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--- en:grundlagenhandbuch:aufschmelzberechnung:aufschmelzmodell_fuer_dispers_verteilte_fuellstoffe [2026/02/02 15:27] – [Melt Temperature Development] pka
+++ en:grundlagenhandbuch:aufschmelzberechnung:aufschmelzmodell_fuer_dispers_verteilte_fuellstoffe [2026/02/05 11:03] (aktuell) – [Temperature of the Solids at the Location of the First Melt] pka
@@ Zeile 28: / Zeile 28: @@
 **Figure:** Temperature trends within a spherically formed particle
-In the area of unsteady heat transmittance, the heat increase starting from single particles [1, 2], or through solid beds [2, 3, 4], is considered. Since according to assumption 2, the melting of single particles should be assumed, it is also sensible to assume single particles in the solid conveying area. The figure shows diagrammatically the temperature trends inside a spherically formed particle. $\bar{T}$ is the average caloric temperature of the particle.
+In the area of unsteady heat transmittance, the heat increase starting from single particles [[en:grundlagenhandbuch:aufschmelzberechnung:literatur|[1, 2]]], or through solid beds [[en:grundlagenhandbuch:aufschmelzberechnung:literatur|[2, 3, 4]]], is considered. Since according to assumption 2, the melting of single particles should be assumed, it is also sensible to assume single particles in the solid conveying area. The figure shows diagrammatically the temperature trends inside a spherically formed particle. $\bar{T}$ is the average caloric temperature of the particle.
 The coupling of the energy theorem and kinetics gives the differential calculus for the spherically symmetric temperature field.
@@ Zeile 38: / Zeile 38: @@
 $$\theta = \frac{T - T_0}{T_0 - T_U} ; \tau = \frac{at}{r_0^2} ; \xi = \frac{r}{r_0} \tag{2}$$
-were introduced. For the case of a unique erratic temperature change from the starting temperature $T_0$ up to the environment temperature $T_U$, the solution of the differential calculus (2) is given by [2]. For sufficient lengths of time t in-side particles of finite expansion, similar temperature profiles can be expected. They are then described by the location function $f(x)$, which with extra time, experiences scaled reductions [2]:
+were introduced. For the case of a unique erratic temperature change from the starting temperature $T_0$ up to the environment temperature $T_U$, the solution of the differential calculus (2) is given by [[en:grundlagenhandbuch:aufschmelzberechnung:literatur|[2]]]. For sufficient lengths of time t in-side particles of finite expansion, similar temperature profiles can be expected. They are then described by the location function $f(x)$, which with extra time, experiences scaled reductions [[en:grundlagenhandbuch:aufschmelzberechnung:literatur|[2]]]:
 $$\theta = g(\tau) \cdot f(\xi) \tag{3}$$
@@ Zeile 82: / Zeile 82: @@
 **Table:** Constants of the determination of the temperature function
-The figure shows the value of the first four roots taken from [2] and the approximate values of those with the equations (11) and (12).
+The figure shows the value of the first four roots taken from [[en:grundlagenhandbuch:aufschmelzberechnung:literatur|[2]]] and the approximate values of those with the equations (11) and (12).
 {{ :en:grundlagenhandbuch:aufschmelzberechnung:en_sigma150_dlg_grundlagenhandbuch_aufschmelzberechnung_004.svg?500%nolink |}}
@@ Zeile 254: / Zeile 254: @@
 Every interval i of the step function should possess, like the flat channel, a linear velocity distribution. If one lets the number of intervals tend towards infinity and the interval widths tend towards zero, it results in the average dissipated output:
-$$(\overline{\tau\dot{\gamma}})_j = \frac{1}{\bar{h}^{1+n}} \int_{-\frac{b_{max}}{2}}^{x_f} \frac{K(T_j)v_0^{1+n}}{h(x)^{1+n}} dx \tag{3}$$
