Time to delta T

[sreid]

Here's another question directed at the Math Gurus. Copper windings are often hit with short bursts of current far above their continous rating. The question is how long can that power burst be applied and not exceed the magnet wire varnish temperature rating.

To get a limiting case, one can assume isothermal conditions [no heat escapes from the wire] and it is easy to derive the rate of temperature rise as

dT/dt = [Watts}/[Copper Mass x Copper Specific Heat]

This assumes the Copper resistance remains constant but the copper resisrance rises significantly with temperature. So the question is, "What's the equation for time to Delta Temperature including the change in resistance with temperature?"

There are two cases: 1) Constant Voltage [the current decreases with rising temperature] and 2) Constant Current [the Wattage increases with rising resistance].

 

[ScottyUK]

The adiabatic equation frequently used in cable sizing calculations might be a starting point. My TGML skills ain't good enough to get it on screen but some Googling will find it.




----------------------------------

If we learn from our mistakes I'm getting a great education!

 

[electricpete]

For the constant current case:
dT / dt = I^2 * R(T) / C
where C is heat capacity, R is resistance

We need to assume a relationship for variation of resistance with temperature.
Assume R(T) = T0 + R1*T

dT / dt = ( I^2 * R0 + I^2 * R1* T(t) ) / C

dT / dt – I^2 * R1 * T(t) / C = I^2 * R0 / C [equation 1]

Let sigma = - I^2 * R1/C
Let mu(t) = exp(sigma *t)

equation 1 can be rewritten as
dT / dt + sigma * T(t) = I^2 * R0 / C [equation 1]

multiply both sides equation 1 by integrating factor mu(t)
mu(t) * dT/dt + sigma * mu(t)*T = mu(t) * I^2 * R0 / C [equation 2]

We have constructed eqation 2 in order to make the left hand side be equal to d/dt [mu(t) * T(t)]. This has not yet been shown, but will now be proven:
d/dt [mu(t) * T(t)] = mu* T' + mu' * T (by product rule)
from the definition of mu, we can see that mu' = sigma * mu, so we have
d/dt [mu(t) * T(t)] = mu* T' + sigma * mu * T [equation 3]
We find the left hand side of equation 2 has the desired form and we substitute equation 3 into the left hand side of equation 2 to give:
d/dt [mu(t) * T(t)] = mu(t) * I^2 * R0 / C

Now we integrate each side:
mu(t) * T(t) = I^2 * R0 / C * integral(mu(t), dt) + C1
(where C1 is an integration constant)

subsitute in definition of mu(t):
exp(sigma *t) * T(t) = I^2 * R0 / C * [integral( exp(sigma *t), dt )] + C1
exp(sigma *t) * T(t) = I^2 * R0 / C * [exp(sigma*t) / sigma ] + C1

Multiply by exp(-sigma*t)
T(t) = I^2 * R0 / C * [1 / sigma] + C1*exp(-sigma*t)

Subsitute in sigma = - I^2 * R1/C
T(t) = I^2 * R0 / C * [1 / sigma ] + C1*exp(-sigma*t)
T(t) = I^2 * R0 / C * [-C/<I^2*R1>] + C1*exp(I^2 * R1*t/C)
T(t) = -R0/R1 + C1*exp(I^2 * R1*t/C)]
Solve for C1 using the relationship T(t=0) = Tinit
Tinit = -R0/R1 + C1*exp(I^2 * R1*0/C)]
Tinit = -R0/R1 + C1
C1 = Tinit + R0/R1

Plug C1 back into the solution:
T(t) = -R0/R1 + (Tinit + R0/R1) *exp(I^2 * R1*t/C)]

That was a lot of work. Could be solved perhaps for t=tfinal as a function of Tinit and T(t=tfinal)

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

 

[electricpete]

Correction

Assume R(T) = T0 + R1*T

of course should have been

Assume R(T) = R0 + R1*T

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

 

[electricpete]

Could be solved perhaps for t=tfinal as a function of Tinit and T(t=tfinal)

The solution for tfinal would be:
tfinal = ln{<Tfinal*R1+R0>/<R0+Tinit*R1>} * C/[I1^2*R1]


=====================================
Eng-tips forums: The best place on the web for engineering discussions.

