In order to guarantee the three goals of improving accuracy, reducing complexity and considering all process zones in the new power model, a model approach was developed using the first law of thermodynamics. To understand the modelling of the model, the enthalpy input into the polymer is shown formulaically in the three process zones, solids conveying zone, melting zone and melt zone.

First of all, a specific enthalpy curve of a semi-crystalline plastic is considered (Picture 1). At the crystallite melting temperature, the plastic has a corresponding specific enthalpy depending on the type. This enthalpy is divided in PAM into $\Delta h_f$ solid enthalpy (enthalpy change to increase the solid temperature) and $\Delta h_a$ melting enthalpy (enthalpy required to dissolve the crystalline regions).

Picture 1: Specific enthalpy curve of a partially crystalline polymer

The specific enthalpy change at a supporting point compared to the filling point of the material results from the multiplication of the solid enthalpy by a percentage factor which is dependent on the filling temperature, the current solid temperature and the crystallite melting temperature. The factor indicates to what percentage of the crystallite melting temperature the solid was heated, starting from the starting temperature of the material. The melting enthalpy is initially not taken into account in the solids conveying zone. The reason is shown in figure 1. For the determination of the enthalpy increase must be referred to the black imaginary straight lines. If the melting enthalpy is taken into account, the straight line would have a too large gradient at low temperatures, resulting in too large a deviation. In the solids conveying range, however, the temperature increase lies precisely in this small value range. Formally expressed, the specific enthalpy change at a support point in the solids conveying range compared to the filling point is given by equation 0-1

$$\Delta h_{FF} = \Delta h_f \cdot \frac{T_{FS} - T_E}{T_K - 0}\tag{Equation 0-1}$$

For the driving power of an element area the difference of the specific enthalpy change between two supporting points is calculated with equation 0-2.

$$\Delta \Delta h_{FF} = \Delta h_{FF_n} - \Delta h_{FF_{n-1}}\tag{Equation 0-2}$$

Using the calculated specific enthalpy difference, the pressure difference and the introduced heat flow, the required drive power required between two support points in the solids transport zone can be calculated from this (equation 0-3).

$$P_{D/FF} = \dot{m} \ast (\Delta \Delta h_{FF} + \frac{\Delta p}{\rho}) - \dot{Q}\tag{Equation 0-1}$$

The melting zone and the melt conveying zone differ from the solids conveying zone in that the material is two-phase. Consequently, a separate consideration of the specific enthalpy change must be carried out for the solids bed and the melt zone.

For the specific enthalpy change of a pure heating of the melt, the following applies in relation to the crystallite melting temperature through the integration of the specific heat capacity via the temperature equation 0-4.

$$\Delta h_{RS} = cp_0 \ast (T_M - T_K) + \frac{m_{cp}}{2} \ast (T_M^2 - T_K^2)\tag{Equation 0-4}$$

In addition to the enthalpy, energy was already introduced into the molten material by heating the melt in order to heat the solid up to the crystallite melting temperature. Formally, the specific enthalpy introduced up to the melting point can be determined according to equation 0-1 with equation 0-5.

$$\Delta h_{FA} = \Delta h_a + \Delta h_f \ast \frac{T_K - T_E}{T_K - 0}\tag{Equation 0-5}$$

From the addition of the equations 0-4 and 0-5, equation 0-6 results for the specific enthalpy increase from the starting point of the filling temperature to the current melt temperature for the melt range:

$$\Delta h_S = \Delta h_a + \Delta h_f \ast \frac{T_K - T_E}{T_K - 0} + cp_0 \ast (T_M - T_K) + \frac{cp_m}{2} \ast (T_M^2 - T_K^2)\tag{Equation 0-6}$$

The solid bed in the melting zone, on the other hand, undergoes a different specific enthalpy increase. This can be determined with equation 0-7, following equation 0-1 in the solids conveying area.

$$\Delta h_{FA} = \Delta h_f \ast \frac{T_{FS} - T_E}{T_K - 0}\tag{Equation 0-7}$$

The enthalpy increases of the solid bed and the melt zone must now be weighted and added up on the basis of the degree of melting in order to obtain the total enthalpy change at a supporting point in the melting zone and the melt conveying zone. For weighting, the total mass flow is multiplied by the respective percentage of the solid or melt. By adding the two specific enthalpy changes, the total enthalpy increase at one support point can be calculated. The differentiation between melting zone and melt area is made by the degree of melting. This is calculated in advance in SIGMA and can also be displayed in visual form.

