Правка разделов «Solution», «Answer»
en/5.4.16.md
+5 −65
| ### Statement | |||
| $5.4.16^*.$ Between two flat parallel plates located at a distance $\delta$ from each other, there is a monatomic gas (the free path of atoms is much longer than $\delta$). Estimate the heat flux density if the temperature of the plates is maintained at T and $T + \Delta T$, respectively, and the unit volume of gas contains $n'$ atoms; $\mu$ is the mass of the atom | |||
| @@ -4,78 +4,18 @@Statement | |||
| ### Solution | |||
| − | |||
| + | In the classical case, the heat flux is determined by the formula: | ||
| − | $$ | ||
| − | J =\frac{\Delta E}{A\cdot t} | ||
| − | $$ | ||
| + | $$q=-\\frac{1}{3}\\lambda\\left\\langle \\upsilon \\right\\rangle \\frac{i}{2}kn\\frac{\\partial T}{\\partial x}S$$ | ||
| − | We know that the change of the energy is given by | ||
| + | However, for the sake of clarity, it would be fair to ignore the coefficient $\\frac{1}{3}$. It should also be noted that for $\\lambda\\gg \\delta$, the mean free path of the molecules is determined by the distance $\\delta$. Also, given the smallness of $\\delta$, the derivative can be considered equal to $-\\frac{\\Delta T}{\\delta}$. The gas is monatomic, meaning $i=3$. We'll take the average velocity to be equal to the root-mean-square value: $\\left\\langle \\upsilon \\right\\rangle = \\sqrt{\\frac{3kT}{\\mu}}$. Hence, the final formula for the flux density is: | ||
| − | $$ | ||
| − | |||
| − | $$ | ||
| + | $$\\omega=\\frac{q}{S}=\\frac{3}{2}kn\\Delta T\\sqrt{\\frac{3kT}{\\mu}}$$ | ||
| − | Where $n$ is the number of moles | ||
| − | |||
| − | We know that Heat flow then | ||
| − | |||
| − | $$ | ||
| − | J = \frac{nR\Delta T}{(\gamma-1)A\cdot t} | ||
| − | $$ | ||
| − | |||
| − | We know that the time is given by because (the free path of atoms is much longer than $\delta$) furthermore $\lambda \ll \delta$ therefore the average collision between the molecules will be $\delta$: | ||
| − | |||
| − | $$ | ||
| − | t = \frac{\delta}{v} | ||
| − | $$ | ||
| − | |||
| − | Likewise: | ||
| − | |||
| − | $$ | ||
| − | J = \frac{nR\Delta Tv}{(\gamma-1)A\cdot\delta} | ||
| − | $$ | ||
| − | |||
| − | Now we have to: | ||
| − | |||
| − | $$ | ||
| − | A\cdot \delta=V | ||
| − | $$ | ||
| − | |||
| − | So from the previous equations we have to: | ||
| − | |||
| − | $$ | ||
| − | J = \frac{nR\Delta Tv}{(\gamma-1)V} | ||
| − | $$ | ||
| − | |||
| − | Another way to write this equation is: | ||
| − | |||
| − | $$ | ||
| − | J = \frac{n'K\Delta Tv}{(\gamma-1)} | ||
| − | $$ | ||
| − | |||
| − | Where $n'$ is the number of the atoms per unit volume: | ||
| − | |||
| − | $$ | ||
| − | v=\sqrt{\frac{3R(T+\Delta T)}{\mu}} | ||
| − | $$ | ||
| − | |||
| − | How $\Delta T \ll T$ is much less than this | ||
| − | |||
| − | $$ | ||
| − | v\approx\sqrt{\frac{3RT}{\mu}} | ||
| − | $$ | ||
| − | |||
| − | Introducing this velocity into the equation for the flow we have to | ||
| − | |||
| − | $$ | ||
| − | J = \frac{n'K\Delta T}{(\gamma-1)}\sqrt{\frac{3RT}{\mu}} | ||
| − | $$ | ||
| − | |||
| #### Answer | |||
| $$ | |||
| − | |||
| + | \\omega=\\frac{3}{2}kn\\Delta T\\sqrt{\\frac{3kT}{\\mu}} | ||
| $$ | |||
| ещё строк без изменений 4 | |||