 

[electricpete]

I should have mentioned that the form of the resistance variation: R = R0 + R1 * T implies that T represents temperature rise above the reference temperature associated with resistance R0. Sorry, for the clumsy notation. I will see if I can re-format it a little. Perhaps I can put it in a format where it is easier to compare to other published forms of the same relationship mentioned by Scotty (I see the result published in several books without derivation).

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

 

[sreid]

Electricpete and ScottyUK,

I had never thought to use wire hesting equations as I didn't know they existed, I have always used code tables. I did think that this problem must have been solved [many times probably]but I've never seen it. And I have a real need in that Voice Coils are often used with brief power bursts.

Having said that, thanks again, Electricpete for your timely response. I plan to use the equation in a document and planned to call it "Electricpete's Equation" unless you wanted a different name. I still may [attribution], this is the first and only place I've seen it.

 

[electricpete]

In the analysis above I have written the relationship:
R = R0 + R1 * T
where as stated before T must represent the temperature rise above reference temperature.

Let us compare to more traditional notation.
R = R20 * [1 + alpha * (TC – 20) ]
which can be rewritten as
R = R20 + R20* alpha *(TC – 20) ]

where R20 is resistance at 20C, alpha is temperature coefficient expressed as ohms/(ohm-C), TC is temperature in degrees C and 20 is of course our reference temeprature.

By comparing these two equations, we can see they corresponds if we subsitute:
R0 -> R20
R20 * alpha -> R1
T -> (TC – 20)

If you make these substitutions into my solution above for Tfinal, (and divide the numerator and denominator inside brackets by alpha for simplificatio), you come up with:
tfinal := ln {(TCfinal-20+1/alpha)/(TCinit-20+1/alpha} * C / [I1^2*alpha*R20] {equation ep1}

In the event that the substitution of symbols at the end of the proof bothers you (it shouldn't), I have reworked the problem from the beginning using TC, R20, and alpha. The algebra is missing (since Maple did all the work), but the important think is that the answer is the same as equation ep1 immediately above, which imo validates my conclusion (no algebra errorrs along the way). As long as the differential equation is set up right, Maple will give us the right answer to the question we ask it. Here is the Maple solution which recreates the same equation ep1:

http://home.comcast.net/~electricpete/eng-tips/AdiabaticTemperatureRise.pdf

=============
Now here is an excerpt from Electric Cables Handbook Chapter 9, pages 152-153:

http://home.comcast.net/~electricpete/eng-tips/ElecCableHdbk.pdf

Equation 9.1 is
I^2 = [K^2 * S^2 / T ] * ln {(Theta1 + beta) / (Theta0 + beta)}

Let's make some substitutaions from their notation to my notation:
T -> tfinal
Theta1 -> Tfinal
Theta0 -> Tinit
beta -> 1/alpha

I^2 = [K^2 * S^2 / tfinal ] * ln {(Tfinal + 1/alpha) / (Tinit + 1/alpha)} (EQ 9.1A)
Note the quantities inside the brackets seem to be missing the –20C compared to my equation. But the physical difference of importance is the difference to the reference temperature of 20C (which is the reference temperature they are using as shown on page 153). Their expression Tfinal without any reference temeprature suggests a temperature difference from 0C. But 0C has no physical significance in this problem.

