Unterschiede

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--- en:grundlagenhandbuch:drehmoment_und_antriebsleistung:urspruenglicher_ansatz [2026/02/07 15:37] – [Original Model] deppe2
+++ en:grundlagenhandbuch:drehmoment_und_antriebsleistung:urspruenglicher_ansatz [2026/04/28 10:26] (aktuell) – gelöscht deppe2
@@ Zeile 1: / Zeile 1: @@
-====== Original Model ======
-===== Original Model =====
-The total power consumption is made up of solid friction and mixing friction in the feeding section, the power in the intermeshing zone of both screws, the power in the radial clearance over the screw tips, the power in the melt layer between solid material and the barrel wall, the power in the melt pool and in the melt section. In the calculation of the processes, the power consumption is considered to be negligibly tiny until the place of the first melt pool formation. As co-rotating twin screws, which will be used as a plasticating unit, usually possess a partly filled feeding zone, this is justifiable.
-In the estimation of the power consumption in [[en:grundlagenhandbuch:drehmoment_und_antriebsleistung:literatur|[1]]] equations were represented, where the power is calculable in the general form:
-$$P = \int_0^z \int_{-\frac{b_{max}}{2}}^{+\frac{b_{max}}{2}} (\tau_{0x} r_{0x} + \tau_{0z} r_{0z}) dx \, dz \tag{1}$$
-With the wall shear stress:
-$$\tau_{0x} = K \left[\left(\frac{\partial v_x}{\partial y}\right)^2 + \left(\frac{\partial v_z}{\partial y}\right)^2\right]^{\frac{n-1}{2}} \frac{\partial v_x}{\partial y} \tag{2}$$
-$$\tau_{0z} = K \left[\left(\frac{\partial v_x}{\partial y}\right)^2 + \left(\frac{\partial v_z}{\partial y}\right)^2\right]^{\frac{n-1}{2}} \frac{\partial v_z}{\partial y} \bigg|_{y=h} \tag{3}$$
-if the shear rates at the wall for conveying elements with the represented approximation equations in the table are used.
-**Table:** Estimation of the wall shear speeds (conveying elements: [[en:grundlagenhandbuch:drehmoment_und_antriebsleistung:literatur|[1]]]) \\
-^ I., III. ^ II. ^
-| $ \left.\frac{\partial v_x}{\partial }\right|_{y=h} = \frac{v_{0x}}{h}\,C_x $ | $ \left.\frac{\partial v_x}{\partial y}\right|_{y=h} = \frac{v_0}{h}\,C_x $ |
-| $ \left.\frac{\partial v_z}{\partial }\right|_{y=h} = \frac{v_{0z}}{h}\,C_z $ | $ \left.\frac{\partial v_z}{\partial y}\right|_{y=h} = \frac{v_0}{h}\,C_z $ |
-==== I. Conveying Elements ====
-$$C_x = \left[(1.368 + 2.634n^{0.1})e^{(n-1)^{\frac{1}{n}}}\right]$$
-$$0.55 \leq \pi_V \leq 1.00 \quad\quad C_z = 1 + 3n^{-0.2131}(1 - \pi_V)$$
-$$1.00 < \pi_V \leq 1.25 \quad\quad C_z = 1 + 3(1 - \pi_V)$$
-$$1.25 < \pi_V \leq 2.00 \quad\quad C_z = C_1 - C_2 \pi_V^{C_3}$$
-==== II. Kneading Blocks ====
-$$C_x = \frac{1}{\pi} + \sqrt{2}$$
-$$C_z = \frac{\pi + 1}{\pi} \cdot \frac{\dot{V}}{v_0 b_{max} h k}$$
-==== III. Reconveying Elements ====
-$$C_x = 1.75 + 1.75n$$
-$$0.00 \leq \pi_V \leq 1.00 \quad\quad C_z = (0.42 + 0.8828n)\pi_V + (1.75 + 1.9692n)$$
-For re-conveying elements, the two equations shown in table can be used. These approximation equations are based on a numerical solution of a system of differential equations, the systems of differential equations table [[en:grundlagenhandbuch:drehmoment_und_antriebsleistung:literatur|[1]]]. The same prerequisites are used as were for the equations of the conveying elements [[en:grundlagenhandbuch:drehmoment_und_antriebsleistung:literatur|[1]]].
