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--- en:grundlagenhandbuch:geometriegroessen_des_gleichdrall-doppelschneckenextruders [2026/05/21 22:28] – [Free Cross Section and Average Channel Depth] neelest
+++ en:grundlagenhandbuch:geometriegroessen_des_gleichdrall-doppelschneckenextruders [2026/06/16 14:37] (aktuell) – [Geometry of Tightly Intermeshing, Co-rotating Twin Screw Extruders] deppe2
@@ Zeile 9: / Zeile 9: @@
   * Modular setup of both screw and barrel (see figure)
-{{ :grundlagenhandbuch:geometriegroessen_des_gleichdrall-doppelschneckenextruders:eingesetzte_schneckenelemente:en_sigma150_dlg_grundlagenhandbuch_geometriegroessen_000.svg?700%nolink |}}
+{{ :en:grundlagenhandbuch:de_sigma150_dlg_grundlagenhandbuch_geometriegroessen_000.svg?nolink&700 |}}
 **Figure:** Modular setup of both screw and barre
@@ Zeile 244: / Zeile 244: @@
 with:
-$$A_{ABC} = \frac{\Omega}{4} \cdot a^2\tag{27}$$
+$$A_{ABC} = \frac{\Omega}{4} \cdot a^2 \tag{27}$$
-$$A_{MB/C} = \left[\Omega + \frac{\pi}{2} - \alpha - \arcsin\left(\frac{R_a}{R_i} \cdot \cos(\alpha)\right)\right] \cdot \frac{R_i^2}{2}\tag{28}$$
+$$A_{MB'C} = \left[\Omega + \frac{\pi}{2} - \alpha - \arcsin\left(\frac{R_a}{R_i} \cdot \cos(\alpha)\right)\right] \cdot \frac{R_i^2}{2} \tag{28}$$
-$$A_{MB/C/} = \frac{1}{2}\left\{\frac{a \cdot \sin(\Omega/2)}{\cos(\alpha - \Omega)} + \frac{R_i \cdot \cos\left[\alpha + \arcsin\left(\frac{R_a}{R_i} \cdot \cos(\alpha)\right)\right]}{\cos(\alpha)}\right\} \cdot R_a \cdot \cos(\alpha)\tag{29}$$
+$$A_{MB'C'} = \frac{1}{2}\left\{a \cdot \sin(\Omega/2) + R_i \cdot \cos\left[\alpha + \arcsin\left(\frac{R_a}{R_i} \cdot \cos(\alpha)\right)\right]\right\} \cdot \frac{r - R_i \cdot \cos\left(\Omega + \phi + \arcsin\left[\frac{R_a}{R_i} \cos(\alpha)\right]\right)}{\cos(\alpha - \Omega)} + R_a \cdot \cos(\alpha) \tag{29}$$
-$$A_{ABC/} = \frac{1}{2} \cdot \frac{a \cdot \sin(\Omega/2)}{\cos(\alpha - \Omega)} \cdot a \cdot \cos(-\alpha + \Omega/2)\tag{30}$$
+$$A_{ABC'} = \frac{1}{2} \frac{a \cdot \sin(\Omega/2)}{\cos(\alpha - \Omega)} \cdot a \cdot \cos(-\alpha + \Omega/2) \tag{30}$$
-$$A_{ABC} = \frac{\Omega}{4} \cdot a^2 \tag{31}$$
+$$A_{B'B'B''} = \frac{1}{2}r^2 \sqrt{1 - \frac{\left[r - R_i \cdot \cos\left\{\Omega + \phi + \arcsin\left(\frac{R_a}{R_i} \cos(\alpha)\right)\right\}\right]^2}{(R_i + r)^2}}$$
-$$A_{MB'C} = \left[\Omega + \frac{\pi}{2} - \alpha - \arcsin\left(\frac{R_a}{R_i} \cdot \cos(\alpha)\right)\right] \cdot \frac{R_i^2}{2} \tag{32}$$
-$$A_{MB'C'} = \frac{1}{2}\left\{a \cdot \sin(\Omega/2) + R_i \cdot \cos\left[\alpha + \arcsin\left(\frac{R_a}{R_i} \cdot \cos(\alpha)\right)\right]\right\} \cdot \frac{r - R_i \cdot \cos\left(\Omega + \phi + \arcsin\left[\frac{R_a}{R_i} \cos(\alpha)\right]\right)}{\cos(\alpha - \Omega)} + R_a \cdot \cos(\alpha) \tag{33}$$
