Revisit: 3 phase 208, tapping for single phase loads, current on retur
Revisit: 3 phase 208, tapping for single phase loads, current on retur
(OP)
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?
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?






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