Heat Transfer in Polymer Extrusion
The extruder barrel is a multi-zone heat exchanger wrapped around a screw. Heat flows simultaneously from barrel heaters into the polymer melt and from the hot melt outward through the barrel wall to cooling jackets. The concepts of thermal resistance from Chapters 2 and 3 do all the work here, applied to the geometry and materials of real processing equipment.
Close your notes. Try each question from memory, then reveal the answer. The ones you cannot recall are exactly what this chapter reinforces.
RC-1. Write Fourier's law for steady 1-D conduction through a flat wall. Define every symbol and give its units.
RC-2. Two flat walls in series: wall 1 has resistance $R_1$ and wall 2 has resistance $R_2$. Write the total resistance and the heat flow rate if the outer temperatures are $T_A$ and $T_B$.
RC-3. What is the convective resistance $R_{\text{conv}}$? What physical quantity does Newton's law connect?
5.1 The Extrusion Barrel
Imagine pushing cold polymer pellets into one end of a heated steel tube and pulling out a smooth, molten ribbon from the other. That is extrusion in one sentence. The screw inside the barrel conveys, melts, and pressurises the polymer. The barrel wall is where the thermal action happens: heater bands on the outside push energy in during start-up, and water-cooled jackets pull energy out during steady operation, keeping each zone at its target temperature.
The central question is: how much heat can the barrel wall transfer, and what limits it? The answer is the same one you used in Chapter 2 for a house wall in winter, but applied to steel, a polymer melt, and water — three materials in series.
Before reading on — the governing equation for this chapter is $1/U = 1/h_i + L/k + 1/h_o$. Does that structure look familiar? It is identical to adding resistors in series in an electrical circuit: $R_{\text{total}} = R_1 + R_2 + R_3$. The analogy is exact — temperature difference plays the role of voltage, heat flow rate plays the role of current, and the largest resistance controls everything. You also saw this logic in chemical kinetics: when one step is much slower than the others, it sets the rate of the whole reaction. Same mathematics — three completely different physical contexts.
5.2 Barrel Zones and Thermal Goals
A single-screw extruder barrel is divided into three to five independently controlled temperature zones along its length. Each zone has its own heater band and thermocouple. Typical zone temperatures for HDPE:
| Zone | Location | Typical Setpoint (HDPE) | Purpose |
|---|---|---|---|
| Zone 1 (Feed) | Near hopper | 150–180 °C | Begin softening without premature melt |
| Zone 2 (Transition) | Middle | 180–210 °C | Complete melting |
| Zone 3 (Metering) | Near die | 200–230 °C | Homogenise melt, build pressure |
| Die zone | At die | 210–240 °C | Shape and surface quality |
During steady-state operation, viscous dissipation (friction between the screw, melt, and barrel) generates heat. The barrel cooling jackets must remove this heat to hold the zones at setpoint. The problem becomes: how much heat can a given barrel wall cross-section transfer per unit time?
5.3 The Overall Heat Transfer Coefficient $U$
When heat passes from a polymer melt through a barrel wall into a cooling jacket, it crosses three resistances in series:
- Melt-side convection from the bulk melt to the inner barrel surface: resistance $1/(h_i A)$
- Conduction through the steel barrel wall of thickness $L$ and conductivity $k$: resistance $L/(kA)$
- Coolant-side convection from the outer barrel surface to the cooling water: resistance $1/(h_o A)$
By the series resistance rule from Chapter 2:
$$\frac{1}{UA} = \frac{1}{h_i A} + \frac{L}{kA} + \frac{1}{h_o A}$$
Dividing through by area $A$ gives the overall heat transfer coefficient $U$ (units W/m²K):
$$\frac{1}{U} = \frac{1}{h_i} + \frac{L}{k} + \frac{1}{h_o}$$
The heat flow across the entire barrel wall is then:
$$\dot{Q} = U A \Delta T_{\text{overall}}$$
where $\Delta T_{\text{overall}} = T_{\text{melt}} - T_{\text{coolant}}$.
$U$ condenses all three resistances into a single coefficient that describes the overall ease of heat transfer. A small $U$ means the wall is thermally stiff; the melt cannot be cooled easily, regardless of how cold the water is.
