The Heat Conduction Equation
From Fourier's law to a single equation that governs all of conduction, in any geometry, with any heat source.
The Burning Question
Imagine you are designing a ceramic heater nozzle for a high-temperature FDM printer (the kind that processes PEEK at 400°C. The resistive heating element is embedded inside the ceramic block. When the printer is running, the element dumps 50 W into a block the size of a sugar cube.
Here is the puzzle: the surface of the block never reaches the same temperature as its core. You know the power going in. You know the convective cooling at the surface. But to find the temperature at any point inside, especially the hottest point where the polymer melt lives, you need more than Fourier's law. You need to know how temperature is distributed in space, not just across a simple slab.
Chapter 2 gave you a powerful shortcut, thermal resistance, that works beautifully for layered geometries where you only care about the heat flow rate between fixed surfaces. But that approach breaks down the moment you have:
Internal heat generation (Joule heating, chemical reaction, viscous dissipation).
Non-uniform temperature within a single material (temperature profile, not just temperature difference).
Time-varying conditions (the block is heating up, not yet at steady state).
To handle all three, we need to write an energy balance on an infinitesimally small volume element and take the limit as the volume shrinks to zero. The result is a partial differential equation (the heat conduction equation) that governs temperature at every point in space and time. Once you have this equation and a set of boundary conditions, physics does the rest.
One equation, derived in this chapter, predicts the temperature distribution in a PEEK nozzle, an injection-molded part cooling in a mold, a polymer film being laser-welded, and every other conduction problem in this textbook. What changes from problem to problem is not the equation, but the geometry and boundary conditions.
Deriving the Heat Conduction Equation
The Cartesian Form
Consider a small rectangular block of material with sides $\Delta x$, $\Delta y$, $\Delta z$. Heat can flow in and out through every face. The block may also generate heat internally at a volumetric rate $\dot{q}$ (W/m³). Our energy accounting is simple:
We apply Fourier's law to find the heat entering and leaving each pair of faces. In the $x$-direction, heat enters the left face at rate $q_x = -kA_x \frac{\partial T}{\partial x}\big|_x$ and leaves the right face at rate $q_{x+\Delta x} = -kA_x \frac{\partial T}{\partial x}\big|_{x+\Delta x}$. The net heat from the $x$-faces alone is $q_x - q_{x+\Delta x}$, which equals $\frac{\partial}{\partial x}\!\left(k\frac{\partial T}{\partial x}\right)\!\Delta x \cdot A_x$ in the limit $\Delta x \to 0$. Repeating for $y$ and $z$, adding generation $\dot{q}\,\Delta x\,\Delta y\,\Delta z$, and noting that storage equals $\rho c_p \frac{\partial T}{\partial t}\,\Delta V$, then dividing through by $\Delta V = \Delta x\,\Delta y\,\Delta z$, we arrive at:
If the material has uniform, constant conductivity $k$, we can factor it out and define the thermal diffusivity $\alpha \equiv k/(\rho c_p)$:
Setup. A volume element at position $(x, y, z)$ with dimensions $\Delta x \times \Delta y \times \Delta z$. Area perpendicular to $x$-axis: $A_x = \Delta y \Delta z$. Fourier's law gives the flux component $q_x'' = -k \partial T/\partial x$, so the heat flow rate into the left face is $\dot{Q}_{x,in} = -k \frac{\partial T}{\partial x}\big|_x \cdot \Delta y \Delta z$.
Net $x$-direction contribution. The heat leaving the right face is $\dot{Q}_{x,out} = -k \frac{\partial T}{\partial x}\big|_{x+\Delta x} \cdot \Delta y \Delta z$. Net influx:
$$\dot{Q}_{x,in} - \dot{Q}_{x,out} = \left(-k\frac{\partial T}{\partial x}\bigg|_x + k\frac{\partial T}{\partial x}\bigg|_{x+\Delta x}\right)\Delta y\Delta z$$Dividing by $\Delta x$ and taking $\Delta x \to 0$, this becomes $\frac{\partial}{\partial x}\!\left(k\frac{\partial T}{\partial x}\right)\Delta x \Delta y \Delta z$.
Add all three directions and generation. By symmetry, $y$ and $z$ contribute analogously. Internal generation contributes $\dot{q}\,\Delta x\Delta y\Delta z$ (W).
