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The temperature in the soil evolves based on the surface temperature and the average temperature at depth. The thermal conductivity and specific heat capacity of the soil determine the depth at which the average temperature is reached.

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Concept

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If the solution for the ground temperature ($T$) is graphed over the depth ($z$) and the time ($t$) using the average environmental temperature ($T_m$), the annual temperature variation in the soil ($\Delta T_s$), the time phase shift ($t_0$), the average depth of variation ($d_m$), and the annual angular frequency ($\omega_m$), the result is:

This yields a curve that, at the surface ($z=0$), exhibits the maximum of summer and the minimum of winter temperatures. The temperature then converges to the average temperature with constant depth. Additionally, there is an inertia effect in the system:

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Parameters

Variables

$q$

q

Heat flow rate

W/m^2

$dQ$

dQ

Heat transported

J

$S$

S

Section

m^2

$dt$

dt

Time variation

s

Calculations

• Alternative method
• Traditional method
• Strategy
• 2D
• 3D

Equation

Number

First, select the equation: to , then, select the variable: to

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Calculations

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Equation

Solved

Translated

Calculations

Symbol

Equation

Solved

Translated

Variable Given Calculate Target : Equation To be used

Equations

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Equation

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The heat flow rate ($q$) is defined in terms of the heat transported ($dQ$), the time variation ($dt$), and the section ($S$) as follows:

$ q \equiv \displaystyle\frac{1}{ S }\displaystyle\frac{ dQ }{ dt }$

Conditions

Variables

$q$

Heat flow rate

$W/m^2$

10178

$dQ$

Heat transported

$J$

10159

$S$

Section

$m^2$

5205

$dt$

Time variation

$s$

10160

Relation

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Equation

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For a conductor with values of the conductor length ($L$) and the section ($S$), the flow of the heat transported ($dQ$) is described under the time variation ($dt$) and the thermal conductivity ($\lambda$) as follows:

In the infinitesimal case, where the conductor length ($L$) reduces to ($$) and the temperature difference in the conductor ($\Delta T_0$) becomes the temperature variation ($dT$), the equation for the heat flow rate ($q$) simplifies to:

$ q = - \lambda \displaystyle\frac{dT}{dz}$

Conditions

Variables

Relation

Development

Since the heat transported ($dQ$) is a function of the conductor length ($L$), the section ($S$), the time variation ($dt$), the temperature difference in the conductor ($\Delta T_0$), and the thermal conductivity ($\lambda$) according to the following equation:

$ q = \displaystyle\frac{ \lambda }{ L } \Delta T_0 $

With the equation for the heat flow rate ($q$) defined as:

$ q \equiv \displaystyle\frac{1}{ S }\displaystyle\frac{ dQ }{ dt }$

In the infinitesimal case, where the conductor length ($L$) reduces to ($$) and the temperature difference in the conductor ($\Delta T_0$) becomes the temperature variation ($dT$), the equation simplifies to:

$ q = - \lambda \displaystyle\frac{dT}{dz}$

[1] "Théorie analytique de la chaleur" (The Analytical Theory of Heat), Joseph Fourier, Cambridge University Press (2009) (original 1822)

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Equation

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The definition of the heat flow rate ($q$) is established using the thermal conductivity ($\lambda$) and the temperature variation ($dT$) as a function of the distance traveled ($dz$) through the following equation:

By studying heat flow, we obtain the equation for the absolute temperature ($T$) as a function of the position along an axis ($z$), the time ($t$), and the thermal conductivity ($\lambda$), which becomes the specific heat ($c$). The equation for the density ($\rho$) simplifies to:

$ \displaystyle\frac{\partial T}{\partial t}=\displaystyle\frac{\lambda}{\rho c}\displaystyle\frac{\partial^2 T}{\partial z^2}$

Conditions

Variables

Relation

Development

The quantity of the heat transported ($dQ$) through ($$) can be calculated using the heat flow rate ($q$) and the time variation ($dt$) with the section ($S$) through the following equation:

$dQ = -\displaystyle\frac{\partial}{\partial z}(q S dt) dz$

Since the heat flow rate ($q$) with the temperature variation ($dT$) and the thermal conductivity ($\lambda$) is defined as:

$ q = - \lambda \displaystyle\frac{dT}{dz}$

Therefore,

$dQ = \displaystyle\frac{\partial}{\partial z}(\lambda \displaystyle\frac{\partial T}{\partial z}) S dz dt$

On the other hand, we can relate the heat supplied to liquid or solid ($\Delta Q_s$) with the mass ($M$), the specific heat ($c$), and the temperature variation in a liquid or solid ($\Delta T_s$) through the equation:

$ \Delta Q = M c \Delta T$

In this case, with the volume Variation ($dV$), the equation becomes:

$dQ=\rho dV c dT = \rho c S dz dT$

And finally, we have:

$ \displaystyle\frac{\partial T}{\partial t}=\displaystyle\frac{\lambda}{\rho c}\displaystyle\frac{\partial^2 T}{\partial z^2}$

