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--- en:grundlagenhandbuch:drehmoment_und_antriebsleistung:urspruenglicher_ansatz:aufschmelzbereich [2025/06/20 09:24] – deppe2
+++ en:grundlagenhandbuch:drehmoment_und_antriebsleistung:urspruenglicher_ansatz:aufschmelzbereich [2026/01/27 15:16] (aktuell) – gelöscht deppe2
@@ Zeile 1: / Zeile 1: @@
-====== Melting Section ======
-===== 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*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 [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:
-$$P_3 = \frac{K(T_{Fl})by\Delta z v_{rel}^{n-1}}{n} (v_{0x}^2 + (v_{0z} - v_{Fz})v_{0z})k \tag{4}$$
-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: $\eta \gamma^2 t$, respectively proportional to the product: $t \gamma t$.
-The screw torque yields from the total power consumption using the:
-$$M_d = \frac{P_{total}}{4n_D \chi} \tag{3}$$
-The screw torque is related to that of one screw.