Mould Cooling

Consider a moulding in the form of a flat plate of thickness 2L between the water cooled platens of a moulding machine (as in the simple sketch above). Because the thermal conductivity of the mould material is very much higher than that of the polymeric material (~ 1000 times greater) it is safe to assume that the heat flux across the mould/coolant interface will be very much higher than the heat flux across polymer/mould interface, especially if the coolant flow is turbulent rather than laminar - i.e. the Reynolds Number,  ,  is greater than 2100.  Where D is a characteristic dimension, i.e. the diameter of a tube, V is the velocity of the coolant, r is the density of the coolant and m is its viscosity.

We can subsequently assume that the mould temperature, T₁, will always be that of the coolant and as such we can use Newton’s Law of Convection to describe the heat transfer between the polymer and coolant.

                                                                                                           (1)

Where Q is the heat flux, A is the surface area of the moulding, normal to the direction of heat flux, i.e., through its thickness, T is the temperature at the geometric center of the moulding (along its center line) at some point in time, t, and h is the heat transfer coefficient.

The heat transfer coefficient, h, is difficult to calculate because it is a function of many variables, such as the nature of the mould surface, the intimacy of contact at the interface and the viscosity of the coolant to name but a few. It is not a constant valid under all conditions of heat transfer. We can assume that values for h are available from various sources in the literature.  (One sensible value given in the literature for polymer in contact with steel is 500 W m^-2 K^-1).

From Fourier’s Equation we get :

                                                                                                                (2)

Where m is the mass of the moulding and C_p is its heat capacity.

If we equate (1) and (2) above, we get :

                                                                                                  (3)

If we let the temperature difference (T - T₁) = q

Then                                                                                                    (4)

Or                                                                                                        (5)

However, when t = 0, i.e., at the start of cooling, q = q ₀ = (T ₀ – T₁)

Where T ₀ is the initial temperature of the melt.

Integrating                                                                                           (6)

We get                                                                                                      (7)

The exponent in equation (7) can be rewritten

                                                                                      (8)

where r and v are the density and volume of the moulding respectively and k is the thermal conductivity.

So that                                                                                                  (9)

Where                                                                                    (the thermal diffusivity).

is the Biot Number (Bi) which describes transient, or unsteady heat conduction. The Biot Number can be thought of in this case as the ratio of heat transferred across the moulding/mould interface relative to the amount of heat conducted through the moulding to the mould. It is a measure of the relative resistance to heat flow – a low Biot Number means that the resistance is predominantly at the interface and a high Biot Number suggests that the resistance is within the moulding itself.

                                                                                                (10)

We can now plot the temperature gradient vs. Fo for various values of heat transfer coefficient, h, hence various values of Bi.

From this graph – if you know T₀ and T₁ and you know the heat transfer coefficient for your polymeric material and your mould material – you can calculate your cooling time, T.

 

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The exact result of the analysis does, of course, depend on the input data available, some of which is difficult to obtain.

Using the data for High Density Polyethylene (Dow KT 10000) gleaned from various sources and using the above 
analysis, the cooling time for a component 4 mm thick ( i.e., ) will be as follows.

The initial melt temperature, T₀ = 250°C

The mould temperature , T₁ = 40°C

High Density Polyethylene Crystallisation Temperature, T = 90°C

We can assume that the melt will super cool to a value of about 90°C, i.e. about 40°C below its crystalline melting temperature of 130°C. We can determine our own degree of super cooling because those doing the DSC usually run the temperature up from 25°C to 200°C then back down again to 25°C. This will provide the melting and the crystallisation data.

Two values are available for the heat capacity, C_p, 2.31 x 10³ and 2.23 x 10³ J kg^-1 K^-1

Dow’s own value for the density, r , of  KT 10000 is 964 kg m^-3

Dow's value for the thermal conductivity, k, of HDPE is 0.50 W m^-1 K^-1

As above, a value quoted in the literature for the heat transfer coefficient, h, for polyolefines in conjunction with steel is 500 W m^-2 K^-1.

From the graph above for Bi = 2,  when = 0.238,  the Fourier Number, Fo = 0.70

Most materials manufacturers supply data on the thermal diffusivity of their plastics but in the absence of any information a value of 1 x 10^-7 m² s^-1 can be used as a first approximation.

In this case the time taken for the center of the moulding to reach the crystallisation temperature of HDPE (the cooling time) will be:

But, using the thermal data supplied by Dow, the thermal diffusivity becomes:

Therefore the time taken for the center of the moulding to cool sufficiently will be:

These two values of cooling time can be regarded as the two extremes and accurate measurement of the thermal data will give a more accurate cooling time.