Wound rotor motor parameters 1

[rockman7892]

I have recently been studying a wound rotor motor that we are getting ready to put in service, and was trying to get a better understanding for how certain parameters of the motor were calculated. The motor is a 6500hp 4000V 6-pole motor. The rotor resistance is controlled by a rheostat which is listed as having an external resistance of. 2.149 ohms/phase.

The rotor is listed as having a voltage of 3370V. Is this rotor voltage determined by the equation Erot = s * Estator where s represents the slip in the motor? I would assume that the 3370V is listed for the motor operating at full load and is a function of slip at this full load. Does this mean then that this 3370V is the rotor voltage when the motor has a slip of .84%?

The motor datasheet has the rotor connection as a wye connection. To determine the rotor current do you simply take the rotor voltage of 3370/1.73= 1947.9V and then divide this number by the resistance listed for the rheostat of 2.149ohms/phase to get 906.4A? The motor datahseet lists the rotor current as 875A so I was trying to figure out why my calculation is off.

I am also trying to see mathmatically how the rotor resistance effects current and torque. I know that as the rotor resistance increases the motor starting current decreases. In this case the motor current is being controlled by the rotor currrent as defined by R2(1-s/s). This shows that as the rotor resistance R2 increases the motor current is lower and as R2 decreases the motor current increases. Do I have this correct.

Lastly I am trying to find how the rotor resistance effects motor torque. The one equation for motor torque that I am aware of is defined by:

Tind = (3*V^2*R2/s)/Wsync[(Rth+R2/s)^2+(Xth+X2)^2]

From this equation the rtor R2 component is both in the numerator and denominator however the lower term is squared. I would think then that an increase in R2 would cause a decrease in the induced torque. How is the induced torque in this quation effected by the rotor resistance?

I may be mistaking in that the torque stays the same for the motor no matter the resistance, but the max (breakdown) torque occurs at lower speeds? How is this effected by the rotor current?

 

[LionelHutz]

Also I thought that increasing R2 brought Tmax closer to s=1 which you are showing that it does. Is there a point where if you bring R2 too high that the torque starts to decrease? Maybe that is what I am missing.

Yes, if you keep increasing the external resistance then the breakdown torque peak moves to a "negative" speed and you begin to operate on the right side of the breakdown peak. This is typically the area where these motors are operated. You want to start with a resistor that moves the rated speed point on the torque curve to 0 speed. This requires a higher resistance than moving the breakdown torque peak to 0 speed.

I assume you have a liquid rheostat? What happens is the resistance is slowly lowered once the stator is connected to line power. As long as the resistance is lowered slowly enough, the motor will only draw the current required to start and could possibly be started with less than rated current. The motor basically operates to the right side of the breakdown torque during the whole start. In the best case, it operates within the running part of the curve (100% to 0% torque) during the whole start. Otherwise, if the load requires more than rated torque, the motor will "slip" back on the torque curve until it reaches a high enough torque to keep accelerating.

 

[rockman7892]

Thanks to everyone who contibuted above.

We filled our new Rheostat this week with the water and electrolyte no problems.

So if I understand the process correctly when the Rheostat starts the rotor leads are touching a small portion of the water and thus have a certain resistane. The water level then raises and as it raises and the rotor leads are more submersed the resistance is lowered (effect of electrolyte) until finally at a certain level the rotor leads are shorted.

So what determines what the initial resistance is? The initial amount of solution covering the rotor leads? Would this value be the 0.43 value or the 2.1493 value from above. If we were saying that 0.43 ohms gave maximum torqe then how are we sure that we do not go above this, and why is the unit then rated for 2.1493 ohms?

 

[7anoter4]

It depends upon the load required torque curve .Usually at the start it is less than 1/2 of rated and it goes down a while with rpm increasing and approximately from 1/2 rated rpm goes straight to the rated torque.
In these conditions in order to maintain the motor torque curve above the load curve you have to reduce the Rext as follows:
At rpm[p.u.] Rext Load Torque[p.u.] Motor Torque[p.u.] current [p.u.]
0 100% 0.5 1 1
0.5 50% 0.3 0.5 jump to 1 0.5-1
0.68 25% 0.5 0.7-1.3 0.7-1.2
0.8 10% 0.7 0.8-1.8 0.8-1.7
0.9 5% 0.8 1-1.7 0.8-1.7
0.95 0% 0.9 1 1

 

[rockman7892]

We finally started this motor, and were very sucessful with its start. The complete sarting time for the Rheostat/Motor combination ended up being 42s which was a factor of how long the Rheostat taks to step through its varying resistance levels.

Full load current on this motor is 829A and when starting the highest reading I saw on the motor was 1060A right at the begining of the start. This current was only for a brief amount of time and then quickly dropped down to below full load (829A) for the remainder of the start. To me this looked very good and the relay indicated that only 1% of the motor thermal capacity was used during the start. This would almost indicate to me that this motor could be started time and time again (But obviously we would never attempt)

At first this starting time of 42s seemed long to me but after seeing the motor start several times it appears to work great. Is this long starting time typical for this type of application and are there any drawbacks due to this? I have heard of another location with a similar size motor and arrangement that has a starting time of 20s or so.

