Ohm's law question

[GeorgeB]

A customer and good friend of mine makes equipment for the steel industry. One of his systems uses resistive heaters, high power, in a 2400F furnace.

His element manufacturer had some quality control issues, and to perform incoming inspection, he uses an AC welder, Lincoln AC-225-S which has 100% duty cycle at one setting (often used, but not by him, to thaw pipes).

A year or so back he decided to quantify his inspection. The data are puzzling.

Heater has 30.3 (+/-0.1V) volts across it, measured with 4 different "true RMS" meters, 3 Fluke, 1 Oriental something. Note that the operating supply is 277V so this monster just gets a lot warm in free air during receiving inspection. The element is designed for negligible temperature coefficient of resistance. It's not changing NOTICEABLY during the measurement.

Leads have 60 amps (+/- 0.2A) thru them, measured with 4 different clamp meters.

Heater has 0.450 ohms resistance; his specification to the element manufacturer is 0.44 to 0.46. This is approximately measured by 2 of the Fluke meters and for more resolution, a 4 wire low resistance meter. He has a 0.2% 0.565 ohm precision resistor he uses to check his meters.

So what's the question? R=E/I, right (DC or resistive AC circuit). Run that calculation, and we get 0.505 ohms, far different (over 10%) from the 0.45 measured with DC instrumentation. Obviously the transformer welder is quite inductive. Is that causing my readings issue?

It's been 50 years since I studied this in EE school and I never used it afterwards ... all I remember is that voltage and current are in phase in a resistor. But there is just 1 loop containing a simplified ideal source, ideal inductor, and ideal resistor. I know that if I put a 1 "ohm" capacitor in series with a 1 ohm resistor across an AC source, there will be something different than 2 resistors (a lossy voltage divider) or 2 capacitors (a lossless voltage divider).

@gar or anyone else, shed your light on my ignorance please.

 

[wwhitney]

For AC linear circuits, Z = E/I, and Z = sqrt(R² + X²). So if your resistance meters are truly measuring R, what you've shown (for the simplest model) is that the welder - heater circuit has a reactance X = sqrt(0.505² - 0.450²) = 0.23 ohms.

Cheers, Wayne

 

[LarryFine]

Maybe this statement is less than accurate:

The element is designed for negligible temperature coefficient of resistance. It's not changing NOTICEABLY during the measurement.

 

[gar]

202021-1726 EDT

GeorgeB:

"Heater has 0.450 ohms resistance; his specification to the element manufacturer is 0.44 to 0.46. This is approximately measured by 2 of the Fluke meters and for more resolution, a 4 wire low resistance meter. He has a 0.2% 0.565 ohm precision resistor he uses to check his meters."

At this resistance level you want to use the 4 wire method. This means current is injected and measured to the resistance. Contact resistance in injecting current is not a problem. The purpose of the 4 wire method is to remove contact resistance problems. The voltage leads must contact the resistance at points inside of where current injection occurs, and be at the two points of the resistance where you want the resistance to be measured.

Assuming the 4 wire meter is correctly calibrated, then what resistance did you measure? This was possibly at a very low power level in the resistance compared to its full power operating point.

Most resistance materials have a positive temperature coefficient. Thus, at full power I might expect resistance to be somewhat higher than at room temperature.

Your high power measurement with welder AC current supply gives an 0.505 ohms result. This is ( (0.505/0.450)-1)*100 = +12% change. In the ball park of what I see on typical space heater.

You may have some instrumentation errors, but off hand I might expect this resistance change.

Use a Variac and a low voltage transformer, 10 V at 20 A with 120 V in, and do a 4 terminal resistance check at about 10 V, and at 1 V. What are these resistance values?

.

 

[GeorgeB]

Maybe this statement is less than accurate:

All things (well, MOST things) are possible.

He has the Fluke 80i-1000s to his scope and sees constant current from cold during the 1st few seconds, continuing for the several minutes of scratching his head. Of course, this is a WELDER, and the inherent design likely attempts to do that too.

I agree with @gar "Use a Variac and a low voltage transformer, 10 V at 20 A with 120 V in, and do a 4 terminal resistance check at about 10 V, and at 1 V. What are these resistance values?"; I've offered to bring the low voltage transformer, variac, and a six pack of his favorite brew to play, but he is content in his confusion. "you can lead a ...."

Thanks!

 

[winnie]

For AC linear circuits, Z = E/I, and Z = sqrt(R² + X²). So if your resistance meters are truly measuring R, what you've shown (for the simplest model) is that the welder - heater circuit has a reactance X = sqrt(0.505² - 0.450²) = 0.23 ohms.

Cheers, Wayne

0.23 ohm of reactance at 60Hz requires about 600uH of inductance. A hand-held LCR meter could measure this with a 1000Hz test frequency. The measurement accuracy might not be very good at a 100Hz test frequency.