Three resistances add in series; U is limited by the largest term
In polymer extrusion: melt side almost always dominates
5.4 Which Resistance Dominates?
The overall $U$ is always smaller than the smallest individual coefficient, just as the slowest link limits a supply chain. For polymer extrusion, the three terms compare very differently:
| Resistance | Typical value (m²K/W) | Physical reason |
|---|---|---|
| Melt-side: $1/h_i$ | 0.005–0.050 | Polymer melts are viscous and have low $k$; convection is weak |
| Steel wall: $L/k$ | 0.0002–0.002 | Steel $k \a\approx 16\text{--}50$ W/mK; wall is thin; resistance is tiny |
| Coolant-side: $1/h_o$ | 0.001–0.010 | Water at moderate velocity; $h_o \a\approx 200\text{--}1000$ W/m²K |
In practice, the melt-side resistance almost always dominates. Improving the barrel wall material from carbon steel to copper, for example, barely changes $U$ because the wall resistance was never the bottleneck. A better return on investment is higher screw speed (more melt mixing, higher $h_i$) or a screw design with better melt circulation.
5.5 The Melt Energy Balance
Once you know $U$, you can compute the heat transferred per unit time. To link that to production rate, you also need the energy balance on the melt itself. If the polymer enters a barrel zone at temperature $T_1$ and exits at $T_2$:
$$\dot{Q} = \dot{m}\,c_p\,(T_2 - T_1)$$
where $\dot{m}$ is the mass flow rate (kg/s) and $c_p$ is the specific heat capacity of the polymer melt (J/kgK). This equation connects heat transfer engineering to process throughput.
Typical polymer melt properties
- HDPE melt: $c_p \a\approx 2300\ \text{J/kgK}$, $k \a\approx 0.44\ \text{W/mK}$
- PLA melt: $c_p \a\approx 1800\ \text{J/kgK}$, $k \a\approx 0.13\ \text{W/mK}$
- ABS melt: $c_p \a\approx 1400\ \text{J/kgK}$, $k \a\approx 0.17\ \text{W/mK}$
- Nylon 6 melt: $c_p \a\approx 2000\ \text{J/kgK}$, $k \a\approx 0.25\ \text{W/mK}$
The low $k$ values mean polymer melts are poor thermal conductors — heat transport within the melt depends heavily on convective mixing, not conduction.
5.6 Die and Post-Die Cooling
After the barrel, the melt enters the die where it is shaped. The die body is also heated to prevent premature solidification. Once the extrudate exits the die, it must be cooled rapidly to lock in its shape. Three post-die cooling methods are common in polymer processing:
Water bath cooling (most common for wire, pipe, and sheet): $h \a\approx 500\text{--}3000\ \text{W/m}^2\text{K}$. Very effective; the high $h$ of water makes this the fastest option.
Air quench (film extrusion, blow moulding): $h \a\approx 20\text{--}80\ \text{W/m}^2\text{K}$. Gentler, gives more control over crystallinity. The low $h$ means longer cooling times.
Chill roll contact (cast film, sheet): $h_{\text{contact}} \a\approx 200\text{--}1000\ \text{W/m}^2\text{K}$. The good metal-to-metal contact provides effective cooling with precise surface finish control.
In each case, the same $U$ framework applies, with the appropriate $h$ values for the contact medium. The polymer thermal conductivity appears as the internal conduction resistance when the extrudate is thick.
5.7 Interactive: Barrel Wall Resistance Explorer
Adjust the three thermal resistances to see how they combine into $U$ and what fraction each contributes to the total. Notice how the overall $U$ is always dominated by the largest resistance.
5.8 Worked Examples
Example 5.1 — Overall $U$ for an HDPE Extruder Barrel
An HDPE extruder barrel has a steel wall 10 mm thick ($k = 10\ \text{W/mK}$). The melt-side convective coefficient is $h_i = 200\ \text{W/m}^2\text{K}$ and the cooling-water side is $h_o = 250\ \text{W/m}^2\text{K}$. Compute $U$.
Solution:
$$\frac{1}{U} = \frac{1}{h_i} + \frac{L}{k} + \frac{1}{h_o} = \frac{1}{200} + \frac{0.010}{10} + \frac{1}{250}$$ $$= 0.0050 + 0.0010 + 0.0040 = 0.0100\ \text{m}^2\text{K/W}$$ $$U = 100\ \text{W/m}^2\text{K}$$
Note that the wall resistance (0.0010) is only 10% of the total. Even replacing the steel with copper ($k = 400\ \text{W/mK}$) would change the wall term from 0.0010 to 0.000025 — negligible impact on $U$.
