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Mathematical relationship of a thermocouple

  1. Jul 10, 2016 #1
    1. The problem statement, all variables and given/known data
    The graph shows the response of a bare thermocouple
    which has been subjected to a step change in temperature from 50°C to
    10°C. Assuming that the bare thermocouple behaves as a single transfer
    lag system, determine the mathematical relationship between the
    temperature (T) and time (t).

    upload_2016-7-10_12-18-20.png upload_2016-7-10_12-18-20.png

    2. Relevant equations
    To=Ti*e-(t/τ) Eq1.

    OR

    f(T)=c*e-(kt) where k= LnT1-LnT2/t1-t2 =Ln5/-20=-0.08047 Eq2.
    3. The attempt at a solution
    Using Eq2., where t=0 :

    50e-0.08047(0)=50

    t=20:

    50e-0.08047(20)=10

    So far so good looking at the graph but for other values of time such as t=6 :

    50e-0.08047(6)= 30.85, which doesn't look right on the graph. Have I got the mathematical relationship wrong? Or is my time constant wrong?
     
  2. jcsd
  3. Jul 10, 2016 #2
  4. Jul 10, 2016 #3

    NascentOxygen

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    There are a number of ways for determining ##\mathsf τ##, the time constant. If you know the initial and final levels, then you can use the time taken to fall a certain % of their difference. Or you can by eye draw a tangent to the curve and passing through the starting point, and use this slope to determine ##\mathfrak τ##.

    Ignoring all but just 2 points like you are trying to means you are not making use of the error-averaging properties of multiple data points.

    But I think the best way is to transfer plenty of data points to semilog graph paper, where an exponential relation becomes a straight line. An attraction of the semilog plot is that you can see at a glance the range where your curve is a close fit to an exponential, and more importantly, where it is not a close fit.
     
  5. Jul 10, 2016 #4

    rude man

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    This is not a 'relevant equation'. Insert t = infinity and what T do you get? You get T = 0 whereas obviously you are supposed to get around 10 deg C.

    So how about T = Tie-t/τ + T instead?
    Read Ti and T from the graph to compute τ.
    Forget any 'curve fitting'. You were told to assume a 'single transfer lag' system. Use just the initial and final T points.
     
  6. Jul 10, 2016 #5
    "So how about T = Tie-t/τ + T∞ instead?
    Read Ti and T∞ from the graph to compute τ."

    Sorry but how would I do that, I dont understand the method,sorry.
     
  7. Jul 10, 2016 #6
    Think I have it :

    T=T+a.e^(-t/τ) (where T = limiting value which is 10°C)

    Choosing 2 points on the graph that are easily identifiable:

    (x=time,y=Temp) : (16,11) and (8,16)

    11=10+a*e^-(16/τ) (1)
    16=10+a*e^(-8/τ) (2)

    Simplifying by subtracting 10 from each equation and then dividing (1) by (2) :

    0.1667=e^(-8/τ)

    Solving for τ = 4.4654

    Re-substituting τ into (1) and solving for a, a=35.985

    So the full equation becomes T=10+35.985*e^-(t/4.4654)

    Still not entrely accurate for some values fo T at given times
     
  8. Jul 10, 2016 #7

    rude man

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    I said to use the initial and final T points. You picked inbetween points that are hard to resolve and give a poor exponential fit for the empirical data you're looking at so you got way the wrong value for "a".
    One look at the graph should tell you "a" (= Ti) very precisely! Then, compute τ.
     
  9. Jul 12, 2016 #8
    Im having trouble with the math:

    T=T∞+a.e^(-t/τ)

    T∞=10

    a=50

    T=50 at 0s

    50=10+50e^(0/τ) is unsolvable

    T=10 at 20s

    10=10+50e^-(20/τ) is unsolvable
     
  10. Jul 12, 2016 #9

    rude man

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    How about t = 4?
     
  11. Jul 12, 2016 #10
    10+50e^(-4/τ)=50 , τ=-4/(ln(4/5))=17.92

    why t=4??

    Im a bit lost now. I dont think this equation is working.
     
  12. Jul 12, 2016 #11

    NascentOxygen

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    a is not 50
     
  13. Jul 12, 2016 #12

    rude man

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    Because that gives you a good measure of τ. You're not going to get τ at Ti or Tf; the curve could drop from 50 towards 10 very slowly or very quickly or anything inbetween. And τ tells you how fast or how slowly the curve drops, right?
    It is if you use it right!
    The equation is T = Tf + a e-t/τ. When t=4 what is T?
     
    Last edited: Jul 12, 2016
  14. Jul 12, 2016 #13

    rude man

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    Good thing to keep in mind!
     
  15. Jul 13, 2016 #14
    Sorry I need some clarity here because it is getting too confusing, I need more methodology here and the correct syntax. According to my notes;

    T = Tf + a e^(-t/τ)

    T= output value
    τ= system time constant in seconds (a*0.368)?? i.e one time constant equals 36.8% temperature input
    a= magnitude of the step change (50-10=40)??
    t=elapsed time
    Tf or 'C' in my notes= final value of a when t >> 5τ (10)??

    The equation is T = Tf + a e-t/τ. When t=4 what is T? On the graph its looks to be about 24°C although im sure this is not what your asking.

    Do I have the right values for all the variables in the equations?

    Initial and final T points are 50 and 10 respectively?
     
  16. Jul 13, 2016 #15

    rude man

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    That statement makes no sense to me. Just consider τ a constant with the dimension of time for now.
    right but it's not a step change. It's the difference between T(0) and T(∞).
    right
    That's exactly what i'm asking and that's correct. It's T(t=4) = 24 .
    Absolutely!
    You're getting there.
     
  17. Jul 14, 2016 #16
    Okey doke here we go T = Tf + a e-t/τ. I can see what you mean when you say you have to use a figure inbetween the initial and final points

    T(4) = 10 + 40 e^-4/τ

    24=10+40e^-4/τ

    solving for τ :

    τ= -4/ln(7/20) = 3.81

    T(t)=10+40e^-t/3.81

    T(20)= 10+40e^-20/3.81=10.21 (close to final value although it never touches in reality)

    T(0)=50 as e^0 =1
     
  18. Jul 14, 2016 #17

    rude man

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    Yes!
    You can also check you equation fo T(∞) of course.
     
  19. Jul 14, 2016 #18
    Of course thanks for your help.
     
  20. Jul 14, 2016 #19

    rude man

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    Congrats for sticking it out this long. 9 out of 10 give up long before this.
    The former president of Sony once said that what makes a good scientist is curiosity and not quitting until the answer is found. So true.
     
  21. Mar 21, 2017 #20
    Hi, I've been following this up until we got to solving for 4 seconds at 24

    24=10+40e^-4/τ

    and then became totally lost when solving for τ

    τ= -4/ln(7/20) = 3.81

    I'm a little lost on how the 3.81 came about and not sure if its some basic maths I'm missing here, or confusion over the written format of some symbols can anyone help with this

    thanks
     
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