Chapter 6

Heat Transfer in Injection Molding

Injection molding is one of the highest-volume polymer manufacturing processes in the world. Of the total cycle time, roughly 70–80% is spent cooling the part to a temperature where it can be ejected without warping. Heat transfer analysis is therefore the single most important lever for improving throughput. This chapter develops the tools to quantify it.

✍

Cold Recall — Before You Read

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 the expression for $T_{\max}$ inside a slab of thickness $2L$ with uniform internal heat generation $\dot{q}$ (W/m³), cooled symmetrically on both faces to temperature $T_w$. What does this expression tell you about which material properties matter most?

RC-2. In Chapter 5 you built the formula $1/U = 1/h_i + L/k + 1/h_o$. Identify what each of the three terms represents physically. What determines which term dominates?

RC-3. State Newton's law of cooling. What are the units of the convective coefficient $h$? Name one practical way to increase $h$ in a mold cooling channel.

6.1 The Cycle Time Problem

A nylon bracket injection mold runs at 4 shots per minute in your factory. Your competitor runs the same part at 6 shots per minute. Their cost per part is 33% lower, and the difference is almost entirely in cycle time — specifically, in how fast they cool the mold.

Injection molding cycle time has four phases: mold close and injection (∼5% of cycle), packing and hold (∼15%), cooling (∼70%), and mold open/part eject (∼10%). If you want to speed up the process, cooling is the only phase large enough to make a significant difference. That is a heat transfer problem.

This chapter develops the tools to answer: how much heat must be removed from the mold per shot, how fast can the mold wall remove it, and which material properties determine whether the cycle time is 5 seconds or 30 seconds.

Pattern Check

Before reading on — the key result of this chapter is $t_{\text{cool}} \propto s^2/\alpha$. Does that ratio look familiar? From Chapter 8 you know the Fourier number $Fo = \alpha\,t/s^2$: it is the dimensionless time for transient conduction in a slab. This chapter just inverts it: to find how long cooling takes, solve $Fo \sim 1$ for $t$, giving $t \sim s^2/\alpha$. The $s^2$ is the universal signature of diffusion — it appears in heat conduction, mass diffusion, and even random walks. Halve the wall thickness, quarter the cooling time. That scaling is worth memorising cold.

6.2 The Injection Molding Cycle

Molten polymer enters the cold mold cavity under high pressure (typically 50–200 MPa). The polymer begins to solidify immediately on contact with the cold mold wall. A frozen skin forms and thickens inward as the hot core continues to transfer heat outward through the solidified layer and into the mold steel.

The part can be ejected only when the entire cross-section has cooled below the ejection temperature $T_e$ — the temperature at which the polymer is stiff enough not to distort under ejection force. Typical values:

Polymer	Melt injection temp (°C)	Ejection temp (°C)	$\Delta T$ removed (°C)
Nylon 6/6	280	80	200
HDPE	220	60	160
ABS	240	75	165
PLA	200	60	140
Polycarbonate	300	120	180

Notice that a large $\Delta T$ must be removed each shot. For a 200-g nylon part with $c_p = 2000\ \text{J/kgK}$: the sensible heat alone is $Q = 0.2 \times 2000 \times 200 = 80\ \text{kJ}$ per shot. For semi-crystalline polymers (HDPE, nylon), the latent heat of crystallisation adds another 10–30% on top.

6.3 Heat Content of the Shot

The total heat to be removed from one part (one shot) is:

$$Q_{\text{shot}} = m\,c_p\,(T_{\text{melt}} - T_e)$$

For semi-crystalline polymers, add the latent heat of crystallisation $H_c$ (J/kg):

$$Q_{\text{shot}} = m\bigl[c_p(T_{\text{melt}} - T_e) + H_c\bigr]$$

Typical $H_c$ values: HDPE 180 kJ/kg; nylon 6/6 80 kJ/kg; PLA 40 kJ/kg; ABS 0 (amorphous, no latent heat).

