What is the temperature at thermal junction T if the EMF output is 1.8mV?

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SUMMARY

The discussion focuses on calculating the temperature at thermal junction T of a thermocouple based on an EMF output of 1.8mV. The relevant equation used is EAB = αT + βT², where α and β were determined to be 4.4333 x 10-5 and -1.3333 x 10-8, respectively. Participants emphasized the importance of correctly summing voltage drops around the circuit, including the digital voltage meter (DVM), and distinguishing between voltage polarities for different materials. The final temperature calculation must adhere to the established voltage loop equations.

PREREQUISITES
  • Understanding of thermocouple principles and EMF output
  • Familiarity with the equation EAB = αT + βT²
  • Knowledge of voltage drop calculations in electrical circuits
  • Ability to interpret circuit schematics
NEXT STEPS
  • Study the derivation and application of thermocouple equations
  • Learn about voltage drop calculations in circuits with dissimilar metals
  • Explore the use of digital voltage meters (DVM) in temperature measurement
  • Investigate common errors in thermocouple temperature calculations
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Students in electrical engineering, professionals working with thermocouples, and anyone involved in temperature measurement and circuit analysis will benefit from this discussion.

FurryMcFlurry
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Hi guys! First time posting here so please be nice. :)

Homework Statement


A thermocouple T is connected in the following circuit with two different types of metal wires to measure a temperature ranging from 0°C to 500°C:

[img=http://s12.postimg.org/9y4nuly15/mechatronics_thermocouple.png]

This thermocouple's EMF output EAB is shown in the following graph:

[img=http://s12.postimg.org/hfdv9tnk9/mechatronics_thermocouple_graph.png]

Suppose that the measurement from the digital voltage meter (DVM) is 1.8mV, estimate the measured temperature at thermal junction T (Answer must match to 1 decimal place).

Homework Equations



EAB = αT + βT2

The Attempt at a Solution


What I've done is that I've taken both the points on the graph (T = 100°C and T = 200°C) and solved for values of α and β by using simultaneous equations. From here, I was able to get α = 4.4333 x 10-5 and β = -1.3333..3x10-8

4. Where I'm stuck
My current stab at the solution is basically the substitution of EAB = 1.8mV (the measurement from the DVM) into the equation and substituting the values of α and β into the equation, and then solving for the roots of T. My values as of such have been completely unrealistic (x10-4). I can't shake off the feeling that that step is wrong... I've been looking through my lecture notes (which are, by the way, are very much incomplete and only mentions that one equation and doesn't have any examples), and I haven't been able to find anything that could help me. Any help would be very much appreciated. :smile:
 

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You need to sum voltage drops around the entire circuit and set = 0. This includes the DVM.

Write this equation.

One question: Does EAB mean EB - EA or EA - EB?
 
Last edited:
Thanks for the reply!

This wasn't mentioned in the question (and was never explained in the lectures for that matter) so I'm assuming that EAB means EB-EA.

Summing the loops, I've gotten the voltage loop (clockwise) as:
VT=50°C + 1.8mV + VT=150°C - VT = 0

After substituting T=50°C and T=150°C into the equation, I got ET=50°C = 2.2mV and ET=150°C = 6.3mV.

Substituting it all into the loop, I get VT = 1.8mV + 2.2mV + 6.3mV = 10.3mV.

Would this be correct?
 
FurryMcFlurry said:
Thanks for the reply!

This wasn't mentioned in the question (and was never explained in the lectures for that matter) so I'm assuming that EAB means EB-EA.

Summing the loops, I've gotten the voltage loop (clockwise) as:
VT=50°C + 1.8mV + VT=150°C - VT = 0
There should be two V150 terms in your loop, not just one. A quick look at the schematic diagram tells you that.

Why not start your voltage loop at the low side of the dvm:

(EB - EA)150 + (EA - EB)T + (EB - EA)150 + (EA - EB)50 = 1.8 mV.

Notice that you need to distinguish between V and -V for each drop across dissimilar metals. I think you missed that fact. Your V50 and V150 have the same polarity which is incorrect, again as can be seen on the schematic diagram.



/QUOTE]
 

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