Thermal Resistance and Steady-State Conduction
Heat resists its own flow. The same mathematics that describes electrical circuits applies to heat moving through walls, pipes, and polymer laminates.
The Heat Break Problem
Every FDM printer has a component called a heat break. It is a thin-walled tube, usually made of stainless steel or PTFE, that connects the 220°C heater block to the 25°C cold end of the hotend. Its job is simple: block heat. If too much heat travels upward, the filament softens too early, jams the drive gears, and ruins the print.
The heat break designer faces a clear optimization problem. The tube must be long enough, thin-walled enough, and made of a material with low enough thermal conductivity to create sufficient thermal resistance. But it cannot be so fragile that it buckles under the filament pushing through it.
This chapter gives you the tools to solve that problem exactly. Thermal resistance is the central idea. It quantifies how much a material opposes heat flow, the same way electrical resistance quantifies opposition to current. Once you can write down a thermal resistance, you can analyze any combination of walls, cylinders, coatings, and convective boundaries as a circuit.
Thermal Resistance
The Slab
Start with the simplest geometry: a flat slab of thickness $L$, cross-sectional area $A$, and thermal conductivity $k$. One face is held at temperature $T_1$, the other at $T_2 < T_1$. At steady state, Fourier's law gives the total heat flow rate:
Now look at this equation and compare it to Ohm's law: $I = V/R$. Current equals voltage divided by resistance. Rearrange Eq. 2.1 the same way:
$$\dot{Q} = \frac{T_1 - T_2}{L/(kA)}$$The denominator $L/(kA)$ plays the role of resistance. Define it as the conductive thermal resistance of a flat slab:
With this definition, heat flow becomes simply:
$$\dot{Q} = \frac{\Delta T}{R_{cond}}$$This is the thermal analogue of Ohm's law. Temperature difference drives heat flow; resistance opposes it. A thick wall has high resistance. A highly conductive material has low resistance. A large area lowers resistance because more pathways are available in parallel.
A typical stainless steel heat break has $L = 22$ mm, $k = 15$ W/m·K, and an inner diameter of 2 mm (so $A \approx 3.1 \times 10^{-6}$ m²). Its thermal resistance is: $R = 0.022 / (15 \times 3.1 \times 10^{-6}) \approx 470$ K/W. At a temperature difference of 195°C, it conducts about 0.41 W upward — small enough that the cooling fans can easily handle it.
Explore: How Geometry and Material Affect Resistance
Move the sliders to see how thickness, conductivity, and area each affect the thermal resistance and the resulting heat flux. The slab colour shifts from blue (low flux) to red (high flux).
Series Resistance Networks
A printer hotend is not a single material. It is a stack: heater block (aluminium), heat break (stainless steel), heat sink (aluminium), with some thermal paste at each interface. Heat must flow through all of them in sequence. The total resistance is the sum of the individual resistances.
The heat flow rate through the entire stack is the same in every layer — because at steady state, energy cannot accumulate anywhere:
$$\dot{Q} = \frac{T_{hot} - T_{cold}}{R_{total}}$$The temperature drop across each layer is proportional to its resistance. A thick, poorly conducting layer drops most of the temperature. A thin, highly conducting layer drops almost none.
A co-extruded barrier film has three layers: HDPE outer (2 mm, k=0.44), EVOH barrier (0.1 mm, k=0.34), and HDPE inner (2 mm). The total resistance per unit area is: $R'' = L_1/k_1 + L_2/k_2 + L_3/k_3 = 0.002/0.44 + 0.0001/0.34 + 0.002/0.44 = 4.55 + 0.29 + 4.55 = 9.4$ mm·K/W. The thin EVOH layer contributes only 3% of the total resistance — it is there for its barrier properties, not its insulation.
Layers that heat must pass through one after another are in series: add resistances. Paths that allow heat to flow simultaneously side-by-side are in parallel: add conductances ($1/R$), not resistances. A wall with a metal window frame and glass panels is a parallel circuit.
Given: A hotend assembly has three components in series.
- Aluminium heater block: $L=25$ mm, $k=160$ W/m·K, $A=4\times10^{-4}$ m²
- Stainless steel heat break: $L=22$ mm, $k=15$ W/m·K, $A=3.1\times10^{-6}$ m²
- Aluminium heat sink: $L=30$ mm, $k=160$ W/m·K, $A=1\times10^{-3}$ m²
Find: Total thermal resistance of the stack.
