Temperature in a Copper Coil by Measuring Resistance

Any two lead hand held meter will not be accurate enough for a temperature calculation. You must use at least a three but better four lead resistance measurement and not move the location of the wires from reading to reading. If you look at the resistivity an temperature coefficients you will see that copper has both a low resistance and that the temperature coefficient will make less than a 0.4% change in the true value. Errors in measurement from contact area to measurement length changes can quickly mask any actual changes purely from temperature change.

This approach sounds rather a bizarre method to measure the temp in the coil of copper.

Why not just measure the temperature with one of the proven methods, such as a thermocouple or an infra red scanner?

Um...does 'cause I was told to count?

When I asked the same question, I was told that UL requires it to be done this way. When I first learned about UL I thought they were a great thing; I have seen sub standard power strip extension cords cause fires due to poor design. But after seeing them in action here I personally feel that their first priority is making executives rich and public safety is an excuse to gouge manufacturers (who are just trying to make their executives rich).

Drew K

I believe UL prefers it because it is a direct temperature measurement of the object (copper tubing) and not a temperature from something (probe) attached to one part of the object. This is a very subtle difference in metrology that rarely makes a difference but when it does...

I don't know the whole story of your measurement needs but it is obvious that you are into the realm where high accuracy of measurement techniques must be substantiated. Question the rules, politely, to find out their reasons.

The four wire method as bizarre as it may sound, is probably the most accurate way to determine the resistance of a length of wire, and by proxy its temperature, as long as you have a base line temperature/resistance reading.

You can obtain the baseline either by measurement at a precisely known temperature, or by knowing the physical properties of the coil; i.e., its exact length, cross-sectional area, temperature and the material.

There are meters called ductors or micro-Ohmmeters that take the guess work out.

Yes... and I have seen the method used in electroplating systems in the printing industry where the resistance of the item being plated is used as an input to a controller to control bath tempertatures.

Seems rather like overkill for a coil of copper.

Is this an electrical coil or a fluid carrying coil?

If it is an electrical coil with current flowing through it when you are trying to measure the resistance, then you can't use an ohmmeter. An ohmmeter measures the resistance by measuring voltage across a resistor with a given current flowing through it. The resistance, of course, is the ratio. If there is current in addition to that provided by the ohmmeter, then you will get the wrong answer. You can't measure resistance in an energized circuit.

The resistance of copper is very low and it's temperature coefficient of resistance is also low. (It is a very poor material to use as a thermistor.) A problem with measuring low resistance with an ohmmeter is that the contact resistance of the probes is in series with the unknown resistance. Very low resistances are generally measured using the four point method. The reference current is injected through two probes and a second set of probes are used to measure the voltage.

Another factor is that if the current through a coil is changing, the voltage is not only due to the resistance but also the inductance times the rate of change of the current. The current has to be steady unchanging to get a resistance reading.

It is not my line of expertise but basic knowledge points to some problems associated with matching temperature to resistance.

If by coil you mean something like a solenoid or transformer then the overall resistance of the coil might predict an average maximum temperature whereas you might want to know the maximum temperature of the inner windings - and that depends a lot on how the heat dissipates - so you probably need to take accurate measurements of temperature and time and plot them against accurate measurements of current and voltage to build up an empirical formula specific to your application.

The experts here in measurement of temperature and resistance can help with the precision.

If by coil you mean something like an electric blanket then I would be keen to hear from the experts here. My electric blanket 'intelligent' controller has just packed up.

Here's the basics, work the rest out yourself

Alas, no post has explained the essentials of the method. Kvsrihdar has explained the resistance to temperature conversion accurately.

Resistance measurement is a standard technique for windings on motors, solenoids etc. While it does not give a "hot-spot"temperature it is a test that is repeatable on every one of a design - the actual hot spot may be buried in the middle of a coil and inaccessible except to a thermocouple buried in a prototype winding. In the case of many windings, the temperature variation within the copper winding is not great and the "hot spot" can be estimated by adding some degrees to the mean temperature got from the resistance test.

When testing motors and similar moving items, the time taken to stop the motor, removing any generated voltages which would upset a DC ohms tester, is such that the winding may have cooled significantly from the hot running state.

However, there is a method which allows the hot temperature to be found from later readings as the winding cools. A stopwatch is started at the moment the power is cut, then resistance readings are noted at regular stopwatch time intervals e.g.

Time (sec), Resistance measured R, Temperature T calculated from R

0, ?, ?, POWER CUT OFF

23, R1, T1, FIRST READING POSSIBLE

25, R2, T2

30, R3, T3

35, R4, T4

etc.

The temperatures corresponding to the measured resistances are calculated and tabulated. Then, using for example, the Microsoft Windows "Calculator" program (in Scientific "View") the Natural logerithms of T1, T2,T3....etc. are listed L1, L2, L3...etc.