This heat must all pass through the mold wall and into the cooling channels before the part can be ejected. $Q_{\text{shot}}$ sets the minimum work the cooling system must do per cycle.

Where does the heat come from? All of it was put in upstream: the extruder barrel heaters melted the pellets, stored sensible and latent heat in the melt, and injected that thermal energy into the mold cavity. The mold cooling system's job is to extract that stored energy fast enough to meet the cycle time target.

6.4 Mold Cooling: The Overall Heat Transfer Coefficient

Heat flows from the solidifying part through the mold steel and into the cooling channel coolant. Using the series resistance model from Chapter 5:

$$\frac{1}{U} = \frac{L_{\text{mold}}}{k_{\text{mold}}} + \frac{1}{h_{\text{coolant}}}$$

Here the contact resistance between the solidified part and the mold wall is often small and is neglected in a first analysis. $L_{\text{mold}}$ is the distance from the cavity surface to the cooling channel, and $k_{\text{mold}}$ is the mold material thermal conductivity.

Mold material	$k$ (W/mK)	Relative cooling speed
Tool steel (P20)	30–37	Standard
Stainless steel 316	16	Slower (−40%)
Aluminium 7075	130–150	Fast (×4)
Beryllium copper	100–130	Fast (×3)
Copper alloy (MoldMAX)	200	Fastest (×5)

In injection molding, the mold wall conductive resistance is more significant than in the extruder barrel because the distance to the cooling channel is typically longer (10–30 mm) and the contact area matters enormously. Channel placement is a critical design parameter.

Coolant-side $h$ for injection mold channels: Water flowing in a 10 mm diameter channel at 2 m/s is turbulent ($Re \approx 20{,}000$) and gives $h \approx 3000\text{--}8000\ \text{W/m}^2\text{K}$. If flow is too slow (laminar), $h$ drops by 10–50$\times$. Always verify turbulent flow in mold channels.

6.5 Estimating Cooling Time

A simple but useful estimate of minimum cooling time comes from dividing the total heat stored in the part by the maximum rate at which the mold can extract it:

$$t_{\text{cool}} \approx \frac{Q_{\text{shot}}}{\dot{Q}_{\text{max}}} = \frac{m\,c_p\,(T_{\text{melt}} - T_e)}{U\,A\,(T_{\text{melt}} - T_{\text{coolant}})}$$

This is an optimistic lower bound because it assumes the full $\Delta T$ is available throughout cooling. In practice, as the part cools, $\Delta T$ decreases and the heat extraction rate falls. The true cooling time is 1.5–2$\times$ this estimate for most parts.

A more rigorous approach uses the transient conduction solution (Chapter 8) for a slab of half-thickness $s$ cooled on both faces:

$$t_{\text{cool}} = \frac{s^2}{\pi^2\,\alpha}\ln\!\left(\frac{4}{\pi} \cdot\frac{T_{\text{melt}} - T_{\text{mold}}}{T_e - T_{\text{mold}}}\right)$$

where $\alpha = k/(\rho c_p)$ is the thermal diffusivity of the polymer and $T_{\text{mold}}$ is the mold wall temperature. This formula (Ballman–Shusman approximation) shows that cooling time scales as $s^2 / \alpha$. Halving the wall thickness cuts cooling time by 4$\times$ — wall thickness is by far the strongest lever available to the part designer.

Design rule of thumb: In injection molding, cooling time scales as (wall thickness)². A 3 mm wall takes 4$\times$ as long to cool as a 1.5 mm wall. This is why thin-wall injection molding is such a powerful cycle-time reduction strategy.