| ### Statement | ### Statement | ||
| $5.4.16^*.$ Between two flat parallel plates located at a distance $\delta$ from each other, there is a monatomic gas (the free path of atoms is much longer than $\delta$). Estimate the heat flux density if the temperature of the plates is maintained at T and $T + \Delta T$, respectively, and the unit volume of gas contains $n'$ atoms; $\mu$ is the mass of the atom | $5.4.16^*.$ Between two flat parallel plates located at a distance $\delta$ from each other, there is a monatomic gas (the free path of atoms is much longer than $\delta$). Estimate the heat flux density if the temperature of the plates is maintained at T and $T + \Delta T$, respectively, and the unit volume of gas contains $n'$ atoms; $\mu$ is the mass of the atom | ||
| @@ -4,78 +4,18 @@Statement | |||
| ### Solution | ### Solution | ||
| In the classical case, the heat flux is determined by the formula: | |||
| $$ | $$q=-\\frac{1}{3}\\lambda\\left\\langle \\upsilon \\right\\rangle \\frac{i}{2}kn\\frac{\\partial T}{\\partial x}S$$ | ||
| J =\frac{\Delta E}{A\cdot t} | |||
| $$ | |||
| We know that the change of the energy is given by | However, for the sake of clarity, it would be fair to ignore the coefficient $\\frac{1}{3}$. It should also be noted that for $\\lambda\\gg \\delta$, the mean free path of the molecules is determined by the distance $\\delta$. Also, given the smallness of $\\delta$, the derivative can be considered equal to $-\\frac{\\Delta T}{\\delta}$. The gas is monatomic, meaning $i=3$. We'll take the average velocity to be equal to the root-mean-square value: $\\left\\langle \\upsilon \\right\\rangle = \\sqrt{\\frac{3kT}{\\mu}}$. Hence, the final formula for the flux density is: | ||
| $$ | $$\\omega=\\frac{q}{S}=\\frac{3}{2}kn\\Delta T\\sqrt{\\frac{3kT}{\\mu}}$$ | ||
| $$ | |||
| Where $n$ is the number of moles | |||
| We know that Heat flow then | |||
| $$ | |||
| J = \frac{nR\Delta T}{(\gamma-1)A\cdot t} | |||
| $$ | |||
| We know that the time is given by because (the free path of atoms is much longer than $\delta$) furthermore $\lambda \ll \delta$ therefore the average collision between the molecules will be $\delta$: | |||
| $$ | |||
| t = \frac{\delta}{v} | |||
| $$ | |||
| Likewise: | |||
| $$ | |||
| J = \frac{nR\Delta Tv}{(\gamma-1)A\cdot\delta} | |||
| $$ | |||
| Now we have to: | |||
| $$ | |||
| A\cdot \delta=V | |||
| $$ | |||
| So from the previous equations we have to: | |||
| $$ | |||
| J = \frac{nR\Delta Tv}{(\gamma-1)V} | |||
| $$ | |||
| Another way to write this equation is: | |||
| $$ | |||
| J = \frac{n'K\Delta Tv}{(\gamma-1)} | |||
| $$ | |||
| Where $n'$ is the number of the atoms per unit volume: | |||
| $$ | |||
| v=\sqrt{\frac{3R(T+\Delta T)}{\mu}} | |||
| $$ | |||
| How $\Delta T \ll T$ is much less than this | |||
| $$ | |||
| v\approx\sqrt{\frac{3RT}{\mu}} | |||
| $$ | |||
| Introducing this velocity into the equation for the flow we have to | |||
| $$ | |||
| J = \frac{n'K\Delta T}{(\gamma-1)}\sqrt{\frac{3RT}{\mu}} | |||
| $$ | |||
| #### Answer | #### Answer | ||
| $$ | $$ | ||
| \\omega=\\frac{3}{2}kn\\Delta T\\sqrt{\\frac{3kT}{\\mu}} | |||
| $$ | $$ | ||
| ещё строк без изменений 4 | |||