1-D Steady-State Conduction: Cylinders, Spheres, and Fins
Pipes, pellets, and heat sinks all share one geometry: curved surfaces where heat flows radially. The mathematics changes from Chapter 2's flat-wall formula, but the physical reasoning is identical. A shell balance on a thin cylindrical or spherical shell gives the governing equation, and the resulting resistance formulas unlock dozens of practical problems in polymer processing and materials science.
The Heat Break Problem
Every FDM printer contains a small but carefully engineered cylinder called the heat break. It connects the heater block (250°C for ABS) to the cold zone where filament is solid and feeding smoothly (30°C). If too much heat travels up that cylinder, the filament softens too early and jams. If the cylinder is too thick or made of the wrong material, the printer fails.
Engineers solve this by choosing a material with very low thermal conductivity (titanium, k = 22 W/mK, compared to steel at 50) and making the wall as thin as manufacturing will allow. The goal is to maximise the thermal resistance R of a hollow cylinder. The formula they use is not the flat-wall resistance from Chapter 2. It involves a logarithm, because the heat flow area changes with radius.
The same geometry appears constantly in polymer science: a nylon pipe carrying hot water, a PVC-insulated electrical cable, a hollow fibre membrane in a dialysis unit, a polypropylene pellet cooling in a water bath. In each case, heat flows radially through a curved shell, and the question is always the same: how much, and how fast?
The heat break is a deliberate application of high-R cylindrical design. A titanium heat break with a 0.5 mm wall and 4 mm inner diameter has a thermal resistance roughly five times larger than a stainless steel version of the same dimensions. That difference determines whether the printer clogs or runs cleanly.
Cylindrical Conduction
Consider a long hollow cylinder with inner radius r1, outer radius r2, and length L. The inner surface is held at temperature T1 and the outer at T2. We want the steady-state heat flow rate q.
We apply a shell balance to a thin cylindrical shell at radius r with thickness dr. At steady state with no heat generation, energy in equals energy out, so the radial heat flow rate qr must be the same at every radius. Fourier's law in cylindrical coordinates gives:
Since qr is constant, we rearrange and integrate from r1 to r2:
Separate variables in Equation 4.1:
$$q_r \frac{dr}{r} = -k \cdot 2\pi L \, dT$$Integrate both sides, left from r1 to r2, right from T1 to T2:
$$q_r \int_{r_1}^{r_2} \frac{dr}{r} = -k \cdot 2\pi L \int_{T_1}^{T_2} dT$$ $$q_r \ln\!\left(\frac{r_2}{r_1}\right) = -k \cdot 2\pi L \,(T_2 - T_1)$$Since T1 > T2, let ΔT = T1 − T2 > 0:
$$q_r = \frac{2\pi k L \,\Delta T}{\ln(r_2/r_1)}$$Writing this as q = ΔT / R:
$$\boxed{R_{\text{cyl}} = \frac{\ln(r_2/r_1)}{2\pi k L}}$$Compare this to the flat-wall result: Rslab = L/(kA). Both are ratios of a geometric length to conductance. The difference is that Rcyl uses ln(r2/r1) as the effective "thickness" because the area changes continuously with radius. When the wall is very thin (r2 − r1 << r1), ln(r2/r1) ≈ (r2−r1)/r1 and the formula reduces to the flat-wall result, as it should.
For a multilayer cylinder with shells of conductivities k1, k2, ... and radii r1 < r2 < r3 ..., the total resistance is Rtot = Rcyl,1 + Rcyl,2 + ... just as for flat walls. A convective resistance at the outer surface adds 1/(hA) = 1/(h · 2πroutL).
Spherical Conduction
The same shell-balance logic applies to a hollow sphere. The heat flow area at radius r is 4πr2. Following exactly the same procedure as before gives:
Spherical geometry appears in polymer science wherever particles exchange heat with a surrounding fluid: cooling pellets in a water quench, encapsulated drug particles releasing at body temperature, or polymer microspheres in a suspension polymerisation. For a solid sphere with a known surface temperature, you only need the convective resistance 1/(h · 4πr2) at the outer surface, and the conduction resistance inside becomes zero (since there is no shell to cross).
HDPE pellets emerging from a die face pelletiser at 180°C are quenched in cold water. Modelling them as spheres, the total resistance is Rsph (HDPE shell, if hollow) plus Rconv = 1/(h · 4πr2) at the surface. The Biot number tells you whether the dominant resistance is inside or outside (covered fully in Chapter 8).
Critical Radius of Insulation
Here is a counterintuitive result. Consider a small pipe covered in foam insulation. You expect more foam to mean less heat loss. But if the pipe is thin enough, adding insulation initially increases the heat loss, because the gain in cylindrical resistance is more than offset by the gain in convective area at the outer surface.