Solution:
Each resistance: $R_i = L_i/(k_i A_i)$
$$R_{block} = \frac{0.025}{160 \times 4\times10^{-4}} = 0.39 \text{ K/W}$$ $$R_{break} = \frac{0.022}{15 \times 3.1\times10^{-6}} = 473 \text{ K/W}$$ $$R_{sink} = \frac{0.030}{160 \times 1\times10^{-3}} = 0.19 \text{ K/W}$$ $$R_{total} = 0.39 + 473 + 0.19 \approx 474 \text{ K/W}$$The heat break carries 99.9% of the total resistance. The aluminium components are essentially invisible. This is the point: one well-chosen material dominates the series circuit.
Parallel Resistance
When two paths for heat flow exist side by side, they act like resistors in parallel. The total conductance is the sum of the individual conductances:
A practical example: an injection mold has steel walls with embedded cooling channels. The wall conducts heat by one path (through solid steel); the water in the channels conducts it by another (through water and then convection). The total heat removal is the sum of both paths.
Parallel paths always reduce the total resistance. The path with the smallest resistance dominates. If one path is 10 times lower resistance than another, it carries roughly 10 times more heat.
Cylindrical Geometry
Polymer pipes, extruder barrels, and wire coatings are cylinders, not flat slabs. The geometry changes the resistance formula because the area through which heat flows is not constant: it is $2\pi r L$ at radius $r$, and that increases as $r$ increases.
The result of accounting for this varying area is a logarithmic resistance:
Notice that the resistance depends on the ratio $r_2/r_1$, not the absolute radii. Doubling both radii does not change the resistance. Adding a thin coating to a large pipe barely changes it. But a thick coating on a thin wire changes it substantially.
Consider a hollow cylinder of inner radius $r_1$, outer radius $r_2$, and length $L$. At radius $r$, the area through which heat flows radially is $A(r) = 2\pi r L$. At steady state, the total heat flow $\dot{Q}$ must be the same at every radius (energy cannot accumulate). Apply Fourier's law at radius $r$:
$$\dot{Q} = -k \cdot 2\pi r L \cdot \frac{dT}{dr}$$Separate variables:
$$dT = -\frac{\dot{Q}}{2\pi k L}\frac{dr}{r}$$Integrate from $r_1$ (where $T = T_1$) to $r_2$ (where $T = T_2$):
$$T_2 - T_1 = -\frac{\dot{Q}}{2\pi k L}\ln\!\frac{r_2}{r_1}$$Rearranging and defining $R_{cyl}$:
$$\dot{Q} = \frac{T_1 - T_2}{\ln(r_2/r_1)/(2\pi k L)} \equiv \frac{\Delta T}{R_{cyl}}$$The $\ln(r_2/r_1)$ term comes directly from integrating $dr/r$. In a flat slab, area is constant so you integrate $dx$ and get $L$. In a cylinder, area grows with $r$, so you integrate $dr/r$ and get a logarithm.
A twin-screw extruder barrel has an inner radius of 25 mm, outer radius 50 mm, and is made of tool steel ($k = 42$ W/m·K). For a 1-metre section: $R = \ln(50/25)/(2\pi \times 42 \times 1) = 0.693/264 = 0.0026$ K/W. The barrel offers almost no thermal resistance — heat from the band heaters reaches the polymer melt very quickly.
Convection Resistance
Newton's law of cooling tells us $\dot{Q} = hA(T_s - T_f)$. Rearranged, this looks exactly like the thermal resistance form:
This is the resistance the fluid layer offers to heat leaving a surface. A high convective coefficient (vigorous flow, or water instead of air) means low convective resistance. Still air ($h \approx 5$ W/m²·K) has high resistance. Water cooling ($h \approx 1000$ W/m²·K) has resistance 200 times lower.
Now you can model a complete system: solid wall plus fluid on both sides. The three resistances — convection on the hot side, conduction through the wall, convection on the cold side — are in series.
Radiation can be linearised near a reference temperature $T_{ref}$ to give $R_{rad} \approx 1/(4\varepsilon\sigma T_{ref}^3 A)$. This allows radiation to be included in a network diagram, though the approximation breaks down when the temperature difference is large.
The Overall Heat Transfer Coefficient
When hot fluid flows inside a pipe and cold fluid flows outside, heat must cross three resistances in series: inner convection, pipe wall conduction, outer convection. Engineers combine these into a single number called the overall heat transfer coefficient $U$:
Once you know $U$, the heat flow is simply $\dot{Q} = UA\,\Delta T_{overall}$. This is the standard form used to design heat exchangers, which Chapter 10 covers in depth.
