0
$\begingroup$

I am new here so hope this is the way to ask.

I have got a machine I want to test. I must stop the test when the machine is about 100 degrees celsius. I have got a logging device and logged the cooling down. I logged for about 10 minutes and I want to predict the temperature in 1 hour cooling.

Here is what I did measure.

$$ \begin{array}{c|lcr} T(min) & \text{T(C Measured)} & \text{T(C Calculated)} \\ \hline 0& 95.28& 95.28&\\ 1& 85.39& 85.44&\\ 2& 76.23& 76.74&\\ 3& 69.03& 69.04&\\ 4& 63.78& 62.24&\\ 5& 58.84& 56.22&\\ 6& 54.99& 50.90&\\ \end{array} $$

Ages ago I had Math lessons, and I though well lets see what I can remember. I did a search on the internet and found this site and it looked familiar.

http://www.ugrad.math.ubc.ca/coursedoc/math100/notes/diffeqs/cool.html

It is about a pan of soup cooling down, I should be able to use this method to calculate the cooling down of the machine right? But when I try and use the first few minutes of the data the calculated graph is different from the measured graph. I changed some numbers in the end formula with the method of “gamble and see what happens” and then it gets better, but then it will be a gamble and not a real calculation. Here is my calculation. The last 3 numbers are not close engough to the real measured data and I want to go to T = 60 min.

Calculation:

So I tried to use the first 3 minutes and see if the formula could predict te following minutes, this is in the column “T(C Calculated)”.

T(t) = To + (To-Ta)e(-kt) 69.03 = 10.25 + (95.28-10.25)e(-3k) 69.03 - 10.25 = 85.03e(-3k) 58.78 = 85.03e(-3k) 58.78/85.03 = e(-3k) 85.03/58.78 = e(3k) 1.447 = e(3k) ln(1.447) = ln (e(3k)) 0.3695 = 3k 0.3695/3 = k k = 0.123

T(t) = 10.25 + 85.03e-0.123*t

With the same method I used Calc (excel of open office) to calculate the k values for all the rows of time and I got this table:

Automatic calculated k values:

$$ \begin{array}{c|lcr} T(min) & \text{k} & \text{Delta k} & \text{AVERAGE} \\ \hline 1 & 0.1236510953 & . & . & . \\ 2 & 0.1268262348 & 0.0031751395 & . &. \\ 3 & 0.1230674915 & -0.0037587433 & . & . \\ 4 & 0.1156904727 & -0.0073770187 & . & . \\ 5 & 0.1119172774 & -0.0037731954 & . & \\ 6 & 0.1070226966 & -0.0048945808 & -0.0033256797 & \\ \end{array} $$

And changed the formula to:

T(t) = 10.25 + 85.03e-1*(0.123-(0.003326*t))*t
And got these values:

$$ \begin{array}{c|lcr} T(min)& \text{T(C)} & \text{Difference from measured values} \\ 0 & 95.2799999989 & 1.07968389784219E-009& \\ 1 & 85.3900071572 & -7.15722647726125E-006& \\ 2 & 77.0935046247 & -0.8635046247& \\ 3 & 70.1099181698 & -1.0799181698& \\ 4 & 64.2137289838 & -0.4337289838& \\ 5 & 59.2230004683 & -0.3830004683& \\ 6 & 54.9904554258 & -0.0004554258& \\ \end{array} $$

It is closer to the real measured data but still I do not trust my method enough to predict the coming hour. Can anyone please tell me what I might be doing wrong and what to change in my method? It would be helpful if someone did something similar, maybe the theory is different from the real thing?

1 Answers 1

0

Newton's Law of Cooling is normally not a very good model of how things cool off. If you're circulating your ambient fluid and/or if you have a very thin body, then, yeah, it's pretty good.

But if you put a sphere of warm terra cotta in tub of cold water, then the sphere warms up the surrounding water and the outside of the sphere cools, leaving the inside warmer. The sphere becomes self-insulating and a similar effect is happening with the water.

If you think about submerging a thin steel plate and a sphere of steel of the same mass into ice water, you can see right away that the steel plate will cool down much more quickly, but Newton's Law doesn't take shape into account, and predicts the same temperature curve for both.

I know about the Stefan-Boltzmann law, which is for black body heat, but I don't think that's your situation.