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Homework Help: Question about thermocouples.

  1. Oct 7, 2013 #1
    Hi guys! First time posting here so please be nice. :)

    1. The problem statement, all variables and given/known data
    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:


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


    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).

    2. Relevant equations

    EAB = αT + βT2

    3. 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:

    Attached Files:

  2. jcsd
  3. Oct 7, 2013 #2

    rude man

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    Gold Member

    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: Oct 7, 2013
  4. Oct 7, 2013 #3
    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?
  5. Oct 8, 2013 #4

    rude man

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    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.

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