Exact formula (Ballman–Shusman)

$$t_c = \frac{s^2}{\pi^2\,\alpha}\ln\!\left(\frac{4}{\pi}\cdot\frac{T_m - T_w}{T_e - T_w}\right)$$

Log factor accounts for temperature ratio; typically 1.5–3 for real processes

Design scaling rule

$$t_c \propto \frac{s^2}{\alpha} \quad \Leftrightarrow \quad Fo = \frac{\alpha\,t_c}{s^2} \approx \text{const}$$

$s^2$ is the diffusion signature — halve $s$, quarter $t$

6.6 Thermal Diffusivity and Why It Controls Cycle Time

The cooling time formula contains $\alpha = k / (\rho c_p)$, the thermal diffusivity (m²/s). Higher $\alpha$ means faster heat penetration through the part, hence faster cooling. Thermal diffusivity combines all three relevant material properties:

$$\alpha = \frac{k}{\rho\,c_p}$$

$k$ (W/mK) is the ability to conduct heat; $\rho c_p$ (J/m³K) is the heat stored per unit volume per degree. High $\alpha$ means the material conducts heat quickly relative to how much it stores — it is a “fast” thermal material.

Material	$k$ (W/mK)	$\rho c_p$ (MJ/m³K)	$\alpha$ (m²/s)	Relative to HDPE
Copper	400	3.4	1.2×10⁻⁴	500×
Tool steel	35	3.5	1.0×10⁻⁵	40×
HDPE	0.44	1.9	2.3×10⁻⁷	1×
Nylon 6	0.25	2.0	1.3×10⁻⁷	0.56×
PLA	0.13	2.3	5.7×10⁻⁸	0.25×

This table explains why PLA parts take roughly 4$\times$ longer to cool than HDPE parts of the same thickness: PLA has both lower $k$ and higher $\rho c_p$. A process engineer moving from HDPE to PLA must adjust cycle times accordingly or reduce wall thickness to compensate.

6.7 Cooling System Design Principles

Four variables control how quickly the mold removes heat from a part. In order of their typical engineering leverage:

1. Wall thickness (part design, scales as $s^2$): the strongest lever. Reducing wall thickness from 3 mm to 2 mm cuts cooling time by more than half. Requires close collaboration between part designer and tool designer.

2. Cooling channel proximity (mold design, reduces $L_{\text{mold}}$): channels drilled closer to the cavity surface reduce the mold wall conductive resistance. Conformal cooling channels (produced by metal additive manufacturing) can follow the cavity geometry and deliver dramatic improvements.

3. Coolant flow rate (process parameter, raises $h_{\text{coolant}}$): ensure turbulent flow in all channels (Re > 4000). Turbulent flow gives $h \approx 3000$–8000 W/m²K vs. laminar $h \approx 300$–800 W/m²K.

4. Mold material (tool design, changes $k_{\text{mold}}$): aluminium or beryllium-copper inserts at high-heat areas can reduce local cycle time. However, the mold wall resistance is often not the dominant one, so this has diminishing returns once channels are close.

Conformal cooling — mold channels manufactured by metal 3D printing that precisely follow the cavity contour — represents the largest recent advance in injection mold cooling. Cycle time reductions of 30–50% are achievable for complex geometries where conventional drilled channels cannot reach.

6.8 Interactive: Injection Molding Cycle Time Explorer

Adjust the part and process parameters to see how cooling time, heat per shot, and required coolant flow rate change. Notice how strongly wall thickness dominates everything else.

Wall half-thickness $s$ (mm): 2

$T_{\text{melt}}$ (°C): 230

$T_{\text{mold}}$ (°C): 25

$T_{\text{eject}}$ (°C): 80

$\alpha \times 10^7$ (m²/s): 2.3

HDPE≈2.3 | Nylon≈1.3 | PLA≈0.57

$s^2/\alpha$ (characteristic time)	—
Cooling time $t_c$ (Ballman-Shusman)	—
Log factor $\ln(4/\pi \cdot \Delta T_\text{ratio})$	—
Suggested total cycle time (×1.4)	—

Cooling time sensitivity

Double $s$:

Halve $s$:

Double $\alpha$:

6.9 Worked Examples

Example 6.1 — Heat Content of a Nylon Shot

A nylon 6/6 part has mass 0.20 kg. Injection temperature: 230 °C. Ejection temperature: 80 °C. Specific heat of melt: $c_p = 2000\ \text{J/kgK}$. Latent heat of crystallisation: $H_c = 60\ \text{kJ/kg}$. Find total heat to remove per shot.