+$$(\overline{\tau\dot{\gamma}})_j = \frac{1}{x_f + \frac{b_{max}}{2}} \int_{-\frac{b_{max}}{2}}^{x_f} \frac{K(T_j)v_0^{1+n}}{h(x)^{1+n}} dx \tag{3}$$
 $x_f$ represents the position of the flow-front in the partially filled channel sections, assuming an ideal perpendicular flow-front. With the average effective channel depth
@@ Zeile 294: / Zeile 294: @@
 By equating (4) and (5) one gets:
-$\frac{\partial r}{\partial t} = -\left(\frac{r_G}{r}\right)^2 \frac{\rho_f}{\rho_s} \frac{\partial r_G}{\partial t} = -\left(\frac{r_G}{r}\right)^2 \frac{\rho_f}{\rho_s} \frac{\partial r_G}{\partial_z} \frac{\partial_z}{\partial t} \tag{6}$
+$$\frac{\partial r}{\partial t} = -\left(\frac{r_G}{r}\right)^2 \frac{\rho_f}{\rho_s} \frac{\partial r_G}{\partial t} = -\left(\frac{r_G}{r}\right)^2 \frac{\rho_f}{\rho_s} \frac{\partial r_G}{\partial_z} \frac{\partial_z}{\partial t} \tag{6}$$
 Solve equation (6) with respect to $\partial t$ and include the average flow velocity in the channel using $\left(\bar{v} = \frac{\partial z}{\partial t}\right)$ to get:
-$\partial t = \left(-\left(\frac{r_G}{r}\right)^2 \frac{\rho_f}{\rho_s} \frac{\bar{v}}{\partial z}\right)^{-1} \partial r \tag{7}$
+$$\partial t = \left(-\left(\frac{r_G}{r}\right)^2 \frac{\rho_f}{\rho_s} \bar{v} \frac{\partial r_G}{\partial z} \right)^{-1} \partial r \tag{7}$$
 Now put equation (7) through (3) with:
-$a_s = \frac{\lambda_s}{\rho_s c_p} \tag{8}$
+$$a_s = \frac{\lambda_s}{\rho_s c_p} \tag{8}$$
 With the definition of the constants:
-$\frac{\partial^2 T}{\partial r^2} + \left[\frac{1}{a_s}\left(\frac{r_G}{r}\right)^2 \frac{\rho_f}{\rho_s} \frac{\bar{v}}{\partial z} \frac{\partial r_G}{\partial z} + \frac{2}{r}\right] \frac{\partial T}{\partial r} = 0 \tag{9}$
+$$\frac{\partial^2 T}{\partial r^2} + \left[\frac{1}{a_s}\left(\frac{r_G}{r}\right)^2 \frac{\rho_f}{\rho_s} \bar{v} \frac{\partial r_G}{\partial z} + \frac{2}{r}\right] \frac{\partial T}{\partial r} = 0 \tag{9}$$
 From equation (8):
-$A = \frac{1}{a_s} r_G^2 \frac{\rho_f}{\rho_s} \frac{\bar{v}}{\partial z} \frac{\partial r_G}{\partial z} \tag{10}$
+$$A = \frac{1}{a_s} r_G^2 \frac{\rho_f}{\rho_s} \bar{v} \frac{\partial r_G}{\partial z} \tag{10}$$
-The double integration results in the following equation:
+The double integration of
-$\frac{\partial^2 T}{\partial r^2} + \left[\frac{A}{r^2} + \frac{2}{r}\right] \frac{\partial T}{\partial r} = 0 \tag{11}$
+$$\frac{\partial^2 T}{\partial r^2} + \left[\frac{A}{r^2} + \frac{2}{r}\right] \frac{\partial T}{\partial r} = 0 \tag{11}$$
+results in the following equation:
+$$T(r) = \frac{C_1}{A}e^{-\frac{A}{r}} + C_2 \tag{12}$$
 Whereby C1 and C2 are the integration constants. With the following boundary conditions:
-$T(r = \infty) = T_m \tag{12}$
+$$T(r = \infty) = T_m \tag{13}$$
-$T(r = r_G) = T_{fl} \tag{13}$
+$$T(r = r_G) = T_{fl} \tag{14}$$
 It results in the integration constants:
-$C_1 = A \frac{T_m - T_{fl}}{\exp\left(\frac{A}{r_G}\right) - 1} \tag{14}$
+$$C_1 = A \frac{T_m - T_{fl}}{\exp\left(\frac{A}{r_G}\right) - 1} \tag{15}$$
-$C_2 = T_m + \frac{T_m - T_{fl}}{\exp\left(\frac{A}{r_G}\right) - 1} \tag{15}$
+$$C_2 = T_m + \frac{T_m - T_{fl}}{\exp\left(\frac{A}{r_G}\right) - 1} \tag{16}$$
 The solution of the differential equation results in:
-$\frac{T_m - T(r)}{T_m - T_{fl}} = \frac{1 - \exp\left(\frac{A}{r}\right)}{1 - \exp\left(\frac{A}{r_G}\right)} \tag{16}$
+$$\frac{T_m - T(r)}{T_m - T_{fl}} = \frac{1 - \exp\left(\frac{A}{r}\right)}{1 - \exp\left(\frac{A}{r_G}\right)} \tag{17}$$
 Which is equilibrium of heat flows.
@@ Zeile 357: / Zeile 361: @@
 Now substitute the constant A.
-$A = \frac{1}{a_s} r_G^2 \frac{\rho_f}{\rho_s} \frac{\bar{v}}{\partial z} \frac{\partial r_G}{\partial z} \tag{21}$
+$A = \frac{1}{a_s} r_G^2 \frac{\rho_f}{\rho_s} \bar{v} \frac{\partial r_G}{\partial z} \tag{21}$
 With the constant A':