Conductive Heat Transfer in Cylindrical Coordinates
Reference ID: MET-8DBA | Process Engineering Reference Sheets Calculation Guide
Introduction & Context
Conductive heat transfer in cylindrical coordinates governs the radial temperature profile and energy loss across pipe insulation, reactor liners, and vessel jackets. In process engineering the calculation is indispensable for:
Determining insulation thickness required to meet maximum heat-loss targets.
Sizing steam-tracing systems to maintain fluid viscosity.
Estimating thermal stress gradients that drive fatigue in thick-walled pressure equipment.
Performing energy audits on long-distance hot-oil or bitumen pipelines.
The one-dimensional, steady-state solution presented here assumes constant thermal conductivity and perfect radial symmetry; it is the first-order check performed before invoking more complex 2-D or transient models.
Methodology & Formulas
Geometry conversion
Inner and outer radii are converted to consistent SI units:
\[
r_{1} = \frac{r_{1,\mathrm{mm}}}{1000},\qquad
r_{2} = \frac{r_{1,\mathrm{mm}}+t_{\mathrm{ins},\mathrm{mm}}}{1000}
\]
Heat-flow rate per unit length
Applying Fourier’s law in cylindrical coordinates and integrating over the radial coordinate gives:
\[
\frac{Q}{L}= \frac{2\pi\,k_{\mathrm{ins}}\left(T_{1}-T_{2}\right)}{\ln\left(\dfrac{r_{2}}{r_{1}}\right)}
\]
where
\(k_{\mathrm{ins}}\) is the thermal conductivity of the insulation,
\(T_{1}\) and \(T_{2}\) are the inner and outer surface temperatures, and
\(\ln\) denotes the natural logarithm.
Validity checks
The following criteria should be satisfied to keep systematic error below ≈1 %:
Check
Condition
Remedy if violated
Positive radii
\(r_{2}>r_{1}\)
Increase insulation thickness or verify input.
Positive conductivity
\(k_{\mathrm{ins}}>0\)
Use temperature-corrected \(k\) or change material.
Positive length
\(L>0\)
Reset length to physical segment of interest.
Long-cylinder limit
\(\dfrac{L}{r_{2}}\ge 10\)
Apply end-correction factor or model 2-D conduction.
For each layer i with constant ki, the radial heat equation reduces to:
1/r · d/dr (r · dT/dr) = 0
Solve twice: Ti(r) = C1,i ln r + C2,i
Match two boundary conditions per interface: T and k · dT/dr must be continuous
Close the set with known inner and outer surface temperatures (or convection)
Resulting heat flow Q̇ = 2πL(Tin–Tout)/Σ ln(rj+1/rj)/kj is used for insulation thickness optimisation
Compare characteristic length Lc = (ro–ri)/ln(ro/ri) with the radius ratio.
If ro/ri < 1.3, the error in Q̇ is < 5 % when the wall is treated as planar
For thicker pipes or refractory linings, always keep the logarithmic term to avoid under-predicting heat loss
Add a volumetric source q̇ (W m-3) to the cylindrical heat equation:
1/r · d/dr (r · dT/dr) + q̇/k = 0
General solution T(r) = –q̇ r2/(4k) + C1 ln r + C2
Apply symmetry (dT/dr = 0 at r = 0) for solid cylinders or specify inner/outer convection
Maximum temperature always occurs at the centre for solid rods; check against material limits
Heat flows radially across the layers, so use the radial kr.
For wound glass or mineral fibre, kr is typically 0.3–0.4 × kaxial
Manufacturers list both; picking the higher axial value by mistake can over-predict insulation performance by > 60 %
Worked Example – Heat Loss from a Steam-Tracer Line
A 52.5 mm OD stainless-steel pipe on a chemical plant carries 165 °C steam. To protect personnel, the line is covered with 50 mm of calcium-silicate insulation (k = 0.045 W m⁻¹ K⁻¹). The outer surface of the insulation is held at 35 °C by natural convection to ambient air. Determine the steady-state heat loss per metre of pipe length.
Knowns
Inner radius, r₁ = 52.5 mm = 0.053 m
Insulation thickness, t = 50 mm
Outer radius, r₂ = r₁ + t = 102.5 mm = 0.103 m
Insulation thermal conductivity, k = 0.045 W m⁻¹ K⁻¹
