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Mold Cooling

The following is a contribution of Rowland Evans of reps Ltd., a UK based company selling pulsed water cooling and heating systems. Their website offers a wealth of information of anything to do with mold cooling.

Mold cooling: How Much Water Pressure is needed?

It is the volume of coolant (flow rate) that is important in mold cooling of plastic molds. Pressure is the force that is required to cause the water to flow through the mold's cooling channels at a given flow rate (Imp. g/m – l/m).

A Delta p of between 0.75 – 1.5 bar, (10-22 psi) across the mold, will generate a flow rate of around 3 Imp. g/m 3.6 US g/m 13.6 l/m through a 10-14mm bore cooling channel (which are generally the largest used). This is a good, energy efficient, turbulent flow rate.

Delta p = Supply Pressure – Drain Pressure.

However, this is only the starting point. You now need to know how many cooling channels you have on the mold so that you can work out the flow rate required from your central system at the machine, e.g. 16 cooling channels x 3 Imp. g/m = 48 Imp. g/m. (220 l/m.). Now we need to know what the smallest diameter flow channel is between the main central cooling systems (supply and return) and the machine, this is usually the hose tails on the flexible pipe that connect the machine to the solid pipe work, so that we can estimate the pressure losses to and from the mold. If we use as and example a diameter of 1" (25mm), then these pressure losses would be between 3-5 bar (43-72 psi.). Now to find the static pressure difference at the machine, on this application with no water flowing, we would need to sum these two results. Minimum Delta p = 3.75 bar (54 psi.), maximum effect Delta p = 6.5 bar (94 psi.) static Delta p for 16 cooling channels.

Obviously, if you have more cooling channels on your mold with the same supply limitations to your machines you will require a much greater pressure difference at the machine.

The figures for the pressure difference at the machine should not be confused with the central system supply line pressure measured near the pumping station, because there are pressure losses within the central system (these effects should be minimised). As a rule-of-thumb, a reasonable figure for these losses in a reasonably designed system would be to add between 2-3 bar, (30-45 psi.), this gives us a central system pumping station pressure of between 6-9 bar, (87-130 psi.).

Mold Cooling

There are methods of configuring the central cooling systems pipe work that can reduce these losses to between 1 – 2 bar (14 – 30 psi.), lowering the pumping station pressure to between 5 – 6.2 bar (75 – 90 psi.).

However this is not always practical in existing molding shops. An alternative possibility is to use your existing central system as a "low pressure" supply for a high pressure / high flow rate controlled circulator (Turbulator), located near the molding machine.


The smaller the diameter of cooling channels,the smaller it's surface area, the greater the Delta p that is required to generate a high volume of coolant flow / molding cycle. A high volume of coolant flow / molding cycle is required to promote a good heat extraction rate & there for optimum cooling. From a practical point of view, to avoid excessive pressure requirements cooling channels of less than 5.5mm (0.217) should not be used as they will require a mold Delta p of over 3.5 bar (51 psi). When you take into account supply line and central system pressure losses, this means you will require a central pumping pressure of 9.5 bar (136 psi). This is based on a mold with 20 cooling channels and a supply line of 1" diameter and an averagely designed central cooling system. The same mold with a well designed, low loss central system with 1 ½" supply lines to the mold would only require a central pumping pressure of 6.7 bar (98 psi).

Now we can see that these calculations lead to a very high central system pressure requirement; the higher the pressure the more costly it is to run.

What is the effect of having different sized cooling channels within the same mold? It is obvious from the above examples that the Delta p required to generate a good turbulent flow is very different for different sized cooling channels e.g. for a 6mm (1/4") we require 3 bar (44 psi), and for a 11mm (7/16") we require 1.9 bar (27 psi)

Pressure losses (Delta p) are directly related to the flow rate through a cooling channel of a certain diameter over a certain length, with additions created by changes in direction and geometry. For the purpose of this paper we have factored in some nominal lengths and geometry's for the mold cooling system layout. Of course there are some very bad layouts where our calculation will be inadequate, and there are a few well laid out where our calculation will leave some spare capacity.

Mold Cooling

So, pressure and flow are directly related, therefore insufficient pressure = insufficient flow, and insufficient flow = insufficient pressure. Most molding plants do not have the necessary equipment to measure flow rates accurately. However, you only need two sets of pressure gauges; 0-10 bar (0-150 psi) for high pressure systems, and 0-4 bar (0-50 psi) for low pressure systems, to be able to measure the Delta p at the different points of the system, and therefore equate these readings to flow rate, or at least test the adequacy of your system.

What is the consequence of our simplistic mold with one 6mm (1/4") and one 11mm (7/16") cooling channel, to generate a good turbulent flow. In the 6mm (1/4") channel we will require a Delta p of 3 bar (44 psi) to generate a flow rate of 1.5 g/m (6.2 L/m). On the 11mm (7/16") channel we require 2.5 g/m (11.4 L/m). This will require a Delta p of 1.9 bar (27 psi). The dilemma is that we only have one supply system. If we have a Delta p of 1.9 bar (27 psi) we will not have good turbulent flow in the 6mm (1/4") cooling channel, therefore we need to generate a pressure loss in the 11mm (7/16") channel of 3 bar (44 psi). To do this we need to increase the flow through that channel from 2.5 g/m (11.4 L/m) to at least 3.8 g/m (17.4 L/m), so the flow rate through the mold will not be 1.5 + 2.5 g/m = 4 g/m (18 L/m) but 1.5 + 3.8 = 5.25 g/m (24 L/m).

Now, if we apply this to the mold with 16 cooling channels, 8 + 8, where two cooling channels on each half are 6mm (1/4") and the rest are 11mm (7/16"), we will see that the flow rate (volume) of the coolant required is drastically increased,

(4 x 1.5 = 6) + (6 x 3.8 = 45.6) = 51.6 g/m (235 L/m) @ 3 bar.

If all cooling channels were 11mm (7/16") then the flow rate would be 40 g/m (182 l/m) @ 2 bar, over 30% less flow and pressure required.

The mixed size cooling channel layout is normal within our industry. The result is that the smaller diameter cooling channels are inevitably starved of pressure and flow, the coolant in these channels has laminar flow leading to poor heat extraction.

1.Connecting different diameter cooling channels in series (looping) will have the opposite effect; turbulent flow in the smallest diameter and laminar flow in with the larger diameter. This is also unadvisable, avoid this practice.

Mold Cooling

2.Keep cooling channels within the molding area the same diameter.
3.If this is not possible, the available coolant flow rate at the mold must be sufficient to generate the pressure required for the smaller diameter flow channels.
4.Pulsed cooling techniques will help divert the flow to the restricted flow channels, where the central system is barely adequate.
5.Use the largest diameter pipework to connect your machines to the central system to reduce pressure losses and allow the maximum flow rate.
6.Use a Turbulator flow pressure booster near the machine to allow for a moderate central system pressure/flow rate.
7.Use large bore flexible pipes and snap-on fittings to connect the mold to the machine supply to reduce external pressure losses.
8.Manifold your mold, don't loop it.
9.Use ringmaining techniques on your central cooling system.
10.Do not try to save money on the central system pipework size (the larger the diameter the better) because you will pay for it every day in extended cycle time.