Regular vs differential form of Fourier's law?

Question

I was watching some of Michel Van Bizen's lectures on youtube, (Heat Transfer, video 16-18) and the guy mentioned that if a pipe or rather a bar was connecting a heat source and a cold source, we could use Fourier's law in this form: $$ dQ/dt = k(T_1 - T_2)A/L $$

He also said if A, the cross-sectional area of the bar varies with x, or the constant of conductivity, k, varies with respect to x, then to find the heat flow rate, the new formula would be $$ dQ/dt = kA(dT/dx) $$ Similarly, if the temperature of the heat reservoir varied with time as heat flowed away from it, the new formula would be $$ dQ/dt = kAT(t)/L $$. I'm pretty sure the difference between the formulas has something to do with temperature changing non-linearly and the temperature gradient, but I'm not sure how to get to these equations and the concept behind it all. Can anyone explain the difference between the formulas and clear up my confusion? Please keep the explanations as simple as possible, with minimal multivariable calculus if possible.

Answer 1

Let's start from the beginning and write the balance equation of the internal (thermal) energy $U$, and then introducing Fourier's law for heat flux.

Internal energy balance equation. We can write the time derivative of the thermal internal energy as the total heat flux and production (if some production, like chemical reactions, exists in the system)

$\dfrac{d U}{d t} = Q$.

For a 3D continuous system we can write it in integral form as

$\dfrac{d}{dt} \displaystyle \int_V \rho u = - \oint_{\partial V} \mathbf{q} \cdot \mathbf{\hat{n}}$,

being $\rho$ the density, $u$ the internal thermal energy per unit mass, $\mathbf{q}$ the (vector) heat flux per unit surface, transferring energy from the system to the external environment and $\mathbf{\hat{n}}$ the unit normal vector pointing outwards.

Differential form of this equation reads $\dfrac{\partial }{\partial t} (\rho u) = -\nabla \cdot \mathbf{q}$

Fourier's law. Fourier's law states that the heat flux per unit surface is linear with the temperature gradient $\nabla T$.

Fourier's law in 3D. General form of Fourier's law in 3D physical space reads

$\mathbf{q} = -k \, \nabla T = - k \left[ \dfrac{\partial T}{\partial x} \mathbf{\hat{x}} + \dfrac{\partial T}{\partial y} \mathbf{\hat{y}} + \dfrac{\partial T}{\partial z} \mathbf{\hat{z}} \right]$

for an isotropic medium, i.e. the heat-flux per unit of area $\mathbf{q}$ is locally proportional to the temperature gradient $\nabla T$, through the conductivity $k$. Thus the balance equation becomes

$\dfrac{\partial }{\partial t} (\rho u) = \nabla \cdot ( k \nabla T)$.

If $k$ is uniform in space, and if we can write internal energy as a function of $T$, $u = c T$, with constant density and $c$, the equation becomes

$\rho c \dfrac{\partial T}{\partial t} = k \, \nabla^2 T$.

Fourier's law in 1D. Looking at an elongated element, such as a structural beam/rod with isolated surface or a pipe containing fluid at rest, we can approximately assume that the behavior of this system is approximately 1-dimensional, with the heat conduction occurring only along the axial direction $x$. Fourier's law thus reduce to its $x$-component only

$q_x = - k \dfrac{\partial T}{\partial x}$,

where functions depends only on $x$ and time $t$. Balance equation becomes

$\rho c \dfrac{\partial T}{\partial t} = k \, \dfrac{\partial^2 T}{\partial x^2}$.

Total heat flux reads

$Q = \displaystyle \int_S \mathbf{q} \cdot \mathbf{\hat{n}} = \int_S q_x \mathbf{\hat{x}} \cdot \mathbf{\hat{x}} = q_x \, A$.

