Unterschiede

Hier werden die Unterschiede zwischen zwei Versionen angezeigt.

--- en:grundlagenhandbuch:zylinderwaermestroeme [2025/06/26 11:37] – deppe2
+++ en:grundlagenhandbuch:zylinderwaermestroeme [2026/05/28 09:53] (aktuell) – [Barrel Heat Flows] deppe2
@@ Zeile 4: / Zeile 4: @@
 One possibility to determine the temperature at the inside wall of the barrel is to calculate it starting from the cooling channel. The heat flow per area which can maximally be discharged from the melt through the temperature control of the barrel can be used to define the heat flow more precisely in the melt temperature calculation in channel height direction. A general approach to calculate the specific heat flow that is discharged is shown with two different arrangements of the cooling channel. The first arrangement used in the ZE-UTX series from Kraus-Maffei Berstorff is shown in the following figure.
+{{ :en:grundlagenhandbuch:en_sigma150_dlg_grundlagenhandbuch_zylinderwaermestroeme_001.png?nolink |}}
 **Figure:** Arrangement of the cooling channels (light) and heating cartridge (dark) in a barrel element of a ZE-UTX from Kraus-Maffei Berstorff GmbH; Source: KraussMaffei Berstorff 2002
@@ Zeile 12: / Zeile 14: @@
 The surface of the twin bore is converted into a rectangular prism with the same surface. This rectangular prism has the width x, the height y and the length $L_{Zyl}/i_n$. Thereby also the distances $s_x$ and $s_y$ of the barrel interface of the cooling channels change into $s_x^*$ and $s_y^*$. The rectangular prism should here have the same ratio of width to height as the corresponding proportion of the real twin bore.
+{{ :en:grundlagenhandbuch:en_sigma150_dlg_grundlagenhandbuch_zylinderwaermestroeme_002.png?nolink |}}
 **Figure:** The surrogate arrangement used for the calculation
 The basis for the calculation is a heat balance on the volume between barrel and cooling channel surface. Let the specific heat flow $\dot{q}_{cool}$ (cooling) be constant along the whole cooling channel length.
+{{ :en:grundlagenhandbuch:en_sigma150_dlg_grundlagenhandbuch_zylinderwaermestroeme_003.png?nolink |}}
 **Figure:** Heat flows occurring
@@ Zeile 21: / Zeile 27: @@
 The balance of the heat flows in height direction is
-$$\dot{Q}_{cyl,y} - \dot{Q}_{\xi} = \dot{Q}_{cool,y}$$
+$$\dot{Q}_{cyl,y} - \dot{Q}_{\xi} = \dot{Q}_{cool,y} \tag{1}$$
 After introducing the dimensionless height coordinate
-$$\xi = \frac{y'}{s_y}$$
+$$\xi = \frac{y'}{s_y} \tag{2}$$
 with y' as moving coordinate the following results from the equation above
-$$\frac{A_{quad,y}}{A_{cool,y}} \cdot \frac{dT}{d\xi} = \left(\frac{A_{quad,y}}{A_{cool,y}} - 1\right) \cdot \xi \cdot \frac{dT}{d\xi} = \frac{s_y^* \cdot \dot{q}_{cool}}{\lambda_{cyl}}$$
+$$\frac{A_{quad,y}}{A_{cool,y}} \cdot \frac{dT}{d\xi} = \left(\frac{A_{quad,y}}{A_{cool,y}} - 1\right) \cdot \xi \cdot \frac{dT}{d\xi} = \frac{s_y^* \cdot \dot{q}_{cool}}{\lambda_{cyl}} \tag{3}$$
-whereas $A_{Quad,y}$ is the area of the rectangular prism surrogating the twin bore in the xz-plane and $A_{Kühl,y}$
+whereas $A_{Quad,y}$ is the area of the rectangular prism surrogating the twin bore in the xz-plane
-$$A_{cool,y} = \frac{1}{2}\pi \cdot D_{cool} \cdot L_{cool,x}$$
+$$A_{Quad,y} = y \cdot \frac{L_{Zyl}}{i_n} \tag{4}$$
+and $A_{Kühl,y}$
+$$A_{cool,y} = \frac{1}{2}\pi \cdot D_{cool} \cdot L_{cool,x} \tag{5}$$
 are corresponding to the lower half of the cooling channel surface in the xz-plane.
-Let the temperature $T_{Kühl}$ be at the cooling channel surface, and let the temperature $T_{Z,y}$ be at the barrel surface in the xz-plane. After integration and further mathematical operations the equation is
+Let the temperature $T_{cool}$ be at the cooling channel surface, and let the temperature $T_{Z,y}$ be at the barrel surface in the xz-plane. After integration and further mathematical operations the equation is
-$$T_{Z,y} = T_{cool} - \frac{s_y^* \cdot \dot{q}_{cool}}{\lambda_{Zyl}} \cdot \frac{\ln\left(\frac{A_{cool,y}}{A_{quad,y}}\right)}{1 - \frac{A_{quad,y}}{A_{cool,y}}}$$
+$$T_{Z,y} = T_{cool} - \frac{s_y^* \cdot \dot{q}_{cool}}{\lambda_{Zyl}} \cdot \frac{\ln\left(\frac{A_{cool,y}}{A_{quad,y}}\right)}{1 - \frac{A_{quad,y}}{A_{cool,y}}} \tag{6}$$
 Analogously, the temperature $T_{Z,x}$ at the barrel surface in the zy-plane can be determined.