-$$A_{ABC'} = \frac{1}{2} \frac{a \cdot \sin(\Omega/2)}{\cos(\alpha - \Omega)} \cdot a \cdot \cos(-\alpha + \Omega/2) \tag{34}$$
-$$A_{B'B'B''} = \frac{1}{2}r^2 \sqrt{1 - \frac{\left[r - R_i \cdot \cos\left\{\Omega + \phi + \arcsin\left(\frac{R_a}{R_i} \cos(\alpha)\right)\right\}\right]^2}{(R_i + r)^2}} \tag{35}$$
 $$- \frac{1}{2}R_i^2 \sin\left\{-\Omega - \phi - \arcsin\left[\frac{R_a}{R_i} \cos(\alpha)\right] - \arccos \frac{r - R_i \cdot \cos\left(\Omega + \phi + \arcsin\left[\frac{R_a}{R_i} \cos(\alpha)\right]\right)}{R_i + r}\right\} +$$
@@ Zeile 270: / Zeile 262: @@
 $$- \frac{1}{2}r^2 \arccos \frac{r - R_i \cdot \cos(\phi + \Omega + \arcsin\left[\frac{R_a}{R_i} \cos(\alpha)\right])}{R_i + r}$$
-$$- \frac{1}{2}R_i^2 \left(-\phi - \Omega + \pi - \arcsin\left[\frac{R_a}{R_i} \cos(\alpha)\right] - \arccos \frac{r - R_i \cdot \cos(\phi + \Omega + \arcsin\left[\frac{R_a}{R_i} \cos(\alpha)\right])}{R_i + r}\right)$$
+$$- \frac{1}{2}R_i^2 \left(-\phi - \Omega + \pi - \arcsin\left[\frac{R_a}{R_i} \cos(\alpha)\right] - \arccos \frac{r - R_i \cdot \cos(\phi + \Omega + \arcsin\left[\frac{R_a}{R_i} \cos(\alpha)\right])}{R_i + r}\right)\tag{31}$$
 Since the maximum channel depths of both the pushing flight element and the traditional conveying element are identical, we can calculate the average channel depth using the following equation:
-$$\bar{h} = \frac{A_{free}}{b_{max}}\tag{9}$$
+$$\bar{h} = \frac{A_{free}}{b_{max}}\tag{32}$$
 ==== Intermeshing Region of the Two Screws ====
@@ Zeile 285: / Zeile 278: @@
 This area can be calculated according to Booy  [[en:grundlagenhandbuch:geometriegroessen_des_gleichdrall-doppelschneckenextruders#references|[Boo78]]] using the following equation:
-$$A_{zw} = \left(\frac{D_s}{2}\right)^2 \cdot \sin(\Omega) - \frac{A_{Fr}}{i} \cdot \left(1 - \frac{\phi \cdot i}{\pi}\right)\tag{10}$$
+$$A_{zw} = \left(\frac{D_s}{2}\right)^2 \cdot \sin(\Omega) - \frac{A_{Fr}}{i} \cdot \left(1 - \frac{\phi \cdot i}{\pi}\right)\tag{33}$$
 ==== Shouldered Kneading Blocks ====
@@ Zeile 295: / Zeile 288: @@
 The model for the traditional kneading blocks, is based on the replacement of the discrete geometry (kneading discs) with a continuous geometry. One can find grooves within flights of this assumed continuous geometry. These grooves are characterized by their height $h_{Nut}$, their length $l_{Nut}$ and their width $b_{Nut}$. The width is half the width of the discs of traditional kneading blocks ($b_{Kn}$).
-$$b_{groove} = \frac{1}{2} \cdot b_{Kn} \tag{11}$$
+$$b_{groove} = \frac{1}{2} \cdot b_{Kn} \tag{34}$$
-$$l_{groove} = \frac{e}{\cos \varphi_s}\tag{12}$$
+$$l_{groove} = \frac{e}{\cos \varphi_s}\tag{35}$$
 The height of the groove results from the staggering angle. The staggering angle needs to be larger than the flight angle in order to form a groove. If the staggering angle is smaller than the flight angle then no groove can be calculated.