Example 5.2 — Throughput from an Energy Balance
A cooling zone (barrel area $A = 0.5\ \text{m}^2$) operates with $U = 100\ \text{W/m}^2\text{K}$, melt temperature 180 °C, and coolant temperature 80 °C. The melt (HDPE, $c_p = 2000\ \text{J/kgK}$) drops 25 °C as it passes through this zone. What is the mass throughput $\dot{m}$?
Solution:
Heat removed by the barrel wall: $$\dot{Q} = UA\Delta T = 100 \times 0.5 \times (180 - 80) = 5000\ \text{W}$$
Energy balance on the melt: $$\dot{Q} = \dot{m}\,c_p\,\Delta T_{\text{melt}} \Rightarrow \dot{m} = \frac{5000}{2000 \times 25} = 0.10\ \text{kg/s}$$
The extruder delivers $0.10\ \text{kg/s} = 360\ \text{kg/h}$ of HDPE through this zone.
5.9 Concept Questions
Part A — Conceptual
An engineer proposes replacing the carbon steel barrel ($k = 50\ \text{W/mK}$) with a copper barrel ($k = 400\ \text{W/mK}$) to improve cooling. The wall is 15 mm thick, $h_i = 150\ \text{W/m}^2\text{K}$, $h_o = 300\ \text{W/m}^2\text{K}$. Will this change make a significant difference? Calculate $U$ for both cases.
Why does a polymer melt have a much lower convective coefficient $h_i$ than water flowing at the same velocity? Name two material properties responsible.
An operator increases the cooling water flow rate by 50%, which aapproximately doubles $h_o$ from 300 to 600 W/m²K. The melt-side coefficient is $h_i = 80\ \text{W/m}^2\text{K}$ and the wall resistance is negligible. By what percentage does $U$ increase?
Explain why rapid quenching of a semi-crystalline polymer (like HDPE) in a water bath produces a different microstructure than slow air cooling. Connect your answer to the $U$ and $\dot{Q}$ formulas.
Derive the dominant-resistance approximation algebraically and apply it.
(a) Starting from $1/U = R_i + R_w + R_o$ (where $R_i = 1/h_i$, etc.), factor out $R_i$ to write $U = h_i / (1 + (R_w + R_o)/R_i)$. Define the ratio $ \varepsilon = (R_w + R_o)/R_i$ and show that when $ \varepsilon \ll 1$, $U \approx h_i\,(1 - \varepsilon)$. What does this tell you about the marginal benefit of reducing a small resistance?
(b) For the system: $h_i = 100\ \text{W/m}^2\text{K}$, $L/k = 0.001\ \text{m}^2\text{K/W}$, $h_o = 400\ \text{W/m}^2\text{K}$. Calculate $U$, then compute $ \varepsilon$. Is the dominant-resistance approximation valid here?
(c) An engineer doubles $h_o$ from 400 to 800 W/m²K (50% more pump energy). By what absolute and relative amount does $U$ increase? Then doubles $h_i$ from 100 to 200 W/m²K (new screw design, same pump energy). Compare the two improvements. Which investment was more effective, and why?
(d) At what value of $h_o$ (keeping $h_i = 100$ and $L/k = 0.001$) would the coolant resistance become negligible (say, contributing less than 5% of total resistance)? Show your calculation.
5.10 Problem Set
Part B — Calculations
An extruder barrel wall has the following properties: melt-side coefficient $h_i = 200\ \text{W/m}^2\text{K}$; wall thickness $L = 10\ \text{mm}$; wall conductivity $k = 10\ \text{W/mK}$; coolant-side coefficient $h_o = 250\ \text{W/m}^2\text{K}$. Compute the overall heat transfer coefficient $U$ in W/m²K.
$1/U = 1/h_i + L/k + 1/h_o$
Three thermal resistances add in series — the largest one limits U, just as the slowest step limits a reaction rate.
A barrel cooling zone has area $A = 0.5\ \text{m}^2$ and overall coefficient $U = 100\ \text{W/m}^2\text{K}$. The melt temperature in this zone is 180 °C and the coolant temperature is 80 °C. How much heat (in W) is removed from this zone?
$\dot{Q} = U A \Delta T$
U × area × overall temperature gap = heat flow rate. All three factors are independently adjustable in process design.
