Conduction through composite walls, cylinders and spheres

Thermal resistance networks for composite plane walls, pipes and spheres, including convective films, interface temperatures and the overall coefficient.

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Why it matters

Real walls are never one material: a furnace has firebrick, insulating brick and red brick; a steam line has a steel wall, lagging and a cladding; a cryogenic tank is a sphere with several shells. Thermal resistances let you combine all these layers, plus the convective films on each side, into one simple series calculation that gives the heat loss and every interface temperature — which is how you check that a brick or an insulant stays below its maximum service temperature.

Key ideas

Thermal resistance. For steady conduction with no generation, the heat rate q is the same through every layer, so each layer obeys q = ΔT/R. Resistances in series add; resistances in parallel combine as reciprocals. The same idea covers a convective film, R = 1/(h·A), and (approximately) radiation, R = 1/(h_r·A).

Plane composite wall. Layers in series: R_total = Σ L_i/(k_i·A) + 1/(h_i·A) + 1/(h_o·A). Once q is known, each interface temperature is found by stepping through the resistances: T_next = T_previous − q·R_layer. The temperature drop across a layer is proportional to its resistance, so the insulating layer carries most of the drop.

Parallel paths. When materials sit side by side (for example a stud and the insulation between studs), the paths are in parallel. The usual 1-D parallel-path model assumes surfaces normal to the heat flow are isothermal; it is an approximation because heat also spreads sideways.

Contact resistance. Real interfaces touch only at surface asperities, adding an interfacial resistance R″_tc (m²·K/W) that depends on pressure, roughness and any filler; take it from data when it matters (it is important for metals, usually negligible next to insulation).

Cylinder (radial conduction). The area 2πrL grows with radius, so the heat flux falls as 1/r and the steady temperature profile is logarithmic, not linear. The heat rate q (W) is constant with r; the flux q″ is not. Resistance R = ln(r₂/r₁)/(2πkL). For thin walls (r₂/r₁ close to 1) this approaches the plane-wall value with the log-mean area.

Sphere. Area 4πr² grows faster; the profile varies as 1/r and R = (r₂ − r₁)/(4πk·r₁·r₂).

Log-mean and geometric-mean areas. A curved wall can be written like a plane wall q = k·A_m·ΔT/(r₂ − r₁) with A_m the logarithmic mean area for a cylinder and the geometric mean area for a sphere. This form is common in chemical engineering texts.

Overall coefficient. Writing q = U·A·ΔT_overall, U·A = 1/R_total. For a cylinder U must be quoted on a stated area (inner or outer), since U_i·A_i = U_o·A_o. This leads directly to exchanger design.

Formulas

q = ΔT_overall / R_total q heat rate (W), ΔT_overall between the two fluids or outer surfaces (K), R_total sum of series resistances (K/W). Steady, no generation.

R_wall = L/(k·A) Plane layer; L thickness (m), k (W/m·K), A face area (m²).

R_conv = 1/(h·A) Convective film; h heat transfer coefficient (W/m²·K).

R_cyl = ln(r₂/r₁)/(2π·k·L) Cylindrical shell; r₁, r₂ inner and outer radii (m), L length (m).

T(r) = T₁ − (T₁ − T₂)·ln(r/r₁)/ln(r₂/r₁) Temperature in a cylindrical shell.

R_sph = (r₂ − r₁)/(4π·k·r₁·r₂) Spherical shell.

A_lm = (A₂ − A₁)/ln(A₂/A₁), q = k·A_lm·(T₁ − T₂)/(r₂ − r₁) Cylinder with log-mean area (m²).

A_gm = √(A₁·A₂) = 4π·r₁·r₂ Sphere with geometric-mean area.

1/R_parallel = 1/R_a + 1/R_b Parallel paths.

U·A = 1/R_total Overall heat transfer coefficient U (W/m²·K) on area A.

Worked examples

Example 1 (standard). A furnace wall has 0.20 m firebrick (k = 1.2 W/m·K), 0.10 m insulating brick (k = 0.15 W/m·K) and 0.10 m building brick (k = 0.7 W/m·K). Furnace gases are at 1000 °C with h_i = 40 W/m²·K; ambient air is at 30 °C with h_o = 10 W/m²·K. Find the heat loss per m² and the interface temperatures.

