Cooling Time Calculator
Calculate plastic injection molding cooling time from wall thickness and material thermal diffusivity.
Introduction
Cooling usually consumes more than half of an injection molding cycle, so estimating it well is the difference between a competitive quote and a press that loses money every shot. The Cooling Time Calculator implements the classical one-dimensional transient conduction solution for a plate-like part: enter wall thickness, melt temperature, mold (coolant-side) temperature, and the ejection temperature your resin tolerates, pick the material, and it returns the theoretical seconds needed before the part core is stiff enough to eject. Thermal diffusivity is looked up per resin — 0.096 mm²/s for PP, 0.102 for ABS, 0.113 for PA6 and PC — values that drive the result as strongly as wall thickness does.
How it Works
The implemented equation is t_cool = (s² ÷ (π² × α)) × ln((8/π²) × (T_melt − T_mold) ÷ (T_eject − T_mold)), the first-term Fourier-series solution for centerline cooling of a slab of thickness s with diffusivity α. Two behaviors follow directly from the math: cooling time scales with the square of wall thickness, so going from 2 mm to 3 mm walls more than doubles the estimate, and raising mold temperature stretches cooling logarithmically by shrinking the temperature ratio inside the ln() term.
Usage Scenarios
- Quoting cycle time for a new part: combining the cooling estimate with fill, pack, and mold-open allowances to predict total cycle before the tool exists.
- Justifying a design change: showing that coring out a 4 mm boss to 2.5 mm walls cuts theoretical cooling time by roughly 60% thanks to the s² term.
- Choosing mold temperature for PC, where a hot mold is needed for surface finish but every extra degree narrows the (T_eject − T_mold) gap and lengthens the cycle.
- Comparing resins for a cost-down: PA6's higher diffusivity (0.113 mm²/s) cools meaningfully faster than POM (0.095 mm²/s) at identical geometry.
- Setting a realistic baseline before a cooling-circuit upgrade, to quantify how far the current process sits from the conduction-limited minimum.
FAQ
Why does wall thickness dominate the result?
Conduction through the part scales with s², so cooling time grows with the square of the wall. Halving thickness quarters the theoretical cooling time — the single strongest lever in part design.
What ejection temperature should I use?
Use the heat-deflection-based ejection temperature from the resin datasheet — the temperature where the part is rigid enough to take ejector loads without marking or warping.
Why must eject temperature exceed mold temperature?
The model cools the part asymptotically toward mold temperature; the ln() term requires T_eject > T_mold. If they were equal, theoretical cooling time would be infinite.
How close is the estimate to a real process?
It is a 1-D plate idealization: real parts add ribs, corners, and imperfect cooling circuits. Treat it as the physics floor — production cycles typically run somewhat above it.