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The slab formula for heat conduction works for walls and rods. Apply it to an insulated pipe or a spherical tank and the answer is wrong, because the area that heat flows through keeps changing. This post derives the two radial formulas, works numbers for a sphere and a pipe, and shows the mistake most students make with the temperature inside the shell.
Why does the slab formula fail for shells?
The NCERT Class 11 chapter on thermal properties of matter gives steady conduction through a bar as H = KA(T_C − T_D)/L, where the cross-section A is the same all along. In a shell, heat flows outward through spheres or cylinders whose area grows with r. In the steady state the same heat current H crosses every layer (otherwise some layer would warm up or cool down), so the temperature gradient must be steeper near the inner surface and gentler near the outer one.
The JEE (Main) 2026 syllabus lists heat transfer by conduction, and the JEE (Advanced) 2026 information brochure lists "heat conduction in one dimension". Radial flow counts, because the temperature depends on r alone. The derivation needs a simple integral.
How do you derive the radial formulas?
Write the slab law for a thin layer of thickness dr: H = −K A(r) dT/dr. Then separate and integrate from r₁ (temperature T₁) to r₂ (temperature T₂).
Spherical shell: A = 4πr². So H dr/(4πr²) = −K dT. Integrating: H(1/r₁ − 1/r₂)/(4π) = K(T₁ − T₂), which gives H = 4πK r₁r₂(T₁ − T₂)/(r₂ − r₁).
Cylindrical shell of length L: A = 2πrL. So H dr/(2πrL) = −K dT. Integrating: H ln(r₂/r₁)/(2πL) = K(T₁ − T₂), which gives H = 2πKL(T₁ − T₂)/ln(r₂/r₁).
Each has a thermal resistance, R = ΔT/H: (r₂ − r₁)/(4πK r₁r₂) for the sphere and ln(r₂/r₁)/(2πKL) for the cylinder. Layers in series add their resistances, just like slabs.
Worked example 1: a spherical shell
These are sample numbers. A spherical shell of insulation has inner radius 10 cm and outer radius 20 cm, with K = 0.04 W/(m K). The inner surface is at 100 °C and the outer at 50 °C.
H = 4π × 0.04 × 0.1 × 0.2 × 50 / 0.1 = 4π × 0.04 × 10 ≈ 5.03 W.
What is the temperature at r = 15 cm, halfway through? From the integrated form, T changes in proportion to 1/r. So (T₁ − T)/(T₁ − T₂) = (1/r₁ − 1/r)/(1/r₁ − 1/r₂) = (10 − 6.67)/(10 − 5) ≈ 0.667. T ≈ 100 − 0.667 × 50 ≈ 66.7 °C, not the 75 °C a straight line would give.
Worked example 2: an insulated pipe
Sample numbers again. A pipe carries hot water. Its insulation runs from r₁ = 2 cm to r₂ = 4 cm, K = 0.05 W/(m K), inner surface 90 °C, outer surface 30 °C. Per metre of pipe:
H = 2π × 0.05 × 1 × 60 / ln 2 ≈ 18.85/0.693 ≈ 27.2 W.
At r = 3 cm, T changes in proportion to ln r: T = 30 + 60 × ln(4/3)/ln 2 ≈ 30 + 60 × 0.415 ≈ 54.9 °C, again below the 60 °C midpoint value. In both shapes, most of the temperature drop happens near the inner surface, where the area is smallest.
Which mistakes cost marks?
Using the slab formula with the inner or outer area. One overestimates H, the other underestimates it.
Assuming a linear temperature profile. It is linear in 1/r for a sphere and in ln r for a cylinder.
Mixing up r₂ − r₁ and r₁r₂. Check units: K r₁r₂/(r₂ − r₁) has units of W/K once multiplied by 4π.
Using diameters instead of radii. For the cylinder this does not matter in the log ratio, but it does for the sphere.
How should you practise this?
Comparing two methods side by side helps more than studying each alone. Rittle-Johnson and Star (2007) randomly assigned 70 seventh-grade students to compare alternative solution methods or study them one at a time; the comparing group gained more in procedural knowledge and flexibility. Here, compare the slab, cylinder and sphere side by side: what changes is only A(r). Our post on heat conduction and thermal resistance covers slabs and composite walls.
If a derivation step does not make sense, Fermi's AI tutor notices where you are stuck and decides whether to explain, ask or show.
Derive both formulas from H = −K A(r) dT/dr without notes.
Redo both examples with the inner and outer temperatures swapped and check that only the sign of H changes.
Solve one two-layer problem (two insulation materials on one pipe) by adding resistances.
Before your next session, derive the sphere formula on one side of a page and the cylinder formula on the other, and circle the single line where they differ.
