--- title: "Heat flow through composite plane wall, three layers" subject: "Fundamental of Thermodynamics & Heat Transfer" chapter: 7 key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:composite-plane-wall-three-layers" source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/composite-plane-wall-three-layers site: BCT Engineering --- # Heat flow through composite plane wall, three layers Fundamental of Thermodynamics & Heat Transfer · Chapter 7: Introduction to Heat Transfer - Asked in: 2073 Bhadra, 6 marks - Video: [Heat Conduction through Composite Wall II Heat and Mass Transfer II ](https://www.youtube.com/watch?v=AjVsGYBpBac) — Shailendra's Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/composite-plane-wall-three-layers - Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-7 ## Question Derive a heat flow equation through a composite plane wall consisting of three layers of different materials. ## Short answer For three layers in series, each resistance R_i = L_i/(k_i A). Total resistance is sum, so heat flow Q = (T1 − T4) / [L1/(k1 A) + L2/(k2 A) + L3/(k3 A)]. ## Answer The heat flow through a composite plane wall of three layers is derived below using Fourier's law for each layer. #### Setup Consider a plane wall of area A made of three layers in perfect contact: thicknesses L₁, L₂, L₃ and conductivities k₁, k₂, k₃. The outer surfaces are at T₁ (hot side) and T₄ (cold side); the interface temperatures are T₂ and T₃. Heat flow is steady and one-dimensional, so the same Q passes through every layer. ```text T1 | k1 | T2 | k2 | T3 | k3 | T4 -->| L1 | | L2 | | L3 |--> Q T1 --[R1]-- T2 --[R2]-- T3 --[R3]-- T4 ``` #### Derivation Fourier's law for each layer: Q = k₁A (T₁ − T₂)/L₁ , Q = k₂A (T₂ − T₃)/L₂ , Q = k₃A (T₃ − T₄)/L₃ Write each as a temperature drop: T₁ − T₂ = Q L₁/(k₁A) T₂ − T₃ = Q L₂/(k₂A) T₃ − T₄ = Q L₃/(k₃A) Add the three equations; the interface temperatures cancel: T₁ − T₄ = Q [L₁/(k₁A) + L₂/(k₂A) + L₃/(k₃A)] **Q = (T₁ − T₄)/(L₁/(k₁A) + L₂/(k₂A) + L₃/(k₃A)) = (T₁ − T₄)/(R₁ + R₂ + R₃)** where R = L/(kA) for each layer. The resistances are in series, so they add, as in an electric circuit. The interface temperatures follow from T₂ = T₁ − Q R₁ and T₃ = T₂ − Q R₂. For equal layer areas the wall can also be written with an overall coefficient: U = 1/(L₁/k₁ + L₂/k₂ + L₃/k₃) and Q = U A (T₁ − T₄). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the same Q used for every layer? A. Steady one-dimensional flow with no storage B. Layers have the same thickness C. Layers have the same k D. Temperatures are equal Answer: A. In steady 1-D flow, heat entering a layer equals heat leaving it. (Key point: Assumption) 2. What happens when the three temperature-drop equations are added? A. Q cancels B. T1 and T4 cancel C. Interface temperatures T2 and T3 cancel D. All temperatures cancel Answer: C. The sum gives T1 - T4 = Q (R1 + R2 + R3). (Key point: Cancellation) 3. The heat flow is A. Q = (T1 - T4)(R1 + R2 + R3) B. Q = A (T1 - T4)/(L1 + L2 + L3) C. Q = (T1 - T4)/(L1/k1A + L2/k2A + L3/k3A) D. Q = (T1 - T4)/(k1 + k2 + k3) Answer: C. Series resistances add: Q = ΔT/(R1 + R2 + R3). (Key point: Final equation) 4. How is T2 found afterwards? A. T2 = T4 + Q R1 B. T2 = T1 - Q R3 C. T2 = T1 - Q R1 D. T2 = (T1 + T4)/2 Answer: C. The drop across layer 1 is Q R1. 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