# Pros & Cons of Wye vs. Delta motor windings.

#### Blue-Photon

Joined Jan 2, 2012
4
What are the advantages / disadvantages of connecting a 3Phase motor in Delta vs. Wye?

(I have been searching for answers to this question for some time. All forums I've visited gave only half baked answers, so I'm hoping this one will be a little more professional.)

I have a small (0.75HP) 6 wire motor - currently connected as Y with center floating. The name plate states: Y/Δ, 230/380 VAC, which tells me I can configure it both ways. If possible I'd like to get more power out of it, so the question is will a Delta connection achieve this?

Many people have stated that there is no power (or other) advantage or disadvantage between the two connections, but if that is the case, why would so many (larger) motors use Wye-Delta starters? These motors clearly "kick-in" when they switch from Y to Δ. If there is no advantage, then why aren't all motors simple connected in Y, and left that way (Y/Δ starters are much more complex and expensive)?

Blue-Photon

#### strantor

Joined Oct 3, 2010
5,636

#### jimkeith

Joined Oct 26, 2011
540
There is no power output advantage when connected Δ.

#### Blue-Photon

Joined Jan 2, 2012
4
Hi Strantor,

Thanks for your quick answer. I fully realize and understand what you have stated, but this dosen't quite answer my question.

I understand that Vpp in Y = Vpp in Δ, and in Wye Vpn = Vpp/√3, so the voltage from phase to neutral is reduced by √3.

If what you have said is correct (that this motor is designed for Y380 and Δ230), then connecting in Delta 380 would definately increase the power, but it could also run hot and fry the windings. To my knowledge however there is no Δ230 (readily) available in Europe. We only have Y380 phase to phase (230 phase to neutral), and Δ380.

Blue-Photon

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#### jimkeith

Joined Oct 26, 2011
540
The winding current is unchanged in either connection--the phase current does change by the √3 factor.

#### GetDeviceInfo

Joined Jun 7, 2009
1,878
your question on wye/delta start has more to do with the nature of the load. The motor just converts electrical energy to mechanical energy. Considering Stantors comments, with a Delta configuration you'll have full torque applied instantaneously to the load. If the load has high inertia, this can be very hard on transmission components, and will result in extended starting currents on the motor, causing high heat generation. Motor overload devices are placed inline to protect motors from extended current draws and when properly sized, will prevent such starts. With the Wye connection and the subsequent reduction of voltage on the windings, your available horsepower is cut to 1/3 rating. This greatly reduces the strain on the drive on start up. Once the drive is up to some speed, it is switched over to Delta to obtain full rated HP.

Not all loads are of this nature. In fact most loads in industry can be started without HP reduction. Electronic softstarters are now being integrated with power contactors and the costs are giving them wider usage. I've seen Wye/Delta starters applied to drives down to 1 Hp.

The other concept is as Stantor mentioned. By reconnecting your windings, you have an option of source voltages that you can connect to. Some of the most useful motors in industry are dual voltage motors. These are not Wye/Delta, but the optional voltage is immensly usefull.

I have a small (0.75HP) 6 wire motor - currently connected as Y with center floating. The name plate states: Y/Δ, 230/380 VAC, which tells me I can configure it both ways. If possible I'd like to get more power out of it, so the question is will a Delta connection achieve this?
Yes, but you cannot exceed your current ratings, so again it's more of a question, can you drive a bigger load? Yes, within the ratings of the motor.

With all that being said, you should probably give us full nameplate data, as you may have a multispeed motor.

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#### Blue-Photon

Joined Jan 2, 2012
4
your question on wye/delta start has more to do with the nature of the load. ...
Yes (you will get more power), but you cannot exceed your current ratings, so again it's more of a question, can you drive a bigger load? Yes, within the ratings of the motor.
So if I understood correctly:

If I connect this motor in Δ(380) I will theoretically increase the possible power output.

If however in Y connection it is already reaching its maximum power rating, pushing it to Δ to deliver more power could (will) cause it to overheat.

This makes more sense to me.

#### strantor

Joined Oct 3, 2010
5,636
So if I understood correctly:

If I connect this motor in Δ(380) I will theoretically increase the possible power output.

If however in Y connection it is already reaching its maximum power rating, pushing it to Δ to deliver more power could (will) cause it to overheat.

This makes more sense to me.
yes. you could theoretically wire it in delta at 380V but you would have to derate it.

#### Blue-Photon

Joined Jan 2, 2012
4

Things are much clearer for me now. It's nice to clear out a few cobwebs, since my EE courses date back 30+ years (I'm mechanical not electrical, but don't hold it against me).

Cheers, and happy new year!

