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

astrosander edited Dec 30, 2024, 5:57 AM

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revision #12606 parent #12604 ← older

en/5.4.17.md +15 −26

		@@ -1,60 +1,52 @@
			### Statement

		−	$5.4.16*.$ ~~[Insert problem description here]~~
		−
		−	__Example Statement__:
		−	~~$1.1.1.$ Between two flat parallel~~ plat~~es located at a distance δ from each oth~~er, there
		−	is a monatomic gas (the free path of atoms is much longer than δ). Estimate
		+	$5.4.16^*.$ 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 + ∆T, respectively, and the unit volume of gas contains n' atoms; µ is the
		+	$T + \Delta T$, respectively, and the unit volume of gas contains $n'$ atoms; $\mu$ is the
			mass of the atom

		−
			### Solution

		−	[Your solution should be placed here]
		−
		−	__Example Solution__:
			Heat flow is defined by

		−
		−	$$J =\frac{ΔE}{A*t} $$
		+	$$J =\frac{\Delta E}{A\cdot t} $$

			We know that the change of the energy is given by

		−	$$ΔE = \frac{nRΔT}{\gamma-1}$$
		+	$$\Delta E = \frac{nR\Delta T}{\gamma-1}$$

		−	Where ~~(n)~~ is the number of moles
		+	Where $n$ is the number of moles

			We know that Heat flow then

		−	$$J = \frac{nRΔT}{(\gamma-1)A*t}$$
		+	$$J = \frac{nR\Delta T}{(\gamma-1)A\cdot t}$$

		−	We know that the time is given by because (the free path of atoms is much longer than δ) furthermore ~~λ>>δ t~~herefore the average collision between the molecules will be δ:
		+	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 $\delta$:

		−	$$t = \frac{δ}{v}$$
		+	$$t = \frac{\delta}{v}$$

			Likewise:

		−	$$J = \frac{nRΔTv}{(\gamma-1)A*δ}$$
		+	$$J = \frac{nR\Delta Tv}{(\gamma-1)A\cdot\delta}$$

			Now we have to:

		−	$$A*δ=V$$
		+	$$A\cdot \delta=V$$

			So from the previous equations we have to:

			$$J = \frac{nRΔTv}{(\gamma-1)V}$$

			Another way to write this equation is:

		−	$$J = \frac{n'KΔTv}{(\gamma-1)}$$
		+	$$J = \frac{n'K\Delta Tv}{(\gamma-1)}$$

		−	Where (n') is the number of the atoms per unit volume:
		+	Where $n'$ is the number of the atoms per unit volume:

		−	$$v=\sqrt{\frac{3R(T+Δ)}{\mu}}$$
		+	$$v=\sqrt{\frac{3R(T+\Delta T)}{\mu}}$$

		−	How ~~ΔT<<T~~ is much less than this
		+	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

			$$J = \frac{n'KΔT}{(\gamma-1)}\sqrt{\frac{3RT}{\mu}}$$

		@@ -65,6 +57,3 @@Solution
			#### Answer

			$$\boxed{J = \frac{n'KΔT}{(\gamma-1)}\sqrt{\frac{3RT}{\mu}}}$$
		−
		−
		−

	@@ -1,60 +1,52 @@
	### Statement		### Statement

	$5.4.16*.$ ~~[Insert problem description here]~~		$5.4.16^*.$ 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
	__Example Statement__:
	~~$1.1.1.$ Between two flat parallel~~ plat~~es located at a distance δ from each oth~~er, there
	is a monatomic gas (the free path of atoms is much longer than δ). Estimate
	the heat flux density if the temperature of the plates is maintained at T and		the heat flux density if the temperature of the plates is maintained at T and
	T + ∆T, respectively, and the unit volume of gas contains n' atoms; µ is the		$T + \Delta T$, respectively, and the unit volume of gas contains $n'$ atoms; $\mu$ is the
	mass of the atom		mass of the atom


	### Solution		### Solution

	[Your solution should be placed here]

	__Example Solution__:
	Heat flow is defined by		Heat flow is defined by

			$$J =\frac{\Delta E}{A\cdot t} $$
	$$J =\frac{ΔE}{A*t} $$

	We know that the change of the energy is given by		We know that the change of the energy is given by

	$$ΔE = \frac{nRΔT}{\gamma-1}$$		$$\Delta E = \frac{nR\Delta T}{\gamma-1}$$

	Where ~~(n)~~ is the number of moles		Where $n$ is the number of moles

	We know that Heat flow then		We know that Heat flow then

	$$J = \frac{nRΔT}{(\gamma-1)A*t}$$		$$J = \frac{nR\Delta T}{(\gamma-1)A\cdot t}$$

	We know that the time is given by because (the free path of atoms is much longer than δ) furthermore ~~λ>>δ t~~herefore the average collision between the molecules will be δ:		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 $\delta$:

	$$t = \frac{δ}{v}$$		$$t = \frac{\delta}{v}$$

	Likewise:		Likewise:

	$$J = \frac{nRΔTv}{(\gamma-1)A*δ}$$		$$J = \frac{nR\Delta Tv}{(\gamma-1)A\cdot\delta}$$

	Now we have to:		Now we have to:

	$$A*δ=V$$		$$A\cdot \delta=V$$

	So from the previous equations we have to:		So from the previous equations we have to:

	$$J = \frac{nRΔTv}{(\gamma-1)V}$$		$$J = \frac{nRΔTv}{(\gamma-1)V}$$

	Another way to write this equation is:		Another way to write this equation is:

	$$J = \frac{n'KΔTv}{(\gamma-1)}$$		$$J = \frac{n'K\Delta Tv}{(\gamma-1)}$$

	Where (n') is the number of the atoms per unit volume:		Where $n'$ is the number of the atoms per unit volume:

	$$v=\sqrt{\frac{3R(T+Δ)}{\mu}}$$		$$v=\sqrt{\frac{3R(T+\Delta T)}{\mu}}$$

	How ~~ΔT<<T~~ is much less than this		How $\Delta T \ll T$ is much less than this

	$$v\approx\sqrt{\frac{3RT}{\mu}}$$		$$v\approx\sqrt{\frac{3RT}{\mu}}$$

	Introducing this velocity into the equation for the flow we have to		Introducing this velocity into the equation for the flow we have to

	$$J = \frac{n'KΔT}{(\gamma-1)}\sqrt{\frac{3RT}{\mu}}$$		$$J = \frac{n'KΔT}{(\gamma-1)}\sqrt{\frac{3RT}{\mu}}$$

	@@ -65,6 +57,3 @@Solution
	#### Answer		#### Answer

	$$\boxed{J = \frac{n'KΔT}{(\gamma-1)}\sqrt{\frac{3RT}{\mu}}}$$		$$\boxed{J = \frac{n'KΔT}{(\gamma-1)}\sqrt{\frac{3RT}{\mu}}}$$