Thermocouples

[omdano]

Hello i am a high school student , i need some help in a project i am working on

So ...
If i have a room with the temp of 37c , a thermocouple and a lighter

Joint 1 is the hot one
Joint 2 is the cold one

both of them are 37c

I use the lighter on joint 1 and the heat goes up to 50c
Joint 2's Temp is how much ? is it still 37 ? is it cold because of comparasion?

The thermocouple Circuit is electricity free and in the output there is a Voltage meter .

Thanks for your answer .

 

[aigomorla]

too difficult to messure without telling us what the resistance is on joint2.
What are the two objects connected by joint...
How much work does each object require for it to go up and down 1C.

 

[omdano]

Lets Say it is Cu And Fe thermocouples
as i said , i am not a " good " high school student
consider resistance being 0
i cant however provide the work ...

 

[SOFTengCOMPelec]

If this is a Homework "project", then "omdano" should be doing it.

 

[omdano]

If this is a Homework "project", then "omdano" should be doing it.

It is for a competition , i am trying to make an electricity free cooler

 

[SOFTengCOMPelec]

It is for a competition , i am trying to make an electricity free cooler

Does this help ?

 

[omdano]

Does this help ?

Cant open the link for some reason , i will reply when i manage to open it .

 

[aigomorla]

Does this help ?

no...

cuz what he's asking for is a passive solution im guessing.
Passive - no moving parts... no electricity required... on the principle of radiation or the usages of vacuum heat pipes to move heat away from object.

Again we couldnt tell you exact figures.

Assumption heat source was copper, and you laid it on top of iron... without masses of both objects its again impossible to tell you.
without heat source numbers.. its too difficult to tell you.


If you cant ask enough basic questions to get answers, its not a experiment worth undertaking.

Basically if ur general knowledge doesnt allow you to understand why were asking these questions, its gonna be a failed experiment because you lack the sufficient knowledge to proceed.

It is for a competition , i am trying to make an electricity free cooler

if your thermodynamics is lacking to understand the basic concepts like specific heat, and Watt potential of load... its already a failed experiment.
First off in a experiment u can never assume resistance = 0
because resistance is NEVER 0 unless we go into superconductive substances...

 

[SOFTengCOMPelec]

no...

cuz what he's asking for is a passive solution im guessing.

No....

He is using the lighter flame as the "power source", if I understand him correctly.

Because thermocouples can generate electricity from heat sources, it "can" be " passive" (NOT properly passive, as this would not use a flame either), i.e. using the flame.

i.e. In theory there could be NO external power source, only sources of heat. But I am NOT sure if that would be enough to cause sufficient cooling, without an external power source.

The link was because he/she does not seem to be supplying enough details to properly answer the question(s). It should be useful, because they can be used for "electronic" non-mechanical movement, coolers.

 

[aigomorla]

No....

He is using the lighter flame as the "power source", if I understand him correctly.

Because thermocouples can generate electricity from heat sources, it "can" be " passive" (NOT properly passive, as this would not use a flame either), i.e. using the flame.

The link was because he/she does not seem to be supplying enough details to properly answer the question(s).

the voltage at which the TEC will supply is totally dependent on the the delta of the hotside and cold side.

Then his question would fail b4 it even began because coupler B would be room temp.. and the TEC would generate current while keeping hotside and cold side separate.

The current the TEC would generate would also be so tiny... were talking microamps... and would not even be viable...

 

[SOFTengCOMPelec]

the voltage at which the TEC will supply is totally dependent on the the delta of the hotside and cold side.

Then his question would fail b4 it even began because coupler B would be room temp.. and the TEC would generate current.

I agree.

The information is NOT clear enough (to me, at least), as to if "electricity free" means outside sources, or NO electricity flowing at all.

N.B. I am pretty hopeless with Thermocouples, and would probably get a D-, if I was tested on them at the moment, to a high enough exam standard.

 

[aigomorla]

N.B. I am pretty hopeless with Thermocouples, and would probably get a D-, if I was tested on them at the moment, to a high enough exam standard.

lol

remember when i said when it came to sub 12V i have a full understanding. :biggrin:

been here... done this.... froze water... didnt want to run dielectric grease again....

im pretty sure the OP doesnt imply TEC's.

 

[SOFTengCOMPelec]

lol

remember when i said when it came to sub 12V i have a full understanding. :biggrin:

We might have to go back to stuff like that, depending on how cpus develop in the future. Hopefully not.

im pretty sure the OP doesnt imply TEC's.

It's their use of the word "cooler", which is confusing me and/or making me suggest such things.

It is for a competition , i am trying to make an electricity free cooler

Which makes me think of that way of cooling things down.

I think we (or at least I) need to wait for more details on what is trying to be achieved and how.

 

[sm625]

I would consider myself an expert on thermocouples. First of all Cu-Fe is not a very common thermocouple type used in industry. Iron-Constantin (Type K) is a common one. Chromel–constantan (Type E) is another very common one. The voltages produces by thermocouples are very small. For a type K, its about 4mV per 100 degrees C. The power produced by thermocouples is even smaller. It cannot be used to drive anything, not even an LED. The signal from a thermocouple junction has to be routed through a junction of known temperature (the cold junction) where it becomes normal copper wire. It is then routed into a high impedence signal conditioning circuit that ultimately ends with the signal being converted to digital and then is converted to a temperature. You then have to subtract the temperature at the cold junction to compensate. Thermocouples are non linear, so you have to use a complex equation or a lookup table to convert any given mV value into temperature.

I dont have numbers for a Cu-Fe thermocuple, but I can give you an example using a Type J. If both junctions are at 37C and your meter is on the cold junction, the meter should read 0.0mV because both junctions are the same temperature. If you raise the temperature of the hot junction to 50C then the meter will read

2.585mV (hot junction at 50C) - 1.902mV (cold junction at 37C)

or 0.683mV.

You probably wont face this problem until college chemistry, but note that you cannot subtract the two temperatures to get 13C and then convert to type J units. In that case you would get a meter reading of 0.660mV, which would be incorrect.