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wye vs delta

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

The voltage across the single windings will be sqrt(3) lower. That makes peak torque three times lower. The actual power output before the motor gets too hot depends mostly on the slip. It increases three times and will reduce possible output quite a lot.

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
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

RE: wye vs delta

In a given application that ACTUAL horsepower output will match the load, as long as you don't hit breakdown torque.  There may be slight slowing of motor which reduces load demanded, but that is usually a secondary effect.  The bottom line, ACTUAL horsepower output when connected to a given load is likely not affected much by the change from wye to 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

Since you mentioned impedance effects, I thought you might be interested in knowing actual vs limiting performance.  If I have misunderstood your question please disregard.

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RE: wye vs delta

I have always taken the somewhat simplistic approach that allowable HP was limited by the heating of the motor and that that in turn was limited by the maximum allowable current through the windings.
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
--------------------
"Why not the best?"
Jimmy Carter

RE: wye vs delta

I^2*R losses are tpyically 50 - 75% of the losses. If they were 100% the current-based analysis would be dead on.  Core losses will go the opposite direction (they will be lower in delta than wye). Friction/windage losses will be relatively unchanged. Stray losses I think may go up.

 

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RE: wye vs delta

Whoops.  Core losses would go up in delta compared to wye.

 

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RE: wye vs delta

...so core loss is lower in wye compared to delta. Hence opposite direction of I^2^R loss for a given input current. (I was right about the opposite part).

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RE: wye vs delta

(OP)
If the impedance on the delta motor is prefixed by the manufacturer and it rewired for wye from delta then it would have to change by the square root of three right or a factor of three
I am thinking the motor would only usable at 57.7% of nameplate hp anything above that would cause overheating  

 

RE: wye vs delta

So the I2R losses will be the same with the same current in the windings. That takes care of about 50% to 75% of the losses. Windage and bearing losses will be the same at the same speed.
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
--------------------
"Why not the best?"
Jimmy Carter

RE: wye vs delta

The torque curve of the motor will be lowered to 33.3% of the rated torque curve. So, to get 57.7% of rated power you will need a motor with a breakdown peak greater than 173.3% of rated torque.

 

RE: wye vs delta

Looking at typical torque curves in the Cowern Papers, that is borderline for design "B" motors, possibly the most common motors. If you need full torque at low speed, these motors will stall.

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

(OP)
so did we decide what would be the outcome, i remember this from college we did mathematical models with resistors but i didnt really understand it then i just accepted it and memorized it for the test, now i want to really know it, no more guesing. it seems that it would be 57.8% of nameplate hp although the mathematics indicate it would 33.3% but that would raise input kva by 1.732. now each coil produces 33.3% of total hp balanced load of course so input kva is product of 3 times kw of each coil or 33.3% times 3. or another to look at it is if the total sum of individual phases equals the single phase input equivalent then it wired correctly if the input is 1.732 higher than hp output it is wrong and going to burn up. last ditch effort on this is to set it up on a bench with a vfd and run it up in delta and and a transducer to measure hp, run it up in wye and measure differences.  

RE: wye vs delta

Well, this is being over thought in odd ways.

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

I would say the rotor I^2*R losses increase in roughly the same proportion as stator I^2*R losses.
  
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

So, 3 times as much rotor I^2*r loss at 57% power would mean there is more heating of the rotor than at full load???
 

RE: wye vs delta

No at 57% it would be the same.  My response was a reply to your question: Will the extra slip cause more loss in the rotor than when running at rated speed and torque?
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

Ok sorry I misread your question. Missed the "than" part.

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

And to be honest I'm still not sure what your question was if it was not answered by "rotor I^2*R varies the same as stator I^2*R"

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RE: wye vs delta

Hi Pete;
I think that the OP would like you to simplify

Quote:

And to be honest I'm still not sure what your question was if it was not answered by "rotor I^2*R varies the same as stator I^2*R"
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

ok, I'll wait to see what further clarification is requested by anyone.  My last 3 posts were response to Lionel.

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RE: wye vs delta

Electricpete - Your first answer was not very clear. I read this part;

Quote:

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.

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

It looks correct to me.

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

Whoops, maybe you have a point.  Slip would be 1/sqrt(3) at the 57% point.   Let me think on that.

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RE: wye vs delta

I tracked down the error in my 15 Jun 10 14:58 post... on items 3 and 4 we need to multiply slip fraction and load fraction together to obtain the losses (because rotor losses represent a fraction slip of OUTPUT LOAD).

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

(OP)
So if the motor is 7.5 hp and delta wound ran on 240vac at 18 amps then rewired for wye at 240 vac then what is the hp output and amp draw of that motor

RE: wye vs delta

As others stated:  steady state limit 4.3hp and corresponding current 10.4A (under assumption that I^2*R losses dominate).  Motor will not have much room to accommodate momentary overloads since torque vs speed curve is reduced by factor of 3.  

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RE: wye vs delta

(OP)
Would it be 10 amps at 240vac and 4.33 hp or 2.5hp  

RE: wye vs delta

(OP)
Since torque drops by three would it 2.d hp
 

RE: wye vs delta

(OP)
That is to say 2.5 hp  

RE: wye vs delta

You would need to look at breakdown torque to see if torque  is limiting on steady state basis or more likely limiting for momentarily overload. If you start with 200% breakdown torque at rated voltage, than at 58% voltage (wye) you end up with 1/3 that or 200%/3 = 67% which is not as limiting on a steady state basis as our 58% thermal limit.    But if you are operating very close to the breakdown torque than relatively brief small overload may cause the motor to trip.  

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RE: wye vs delta

So the motor can likely be operated at 4.3hp on a steady state basis, but you are more likely to trip on any small overload than a motor operated a rated load and voltage.

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RE: wye vs delta

(OP)
The thing that's not clear is the torque because if it lowers to 33.3% then would hp be 2.5

RE: wye vs delta

Let's back up to the 7.5hp motor.  The 7.5 hp corresponds to a thermal limit, not a torque limit.  The motor likely has a breakdown torque which is in the neighborhood of 200% of rated torque (see NEMA tables for exact value).  You have lots of margin between torque corresponding to steady state 7.5hp thermal limit and the breakdown torque.

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

Clarification in bold:
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

I'm not sure if it was mentioned but starting would be something to look at as well:
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

troyboy - Reading thru your comments (discussion of impedance in 1st post, discussion of test bed, and recent discussion of torque)  I THINK I see a theme that suggests you believe motor determines the power level.    It is not the case... the power level is determined primarily by the load.  The motor delivers whatever torque is required to drive the load near sync speed (as long as it is within motor capability).     I apologize if I misunderstood your comments.  

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RE: wye vs delta

(OP)
thanks electricpete i got it now breakdown thermal and steady torque are closer as voltage decreases so that stall overlaps at a certain voltage and torque. the bigger the gap between any two of them allows the motor to run on other than manufactured voltages.

RE: wye vs delta

(OP)
hey electricpete i got a motor here that is 3hp with 12 leads it has a nameplate that has delta wye double wye and double delta symbols on it with their respective voltages and amperages written on it. each configuration is 2400watts three phase for 3hp at 850 rpm. how can the different wiring configurations carry the 2400w when their is a difference as we have discussed earlier?

RE: wye vs delta

The voltage and current through each of the individual coils will always be the same.
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
--------------------
"Why not the best?"
Jimmy Carter

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