In the pure melt conveying zone, for example, the melting degree is equal to one, so that the second term of equation 0-8 is omitted.

$$\Delta h_{AS/S} = m_{asv} \ast \Delta h_S + (1 - m_{asv}) \ast \Delta h_{FA}\tag{Equation 0-8}$$

The required driving power of an element area can be determined with the available results via the difference of the specific enthalpy increase of two supporting points (equation 0-8) via equation 0-9.

$$\Delta \Delta h_{AS/S} = \Delta h_{AS/S_n} - \Delta h_{AS/S_{n-1}}\tag{Equation 0-9}$$

$$P_{D_{AS/S}} = \dot{m} \ast (\Delta \Delta h_{AS/S} + \frac{\Delta p}{\rho}) - \dot{Q}\tag{Equation 0-10}$$

In the single-stage compounding process, two polymers are metered into the hopper and then compounded. This means that the components are dosed in a certain ratio in the hopper and simultaneously plasticized. As a result, both materials have the same temperature at one point along the extrusion process. However, the polymers differ in their melting behaviour. At the same temperature, both materials also have different enthalpy levels. The consequence is that the components mixed in the hopper cannot be seen as a unit, but that the enthalpy levels must be considered separately and then added, weighted, to the corresponding mass flow. The advantage of this method is that the model is not only applicable for two polymers, but for any number of them. In addition, the basic model already modelled can be used and only a weighting and addition of both enthalpy levels is required.

In the solids transport zone, the enthalpy level of the individual polymers can be calculated using equation 0-1. These are then weighted with equation 0-11 and added together to obtain the enthalpy level at a supporting point.

$$\Delta h_{FF_{C1}} = \frac{\dot{m}_1}{\dot{m}_{SS}} \ast \Delta h_{FF1} + \frac{\dot{m}_2}{\dot{m}_{SS}} \ast \Delta h_{FF2} + \cdots + \frac{\dot{m}_n}{\dot{m}_{SS}} \ast \Delta h_{FFn}\tag{Equation 0-11}$$

Subsequently, identical to equation 0-2 and equation 0-3, the enthalpy change and from this the required drive power between two adjacent support points can be calculated.

The same method is used in the melting zone and the melt conveying zone. The enthalpy levels of the polymers are considered separately in the first step and then added together. Equations 0-6 and 0-7 of the basic model provide the enthalpy levels of the corresponding melt and solid content for the individual polymers. In equation 0-8 the proportions are weighted with the enthalpy level. For a process with several polymers that melt simultaneously, the weighting of the different mass flows must also be taken into account. Equation 0-12 applies to the specific change in the enthalpy when compounding polymers when mixed in the hopper:

$$\Delta h_{AS/S_{C1}} = \frac{\dot{m}_1}{\dot{m}_{SS}} \left(\dot{m}_1 \ast m_{asv_1} \ast \Delta h_{S_1} + \dot{m}_1 \ast (1 - m_{asv_1}) \ast \Delta h_{FA_1}\right) + \frac{\dot{m}_2}{\dot{m}_{SS}} \left(\dot{m}_2 \ast m_{asv_2} \ast \Delta h_{S_2} + \dot{m}_2 \ast (1 - m_{asv_2}) \ast \Delta h_{FA_2}\right)$$ $$ + \cdots + \frac{\dot{m}_n}{\dot{m}_{SS}} \left(\dot{m}_n \ast m_{asv_n} \ast \Delta h_{S_n} + \dot{m}_n \ast (1 - m_{asv_n}) \ast \Delta h_{FA_n}\right)\tag{Equation 0-12}$$

Subsequently, identical to equation 0-9 and equation 0-10, the specific enthalpy difference and the required drive power between two supporting points can be calculated.

