Unterschiede
Hier werden die Unterschiede zwischen zwei Versionen angezeigt.
| Beide Seiten der vorigen RevisionVorhergehende ÜberarbeitungNächste Überarbeitung | Vorhergehende Überarbeitung | ||
| en:grundlagenhandbuch:aufschmelzberechnung:aufschmelzmodell_fuer_dispers_verteilte_fuellstoffe [2026/02/02 15:45] – [Calculation of the Solid Bed Reduction Along the Melt Path] 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 | ||
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| Zeile 28: | Zeile 28: | ||
| **Figure:** Temperature trends within a spherically formed particle | **Figure:** Temperature trends within a spherically formed particle | ||
| - | In the area of unsteady heat transmittance, | + | In the area of unsteady heat transmittance, |
| The coupling of the energy theorem and kinetics gives the differential calculus for the spherically symmetric temperature field. | The coupling of the energy theorem and kinetics gives the differential calculus for the spherically symmetric temperature field. | ||
| Zeile 38: | Zeile 38: | ||
| $$\theta = \frac{T - T_0}{T_0 - T_U} ; \tau = \frac{at}{r_0^2} ; \xi = \frac{r}{r_0} \tag{2}$$ | $$\theta = \frac{T - T_0}{T_0 - T_U} ; \tau = \frac{at}{r_0^2} ; \xi = \frac{r}{r_0} \tag{2}$$ | ||
| - | 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: |
| $$\theta = g(\tau) \cdot f(\xi) \tag{3}$$ | $$\theta = g(\tau) \cdot f(\xi) \tag{3}$$ | ||
| Zeile 82: | Zeile 82: | ||
| **Table:** Constants of the determination of the temperature function | **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: |
| {{ : | {{ : | ||
| Zeile 298: | Zeile 298: | ||
| 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: | 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: | ||
| - | $$\partial t = \left(-\left(\frac{r_G}{r}\right)^2 \frac{\rho_f}{\rho_s} | + | $$\partial t = \left(-\left(\frac{r_G}{r}\right)^2 \frac{\rho_f}{\rho_s} \bar{v} |
| Now put equation (7) through (3) with: | Now put equation (7) through (3) with: | ||
| Zeile 306: | Zeile 306: | ||
| With the definition of the constants: | 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{\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}$$ |
| From equation (8): | From equation (8): | ||
| - | $$A = \frac{1}{a_s} r_G^2 \frac{\rho_f}{\rho_s} | + | $$A = \frac{1}{a_s} r_G^2 \frac{\rho_f}{\rho_s} \bar{v} \frac{\partial r_G}{\partial z} \tag{10}$$ |
| - | The double integration | + | The double integration |
| $$\frac{\partial^2 T}{\partial r^2} + \left[\frac{A}{r^2} + \frac{2}{r}\right] \frac{\partial T}{\partial r} = 0 \tag{11}$$ | $$\frac{\partial^2 T}{\partial r^2} + \left[\frac{A}{r^2} + \frac{2}{r}\right] \frac{\partial T}{\partial r} = 0 \tag{11}$$ | ||
| + | |||
| + | 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: | 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: | 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: | 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. | Which is equilibrium of heat flows. | ||
| Zeile 357: | Zeile 361: | ||
| Now substitute the constant A. | Now substitute the constant A. | ||
| - | $A = \frac{1}{a_s} r_G^2 \frac{\rho_f}{\rho_s} | + | $A = \frac{1}{a_s} r_G^2 \frac{\rho_f}{\rho_s} \bar{v} \frac{\partial r_G}{\partial z} \tag{21}$ |
| With the constant A': | With the constant A': | ||