Thermal-electrical analogy

Videos

Video 1

Introduction to the thermal-electrical analogy. PDF slides

Video 2

Some examples on how to use the thermal-electrical analogy to model a variety of transfer phenomena. PDF slides

Formulas

The heat flux per square meter of a wall is proportional to the temperature difference across the wall $\Delta T$, and to its transmittance $U$ [W/(m$^2$.K)] :

$$\varphi = U . \Delta T = \dfrac{1}{R} . \Delta T$$

The unit of this heat flux $\varphi$ is [W/m$^2$]. The total heat $\phi$ flowing across a wall of area $S$, in [W], is then:

$$\phi = S . \varphi = S . U . \Delta T$$

We can also define a global heat loss coefficient such as $D = S.U$ [W/K]

The thermal resistance $R_i$ of a component of thickness $e_i$ and heat conductivity $\lambda_i$ is:

$$R_i=\dfrac{e_i}{\lambda_i}$$

A wall made of several material layers behaves like a series circuit of resistances: their values add up.

$$R = \sum_i R_i = \sum_i \dfrac{e_i}{\lambda_i}$$

A building envelope made of several separated parts (opaque wall, windows, roof…) can be modelled as a parallel circuit of resistances: heat flows through each surface $S_i$ (m$^2$) simultaneously with more or less intensity. The heat loss coefficients $D$ of each surface add up:

$$D = S.U = \sum_i S_i.U_i$$

	Variable	Unit
$R$	Thermal resistance	(m$^2$.K)/W	can be summed in series
$U$	Thermal transmittance	W/(m$^2$.K)
$D$	Heat loss coefficient	W/K	can be summed in parallel

Cette méthode peut être appliquée pour représenter les transferts à l’échelle de tout un bâtiment, y compris en incluant des transferts par renouvellement d’air (voir vidéo 2).

Exercise

Calculate the heat loss coefficient of an insulated room.

• Questioning
• Solution