Storage term. The energy stored as sensible heat is $\rho c_p \frac{\partial T}{\partial t}\,\Delta x\Delta y\Delta z$.
Assemble and divide by $\Delta V$.
$$\frac{\partial}{\partial x}\!\left(k\frac{\partial T}{\partial x}\right) + \frac{\partial}{\partial y}\!\left(k\frac{\partial T}{\partial y}\right) + \frac{\partial}{\partial z}\!\left(k\frac{\partial T}{\partial z}\right) + \dot{q} = \rho c_p \frac{\partial T}{\partial t}$$This is Eq. (3.1). Every term has units of W/m³. Note that when $k$ is uniform it factors out, giving $k\nabla^2 T + \dot{q} = \rho c_p \partial T/\partial t$, which divides through to Eq. (3.2).
Other Coordinate Systems
Many polymer geometries are not Cartesian. A pipe is cylindrical; a pressure vessel head is spherical. The same energy accounting in different coordinate systems gives:
Limiting Cases
Equation (3.2) has four well-known special cases. Knowing which one applies immediately simplifies the problem.
Thermal Diffusivity
The thermal diffusivity $\alpha = k/(\rho c_p)$ (m²/s) tells you how fast temperature disturbances spread through a material. Think of it like this: if you touch a cold metal rod versus a cold plastic rod, the metal feels colder even though both are at the same temperature. That is $\alpha$ at work. Metal has a high $\alpha$: temperature changes propagate quickly from the metal surface into your fingertip and away from it. Plastic has a low $\alpha$: temperature spreads slowly, so the surface stays cooler for longer, and your fingertip warms it up instead.
High $\alpha$ means the material reaches thermal equilibrium quickly. Low $\alpha$ means thermal disturbances take a long time to penetrate. For polymer processing, low $\alpha$ is precisely why thick injection-molded parts take so long to cool: the heat generated during solidification is trapped inside and can only escape slowly.
The reason PLA parts cool faster in still air than ABS parts (even though ABS is printed hotter) is partly the difference in thermal diffusivity. PLA has $\alpha \approx 8 \times 10^{-8}$ m²/s vs ABS at about $1.1 \times 10^{-7}$ m²/s. ABS conducts heat away from the hot bead slightly faster, but its higher deposition temperature creates a larger thermal gradient. Both factors interact, and the governing equation captures them together through the $\alpha$ and the boundary conditions.
Widget: Boundary Condition Visualizer
Before we define boundary conditions formally, try changing them below and observe how the temperature profile in a slab changes shape. Use the toggles to switch between an insulated face (zero flux) and a convective face (Newton's law of cooling). The slider controls heat generation inside the slab. The blue curve shows the resulting steady-state temperature profile.
Notice: when one face is insulated, the temperature maximum shifts to that face. When both faces are convective, the profile is symmetric with maximum at the center. Try setting both faces to insulated: the solver correctly shows that no steady state exists when heat is generated with no escape path.
Boundary Conditions
A differential equation alone does not have a unique solution. We need boundary conditions: mathematical statements about what is happening at each surface, plus an initial condition if the problem is transient. There are three standard types.
Type 1: Specified Temperature (Dirichlet)
You just specify the temperature value. In practice, it arises when a surface is in perfect thermal contact with a large reservoir (like cooling water or a phase-change material) that holds the surface temperature constant.
Type 2: Specified Heat Flux (Neumann)
The most important special case is $q_s'' = 0$: the insulated or symmetry condition. A plane of symmetry has zero heat flux by definition: what crosses from the left side must equal what crosses from the right, and in a symmetric problem these are equal and opposite, so the net flux is zero. You often exploit this to solve only half the domain.
The midplane of a symmetric heated build plate (same temperature on top and bottom) is an adiabatic plane. You only need to solve from the plate surface to its center; the symmetry condition dT/dx = 0 at the center replaces the unknown boundary condition there, halving the computational domain.
Type 3: Convection (Robin)
This is the most physically realistic boundary condition for an exposed surface. It says: whatever heat arrives at the surface by conduction must leave by convection. The surface temperature $T_s$ is not given ahead of time; it is an unknown that this equation helps determine. Applying the Type 3 condition at $x = L$ links the interior temperature profile to the ambient fluid.
Steady-State Solutions with Heat Generation
Plane Wall with Uniform Heat Generation
This problem appears constantly in polymer processing. A resistive heating element (film heater, cartridge heater, PTC element) dumps heat uniformly into a polymer block. One face may be insulated (the back of a mold); the other face loses heat to a fluid by convection.
