--- title: "Conduction through a composite cylinder" subject: "Fundamental of Thermodynamics & Heat Transfer" chapter: 7 key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:conduction-composite-cylinder" source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/conduction-composite-cylinder site: BCT Engineering --- # Conduction through a composite cylinder Fundamental of Thermodynamics & Heat Transfer · Chapter 7: Introduction to Heat Transfer - Asked in: 2072 Asoj, 6 marks - Video: [Heat Conduction through Composite Cylinder | Heat Transfer](https://www.youtube.com/watch?v=TwV8hlTM1yA) — Engineering Xplained - 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/conduction-composite-cylinder - Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-7 ## Question Derive an expression for conduction heat transfer through a composite cylinder. ## Short answer For a composite cylinder of n layers, steady radial conduction gives Q = 2πL (T₁ − T_{n+1}) / Σ [ln(r_{j+1}/r_j)/k_j]. Each layer’s thermal resistance is R_j = ln(r_{j+1}/r_j)/(2πk_jL); they sum in series, so total resistance is ΣR_j and Q = ΔT / ΣR_j. ## Answer A composite cylinder is a cylinder made of several concentric layers of different materials in perfect contact (for example a pipe with insulation layers). Heat flows steadily in the radial direction. #### Single cylindrical layer For a cylindrical layer of inner radius r₁, outer radius r₂, length L and conductivity k, with temperatures T₁ at r₁ and T₂ at r₂ (steady, radial, no heat generation), the area changes with radius, A = 2πrL. Fourier's law: Q = −k (2πrL) dT/dr Separate variables and integrate: Q ∫_{r₁}^{r₂} dr/r = −2πkL ∫_{T₁}^{T₂} dT Q ln(r₂/r₁) = 2πkL (T₁ − T₂) **Q = 2πkL (T₁ − T₂)/ln(r₂/r₁)**, so **R_th = ln(r₂/r₁)/(2πkL)** #### Composite cylinder of n layers Let the layers have radii r₁, r₂, ..., r_{n+1} and conductivities k₁, ..., k_n. The inner surface (r₁) is at T₁ and the outer surface (r_{n+1}) at T_{n+1}. The same Q passes through every layer. For layer j: T_j − T_{j+1} = Q ln(r_{j+1}/r_j)/(2π k_j L) Add all layers; the interface temperatures cancel: T₁ − T_{n+1} = (Q/2πL) Σ ln(r_{j+1}/r_j)/k_j **Q = 2πL (T₁ − T_{n+1})/Σ [ln(r_{j+1}/r_j)/k_j]** For three layers: Q = 2πL (T₁ − T₄)/[ln(r₂/r₁)/k₁ + ln(r₃/r₂)/k₂ + ln(r₄/r₃)/k₃]. The layers act as thermal resistances R_j = ln(r_{j+1}/r_j)/(2π k_j L) in series, so Q = ΔT_total/ΣR_j. If convection at the inner and outer surfaces is present, add 1/(h_i 2π r₁ L) and 1/(h_o 2π r_{n+1} L) to the sum. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Integrating Q = -k 2π r L dT/dr gives A. Q ln(r2/r1) = k L B. Q (r2 - r1) = 2π k L (T1 - T2) C. Q ln(r2/r1) = 2π k L (T1 - T2) D. Q = 2π k L (T1 - T2) Answer: C. Separating variables, ∫ dr/r = ln(r2/r1). (Key point: Integration step) 2. R_th for a cylindrical layer is A. ln(r2/r1)/(2π k L) B. 2π k L/ln(r2/r1) C. 1/(h 2π r L) D. (r2 - r1)/(k A) Answer: A. From Q = ΔT/R. (Key point: Single cylinder resistance) 3. Total heat flow through n layers is A. 2πL ΔT/Σ[ln(r_{j+1}/r_j)/k_j] B. ΔT/Σ k_j C. 2πL ΔT/ln(r_{n+1}/r_1) D. 2πL ΔT Σ[ln(r_{j+1}/r_j)/k_j] Answer: A. The same Q passes through each layer, so drops add. (Key point: n layers) 4. How is convection at the surfaces included? A. Multiply by h B. Ignore it C. Add 1/(h_i 2π r1 L) and 1/(h_o 2π r_{n+1} L) in series D. Add h directly Answer: C. Film resistances 1/(hA) with A = 2π r L are added. 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