$1.1.1.$ Between two flat parallel plates located at a distance δ from each other, 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
T + ∆T, respectively, and the unit volume of gas contains n' atoms; µ is the
mass of the atom
### Solution
[Your solution should be placed here]
__Example Solution__:
Heat flow is defined by
$$J =\frac{ΔE}{A*t}$$
We know that the change of the energy is given by
$$ΔE = \frac{nRΔT}{\gamma-1}$$
Where (n) is the number of moles
We know that Heat flow then
$$J = \frac{nRΔT}{(\gamma-1)A*t}$$
We know that the time is given by because (the free path of atoms is much longer than δ) furthermore λ>>δ therefore the average collision between the molecules will be δ:
$$t = \frac{δ}{v}$$
Likewise:
$$J = \frac{nRΔTv}{(\gamma-1)A*δ}$$
Now we have to:
$$A*δ=V$$
So from the previous equations we have to:
$$J = \frac{nRΔTv}{(\gamma-1)V}$$
Another way to write this equation is:
@@ -50,16 +50,20 @@Solution
$$J = \frac{n'KΔTv}{(\gamma-1)}$$
−
Where (n') is the number of the atoms
+
Where (n') is the number of the atoms per unit volume:
+
$$v=\sqrt{\frac{3R(T+Δ)}{\mu}}$$
−
How ΔT<<T is much less than this
−
#### Answer
+
$$v\approx\sqrt{\frac{3RT}{\mu}}$$
−
[Insert a concise answer or boxed result, like this:]
+
Introducing this velocity into the equation for the flow we have to
$1.1.1.$ Between two flat parallel plates located at a distance δ from each other, there
$1.1.1.$ Between two flat parallel plates located at a distance δ from each other, there
is a monatomic gas (the free path of atoms is much longer than δ). Estimate
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 + ∆T, respectively, and the unit volume of gas contains n' atoms; µ is the
mass of the atom
mass of the atom
### Solution
### Solution
[Your solution should be placed here]
[Your solution should be placed here]
__Example Solution__:
__Example Solution__:
Heat flow is defined by
Heat flow is defined by
$$J =\frac{ΔE}{A*t}$$
$$J =\frac{ΔE}{A*t}$$
We know that the change of the energy is given by
We know that the change of the energy is given by
$$ΔE = \frac{nRΔT}{\gamma-1}$$
$$ΔE = \frac{nRΔT}{\gamma-1}$$
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
$$J = \frac{nRΔT}{(\gamma-1)A*t}$$
$$J = \frac{nRΔT}{(\gamma-1)A*t}$$
We know that the time is given by because (the free path of atoms is much longer than δ) furthermore λ>>δ therefore 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 δ) furthermore λ>>δ therefore the average collision between the molecules will be δ:
$$t = \frac{δ}{v}$$
$$t = \frac{δ}{v}$$
Likewise:
Likewise:
$$J = \frac{nRΔTv}{(\gamma-1)A*δ}$$
$$J = \frac{nRΔTv}{(\gamma-1)A*δ}$$
Now we have to:
Now we have to:
$$A*δ=V$$
$$A*δ=V$$
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:
@@ -50,16 +50,20 @@Solution
$$J = \frac{n'KΔTv}{(\gamma-1)}$$
$$J = \frac{n'KΔTv}{(\gamma-1)}$$
Where (n') is the number of the atoms
Where (n') is the number of the atoms per unit volume:
$$v=\sqrt{\frac{3R(T+Δ)}{\mu}}$$
How ΔT<<T is much less than this
How ΔT<<T is much less than this
#### Answer
$$v\approx\sqrt{\frac{3RT}{\mu}}$$
[Insert a concise answer or boxed result, like this:]
Introducing this velocity into the equation for the flow we have to