I conclude they made a mistake. The only logical thing is to assume they have used the following temeprature relationship:
R = R20 ( 1 + alpha TC)
rather than the correct tempertaure relationship
R = R20 ( 1 + alpha (TC-20))

So I am going to correct their mistake and express temperature as a difference to the reference temeprature (expected) rather than as a difference to 0C which has no significance in the problem. That gives us:
I^2 = [K^2 * S^2 / tfinal ] * ln {(Tfinal –20C + 1/alpha) / (Tinit –20C + 1/alpha)} (EQ 9.1B)

Swap the location of tfinal and I^2
tfinal = [K^2 * S^2 / I^2 ] * ln {(Tfinal + 1/alpha) / (Tinit + 1/alpha)}

Now look at S^2 * K^2
{S^2} * {K^2} = {Qc * (1/alpha + 20) / Rho20} * {Area^2}
where Qc is (heat capacity) in Joules * degC /Volume

Break this up into three factors: [Qc*Area], [1/Rho20*Area], and [1/alpha + 20]:

[Qc * Area] = C expressed on a per-length basis

[1 / Rho20 *Area] = 1/R20 where R20 is expressed on a per-length basis

[1/alpha + 20] - the origin of the 20 here again is unclear, but I suspect it arises from the same erroneuous assumption as above regarding form of the resistance vs temperature relationship. I'm going to discard it and stick with 1/alpha (because that is what is required to make the expressions match).

That gives us
S^2 * K^2 = C * (1/R20) * (1/alpha)
where again C and R20 are per-length

Susbsituting S^2 * K^2 = C * 1/R20 * 1/alpha back into (EQ 9.1B) gives us:
tfinal = [C / (R20 *alpha * I^2) ] * ln {(Tfinal –20C + 1/alpha) / (Tinit –20C + 1/alpha)} (EQ 9.1C)


After all this we see that my original equation ep1 posted above matched EQ 9.1C which was obtained from the Cable Handbook equation 9.1 by adding two "corrections" that involve treatment of the 20C reference tmperature. Again by physical considerations, I don't see how it is relevant in this context to list a temperature in C without expressing it as a difference from a temeprature that has physical relevance such as 20C and so the appearance of temperature without a subtracted reference temeprature must be wrong in my opinion (can anyone argue otherwise?). Further, I have seen the derivation of mine from first principles, while theirs was handed down without explanation. Personally, I will choose the one that I have seen proved within my own eyes.

Of course, I could always be wrong. I will be happy if someone can prove me wrong or provide additional insight by explaining the difference in these two sets of results in some other way.


=====================================
Eng-tips forums: The best place on the web for engineering discussions.

 

[sreid]

electricpete,

Wow! Thanks again, you have provided more that I could have ever expected. It is an interesting [and useful IMHO] derivation. I will check the work in detail soon but as you say, if Maple came up with the same answer, that's really all the back check one needs.

I would like to attribute the equations to you as "Electric Pete's Equation(s)" unless you'd prefer something else. Plagerism may be the secret to success but I hate that kind of sleeze.

 

[electricpete]

LOL. I'll take all the credit I can get... deserved or not.

Obviously other people have solved the problem before. Just wish they'd have showed us their derivations so we could reconcile the difference in results.

The difference between 1/alpha and (1/alpha + 20) is small <8% for copper.

The difference ln{()/(Tinit –20C + 1/alpha)} and ln{()/(Tinit + 1/alpha)} would be pretty darned big if Tinit is anywhere near 20C.

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

 

[electricpete]

No, I take it back, that second difference is small also. 1/alpha is up around 250C. So difference of 20 inside the ln would be a small difference (unless we are talking about temperatures far below 0C which is outside my range of interest). So even though I can't reconcile the differences, I'm pretty sure the two equations would come out within 10% of each other for the answer tfinal.

But still the nagging question why the difference...

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

 

[electricpete]

I found another reference which lists exactly the same equation as the Electric Cable Handbook equation 9.1 above.

It is the ABB Handbook of Electrical Devices, Appendix D. However they define their symbols more completely than the Cable Handbook. They are using a reference resistance (R20) corresponding to 20C, and a fractional temperature coefficient of resistance alpha corresponding to 0C. I was assuming that the temperature coefficient would have the same reference temperature as the reference resistance, but that is not the case here.

