Introduction to Heat Transfer
Heat moves. Understanding how it moves — and how fast — is what heat transfer is about. This chapter introduces the three mechanisms and the rate equation for each.
The 3D Printer on Your Desk
Watch a 3D printer make a part. The metal nozzle is at 220 °C. Cold filament enters from above. Molten plastic exits below and lands on a platform held at 60 °C. The part builds up one 0.2 mm layer at a time.
Three things are happening simultaneously. Heat flows through the nozzle metal from heater to plastic — that is conduction. The deposited plastic loses heat to the air around it — that is convection. The hot bed radiates energy toward the cooler room walls — that is radiation.
Every layer's quality depends on the balance of these three flows. Too much cooling and the layer warps before the next one bonds. Too little and the part slumps. This is not a thermodynamics problem. Thermodynamics tells you the plastic will eventually cool to room temperature. It says nothing about how fast, or where the hot spots are, or whether layer 3 has cooled enough before layer 4 arrives.
Fig. 1.1 — All three modes of heat transfer occur simultaneously in a 3D printer.
1.1 Heat Transfer vs. Thermodynamics
Thermodynamics answers: how much? and in which direction? It says a hot object will cool and energy will flow from hot to cold. That is the Second Law.
Heat transfer answers: how fast? It provides rate equations — laws that connect the speed of energy flow to temperature difference, geometry, and material properties.
Think of it this way. Thermodynamics says your coffee will cool to room temperature. Heat transfer tells you it will take about 20 minutes in a ceramic mug, or 5 minutes if you pour it into a thin aluminium cup in a breeze.
Semi-crystalline polymers like nylon and PET crystallise at a rate that depends strongly on cooling speed. Cool too fast and you get an amorphous, transparent film. Cool slowly and you get a semi-crystalline, opaque one with higher stiffness. Thermodynamics tells you the equilibrium crystal structure. Heat transfer tells you whether you will actually reach it before the polymer freezes.
1.2 Conduction
Touch the metal nozzle of your 3D printer and you burn your finger. Touch the surrounding plastic housing — same temperature — and it barely feels warm. Both are at the same temperature. The difference is not temperature but thermal conductivity.
Conduction is heat transfer through a stationary material. At the molecular level, energetic atoms pass vibrations to their neighbours — like a row of billiard balls. No bulk movement of matter occurs. Only energy moves.
Conduction occurs in solids, liquids, and gases. In solids, lattice vibrations and free electrons carry energy. In fluids, molecular collisions do the work.
Most polymers have thermal conductivity of 0.1 to 0.5 W/m·K. Aluminium is 205 W/m·K. This means heat conducts through a polymer about 400 to 2000 times more slowly than through aluminium. A 10 mm thick polymer part takes much longer to cool at its core than a metal part of the same thickness. That delay causes warping and residual stress.
Fourier's Law
Jean-Baptiste Fourier showed in 1822 that the rate of heat flow through a slab is proportional to the area and to the temperature gradient. For heat flow in the $x$-direction:
Reporting heat in watts tells you the total power flowing through a surface, but that number depends on how large the surface is. A bedroom wall and a postage stamp could both carry 10 W while behaving very differently at the surface level. To compare materials on equal footing, we divide by area. The result is the heat flux, $q$: the rate of heat flow per unit area, regardless of the total surface size.
The minus sign is not arbitrary. Heat flows from hot to cold — in the direction of decreasing temperature. If $T$ decreases as $x$ increases, then $dT/dx$ is negative, making $q$ positive. Without the minus sign, the equation would say heat flows toward higher temperature, which is wrong.
Fig. 1.2 — Fourier's law. Heat flows from high temperature (left) to low (right) through a slab of thickness L.
Imagine a thin slab of thickness $\Delta x$ and cross-sectional area $A$, sitting inside a larger wall. At steady state, temperature does not change with time, so no energy can be accumulating inside the slab. Whatever heat enters through the left face must leave through the right face:
$$A \cdot q_x \big|_x \;=\; A \cdot q_x \big|_{x+\Delta x}$$Rearrange by bringing both terms to one side:
$$A \cdot q_x \big|_{x+\Delta x} \;-\; A \cdot q_x \big|_x \;=\; 0$$Now divide every term by $A \cdot \Delta x$. The area $A$ cancels because it appears in every term. Dividing by $\Delta x$ converts the difference in flux across the slab into a rate of change per unit length. Taking the limit as the slab thickness $\Delta x \to 0$ turns that ratio into a derivative:
$$\lim_{\Delta x \to 0}\frac{q_x\big|_{x+\Delta x} - q_x\big|_x}{\Delta x} \;=\; \frac{dq_x}{dx} \;=\; 0$$This says $q_x$ is constant across the slab — the same flux enters and leaves every thin slice. Experiments then show that this flux is proportional to the local temperature gradient:
$$q_x = -k\frac{dT}{dx}$$That is Fourier's law. In Chapter 2, we apply exactly this pattern — write an energy balance on a thin shell, divide by its volume, take the limit — in cylindrical and spherical geometries. The shell balance is the core derivation tool of this subject.
