- #1
skrat
- 748
- 8
Homework Statement
American hardwood 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!
Homework Equations
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?