Fourier's law in 1D steady problem without heat source. Internal energy equation for 1D steady problem without heat source can be written using ordinary derivative (with functions of one variable only, namely $x$) as

$k \, \dfrac{d^2 T}{d x^2} = 0$$\qquad \rightarrow \qquad $$\dfrac{d q_x}{d x} = 0$

and thus $q_x = - k\dfrac{dT}{dx} = \overline{q}_x = \text{const}$, and thus the temperature profile in space is linear

$T(x) = -\dfrac{ \overline{q}_x }{k} x + C = A x + C$.

To complete the differential problem, you need boundary conditions. As an example, if you know the temperature at two locations $T(x_0) = T_0$, $T(x_1) = T_1$, you can write the temperature and the heat flux as

$T(x) = T_0 + (T_1 - T_0 ) \dfrac{x-x_0}{x_1 - x_0}$$\quad ,\qquad$ $q_x = - k \dfrac{T_1-T_0}{x_1-x_0}$$\quad ,\qquad$ $Q = - k A \dfrac{T_1-T_0}{x_1-x_0}$

Answer 2

For clarity, this answer always specifies the spatial and thermal dependencies when they exist (e.g., the temperature $T(x,t)$ as a function of the position $x$ and time $t$).

Fourier's law expresses the phenomenological finding that the instantaneous conduction heat flux $q(x,t)$ (in units of power) perpendicular to a slice of material in the $y$–$z$ plane at location $x$ is proportional to the thermal conductivity $k(x,t)$, the cross-sectional area $A(x,t)$, and the temperature gradient $\frac{dT(x,t)}{dx}$:

$$q(x,t)=k(x,t)A(x,t)\frac{dT(x,t)}{dx}\tag{1},$$

where the parameters are all specific to point $x$ and may depend on the time $t$.

Let's focus on steady-state conduction for a moment. The time dependence disappears, so we have

$$q(x)=k(x)A(x)\frac{dT(x)}{dx} \tag{2}.$$

If no other heat transfer modes (e.g., radiation, convection, heat generation) are present, then the conduction energy flux is the complete energy flux, and conservation of energy requires $q(x)$ to be the same at any $x$: $q(x)=q$. In addition, if the cross section is uniform, then $A(x)=A$, and if a single material is used and the thermal conductivity variation over the relevant temperature range is negligible, then $k(x)=k$. We can then integrate $\frac{dT(x)}{dx}=\frac{q}{kA}$ to obtain

$$T(x)=\frac{q}{kA}x+C \tag{3},$$

where $C$ is a constant. If we have fixed-temperature boundary conditions at two $x$ locations separated by a distance $L$ (say, $T_1$ and $T_1+\Delta T$), then we can insert those to obtain

$$q=kA\frac{\Delta T}{L} \tag{4}.$$

Importantly, this equation isn't a starting point, as might be concluded from Van Bizen's lecture, but rather a substantial simplification of the more general Equation (1).

Let's return to that general case. Since the instantaneous heat flux $q(x,t)$ is a measure of the energy $Q(x,t)$ conducted past point $x$ per unit time, or $\frac{dQ(x,t)}{dt}$, we may integrate

$$q(x,t)=\frac{dQ(x,t)}{dt}=k(x,t)A(x,t)\frac{dT(x,t)}{dx} \tag{5}$$

to obtain

$$Q(x,t)=\int k(x,t)A(x,t)\frac{dT(x,t)}{dx}\,dt, \tag{6} $$

or

$$Q(x,t)=Q(t) = k(x)A(x)\frac{dT(x)}{dx}t \tag{7} $$

for steady-state conduction with no other heat transfer modes present, or

$$Q(t)=kA\frac{\Delta T}{L}t \tag{8} $$

for the simple case of the 1-D bar or rod with uniform cross section and constant thermal conductivity. Equation (4) thus gives the heat flux moving in the $x$ direction, and Equation (8) gives the total energy that's been conducted in that direction.

Does this get at what you're asking about, i.e., the various spatial and temporal dependencies and how to handle them?