Given: A = 1 m², temperatures and properties as above.

  1. Resistances (K/W for 1 m²): 1/h_i = 0.0250; firebrick 0.20/1.2 = 0.1667; insulating brick 0.10/0.15 = 0.6667; building brick 0.10/0.7 = 0.1429; 1/h_o = 0.1000.
  2. R_total = 1.1012 K/W.
  3. q = ΔT/R_total = (1000 − 30)/1.1012 = 880.9 W.
  4. Step through: inner surface = 1000 − 880.9 × 0.025 = 978.0 °C; firebrick/insulation = 978.0 − 880.9 × 0.1667 = 831.2 °C; insulation/building brick = 831.2 − 880.9 × 0.6667 = 243.9 °C; outer surface = 243.9 − 880.9 × 0.1429 = 118.1 °C (check: 30 + 880.9 × 0.1 = 118.1 °C).

Answer: q ≈ 881 W/m²; interface temperatures 978, 831, 244 and 118 °C. The insulating brick at 831 °C on its hot face must be rated above that.

Example 2 (GATE level). Steam at 200 °C flows in a steel pipe (r₁ = 50 mm, r₂ = 55 mm, k = 45 W/m·K) covered with 50 mm of insulation (k = 0.06 W/m·K). Inside h = 500 W/m²·K; outside h = 10 W/m²·K to air at 25 °C. Find the heat loss per metre and the outer surface temperature.

Given: r₁ = 0.050 m, r₂ = 0.055 m, r₃ = 0.105 m, L = 1 m.

  1. Inside film: 1/(h_i·2π·r₁·L) = 1/(500 × 2π × 0.05) = 0.00637 K/W.
  2. Steel: ln(r₂/r₁)/(2π·k·L) = ln(1.1)/(2π × 45) = 0.000337 K/W.
  3. Insulation: ln(0.105/0.055)/(2π × 0.06) = 0.6466/0.3770 = 1.7152 K/W.
  4. Outside film: 1/(10 × 2π × 0.105) = 0.1516 K/W.
  5. R_total = 1.8735 K/W; q = (200 − 25)/1.8735 = 93.4 W per metre.
  6. Outer surface: T₃ = 25 + q·R_o = 25 + 93.4 × 0.1516 = 39.2 °C.

Answer: q ≈ 93.4 W/m; outer surface ≈ 39.2 °C. The steel wall and the steam film are negligible; the insulation is over 90 % of the resistance.

Common mistakes

  • Adding conductivities instead of resistances, or adding resistances of layers that are actually in parallel.
  • Using a plane-wall formula for a thick pipe lagging: ln(r₂/r₁) matters when r₂/r₁ is not close to 1.
  • Using diameters inside ln — harmless for the ratio — but then using diameter instead of radius in 2πrL or 4πr².
  • Forgetting the convective films, or using the inner area for the outer film.
  • Quoting U without saying which area it is based on.
  • Claiming the heat flux is constant through a pipe wall; only the heat rate is.

For GATE CH

Typical questions: heat loss and an interface temperature through a two- or three-layer wall, the thickness of insulation needed to keep an outer surface below a limit, pipe lagging with inside and outside films, spherical shells, and the ratio of resistances to identify the controlling layer. Practise drawing the resistance network first, checking that areas match each resistance, and verifying your answer by recomputing the outer temperature from the other side.

Quick check

  1. Two layers, R₁ = 0.2 K/W and R₂ = 0.6 K/W in series, ΔT = 80 K. What is q?
  2. In steady radial conduction through a pipe wall, which is constant with r: the heat rate or the heat flux?
  3. Write the resistance of a spherical shell.
  4. In Example 1, which layer has the largest temperature drop and why?
  5. If U_o = 50 W/m²·K is based on an outer area of 2 m², what is U_i for an inner area of 1.6 m²?

Answers: 1. 100 W. 2. The heat rate. 3. (r₂ − r₁)/(4πk·r₁·r₂). 4. The insulating brick, because it has the largest resistance. 5. 62.5 W/m²·K.

See it move

All Chemical animations
Conduction Through Composite Walls

Adjust the thickness and thermal conductivity of each layer to see how they affect the total thermal resistance and heat transfer rate.

