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1D Fourier Transient Cooling Time Formula

Calculates the required in-mold cooling time (tc in seconds) for a plastic component center-plane temperature to drop below ejection threshold.

Mathematical Canonical Relation

t_c = \frac{h^2}{\pi^2 \cdot \alpha} \ln\left[ \frac{8}{\pi^2} \cdot \frac{T_{melt} - T_{mold}}{T_{eject} - T_{mold}} \right]

Plain Formula: tc = (h^2 / (pi^2 * alpha)) * ln((8/pi^2) * ((T_melt - T_mold)/(T_eject - T_mold)))

Live Engineering Parameter Solver

Input your production or test values below for instant, verified output.

Verified Engine

Wall Thickness (mm) (mm)

h_mm

Range: 0.5 – 8 mm

Thermal Diffusivity (mm²/s) (mm²/s)

alpha

Range: 0.04 – 0.2 mm²/s

Melt Temp (°C) (°C)

Range: 140 – 360 °C

Mold Temp (°C) (°C)

Range: 10 – 120 °C

Ejection Temp (°C) (°C)

Range: 40 – 200 °C

Calculated Engineering Result:

8.4 seconds

Engineering Verification:Optimized fast cycle cooling time.

When & Where to Apply

Use during mold design and cycle time optimization to estimate required cooling time based on wall thickness and thermal diffusivity.

Common Engineering Pitfalls

Cooling time scales with h^2 — doubling wall thickness quadruples cooling time!
Using thermal conductivity k instead of thermal diffusivity alpha

Physical Assumptions & Validity Limits

1D transient heat conduction through flat plate of uniform wall thickness h
Constant mold wall surface temperature Tw
Constant material thermal diffusivity alpha

Variable Symbols, SI Units & Dimensional Specifications

Symbol	Variable Meaning	SI Unit	Typical Industrial Range
t_c	Cooling time	seconds	—
h	Part wall thickness	mm	2.5
\alpha	Thermal diffusivity	mm²/s	0.085
T_{melt}	Melt injection temperature	°C	230
T_{mold}	Mold wall temperature	°C	40
T_{eject}	Ejection temperature	°C	90

Worked Industrial Numerical Example: Polypropylene Container 3mm Wall Cooling Time

Problem Statement:

Calculate in-mold cooling time for a polypropylene container wall (h = 3.0 mm, thermal diffusivity alpha = 0.085 mm²/s) injected at Tm = 230°C into a mold at Tw = 30°C and ejected at Te = 90°C.

Step-by-Step Numerical Solution:

Calculate temperature ratio term: (8 / pi^2) * ((230 - 30) / (90 - 30)) = 0.81057 * (200 / 60) = 2.7019
Calculate natural log: ln(2.7019) = 0.9939
Calculate diffusion time factor: h^2 / (pi^2 * alpha) = (3.0)^2 / (9.8696 * 0.085) = 10.728 s
Multiply terms: tc = 10.728 * 0.9939 = 10.66 s

Final Answer: tc = 10.7 seconds

💡 Engineering Interpretation: Cooling time is 10.7s. Thinning the wall to 2.0 mm would reduce cooling time to 4.7s, increasing production output by 56%.

Authoritative Technical Reference & Source Provenance

Principles of Polymer Processing by Z. Tadmor & C. Gogos (Wiley, 2006), 2nd Edition, Page 450-458.

Connected Curriculum Lesson

Cooling System Design & Cycle Time Optimization

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