[the value T1 is entered to the calculator, then the "ln" key is pressed - the display shows the Natural logerithm L1 etc].

A plot is then made on graph paper, with time on the X axis and the corresponding L1, L2, L3...etc. values on the Y axis.

If all is well, the L1, L2, L3.... values will form a straight line, sloping down to the right, which can be extended back to zero time on the graph - Y value at this point, L0, is the log of the initial hot temperature, before cooling started. [The L0 value is entered on the calculator, the "Inv" box is clicked with the mouse, then "ln" button clicked - the display shows the inverse of the Natural logerithm L0, which is the initial temperature].

All this calculation is based on the assumption that the rate of heat loss and temperature fall is proportional to difference between copper temperature and "ambient temperature". The same basic formula relates voltage to time when a capacitor discharges through a resistor.

In a real situation, "ambient" is not that straightforward - the copper cools to the general temperature of the motor or other close surroundings, then that cools down to ambient with a different slope. When you have no idea of the cooling rate of the winding under test, it is wise to draw the points to show that the first points really are a straight line.

Drew K has not given the cold value of copper coil, if it is tens of ohms, one can subtract the lead resistance (with ohmmeter probes shorted) from measured values. If it is a few ohms, the 4 wire method is needed, with a bridge or accurate current source and millivoltmeter and good relays to disconnect the metering while the coil is energised.

67model

I had a more pressing project put the coil temperature on the back burner. When I do the resistive method, I will post the data I collect and my calculations here. The formula I am planning on using does not include logarithms.

The formula is:

Tt=Tc+(Rh-Rc)/R×(Tc+234.5)

T(t)=total winding temp

T(c)=cold motor temp

R(h)=Hot motor resistance

R(c)=Cold motor resistance

234.5=constant for copper windings

Drew K

Be very careful with your units. A spreadsheet will not care if the numbers you enter are ohms, milli-ohms, Siemens, Fahrenheit or Centigrade.

What is R in your formula? Is not Tt the same as Th, corresponding to Rh? and ....

Tc = cold winding temperature ; Rc = cold winding resistance

Th = hot winding temperature ; Rh = hot winding resistance

Did you mean Tt = Tc + ( (Tc + 234.5) X ((Rh - Rc) /Rc))

which is the same as...

Tt -Tc = (Tc +234.5) x (Rh - Rc) x (1/Rc)

Further to my previous post #7, no comment has been posted in relation to my view that the coil will have a temperature depending on the heat dissipation, and being a coil the innermost layers will reach a temperature much higher than the outermost layers.

The overall resistance which is an average for the whole coil, will give a temperature that is the average for the whole coil.

Whereas in practice the average is a result of the very hot inner coils in series with much cooler outer coils.

Unless I have missed the point of the question, I guess you need to know the highest temperature within the coil rather than just the average - and whilst there might be a strict correlation between the two - this will only be known by a formula evolved from empirical testing of that particular coil..

In round figures (for easy mental math) starting with a cold copper coil of 100 turns at 100Ω and 20c throughout, will have an average resistance of 170Ω when heated to 200c.

But assuming a temperature gradient exists that is linear over the length and depth of 10c per turn, we are talking of one coil at the cold end of 1Ω at 20c being matched by one coil at 2.4Ω and 328c at the hot end - where 170Ω and 200c are the respective averages. That is a big difference in heat ....!

Heat loss is not linear across the coii and temperature distribution is not linear along the wire. The numbers used here are to demonstrate the idea of my argument. They can be brought to precision by others in possession of better knowledge and exact parameters.

Here's a link to a ready made auto formula: Put in your own numbers and click the 'Blue' value in the formula for the answer you want.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/restmp.html

....or am I wrong in this logic ? Does it matter ?

A useful calculator. But its coefficient of resistance given for copper is 0.39, which is a bit off the standard value for IACS (international Annealed Copper standard) of 0.393 x 10^-2 .

It is useful to remember that the IACS data means that a coil of electrical grade copper which has a resistance of 254.5 ohms at 20 Celsius has a change per degree of 1.000 ohms between 0 and 200 'C (at least). This holds within the range of conductivities found for electrical copper (several % variation from IACS). The only condition is that the copper is free to expand.

So if you have a coil which is 57 ohms at 17 'C you can proceed...

Resistance of "standard" coil at 17 'C = 254.5 -3 = 251.5 ohm

Hence resistance change per 'C for test coil = 57/ 251.5 = 0.2266 ohm per 'C.

Hence temperature of test coil when resistance is R ohms is (R- 57)/0.2266 + 17 'C.

The formula given by Drew K is correct, so long as multiplies & divides are done in right order. Check arithmetic against the resistances given in NBS document for 8 AWG copper wire per km which are :-

'C, ohms

0, 1.899

20, 2.061

25, 2.101

50, 2.304

75, 2.506

100, 2.708

200, 3.518

As you wrote, the "hot spot" temperature is more than the mean given by resistance, but is not enough more that much electrical gear cannot be rated on temperature rise "by resistance".