The total resistance for a pipe of inner radius r1, insulation outer radius r, insulation conductivity kins, and external convection h is:
Minimising Rtot with respect to r (i.e., dR/dr = 0) yields a minimum resistance at the critical radius:
For foam insulation (k ≈ 0.04 W/mK) in still air (h ≈ 8 W/m²K), the critical radius is only 5 mm. Most industrial pipes exceed this radius, so insulation always helps. But a thin electrical wire (r = 1 mm) or a capillary tube can sit well below the critical radius, and wrapping it in polymer insulation actually increases heat transfer to the surroundings.
Students often assume insulation always reduces heat loss. Always compare the bare-pipe radius to rcr = k/h first. If rpipe < rcr, heat loss increases as you add insulation up to r = rcr, then decreases for r > rcr.
Interactive: Fin Efficiency Explorer
Fins are extended surfaces added to a body to increase its heat transfer area. How effective a fin is depends critically on its thermal conductivity. Use the controls below to see how fin efficiency changes with material choice and length.
Extended Surfaces and Fins
A fin is any surface extended from a body to increase the area available for heat transfer. Radiator fins on a car engine, heat sinks on electronics, the corrugated surface of a polymer heat exchanger: all work by the same principle. The catch is that a fin only helps if heat can conduct along it quickly enough to reach the tip. If a fin's thermal conductivity is too low, its tip reaches the ambient temperature well before the end, and extra length adds area but almost no heat transfer.
The Fin Equation
Consider a rectangular fin of cross-sectional area Ac, perimeter P, length L, and conductivity k, with convection coefficient h and ambient temperature T∞. Define the excess temperature θ(x) = T(x) − T∞. An energy balance on a differential element of length dx gives:
Energy in by conduction minus energy out by conduction equals energy out by convection:
$$-kA_c\frac{dT}{dx}\bigg|_x + kA_c\frac{dT}{dx}\bigg|_{x+dx} = h(P\,dx)(T - T_\infty)$$Dividing by dx:
$$kA_c\frac{d^2T}{dx^2} = hP(T - T_\infty)$$Substituting θ = T − T∞:
$$\frac{d^2\theta}{dx^2} = m^2\theta \qquad \text{where } m = \sqrt{\frac{hP}{kA_c}}$$This is a second-order ODE with solution:
$$\theta(x) = C_1 e^{mx} + C_2 e^{-mx}$$With boundary conditions θ(0) = θb (base temperature) and dθ/dx = 0 at x = L (insulated tip), the solution is:
$$\theta(x) = \theta_b \frac{\cosh m(L-x)}{\cosh mL}$$The heat lost by the fin (integrating convection over the entire surface) equals the heat conducted in at the base:
Fin Efficiency
The fin efficiency ηf compares the actual heat transfer to the ideal case where the entire fin is at the base temperature:
When mL is small (short fin, high k, low h), ηf approaches 1: the fin is nearly isothermal and every centimetre contributes fully. When mL is large, tanh(mL) → 1 and ηf → 1/(mL), which falls quickly.
For a PEEK fin (k = 0.25 W/m·K) with h = 50 W/m²·K, t = 4 mm, w = 20 mm: m = 346 m−1. Even a 30 mm fin has mL = 10.4 and efficiency of only 9.6%. An identical aluminium fin (k = 200) has m = 12.3 m−1, mL = 0.37, and efficiency of 95.7%. This is why heat sinks are always made of metal, and why adding polymer ribs to a heated mould does almost nothing.
Worked Examples
Nylon Pipe with Foam Insulation
A nylon pipe (k = 0.25 W/m·K) carries hot water. Inner radius r1 = 20 mm, outer radius r2 = 30 mm. A layer of foam insulation (k = 0.04 W/m·K) covers the pipe from r2 = 30 mm to r3 = 50 mm. The inner water temperature is 70°C, the surrounding air is 20°C, and the convection coefficient at the outer foam surface is h = 8 W/m²·K. Find the heat loss per metre of pipe.
Solution:
Three resistances in series: nylon shell, foam shell, and outer convection.
$$R_{\text{nylon}} = \frac{\ln(r_2/r_1)}{2\pi k_1 L} = \frac{\ln(30/20)}{2\pi \times 0.25 \times 1} = \frac{0.405}{1.571} = 0.258 \;\text{K/W}$$ $$R_{\text{foam}} = \frac{\ln(r_3/r_2)}{2\pi k_2 L} = \frac{\ln(50/30)}{2\pi \times 0.04 \times 1} = \frac{0.511}{0.251} = 2.035 \;\text{K/W}$$ $$R_{\text{conv}} = \frac{1}{h \cdot 2\pi r_3 L} = \frac{1}{8 \times 2\pi \times 0.050 \times 1} = \frac{1}{2.513} = 0.398 \;\text{K/W}$$ $$R_{\text{tot}} = 0.258 + 2.035 + 0.398 = 2.691 \;\text{K/W}$$ $$q = \frac{\Delta T}{R_{\text{tot}}} = \frac{70 - 20}{2.691} = 18.6 \;\text{W/m}$$Notice that the foam dominates: 2.035 out of 2.691 K/W, or 76% of the total resistance. Doubling the foam thickness would halve the heat loss; doubling the nylon thickness would have almost no effect.