Blue Photon

___________

Fear of the Lord is the beginning of wisdom. Pr. 1:7

#### jimkeith

Joined Oct 26, 2011
540
yes. you could theoretically wire it in delta at 380V but you would have to derate it.
This will saturate the magnetics and probably will not provide significant additional torque. In saturation, the current will dramatically increase and overheat the motor. You must derate the voltage in this case--back to 230VAC.

#### darkfeffy

Joined Apr 1, 2009
12

Things are much clearer for me now. It's nice to clear out a few cobwebs, since my EE courses date back 30+ years (I'm mechanical not electrical, but don't hold it against me).

Cheers, and happy new year!

Blue Photon

___________

Fear of the Lord is the beginning of wisdom. Pr. 1:7
Hi Blue-Photon,
I saw your post today while searching for something on the internet. Now, while I appreciate the answers provided by the other members, and even though you might have already moved on from this topic, I will provide you with mine as I wasn't exactly satisfied by the answers I saw.
The situation is a little bit nuanced. Let me explain.
First, we should look at the steady-state torque equation of the induction machine (which results from the circuit model of the machine). I have a photo of the equation here below (I can't find equation insert functions on this website):

In the photo of the torque equation, a, b, c, and d represent circuit parameters which I have not shown in further detail here. 's' is the slip (which is the ratio between the difference of the rotor speed and synchronous speed to the synchronous speed). The slip is high for low speeds (=1 at stall) and low for high speeds (approx =0 when no load is attached to the shaft). V_line-neutral is the equivalent line-to-neutral voltage. Notice that it says "line-to-neutral". This means that no matter what voltage you have at the input or how the stator windings are connected, you must obtain an equivalent line-to-neutral voltage. So theoretically, there is no difference between using a 400V line-line with the windings connected in delta and a 230V line-neutral with the windings connected in star. In both cases, the equivalent line-neutral voltage is 230V. However, if you used a 230V line-line with the windings connected in delta, then that situation is different from the 230V line-neutral with windings connected in star.
From that equation, you can see that the torque depends on the slip and the rms of the line-neutral voltage (squared) . If you plot this torque, then you get a graph as in the second photo here below:

In this second photo, as the rms line-neutral voltage increases, the peak torque increases (and the locked rotor torque increases too), but the shape remains practically the same. Now the reason for using the star-delta starting method is because of that locked rotor torque, which as you can see increases as the voltage increases. There is another expression of the torque which gives us the torque as a function of the stator current. I have not written it here, but it can be used to compute the stator current when the torque is known; this will show that at startup, the current is really high but this high value reduces with decreasing voltage.

Finally, the torque curve is a characteristic of the motor, meaning that in steady state, it is the mechanical load that determines what the machine torque (and therefore speed) will be. I have made a zoom in on one arbitrary operating point. It is shown in the third figure, here below:

Here you can see that as the line-to-neutral voltage increases the speed at which the motor runs increases (or the slip decreases). There are other machine equations (I will spare you the details) that relate the electrical losses to the slip. The higher the slip, the higher the electrical losses. So, it is advantageous to run the motor at a higher voltage because you will get less heating in the windings. This would mean that you could get more power out of the machine (without breaking it) if you ran at a higher voltage. However, there is a caveat: a higher and higher voltages result in higher and higher magnetic flux in the stator and rotor cores (this is Faraday's law or Maxwell's 3rd law), which can saturate the machine (at saturation, the windings will all appear as shorts which will cause a very huge current to flow, possibly damaging the motor). So you also don't want very high voltages for the sake of preventing saturation and avoiding insulation breakdown.

I hope this was a more convincing response.

PS. it seems the inline photo insertion doesn't work. So I have put them in this mail as attached files.

regards

#### darkfeffy

Joined Apr 1, 2009
12
Oops, the last figure is turned upside down, that's weird ...

#### t_n_k

Joined Mar 6, 2009
5,455

#### darkfeffy

Joined Apr 1, 2009
12
Well, perhaps in the future some other person doing random searches on the internet will fall on this same question just like I did. I'd like for them to have my point of view.

#### narenvcm

Joined Jan 19, 2015
1
Delta vs Wye phase connections
The delta and wye motors provide the exact same performance in theory. The delta has a lower resistance, a lower torque constant and a higher current. The end result is that force/torque is exactly the same.

A delta winding is used if more speed is needed, and the winding can not be scaled, and more voltage is not available.

A delta winding should be avoided because it allows currents to circulate in a circle in the delta if the back emf wave form has odd harmonic content. This causes additional losses. Therefore, a scaled wye windings is better for high speed than a delta winding.

For a wye wound motor as compared to a delta wound motor, presuming each phase resistance is the same, between any two phases, a delta wound motor has 1/3 the resistance as compared to the wye.

The wye-wound has higher resistance and higher inductance, better for drives with lower switching frequencies.
Above stated reason is very imp for choosing wye or delta
where we need to start or stop motor frequently, we use delta and elsewhere wye