Mathematical relationship of a thermocouple

[CNC101]

Homework Statement

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).

Homework Equations

T_o=T_i*e-(t/τ) Eq1.

OR

f(T)=c*e-(kt) where k= LnT₁-LnT₂/t₁-t₂ =Ln5/-20=-0.08047 Eq2.

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?

 

[Irene Kaminkowa]

You shoold use many points: 10 to 20 - to restore the equation.
The relevant method is Least Squares: https://en.wikipedia.org/wiki/Least_squares

 

[NascentOxygen]

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 $\mathfrak τ$[/size].

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.

 

[rude man]

Homework Statement

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).

View attachment 103040View attachment 103040

Homework Equations

T_o=T_i*e-(t/τ) Eq1.

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 = T_ie^-t/τ + T_∞ instead?
Read T_i 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.

 

[CNC101]

"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 don't understand the method,sorry.

 

[CNC101]

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

 

[rude man]

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)

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" (= T_i) very precisely! Then, compute τ.

 

[CNC101]

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

 

[rude man]

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

How about t = 4?

 

[CNC101]

10+50e^(-4/τ)=50 , τ=-4/(ln(4/5))=17.92

why t=4??

Im a bit lost now. I don't think this equation is working.

 

[NascentOxygen]

I don't think this equation is working.

a is not 50

 

[rude man]

10+50e^(-4/τ)=50 , τ=-4/(ln(4/5))=17.92
why t=4??

Because that gives you a good measure of τ. You're not going to get τ at T_i or T_f; 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?

I don't think this equation is working.

It is if you use it right!
The equation is T = T_f + a e^-t/τ. When t=4 what is T?

 

[rude man]

a is not 50

Good thing to keep in mind!

 

[CNC101]

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 I am 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?

 

[rude man]

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

That statement makes no sense to me. Just consider τ a constant with the dimension of time for now.

a= magnitude of the step change (50-10=40)??

right but it's not a step change. It's the difference between T(0) and T(∞).

t=elapsed time
Tf or 'C' in my notes= final value of a when t >> 5τ (10)??

right

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 I am sure this is not what your asking.

That's exactly what I'm asking and that's correct. It's T(t=4) = 24 .

Initial and final T points are 50 and 10 respectively?

Absolutely!
You're getting there.

 

[CNC101]

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

 

[rude man]

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

Yes!
You can also check you equation fo T(∞) of course.

 

[CNC101]

Of course thanks for your help.

 

[rude man]

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.

 

[Jayson]

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

 

[rude man]

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

If you're having trouble solving for τ here you need to go back & review high school math. Like Algebra 1 rather than Algebra 2 I think.

 

[Jayson]

If you're having trouble solving for τ here you need to go back & review high school math. Like Algebra 1 rather than Algebra 2 I think.

ok solving for t
24 = 40e^-4/t +10
14/40 = e^-4/t
7/20 = e^-4/t

and then I'm lost as to how e disappears to become

τ= -4/ln(7/20) = 3.81

any help in the right direction much appreciated before I go mad

 

[gneill]

and then I'm lost as to how e disappears to become

τ= -4/ln(7/20) = 3.81

You should recall that e^x and ln(x) are inverses of each other. Do a bit of googling on "e^x and ln(x)".

 

[rude man]

If y = e^x
then taking the natural log of each side,
ln(y) = ln(e^x) = x
This you must assimilate indelibly!

 

[David J]

Hello

I was following this thread and I was just wondering why you used t = 4? In my notes it tells me that

"The important thing to realize is that the time constant is the time taken for 63.2% of the full change to occur"

In the case of this question we started at 50 degrees and ended at 10 degrees, a change of 40 degrees.

63.2 % of 40 is 25.28. 50 - 25.28 = 24.72

I know its not exactly 24 but I was just wondering if there was a connection at all ??

 

[HopelessDropout]

Can anybody help me get through determining the relationship using equation 1 as mentioned at the start?

output change (initial) = input change (final) x (1 - e-t/τ )

Thanks!

 

[jim hardy]

You could start with Newton's Law of Cooling

for example

https://courses.lumenlearning.com/ivytech-collegealgebra/chapter/use-Newtons-law-of-cooling/It's a feature of math that any quantity whose rate of change depends on what is its magnitude will follow an exponential curve.
That should become part of your repertoire.
Here's a tutorial with worked examples and practice problems.

https://courses.lumenlearning.com/waymakercollegealgebra/chapter/exponential-growth-and-decay/

 

[David J]

I struggled with this in the beginning but if you follow all of the posts in this thread the correct method is explained.

I used $T=T\infty+a*e^(\frac{-t}{\tau})$

$T\infty$ is the final limit which in the case of this question is 10

$a$ is the difference between $To$ which is 50 and $T\infty$ which is 10 so a must = ...

the task in hand is to identify the time constant $\tau$

I picked 10 points from the graph given in the question, the first point being $x=2, y=32.5$ and the last point being $x=20, y=10.1$

so I found $T=32.5$ and $t=2$

I then plugged these values into the equation and after evaluating for $\tau$ i had a time constant for my first point which equated to 3.51

I did this for every point and took all of the values and found the mean value. I used this mean value for $\tau$. When i used it in the equation $T=T\infty+a*e^(\frac{-t}{\tau})$ it proved to be very close to $T$

I think as suggested by the lady in post 2, the more points you use the closer you will get. My answer was marked correct

I also created a graph in excel to try and get as close as possible.

I am aware of the course you are doing and my advice is to follow all of the advice in this thread and go into as much detail as you can and you won't go far wrong

 

[HopelessDropout]

Thanks, all above, I think I managed to get there!

 

[ckfo]

Transfer lag of the time constant is 63.25%

https://en.wikipedia.org/wiki/Time_constant

change in temperature - time constant

Then find on the graph where that correlates.

Initial rate of change = change in temperature/change in time

With the values you work out, you will then plot those points on the graph.