Solution:

Sensible heat: $Q_s = 0.20 \times 2000 \times (230 - 80) = 0.20 \times 2000 \times 150 = 60{,}000\ \text{J}$

Latent heat: $Q_L = 0.20 \times 60{,}000 = 12{,}000\ \text{J}$

$$Q_{\text{shot}} = 60{,}000 + 12{,}000 = 72{,}000\ \text{J} = 72\ \text{kJ}$$

At 4 shots/min, the cooling system must handle $72{,}000 \times 4 / 60 = 4800\ \text{W} = 4.8\ \text{kW}$ continuously.

Example 6.2 — Mold Wall $U$ and Heat Removal Rate

A tool-steel mold has a cavity-to-channel distance $L = 20\ \text{mm}$, $k = 40\ \text{W/mK}$. Cooling water gives $h_o = 500\ \text{W/m}^2\text{K}$. The cavity face area is $A = 0.10\ \text{m}^2$. Melt at 230 °C, coolant at 30 °C. Find $U$ and the heat removal rate.

Solution:

$$\frac{1}{U} = \frac{L}{k} + \frac{1}{h_o} = \frac{0.020}{40} + \frac{1}{500} = 0.0005 + 0.0020 = 0.0025\ \text{m}^2\text{K/W}$$ $$U = 400\ \text{W/m}^2\text{K}$$

$$\dot{Q} = UA\Delta T = 400 \times 0.10 \times (230 - 30) = 8000\ \text{W} = 8\ \text{kW}$$

6.10 Concept Questions

Easy

A part designer wants to reduce cycle time. They propose doubling the wall thickness to allow the part to be stiffer. How will this affect cooling time? By what factor?

Easy

Why does ABS not have a latent heat term in the shot heat calculation?

Medium

A mold channel runs at 1 m/s, giving laminar flow and $h_o = 400\ \text{W/m}^2\text{K}$. The operator increases flow velocity to 3 m/s, shifting to turbulent flow and $h_o = 3200\ \text{W/m}^2\text{K}$. If the mold wall resistance is $L/k = 0.0005\ \text{m}^2\text{K/W}$, calculate $U$ before and after. By what factor does $U$ improve?

Medium

A process engineer switches a mold from PLA to HDPE. Both parts have the same geometry (wall thickness 2 mm), same mold temperature (25 °C), and similar melt and ejection temperatures. Which polymer will have the shorter cooling time, and roughly by what factor? Use: $\alpha_{\text{HDPE}} = 2.3\times10^{-7}$, $\alpha_{\text{PLA}} = 5.7\times10^{-8}\ \text{m}^2/\text{s}$.

Hard

From the diffusion equation to the Ballman–Shusman formula. This problem builds the full chain: governing PDE → dimensionless group → closed-form cooling time.

(a) Write the 1-D transient heat conduction equation $\partial T/\partial t = \alpha\,\partial^2 T/\partial x^2$ for a slab of half-thickness $s$. Non-dimensionalise using $\theta = (T-T_w)/(T_m-T_w)$, $\xi = x/s$, and $\tau = \alpha\,t/s^2$. What dimensionless group naturally emerges as the dimensionless time, and what is its physical meaning?

(b) The first-term series solution at the centre of the slab ($\xi=0$) gives $$\theta_c \approx \frac{4}{\pi}\exp(-\pi^2\,Fo), \quad Fo = \frac{\alpha\,t}{s^2}.$$ Define the ejection criterion $\theta_e = (T_e - T_w)/(T_m - T_w)$. Solve for $Fo^*$ at which $\theta_c = \theta_e$, then convert back to real time $t_c = Fo^* \cdot s^2/\alpha$ to derive the Ballman–Shusman formula.