Heat flux through a 3D printer nozzle tip
A brass nozzle tip ($k = 109$ W/m·K) is 8 mm long. The heater block end is at 220 °C and the exit end is at 185 °C. Find the heat flux through the tip.
Solution
The temperature gradient (linear for steady state, constant $k$):
$$\frac{dT}{dx} = \frac{185 - 220}{0.008} = \frac{-35}{0.008} = -4375 \text{ K/m}$$Applying Eq. 1.2:
$$q = -k\frac{dT}{dx} = -(109)(-4375) = 476{,}875 \text{ W/m}^2 \approx 477 \text{ kW/m}^2$$That is nearly half a megawatt per square metre — a large flux. But the tip area is tiny ($\approx 5 \times 10^{-7}$ m²), so the total power is only about 0.24 W through the tip itself. This illustrates a key point: high flux does not necessarily mean large total power.
Let $x=0$ at the heater end (220 °C) and $x = L = 0.008$ m at the exit (185 °C).
For a slab in steady state with constant $k$, $T(x)$ is linear: slope = $(T_2 - T_1)/L$.
1.3 Convection
Blow on a hot spoonful of soup to cool it faster. You are increasing convection. Moving fluid carries thermal energy away from a surface much more effectively than still air does.
Convection occurs when a fluid moves over a surface at a different temperature. The fluid closest to the surface picks up (or delivers) energy, then moves away, replaced by cooler (or hotter) fluid. Energy transport and mass transport are coupled.
Two types:
- Natural convection — density differences due to temperature drive the flow. No pump or fan needed.
- Forced convection — an external device drives the flow. Fans, pumps, atmospheric wind.
Fig. 1.3 — In convection, fluid near the surface picks up heat and is swept away. Velocity is zero at the surface and rises to the bulk value far away.
Newton's Law of Cooling
The rate equation for convection is Newton's law of cooling. It says the heat loss from a surface is proportional to the temperature difference between surface and fluid:
The coefficient $h$ is not a material property. It depends on fluid velocity, geometry, fluid properties, and whether the flow is laminar or turbulent. Much of the later chapters on convection are about finding $h$ for different situations.
| Situation | h (W/m²·K) |
|---|---|
| Air, natural convection | 6 – 30 |
| Air, forced convection | 10 – 200 |
| Water, natural convection | 500 – 1000 |
| Water, forced convection | 600 – 8000 |
| Boiling water | 2500 – 100,000 |
The layer-by-layer temperature in FDM printing is governed largely by convection to ambient air ($h \approx 10$–25 W/m²·K). Some printers use an enclosure with elevated temperature to reduce this cooling — particularly for ABS, which warps if it cools too quickly. A heated enclosure cuts the temperature difference $(T_s - T_f)$ without changing $h$.
1.4 Radiation
Hold your hand near — not touching — a hot toaster. You feel warmth even in still air. That is radiation: electromagnetic energy emitted by every object above absolute zero, travelling at the speed of light, requiring no medium.
The portion of the electromagnetic spectrum between 0.1 μm and 100 μm — from near-ultraviolet to far-infrared — is called thermal radiation. When absorbed by a surface, it converts to heat. At room temperature, objects emit mostly in the infrared. At 1000 °C, some emission shifts into the visible range, which is why hot metal glows.
Fig. 1.4 — Net radiation exchange between two surfaces. Both surfaces emit; the net flow is from hot to cold.
Stefan–Boltzmann Law
The maximum radiation a surface can emit is set by its temperature. A perfect emitter (called a black body) obeys the Stefan–Boltzmann law:
Real surfaces emit less. Their emissivity $\varepsilon$ (0 to 1) scales the emission:
For a small body (area $A_1$, emissivity $\varepsilon_1$, temperature $T_1$) inside a large enclosure at temperature $T_2$:
The heated print bed (typically 60–110 °C) radiates toward the cooler room walls. At 60 °C (333 K) with $\varepsilon \approx 0.9$: $E = 0.9 \times 5.67 \times 10^{-8} \times 333^4 \approx 590$ W/m². Over a 0.04 m² bed, that is about 24 W of radiated power — comparable to the convective loss. Designers of enclosed printers must account for this.