The value of $U$ is always dominated by the largest resistance. If the outer convective resistance is ten times larger than the wall resistance, then doubling the wall thickness changes $U$ by only a few percent. This is a useful diagnostic: improving the wrong resistance wastes effort.
A nylon pipe carries hot process water ($h_i = 800$ W/m²·K) through a water bath ($h_o = 300$ W/m²·K). The pipe wall is 3 mm thick ($k = 0.25$ W/m·K). Per unit area: $1/U = 1/800 + 0.003/0.25 + 1/300 = 0.00125 + 0.012 + 0.00333 = 0.0166$ m²·K/W. So $U = 60$ W/m²·K. The conduction term dominates at 72% of total resistance — a thinner pipe wall or higher-conductivity material would improve cooling most.
Worked Examples
Given: A HDPE pipe (inner radius 20 mm, outer radius 25 mm, $k_{HDPE} = 0.44$ W/m·K) is wrapped in 20 mm of polyurethane foam insulation ($k_{foam} = 0.033$ W/m·K). Inner surface is at 80°C; outer surface of foam at 20°C. Pipe length = 2 m.
Find: Heat loss rate.
Solution: Two cylindrical shells in series.
HDPE wall: $r_1 = 0.020$ m, $r_2 = 0.025$ m
$$R_{HDPE} = \frac{\ln(0.025/0.020)}{2\pi \times 0.44 \times 2} = \frac{0.223}{5.53} = 0.0403 \text{ K/W}$$Foam: $r_3 = 0.025$ m, $r_4 = 0.045$ m
$$R_{foam} = \frac{\ln(0.045/0.025)}{2\pi \times 0.033 \times 2} = \frac{0.588}{0.415} = 1.418 \text{ K/W}$$ $$R_{total} = 0.040 + 1.418 = 1.458 \text{ K/W}$$ $$\dot{Q} = \frac{80 - 20}{1.458} = 41.1 \text{ W}$$Without the foam, $R_{total} = 0.040$ K/W and $\dot{Q} = 1500$ W. The foam reduces heat loss by 97%. The HDPE wall itself is negligible.
Given: A polymer tube ($k = 0.25$ W/m·K, wall 2 mm) separates a hot fluid ($h_i = 500$ W/m²·K) from a cold fluid ($h_o = 200$ W/m²·K).
Find: Overall $U$ per unit area of outer surface.
$$\frac{1}{U} = \frac{1}{h_i} + \frac{L}{k} + \frac{1}{h_o} = \frac{1}{500} + \frac{0.002}{0.25} + \frac{1}{200} = 0.002 + 0.008 + 0.005 = 0.015 \text{ m}^2\text{K/W}$$ $$U = 66.7 \text{ W/m}^2\text{K}$$
The wall conduction term contributes 53% of the total resistance. For a metal tube, the wall term would be below 1%. This illustrates why polymer heat exchangers must use very thin walls or high-conductivity polymer composites to compete.
Concept Check
A HDPE slab (k = 0.44 W/m·K) and a copper slab (k = 400 W/m·K) have the same thickness and area. The ratio of their thermal resistances, R_HDPE / R_copper, is closest to:
You double the thickness of an insulating foam layer. What happens to the thermal resistance and the heat flow rate (all else equal)?
A pipe wall (cylindrical) and a flat slab have the same material, the same inner-to-outer thickness, and the same temperature difference. Which has the higher heat flow rate per unit length?
Key Concept
Thermal resistance for a slab is $R = L/(kA)$: proportional to thickness, inversely proportional to conductivity and area. For a cylinder, it is $R = \ln(r_2/r_1)/(2\pi k L)$: the logarithm arises because area increases with radius. In both cases, $\dot{Q} = \Delta T / R$.
A wall has three layers in series. The middle layer has 10× the resistance of the other two combined. If you want to reduce total heat flow by 30%, which layer should you improve?
In the overall heat transfer coefficient equation $1/U = 1/h_i + L/k + 1/h_o$, which term dominates when the fluid is still air on both sides ($h \approx 5$ W/m²·K) and the wall is 5 mm of aluminium ($k = 160$ W/m·K)?
Key Concept
In a series resistance network, the largest resistance dominates the total. When still air is involved, convection resistance ($1/hA$) almost always dwarfs conduction resistance through metal walls. This is why insulation and airflow management matter more than wall material in most building and industrial applications.
Problem Set — Chapter 2
Part A — Short Answer
A Styrofoam cup and a steel cup each contain the same hot coffee. Which cup feels hotter on the outside, and why? Express your answer using the concept of thermal resistance.
