Motor thermal model : neg. seq. current biasing factor k

obut4 · Feb 23, 2012

In motor protection relays, the thermal curve element (49TC) models the thermal capacity in a motor.
The 49 thermal element takes the load level and negative sequence current into account to creates a thermal model of the motor. Thus the thermal model creates an "equivalent current," Ieq, that best represents the actual motor dynamics:

Ieq =I*sqrt(1+k(I1/I2)^2)

where,

Ieq = equivalent thermal current in pu (unit of thermal pickup current)
I = maximum phase current in pu
I1 = positive sequence fundamental component of current in pu
I2 = negative-sequence fundamental component of current in pu
k = constant to determine additional heating caused by negative-sequence current in pu

The k value is used to calculate the contribution of the negative-sequence current flowing in the rotor due to unbalance. It is defined as:

K= Rr2/Rr2

where:
Rr2 = rotor negative-sequence resistance
Rr1 = rotor positive-sequence resistance.

QUESTION :

I need to understand why k is equal to this ratio?

For me the heating of the motor comes mainly from:
- the stator winding Rs x I^2 with I = stator current
- the rotor heating with positive sequence current Rr1x (I1)^2
- the rotor heating with negative sequence current Rr2x (I2)^2

It means heating is proportional to

Rs x I^2 + Rr1x (I1)^2 + Rr2x (I2)^2

If we divide by Rr1x (I1)^2:

[(Rs/Rr1) x( I / I1)^2] + 1 + k x (I2/I1)^2

1 + k x (I2/I1)^2 is closed to Ieq

But why does [(Rs/Rr1) x( I / I1)^2] not appear in the equivalent current Ieq?

Why is there no reference to the stator resistance Rs?

Thanks for your help,

electricpete · Feb 23, 2012

As clarification?
Ieq =I*sqrt(1+k(I1/I2)^2) should be Ieq =I*sqrt(1+k(I2/I1)^2) ?
K= Rr2/Rr2 should be K= Rr2/Rr1 ?

K gives a weighting factor for heating due to rotor current compared to stator current.
Plug K into the equation
Ieq =I*sqrt(1+R2/R1 * (I2/I1)^2)

I assume I2<< I1, so I ~~ I1.
Then the equation makes a little more sense to me:
Ieq =I1*sqrt(1+R2/R1 * (I2/I1)^2) = sqrt(I1^2 + R2/R1 * I2^2)
Ieq^2 =(I1^2 + R2/R1 * I2^2)

Take P = Ieq^2*R1 and what do you get?
P = R1*I1^2 + R2*I2^2

Where 1 subscripts are + seq and 2 are – seq, and we focus on rotor heating exclusively here.
That looks pretty reasonable to me. The only weird part was assuming I~~~I1.

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(2B)+(2B)' ?

electricpete · Feb 23, 2012

Major Correction:

K gives a weighting factor for heating due to rotor current compared to stator current.

should've been:

K gives a weighting factor for heating due to negative sequence rotor current compared to positive sequence rotor current.

=====================================
(2B)+(2B)' ?

electricpete · Feb 23, 2012

It should also be mentioned that there is a physical reason that effective rotor resistance will be higher for negative sequence than postive sequence. The negative sequence rotates backwards while rotor is rotating forward. Therefore the associated currrent has a frequency near twice line frequency, so there will be a very pronounced deep bar effect (similar to skin effect), which increases the effective resistance to negative sequence.

=====================================
(2B)+(2B)' ?

electricpete · Feb 23, 2012

The only weird part was assuming I~~~I1.

I think the author could have used I1 to be more exact, but chose I = max phase current as a conservatism/simplification.

Let’s say you are given 3 currents, then you could compute I1 if you know phase angles.

Or if you just take the highest one, you relieve yourself of the burden to do the calculation, and you know the result will be conservative (I = max phase current ? I1 = positive sequence current).

=====================================
(2B)+(2B)' ?

obut4 · Feb 23, 2012

Thanks for your answers.

Sorry for the mistakes that you have perfectly corrected.

When you say:

"Take P = Ieq^2*R1 and what do you get?
P = R1*I1^2 + R2*I2^2"

I assume R1, R2 are the rotor positive/negative sequence resistance.

P seems to depend only on the rotor resistances. What about the stator resistance? Can it be neglected?

P = Rs*I^2 + R1*I1^2 + R2*I2^2 ????

electricpete · Feb 23, 2012

I assume R1, R2 are the rotor positive/negative sequence resistance.

That’s what I had in mind... the same as your Rr1 and Rr2.

P seems to depend only on the rotor resistances. What about the stator resistance? Can it be neglected?

It’s a good question. I have some ideas, related to the assumptions and the purposes of the calculation and again simplifications that result in convervatism, but let me sit on those for a moment.

Do you have a link to the source of this equation?

=====================================
(2B)+(2B)' ?

obut4 · Feb 27, 2012

My sources for equivalent current calculation are GE and Basler.

If you have more ideas about the K factor, i 'am interested.

7anoter4 · Feb 27, 2012

Try this link:

http://www.google.co.il/url?sa=t&rc...urce=web&cd=1&sqi=2&ved=0CB8QFjAA&url=http://

http://www2.selinc.com/techpprs/thr...9KjaDQ&usg=AFQjCNHlv7rsJnuO1QXKtDIHz_SP0YsHJA

electricpete · Mar 3, 2012

I think it is just a simplification for conservatism.

If we refer the rotor to the stator, then the total I^2*R heat produced in the motor is:
I^2*Rstator + I^2 * Rrotor

where the current I is the same same for both terms assuming we have adjusted Rrotor to be referred to the stator.

To calculate total heating(*), the correction for negative sequence SHOULD apply only to the second term. But if you apply it to both terms, then you have a conservative result (predicts higher than actual heating) and a simpler calculation. I think that’s what they did.

* Also note that for large motors during starting, the heating produced in the rotor may be more relevant to the thermal limit than the heating produced in stator because large motors tend to be rotor-limited during starting.

=====================================
(2B)+(2B)' ?

electricpete · Mar 3, 2012

I should add, there is a normalization inherent in translating I^2*(Rstat+Rrot) to temperarature rise. (the unknown quantity represents a thermal resistance). In particular, at steady state full load current we expect full load temperature rise. So identifying the specific values of effective resistance Rstator+Rrotor is not critical as long as the model is calibrated (by computing thermal resistance) to give full load temperature rise at full load current. (if you input R1+R2 higher than actual, it would estimate thermal resistance lower than actual to give the same temperature rise).

The actual model of course also includes a thermal capacity.

=====================================
(2B)+(2B)' ?

electricpete · Mar 3, 2012

The above discussion is based on a "single degree of freedom" model characterized by single internal temperautre and single thermal capacity. There are of course more complicated models possible (like a 2dof model with two thermal capacities and two temperatures... perhaps one for rotor and one for stator), but it doesn't look like Basler uses that.

=====================================
(2B)+(2B)' ?

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