Thermal conductance of an angular section of a disc

[BrenoVA]

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.

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?

 

[Chestermiller]

In cylindrical coordinates, the Laplace equation is satisfied by $$T=T_0+(T_1-T_0)\frac{\theta}{\alpha}$$

 

[TSny]

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.

 

[Chestermiller]

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 $$Q=\int_{R_0}^{R_1}{qhdr}$$

 

[BrenoVA]

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...