I'm pretty sure that fact (using reference resistance at 20C and coefficient of resistance at 0C) will reconcile the differences. If I get a chance I will write it out to show the expressions are equivalent.


=====================================
Eng-tips forums: The best place on the web for engineering discussions.

 

[electricpete]

I have reconciled the differences in the equations and now I am sure my equation ep1 above is correct if the alpha is based on 20C (call it alpha20). Also I derived an alternate form below which matches the Cable Handbook if alpha is based on 0C (call it alpha0).

Attached is an excerpt from above-referenced ABB handbook. On the right side of the page you see the same equation as in the cable handbook. There parameter B (similar to Beta in cable handbook) is listed as "the reciprocal of temperature coefficient of resistivity at 0C", while the resistivity itself is given at 20C. The only reason I provided this attachment is to establish that the result is based on use of resistance standard at 20C and temperature coefficient at 0C.

Start with the expression I derived as posted 21 Feb 09 12:41

tfinal = ln{<Tfinal*R1+R0>/<R0+Tinit*R1>} * C/[I1^2*R1] [eq A]

As previously stated this result was derived using the following relationship:
R(T) = R0 + R1 * T [equation B]
where we have not yet defined R0, R1, T, but the result is valid for any collection of definitions of R0, R1, T that give the correct value R(T).

Let us start by defining T and R0 in a very natural way:
T is temperature in degrees C
R0 is resistance at 0 degrees C

Now we wish to determine the proper values of R1 assuming we are using 20C as the temperature for the reference resistance (call it R20) and 0C as the reference temperature for temperature coefficient (call it alpha0.... not much different than alpha20).

Using R0 and alpha0 as a reference, the general relationship for resistance would be:
R(T) = R0 * (1 + alpha0 * T) = R0 + R0 * alpha0 * T [equation C]

We can use this equation C to find R0 in terms of R20 and alpha0:
R20 = R0 * (1 + alpha0 * 20)

Solve for R0
R0 = R20 / ( 1 + alpha0 * 20)

We can also compare equation B to equation C to solve for R1
R(T) = R0 + R1 * T [equation A]
R(T) = R0 + R0 * alpha0 * T [equation B]

By inspection of the above equations A and B, we can see that
R1 = alpha0 * R0

And plugging in the equation for R0 into the result for R1 we have:
R1 = alpha0 * R20 / ( 1 + alpha0 * 20)
Divide numerator and denominator by alpha0
R1 = R20 / ( 1/alpha0 + 20)


So now we have expressions for R0 and R1 in terms of R20 and alpha0. Let us plug them back into equation A.

Start with original equation A
tfinal = ln{<Tfinal*R1+R0>/<Tinit*R1+R0>} * C/[I1^2*R1]

Divide numerator and denominator inside parantheses by R1
tfinal = ln{<Tfinal+R0/R1>/<Tinit + R0/R1>} * C/[I1^2*R1]

Plug our new definitions of R0 and R1 into the above equation:
R1 => R20 / ( 1/alpha0 + 20)
R1/R0 => alpha0
tfinal = ln{<Tfinal+1/alpha0>/<Tinit + 1/alpha0>} * C * ( 1/alpha0 + 20) /[ R20 * I1^2] [equation D]

Now if we we revisit the previous equation from the Electric Cables Handbook Chapter 9, pages 152-153 without making any "corrections" this time, we see they it is the same as equation D above.

Equation 9.1 was
I^2 = [K^2 * S^2 / T ] * ln {(Theta1 + beta) / (Theta0 + beta)}

Let's make some substitutions from their notation to my notation:
T -> tfinal
Theta1 -> Tfinal
Theta0 -> Tinit
beta -> 1/alpha0

I^2 = [K^2 * S^2 / tfinal ] * ln {(Tfinal + 1/alpha0) / (Tinit + 1/alpha0)} (EQ 9.1A)

Swap the location of tfinal and I^2
tfinal = [K^2 * S^2 / I^2 ] * ln {(Tfinal + 1/alpha0) / (Tinit + 1/alpha0)} (EQ 9.1B)

Now look at K^2 * S^2
{K^2} {S^2} = {Qc * (1/alpha + 20) / Rho20} * {Area^2}
where Qc is (heat capacity) in Joules * degC /Volume

Break this up into three factors: [Qc*Area], [1/Rho20*Area], and [1/alpha + 20]:

[Qc * Area] = C expressed on a per-length basis

[1 / Rho20 *Area] = 1/R20 where R20 is expressed on a per-length basis

[1/alpha + 20] = no clarification needed.

