Thermocouple basics 1

[omby]

I wish to test a thermocouple loop by injecting a millivoltage from in the field where the thermocouple terminates to compensating cable, back to a substation and a thermocouple/mA convertor card with cold junction compensation. Lets say the field temp is 30 degrees c and the substation temp is 10 degrees c.

To simulate 1000 degrees c, it as simple as injecting the mV equivelent of 990 degrees c onto the comp cable(the cjc on the card adding the 10 degs c substation temp to the mA o/p)?

or...

Do I have to take into account the 30 degree c field temp from where I am injecting, and the fact that I am injecting a signal on compensating cable.

 

[IRstuff]

You have also to be concerned about any thermocouple effects of the wires you insert in the loop itself.

TTFN

 

[omby]

This is one point causing confusion.

I always thought that due to the law of intermediate metals that because the internal wiring of the injector, leads, and point at where I am injecting in the field are all at the same 30 degrees c, that yes, other hot and cold junctions are made, but the overall error is zero because they are all at the same temp?

The other point of question is the fact if you short the comp cable in the field you will read ambient field temp at the convertor card in the substation, ie 30 degs c. Will this 30 degrees come into play whan you effectively short the compensating cable with the mV injector and inject the mV equivelent of 990 degc c to simulate 1000?

 

[itsmoked]

I think you are correct omby.

Just try to keep the two "source leads" isothermal or everything will go to hell in a hand basket.

Apply the mVs to the point the T/C would be connected. Everything else should be in place and functioning, i.e. the converter card and the ice ref, etc. Use the mV given by a table for the T/C of interest. The ice ref and your reading instrument should take all the other things into account.

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[danw2]

Omby, you are correct.

1) If the intermediate junction points are held isothermal (same temperature) across both junction points, the result is zero error.

2) The cold junction compensation on the card algebraically adds its measured temperature of the card's connection block to the reading received from the thermcouple, which is the emf of the hot end minus the cold end.

If one shorts the input to the card with a copper wire, the result of shorting is that the device will dislay the temperature of the connection block. The cold junction contributes the connection block temperature and the copper wire contributes 0 mV, the sum of which is the connection block temperature.

If the connection block temperature corresponds to an arbitrary 2.000 mV from the standard thermocouple tables, then the applied calibration voltage has to be 2.000 mV less than the applied or injected signal, because the 2.000 mV (or equivalent temperature) is added in by the input card's cold junction.

The mV values from the thermocople tables assume ice point reference, which is in modern electronics is
TC hot end minus TC cold end plus cold junction.

The hot end minus cold end concept is important, because a T/C in a 1,000° furnace connected to your input card does not generate the mV table equivalent of 1,000°. That T/C generates the mV table equivalent of 990°, because the 1,000° hot end minus the 10° cold end is 990°.

Dan

 

[omby]

Thanks Dan, I was reasonably sure I was right on that, but after discussing it will a colleague, I had sown seeds of doubt in my mind.

I am happy that in the field, my test equipment will be isothermal, but what about the compensating cable?
From the original example I gave the end I am injecting a mV from in the field is 30 degrees c, whereas the substation end is 10 degrees c.

If I inject the equivelent mV for 990 degs c (to give me a reading of 1000 on the convertor card o/p), will I have to make any adjustments to this figure to take into account the 20 degree difference between the two ends of the compesating cable??

 

[danw2]

I think you got the 20 deg difference number from:

hot end = 1000
intermediate junction block = 10
input card = 10

1000 - 10 - 10 = 980.

No, you don't include the intermediate junction block temperature, because its isothermal properties are self cancelling, meaning that the error on one leg is cancelled by the error in the other leg.

You were right the first time
hot end = 1000
ignore intermediate junction block temperature
cold end = 10
injection value = 1000 - 10 = 990

If I misinterpreted, try again.

I'm assuming that you have thermocouple wire or TC extension wire from for all sections of wiring, from the TC element all the way back to the input card, right?

Dan

 

[omby]

Thanks Dan.

Thermocouple extension cable (compensating cable) is used from the t/c head to the tc/mA converter card.

The 20 degrees is the difference in temp of the compensating cable between the field connection (30 degs) and the substation (10 degs).

If I inject in the substation, everything is at 10 degs c and the card will compensate for this, and add on 10 degs to the 990 I am injecting.

If I am out in the field and injecting down the compensating cable, will connecting an injector turn the compensating cable into a thermocouple which will add 20 degs (difference between field 30 degs and sub 10 degs)to the 990 I am injecting which in turn will have 10 degrees cjc added to it to give a final o/p temp at the card of 1020 degs c??

Basically, does connecting an injector to compensating cable and injecting at a point where the ambient temp differs from the cj, introduce an error by virtue of the fact that the compensating cable produces a mV when one end differs in temp from the other?

omby

 

[waross]

Hi omby;
Rather than subtract 10 C or 30 C look up the milivolt equivlent of 10 C and subtract that from the milivolt equivalent of 1000 C. That will give you the correct figure for direct injection at the instrument in an ambient of 10 C.
For a field test at 30 C ambient, subtract the milivolt eqivalent of 30 C from the milivolt equivalent of 1000 C, and use that for your test.
This method will compensate for, or avoid, any non-linearity in the tables.
respectfully

 

[omby]

Waross, I think you missunderstand my problem.