The barrel zone in P2 removes $\dot{Q} = 5000\ \text{W}$ from the melt. The polymer is a nylon melt with $c_p = 2000\ \text{J/kgK}$. The melt temperature drops by 25 °C as it passes through this zone. What is the polymer mass flow rate $\dot{m}$ in kg/s?
$\dot{Q} = \dot{m}\,c_p\,\Delta T$, rearranged for $\dot{m}$
Energy balance: heat removed = mass flow × heat capacity × temperature drop. This links the thermal calculation directly to production throughput.
A barrel wall has: melt-side $h_i = 50\ \text{W/m}^2\text{K}$ (viscous melt, poor mixing); wall $L = 20\ \text{mm}$, $k = 20\ \text{W/mK}$; coolant-side $h_o = 200\ \text{W/m}^2\text{K}$. Compute each individual resistance per unit area (m²K/W) and state which one dominates. What percentage of the total resistance does it represent?
$R_i = 1/h_i$, $R_w = L/k$, $R_o = 1/h_o$, % = 100 R_i / R_{\text{total}}$
Compute each resistance, sum them, and express each as a percentage — the biggest % tells you where to invest engineering effort.
A barrel zone must remove $\dot{Q} = 6000\ \text{W}$ of viscous heat. The coolant is water with $c_p = 4000\ \text{J/kgK}$, and the maximum allowable coolant temperature rise is $\Delta T_{\text{coolant}} = 5\,°\text{C}$ (to keep the coolant temperature reasonably uniform). What mass flow rate of cooling water is required (kg/s)?
$\dot{Q} = \dot{m}_{\text{water}}\,c_p\,\Delta T$, rearranged for $\dot{m}$
Same energy balance as P3, now applied to the coolant. The tighter the allowed temperature rise, the more water you need to push through.
In your unit operations lab, a single-screw extruder running HDPE at 200 kg/h is supposed to maintain Zone 3 at 220 °C. The temperature controller shows the zone is consistently running 15 °C above setpoint even with the cooling jackets fully open.
(a) List the three resistances in the cooling circuit and, without calculating, identify which one you suspect is most likely limiting the cooling. Justify your reasoning from the physical properties of HDPE melt vs. water.
(b) The cooling water inlet is 20 °C and the outlet temperature is measured at 45 °C. The water flow rate is 8 L/min ($\dot{m} = 0.133\ \text{kg/s}$, $c_p = 4200\ \text{J/kgK}$). Compute the actual heat removal rate $\dot{Q}_{\text{water}}$.
(c) Estimate the expected $\dot{Q}$ from viscous dissipation: HDPE melt specific heat $c_p = 2300\ \text{J/kgK}$, throughput $\dot{m} = 0.056\ \text{kg/s}$ (200 kg/h), melt temperature excess $\Delta T_{\text{melt}} = 15\,°\text{C}$. Compare with your answer to (b) and diagnose the problem.
(d) Propose two specific, actionable changes to fix the overheating. For each, state which resistance in the $1/U$ equation you are targeting.
5.11 Chapter Summary
Key Equations
| Quantity | Formula | Notes |
|---|---|---|
| Overall heat transfer coefficient | $1/U = 1/h_i + L/k + 1/h_o$ | Flat geometry; add more terms for more layers |
| Heat flow | $\dot{Q} = UA\Delta T_{\text{overall}}$ | $\Delta T$ from melt to coolant bulk |
| Melt energy balance | $\dot{Q} = \dot{m}\,c_p\,\Delta T_{\text{melt}}$ | Links heat transfer to throughput |
| Dominant resistance | $R_{\text{dominant}} = \max(1/h_i,\ L/k,\ 1/h_o)$ | Controls $U$; target this one first |
Key Ideas
- The overall heat transfer coefficient $U$ combines all resistances in series: melt convection, wall conduction, and coolant convection.
- In polymer extrusion, the melt-side resistance always dominates because polymer melts have low $k$ and high viscosity.
- Improving the wall material (carbon steel → copper) barely changes $U$ when the melt side dominates. Target the largest resistance.
- The melt energy balance $\dot{Q} = \dot{m}\,c_p\,\Delta T$ connects heat transfer engineering to production throughput.
- Post-die cooling method determines cooling rate: water bath $\gg$ chill roll $\gg$ air quench; this controls polymer crystallinity and final properties.