Step 1: Draw the resistance network. Three elements in series: Rnylon + Rfoam + Rconv.
Step 2: Apply Rcyl = ln(rout/rin) / (2πkL) to each shell.
Step 3: Rconv = 1/(h · 2πroutL). Then q = ΔT / Rtot.
PEEK Fin Efficiency and the Case for Metal
A rectangular fin has these dimensions: length L = 30 mm, thickness t = 4 mm, width w = 20 mm. Convection coefficient h = 50 W/m²·K. Compare the heat transfer and efficiency of a PEEK fin (k = 0.25 W/m·K) to an identical aluminium fin (k = 200 W/m·K). Base temperature 120°C, ambient 20°C (θb = 100 K).
Solution:
Fin perimeter P = 2(w + t) = 2(0.020 + 0.004) = 0.048 m. Cross-section Ac = wt = 8×10−4 m².
PEEK:
$$m = \sqrt{\frac{50 \times 0.048}{0.25 \times 8\times10^{-4}}} = \sqrt{\frac{2.4}{2\times10^{-4}}} = \sqrt{12000} = 346 \;\text{m}^{-1}$$ $$mL = 346 \times 0.030 = 10.4 \quad \Rightarrow \quad \tanh(mL) \approx 1$$ $$\eta_f = \frac{\tanh(10.4)}{10.4} = \frac{1}{10.4} = 9.6\%$$ $$q_{\text{PEEK}} = \sqrt{50 \times 0.048 \times 0.25 \times 8\times10^{-4}} \times 100 \times 1 = 0.69 \;\text{W}$$Aluminium:
$$m = \sqrt{\frac{50 \times 0.048}{200 \times 8\times10^{-4}}} = \sqrt{\frac{2.4}{0.16}} = \sqrt{15} = 3.87 \;\text{m}^{-1}$$ $$mL = 3.87 \times 0.030 = 0.116 \quad \Rightarrow \quad \tanh(0.116) = 0.115$$ $$\eta_f = \frac{0.115}{0.116} = 99.3\%$$ $$q_{\text{Al}} = \sqrt{50 \times 0.048 \times 200 \times 8\times10^{-4}} \times 100 \times 0.115 = \sqrt{384} \times 100 \times 0.115 = 22.6 \;\text{W}$$The aluminium fin transfers 33 times more heat with the same geometry and the same convection conditions. The PEEK fin operates deep in the "thermally long" regime where the tip is already at ambient and extra length is wasted area. This is the quantitative argument for metal in any fin application.
Concept Questions
A copper pipe (k = 400 W/m·K) and a nylon pipe (k = 0.25 W/m·K) have identical inner and outer radii and carry the same temperature fluid. All else equal, which has higher thermal resistance Rcyl?
You double the outer radius of a cylindrical pipe wall while keeping the inner radius and conductivity fixed. How does Rcyl change?
For a cylindrical pipe, you double the pipe length L while keeping all other parameters fixed. What happens to the heat loss rate q?
Concept: Cylindrical Thermal Resistance
Rcyl = ln(r2/r1) / (2πkL). Resistance is inversely proportional to k and to L, and grows logarithmically with the radius ratio. The logarithm is the key difference from flat-wall resistance: doubling the outer radius does not double the wall thickness in the thermal sense because outer layers have more area.
A PEEK fin has mL = 10. What is the approximate fin efficiency?
To improve a fin's efficiency, which change is most effective?
Concept: Fin Efficiency and Material Choice
Fin efficiency η = tanh(mL)/(mL). For efficiency close to 1, mL must be small (mL < 0.3 is a practical guideline). Since m = √(hP/kAc), decreasing k increases m and destroys efficiency. Polymer fins are almost always impractical for heat transfer; their value is structural or chemical, not thermal.
Problem Set
Part A: Conceptual
Explain in physical terms why the cylindrical resistance formula uses ln(r2/r1) rather than a simple difference (r2−r1). What happens to the formula as the wall thickness becomes very small compared to the radius?
The critical radius of insulation is rcr = kins/h. Explain why increasing the external convection coefficient h makes the critical radius smaller. Under what practical conditions is this important?
A fin with mL = 0.1 and a fin with mL = 5 are added to the same heated surface. Compare their efficiencies and comment on whether either fin is a good engineering decision. What physical properties distinguish them?
