Unterschiede

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--- en:grundlagenhandbuch:zylinderwaermestroeme [2026/04/15 09:24] – [Barrel Heat Flows] neelest
+++ en:grundlagenhandbuch:zylinderwaermestroeme [2026/05/28 09:53] (aktuell) – [Barrel Heat Flows] deppe2
@@ Zeile 27: / 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.
@@ Zeile 45: / Zeile 49: @@
 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 [[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 $T_{cool}$ 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.
@@ Zeile 85: / 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 99: / 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 125: / Zeile 129: @@
 Therefore the temperature $T_r$ at the barrel surface is
-$$T_{cyl,r} = T_{cool} - \frac{\dot{q}_{cool} \cdot s_r}{\lambda_{cyl}} \cdot \frac{\ln\left(\frac{A_{cool,r}}{A_{tube}}\right)}{1 - \frac{A_{tube}}{A_{cool,r}}}$$
+$$T_{cyl,r} = T_{cool} - \frac{\dot{q}_{cool} \cdot s_r}{\lambda_{cyl}} \cdot \frac{\ln\left(\frac{A_{cool,r}}{A_{tube}}\right)}{1 - \frac{A_{tube}}{A_{cool,r}}} \tag{22}$$
 ===== 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