 

[rockman7892]

I was talking with someone regarding this starting time and they told me that they believed the normal starting time was between 20-30 sec, but could not give me a good explanation of why. It was just stated that in his experience, this is what he has seen?

Does anyone else have similar numbers or starting time information to share regarding how long the starting sequence should be set for?

 

[LionelHutz]

Speed up the process. When the current draw during acceleration is or rises to around say 800-1000A then you've found the correct time.

If you don't care about the acceleration time then what you have sounds fine.

 

[7anoter4]

LionelHutz is right. I think that the load torque curve as I drew it in the above post is very "optimistic". Actually, the load torque curve is located higher so the acceleration torque [Ta]
[equal to the difference between Motor Torque and Load Torque] is small. Then the acceleration time [ta] will be elevated since ta=Wk^2*Delta N/308/Ta.[Ta small= ta big].
If, for instance, you'll change the Rext from 100% to 50% at 25%velocity [300 rpm] instead of 50% [600 rpm] and further you'll change to 25%Rext at 50%velocity [600 rpm], to 10% at 68%velocity, to 5%Rext at 80% and to 0 % at 90%, you'll get only 12.5 seconds total acceleration time.
But, in this arrangement the current will be from 1.5 to 5.5[p.u].If the stall time [state by manufacturer] would be 10-13 second, this could be fair.

 

[LionelHutz]

Your numbers seemed odd to me unless you just posted them differently. Ideally, you always operate the motor to the right of the breakdown torque. If so, the current and torque start out high on each step and drop as the motor accelerates. Then, at the next step, the torque and current jump up again and drop off again as the motor accelerates to the next speed.

Like from 0 -> 50% speed I'd expect
torque to go 100% -> 50%
current to go 100% -> 50%

Then, when the next step is energized the torque and current would jump up again and the process would repeat.

But then, if the torque at 50% speed is 30%, the motor will continue to accelerate past 50% speed until the motor torque and load torque are equal.

This motor has a liquid rheostat. Ideally, you want the rheostat resistance to lower directly proportional to the motor speed. Then, the motor would remain at 100% torque the whole start - it would start the load as quickly as possible without going above rated current.

This is not possible with a feedback loop though, so the rheostat speed can just be set to keep the motor current from going above (or at least not much above) the rated current.

 

[7anoter4]

If the Rext is a liquid rheostat than one can change the resistance continuously, indeed. I did the last appreciation based on the manufacturer attached curves for Rext =100%,50%,25%,10%,5% and 0.The acceleration time I calculated based on relation: ta=Wk^2*DeltaN/308/Ta. where :
Wk^2= 31750 Lbft² [rotor+load] ; Delta N= Delta N[p.u.]*1200 rpm. Ta=(av.motorTq[p.u.]-av.loadTq[p.u.])*Tn
Where Tn[rated Torque]= 29611 lb.ft .
I took the Rext switching rpm deliberately at 50%-600 rpm- [first time] and 25%-300 rpm [second time] and so on.
The Load Torque Curve is calculated so that total ta=40 sec. [in first Rext switching proposal].
Here is the result table :
Rext Delta N average average
switching [p.u.] Rext.[%] loadTq[pu] motorTq[pu] ta[sec]
rpm[p.u.]
0-0.25 0.25 100% 0.775 0.9 8.355
0.25-0.50 0.25 50% 0.565 1.25 1.524
0.50-0.68 0.18 25% 0.725 1.3 1.308
0.68-0.8 0.12 10% 0.825 1.9 0.466
0.8-0.9 0.10 5% 0.875 2.2 0.315
0.9-0.98 0.08 0% 0.95 1.8 0.393
Total sec.= 12.363
The attached sketch may clear more the explanation.

 

[rhatcher]

How can the load torque be different from the motor torque? The motor will accelerate as fast as the torque that it produces allows.

 

[LionelHutz]

rhatcher - the load torque is the torque the motor must produce just to keep the load turning. The motor torque is the torque the motor produces. The difference between the motor torque and the load torque is the accelerating torque.


Yes, the curves help and that is exactly the type of acceleration that occurs. I'll just note that you want to avoid a situation like the last step where there is a large current and torque peak. You would try to accelerate closer to full load before shorting.

I just realized the last line in the last post should have read "This is not possible without a feedback loop".

 

[rhatcher]

Thanks LionelHutz, I was having a "no brain" moment when I pondered that question. My thoughts were: "How can the torque output of the motor not equal the torque applied to the load?" Your explanation of what 7anoter4 meant by "load torque" is the textbook definition of that term and I agree that is what he meant.

That being said, the shape of that torque curve is unusual. Definitely not variable torque in the normal sense and also not quite constant torque. Do you think that it is a measured value from a load cell or a calculated value based on load intertia, motor toqrue, and acceleration time?

Also, I am thinking that the OP's description of the resistor(s) as a rheostat implies that it is continually adjustable from 0-100%. That would imply a curved acceleration rate based on the rate that the rheostat changes rather than the steps that the graph shows. If so, it would allow the OP a lot of flexibilty in determining acceleration time.