Injection Moulding Cooling Time From Wall Thickness

How long a moulded part takes to cool, which on most parts is most of the cycle and most of the cost.

Run the calculator

Example

You enter

You get

Details, formula, and sources

The plate solution divides the wall thickness squared by the thermal diffusivity and multiplies by a logarithmic temperature term, and the SQUARE is the fact that governs everything downstream: doubling a wall quadruples the cooling, and a 25 percent increase costs 56 percent more time. This is why plastic parts look the way they do -- uniform, thin walls with ribs and coring rather than solid sections -- and the reason is economic rather than aesthetic. The consequence a designer rarely sees is that thickness is decided long before anyone counts cycles. A section thickened for stiffness during design is a permanent cost on every part the tool ever makes, and on a production part it is a large number of machine-hours a year that no process adjustment recovers. Coring is the counter-move: splitting a thick section into two thinner walls returns the cooling to the thin-wall figure, because it is the distance heat must travel that matters and not the amount of material. That change is free per part and it is available only before the tool is cut. Mould temperature is the weaker lever and it works differently. It enters through a logarithm rather than a square, so a large temperature change buys a small time change -- and running the mould colder to save cycle time costs crystallinity in semi-crystalline materials, adds moulded-in stress, and degrades surface finish. It is the adjustment available after the tool exists, which is why it is the one reached for, and it is the one that quietly moves part properties. This is a one-dimensional plate solution with constant properties. It does not handle three-dimensional heat flow, corners, or ribs (which cool from more than one direction and are why real parts beat this figure at features and miss it at thick sections), does not model the actual mould cooling circuit, its layout, flow rate or turbulence, does not account for crystallisation heat in semi-crystalline materials, and does not compute the injection, hold, or mould movement portions of the cycle. Thermal diffusivity is entered because it varies with temperature and with the material; for common thermoplastics it runs roughly 0.00013 to 0.00023 in2/s, which is what the conductivity, density and specific heat of ABS, polypropylene, polycarbonate and HDPE give -- a figure several times larger than that shortens every answer here in proportion, so it is worth checking against the moulder's own cycle records. The resin supplier's data, a mould cooling analysis, and the moulder's own cycle records govern.

the one-dimensional plate solution t = h^2 / (pi^2 alpha) x ln[(4/pi) x (T_melt - T_mould) / (T_eject - T_mould)]; cooling goes with the SQUARE of the wall and only the LOGARITHM of the temperatures.

The classical plate cooling solution for injection moulding, with thermal diffusivity ENTERED. Note that alpha = k/(rho x cp) gives roughly 0.00013 in2/s for ABS, 0.00020 for polypropylene, 0.00021 for polycarbonate and 0.00023 for HDPE -- several times lower than figures sometimes quoted, and a diffusivity too high shortens every answer in proportion.

One published closed-form solution.

Estimate. AHJ and licensed professional govern.

Field names used by the API: wall_thickness_in, alpha_in2_s, melt_temp_f, mould_temp_f, eject_temp_f, alt_wall_thickness_in, non_cooling_cycle_s, annual_parts, cooling_time_s, alt_cooling_time_s, time_ratio, cored_time_s, cycle_time_s, annual_machine_hours

Related tools