(c) For HDPE: $T_m = 230\,°\text{C}$, $T_w = 25\,°\text{C}$, $T_e = 80\,°\text{C}$, $s = 2\ \text{mm}$, $\alpha = 2.3\times10^{-7}\ \text{m}^2/\text{s}$. Calculate $\theta_e$, then $Fo^*$, then $t_c$ numerically.

(d) A quick engineering rule sets $Fo \sim 1$ to estimate $t \sim s^2/\alpha$. Using your value of $Fo^*$ from (c), by what factor does the exact result differ from this rough estimate? Why is the rough estimate still valuable despite the factor?

(a) Non-dimensionalisation. Substituting $\theta$, $\xi$, $\tau$ into the PDE: $$\frac{\partial\theta}{\partial\tau} = \frac{\partial^2\theta}{\partial\xi^2}.$$ The only dimensionless group is the Fourier number $Fo = \tau = \alpha\,t/s^2$, which is the ratio of heat diffused so far ($\alpha\,t$) to the distance the heat must travel ($s^2$). When $Fo \sim 1$ the temperature profile has penetrated the full thickness.

(b) Deriving the Ballman–Shusman formula. Set $\theta_c = \theta_e$: $$\theta_e = \frac{4}{\pi}\exp(-\pi^2\,Fo^*) \implies Fo^* = \frac{1}{\pi^2}\ln\!\left(\frac{4/\pi}{\theta_e}\right) = \frac{1}{\pi^2}\ln\!\left(\frac{4}{\pi}\cdot\frac{T_m-T_w}{T_e-T_w}\right).$$ Converting back: $t_c = Fo^* \cdot s^2/\alpha$, which gives $$\boxed{t_c = \frac{s^2}{\pi^2\alpha}\ln\!\left(\frac{4}{\pi}\cdot\frac{T_m-T_w}{T_e-T_w}\right).}$$ This is the Ballman–Shusman formula — it emerges directly from non-dimensionalising the diffusion equation and inverting the leading-term Fourier series.

(c) Numerical check for HDPE. $$\theta_e = \frac{80-25}{230-25} = \frac{55}{205} \approx 0.268.$$ $$Fo^* = \frac{1}{\pi^2}\ln\!\left(\frac{4}{\pi\times0.268}\right) = \frac{1}{9.870}\ln(4.775) \approx \frac{1.563}{9.870} \approx 0.158.$$ $$t_c = 0.158 \times \frac{(0.002)^2}{2.3\times10^{-7}} = 0.158 \times 17.39 \approx \mathbf{2.75\ \text{s}}.$$ (The interactive widget gives ≈ 2.8 s — excellent agreement.)

(d) Rough estimate vs exact. The crude rule $Fo \sim 1 \Rightarrow t \sim s^2/\alpha = 17.4\ \text{s}$ overestimates by $1/Fo^* \approx 6\times$. The exact prefactor is $Fo^* \approx 0.16$, not 1. Despite the factor-of-6 error, the rough rule is useful because: (i) it gives the correct scaling — $t \propto s^2/\alpha$ is exact; (ii) a factor of 6 is irrelevant when comparing materials or wall thicknesses (ratios cancel); and (iii) it requires no temperatures, making it a fast first-pass screen before detailed optimisation.

6.11 Problem Set

Easy

A nylon part has mass $m = 0.20\ \text{kg}$. Melt injection temperature: $T_{\text{melt}} = 230\,°\text{C}$. Ejection temperature: $T_e = 80\,°\text{C}$. Specific heat of melt: $c_p = 2000\ \text{J/kgK}$. Ignore latent heat. How much heat (J) must be removed from the part before it can be ejected?

Q = J

Formula

$Q = m\,c_p\,(T_{\text{melt}} - T_e)$

Physical meaning

total thermal energy stored in the melt that the mold must extract before the part can be ejected

Easy

A tool-steel mold has cavity-to-channel distance $L = 20\ \text{mm}$, conductivity $k = 40\ \text{W/mK}$. The cooling channel gives $h_o = 500\ \text{W/m}^2\text{K}$. Compute $U$ (W/m²K).