At steady state, $\partial T/\partial t = 0$. With only one spatial coordinate and constant $k$, the Poisson equation (3.5) reduces to:
$$\frac{d^2T}{dx^2} = -\frac{\dot{q}}{k}$$Integrating twice:
$$T(x) = -\frac{\dot{q}}{2k}x^2 + C_1 x + C_2$$Now apply BCs. Left face insulated at $x=0$: $dT/dx = 0 \Rightarrow C_1 = 0$. Right face convection at $x = L$: $-k\,dT/dx\big|_L = h(T_L - T_\infty) \Rightarrow \dot{q}L = h(T_L - T_\infty)$, so $T_L = T_\infty + \dot{q}L/h$. Applying $T(L) = T_L$ to find $C_2$:
This result is illuminating: the surface temperature $T_\infty + \dot{q}L/h$ depends only on the convection coefficient and the total heat generated. The extra term $\dot{q}L^2/(2k)$ at $x=0$ represents how much hotter the interior gets above the surface due to limited thermal conductivity. Low $k$ (as in polymers) makes this second term large: the core runs far hotter than the surface.
Cylinder with Uniform Heat Generation
For a long cylinder of radius $R$ with uniform heat generation, the cylindrical form of Eq. (3.5) with radial symmetry reduces to:
$$\frac{1}{r}\frac{d}{dr}\!\left(r\frac{dT}{dr}\right) = -\frac{\dot{q}}{k}$$Integrating once: $r\,dT/dr = -\dot{q}r^2/(2k) + C_1$. BCs: symmetry at $r=0$ gives $dT/dr = 0 \Rightarrow C_1 = 0$. Convection at $r = R$: $T_s = T_\infty + \dot{q}R/(2h)$. Integrating again and applying $T(R) = T_s$:
Why $4k$ for a cylinder versus $2k$ for a plane wall? In the plane-wall case, heat can only escape in one direction. In the cylinder, heat escapes radially in all directions, and the curved surface area increases with $r$, giving the heat more "room" to escape. This dilutes the internal temperature rise by a factor of 2.
Worked Examples
Problem. A PEEK heater block (k = 0.25 W/m·K) is 20 mm thick and has a resistive heater embedded uniformly, generating $\dot{q} = 2 \times 10^5$ W/m³. The back face is insulated against the frame. The front face is exposed to forced airflow ($h = 200$ W/m²·K, $T_\infty = 25°C$). Find the surface temperature and the maximum temperature inside the block.
Step 1: surface temperature from energy balance.
$$T_s = T_\infty + \frac{\dot{q}L}{h} = 25 + \frac{(2\times10^5)(0.02)}{200} = 25 + 20 = 45\,°C$$Step 2: maximum temperature at insulated face ($x = 0$).
$$T_{max} = T_s + \frac{\dot{q}L^2}{2k} = 45 + \frac{(2\times10^5)(0.02)^2}{2(0.25)} = 45 + 160 = 205\,°C$$Interpretation. The surface is only 45°C, but 20 mm into the block the temperature is 205°C. That is a 160°C gradient across just 2 cm of polymer. This is why heat-resistant polymers are essential in heater blocks: the polymer near the heater operates far above the ambient temperature. PEEK (continuous-use limit ~260°C) is marginal here; PEI/Ultem would be needed at higher power densities.
$T_s = T_\infty + \dot{q}L/h$: all generated heat exits through the convective face at steady state.
$T_{max} = T_s + \dot{q}L^2/(2k)$ from Eq. (3.11) at $x=0$.
Problem. A 2 mm thick polymer film ($k = 0.30$ W/m·K) is heated by laser irradiation, which deposits $\dot{q} = 6 \times 10^6$ W/m³ uniformly throughout. Both surfaces are held at $T_s = 60°C$ by cooling rollers. Find the temperature at the center of the film.
Identify the case. Both surfaces are at the same temperature (symmetric, Type 1 BCs). The problem is symmetric about the midplane at $x = L/2$.
