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Wound rotor motor parameters

Wound rotor motor parameters

Wound rotor motor parameters

(OP)
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?

RE: Wound rotor motor parameters

you should be able to find typical curve of torque vs speed with varying rotor resistance in any motor textbook.  The peak torque shifts to lower speeds as rotor resistance increases but the peak torque remains the same magnitude.   

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RE: Wound rotor motor parameters

(OP)
electricpete

Yes I understand how the torque vs speed curve shifts with varying resistance and I've seen several examples of these curves.  I am now trying to understand from a motor parameter and motor model/equation point of view why this shift takes place.

RE: Wound rotor motor parameters

You can understand it with some algebraic manipulations.  

First thing you have to figure out what is the expression for the slip where peak (breakdown) torque occurs.
You could take your expression for torque T(s) and find the maximum based on the point where the derivative of 0.  
i.e sTmax is the  s which satisfies dT/ds = 0
You can easily solve that with a computer algebra program such as Maple.  By hand it is a little tough.  Fitzgerald has a little bit of a clever short cut which does not require so much algebra to determine sTmax.

We start with Tmech = Pmech / wm = [(1-s)*Pgap] / [(1-s)*wsync]  = Pgap / wsync
Tmech = Pgap / wsync
Since wsync is a constant, the torque is maximized at whatever slip that Pgap is maximized.  We can figure that out from simple maximum power transfer principles.  The maximum power transfer into the element R2/s (i.e. airgap power)  occurs when the impedance supplying that element is equal to R2/s

R2/sTmax = sqrt(R1th^2 + (X1th + x2)^2)
sTmax = sqrt(R1th^2 + (X1th + x2)^2) / R2
From the equation above it is obvious that if R2 goes up sTmax goes down and vice versa.   We proved part of what we started.

Now plugging sTmax back into your equation for torque is again messy but if you go through the details you will see R2 cancels out and the magnitude of the peak torque is independent of R2

Now a little bit of graphical intuition.   If the peak breakdown torque shifts towards the left (lower speeds), that roughly means the torque speed curve shifts up for all speeds below the speed of max breakdown torque (hence good to have higher resistance for starting).  If the peak breakdown torque shifts to the right (higher speeds), that roughly means the torque speed curve shifts down for all speed above the speed of max breakdown torque.  
 

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RE: Wound rotor motor parameters

Sorry, a little garbled at the end. Correction in bold.

Quote:

Now a little bit of graphical intuition.   If the peak breakdown torque shifts towards the left (lower speeds), that roughly means the torque speed curve shifts up for all speeds below the speed of max breakdown torque (hence good to have higher resistance for starting).  If the peak breakdown torque shifts to the right (higher speeds), that roughly means the torque speed curve shifts up for all speed above the speed of max breakdown torque and down for all speeds below the speed of max breakdown torque.

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RE: Wound rotor motor parameters

(OP)
Pete

I follow your calculations above and understand how you derived the fact that the max torque is independent of R2.  I do not understand however from these equations why the maximum torque is shifted to the left with higher resistance?  Does this shift have something to do with the slip in the above equtions?

RE: Wound rotor motor parameters

I see why you can't follow it - the result was inverted. Let's try again:
Maximum airgap power transfer when source impedance is R2/s:
R2/sTmax = sqrt(R1th^2 + (X1th + X2)^2)

Solve for sTmax
sTmax = R2/sqrt(R1th^2 + (X1th + X2)^2)

sTmax is the slip at which max torque occurs and it is proportional to R2.  

When R2 increases, sTmax increases, the speed at which max torque occurs decreases.

When R2 decreases, sTmax decreases, the speed at which max torque occurs increases.

