Heat loss through a wall
Heat Loss Through a Wall
Calculates steady-state conductive heat loss through a flat composite wall using Fourier's law.
Inputs
A_wall:=12m^2=129.17 ft^2
Wall surface area
Indoor air temperature
T_inside:=21degC=69.8 °F
Outdoor air temperature
T_outside:=-5degC=23 °F
Wall Layer Properties
- Layer 1: Brick outer leaf
- Layer 2: Mineral wool insulation
- Layer 3: Plasterboard inner leaf
k_brick:=(0.77W)/(m·K)=0.77 kg·m/K/s^3
Thermal conductivity — brick
Thickness — brick layer
d_brick:=0.102m=4.02 in
k_insul:=(0.04W)/(m·K)=0.04 kg·m/K/s^3
Thermal conductivity — mineral wool
Thickness — insulation layer
d_insul:=0.100m=3.94 in
k_plaster:=(0.25W)/(m·K)=0.25 kg·m/K/s^3
Thermal conductivity — plasterboard
Thickness — plasterboard layer
d_plaster:=0.013m=0.51 in
Surface Resistances
Internal and external surface film resistances per unit area (ISO 6946).
R_si:=0.13m^2·K/W=0.13 K·s^3/kg
Internal surface resistance
R_se:=0.04m^2·K/W=0.04 K·s^3/kg
External surface resistance
Calculations
R_brick:=(d_brick)/(k_brick)=0.13 K·s^3/kg
Thermal resistance — brick (m²·K/W)
R_insul:=(d_insul)/(k_insul)=2.5 K·s^3/kg
Thermal resistance — insulation (m²·K/W)
R_plaster:=(d_plaster)/(k_plaster)=0.05 K·s^3/kg
Thermal resistance — plasterboard (m²·K/W)
Total thermal resistance (m²·K/W)
R_total:=R_si+R_brick+R_insul+R_plaster+R_se=2.85 K·s^3/kg
U_value:=(1)/(R_total)=0.35 kg/K/s^3
U-value (W/m²·K)
Temperature difference across wall
delta_T:=T_inside-T_outside=26 K
Results
Heat loss rate through wall
Q_loss:=U_value·A_wall·delta_T=0.15 hp
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