1.5 When All Three Act Together
In practice, two or three modes operate simultaneously. The 3D printer is the proof. The approach is always the same: write an energy balance and include every relevant term.
Combined convection and radiation
A PLA part ($\varepsilon = 0.9$) cools on a bench. The part surface is at 65 °C. The room is at 25 °C. The convective coefficient is $h = 12$ W/m²·K. Find the total heat flux leaving 1 m² of the part surface.
Solution
Convective flux:
$$q_\text{conv} = h(T_s - T_\infty) = 12(65-25) = 480 \text{ W/m}^2$$Radiation flux ($T_s = 338$ K, $T_\infty = 298$ K):
$$q_\text{rad} = \varepsilon\sigma(T_s^4 - T_\infty^4) = 0.9 \times 5.67\times10^{-8}(338^4 - 298^4) = 215 \text{ W/m}^2$$Total:
$$q_\text{total} = 480 + 215 = 695 \text{ W/m}^2$$Radiation contributes 31% of the total. Ignoring it would under-estimate heat loss by almost a third — a significant error if you are designing cooling time for a batch process.
Concept Check
Each question below has a pool of related variants. If you answer wrong, a new version appears. After the pool is exhausted, the worked solution is shown.
A hot PLA part sits on a bench in still air. Which mode(s) of heat transfer are active?
A metal rod is held with one end in a flame and the other in open air. No fluid flows over the rod. Which mode carries heat along the length of the rod?
Heat from the Sun reaches Earth across 150 million km of space. Which mode is this?
The three modes: Conduction — through stationary matter (solid or fluid at rest). Convection — requires a moving fluid. Radiation — electromagnetic energy, requires no medium. The key question: is there bulk fluid motion? If yes, convection is active. Is there a solid or stationary path? Conduction. Is either surface above absolute zero? Radiation.
In Fourier's law $q = -k\,dT/dx$, why is there a minus sign?
A slab has $T = 80$ °C at $x = 0$ and $T = 20$ °C at $x = L$. The temperature gradient $dT/dx$ is:
Which of these is the correct expression for heat flux through a flat slab of thickness $L$ and conductivity $k$, with temperatures $T_1$ (hot) and $T_2$ (cold)?
The minus sign in $q = -k\,dT/dx$ ensures that heat flux is positive when heat flows in the +x direction (where T is decreasing in +x). For a slab in steady state, $dT/dx = (T_2 - T_1)/L$, and $q = -k(T_2-T_1)/L = k(T_1-T_2)/L$. Since $T_1 > T_2$, $q > 0$.
A PLA part is cooled in a water bath rather than still air. The main reason water cools the part much faster is:
Doubling the temperature difference $(T_s - T_f)$ in Newton's law of cooling (while keeping $h$ and $A$ fixed) will:
$Q = hA(T_s - T_f)$. Both $h$ and $(T_s - T_f)$ matter. In practice, changing the coolant (air → water) changes $h$ by 50–200×, which swamps any practical change in temperature difference. That is why cooling medium matters far more than cooling temperature in most industrial processes.
A surface at 200 °C emits more radiation than one at 100 °C. The ratio of their emissive powers (assuming equal emissivity) is closest to:
Why must you use temperature in Kelvin (not Celsius) in the Stefan–Boltzmann law?
Radiation emission is $E = \varepsilon\sigma T^4$. The $T^4$ dependence means radiation grows rapidly with temperature — doubling T (in Kelvin) increases emission by $2^4 = 16$. This is why radiation dominates at high temperatures (furnaces, re-entry vehicles) but is minor at room temperature. Always use Kelvin.
You want to know how long it takes a 10 mm thick nylon part to cool to 40 °C after molding. This is a problem in:
The Second Law of Thermodynamics says heat flows from hot to cold. What additional information does Fourier's law provide?
Thermodynamics: direction and totals. Heat transfer: rates and distributions. The key addition of heat transfer is the rate equation — Fourier's law says $q = -k\,dT/dx$, connecting flux to gradient. Without this, you cannot calculate cooling time, temperature profiles, or wall thickness requirements.
Problem Set — Chapter 1
Part A — Short Answer
The metal frame of a 3D printer feels much colder than the plastic enclosure panels, even though both are at room temperature. Explain why using only one term from this chapter.