At the beginning of the process, first one material is plasticized and later in the direction of extrusion two or more materials are plasticized simultaneously. As a result, different numbers of polymers are present at different support points in the extrusion process, which determine the enthalpy level. For this reason, the process must be regarded as a separate application. For the modelling of a two-stage process, it is again possible to fall back on the models already created. Before the calculation, however, a case distinction must be made to check whether one or more polymers are processed in the extruder. Up to the second material stage, the enthalpy increase compared to the filling temperature can be calculated strictly according to the basic model from chapter 5. If a support point meets an additional material stage or material addition, the model from the single-stage compounding process must be used. Please note that the individual mass flows are not related to the total mass flow of all polymers, but only to the mass flow of the current support point.

During compounding, fillers such as chalk or talcum are often incorporated to change material properties. For this purpose, the fillers are added to the plasticized melt. A modification of the models presented in this paper is not necessary. However, two cases are to be considered critically. In the case of extremely high filler contents, additional power is required due to the newly generated frictional forces. However, these are special processes and will not be considered at first. The second critical case is the support point where the fillers are incorporated. Through the addition, the melt experiences a significant reduction in temperature. The consequence is that the enthalpy difference with the previous supporting point becomes negative. As a result, the total power requirement between the two support points becomes less than zero. This process is physically impossible because theoretically no power can be obtained from the extruder. The solution is to set the power requirement at this point to zero. This is acceptable because conveying elements are used at the filler addition point, which generally have a low power requirement.

In the melt extruder, the specific increase in enthalpy due to melting of the solid is eliminated. The melt conveyed in the extruder only undergoes a change in enthalpy due to a temperature variation. For this reason, the calculation of the enthalpy increase up to the melting point can be neglected and the reference point for the enthalpy change is not the filling temperature but the crystallite melting temperature. With later differentiation of the enthalpy changes this portion would be shortened out again anyway. The specific enthalpy change of a melt at a support point in the melt extruder in relation to the crystallite melting temperature can be calculated using Equation 0-6 with Equation 0-13.

$$\Delta h_{SE} = cp_0 \ast (T_M - T_K) + \frac{cp_m}{2} \ast (T_M^2 - T_K^2)\tag{Equation 0-13}$$

For the driving power, the difference of the specific enthalpies is calculated again (equation 0-14):

$$\Delta \Delta h_{SE} = \Delta h_{SE_n} - \Delta h_{SE_{n-1}}\tag{Equation 0-14}$$

Equation 0-15 applies to the drive power:

$$P_{D_{SE}} = \dot{m} \ast (\Delta \Delta h_{SE} + \frac{\Delta p}{\rho}) - \dot{Q}\tag{Equation 0-15}$$

In the case of a twin-screw extruder, which functions as a melt extruder and is equipped with a melt consisting of several polymers, the model must be modified for a melt extruder. The procedure is similar to that for compounding. The individual enthalpy levels of the components are determined in a first step and then weighted with the aid of the mass flows. Formally, equation 0-16 is obtained.

$$\Delta h_{SE_M} = \frac{\dot{m}_1}{\dot{m}_{ges}} \left(cp_{01} \ast (T_M - T_K) + \frac{cp_{m1}}{2} \ast (T_M^2 - T_K^2)\right) + \frac{\dot{m}_2}{\dot{m}_{ges}} \left(cp_{02} \ast (T_M - T_K) + \frac{cp_{m2}}{2} \ast (T_M^2 - T_K^2)\right) $$ $$+ \cdots + \frac{\dot{m}_n}{\dot{m}_{ges}} \left(cp_{0n} \ast (T_M - T_K) + \frac{cp_{mn}}{2} \ast (T_M^2 - T_K^2)\right)\tag{Equation 0-16}$$

Subsequently, as in the case of a melt extruder, which is loaded with a polymer, the difference of the enthalpy change can be formed and from this the drive power between two support points.

With the relationships shown, the individual required drive powers between two support points or the individual elements can be determined and visualized in SIGMA. Finally, the individual calculated element powers must be added up to obtain the total power. The sole consideration or difference of the initial and final enthalpy to determine the power leads to errors in various cases. This is the case, for example, if the melt temperature drops once during the extrusion process due to fillers and the melt is then heated again. The subsequent heating of the melt must be done by new energy from outside. This requires a further input of power, which would be neglected if the initial and final enthalpy were simply compared.

The new power model was implemented and verified in SIGMA. Identical to other models, the new power model was compared with the values of the experimental investigations to validate the model. The deviations of different process points and material combinations are shown in the following picture.

Figure: Comparison of experimental investigations and simulations