wye vs delta
wye vs delta
(OP)
if a delta motor is rewired for wye the run amps are lowered and the hp output is lowered. the impedence of the windings has a part to play in that happening, but what is the expected hp output of that motor? is it 1/3 hp of nameplate or is the square root of three less?





RE: wye vs delta
There is no simple 'divide by three' or sqrt(3) answer to your question.
Ask the manufacturer or run a test.
Gunnar Englund
www.gke.org
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
RE: wye vs delta
The current demanded for a given application could go up or down as result of efficiency and power factor effects, but it may not be a huge change because you are delivering roughly the same power.
The THERMAL/LIMITING steady state horsepower output may change. I agree best to talk to OEM to determine that limit.
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RE: wye vs delta
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RE: wye vs delta
If maximum rated current through a winding is 10 Amps, then maximum motor current on a wye connection will be 10 Amps. The maximum motor current on a delta connection will be 17.3 amps.
Reality check:
With maximum allowable current in the windings the KVA in will drop by a factor of 1/1.73 KVA in is a fair indication of kW out plus losses.
Limit: The breakaway torque will drop by a square factor. If breakaway torque becomes less than full load torque, the motor will stall.
The motor may develop 1/1.73 HP or the motor may stall.
Am I missing something Gunnar?
Bill
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"Why not the best?"
Jimmy Carter
RE: wye vs delta
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RE: wye vs delta
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RE: wye vs delta
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RE: wye vs delta
I am thinking the motor would only usable at 57.7% of nameplate hp anything above that would cause overheating
RE: wye vs delta
I'll go with 57.8% with the disclamer that the motor may not have enough torque to start. If it comes up to rated speed it will be usable in most cases at 57.8%.
Bill
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"Why not the best?"
Jimmy Carter
RE: wye vs delta
RE: wye vs delta
I had a nagging feeling that a factor of root-three was missing somewhere. I think that you have identified it, Lionel. Thanks.
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: wye vs delta
RE: wye vs delta
As Gunner posted, the torque @ rated slip drops to 33.3%.
As Bill and Electricpete posted, the per phase current capacity reduces to 57%. In theory, this means the maximum running torque reduces to 57%. This is possible by allowing a higher than rated slip. However, the effect this higher slip has on the motor is unknown.
As I posted, you also need to have a motor with a breakdown torque peak greater than 173% to get to the above 57% torque.
In summary;
You can easily get 33% torque from the motor.
You can also theoretically get 57% torque from a properly selected motor except the effects of the approximately 3 times extra slip are unknown.
So, this brings us to the big question: Will the extra slip cause more loss in the rotor than when running at rated speed and torque?
RE: wye vs delta
Whether you want to look at it as result of increased slip or decreased flux density is perhaps a matter of opinion.
The resistance (referred to stator) associated with rotor I^2*R losses is: R2
The resistance (referred to stator) associated with real power output is R2*(1-s)/s which is for all practical purposes still very close to R2/s
Therefore output power is 1/s times as high as rotor losses. Turn it around and you say rotor losses are a fraction s of output power. 1/sqrt(3) voltage will mean 3x the slip (slip~V^2) for a given output and therefore 3x the rotor I^2*R losses for a given output. Since stator current increased by roughly sqrt(3), stator I^2*R losses also increased by the same factor of 3.
Neglected 2nd order effects such as magnetizing current, change in rotor resistance with frequency etc.
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RE: wye vs delta
RE: wye vs delta
If we reduce voltage by sqrt(3) and leave the motor connected to a load that draws original rated torque (rated power assuming speed change is negligible), then current will increase by roughly sqrt(3) and I^2*R losses will increase by 3 (compared to full-voltage rated load conditions.) To compensate, we derate load capability to 1/sqrt(3). Under reduced voltage, derated-output conditions, the I^2*R losses would be roughly the same as under full-output rated voltage conditions.