@@ Zeile 303: / Zeile 296: @@
 For staggering $\Phi \leq \alpha \leq 45°$ we use a linear relationship between staggering angle and depth of the groove. After conducting experimental investigations we found that it is necessary to apply a factor of 0.5 to that relationship.
-$$h_{groove} = 0.5 \cdot \left(\frac{\bar{h}}{\frac{1}{4} \cdot \pi - \Phi} \cdot \alpha - \frac{\Phi \cdot \bar{h}}{\frac{1}{4} \cdot \pi - \Phi} + s_R\right)\tag{13}$$
+$$h_{groove} = 0.5 \cdot \left(\frac{\bar{h}}{\frac{1}{4} \cdot \pi - \Phi} \cdot \alpha - \frac{\Phi \cdot \bar{h}}{\frac{1}{4} \cdot \pi - \Phi} + s_R\right)\tag{36}$$
 {{ :en:grundlagenhandbuch:geometriegroessen_des_gleichdrall-doppelschneckenextruders:en_sigma150_dlg_grundlagenhandbuch_geometriegroessen_016.svg?600%nolink |}}
@@ Zeile 311: / Zeile 304: @@
 The channel model for conveying shouldered kneading blocks is shown in the figure. We get the following equilibrium of flow rates within the control volume A-B-C:
-$$\dot{V} = k \cdot \dot{V}_{channel} \pm \dot{V}_{gap} \pm l \cdot \dot{V}_{groove}\tag{14}$$
+$$\dot{V} = k \cdot \dot{V}_{channel} \pm \dot{V}_{gap} \pm l \cdot \dot{V}_{groove}\tag{37}$$
 Where $l$ is the number of grooves between the points A and B. The number of the grooves corresponds to the number of kneading discs per rotation, hence follows:
-$$l = \frac{b_{Kn}}{2 \cdot \pi} \cdot (2 \cdot \pi - \Omega) \cdot \cos \varphi_s\tag{15}$$
+$$l = \frac{b_{Kn}}{2 \cdot \pi} \cdot (2 \cdot \pi - \Omega) \cdot \cos \varphi_s\tag{38}$$
 With $b_{Kn}$ being the width of the kneading discs.
@@ Zeile 321: / Zeile 314: @@
 The width of the flights $b_{Steg}$ is shortened by the total sum of all groove widths $l$.
-$$b_{threads} = (2 \cdot \pi - \Omega) \cdot D_s \cdot \cos \varphi_s - l \cdot b_{groove}\tag{16}$$
+$$b_{threads} = (2 \cdot \pi - \Omega) \cdot D_s \cdot \cos \varphi_s - l \cdot b_{groove}\tag{39}$$
 ==== Eccentric Kneading Blocks ====
@@ Zeile 337: / Zeile 330: @@
 Maximum diameter ratio of three-way profile ($i=3$)
-$$D_{v,3} = D_{v,3,max} = \frac{1}{2 \cdot \cos\left(\frac{\pi}{2 \cdot 3}\right) - 1} = 1.366  \tag{Equation 3-13}$$
+$$D_{v,3} = D_{v,3,max} = \frac{1}{2 \cdot \cos\left(\frac{\pi}{2 \cdot 3}\right) - 1} = 1.366  \tag{40}$$
 Outer diameter of three-course profile
-$$D_{a,3} = \frac{2 \cdot a}{1 + D_{v,3}} \tag{Equation 3-14}$$
+$$D_{a,3} = \frac{2 \cdot a}{1 + D_{v,3}} \tag{41}$$
 Eccentricity $e$
-$$e = \frac{D_{a,3} - D_{i,3}}{2} \tag{Equation 3-15}$$
+$$e = \frac{D_{a,3} - D_{i,3}}{2} \tag{42}$$
 Inside diameter of three-course profile
-$$D_{i,3} = 2 \cdot a - D_{a,3} \tag{Equation 3-16}$$