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Equation

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If we consider that the surface temperature experiences rapid daily fluctuations and a slow annual variation, and that the inertia of the system prevents the daily fluctuations from affecting the soil, we can estimate the ground temperature ($T$) as a function of the depth ($z$) and the time ($t$) throughout the year as the solution to the following equation:

where the thermal conductivity ($\lambda$), the specific heat ($c$), and the density ($\rho$). Therefore, the solution is obtained with the average environmental temperature ($T_m$), the annual temperature variation in the soil ($\Delta T_s$), the average depth of variation ($d_m$), the time phase shift ($t_0$), and the annual angular frequency ($\omega_m$) as follows:

$ T = T_m - \displaystyle\frac{1}{2} \Delta T_s e^{- z / d_m }\cos( \omega_m ( t - t_0 ) - z / d_m )$

Conditions

Variables

Relation

The value of the time phase shift ($t_0$) is usually used to adjust the solution to the corresponding hemisphere. In this regard, in the northern hemisphere, this phase difference is almost zero, while in the southern hemisphere, it is approximately half a year.

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Equation

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The solution for the ground temperature ($T$) in the depth ($z$) and the time ($t$) is obtained using the average environmental temperature ($T_m$), the annual temperature variation in the soil ($\Delta T_s$), the time phase shift ($t_0$), the average depth of variation ($d_m$), and the annual angular frequency ($\omega_m$), resulting in:

This solution solves the conduction equation when the average depth of variation ($d_m$) is equal to the thermal conductivity ($\lambda$), the specific heat ($c$), and the density ($\rho$) through:

$ d_m = \sqrt{\displaystyle\frac{2\lambda}{c\rho\omega_m}}$

Conditions

Variables

Relation

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Equation

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To solve for the ground temperature ($T$) in the depth ($z$) and the time ($t$), we use the average environmental temperature ($T_m$), the annual temperature variation in the soil ($\Delta T_s$), the time phase shift ($t_0$), the average depth of variation ($d_m$), and the annual angular frequency ($\omega_m$), resulting in:

This allows us to calculate the heat flux at the surface in the case where the time phase shift ($t_0$) is assumed to be zero and with the thermal conductivity ($\lambda$) and the average depth of variation ($d_m$) using:

$ q = \displaystyle\frac{ \lambda \Delta T_s }{2 d_m }(\cos \omega_m ( t - t_0 ) - \sin \omega_m ( t - t_0 ))$

Conditions

Variables

Relation

Development

As the heat flow rate ($q$) along with the thermal conductivity ($\lambda$), the ground temperature ($T$), and the depth ($z$) results in

$ q = - \lambda \displaystyle\frac{dT}{dz}$

For the solution of the ground temperature ($T$) with the depth ($z$) and the time ($t$), we use the average environmental temperature ($T_m$), the annual temperature variation in the soil ($\Delta T_s$), the time phase shift ($t_0$), the average depth of variation ($d_m$), and the annual angular frequency ($\omega_m$):

$ T = T_m - \displaystyle\frac{1}{2} \Delta T_s e^{- z / d_m }\cos( \omega_m ( t - t_0 ) - z / d_m )$

This is obtained at the surface ($z=0$) and without phase shift ($t_0=0$):

$ q = \displaystyle\frac{ \lambda \Delta T_s }{2 d_m }(\cos \omega_m ( t - t_0 ) - \sin \omega_m ( t - t_0 ))$

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Equation

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To solve for the ground temperature ($T$) in the depth ($z$) and the time ($t$), we use the average environmental temperature ($T_m$), the annual temperature variation in the soil ($\Delta T_s$), the time phase shift ($t_0$), the average depth of variation ($d_m$), and the annual angular frequency ($\omega_m$), resulting in:

This allows us to calculate the heat flux at the surface in the case where the time phase shift ($t_0$) is assumed to be zero and with the thermal conductivity ($\lambda$) and the average depth of variation ($d_m$) using:

$ \Delta T_s = \displaystyle\frac{ \alpha }{ \alpha + \lambda / d_m } \Delta T_m $

Conditions

Variables

Relation

Development

The heat flow rate ($q$) as a function of the thermal conductivity ($\lambda$), the annual temperature variation in the soil ($\Delta T_s$), the average depth of variation ($d_m$), the annual angular frequency ($\omega_m$), the time ($t$), and the time phase shift ($t_0$) is represented by:

$ q = \displaystyle\frac{ \lambda \Delta T_s }{2 d_m }(\cos \omega_m ( t - t_0 ) - \sin \omega_m ( t - t_0 ))$

This represents the total flow passing through the area above the ground and the ground itself. In the first case, the flow is equal to the transmission coefficient ($\alpha$) due to the temperature difference between the environment and the ground surface. For the situation where the time ($t$) is equal to the time phase shift ($t_0$), the flow in the area above the ground can be described as:

$\alpha \left(\displaystyle\frac{\Delta T_m}{2}-\displaystyle\frac{\Delta T_s}{2}\right) = \displaystyle\frac{\lambda \Delta T_s}{2 d_m}$

Solving for the annual temperature variation in the soil ($\Delta T_s$), we obtain:

$ \Delta T_s = \displaystyle\frac{ \alpha }{ \alpha + \lambda / d_m } \Delta T_m $

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