Edits to “Statement”, “Solution”, “Answer”

naz edited
revision #18483 parent #18482 ← older newer →
@@ -1,50 +1,61 @@
### Statement
−$5.3.13.$ [Insert the problem statement]
+$5.3.13.$ The thermal conductivity of gases $A_1$ and $A_2$ is equal to $k_1$ and $k_2$, respectively. Determine the thermal
+conductivity of a mixture in which $A_1$ gas molecules are α times larger than $A_2$ gas molecules. The
+temperature of the gases is the same, the gases are monatomic. The molar mass of gases is $µ_1$ and $µ_2$,
+respectively.
### Solution
−Определим коэффициенты теплопроводности газов в смеси обозначим их $\kappa'_{1}$ и $\kappa'_{2}$.Так как газы одноатомные и температуры одинаковые можно считать, что:
+Let's determine the thermal conductivity coefficients of gases in a mixture, denote them $\kappa'_{1}$ and $\kappa'_{2}$. Since the gases are monatomic and the temperatures are the same, we can assume that:
$$\kappa\sim \frac{1}{\sigma\cdot \sqrt{\mu}}$$
−Значит после всех преобразований можно получить:
+This means that after all the transformations we can obtain:
+
$$\frac{\kappa_{1}}{\kappa_{2}}=\sqrt{\frac{\mu_{2}}{\mu_{1}}}\cdot \frac{r_{2}^2}{r_{1}^2}$$
−Где $r_{1}$ и $r_{2}$ радиусы атомов газа.
+Where $r_{1}$ and $r_{2}$ are the radii of the gas atoms.
+
$$r_{2}= r_{1}\cdot (\frac{\kappa_{1}}{\kappa_{2}}\cdot \sqrt{\frac{\mu_{1}}{\mu_{2}}})^{\frac{1}{2}}$$
−Концентрации газов в смеси относятся как $$\frac{n_{1}}{n_{2}}=\alpha$$
−Число столкновений молекул газа 1 и газа 2 в смеси:
+The concentrations of gases in a mixture are related as $$\frac{n_{1}}{n_{2}}=\alpha$$
−$$\nu_{1}=4\cdot \pi\cdot r_{1}^{2}\cdot n_{1}\cdot <\upsilon_{кв_{1}}> + \frac{n_{1}}{\alpha}\cdot \pi\cdot (r_{1} + r_{2})^{2}\cdot <\upsilon_{кв_{1}}>$$
−$$\nu_{2}=4\cdot \pi\cdot r_{2}^{2}\cdot n_{2}\cdot <\upsilon_{кв_{2}}> + n_{2}\cdot \alpha\cdot \pi\cdot (r_{1} + r_{2})^{2}\cdot <\upsilon_{кв_{2}}>$$
+The number of collisions of molecules of gas 1 and gas 2 in the mixture:
−Запишем отношения теплопроводностей "до" к "после", после преобразований получим:
+$$\nu_{1}=4\cdot \pi\cdot r_{1}^{2}\cdot n_{1}\cdot <\upsilon_{sq_{1}}> + \frac{n_{1}}{\alpha}\cdot \pi\cdot (r_{1} + r_{2})^{2}\cdot <\upsilon_{sq_{1}}>$$
+$$\nu_{2}=4\cdot \pi\cdot r_{2}^{2}\cdot n_{2}\cdot <\upsilon_{sq_{2}}> + n_{2}\cdot \alpha\cdot \pi\cdot (r_{1} + r_{2})^{2}\cdot <\upsilon_{sq_{2}}>$$
+
+
+Let's write down the ratio of thermal conductivities "before" to "after", after transformations we get:
+
$$\frac{\kappa_{1}}{\kappa'_{1}}=\frac{4\cdot \pi\cdot r_{1}^{2}\cdot n_{1} + \frac{n_{1}}{\alpha}\cdot \pi\cdot (r_{1} + r_{2})^{2}}{4\cdot \pi\cdot r_{1}^{2}\cdot n_{1}}$$
−Подставим выражение для $r_{2}$, закончим преобразования и выразим новую теплопроводность газа 1:
+We substitute the expression for $r_{2}$, complete the transformations and express the new thermal conductivity of gas 1:
+
$$\kappa'_{1} = \frac{\kappa_{1}}{1 + \frac{1}{4\cdot \alpha}\cdot [1 + (\frac{\kappa_{1}}{\kappa_{2}}\cdot \sqrt{\frac{\mu_{1}}{\mu_{2}}})^\frac{1}{2}]^{2}}$$
−Аналогично для газа 2:
+Similarly for gas 2:
$$\kappa'_{2} = \frac{\kappa_{2}}{1 + \frac{\alpha}{4}\cdot [1 + (\frac{\kappa_{2}}{\kappa_{1}}\cdot \sqrt{\frac{\mu_{2}}{\mu_{1}}})^\frac{1}{2}]^{2}}$$
−Пусть создан градиент температур $\frac{\partial T}{\partial x}$ внутри смеси. Поток тепла через смесь равен:
+Let a temperature gradient $\frac{\partial T}{\partial x}$ be created within the mixture. The heat flux through the mixture is:
+
$$q_{12} = \kappa_{12}\cdot \frac{\partial T}{\partial x}\cdot S = q_{1} + q_{2} = \kappa'_{1}\cdot \frac{\partial T}{\partial x}\cdot S + \kappa'_{2}\cdot \frac{\partial T}{\partial x}\cdot S$$
−Отсюда:
+
+From here:
$$\kappa_{12} = \kappa'_{1} + \kappa'_{2}$$
$$\kappa_{12} = \frac{\kappa_{1}}{1 + \frac{1}{4\cdot \alpha}\cdot [1 + (\frac{\kappa_{1}}{\kappa_{2}}\cdot \sqrt{\frac{\mu_{1}}{\mu_{2}}})^\frac{1}{2}]^{2}} + \frac{\kappa_{2}}{1 + \frac{\alpha}{4}\cdot [1 + (\frac{\kappa_{2}}{\kappa_{1}}\cdot \sqrt{\frac{\mu_{2}}{\mu_{1}}})^\frac{1}{2}]^{2}}$$
#### Answer
−[Insert a concise answer or boxed result]
+$$\kappa_{12} = \frac{\kappa_{1}}{1 + \frac{1}{4\cdot \alpha}\cdot [1 + (\frac{\kappa_{1}}{\kappa_{2}}\cdot \sqrt{\frac{\mu_{1}}{\mu_{2}}})^\frac{1}{2}]^{2}} + \frac{\kappa_{2}}{1 + \frac{\alpha}{4}\cdot [1 + (\frac{\kappa_{2}}{\kappa_{1}}\cdot \sqrt{\frac{\mu_{2}}{\mu_{1}}})^\frac{1}{2}]^{2}}$$