Equations used
  • Q = (T1 - T2) / R_total — Q heat transfer rate, T1 and T2 temperatures on either side, R_total total thermal resistance
  • R = L / (k * A) — R thermal resistance, L thickness, k thermal conductivity, A cross-sectional area

Interview questions

All Heat Transfer interview questions

Try answering each one aloud before you open it.

  1. 1.What is thermal conduction and how does it occur in composite walls?Concept

    Thermal conduction is the transfer of heat through a material without any movement of the material itself. In composite walls, which are made up of different layers of materials, heat conduction occurs through each layer. The rate of heat transfer depends on the thermal conductivity of each material, the thickness of each layer, and the temperature difference across the wall.

  2. 2.Explain the concept of thermal resistance in the context of heat conduction through composite walls.Concept

    Thermal resistance R is the temperature difference needed to drive unit heat rate, R = ΔT/q (K/W); for a plane layer R = L/(kA) and for a convective film R = 1/(hA). In steady conduction without generation the same q passes through every layer in series, so the resistances add and q = ΔT_overall/ΣR, exactly like current through series resistors. Layers side by side are parallel paths and combine as reciprocals. The temperature drop across each layer is proportional to its resistance, which is how interface temperatures are found.

  3. 3.Why is it important to consider the thermal conductivity of materials when designing composite walls?Application

    The thermal conductivity of a material determines how easily heat can pass through it. In designing composite walls, selecting materials with appropriate thermal conductivities is crucial to achieve the desired insulation or heat transfer properties. High thermal conductivity materials are used where heat transfer is needed, while low thermal conductivity materials are used for insulation.

  4. 4.What happens to the overall thermal resistance if one layer in a composite wall is replaced with a material of higher thermal conductivity?Application

    If one layer in a composite wall is replaced with a material of higher thermal conductivity, the thermal resistance of that layer decreases. As a result, the overall thermal resistance of the composite wall decreases, allowing more heat to pass through the wall.

  5. 5.Explain how heat conduction through a cylindrical wall differs from that through a flat wall.Concept

    In a flat wall the area normal to the heat flow is constant, so with constant k the temperature profile is linear and R = L/(kA). In a cylindrical wall heat flows radially through an area 2πrL that grows with radius; the heat rate is the same at every radius but the heat flux falls as 1/r, so the temperature profile is logarithmic and R = ln(r₂/r₁)/(2πkL). For a thin pipe wall (r₂/r₁ near 1) the two results converge if the log-mean area is used.

  6. 6.Why are spherical shapes often used in applications requiring minimal heat loss?Application

    Spherical shapes are used in applications requiring minimal heat loss because they have the smallest surface area for a given volume. This minimizes the area through which heat can be lost, making spheres more efficient for insulation purposes compared to other shapes.

  7. 7.What is the effect of increasing the thickness of insulation on the outer surface of a cylindrical pipe?Application

    Adding insulation increases the conduction resistance ln(r₂/r₁)/(2πkL) but also increases the outer surface area, which lowers the outside convective resistance 1/(h·2πr₂L). For large pipes, where the outer radius already exceeds the critical radius k/h, more insulation always reduces heat loss, with diminishing returns because of the logarithm. For thin wires or small tubes with radius below k/h, a thin layer of insulation actually increases heat loss until the radius exceeds k/h. The economic thickness balances insulation cost against the value of heat saved.

  8. 8.Calculate the total thermal resistance of a composite wall consisting of three layers with thermal resistances of 0.2, 0.3, and 0.5 K/W.Numerical

    The total thermal resistance of a composite wall is the sum of the thermal resistances of each layer. Therefore, the total thermal resistance is 0.2 + 0.3 + 0.5 = 1.0 K/W.

  9. 9.A spherical tank has an inner radius of 1 m and an outer radius of 1.2 m. If the thermal conductivity of the material is 0.5 W/m·K, calculate the thermal resistance of the tank wall.Numerical

    For a spherical shell R = (1/r₁ − 1/r₂)/(4πk) = (r₂ − r₁)/(4πk·r₁·r₂). Substituting: R = (1/1 − 1/1.2)/(4π × 0.5) = 0.1667/6.283 = 0.0265 K/W.

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