【正文】 a lot of parameters involved. The parts that are thicker need more cooling than the thinner parts. To analyze where you need to put which cooling we need to make a heat balance. This heat balance encounters all the possible heat transactions of the product, the mould and the cooling channels. The formula is written beneath. (1) Where is the heat flux from moulding (kJ/m), the heat exchange with the environment, (kJ/m) and the heat exchange with the coolant (kJ/m) The heat exchange with the environment takes only around five percent of the total heat exchanging. This shows that it is necessary to put some extra cooling. Because the exchange with the environment takes only five percent we can neglect it in relation of the total heat exchange. Then we bee the next formula. (2) Now we need to calculate the heat exchange with the cooling and with the mould itself. The heat flux exchange between the cooling and the molten polymer can be expressed as (Rao et al 2002) (3) Where is the mould temperature (176。C), the demoulding temperature (176。C), the specific heat of the melt (kJ/kg176。C), the latent heat of fusion of polymer (kJ/kg), the melt density (), s the part thickness (mm) and a the distance between two cooling channels. The find the heat flux exchange during the cooling time (s) we use the next formula : （4）a lWhere is the thermal conductivity of mould steel (W/m176。C), the shape factor, α the heat transfer coefficient, the mould wall temperature (176。C) and the temperature of coolant (176。C). The shape factor Se also has a big influence on the cooling quality of the mould. This factor is given as follows for circular cooling channels. (Thorne 1997) （5）With b the distance from the mould surface to the center of the cooling channel (mm) and the diameter of the cooling channel (mm) The heat transfer coefficient for water is as follows: （6）With Re the Reynolds number. The Reynolds number decides if the flow is laminar or turbulent. （7）With u the velocity of the cooling (m/s) and ν the kinematical viscosity of the medium water (㎡/s). As mentioned above the necessary cooling level is not everywhere the same. That’s why the mould needs to be divided in several zones. These zones can be obtained by a simulation program. For the practical use of the formulas we will calculate the necessary cooling time in function of the distance between two cooling lines a, the distance from the mould surface to the cooling channel b and the diameter of the cooling channels . If we put this into one formula we get the next : （8）In this formula we need to make sure that is as minimum as possible to decrease the total cycle time. The formula needs to be used on different temperature zones of the mould. To solve the problem we can use a mathematical program such as Matlab. Out of the program we get the results for the time of cooling for different values of the measurements a, b and of the cooling. Not all the solutions are useful. For the variables a, b and limits should be inserted. The reason why we need boundaries is because the dimensions of the cooling could be oversized. If this happens the strength of the mould can decrease a lot. It is also possible that the design is not realistic anymore. To avoid this there are made some upper and lower boundaries. （9）The boundaries of the variable parameters are found by experience and engineering analyses. They are shown beneath. （10）It is not only necessary to take into account the boundaries. Because the part goes out of the mould in one time, all the cooling times of the different cooling zones need to be the same. Once everything is calculated with Matlab or another mathematical program the position and size of the cooling is known. [6] 4. THE PRODUCT The product we will examine the previous study on is a screw for an LPG tank of a car, illustrated in Figure 4. Now the injection point is at the edges, what gives a bad result. Figure 4: The product a screw for an LPG tank.5. PRACTICAL GATE LOCATION The product shows a few problems. Some of these problems can be solved by choosing a better location for the gate. The injection point of the product is now positioned on the side of the product. From there the piece gets filled perpendicular on the radius and parallel with the top of the piece. First we will examine the possible hesitation problem. If we take a look to the piece from the side we can see that the screw is not parallel with the injection flow. The flow with the less resistance is going straight on. To fill the screw there is more resistance. So the filling of the screw will go much slower than the rest of the piece. It is possible that the screw doesn’t get filled properly. Figure 5 (left) shows the actual injection point location. If we try to look for a better way of injection we can move the injection point in the centre of the product. The advantage is that all the flow lines will be symmetric. Now we can inject from the top or bottom. The screw side is preferred because otherwise the injection point will be visible on top of the product. Figure 5 (right) shows the suggested gate location Figure 5: Hesitation.At the other hand it will be examined the possible racetrack problem. Now we will look to the piece from the top. The injection point is still at the edge of the radius. When we inject over there the outer and the inner flowlines will be as shown on figure 6 (left). As you can see the flowlines e together at the other side of the product. A pression of air will occur at that