U = W/m²K

Formula

$1/U = L/k + 1/h_o$

Physical meaning

mold-wall conduction resistance plus coolant convection resistance in series — the larger term dominates and sets the overall bottleneck

Medium

Using $U = 400\ \text{W/m}^2\text{K}$ (from P2), cavity face area $A = 0.10\ \text{m}^2$, melt temperature 230 °C, and coolant temperature 30 °C, compute the rate of heat removal $\dot{Q}$ (W) from the cavity face.

Q̇ = W

Formula

$\dot{Q} = U\,A\,\Delta T$

Physical meaning

heat removal rate = overall conductance × area × driving temperature gap; larger mold area or colder coolant both help

Medium

A part ($m = 0.20\ \text{kg}$, $c_p = 2000\ \text{J/kgK}$) must drop $\Delta T = 100\,°\text{C}$ before ejection. The mold can remove heat at $\dot{Q} = 8000\ \text{W}$. What is the minimum cooling time (s)?

t_cool = s

Formula

$t = Q_{\text{total}} / \dot{Q}$, with $Q_{\text{total}} = m\,c_p\,\Delta T$

Physical meaning

cooling time = total heat stored ÷ rate of removal; a heavier part or weaker mold cooling both extend the cycle

Medium

Two polymers have the same density and specific heat but different conductivities: Material A has $k_A = 0.50\ \text{W/mK}$ and Material B has $k_B = 0.10\ \text{W/mK}$, both with $\rho = 1000\ \text{kg/m}^3$ and $c_p = 2000\ \text{J/kgK}$. Both are molded into parts with the same wall thickness. By what factor is Material B’s cooling time longer than Material A’s?

t_B/t_A = (ratio)

Formula

$\alpha = k/(\rho\,c_p)$ and $t_{\text{cool}} \propto s^2/\alpha$, so $t_B/t_A = \alpha_A/\alpha_B$

Physical meaning

diffusivity bundles all three material properties into one; low $\alpha$ (e.g. PLA) means slow heat diffusion and a long cooling penalty

Bridge → Lab

In your unit operations injection molding lab, a polycarbonate housing (wall thickness 3.5 mm) is warping on ejection. The mold temperature is set to 90 °C (required for PC to fill properly), and the mold uses water at 80 °C. The melt temperature is 310 °C and the target ejection temperature is 130 °C. Current cycle time is 45 seconds. The lab supervisor asks you to determine if the warping is a thermal problem or a packing/flow problem.

(a) Estimate the minimum cooling time for this part using $\alpha_{\text{PC}} = 1.2\times10^{-7}\ \text{m}^2/\text{s}$, $s = 1.75\ \text{mm}$ (half-thickness). Use the Ballman–Shusman formula with $T_{\text{mold}} = 90\,°\text{C}$.

(b) Compare your estimate to the 45-second cycle. Is the cooling time consistent with adequate cooling, or is the part being ejected too early?

(c) The operator suggests reducing the mold temperature to 40 °C to speed cooling. What are the risks of this approach for a polycarbonate part? Suggest a safer alternative that addresses the root cause.

(a) Ballman–Shusman cooling time:
$s = 0.00175\ \text{m}$, $\alpha = 1.2\times10^{-7}\ \text{m}^2/\text{s}$
$s^2/\alpha = (0.00175)^2 / (1.2\times10^{-7}) = 3.0625\times10^{-6}/1.2\times10^{-7} = 25.5\ \text{s}$
$\Delta T_{\text{ratio}} = (310 - 90)/(130 - 90) = 220/40 = 5.5$
$\ln(4/\pi \times 5.5) = \ln(7.00) = 1.946$
$t_c = (25.5/\pi^2) \times 1.946 = (25.5/9.87) \times 1.946 = 2.583 \times 1.946 \approx 5.0\ \text{s}$

Wait — this seems short. Let me recalculate: $s^2/(\pi^2\alpha) = 25.5/9.87 = 2.58\ \text{s}$; then $\times \ln(4/\pi \times 5.5) = \times 1.946$, giving $t_c \approx 5.0\ \text{s}$.