Spherical geometry appears in drug delivery. A polymer microsphere (radius 0.5 mm, k = 0.20 W/m·K) releases drug by diffusion in body tissue (h ≈ 500 W/m²·K). Which resistance dominates: conduction through the shell or convection at the surface? How would your answer change if the sphere were 10 times larger?
An engineer proposes adding PEEK ribs (fins) to the outside of a heated injection mould to improve cooling. Using what you know about fin efficiency, argue for or against this design. If the ribs could instead be made of any material, what conductivity would you specify as a minimum, and why?
Part B: Calculation
A nylon pipe (k = 0.25 W/m·K) has inner radius r1 = 20 mm and outer radius r2 = 40 mm. It carries hot water at 80°C; the outer pipe surface is held at 30°C. Calculate the heat loss per metre of pipe length.
Hint 1: Use Rcyl = ln(r2/r1) / (2πkL). Note that r2/r1 = 2, so ln(r2/r1) = ln 2 = 0.693.
Hint 2: With L = 1 m: Rcyl = 0.693 / (2π×0.25×1) = 0.441 K/W. Then q = ΔT/R = 50/0.441.
A foam insulation (kins = 0.040 W/m·K) is applied to a pipe. The external convection coefficient is h = 8 W/m²·K. Calculate the critical radius of insulation. If the bare pipe has an outer radius of 3 mm, will adding foam initially increase or decrease heat loss?
Hint 1: rcr = kins/h. This is the radius at which the total resistance (insulation + convection) is at its minimum.
Hint 2: Compare rpipe = 3 mm to rcr. If rpipe < rcr, adding foam increases heat loss initially.
A hollow HDPE sphere (k = 0.44 W/m·K) has inner radius r1 = 40 mm and outer radius r2 = 50 mm. The inner surface is at 60°C (heated water inside). The outer surface faces air at 20°C with convection coefficient h = 20 W/m²·K. Find the steady-state heat loss from the sphere.
Hint 1: Two resistances in series: Rsph = (1/r1−1/r2) / (4πk) and Rconv = 1/(h×4πr2²).
Hint 2: Rsph = (1/0.040−1/0.050)/(4π×0.44) = 5/(5.53) = 0.904 K/W. Rconv = 1/(20×4π×0.050²) = 1.59 K/W. q = 40/2.49.
A PEEK rectangular fin (k = 0.25 W/m·K) has length L = 30 mm, thickness t = 4 mm, and width w = 20 mm. The base temperature is 120°C, ambient is 20°C, and the convection coefficient is h = 50 W/m²·K. Calculate the fin efficiency (as a percentage). Assume an insulated tip.
Hint 1: Compute m = √(hP/(kAc)) where P = 2(w+t) = 0.048 m and Ac = wt = 8×10−4 m².
Hint 2: m = √(50×0.048/(0.25×8×10−4)) = √12000 = 346 m−1. Then mL = 346×0.030 = 10.4. For large mL, tanh(mL) ≈ 1, so η = 1/mL.
A two-layer cylindrical pipe carries hot fluid at 70°C. The inner HDPE layer (k = 0.44 W/m·K) runs from r1 = 20 mm to r2 = 30 mm. The outer foam insulation (k = 0.040 W/m·K) runs from r2 = 30 mm to r3 = 50 mm. The outer foam surface is held at 20°C. Find the heat loss per metre of pipe.
Hint 1: Two Rcyl in series. RHDPE = ln(30/20)/(2π×0.44×1) and Rfoam = ln(50/30)/(2π×0.04×1).
Hint 2: ln(1.5) = 0.405, so RHDPE = 0.147 K/W. ln(5/3) = 0.511, so Rfoam = 2.03 K/W. Foam dominates. q = 50/2.18.
Chapter Summary
Cylindrical and spherical geometries produce resistance formulas with logarithmic and reciprocal terms because the conduction area changes with radius. For cylinders, Rcyl = ln(r2/r1)/(2πkL); for spheres, Rsph = (1/r1−1/r2)/(4πk). Both reduce to the flat-wall formula when the wall is thin relative to the radius. Cylindrical resistances add in series exactly as flat-wall ones do, with a convective term 1/(h×2πrL) at each fluid-solid boundary.
Insulation on a small-radius pipe may increase heat loss until the outer radius reaches the critical value rcr = kins/h. For most industrial pipes this is not a concern, but for thin wires and capillaries the effect is real.
Fins are most effective when mL is small, which requires high conductivity, large cross-sectional area, and short length. Polymer fins have m values in the hundreds of inverse metres, making them thermally useless even at short lengths. Any application that genuinely needs fin-enhanced heat transfer requires a metal. The next chapter applies these tools to polymer extrusion, where cylindrical geometry and heat generation combine in a process you can see on any production floor.