Unterschiede

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--- en:grundlagenhandbuch:zylinderwaermestroeme [2026/05/28 09:52] – deppe2
+++ en:grundlagenhandbuch:zylinderwaermestroeme [2026/05/28 09:53] (aktuell) – [Barrel Heat Flows] deppe2
@@ Zeile 89: / 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 103: / 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 129: / Zeile 129: @@
 Therefore the temperature $T_r$ at the barrel surface is
-$$T_{cyl,r} = T_{cool} - \frac{\dot{q}_{cool} \cdot s_r}{\lambda_{cyl}} \cdot \frac{\ln\left(\frac{A_{cool,r}}{A_{tube}}\right)}{1 - \frac{A_{tube}}{A_{cool,r}}}$$
+$$T_{cyl,r} = T_{cool} - \frac{\dot{q}_{cool} \cdot s_r}{\lambda_{cyl}} \cdot \frac{\ln\left(\frac{A_{cool,r}}{A_{tube}}\right)}{1 - \frac{A_{tube}}{A_{cool,r}}} \tag{22}$$
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