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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

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

For steady problems we can use ordinary derivative assuming all the functions depend only on one variable, namely $x$ coordinate.

and the total heat flux is

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

Fourier's law in 1D steady problem without heat source. Internal energy equation for 1D steady problem without heat source can be written as

$\dfrac{d q_x}{d x} = 0$

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

$T(x) = T_0 + (T_1 - T_0 ) \dfrac{x-x_0}{x_1 - x_0}$
$q_x = - k \dfrac{T_1-T_0}{x_1-x_0}$

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$ $Q = - k A \dfrac{T_1-T_0}{x_1-x_0}$

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 $k$.

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

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

where we can use ordinary derivative assuming all the functions depend only on one variable, namely $x$ coordinate, and the total heat flux is

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

Fourier's law in 1D steady problem without heat source. Internal energy equation for 1D steady problem without heat source can be written as

$\dfrac{d q_x}{d x} = 0$

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

$T(x) = T_0 + (T_1 - T_0 ) \dfrac{x-x_0}{x_1 - x_0}$
$q_x = - k \dfrac{T_1-T_0}{x_1-x_0}$

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

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

For steady problems we can use ordinary derivative assuming all the functions depend only on one variable, namely $x$ coordinate.

and the total heat flux is

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

Fourier's law in 1D steady problem without heat source. Internal energy equation for 1D steady problem without heat source can be written as

$\dfrac{d q_x}{d x} = 0$

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

$T(x) = T_0 + (T_1 - T_0 ) \dfrac{x-x_0}{x_1 - x_0}$
$q_x = - k \dfrac{T_1-T_0}{x_1-x_0}$

Total heat flux $Q$ across a surface can be written as

$Q = \displaystyle \oint_S \mathbf{q} \cdot \mathbf{\hat{n}}$,

where $\mathbf{\hat{n}}$ is the unit normal vector orthogonal to the surface $S$, and $\mathbf{q}$ is the (vector) heat flux per unit surface.

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 $k$.

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

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

where we can use ordinary derivative assuming all the functions depend only on one variable, namely $x$ coordinate, and the total heat flux is

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

Fourier's law in 1D steady problem without heat source. Internal energy equation for 1D steady problem without heat source can be written as

$\dfrac{d q_x}{d x} = 0$

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

$T(x) = T_0 + (T_1 - T_0 ) \dfrac{x-x_0}{x_1 - x_0}$
$q_x = - k \dfrac{T_1-T_0}{x_1-x_0}$

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 $k$.

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

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

where we can use ordinary derivative assuming all the functions depend only on one variable, namely $x$ coordinate, and the total heat flux is

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

Fourier's law in 1D steady problem without heat source. Internal energy equation for 1D steady problem without heat source can be written as

$\dfrac{d q_x}{d x} = 0$

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

$T(x) = T_0 + (T_1 - T_0 ) \dfrac{x-x_0}{x_1 - x_0}$
$q_x = - k \dfrac{T_1-T_0}{x_1-x_0}$