A polymer composite is developed with carbon fibre reinforcement, raising its through-thickness conductivity from 0.25 to 1.8 W/m·K. In an application where the polymer wall is sandwiched between two convective boundaries (h = 50 W/m²·K on each side), estimate whether this improvement meaningfully changes the overall heat transfer coefficient. Show your reasoning without a full calculation.
An engineer argues that doubling the insulation thickness always halves the heat loss from a cylindrical pipe. A colleague disagrees. Who is right, and why? Under what condition, if any, is the engineer's statement approximately correct?
The critical radius of insulation is the outer radius at which adding more insulation actually increases heat loss from a cylinder. Derive the expression for this critical radius and explain physically why it exists. (Hint: the outer convective resistance decreases as the outer radius grows.)
A multilayer co-extruded film has 5 alternating layers of HDPE and EVOH. Assume each HDPE layer is 100 µm thick (k = 0.44 W/m·K) and each EVOH layer is 10 µm thick (k = 0.34 W/m·K). The outer surfaces are at 80°C and 20°C. A packaging engineer proposes replacing all five layers with a single 520 µm HDPE layer of the same total thickness. Will the heat flow rate increase or decrease, and by roughly how much?
Part B — Calculations
A 3 mm thick PTFE sheet ($k = 0.25$ W/m·K, area = $0.5$ m²) separates two fluids. The hot-side surface is at 120°C and the cold-side surface is at 40°C. Find the steady-state heat flow rate (W).
Find $R = L/(kA) = 0.003/(0.25 \times 0.5)$.
Then $\dot{Q} = \Delta T / R = (120-40)/R$.
A nylon pipe ($k = 0.25$ W/m·K, inner radius 15 mm, outer radius 20 mm, length 3 m) carries water at 90°C. The outer surface is at 30°C. Find the heat loss rate (W).
Use $R_{cyl} = \ln(r_2/r_1)/(2\pi k L)$ then $\dot{Q} = \Delta T / R$.
$r_1=0.015$, $r_2=0.020$, $\ln(0.020/0.015) = 0.2877$. $R = 0.2877/(2\pi \times 0.25 \times 3) = 0.0611$ K/W. Then $\dot{Q} = 60/0.0611 \approx 983$ W.
A co-extruded packaging film has three layers: HDPE (1.5 mm, $k=0.44$), EVOH (0.08 mm, $k=0.34$), HDPE (1.5 mm, $k=0.44$). The total area is $2$ m². Outer surfaces are at 60°C and 20°C. Find the total heat flow rate (W).
Three layers in series. $R_i = L_i/(k_i A)$. Add resistances, then $\dot{Q} = \Delta T / R_{total}$.
A polymer tube ($k=0.25$ W/m·K, wall 4 mm) separates hot process water ($h_i = 600$ W/m²·K) from cooling water ($h_o = 400$ W/m²·K). Find the overall heat transfer coefficient $U$ (W/m²·K) per unit area.
$1/U = 1/h_i + L/k + 1/h_o = 1/600 + 0.004/0.25 + 1/400 = 0.00167 + 0.0160 + 0.00250 = 0.02017$ m²·K/W. So $U = 49.6$ W/m²·K.
An HDPE pipe ($k=0.44$ W/m·K, inner radius 20 mm) carries a hot fluid at 80°C. Air at 25°C flows over the outside ($h_o = 10$ W/m²·K). Find the critical radius of insulation $r_c$ (mm) if the insulating foam has $k_{ins} = 0.04$ W/m·K. (Use the formula $r_c = k_{ins}/h_o$.)
$r_c = k_{ins}/h_o = 0.04/10$. Convert to mm.
Chapter Summary
The central idea of this chapter is that heat flow through any geometry can be expressed as a temperature difference divided by a thermal resistance. For a flat slab, $R = L/(kA)$. For a hollow cylinder, $R = \ln(r_2/r_1)/(2\pi kL)$. For a convective boundary, $R = 1/(hA)$. Geometries in series add their resistances; paths in parallel add their conductances.
The dominant resistance in a series network controls heat flow. Identifying it is the first step in any engineering analysis: improving a small resistance does almost nothing; improving the dominant one yields proportional gains.
The overall heat transfer coefficient $U$ packages a series network (inner convection, wall conduction, outer convection) into a single number. It is the standard tool for heat exchanger analysis in Chapter 10.
Chapter 3 takes the next step: instead of assuming a linear temperature profile (as we have done throughout this chapter), we derive the equation that governs the temperature field everywhere inside a conducting solid, including when heat is generated internally and when conditions are not steady.