1.2. Heat Transfer in Composite Walls

The image below shows an example of a composite wall used for heat transfer simulation:

Composite Wall

Example of heat transfer simulation in a composite wall:

from HeatTransfer import CompositeWall

# Create a composite wall with external and internal convection coefficients
wall = CompositeWall.Object(he=23, hi=8, Ti=20, Te=-10, A=10)

# Add layers to the wall using material names
wall.add_layer(thickness=0.20, material='Parpaings creux')  # Hollow concrete blocks
wall.add_layer(thickness=0.05, material='Polystyrène')  # Polystyrene
wall.add_layer(thickness=0.02, material='Plâtre')  # Plaster

# Calculate heat transfer and temperatures at each layer interface
wall.calculate()
wall.df
print(f"df = {wall.df}")
print(f"R_total = {wall.R_total} m².°C/W")

Expected result:

Thickness (m)

Material

Conductivity (W/m.°C)

Resistance (m².°C/W)

Entry Temperature (°C)

Exit Temperature (°C)

Q (W)

A (m²)

NaN

Outdoor air

NaN

0.043478

-10.000000

-9.353644

148.661889

10

0.20

Hollow concrete blocks

1.40

0.142857

-9.353644

-7.229903

148.661889

10

0.05

Polystyrene

0.03

1.666667

-7.229903

17.547079

148.661889

10

0.02

Plaster

0.50

0.040000

17.547079

18.141726

148.661889

10

NaN

Indoor air

NaN

0.125000

18.141726

20.000000

148.661889

10

List of Available Materials

Anglais

Français

Conductivité thermique (W/m.°C)

Glass wool

Laine de verre

0.034

Expanded cork agglomerated with pitch

Liège expansé aggloméré au brai

0.048

Pure expanded cork

Liège expansé pur

0.043

Hollow concrete blocks

Parpaings creux

1.4

Hard limestone (marble)

Pierre calcaire dure (marbre)

2.9

Soft limestone

Pierre calcaire tendre

0.95

Granite

Pierre granit

3.5

Expanded polystyrene

Polystyrène expansé

0.047

Polystyrene

Polystyrène

0.03

Extruded polystyrene

Polystyrène extrudé

0.035

Polyurethane foam

Mousse de polyuréthane

0.03

Plaster

Plâtre

0.5

Glass

Verre

1.0

Air

Air

None

Explanation of the Equations Used

The composite wall heat transfer model uses the following equations to calculate total thermal resistance, heat flux, and temperatures at layer interfaces:

  1. Convective thermal resistance: - External resistance:

    \[R_e = \frac{1}{h_e}\]
    • Internal resistance: .. math:

      R_i = \frac{1}{h_i}
      
  2. Thermal resistance of layers: - For each layer, thermal resistance is calculated as follows:

    \[R_{\text{layer}} = \frac{\text{thickness}}{\text{conductivity}}\]
  3. Total thermal resistance: - The total thermal resistance of the composite wall is the sum of convective resistances and layer resistances:

    \[R_{\text{total}} = R_e + R_i + \sum R_{\text{layers}}\]
  4. Heat transfer coefficient: - The heat transfer coefficient is the inverse of total thermal resistance:

    \[U = \frac{1}{R_{\text{total}}}\]
  5. Heat flux: - Heat flux through the composite wall is calculated using Fourier’s law:

    \[Q = U \cdot A \cdot (T_i - T_e)\]

    where ( A ) is the wall surface area, ( T_i ) is the indoor temperature, and ( T_e ) is the outdoor temperature.

  6. Temperatures at layer interfaces: - The external wall temperature after convective resistance is calculated as follows:

    \[T_{\text{external wall}} = T_e + \frac{Q \cdot R_e}{A}\]
    • Les températures aux interfaces des couches sont ensuite calculées en utilisant le flux thermique et les résistances thermiques : .. math:

      T_{\text{interface}} = T_{\text{précédente}} + \frac{Q \cdot R_{\text{couche}}}{A}
      

Ces équations permettent de déterminer la distribution de température à travers le mur composite et le flux thermique total traversant le mur.