-Jon

 

[synchro]

He has the Fluke 80i-1000s to his scope and sees constant current from cold during the 1st few seconds, continuing for the several minutes of scratching his head. Of course, this is a WELDER, and the inherent design likely attempts to do that too.

If the scope has at least 2 channels, you could put the current probe on one channel and the the voltage across the element on the other channel. If the discrepancy you are seeing in the impedance is due to a series inductance of the heating element, then the current should lag the voltage by arccos(0.45/0.505) ≈ 27 degrees at 60 Hz, or (27°/360°) x 1/(60 sec^-1) = 1.25 milliseconds on your scope.

It shouldn't matter if the applied voltage or the load current varies as long as you are measuring them at the same time, and then you take their ratio to determine the magnitude of the impedance.

 

[winnie]

But there is just 1 loop containing a simplified ideal source, ideal inductor, and ideal resistor. I know that if I put a 1 "ohm" capacitor in series with a 1 ohm resistor across an AC source, there will be something different than 2 resistors (a lossy voltage divider) or 2 capacitors (a lossless voltage divider).

Re-reading what you wrote, I think the above contains an incorrect assumption.

IMHO the heater is very unlikely to be an ideal resistor. This is a rather _huge_ heater, 0.45 ohm at 277v is something like 150kW. I think it very likely that it has some inductance.

Synchro has described a good way to measure the inductance with the instruments your friend has on hand.

Another possibility is pointed to by the discussion of doing a 4 wire measurement. Was the test with the welder a proper '4 terminal' test, with the voltage measurement made on the heater terminals? Or were the welding cables and connectors part of the voltage drop being measured? If the welding cables are just clipped on to the heater terminals, I could easily imaging 20 or more milliohms of contact resistance there.

-Jon

 

[GeorgeB]

OP here; thanks for all of the input. My friend/customer who designed and builds the system truly feels there is no inductance, but has decided to just take the readings on the last 2 as a reference and use them in the future (he sells one of these ever 3 or 4 years, replacement elements every year or so).

Two suggestions here seem reasonable, L in the element and resistance change with temperature. He has several further tests he COULD do and I have equipment to assist ... a low voltage transformer to measure with AC at around 4 amps, and a DC supply ... we'd probably use a car battery at 12V, 24 amps. My DAS will let us see sub-millisecond response from power application ... but he's ok as it is. And I've offered to do it for a cold beer.

Thanks folks!

 

[Besoeker3]

If the scope has at least 2 channels, you could put the current probe on one channel and the the voltage across the element on the other channel.

Not usually to do that easily. Isolated probes are a possibility but not common.

 

[synchro]

If the scope has at least 2 channels, you could put the current probe on one channel and the the voltage across the element on the other channel.

Not usually to do that easily. Isolated probes are a possibility but not common.

It may not be necessary to have isolated probes if the scope is battery powered and/or the applied voltage source to the element is floating (such as an ungrounded transformer output). I still have a Lincoln AC225 welder like the OP mentioned which I bought in high school and its secondary winding is floating. I rarely use it now because I have a TIG welder.
If the voltage source applying the AC power was not floating then a differential probe could be used, or if there is another scope channel an A-B measurement could be made.

 

[Besoeker3]

It may not be necessary to have isolated probes if the scope is battery powered and/or the applied voltage source to the element is floating (such as an ungrounded transformer output). I still have a Lincoln AC225 welder like the OP mentioned which I bought in high school and its secondary winding is floating. I rarely use it now because I have a TIG welder.
If the voltage source applying the AC power was not floating then a differential probe could be used, or if there is another scope channel an A-B measurement could be made.

Yes, I understand. I was just offering the more usual approach. If even a scope needed in the first place.

 

[zbang]

My friend/customer who designed and builds the system truly feels there is no inductance, but has decided to just take the readings on the last 2 as a reference and use them in the future (he sells one of these ever 3 or 4 years, replacement elements every year or so).

I'd take inductance readings on every element as they go by (and cold/warm resistance). Unless he's making the elements himself, the manufacturer could change construction over time and no one would notice.

 

[gar]

202024-1020 EDT

See https://en.wikipedia.org/wiki/Nichrome for change with temperature, near bottom.

600 microH is a substantial coil. See the ARRL Lightning Calculator.

Need more information of how this resistor is made. I suspect it might be a hairpin loop. It is not small. 2400 F is quite possibly high for Nichrome. At 30 V vs 277 the power dissipation is about 1%. Thus, at 30 V not much temperature rise compared to its normal operating point.

I suspect error in the way the measurement of resistance is made is the problem, rather than the resistor.

Over a wide temperature range I do not believe there are any low temperature coefficient resistive materials. On the other hand my brother-in-law had to develop an almost zero temperature coefficient material, mechanical, for spacecraft antennas. But the temperature range was from full darkness in space to full brightness of the sun as the satellite circled the earth, not 2400 F.

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