(b) Comparison: The minimum cooling time estimate is about 5 seconds. The cycle is running at 45 seconds — far longer than the thermal minimum. The cooling time itself is not likely the cause of warping. The part has plenty of time to cool thermally at 45 seconds.

(c) Root cause analysis: With 45 seconds and 5-second thermal minimum, warping is likely caused by one of:

Residual stress from packing: over-packing or under-packing creates a frozen-in stress field that relaxes on ejection.
Non-uniform cooling: one side of the part cooling faster than the other creates a temperature gradient that drives bending on ejection.
Ejection before $T_g$ fully reached: PC $T_g \approx 147\,°\text{C}$; if any interior region is above 130 °C, it can relax under ejection force.

Lowering mold temperature from 90 to 40 °C risks poor surface finish and short shots in PC (it needs a warm mold to fill long thin sections), and may increase residual stress by increasing the $\Delta T$ through the part cross-section.

Safer alternative: Check that both mold halves (core and cavity) are at the same temperature. Temperature imbalance between the two halves is the most common cause of warping in thick PC parts.

6.12 Chapter Summary

Key Equations

Quantity	Formula	Notes
Heat per shot (sensible)	$Q = m\,c_p(T_{\text{melt}} - T_e)$	Add $m H_c$ for semi-crystalline
Mold overall U	$1/U = L_{\text{mold}}/k + 1/h_o$	Neglects part-mold contact R
Heat removal rate	$\dot{Q} = U\,A\,(T_{\text{melt}} - T_{\text{coolant}})$	Driving force = full melt-to-coolant ΔT
Minimum cooling time	$t_{\text{cool}} \approx Q / \dot{Q}$	Lower bound; actual ×1.5–2
Ballman–Shusman	$t_c = \dfrac{s^2}{\pi^2\alpha}\ln\!\left(\dfrac{4}{\pi}\cdot\dfrac{T_m - T_w}{T_e - T_w}\right)$	$s$ = half-thickness; $\alpha$ = polymer diffusivity
Thermal diffusivity	$\alpha = k/(\rho\,c_p)$	m²/s; higher = faster cooling

Key Ideas

Cooling occupies 70–80% of the injection molding cycle; it is the primary lever for cycle time improvement.
Cooling time scales as $s^2/\alpha$: wall thickness squared is the strongest design variable.
Thermal diffusivity $\alpha = k/(\rho c_p)$ determines how fast heat conducts out of the polymer. PLA is about 4$\times$ slower than HDPE.
The mold wall overall U combines conductive resistance (mold steel) and convective resistance (coolant channel). Turbulent coolant flow is critical.
Semi-crystalline polymers release latent heat of crystallisation during cooling, adding significantly to the heat per shot.
Conformal cooling channels reduce $L_{\text{mold}}$, cutting the conductive resistance and enabling 30–50% cycle time reductions.

💡 Memory anchor: “In injection molding, cooling time scales as $s^2/\alpha$. Halve the wall thickness, quarter the cooling time. $\alpha = k/(\rho c_p)$ — high $k$, fast cooling. That is the whole game.”

Heat Transfer in Injection Molding

6.1 The Cycle Time Problem

6.2 The Injection Molding Cycle

6.3 Heat Content of the Shot

6.4 Mold Cooling: The Overall Heat Transfer Coefficient

6.5 Estimating Cooling Time

6.6 Thermal Diffusivity and Why It Controls Cycle Time

6.7 Cooling System Design Principles

6.8 Interactive: Injection Molding Cycle Time Explorer

6.9 Worked Examples

Example 6.1 — Heat Content of a Nylon Shot

Example 6.2 — Mold Wall $U$ and Heat Removal Rate

6.10 Concept Questions

6.11 Problem Set

6.12 Chapter Summary

Key Equations

Key Ideas

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