplying or returning for your understanding) is different at different times.

Now if you take any one 'instant' by drawing a vertical line (perpendicular to time axis) and review the 'instantaneous' value at that 'instant' you will find the sum of 'amplitude' of any two phases is equal and opposite to that of the third phase.

This should explain why you keep reading 10A constant in each phase, while in reality at any instant some of that 10A is 'supply' and some is 'return' and net current of the 3 phase currents is zero and hence you do not see or need a net return wire (or neutral) in a balanced 3 phase system.

 

[peebee]

One caveat to all the above -- If you have cheap power supplies that draw a lot of harmonic current (and I think that's a reasonably good bet), then all the above calcs go out the window. The above discussions all assume you have a pretty clean sinusoidal current draw.

If you have some displacement power factor, you can easily calculate how the currents add up as indicated above. But if you have distortion power factor, it may be difficult or impossible to determine how the currents will add up.