-$$T_{Z,x} = T_{cool} - \frac{s_x^* \cdot \dot{q}_{cool}}{\lambda_{cyl}} \cdot \frac{\ln\left(\frac{A_{cool,x}}{A_{quad,x}}\right)}{1 - \frac{A_{quad,x}}{A_{cool,x}}}$$
+$$T_{Z,x} = T_{cool} - \frac{s_x^* \cdot \dot{q}_{cool}}{\lambda_{cyl}} \cdot \frac{\ln\left(\frac{A_{cool,x}}{A_{quad,x}}\right)}{1 - \frac{A_{quad,x}}{A_{cool,x}}} \tag{7}$$
 With the areas
-$$A_{quad,x} = x \frac{L_{cyl}}{i_n}$$
+$$A_{quad,x} = x \frac{L_{cyl}}{i_n} \tag{8}$$
 and
-$$A_{cool,y} = \frac{1}{2}\pi \cdot D_{cool} \cdot L_{cool,x}$$
+$$A_{cool,y} = \frac{1}{2}\pi \cdot D_{cool} \cdot L_{cool,x} \tag{9}$$
 As a temperature assumed inconsistent at the area of the twin bore is impracticable for the further calculations the average barrel temperature $T_Z$ is determined as follows:
-$$T_Z = \frac{A_{quad,x}}{A_{quad,x} + A_{quad,y}} \cdot T_{Z,x} + \frac{A_{quad,y}}{A_{quad,x} + A_{quad,y}} \cdot T_{Z,y}$$
+$$T_Z = \frac{A_{quad,x}}{A_{quad,x} + A_{quad,y}} \cdot T_{Z,x} + \frac{A_{quad,y}}{A_{quad,x} + A_{quad,y}} \cdot T_{Z,y} \tag{10}$$
-A decrease of the heat transfer coefficient is related to a decrease of the specific heat flow at the cooling channel surface $\dot{q}_{cool}$. The specific heat flow at the cooling channel surface can be determined according to Reiners [Rein87]. This is done with the product from the heat transfer coefficient of the cooling medium $\alpha_{Med}$ and the temperature difference between cooling medium and cooling channel surface/cooling medium-interface $(T_{Kühl}-T_{Med})$
+A decrease of the heat transfer coefficient is related to a decrease of the specific heat flow at the cooling channel surface $\dot{q}_{cool}$. The specific heat flow at the cooling channel surface can be determined according to Reiners [[en:grundlagenhandbuch:zylinderwaermestroeme#literatur |[Rei87]]]. This is done with the product from the heat transfer coefficient of the cooling medium $\alpha_{Med}$ and the temperature difference between cooling medium and cooling channel surface/cooling medium-interface $(T_{cool}-T_{med})$
-$$\dot{q}_{cool} = \alpha_{med}(T_{cool} - T_{med})$$
+$$\dot{q}_{cool} = \alpha_{med}(T_{cool} - T_{med}) \tag{11}$$
-With inserting $T_{Kühl}$ in the equation mentioned above the following results:
+With inserting $T_{cool}$ in the equation mentioned above the following results:
-$$\dot{q}_{cool} = \frac{\alpha_{med}(T_{Z,y} - T_{med})}{1 - \frac{Bi_y}{A_{quad,y}} \cdot \ln\left(\frac{A_{cool,y}}{A_{quad,y}}\right) \cdot \frac{1}{1 - \frac{A_{quad,y}}{A_{cool,y}}}}$$
+$$\dot{q}_{cool} = \frac{\alpha_{med}(T_{Z,y} - T_{med})}{1 - \frac{Bi_y}{A_{quad,y}} \cdot \ln\left(\frac{A_{cool,y}}{A_{quad,y}}\right) \cdot \frac{1}{1 - \frac{A_{quad,y}}{A_{cool,y}}}} \tag{12}$$
 Another widely-used arrangement of the cooling channels is shown in the following figure. In this case the cooling channels are arranged parallel to the extrusion direction.
+{{ :en:grundlagenhandbuch:en_sigma150_dlg_grundlagenhandbuch_zylinderwaermestroeme_004.png?nolink |}}
 **Figure:** Barrel geometry with parallel cooling channels in extrusions direction [picture Krauss-Maffei Berstorff, 2007]
 Considering the surrogate arrangement shown in the following figure the interface temperature can be determined analogously to the procedure explained above.
+{{ :en:grundlagenhandbuch:en_sigma150_dlg_grundlagenhandbuch_zylinderwaermestroeme_005.png?nolink |}}
 **Figure:** The surrogate arrangement for the barrel geometry with parallel cooling channels in extrusion direction used for the calculation
@@ Zeile 75: / Zeile 89: @@
 In this case, the twin bore is not replaced by a rectangular prism with the same surface but by a pipe with the same surface.