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RE: wye vs delta
The point of my first post was rotor I^2*R varies the same as stator I^2*R. Maybe that was already obvious... not sure.
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RE: wye vs delta
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RE: wye vs delta
I think that the OP would like you to simplify to something simpler like;
Yes,
No,
3,
or Root three.
Numbers don't talk to everyone as they do to you. A relationship that is obvious to you may not be obvious to some others.
Yours
Bill
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: wye vs delta
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RE: wye vs delta
And believed you meant this compared to rated voltage. In other words, the rotor losses at 3 times slip would be 3 times greater then when running at rated voltage. So, 3 times the slip and 3 times the losses at 57% load would have rotor losses that are higher than at rated voltage and load.
RE: wye vs delta
1 - Let's say at full load, full voltage we have slip of "Sfl" and rotor losses of "Wfl"
2 - For conditions of full load, 57% voltage, we would have slip of 3*Sfl and rotor losses of 3*Wfl.
3 - For conditions of 57% load, full voltage, we would have slip of 1/3 * Sfl and rotor losses of 1/3 * Wfl.
4 - For conditions of 57% load, 57% voltage, we would have slip of 1*Sfl and rotor losses of 1*Wfl.
Comparing items of "a given output" (as in my quote):
Item 2 has 3x the slip and 3x the rotor losses of item 1.
Item 4 has 3x the slip and 3x the rotor losses of item 3.
Am I missing something?
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RE: wye vs delta
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RE: wye vs delta
So here is corrected version of my 15 Jun 10 14:58 post:
1 - Let's say at full load, full voltage we have slip of "Sfl" and rotor losses of "Wfl"
2 - For conditions of full load, 57% voltage, we would have slip of 3*Sfl and rotor losses of 3*Wfl.
3 - For conditions of 57% load, full voltage, we would have slip of 1/sqrt(3) * Sfl and rotor losses of 1/sqrt(3) * [57%*Wfl] =Wfl/3
4 - For conditions of 57% load, 57% voltage, we would have slip of sqrt(3)*Sfl and rotor losses of sqrt(3)*[57%*Wfl] = Wfl
So it is correct comparing 1 to 2 and 3 to 4 that the 57% voltage case has 3 times the rotor I^2*R losses (as well as 3x the stator I^2*R losses) as the full voltage case of the same load.
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RE: wye vs delta
RE: wye vs delta
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RE: wye vs delta
RE: wye vs delta
RE: wye vs delta
RE: wye vs delta
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RE: wye vs delta
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RE: wye vs delta
RE: wye vs delta
As you decrease voltage, the steady state thermal capacity goes down roughly as 1/V and the torque goes down roughly as 1/V^2. So the margein between steady state thermal limit and breakdown torque decreases as you decrease voltage and at some point go to zero. If we assume you had initial breakdown torque of 200%, you would reach the point of zero margin at around 50% voltage, at which point you have a thermal rated torque 50% of original and a breakdwon torque 50% of original.
However at 58% voltage, you still have margin. In this case torque is not limiting on a steady state basis... thermal performance is. How much margin between steady state torque and breakdown torque is required will depend on the application and user preferences.
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RE: wye vs delta
As you decrease voltage, the steady state thermal capacity goes down roughly as 1/V and the breakdown torque goes down roughly as 1/V^2.
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RE: wye vs delta
1 - does motor torque speed curve remain above load torque speed curve at all speeds.
2 - does the start take too long based on load inertia, load torque, and reduced voltage.
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RE: wye vs delta
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RE: wye vs delta
RE: wye vs delta
RE: wye vs delta
When the windings are changed from a series connection to a parallel connection, the rated voltage across the group will double and the current will be half
When the connections are changed from delta to wye, the rated voltage increases by a factor of 1.73 and the current decreases by the same factor. The voltage applied to each individual winding stays the same and the rated current of each individual winding stays the same. Hence the same KVA for all the various connections.
Bill
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"Why not the best?"
Jimmy Carter