+$$D_{i,3} = 2 \cdot a - D_{a,3} \tag{43}$$
@@ Zeile 359: / Zeile 352: @@
 This profile shows that the comb angles are zero. Therefore the flank angles of the standard profile are the same:
-Comb angle: $\angle AOB = \angle BOC = \angle COD = \Omega_3 = 60° \tag{Equation 3-17}$
+Comb angle: $$\angle AOB = \angle BOC = \angle COD = \Omega_3 = 60° \tag{44}$$
-Flank angle: $\phi_{k,3} = \phi_{g,3} = 0 \tag{Equation 3-18}$
+Flank angle: $$\phi_{k,3} = \phi_{g,3} = 0 \tag{45}$$
-Outside radius: $R_{a,3} = \frac{D_{a,3}}{2} \tag{Equation 3-19}$
+Outside radius: $$R_{a,3} = \frac{D_{a,3}}{2} \tag{46}$$
-Inside radius: $R_{i,3} = \frac{D_{i,3}}{2} = \left(a - \frac{D_{a,3}}{2}\right) \tag{Equation 3-20}$
+Inside radius: $$R_{i,3} = \frac{D_{i,3}}{2} = \left(a - \frac{D_{a,3}}{2}\right) \tag{47}$$
 The coordinates of the points in the three-course standard profile can be calculated by these geometry sizes. Because the profile is symmetrical, only half the contour is selected as the calculation object.
-A: $x_A = R_{i,3}$; $y_A = 0$
+A: $x_A = R_{i,3}; y_A = 0\tag{48}$
-B: $x_B = R_{a,3} \cdot \cos 60° = \frac{1}{2}R_{a,3}$; $y_B = R_{a,3} \cdot \sin 60° = \frac{\sqrt{3}}{2}R_{a,3}$
+B: $x_B = R_{a,3} \cdot \cos 60° = \frac{1}{2}R_{a,3}; y_B = R_{a,3} \cdot \sin 60° = \frac{\sqrt{3}}{2}R_{a,3}\tag{49}$
-C: $x_C = R_{i,3} \cdot \cos 120° = -\frac{1}{2}R_{i,3}$; $y_C = R_{i,3} \cdot \sin 120° = \frac{\sqrt{3}}{2}R_{i,3}$
+C: $x_C = R_{i,3} \cdot \cos 120° = -\frac{1}{2}R_{i,3}; y_C = R_{i,3} \cdot \sin 120° = \frac{\sqrt{3}}{2}R_{i,3}\tag{50}$
-D: $x_D = -R_{a,3}$; $y_D = 0$
+D: $x_D = -R_{a,3}; y_D = 0\tag{51}$
 ==== Screw Mixing Elements ====
@@ Zeile 393: / Zeile 386: @@
 From the control volume ABC we get the equilibrium of flow rates:
-$$\dot{V} = k \cdot \dot{V}_{channel} \pm \dot{V}_{gap} \pm l \cdot \dot{V}_{groove}\tag{21}$$
+$$\dot{V} = k \cdot \dot{V}_{channel} \pm \dot{V}_{gap} \pm l \cdot \dot{V}_{groove}\tag{52}$$
 Where $l$ is the number of grooves between the points A and B. $l$ is calculated using the following equation:
-$$l = \frac{l_{Nut}}{2\pi} \cdot (2\pi - \Omega) \cdot \cos \varphi_s\tag{22}$$
+$$l = \frac{l_{Nut}}{2\pi} \cdot (2\pi - \Omega) \cdot \cos \varphi_s\tag{53}$$
 ==== Blister Elements ====
@@ Zeile 422: / Zeile 415: @@
 These sections are characterized by the following dimensionless numbers:
-$$k_1 = \frac{D_a}{D_z}\tag{23}$$
+$$k_1 = \frac{D_a}{D_z}\tag{54}$$
-$$k_2 = \frac{D_i}{D_a}\tag{24}$$
+$$k_2 = \frac{D_i}{D_a}\tag{55}$$
-$$cl = \frac{a}{D_z/2}\tag{25}$$
+$$cl = \frac{a}{D_z/2}\tag{56}$$
 ==== Thoothed Mixing Elements ====