That gives us
S^2 * K^2 = C * (1/R20) * [1/alpha + 20]
where again C and R20 are per-length

Susbsituting K^2 * S^2 = C * 1/R20 * [1/alpha + 20] back into (EQ 9.1B) gives us:
tfinal = [C * (1/alpha + 20)]/ [ I^2*R20 ] * ln {(Tfinal + 1/alpha0) / (Tinit + 1/alpha0)} (EQ 9.1C)

After all this we see that my equation D which was fully derived above matches equation 9.1C which came from the Cable Handbook and so the two solutions agree.

Assuming there were no difference between alpha0 and alpha20 (i.e. assuming the same alpha could be used in the two different equations T = R0*[1+ alpha * T] and T = R20[1+ alpha * <T-20>].... not a 100% correct assumption even for a linear relationship) then both of these equations should give the same solution as the equation ep1 that I previously posted 22 Feb 09 12:57. Strictly speaking, if the alpha is associated with 20C (as is most common), then equation ep1 is more accurate. If the alpha is associated with 0C, then EQA and EQ 9.1C are more accurate. I see various definitions of alpha used in different references. My impression is that 0C is used as a base point when we are using a higher-order approximation (either T = a + b T + c T^2..etc or the Calendar /Van-Dusen equation). If we want to use a single parameter for a linear approximation, then it is most often based on 20C or else it is based on connecting the endpoints of the interval 0C-100C and exact at one point in the center of that interval... which might happen to be close to 20C).

I am certainly splitting hairs to worry about the difference between alpha0 and alpha20. Either way, there is some error from using a linear approximation on a curve that deviates from linear if we go far above 100C, and likely other larger errors in assumptions for a calculation of this type.


=====================================
Eng-tips forums: The best place on the web for engineering discussions.

 

[electricpete]

(i.e. assuming the same alpha could be used in the two different equations T = R0*[1+ alpha * T] and T = R20[1+ alpha * <T-20>].... not a 100% correct assumption even for a linear relationship)

It may not be immediately obvious that these two relationships require different alpha. But we can see it if we rewrite them as follows:
R = R0 + R0 * alpha0 * T
R = R20 + R20 * alpha20 * <T-20>.

Look at the slope of the curve (slope = dR/dT)
For the first curve, dR /dT = R0 * alpha0
For the first curve, dR /dT = R20 * alpha20

For the slopes to be the same:
R0 * alpha0 = R20 *alpha20
alpha0/alpha20 = R20 / R0

We know that R0 and R20 are different. So these two curves cannot represent the same behavior unless the alpha0 and alpha20 are different.

We can see now that the selection and definition of alpha depends on the reference temperature and reference resistance. With this in mind, the very obvious approach would be to select the same reference temperature for R as we do for alpha. The immense confusion that I suffered in this thread results from the bizarre choice to use alpha based on 0C and R based on 20C which is inherent in the result published in the Cable Handbook and the ABB Electrical Equipment Handbook. The Cable Handbook did not even bother to highlight this bizarre choice. The ABB Handbook does spell it out as shown in the attached (I remembered to attach it this time) where you can see rho is based on 20C and B = 1/alpha is based on 0C.


=====================================
Eng-tips forums: The best place on the web for engineering discussions.

 

[electricpete]

Here is the ABB attachment.

=====================================
Eng-tips forums: The best place on the web for engineering discussions.