Set up. With origin at the left surface ($x = 0$), the temperature distribution is:
$$T(x) = T_s + \frac{\dot{q}}{2k}x(L - x)$$Evaluate at center $x = L/2 = 0.001$ m.
$$T_{center} = 60 + \frac{6\times10^6}{2(0.30)}(0.001)(0.002 - 0.001) = 60 + \frac{6\times10^6}{0.60} \times 10^{-6} = 60 + 10 = 70\,°C$$Interpretation. The center is only 10°C hotter than the surfaces, a modest gradient for $6\times10^6$ W/m³. That seems surprising, but note that the film is very thin (2 mm). The factor $(L/2)^2 / k$ is small because $L$ is small. This is why thin films can be laser-processed without overheating their core: the short conduction path lets heat escape quickly.
Both surfaces at $T_s$ with uniform generation: $T(x) = T_s + \frac{\dot{q}}{2k} x(L-x)$. Maximum at $x = L/2$.
$T_{max} = T_s + \frac{\dot{q}L^2}{8k}$. Notice the factor of $8k$ vs $2k$ for the insulated-face case. Symmetric problem is always cooler at center.
Concept Check
A polymer slab has heat generation $\dot{q} > 0$ and its left face is insulated. Where does the maximum temperature occur at steady state?
You double the heat generation rate $\dot{q}$ in a slab. How does the temperature excess at the center above the surface temperature change?
Which equation governs the steady-state temperature distribution in a polymer slab with no internal heat generation and no time variation?
Key Insight: Limiting Cases of the Heat Equation
The full heat equation $\nabla^2 T + \dot{q}/k = (1/\alpha)\,\partial T/\partial t$ contains three mechanisms: conduction (Laplacian), generation ($\dot{q}$), and storage (time derivative). Switch off generation and time variation, and you have Laplace's equation: a linear temperature profile in 1D. Switch off only the time derivative but keep generation, and you have Poisson's equation: a parabolic profile in 1D. This hierarchy helps you immediately identify the right equation for any problem.
A thick polymer part is cooling in a mold. What boundary condition most accurately represents the midplane of the part?
A polymer film surface is cooled by flowing water at $T_\infty = 15°C$ with $h = 2000$ W/m²·K. The film has $k = 0.30$ W/m·K. Which boundary condition type applies at this surface, and which parameter(s) does it involve?
Boundary Condition Guide
Match the physical situation to the right BC type. Surfaces in contact with reservoirs (boiling water, phase-change materials) approximate Type 1. Insulated or symmetric surfaces use Type 2 with $q'' = 0$. Any surface exposed to a flowing fluid (air, water, polymer melt) uses Type 3. When in doubt, sketch the interface and ask "do I know the temperature, the flux, or the convection coefficient?"
Problem Set
Part A is conceptual. Part B is numerical. All answers should be entered without units; units are indicated. Type your answer; no scroll-bar inputs are needed.
Part A: Conceptual Problems
Explain in physical terms why a plane wall with heat generation and one insulated face has a parabolic temperature profile, while a plane wall with no generation and the same BCs has a linear temperature profile.
A polymer rod has uniform heat generation $\dot{q}$. Its surface temperature is found from the energy balance $T_s = T_\infty + \dot{q}R/(2h)$. Why does this formula involve only $h$ and not the thermal conductivity $k$ of the polymer?
Two polymer slabs A and B have identical thickness, the same $\dot{q}$, and the same convective BCs on both faces. Slab A has $k = 0.25$ W/m·K; Slab B has $k = 0.50$ W/m·K. Which slab has a higher center temperature, and by how much (as a factor)?
Explain why the Fourier number $\text{Fo} = \alpha t / L^2$ is dimensionless and what physical ratio it represents. What does a Fourier number much less than 1 tell you about the state of a polymer part cooling in a mold?
The heat equation in cylindrical coordinates has a $1/r$ factor in front of the radial derivative: $(1/r)\,d/dr(r\,dT/dr) = -\dot{q}/k$. A student claims this factor is just a mathematical artefact. Explain physically where it comes from and why it is essential to get the correct temperature distribution in a rod.
The cylindrical shell at radius $r$ and thickness $dr$ has area $2\pi r L$, which grows with $r$. What does that mean for how much heat crosses each shell face?
Part B: Calculation Problems
A PEEK heater block ($k = 0.25$ W/m·K, $L = 20$ mm) has uniform generation $\dot{q} = 2 \times 10^5$ W/m³. Its back face is insulated; its front face is cooled by air ($h = 200$ W/m²·K, $T_\infty = 25°C$). Find the maximum temperature in the block (°C).