The conductivity of copper and the heat transfer wire to wire is so much more than surface heat transfer by natural convection and radiation that most of the temperature rise is between ambient and coil surface.

We are "in the dark" because DrewK has not told us if his coil has natural or forced cooling and whether it is a 10 MW motor winding or motor coil for a watch.

Thanks Randall. Very interesting. I take the precise values you have given at face value. It is your point about the internal heat distribution being within acceptable rated limits when assessing temperature rise when using the precise resistance as a guide.

I have no personal experience of the various parameters working with coils, but I posed the question about heat dissipation from a past experience when using an air compressor for hours on end to drive a pneumatic needle-hammer gun to remove rust from my narrowboat.

It just happened that the only air hose I had was a few feet short - so I moved the compressor nearer to the boat. Then I plugged it into the mains with an extension lead by unwinding a couple of turns off a 50m portable cable drum.

A few hours later the air pressure dropped because the compressor had stopped. On checking it turned out to be an earth leakage fault within the cable on the drum.

That was when I found out the cable drum was 'red hot' inside when I tried to unwind the cable - after a few turns it was a solid mass of soft gooey plastic - that formed a solid lump when cold.

...obvious with hindsight being an overheating coil and I should have known* - but a bit of a surprise at the time because there were no obvious signs of external heat - warming up to the touch yes - but not with a temperature gradient to the extent you would expect it to melt the plastic insulation inside the coils of the drum - but it did!

Lesson learned!

Thanks for your help on this.

* Hi Savvi. We need a 'red faced' emoticon.

You give a prime example of the temperature gradients within a coil being unacceptable!

Machine windings usually have enamel insulation which is far thinner than the insulation on 240V cable. The space between wires in such a cable is about equal to wire diameter & the oversheath as much again.

A PVC 3 core x 1 sq.mm, 5 metre cable drum I have ( about 7" OD and 3" ID x 3") has marked ratings of 5 amps wound-up and 10 amps fully extended. This suggests that even at 5 amps temperature rise inside is 40 'C at 30'C ambient. The Wiring Regulations for buildings have half the amp rating buried in thermal insulation compared to clipped on a wall for PVC flat cables.

I just looked up thermal conductivities; theoretical transfer from face to face of a copper cube with 1 cm sides,in vacuum, is 3W/'C. Flexible PVC is about 0.0015 W/'C, say 1/2000 the thermal conductivity for PVC compared to copper.

67model

Thanks 67model. Apart from not thinking when I overheated my cable drum at a current well in excess of any sensible rating for the cable - circa 12amps wound up - the rise in temperature from ambient was enough to 'melt' the plastic.

Actually it did not melt in the sense that it got runny - because it retained it's insulation properties except for the earth-leakage fault - the coils just fused together in a solid lump when cold.

The temperature must have been high enough in the middle to melt the pvc, and since it was not hot on the day, the outer coils were 'cold' which implies there must be a temperature gradient - and a resistance gradient as well - that would be 'lost' in an average resistance reading.

You have said difference in conductivity explains why my plastic insulated coil got hot compared to a coil with enamel insulation, and no argument with that, but somewhere in all this is the basis of my view that the temperature of a coil will be higher (in parts) than the resistance reading will predict.

Unless in common practice (amongst coil making experts) it is safe to assume there is no temperature and resistance gradient to take account of, and the change of temperature derived from the resistance is acceptable for proving compliance with temperature specifications and/or standards.

To my mind, spending a fortune on test meters to measure redistance with an accuracy to the nth degree is not going to be much use if there is an unknown temperature gradient.

We had a similar event in Turkey, a 6 plug 220V power strip caused a fire because the 3 meter extension was left bound with the twist-tie when the strip was used. I have heard somewhere that you should never use an extension cord unless you are unwinding it to prevent overheating.

Drew K

Sorry guys, I have been out with a flu.

If I can get caught up today I will take data for the thermal test today or tomorrow. This test is on a small shaded pole motor with less than .2 A when running. The only reason we are using this method is for underwiter labatories test requirements. We use data from the thermocouple datalogger for our development because we know the wire does not get hot enough to cause damage and are more concerned with the heat inside the splash cover.

The motor I am testing now does not have any kind of extra cooling. We only use forced air cooling on 80 laminate or a few specific motors we know are hot from design.

Drew K

Hi..i am a new user here. I think UL prefers it because it is a direct temperature measurement of the object and not a temperature from something attached to one part of the object. This is a very subtle difference in metrology that rarely makes a difference but when it does.

Welcome to the forum, we are an eclectic group from all over the world and welcome your contribution. I am not admin just a long time member.

I appreciate your input, but I completed that job ages ago, I look forward to future contributions.

Drew K