Your finger is at 37 °C. Both surfaces are at ~22 °C. The sensation of "cold" is energy leaving your finger. Which property of the material controls how fast that energy leaves?
A FDM printer uses ABS filament, which must stay above its glass transition temperature (~105 °C) during printing to prevent delamination. The manufacturer recommends an enclosed build chamber at 70–90 °C. Explain how the enclosure affects each of the three modes of heat transfer from the printed part.
Bird, Stewart, and Lightfoot (BSL) begin with a shell momentum balance to derive Newton's law of viscosity. In this textbook, we state Fourier's law first, then show its shell balance derivation in a collapsible box. What is the argument for each approach? Which do you think is better for your learning style, and why?
A student says: "I can make my 3D-printed part cool faster by just reducing the bed temperature from 60 °C to 30 °C." Using Newton's law of cooling and the Stefan-Boltzmann law, evaluate this claim. Under what conditions is the student correct? What other changes would be more effective?
In a polymer nanopore sequencing device, a single DNA or polymer chain threads through a ~2 nm pore in a membrane. An electric current drives the process, generating Joule heat in the pore at a rate of about 10 pW. Identify each relevant mode of heat transfer, state the "hot body" and "cold body" for each, and justify which mode dominates at this scale.
Part B — Calculations
A HDPE slab ($k = 0.44$ W/m·K) is 25 mm thick. Hot side: 200 °C. Cold side: 30 °C. Find the steady-state heat flux (W/m²).
Use $q = k(T_1-T_2)/L$. Plug in $k=0.44$, $T_1=200$, $T_2=30$, $L=0.025$ m.
A printed PLA part ($\varepsilon = 0.92$) has a surface temperature of 80 °C. Room walls are at 22 °C. Find the radiation heat flux leaving the surface (W/m²).
$T_s = 80+273 = 353$ K, $T_{room} = 22+273 = 295$ K. Then use Eq. 1.6 per unit area.
$q = \varepsilon\sigma(T_s^4 - T_{room}^4) = 0.92 \times 5.67\times10^{-8} \times (353^4 - 295^4)$
A 3D printed part loses heat from a 0.08 m² surface. The part is at 70 °C, air is at 25 °C, and $h = 18$ W/m²·K. How many watts does the part lose by convection?
A nylon part ($\varepsilon = 0.88$) at 120 °C is surrounded by air at 30 °C with $h = 15$ W/m²·K. What fraction (%) of the total heat flux leaving the surface is due to radiation?
$q_{conv} = h(T_s-T_f) = 15(120-30) = 1350$ W/m². For radiation: $T_s=393$ K, $T_f=303$ K.
Fraction = $q_{rad}/(q_{conv}+q_{rad}) \times 100$%.
A 3D printer insulation panel (polystyrene foam, $k = 0.033$ W/m·K) separates an enclosure at 80 °C from outside air at 22 °C. The total allowable heat loss through the panel is 30 W/m². What is the minimum panel thickness (mm)?
$q = k\Delta T/L \leq 30$ W/m². Solve for $L \geq k\Delta T/30 = 0.033 \times (80-22)/30$. Convert to mm.
Chapter 1 Summary
Key Ideas
- Heat transfer answers "how fast" — thermodynamics gives direction and totals.
- Three modes: conduction (no bulk motion), convection (moving fluid), radiation (electromagnetic, no medium needed).
- Polymers conduct poorly (k ≈ 0.1–0.5 W/m·K). Heat transfer rates control processing outcomes — crystallinity, warping, residual stress.
- Water vs. air: both are convection. Water wins because h is 50–200× higher.
- Radiation scales as T⁴ — minor at room temperature, dominant at high temperatures. Always use Kelvin.
- Shell balance is the core derivation tool — it will reappear in every geometry in Chapter 2 onward.
Key Equations
- Fourier (flux): $q = -k\,dT/dx$
- Fourier (slab): $q = k(T_1-T_2)/L$
- Newton's cooling: $Q = hA(T_s-T_f)$
- Black body: $E_b = \sigma T^4$
- Real surface: $E = \varepsilon\sigma T^4$
- Net radiation: $Q = A_1\varepsilon_1\sigma(T_1^4-T_2^4)$
Next — Chapter 2
We derive the general heat conduction equation using a shell balance, then solve it for flat slabs, cylinders, and spheres. Thermal resistance — the thermal analogue of electrical resistance — becomes the main tool for multi-layer problems.