Originally posted by len1
Here is an explanation of what I had in mind
____________V2(T0)_____________________
V(T)
____________V3(T0)_____________________
The high temperature thermocouple is at the temperature of interestT, the contact to the cheaper wire (which must be the same at both contacts is at a
location where the temperature is T0.
All junctions generate voltage, and as most multimeter type thermometers assume a linear relationship, Ill assume that too (but its inessential to the
result, just makes thing seasier to see).
V(T) = a T
V2(T0) = b T
V3(T0) = c T = (a - b) T0
the last equation arises from conservation of energy. Then the voltage measured on the multimeter is
a T + b T0 - (a - b)T0 = a (T - T0)
(note that b wil cancel out even if its a function of T)
i.e. the tempearture at the measurement point is the sum of the temperatures measured by the two thermocouples, one with short leads located at the
measurement point, the other at the point of contact with the copper leads. The later could be located on the outer iron casing of the tube, with
holes cut for the high T thermocouple leads, which would need to be about 6 cm long.
There is also the ambient temperature. Most thermometers take this into account by adding it onto the temperature measured by the thermocouple (else
at ambient we get zero). So the relation is actualy
V(T) = a T - a Tambient
To avoid adding this twice from the two multimeters the formula becomes
Tactual = T + T0 - Tambient
Tambient you can get by noting the thermometer reading prior to measurements.
I didnt quite understand why you want a temperature gradient across the tube, or a clam type arrangement.
PS Do you have suppliers for the molyb elements and alumina tube? Len |