Thermal conductance of an angular section of a disc

  • Thread starter Thread starter BrenoVA
  • Start date Start date
  • Tags Tags
    Heat conduction
BrenoVA
Messages
2
Reaction score
0
Homework Statement
Calculate the conductance of an angular section of a disc using the integral form of Fourier's Law.
Relevant Equations
The integral form of Fourier's Law considering 1D angular direction on a cylindrical coordinate system.
I'm trying to calculate the thermal conductance of the disc segment shown in the figure below.
disk_thermal_prob.png


We consider a thickness ##d##, only a non-zero heat flux only in the angular direction (no thermal loss through the sides or above and below), no heat generation and steady state condition. Also, the thermal conductivity k is constant in the material.

According to Fundamentals of Heat and Mass transfer by Incropera et. al., 6th edition, section 3.2, we can use the integral form of Fourier's Law, which for a Cartesian coordinate system as:
q_x\int_{x_0}^x\frac{dx}{A(x)}=-\int_{T_0}^{T_1}k(T)dT

I am not sure how to determine the equivalent equation for the angular coordinate and then solve it, thus finding the thermal conductance C=-\frac{q}{\Delta T}.

My attempt
Note that the cross-section is constant with respect to \theta\in[0, \alpha] and is equal to A=d(R_1-R_0). Then, since we have rd\theta instead of dx, I believe we integrate

q_\theta \int_0^\alpha\int_{R_0}^{R_1}\frac{r}{d(R_1-R_0)}drd\theta = q_\theta \frac{\alpha}{d} \frac{(R_1^2-R_0^2)}{2(R_1-R_0)} = q_\theta \frac{\alpha}{d}\frac{(R_1+R_0)}{2}=-k\Delta T

Therefore, the thermal conductance is
C=k\frac{d}{\alpha}\frac{2}{R_1+R_0}

The units are wrong and it doesn't match a reference I have (which has a poor mathematical explanation) that yields
C=k\frac{d}{\alpha}\ln\left(\frac{R_1}{R_0}\right)


I guess I'm incorrectly applying the surface integrals?
 
Physics news on Phys.org
In cylindrical coordinates, the Laplace equation is satisfied by $$T=T_0+(T_1-T_0)\frac{\theta}{\alpha}$$
 
BrenoVA said:
According to Fundamentals of Heat and Mass transfer by Incropera et. al., 6th edition, section 3.2, we can use the integral form of Fourier's Law, which for a Cartesian coordinate system as:
q_x\int_{x_0}^x\frac{dx}{A(x)}=-\int_{T_0}^{T_1}k(T)dT
This formula does not readily go over to your situation in cylindrical coordinates.

Instead, I suggest starting with the differential form of Fourier's law to express the rate of heat flow, ##dq##, through the blue shaded area, ##dA##, shown below. I use ##h## for the thickness since the symbol ##d## can get confused with the differential operator.

1731695964096.png
 
  • Like
Likes BrenoVA and SammyS
Continuing the analysis I started in post #2, the magnitude of the (circumferential) heat flux vector is given by $$q=-k\frac{1}{r}\frac{\partial T}{\partial \theta}=k\frac{1}{r}\frac{T_0-T_1}{\alpha}$$The total rate of heat flow across each cross section is then given by $$Q=\int_{R_0}^{R_1}{qhdr}$$
 
I finally see the multiple sources of my confusion now, thank you all.

Firstly, the gradient operator in cylindrical coordinates on the scalar field ##T## is:
$$\nabla T = \frac{\partial T}{\partial r}\hat{\boldsymbol{r}} + \frac{1}{r}\frac{\partial T}{\partial \theta}\hat{\boldsymbol{\theta}} + \frac{\partial T}{\partial z}\hat{\boldsymbol{z}}$$
I had forgotten about the ##\frac{1}{r}##.

Secondly, following @TSny, the differential area perpendicular to the direction of the heat flux ##q_\theta## here is ##dA = drdz##

Finally, since ##\frac{\partial T}{\partial r}=\frac{\partial T}{\partial z} = 0## and since we have that ##T## varies linearly with ##\theta##, as written by @Chestermiller, then ##\frac{\partial T}{\partial \theta}=\frac{\Delta T}{\Delta\theta}=\frac{\Delta T}{\alpha}##, and finally:
$$Q=\oint_{S}\nabla T \cdot d \boldsymbol{S} = \int_0^h\int_{R_0}^{R_1}-k\frac{1}{r}\frac{\partial T}{\partial \theta}drdz=-k\frac{\partial T}{\partial \theta}\int_0^hdz\int_{R_0}^{R_1}\frac{1}{r}dr=-k\frac{\Delta T}{\alpha}h\ln\left(\frac{R_1}{R_0}\right)$$

Therefore
$$C=\frac{h}{\alpha}\ln\left(\frac{R_1}{R_0}\right)$$

One thing I noticed is that I gotta refresh my memory on what surface integrals are...
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top