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RE: Wound rotor motor parameters

Since electripete did not refer to a part of your question [quote:]
"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%?" I will try to explain as follows:
Erotor is not equal to Estator*s but E'stator*s as
E'stator=Estator/(mstat/mrotor*windingfactorstator/windingfactorrotor*no.turnstator/no.turnsrotor) where:
mstator =no.of stator phases ; mrotor =no.of rotor phases
The winding factor depends upon how the coils are spread along the stator or rotor, the shorting of the turn and sometime upon rotor slot inclining.
Usually winding factor=0.9-0.7 .
Now, I think the voltage you have got is the standstill voltage[s=1].
So kwr/kws=3370/6000*sqrt(3)=.973
 

RE: Wound rotor motor parameters

Sorry, kwr/kws=3370/4000=0.84

RE: Wound rotor motor parameters

(OP)
Pete

I understand your new equation now since it solves for the slip at which torqe is at maximum.  I need to take a little more time to digest it and go through the derivation again but I get the general principal.  Thanks!

7anoter4

I have always seen and thought that the rotor induced voltage was given from the equation that I listed above.  Is this not the case, or is what I listed usually just a generalization based on being proportional to the slip?  

If the voltage I have is for standstill (s=1) then this would be independednt of any slip and appears that your equation would overide the one that I have.

Was I correct in figuring out the rotor current?

RE: Wound rotor motor parameters

I think the total rotor resistance and the reactance has to be taken into consideration.
Let's say rpm[mec]=1178 wsync=1200 then s=0.0183
Since, as electricpete explained, Pgap=3*Irotor^2*Rrotor/s if we suppose s=.0183 Pgap=Pmec/(1-s)*746=4939391 w and if Irotor=875 we will get Rrotor=0.03935 ohm let's say Xrot=10*Rrotor=0.394 ohm[at start] and negligible at rated rpm.
The wound rotor resistance usually does not present big difference from start up to rated rpm.
If Erotor=3370V and Rtot=Rrotor+Rext=2.188 then Istart=3370/sqrt(3)/sqrt(2.188^2+0.394^2)=875 A
Pgapstart=3*875^2*2.188=5025562.5 w
Pgapstart/Pgaprated=1.0174
Further, as electricpete stated, Tstart=Pgapstart/wsync then the start Torque will be 1.0174*Trated, too.
 

RE: Wound rotor motor parameters

(OP)
7anoter4

I am trying to follow your calculation.  It looks like you are saying that the rotor voltage of 3370V is at starting when s=1.  However in your calculation you are using a slip of near full speed and no where near starting speed.  So are you using a full speed slip to calculate parameters at near zero speed slip?

Once we are running at speed (1178 rpm) then the rotor current is not 875A anymore is it?

I'm just trying to follow the various steps in your calculation but do not understand what you did to get .03935ohms in the first calculation

RE: Wound rotor motor parameters

(OP)
Also the last part of your calculation does not make sense for the Tstart.  This calculated Tstart parameter is only 1% greater then the full load torque.  I would imagine that the Tstart torque should be equal to the breakdown torque since we are adding resisance to shift Tmax to a higher slip (starting).  I would expect the Tstart to be close to the breakdown torque somewhere in the area of 250%.

I will post motor datasheets tomorrow.   

RE: Wound rotor motor parameters

(OP)
Attached are motor performance curves.

The one thing I dont understand on the attached speed vs torque curves is how the maximum torque is less for 100% external resistance than it is for 10% external resistance.  I understood that all above equations kept the maximum torque the same, and only shifted the slip point where it occured.  Why then on the attached curves is the greater amount of external resistance (100%) have the lowest avaliable torque?

RE: Wound rotor motor parameters

Your formula for the rotor voltage is computing it directly from the stator voltage and slip. This isn't true because in the basic sense the rotor voltage comes from the transformer ratio of the motor. When the rotor is not rotating think of the motor as having two windings and the winding ratio will determine the secondary or rotor voltage. You are missing this winding ratio factor in your calculation.

Your determination of current would be correct for the case where the motor is not rotating and that resistor is connected. Once the motor begins to rotate then the current will change. The rated rotor current really has nothing to do with the external connected resistance. The rated rotor current is the current that flows through the rotor when the motor is running at rated speed and rated load with the rotor shorted.

