Cylinder with heat generation, Separation of variables

[16universes]

First, I would like to thank everyone's help with this problem. I spent some time thinking about it, and realized that I probably made it more complex than it really was. After discussing the problem with my advisor, he suggested the following solution:

Let $T$ be the temperature of the thermistor, and $R = R_0 + \alpha T$ be the resistance of the thermistor as a function of its temperature. Note that $\alpha < 0$. Heat generated by the thermistor at constant current is $I^2 (R_0 + \alpha T)$. Assume heat loss due to the airflow is $h_c (T - T_0)$, where $h_c$ is the heat transfer coefficient.

Combining everything together:

$$C_p \frac{dT}{dt} = I^2 (R_0 + \alpha T) - h_c (T-T_0)$$

where C_p is the heat capacity of the thermistor. If we define $m \equiv I^2 \alpha - h_c$ and $b \equiv I^2 R_0 + h_c T_0$, then this can be rewritten as

$$C_p \frac{dT}{dt} = mT + b$$

Note that $m < 0$ and $b > 0$.

This has the following solution:

$$T(t) = k_1 e^{\frac{m}{C_p}t} - \frac{b}{m}$$

Again, noting that $m<0$ and $b>0$, the temperature asymptotically approaches a positive value. Furthermore, according to www.engineeringtoolbox.com, $h_c$ can be approximated as $h_c \approx 10.45 - v + 10v^{1/2}$, so we can also determine $T$ as a function of airflow velocity.

I'm happy with this solution. I was way overthinking it, but this was quite elegant. What do you guys think?

 

[Orodruin]

$$C_p \frac{dT}{dt} = mT + b$$
...
I'm happy with this solution. I was way overthinking it, but this was quite elegant. What do you guys think?

...
You could then just as well formulate the problem as an ODE directly:
$$
\partial_t T = - \kappa T + \beta
$$
for some constants $\kappa, \beta$.
...
The stationary solution to this equation is given by letting the LHS = 0 yielding
$$
\kappa T = \beta
$$
or solving a 4th order equation if you add the radiation term.
If you linearize the problem around this solution you will find a solution that exponentially approaches the stationary solution.

This is what I think ;)

 

[16universes]

Ha I apologize. I was still caught up "seeing" the problem differently. It took some time to change how I was looking at it. But yes, you were correct. Thanks