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#020 Example of Calculating the Cooling Time (1)

Category : Cost
November20, 2009

Problem:

How long is the necessary cooling time for an injection molded product made of ABS plastic (natural material) with a wall thickness of 1.5 mm? However, assume that the cavity surface temperature is 50°C, the temperature of the molten plastic is 230°C, and the molded product releasing temperature is 90°C.

Sample answer:

The cooling time tla required until the average temperature of the molded product becomes 50°C is calculated using the following equation.

tla = s2 / (π2•α) ln(8 / π2•(θr - θm) / (θe - θm)),  where,

tla is the cooling time (sec) related to the average temperature of the wall thickness; s is the wall thickness 1.5 (mm) of the molded product; α is the heat diffusion rate of the plastic at the cavity surface temperature, α = λ/(c•ρ); λ is the coefficient of thermal conductivity of the plastic (kcal/m•.h•°C); c is the specific heat of the plastic (kcal/kg•°C); ρ is the density of the plastic (kg/m3); and with the ABS plastic (natural material) at a cavity surface temperature of 50°C, α = 0.0827 mm2/sec; θr is the temperature of the molten plastic (230°C); θe is the temperature for taking out the molded product (90°C); and θm is the cavity surface temperature (50C). Substituting these values into the above equation, we get:

tla = 1.52 / (3.142•0.0827) ln(8 / 3.142•(230 - 50) / (90 - 50)) = 3.57 (sec)

The cooling time required until the average temperature of the molded product becomes 50°C is 3.57 sec.

*Referencee:"Molds for Injection Molding" by Keizo Mitani, Sigma Publications, 1997 (in Japanese)

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