If you wanted to look at it from a very simplified perspective.

The rotor voltage is inversely proportional to the speed. The rotor voltage is the rotor rated voltage at 0 speed and 0 volts at synchronous speed.

The rotor current is proportional to load. Of course, this isn't true for low loads since there will be a magnetizing current but is a good approximation near full load.
 

RE: Wound rotor motor parameters

If Rext will increase up to 0.43 ohm the start torque will grow up to the maximum[2.5*Trated].Further the start torque will decrease as Rext will increase. See the sketch:

RE: Wound rotor motor parameters

(OP)
7anoter4

I do not see where you are getting the .43ohms from?  It looks like the max torque occurs at 25% resistance which is .25*2.1493=.53 but this is not matching the .43 ohms that you have.  Where do you get this value.

So you are saying that although we have an external resistance of 2.1493ohms avaliable, we only use a maximum of .43ohms to have an effective torque at start?  I always thought that Tmax was independent of R2 and only dependent on voltage and other motor parameters?  Why do you show Tmax changing above as a function of R2.

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.

 

RE: Wound rotor motor parameters

The specs in the attached are motor terminal voltage and current, not rotor. As indicated, your rotor voltage is formed by the turns ratio and by %slip.

The current in your rotor is determined by the impedence of the rotor circuit. The reactive component is influenced by %slip or frequency difference. Notice in the chart how the pf relates to slip.

As you adjust your external resistance, you are changing your pf, which also changes with the slip. Notice in the chart how current and voltage move from highly reactive (shorted rotor) to resistive (100% resistance).

RE: Wound rotor motor parameters

Fist of all you are right:-the breakdown torque does not change with Rext. Only sTmax is changing with Rrotor. But what you do need is the Starting Torque and this will be equal to Breakdown Torque only in the case that sTmax=1.
I don't intend to demonstrate this but I jump directly to:
sTmax=Rtotref/sqrt (Rs^2+ (Xs+Xrotref)^2)
where  Rtotref=(Rrotorwinding+Rext) referred to stator [R'r]
            Xrotref=Rotor reactance referred to stator.[X'r]
I think the rotor voltage 3370 V is the secondary voltage in rotor open circuit when the primary [the stator] is supplied by rated voltage[4000 V].So the voltage drop on the stator impedance may be neglected and the rotor voltage drop is zero.
If we'll take krs= Erotor /Es=3370/4000=0.8425 then I'rotrated=875*krs and
Rtotref= (Rrotorwinding+Rext)/krs^2
Xrotref=Xrot/krs^2
The starting torque=break down torque if
 Rtotref= sqrt(Rs^2+(Xs+Xrotref)^2)
If Rs=0.0192 ohm and Xs= 14*Rs =0.2688 ohm and Xrot=13* Rrotorwinding=14*.02=0.26 ohm
Xrotref=0.26/.8425^2=0.3663
Rrotref=sqrt(0.0192^2+(0.2688+.3663)^2)=0.6354
Rrot=Rrotref*krs^2=0.45 ohm
Rext=Rrot-Rrotwinding=0.45-.02=0.43 ohm.
The factor [13or14] to multiply the Rs or Rrotor in order to get the reactance[ Xs and Xrot] it is resulted from recalculating process of  rated torque and breakdown torque.
The calculation was based on Steinmetz diagram.
In this calculation the Io[no load current] was neglected.  
 

RE: Wound rotor motor parameters

Quote:

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.

 

RE: Wound rotor motor parameters

(OP)
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?

RE: Wound rotor motor parameters

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

RE: Wound rotor motor parameters

(OP)

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.

RE: Wound rotor motor parameters

(OP)
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?

 

RE: Wound rotor motor parameters

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.

RE: Wound rotor motor parameters

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.
 

RE: Wound rotor motor parameters

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.
 

RE: Wound rotor motor parameters

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.

RE: Wound rotor motor parameters

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

RE: Wound rotor motor parameters

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".

RE: Wound rotor motor parameters

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.

 

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