Find $T_s$ from energy balance: $T_s = T_\infty + \dot{q}L/h$.
Add the interior rise: $T_{max} = T_s + \dot{q}L^2/(2k)$. Maximum is at the insulated face.
A 2 mm thick polymer film ($k = 0.30$ W/m·K) is symmetrically heated by a laser depositing $\dot{q} = 6 \times 10^6$ W/m³ throughout. Both surfaces are maintained at 60°C by cooling rolls. Find the temperature at the center of the film (°C).
Symmetric case: $T_{max} = T_s + \dot{q}L^2/(8k)$ where $L = 0.002$ m.
$T_{max} = 60 + (6\times10^6)(0.002)^2/(8 \times 0.30) = 60 + 10 = 70°C$.
An HDPE sheet ($\alpha = 1.5 \times 10^{-7}$ m²/s, thickness $L = 5$ mm) is suddenly exposed to a heat source. After $t = 30$ s of heating, compute the Fourier number $\text{Fo} = \alpha t / L^2$. (Give your answer to two decimal places.)
$\text{Fo} = \alpha t / L^2 = (1.5\times10^{-7})(30)/(0.005)^2$.
Numerator: $4.5\times10^{-6}$. Denominator: $2.5\times10^{-5}$. Fo = 0.18.
A polymer rod ($k = 0.30$ W/m·K, radius $R = 5$ mm) generates heat uniformly at $\dot{q} = 1 \times 10^6$ W/m³ (Joule heating from a resistive fiber). Air convects at $h = 150$ W/m²·K, $T_\infty = 30°C$. Find the surface temperature of the rod (°C). Round to the nearest degree.
$T_s = T_\infty + \dot{q}R/(2h)$. The $2h$ comes from the cylindrical energy balance (area per unit length $= 2\pi R$, volume per unit length $= \pi R^2$, ratio $= R/2$).
Using the same polymer rod as P4a, find the temperature at the center of the rod (°C). Round to the nearest degree.
$T_{center} = T_s + \dot{q}R^2/(4k)$. Use your answer from P4a as $T_s$.
$T_{center} = 47 + (10^6)(0.005)^2/(4\times0.30) = 47 + 20.8 \approx 68°C$.
A symmetric polymer slab has thickness $2L$ and uniform generation $\dot{q}$. Both outer surfaces experience convection with coefficient $h$ and ambient temperature $T_\infty$. Write the expression for the temperature excess at the center above $T_\infty$ (in terms of $\dot{q}$, $L$, $k$, and $h$). Then identify which term becomes dominant when $k$ is very small (typical for a polymer).
$T_{center} - T_\infty = \underbrace{\dot{q}L/h}_{\text{convective resistance}} + \underbrace{\dot{q}L^2/(2k)}_{\text{conductive resistance}}$. For polymers with small $k$, the second term dominates.
Chapter Summary
This chapter derived the governing equation for heat conduction from first principles. An energy balance on an infinitesimal volume element leads directly to Eq. (3.1), a single partial differential equation that governs temperature at every point in space and time. The central insight is that all conduction problems, regardless of geometry or complexity, obey this same equation. What distinguishes one problem from another is only the coordinate system, the value of $\dot{q}$, and the boundary conditions.
Thermal diffusivity $\alpha = k/(\rho c_p)$ determines how fast temperature disturbances propagate. Polymers have $\alpha$ roughly three orders of magnitude smaller than copper, which is why polymer parts heat and cool slowly and why temperature gradients inside polymer components are much steeper than in metal ones.
There are three standard boundary conditions: Type 1 (specified surface temperature), Type 2 (specified heat flux, with the adiabatic/symmetry case as the key special case), and Type 3 (convective coupling to an ambient fluid). Identifying the correct BC for each surface is often the most critical step in setting up a problem. The symmetry condition $\partial T/\partial x = 0$ is a Type 2 BC that halves the computational domain for symmetric problems.
For steady-state problems with uniform heat generation, the temperature profiles are parabolic. The surface temperature is set entirely by the energy balance between generation and convective loss; the thermal conductivity $k$ only affects how much hotter the interior runs relative to the surface. For low-$k$ polymers, this interior temperature rise can be substantial.
In the next chapter, we solve the heat equation analytically for a broad set of geometries: extended surfaces (fins), composite cylinders, and the critical radius of insulation, developing further the tools for practical polymer process design.