Determine the flow and amplitude equation for thermal energy (with Del operator)

Question

It is a question vector calculus and Maxwell's laws. I put it this way. Let's say, we are working in a $3$-Dimensional space ( e.g $x\cdot y\cdot z = 4\cdot3\cdot2$, a certain room/class of that size ) .

Within this room, the heat obey a certain equation ( for e.g. $T = 25 + 5z$ ) .We know that heat flows from higher temperature regions to lower temperature regions. With this information in mind

How could I be able to determine the amplitude and the direction of travel of the thermal energy with the Del operator ( gradient ) ? .

I'm not looking for a definitive response, but an equation that could give me potentially the result for the amplitude and the direction. I also want to know if thermal energy follows a loop pattern within my room (and be able to explain it mathematically by using the del operator once again of course)?

You can find the lecture source here.

Thank you.

Answer 1

(Thermal) Energy is scalar so it does NOT have orientation. Thermal field(If such a field is defined in Physics) has both orientation and magnitude.

In Electromagnetics :

Orientation: $E= \nabla\cdot V$

Magnitude (If energy is stored in field) :

$$W= \dfrac{1}{2}\large\int_{\mathbb{R}^{3}} |E|^2 \ dt$$

I think you may plug $T$ for $V$ in Thermodynamics.

Answer 2

Hello I try to present what I know the best.

We know that heat flow takes place in three forms :

1. Conduction.
2. Convection.
3. Radiation.

For one dimensional heat flow, we have $q= k \dfrac{T_2-T_1}{L}$ , where $T_1$ and $T_2$ are terminal temperatures and $L$ is the length of the material. Switching to $3$-D spaces as you said, we need to look at the fourier system of the heat transfer before proceeding further. $\dfrac{q_x}{A}= - K \dfrac{dT}{dx}$. So integrating we obtain $$\dfrac{q_x}{A}\large \int_{0}^{L} \ dx= -k \int_{T_1}^{T_2} \ dT.$$

So given that the heat flow ( rate of conduction ) in the $3$-D space $(x,y,z)$ is given by the following equation $$ q= - k \nabla T = - k \bigg( \hat{i} \dfrac{\partial T}{\partial x}+\hat{j} \dfrac{\partial T}{\partial y}+\hat{k} \dfrac{\partial T}{\partial y}\bigg).$$The negative sign, there indicates the transfer of heat from one place to another. So you can simply substitute the heat equation in the place of $T$ and compute the partial derivatives to get the rate of flow of heat.

I can write up the cylindrical and spherical coordinate version too . It can be like this .

1. For cylindrical coordinate system $(r,\phi)$ we have $$ \large q= - k \nabla T = - k \bigg(\hat{i} \dfrac{\partial T}{\partial r}+\hat{j} \dfrac{1}{r} \dfrac{\partial T}{\partial y}+\hat{k} \dfrac{\partial T}{\partial Z}\bigg).$$ and individual heat flows are given by $$\large q_r = - K \dfrac{dT}{dx}, \\ \large q_{\phi} = - \dfrac{K}{r}\dfrac{dT}{d \phi},\\ \large q_{Z} = - K\dfrac{dT}{dZ}$$ where $\large q_r$ represents the flow in $r$-th direction. Here we include an extra dimension $Z$ to make $2$-D into $3$-D.
2. For Spherical coordinate system $(r,\theta,\phi)$ we have $$ \large q= - k \nabla T = - k \bigg(\hat{i} \dfrac{\partial T}{\partial r}+\hat{j} \dfrac{1}{r} \dfrac{\partial T}{\partial \theta}+\dfrac{\hat{k}}{r\sin \theta} \dfrac{\partial T}{\partial \phi}\bigg).$$

I hope this will be of some use to you. Feel free to ask if you want to hear more on anything.

Thank you.