@sophiecentaur
sophiecentaur said:
Trying to work it our for myself got me in a muddle.
I must've been lucky in 1965. Our grad student AC power instructor spent twenty minutes at the blackboard drawing phasors and explaining it to us boys.
It has to be presented right the first time so it'll be intuitive thereafter and he did a great job.the key to it is rigor in naming, a la Lavoisier.
We must distinguish between
current in an individual phase winding
and
current in an individual line ,
call them I
phase and I
line
also between voltage across a phase winding V
phase and voltage between two lines V
line
when we do that we see
for delta connection
V
line = V
phase
and
I
line = sum of two I
phase 's.plain trigonometry and single step thinking will get I
line to I
phase relationship...
take a delta connected machine, motor or transformer - in any of the three single windings power is VI cosθ and for simplicity assume unity pf (resistor bank?) .
To be more specific in terminology power in each phase is V
phase I
phase.
Total power is 3X V
phase I
phase.
Now what about the wires carrying power to(or from) the device? That's where we'd hook up measuring instruments.
Each wire carries the current for two phase windings.
Call that the line current .
Phase currents are 120 deg out of phase with one another
and if you add two equal phasors head to tail at 120 deg, their sum is √3 not twice their individual magnitudes.
SO I
line is √3I
phase
AHA !
With the delta connection ,
Current in the Line is greater than current in the phases by that ubiquitous √3 !
I
line = I
phase X √3
and I
phase = I
line/√3
So - were i to read ammeters connected in
series with the
lines
and voltmeters connected
between the lines ,
and multiply those two numbers,
wth delta connection i'll get a result
V
phase X (I
phase X √3) because I
line >I
phase by √3
which is neither total power nor power in a single phase
Total power is
V
phase X I
phase X 3,
Since with delta connection, V
phase = V
line
we can write for delta connection
total power = V
line X I
phase X 3...
and since with delta connection I
phase = Iline/√3
we can write for delta connection
total power = Vline X I
line/√3 X 3 = Vline X I
line X√3
plod through it once drawing those phasors and it's intuitive ever after.
Wye connection?
Since for wye connection
I
phase and I
line are equal
it's V
phase and V
line that differ by √3
step by step plodding will get to the exact same expression
KVA3phase = Vline Iline√3
That it's not intuitive to everybody suggests to me that our grad student Charlie Gross was an exceptional teacher. He's at Auburn now and has written several textbooks.
Sorry Sophie I'm too awkward with latex and graphics to draw a picture
the key is that √3 line to phase ratio for either voltage or current depending on Δ-Y
old jim