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?

 

[electricpete]

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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[rockman7892]

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.

 

[electricpete]

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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[electricpete]

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

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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[rockman7892]

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?

 

[electricpete]

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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[7anoter4]

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

 

[7anoter4]

Sorry, kwr/kws=3370/4000=0.84

 

[rockman7892]

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?

 

[7anoter4]

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.

 

[rockman7892]

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

 

[rockman7892]

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.

 

[rockman7892]

Attached are motor parameters

 

[rockman7892]

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?

 

[LionelHutz]

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.

 

[7anoter4]

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:

 

[rockman7892]

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.

 

[automatic2]

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).

 

[7anoter4]

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.