Rate of gas leakage through small hole

In summary, the conversation discusses a problem involving the flow of gas through a small hole in a container. The approach of calculating the final mass of gas inside the container and the change in mass of the gas is questioned, and an alternative approach is suggested. However, it is noted that the given formula does not provide enough information to solve the problem. The participants suggest calculating the flow rate and density as functions of time, and using calculus to solve for the initial flow rate. The next part of the problem involves calculating the rate of fall of pressure in a cylinder filled with helium, assuming isothermal expansion. The potential effect of the hole size on the solution is also mentioned.
  • #1
rohanlol7
67
2

Homework Statement


Here http://imgur.com/a/4LRM6

upload_2016-8-13_13-21-9.png


2. Homework Equations

The equation is given int he question

The Attempt at a Solution



gas will stop flowing when the pressure inside the gas is equal to that of the surroundings. I calculated the final mass of gas inside the cube. Then i calculated the change in mass of the gas and i got that to be c^2*L^3/3*(1/p-1/Po) then i divided by t and converted L/t=c. However i don't think this is correct as i don't see why the rate of flow of gas should be constant and also my answer seems to be independant of a which seems counter intuitive.
 
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  • #2
Your approach looks questionable.
You're not interested in what happens after pressures equalize. You're interested in the outflow of gas due to the delta in the partial pressure between the inside & outside. The assumption is that outside the box the partial pressure of the gas is zero. So I would consider the component of gas molecules' velocity normal to the side containing the hole, then velocity times area times density = outflow rate.
 
  • #3
rude man said:
Your approach looks questionable.
You're not interested in what happens after pressures equalize. You're interested in the outflow of gas due to the delta in the partial pressure between the inside & outside. The assumption is that outside the box the partial pressure of the gas is zero. So I would consider the component of gas molecules' velocity normal to the side containing the hole, then velocity times area times density = outflow rate.
I agree my approach was dumb. I figured out that the rate flow is proportional to the density, however I can't seem to figure out what to do from there since the density is changing. I can figure out the change in density of the gas in a time t and then I'm stuck
 
  • #4
On the oth
rohanlol7 said:
I agree my approach was dumb. I figured out that the rate flow is proportional to the density, however I can't seem to figure out what to do from there since the density is changing. I can figure out the change in density of the gas in a time t and then I'm stuck
other hand I was able to derive an equation for the no of moles of gas left in the cube as a function of time. ( I got an exponential decay) but I guess this doesn't quite help me
 
  • #5
rohanlol7 said:
On the oth

other hand I was able to derive an equation for the no of moles of gas left in the cube as a function of time. ( I got an exponential decay) but I guess this doesn't quite help me
EDIT: thinking more about it I guess since the size of the box was given we should compute density ρ and flow rate as functions of time.

But the approach is the same: compute the normal velocity as a function of ρ, then compute flow rate, then ρ decreases with flow rate which reduces flow rate etc. Calculus involved.

Also I would forget no. of moles. Stick with kg since you're given the expression for the molecular velocity as a function of density (assumed in in kg/m3.)
 
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  • #6
Looking at this some more, it appears that not enough info is given.

It's obvious that you need to solve for initial density ρ0 and molecular velocity c0 somehow. But even if initial pressure p0 were given, which it isn't, that does not suffice to determine ρ0 and c0, obviously. And even if temperature were given, that would give us number of moles but not density.

So i would proceed by ignoring the given formula, assuming initial values ρ0 and c0, then solving for flow rate dm/dt which will be a function of t.
 
  • #7
rude man said:
since the size of the box was given we should compute density ρ and flow rate as functions of time.
I think the size of the box was specified as l merely in order to write l>>√a. I don't think it is supposed to feature in the answer.
rude man said:
Looking at this some more, it appears that not enough info is given.
... unless we are to assume current pressure p etc. It is not in terms of initial state, merely current state.
 
  • #8
rude man said:
EDIT: thinking more about it I guess since the size of the box was given we should compute density ρ and flow rate as functions of time.

But the approach is the same: compute the normal velocity as a function of ρ, then compute flow rate, then ρ decreases with flow rate which reduces flow rate etc. Calculus involved.

Also I would forget no. of moles. Stick with kg since you're given the expression for the molecular velocity as a function of density (assumed in in kg/m3.)
I think that's the right approach to the problem
 
  • #9
However all these approaches seem fine but, there is a next part to the question and this makes no sense unless what its asking me to calculate is the INITIAL flow rate!
A cylinder that contains helium has a diameter of 300 mm and a length of 1.5 m. The pressure of the gas in the cylinder is 10 atmospheres above the ambient air pressure. There is a small hole in the piston, diameter 2 mm. Calculate the rate of fall of pressure in the cylinder assuming that the expansion is isothermal.
 
  • #10
and the funnier thing is that the answer kind of needs to be independent of the velocity of the particles which is weirder
 
  • #11
Sorry for the delay. rude man and I were having a side discussion. It bothers me that the hole might not be sufficiently small to ignore bulk flow (Bernoulli style) and just treat it as pure diffusion. However, it looks like that is what is expected here.

I started by checking my ideas by deriving the given equation from first principles. Consider a gas of atoms mass m between two parallel plates area A a small distance h apart. Consider one moving at speed v at an angle theta to the normal to the plates.
The probability density of such an angle is sin(theta).
The momentum change normal to the plate on striking it is 2 mv cos(theta).
The frequency with which the atom strikes the same plate is v cos(theta)/2h.
The total number of atoms is ##\rho Ah/m##.
Putting all this together, the force on a plate is ##\int _0^{\pi/2}\rho \frac{Ah}m\frac{v\cos(\theta)}{2h}\sin(\theta)2mv\cos(\theta).d\theta = \frac 13\rho Av^2##.
Since c is the rms speed, the pressure is ##\frac 13\rho c^2##.
So far so good.

The escape rate through a hole area A would be the rate at which atoms would strike that area. This is the same integral as above, but leaving out the factor for momentum change per atom strike.

By this method, I get a mass loss rate of ##\frac 18\rho Av##.
But this is awkward since the average v is not c. So we need to plug in the distribution of speeds in an ideal gas.

@rude man , do you see any flaw in that?
 

Related to Rate of gas leakage through small hole

1. What factors affect the rate of gas leakage through a small hole?

The rate of gas leakage through a small hole is affected by the size of the hole, the pressure difference between the two sides of the hole, the type of gas, and the temperature.

2. How can the rate of gas leakage through a small hole be calculated?

The rate of gas leakage through a small hole can be calculated using the Hagen-Poiseuille equation, which takes into account the factors mentioned above.

3. How does the rate of gas leakage through a small hole change with temperature?

The rate of gas leakage through a small hole increases with an increase in temperature, as the molecules of the gas have more energy and can escape through the hole more easily.

4. Can the rate of gas leakage through a small hole be reduced?

The rate of gas leakage through a small hole can be reduced by decreasing the pressure difference between the two sides of the hole, decreasing the size of the hole, or using materials that are less permeable to gases.

5. Is the rate of gas leakage through a small hole a constant value?

No, the rate of gas leakage through a small hole is not a constant value. It can vary depending on the factors mentioned above and can also change over time as the hole may become larger or smaller due to wear and tear.

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