# Mathemathical physics - tensors

1. May 11, 2014

### skrat

1. The problem statement, all variables and given/known data
American hard wood quebracho has thermal conductivity $14W/mK$ in axial direction, $10W/mK$ in radial direction and $11W/mK$ in $z$ direction. From a large block of wood we cut out a 10 cm long stick with radius $4mm$ in direction that forms the same angles with all eigen axes. We isolate the stick and put it on ice, so that the temperature difference between the ends of the stick is $10K$.

Calculate the heat flux through the stick!

2. Relevant equations

3. The attempt at a solution

$\vec{j}=-\lambda \nabla T$ so $\nabla T=-\lambda ^{-1}\vec{j}$

For the whole block of wood $\lambda =\begin{bmatrix} \lambda _\varphi & & \\ & \lambda _R & \\ & & \lambda _z \end{bmatrix}$ and since this is a diagonal tensor than $\lambda ^{-1}=\begin{bmatrix} \frac{1}{\lambda _\varphi} & & \\ & \frac{1}{\lambda _R} & \\ & & \frac{1}{\lambda _z} \end{bmatrix}$

Also, vector j in the same system is $\vec{j}=j(sin(45°)cos(45°),sin(45°)sin(45°),cos(45)=j(1/2,1/2,\sqrt{2}/2)$

So $\nabla T=-\begin{bmatrix} \frac{1}{\lambda _\varphi} & & \\ & \frac{1}{\lambda _R} & \\ & & \frac{1}{\lambda _z} \end{bmatrix}j\begin{bmatrix} \frac{1}{2}\\ \frac{1}{2}\\ \frac{\sqrt{2}}{2} \end{bmatrix}=-j\begin{bmatrix} \frac{1}{2\lambda _\varphi }\\ \frac{1}{2\lambda _R}\\ \frac{\sqrt{2}}{2\lambda _z} \end{bmatrix}$ in the system of this big block of wood.

Now I would have to rotate $\nabla T$ into the system of the stick! I think this is how this is done (for c=cos and s=sin) $\nabla T^{'}=R_xR_y\nabla T=\begin{bmatrix} 1 & 0 & 0\\ 0& c & -s\\ 0& s & c \end{bmatrix}\begin{bmatrix} c & 0 &-s \\ 0& 1 & 0\\ s& 0 & c \end{bmatrix}(-j)\begin{bmatrix} \frac{1}{2\lambda _\varphi }\\ \frac{1}{2\lambda _R}\\ \frac{\sqrt{2}}{2\lambda _z} \end{bmatrix}=-j\begin{bmatrix} c\frac{1}{2\lambda _\varphi }-s\frac{\sqrt{2}}{2\lambda _z} \\ c\frac{1}{2\lambda _R}-s(s\frac{1}{2\lambda _\varphi }+c\frac{\sqrt{2}}{2\lambda _z}) \\ s\frac{1}{2\lambda _R}+c(s\frac{1}{2\lambda _\varphi }+c\frac{\sqrt{2}}{2\lambda _z}) \end{bmatrix}$

Now we know that $\Delta T=\int _{-h/2}^{h/2}\nabla T_z^{'}dz^{'}=-j\int _{-h/2}^{h/2}(\frac{1}{2\lambda _R}+c(s\frac{1}{2\lambda _\varphi }+c\frac{\sqrt{2}}{2\lambda _z})) dz^{'}=-j(s\frac{1}{2\lambda _R}+c(s\frac{1}{2\lambda _\varphi }+c\frac{\sqrt{2}}{2\lambda _z}))h$

The rotation angle was of course for 45°, therefore $\Delta T=-j(s\frac{1}{2\lambda _R}+c(s\frac{1}{2\lambda _\varphi }+c\frac{\sqrt{2}}{2\lambda _z}))h=-j(\frac{\sqrt{2}}{4\lambda _R}+\frac{1}{4\lambda _\varphi }+\frac{1}{4\lambda _z })$

therefore $j=\frac{-4\Delta T}{(\sqrt{2}/\lambda _R+1/\lambda _\varphi +1/\lambda _z)h}$ where of course $\Delta T<0$

This should be the answer IF i am not mistaken? What do you think?

2. May 11, 2014

### Staff: Mentor

What is the difference between the axial direction and the z direction?

Chet

3. May 12, 2014

### skrat

Hmmm, let me try to translate that part again...

