wye vs delta ?

trying to get my head around the difference as it pertains to current

for instance

if i have a 10 alternator (delta configuration)
and reconnect the stator to wye, then the voltage will now be 10v * 1.73
or 17.3 volts, correct?

the question then becomes,

if the same alternator connect in delta produces 10amps at 10 volts
reconnecting it in wye gets me to 17.3 volts, but at what amperage?

is the amperage limited to the original phase current of 3.3 amps?
(or do i calculate the amps on the inverse of 1.73 factor of .578?)
which would be 10amps * .578 = 5.78amps?

any help with this would be appreciated

(the 10volt/10amp alternator is for illustrative purposes only, and no i don't have one)

thanks
bob g

Huh?

If 2 + 2 = 4 then whats a refrigerator / an apple pie?


Current is based on demand. More current is generally available in a Delta configuration, however the used current is based on how many lights you turn on. (for example)

I don't feel like rambling right now so try this.

http://forums.mikeholt.com/index.php?

Mike Holt wrote most of the text books I had to read during my apprenticeship. Love 'm or hate 'm he must be ok if he's still around.

mobile_bob said:

if i have a 10 alternator (delta configuration)
and reconnect the stator to wye, then the voltage will now be 10v * 1.73
or 17.3 volts, correct?

the question then becomes,

if the same alternator connect in delta produces 10amps at 10 volts
reconnecting it in wye gets me to 17.3 volts, but at what amperage?

is the amperage limited to the original phase current of 3.3 amps?
(or do i calculate the amps on the inverse of 1.73 factor of .578?)
which would be 10amps * .578 = 5.78amps?

Click to expand...

As a general rule transformers, generators etc are rated watts or VA, both indicate the total power possible. If you increase the Votage capability of a device, then the available current has to be reduced in order to maintain the Wattage or VA rating.
M.

i understand that there must be a reduction in current for a rise in voltage

by reconnecting the stator from delta to Wye, the voltage goes up by a factor of
1.73

so then after doing so, do i use the inverse (.578) to determine the ampacity of the alternator ?

i guess the real question is

"is the 57.8% a correct factor?"

(i am speaking of ampacity, or capacity of the Wye connection, and realize that load will determine how much amps i pull up
to this capacity)

bob g

mobile_bob said:

is the amperage limited to the original phase current of 3.3 amps?

Click to expand...

No, the 3.3 amp number isn't correct. I assume you got that by dividing 10 amps by 3, which isn't correct. In the original Delta configuration, the current in each winding is 10/1.73 = 5.78A

(or do i calculate the amps on the inverse of 1.73 factor of .578?)
which would be 10amps * .578 = 5.78amps?

Click to expand...

Yes, that's correct. The Wye current rating would be 5.78 amps per phase, and the voltage would be 17.3V line-to-line.

Guys:
thank you

took a bit today, finally sketched it out vectorally on a piece of paper
and compared the result with the math,, matched close enough for me.

the balance of power makes sense, in theory not figureing normal losses
VA = VA whether delta or wye.

makes sense

bob g