The converter card always adds the 10 degs ambient cj temp to the mV I inject. I am happy with this, and inject the equivelent mV of 990 degs at the card to give me an o/p equivelent to 1000 degs.

When I connect my mV injector to the compensating cable in the field -which has similar properties to the thermocouple itself- am I creating a 30 deg hj in the field which with the cj at 10 degs introduces an error of 20 degs to the 990 I am injecting?

 

[waross]

Hi omby;
I think that I do understand your problem and have expressed myself poorly.
If you are injecting a signal in an area at an ambient temperature of 30 C you should use 30 C compensation.
The leads, passing through a 20 degree temperature difference will introduce an error as you suspected.
As I understand thermocouple theory, part of the voltage is generated in the junction and part of the voltage is generated by the temperature gradient in the connection wires. I don't want to discuss thermo-couple theory.
I want to suggest a test method that will work no matter what the fact or theory of thermo-couples.

If you are injecting a signal in an ambient temperature of 30 C., Don't look up the mV for 970 C. (1000 C-30 C)
Look up the mV for 1000 C.
Look up the mV for 30 C.
Subtract the mVs and use the difference as your test voltage.
Respectfully

 

[cbarn24050]

Thermocouple voltages are only produced at JUNCTIONS of diisimilar metals.

 

[cbarn24050]

I withdraw that staement as it is not correct.

 

[itsmoked]

cbarn24050; Why would you withdraw a correct statement?


Why would you get an error waross because of a gradient?

If it matters what gradient T/C wires "run thru" than the entire basis of TC would be useless!! ITDOESNOTMATTER.

Same with junction of the same material.. If they mattered how could you ever use a TC with out measuring the temperatures of the temperatures of the temperatures....

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[waross]

Hi Keith
The theory that the voltage was only generated in the junction was the first thermocouple theory that I was exposed to. Also the second, third, and possibly the fourth.
Later, I was exposed to a thermocouple theory that states that a thermo-couple voltage is the sum of the junction voltage and the difference between the thermal gradient voltages developed in the extension wires.
The theory states that different materials have different voltage gradients for a given temperature gradient.
I always wondered why the extra money was spent on special extension wires instead of just using copper extension wires. By the thermo-couple laws, everything should cancel out and it should work fine. It does, but it develops a voltage proportional to the difference between the temperature of the measured variable and the temperature of the connection with the copper wires. The difference between that temperature and the control room temperature is lost.
That's why I didn't want to discuss thermo-couple theory.

The method that I have suggested should give accuracy no matter what the correct theory may be.

I will suggest that manufacturers of thermocouple extension wires pay attention to and work to "Limits of errors" in the construction of extension wire because of the thermal gradient voltages.
Respectfully

 

[omby]

Hi Waross,

As I understood things.....
the junction of two dissimilar metals would produce a voltage equivelent to the difference in temp between the hot and cold end of the wires.

The linearity and amount of voltage produced is dependant on the types of wire used, hence the different temp ranges given to commercially available t/c types.

The extension (compensating) cable has similar properties to the thermocouple wires (within its given ambient temp parameters) and is used to move the 'cold' end of the thermocouple to a convenient point ie a sub or panel house (for cold juction compensation of some form).

As you say, if copper leads were used the temp difference would be between the hot end, and the temp at the t/c and copper junction. So a steady 1000 degrees at the hot end would vary by ambient temp if the tc/copper junction was outside.

This brings me to the point of my original post, that if tc extension leads have similar properties to the thermocouple itself, then would connecting an injector in the field produce a voltage in addition to the o/p of the injector if the ends of the extension were at different temps.
I have asked a few people and have had mixed responses so far.
Normal maintenance never actualy brings us to simulate from in the field but I was working on a new installation and it seemed a good way to check the whole loop back from the tc head (the thermocouple being the last thing to be fitted in the loop). It seems to have caused some confusion though...

regards
omby

 

[waross]

Hi omby;
Theory #1
The voltage is generated at the junction.
Theory #2
The voltage is generated in the wires.
Theory #3
All of the above. (This is the theory that makes the most sense to me).
The method that I suggested will compensate correctly whatever theory is correct.
DON'T subtract the temperatures.
Look up the millivolts for the temperatures that you want to correct for and subtract the millivolts.
respectfully

 

[Bomar2]

I was reading documents on Omega's web site and came across this statement:

Temperature changes in the wiring between the input and output ends do not affect the output voltage, provided that the wiring is of thermocouple alloy or a thermoelectric equivalent (Figure 1a). For example, if a thermocouple is measuring temperature in a furnace and the instrument that shows the reading is some distance away, the wiring between the two could pass near another furnace and not be affected by its temperature, unless it becomes hot enough to melt the wire or permanently change its electrothermal behavior.

from the Practical Guidelines for Temperature Measurement

at

http://www.omega.com/temperature/Z/PracticalGuidelinesforTemperatureMeasurement.html

 

[itsmoked]

Yes. They got it correct.
Thanks Bomar2.

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[jgv]

You can check the information at:

http://www.omega.com/

about how you can calibrate a termopar at the field