"Thermal conductivity in the direction of fibers is $14 W/mK$, perpendicular to the fibers $10W/mK$ and $11W/mK$ parallel to the fibers. That would be the exact translation and the way I understood this is that in polar coordinates the first axis in the direction of $\varphi$, second one is radial and the third one is vertical ($z$).

4. May 12, 2014

### Staff: Mentor

So the fibers are oriented circumferentially, correct?

Chet

5. May 12, 2014

### skrat

That's how I understand it, yes.

6. May 12, 2014

### Staff: Mentor

If the rod makes the same angle with all 3 principal axes, the three direction cosines must be equal. So what is the direction cosine of the rod with respect to each of the three axes? From this, what is the equation for a unit vector in the direction of the rod in terms of the unit vectors in the three principal directions? If the sides of the rod are insulated, what is the direction of the temperature gradient (neglecting end effects)?

Chet

7. May 12, 2014

### skrat

Unit vector in the direction of the rod is $\vec{s}=\frac{1}{\sqrt{3}}(1,1,1)$ meaning the angle with the principal axes is in fact $arccos(\frac{\vec{s}\cdot (0,0,1)}{1})=arcos(\frac{\sqrt{3}}{3})=54.74°$ and not 45° as I originally thought.

The temperature gradient has the same direction as the rood.

8. May 12, 2014

### Staff: Mentor

So, let the magnitude of the temperature gradient along the rod be dT/dl. Now you know the magnitude and direction of the temperature gradient. So, if you dot the temperature gradient with the thermal conductivity tensor, you get the heat flux vector.

Chet

9. May 12, 2014

### skrat

so $\nabla T=\frac{dT}{dl}$ meaning $j_z=\lambda _z\frac{\sqrt{3}}{3}\Delta T$ is the answer?

10. May 12, 2014

### Staff: Mentor

The problem statement is very ambiguous. Do they mean that the direction of the heat flux is along the axis of the rod, or do they mean that the direction of the temperature gradient is along the axis of the rod? Your guess is as good as mine. If the direction of the temperature gradient is along the direction of the rod, then

$j_z=\lambda _z\frac{\sqrt{3}}{3}\frac{dT}{dl}$
$j_r=\lambda _r\frac{\sqrt{3}}{3}\frac{dT}{dl}$
$j_θ=\lambda _θ\frac{\sqrt{3}}{3}\frac{dT}{dl}$

But, for this case, there will be heat flux normal to the surface of the rod (even though the temperature gradient is oriented along the rod axis).

If the direction of the heat flux is along the axis of the rod (so that no heat leaves through the cylindrical surface of the rod), the one must determine dT/dr, dT/dz, and dT/dθ such that the heat flux components normal to the surface of the rod are zero.

Do you have any idea which they mean?

Chet

11. May 13, 2014

### skrat

hmmm, I asked today to get a detailed explanation and as it turns out, the heat flux is in the direction of the rod and not the gradient.

Knowing that, my solution sounds something like:

$\nabla T=-\lambda ^{-1}\vec{j}$ where $\vec{j}=j\frac{1}{\sqrt{3}}\begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix}=j\hat{n}$

so $\nabla T=-\lambda ^{-1}\vec{j}=-\begin{bmatrix} \lambda _1 & & \\ & \lambda _2 & \\ & & \lambda _3 \end{bmatrix}j\frac{1}{\sqrt{3}}\begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix}=-j\frac{1}{\sqrt{3}}\begin{bmatrix} \frac{1}{\lambda _1 }\\ \frac{1}{\lambda _2 }\\ \frac{1}{\lambda _3 } \end{bmatrix}$

We also know that $\nabla T\cdot \hat{n}=-j\frac{1}{\sqrt{3}}\begin{bmatrix} \frac{1}{\lambda _1 }\\ \frac{1}{\lambda _2 }\\ \frac{1}{\lambda _3 } \end{bmatrix}\frac{1}{\sqrt{3}}\begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix}=-\frac{j}{3}(\lambda _{1}^{-1}+\lambda _{2}^{-1}+\lambda _{3}^{-1})=\frac{\Delta T}{l}$

From where $j=\frac{3\Delta T}{l(\lambda _{1}^{-1}+\lambda _{2}^{-1}+\lambda _{3}^{-1})}$

12. May 13, 2014

### Staff: Mentor

Yes. This looks OK to me.

Chet