Help with finding time constant from graph

Question

Given the following for oven temperature decay,

                                                           

I have to find the time constant and temperature outside the oven. The answers given in the book are:

$\tau=35.8 \text{ sec}$, Outside Temp = $40^{\circ}C$.

The instructor stated that these values are incorrect and wants us to find the correct solution. I got the following values for $\tau$ but not sure if I am approaching this problem correctly:

$\dfrac{d\theta}{dt} \big|_{t=0}=\dfrac{1}{\tau}=\dfrac{12^{\circ}C}{470\text{ sec}}$

$\tau=39.175 \text{ sec}$

This is close to what the book has, but I feel it might be wrong based on the total time it takes for the system to reach its final value (~4250 sec). I also tried normalizing by the total $\Delta T$ and I get $\tau = 3,113.75 \text{ sec}$, which just seems way too high.

What is the best method of approaching such a problem?

Answer 1

Supposing that the low of cooling is : $$\theta(t)=\theta_{out}+\beta\exp\left(-\frac{t}{\tau}\right)$$ $t$ is the time, $\theta$ is the temperature, $\theta_{out}$ is the outside temperature, $\tau$ is the time constant and $\beta$ is a coefficient.

$$\ln\left(\theta(t)-\theta_{out}\right)=\ln(\beta)-\frac{t}{\tau}$$ Thus, the graphical representation of $\quad\ln\left(\theta(t)-\theta_{out}\right)\quad$ vs. $t$ should be $\underline{\text{linear}}$.

This is drawn in BLACK on the figure below in case of $\theta_{out}=40.\quad$ Obviously, the curve is far from a straight line. This confirmes that $\theta_{out}=40$ is false.

The drawings for various values of $\theta_{out}$ show that the best linear fit is obtained with $\theta_{out}\simeq 0.$

The case $\theta_{out}\simeq 0$ is drawn in red. In order to compare, the straight line is drawn in black dots.

The approximate values of $\beta$ and $\tau$ can be computed thanks to linear regression ($\theta_{out}= 0)$ according to the equation : $$\ln\left(\theta(t)\right)=\ln(\beta)-\frac{t}{\tau}$$ Which leads to : $\beta\simeq 120$ and $\tau\simeq 3900$.

This is the best fit with criteria of least mean square $\underline{\text{relative}}$ deviation (because linear regression vs. the log function).

The result would be slightly different if the criteria was the least mean square deviation. This requires a non-linear method of calculus. But, in practice, the least relative deviation is often preferred.

Answer 2

In the same spirit as JJacquelin's answer.

Consider that $\theta_{out}$ is fixed. Then, le linear regression gives $\beta$ and $\tau$ which are implict functions of $\theta_{out}$. Recompute the values of the $\theta$'s and let $SSQ$ be the sum of the squares of the residuals.

Using the same values as JJacquelin for his plots, you would get the following results $$\left( \begin{array}{cc} \theta_{out} & SSQ \\ -40 & 24.492 \\ -30 & 18.867 \\ -20 & 13.258 \\ -10 & 8.173 \\ 0 & 4.647 \\ 10 & 4.843 \\ 20 & 13.330 \\ 30 & 39.862 \\ 40 & 104.916 \end{array} \right)$$ Use the four values corresponding to $\theta_{out}=-10,0,10,20$ and compute the cubic going through the points. You should get $$SSQ=0.000761715\, \theta_{out}^3+0.0186079\, \theta_{out}^2-0.242684 \,\theta_{out}+4.6471 \tag 1$$ Compute the derivative $$\frac{d SSQ}{d \theta_{out}}= 0.00228515 \, \theta_{out}^2+0.0372157 \, \theta_{out}-0.242684$$ Set it equal to zero since you look for the minimum. This will give you two solutions $\theta_{out}\approx -21.3$ (to be discarded since outside the range) and $\theta_{out}\approx 5.0$.

For $\theta_{out}=5$, using $(1)$ you predict $SSQ=3.994$.

Repeat the linear regression using $\theta_{out}=5$ and you will get $SSQ=4.073$ (quite close, isn't it ?) and the required parameters.

A full nonlinear regression would have given $\theta_{out}=4.967$ and $SSQ=4.073$ and the best parameters would have been $\beta=116.60$ and $\tau=3608$ almost the same as those given by JJacquelin.