-$$\frac{O_{Acht}}{O_{Rohr}} = 1$$
+$$\frac{O_{Acht}}{O_{Rohr}} = 1 \tag{13}$$
 Hence, the diameter of the pipe $r_i$ results.
-$$r_i = -\frac{L_{cyl}}{2} + \sqrt{\left(\frac{L_{cyl}}{2}\right)^2 + U}$$
+$$r_i = -\frac{L_{cyl}}{2} + \sqrt{\left(\frac{L_{cyl}}{2}\right)^2 + U} \tag{14}$$
 whereas
-$$U = \frac{\frac{1}{2}(2\pi - \Omega) \cdot D_Z^2 + a \cdot D_Z \cdot \sin\left(\frac{\Omega}{2}\right) + (2\pi - \Omega) \cdot D_Z \cdot L_{cyl}}{2\pi}$$
+$$U = \frac{\frac{1}{2}(2\pi - \Omega) \cdot D_Z^2 + a \cdot D_Z \cdot \sin\left(\frac{\Omega}{2}\right) + (2\pi - \Omega) \cdot D_Z \cdot L_{cyl}}{2\pi} \tag{15}$$
 The surrogate arrangement is divided into an amount of elements which correspond to amount of the cooling channels.
@@ Zeile 89: / Zeile 103: @@
 The distance $s_r^*$ is then calculated with
-$$s_r = r_a - r_i$$
+$$s_r = r_a - r_i \tag{16}$$
 After introducing the dimensionless coordinate in radial direction
-$$\zeta = \frac{r'}{s_r}$$
+$$\zeta = \frac{r'}{s_r} \tag{17}$$
 with r' as moving coordinate the heat balance in radial direction is
-$$\dot{Q}_{cyl,r} - \dot{Q}_{\zeta} = \dot{Q}_{cool,r}$$
+$$\dot{Q}_{cyl,r} - \dot{Q}_{\zeta} = \dot{Q}_{cool,r} \tag{18}$$
 After inserting the following equation the result is
-$$\frac{A_{tube}}{A_{cool,r}} \cdot \frac{dT}{d\zeta} = \left(\frac{A_{tube}}{A_{cool,r}} - 1\right) \cdot \zeta \cdot \frac{dT}{d\zeta} = \frac{\dot{q}_{cool} \cdot s_r}{\lambda_{Zyl}}$$
+$$\frac{A_{tube}}{A_{cool,r}} \cdot \frac{dT}{d\zeta} = \left(\frac{A_{tube}}{A_{cool,r}} - 1\right) \cdot \zeta \cdot \frac{dT}{d\zeta} = \frac{\dot{q}_{cool} \cdot s_r}{\lambda_{Zyl}} \tag{19}$$
 with the area of the section of the surrogate pipe
-$$A_{tube} = \frac{2\pi \cdot r_i \cdot L_{cyl}}{i_{cool}}$$
+$$A_{tube} = \frac{2\pi \cdot r_i \cdot L_{cyl}}{i_{cool}} \tag{20}$$
 And the cooling channel area
-$$A_{cool,r} = \frac{\pi \cdot D_{cool} \cdot L_{cyl}}{2}$$
+$$A_{cool,r} = \frac{\pi \cdot D_{cool} \cdot L_{cyl}}{2} \tag{21}$$
 Here the length of the cooling channel in z-direction $L_{Kühl,z}$ is assumed to correspond to the length of the barrel element $L_{Zyl}$.
@@ Zeile 115: / Zeile 129: @@
 Therefore the temperature $T_r$ at the barrel surface is
-$$T_{cyl,r} = T_{cool} - \frac{\dot{q}_{cool} \cdot s_r}{\lambda_{cyl}} \cdot \frac{\ln\left(\frac{A_{cool,r}}{A_{tube}}\right)}{1 - \frac{A_{tube}}{A_{cool,r}}}$$
+$$T_{cyl,r} = T_{cool} - \frac{\dot{q}_{cool} \cdot s_r}{\lambda_{cyl}} \cdot \frac{\ln\left(\frac{A_{cool,r}}{A_{tube}}\right)}{1 - \frac{A_{tube}}{A_{cool,r}}} \tag{22}$$
-Excerpts from A. Thümen's dissertation:
+===== Literatur =====
-[Thüm08] A. Thümen: „Untersuchung und Beschreibung des dispersen Aufschmelzens in Gleichdrall-Doppelschneckenextrudern", Dissertation, Universität Paderborn, 2008
+[Thü08] A. Thümen: „Untersuchung und Beschreibung des dispersen Aufschmelzens in Gleichdrall-Doppelschneckenextrudern", Dissertation, Universität Paderborn, 2008
 [Rein87] U. Reiners: „Wärmeübertragung durch Spritzwasserkühlung heißer Oberflächen im Bereich stabiler Filmverdampfung", Dissertation, TU Clausthal, 1987