Thermal resistance of brick wall with plaster

Thermal resistance is the opposition a layer offers to heat flow. By analogy with Ohm's law (I = V/R), heat flow Q (like current) is driven by a temperature difference ΔT (like voltage) against a thermal resistance R_th:

Q = ΔT/R_th, so R_th = ΔT/Q (unit K/W or °C/W)

Plane brick wall with plaster on both sides and convection

Let the wall have area A, brick thickness L_b (conductivity k_b) and a plaster layer of thickness L_p (conductivity k_p) on each face. Hot fluid at T_∞1 (film coefficient h₁) is on the inside, cold fluid at T_∞2 (film coefficient h₂) on the outside. Steady, one-dimensional heat flow; the same Q passes through all layers.

 T_inf1 --[ 1/h1A ]-- plaster --[ brick ]-- plaster --[ 1/h2A ]-- T_inf2
                      Lp/kpA     Lb/kbA     Lp/kpA

Resistances in series

• Inside convection: R₁ = 1/(h₁A)
• Inside plaster: R₂ = L_p/(k_pA)
• Brick: R₃ = L_b/(k_bA)
• Outside plaster: R₄ = L_p/(k_pA)
• Outside convection: R₅ = 1/(h₂A)

Derivation

Each element carries the same Q, so each temperature drop is Q × (its resistance):

T_∞1 − T_s1 = Q R₁, T_s1 − T_a = Q R₂, T_a − T_b = Q R₃, T_b − T_s2 = Q R₄, T_s2 − T_∞2 = Q R₅

Add all five equations:

T_∞1 − T_∞2 = Q (R₁ + R₂ + R₃ + R₄ + R₅)

R_total = 1/(h₁A) + L_p/(k_pA) + L_b/(k_bA) + L_p/(k_pA) + 1/(h₂A)

Q = (T_∞1 − T_∞2)/R_total

Written per unit area: R_total·A = 1/h₁ + 2L_p/k_p + L_b/k_b + 1/h₂, and the overall coefficient is U = 1/(R_total·A), so Q = U A (T_∞1 − T_∞2).