Calculating heat transfer as if it were electric current

## 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)] :

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:

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:

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

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:

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.

The walls of a room are made of:

• 44 m$^2$ of concrete wall ($e_c=15$ cm, $\lambda_c=2,3$ W/(m.K)) with an insulation layer ($e_{i}=10$ cm, $\lambda_{i}=0,04$ W/(m.K))
• 8 m$^2$ of double glazing ($U_v=3,3$ W/(m$^2$.K))
• The indoor heat transfer coefficient is $h_i = 0,11$ (m$^2$.K)/W, the outdoor one is $h_e = 0,07$ (m$^2$.K)/W

The room is ventilated with an air renewal rate of 9 m$^3$/h.

Calculate the heating power that should be prescribed to maintain an indoor temperature of 19°C, if the outdoor temperature is 2°C.

The total heat loss of the room is the sum of three parts: heat flux through the concrete+insulation wall, heat flux through the windows and air renewal.

1. Insulated concrete wall

The thermal resistance is the sum of each layer’s resistance, and the surface resistances:

$R_1 = h_i + \dfrac{e_c}{\lambda_c} + \dfrac{e_{i}}{\lambda_{i}} + h_e = 0,11 + \dfrac{0,15}{2,3} + \dfrac{0,10}{0,04} + 0,07 = 2,75$ (m$^2$.K)/W

The heat loss is the reciprocal of this total resistance, times the area of the wall:

$D_1 = \dfrac{S_1}{R_1} = \dfrac{44}{2,75} = 16,03$ W/K

2. Windows

The $U_v$ transmittance that describes the glazing probably doesn’t include the heat transfer coefficients $h_e$ and $h_i$. They should be included in the equation of the total glazing heat loss coefficient:

$D_2 = \dfrac{S_2}{h_i+\frac{1}{U_v}+h_e} = \dfrac{8}{0.11+\frac{1}{3,3}+0,07} = 16,56$ W/K

3. Air renewal

If the air renewal rate is given in the unit (m$^3$/h), the heat loss associated to it can be calculated easily:

$D_3 = 0,34 \, Q_v = 3,06$ W/K

Summary

The total heating power lost by the room is the sum of these three types of heat loss, times the indoor-outdoor temperature difference:

$\phi = \underbrace{(D_1+D_2+D_3)}_{W/K}.\underbrace{\Delta T}_{K} = (16,03 + 16,56 + 3,06 ). (19-2) = 606$ W