-The estimation of the power consumption must differ in the 3 function sections in the figure on the barrel wall and in the melting section.
-**Figure:** Model for the power calculation.
-The calculation of the power consumption within each section is explained hereafter (for conveying elements and kneading blocks).
-===== Melting Range =====
-The power consumption can be calculated in zones of constant geometry using the following equation:
-$$P_1 = \left\{\int_0^{l_n} \left[C_z^2 + \tan^2(\varphi_s) C_x^2\right]^{\frac{1}{2}} C_z + v_{0x}^{l_n} \left[C_z^2 \cot^2(\theta_G) + C_x\right] + C_x \right\} \frac{|k(T)b(1-y)\Delta z_k}{h} \tag{1}$$
-Depending on the value of y a pure melt section, consider equation (1) (y = 0) a melt pool in the melting section (0 < y < 1).
-In addition the radial clearance is considered through:
-$$P_2 = \frac{K(T_Z)e_{max}\Delta z v_{0z}^{1+n}}{s_R^n} \{1 + \tan^2(\varphi_s)\}^{\frac{n+1}{2}} k \tag{2}$$
-Equation (2) assumes a pure drag flow over the radial clearance.
-===== Melting Section =====
-According to the figure the melting section is divided up into two areas:
-Melt film on the barrel wall with an underlying solid bed of the width: $b \cdot y$
-The principle is again the same as that in equation $$P = \int_0^z \int_{-\frac{b_{max}}{2}}^{+\frac{b_{max}}{2}} (\tau_{0x} r_{0x} + \tau_{0z} r_{0z}) dx \, dz$$ For the melt film, the shear stresses are to be replaced according to [[en:grundlagenhandbuch:drehmoment_und_antriebsleistung:literatur|[3]]] as follows:
-$$\tau_{0x} = K(T_{Fl}) \left(\frac{v_{rel}}{\delta}\right)^{n-1} \frac{v_{0x}}{\delta} \tag{1}$$
-$$\tau_{0z} = K(T_{Fl}) \left(\frac{v_{rel}}{\delta}\right)^{n-1} \frac{v_{0z} - v_{Fz}}{\delta} \tag{2}$$
-With this the relative speed of the melt film $v_{rel}$ will be formed using:
-$$v_{rel} = \sqrt{(v_{0z} + v_{Fz})^2 + v_{0x}^2} \tag{3}$$
-This results in the consumption:
-$$P_3 = \frac{K(T_{Fl}) \overline by\Delta z v_{rel}^{n-1}}{n} (v_{0x}^2 + (v_{0z} - v_{Fz})v_{0z})k \tag{4}$$
-A prerequisite of this way of modelling is the assumption of a pure drag flow in the melt film.
-===== Melt Layer =====
-Depending on the value of y, bearing in mind equation
-$$P_1 = \left\{v_{0z}^{1+n} \left[C_z^2 + \tan^2(\varphi_s) C_x^2\right]^{\frac{n-1}{2}} C_z + v_{0x}^{1+n} \left[C_z^2 + \cot^2(\varphi_z) + C_x\right]^{\frac{n-1}{2}} C_x\right\} \frac{K(T_{Fl})b(1-y)\Delta z}{n} k$$
-a pure melt section ($y = 0$) or a melt pool in the melting section ($0 < y < 1$).
-The individual powers in the different function ranges and zones yield the power consumption in the processing unit of the machine.
-$$P = \sum(P_1)_i + \sum(P_2)_i + \sum(P_3)_i \tag{1}$$
-The specific energy yield is a frequently used parameter for the interpretation of a processing unit. It is calculated by using the ratio of power consumption to mass flow:
-$$S_{Ve} = \frac{P}{\dot{m}} \tag{2}$$
-And it is proportional to the product: $\overline{\eta \dot{\gamma}^2} t$, respectively proportional to the product: $\overline{\tau \dot{\gamma}} t$.
-The screw torque yields from the total power consumption using the:
-$$M_d = \frac{P_{total}}{4n_D x} \tag{3}$$
-The screw torque is related to that of one screw.