Measuring the True Density of Spheres in a Container

Quality27

Involved In Discussions
#1
Hello all,

I have been thinking about an issue I came across at work with measuring the true density of spheres in a container. For example, we will have 60+/- 4mm diameter spheres randomly filled in a 1 m^3 volume container. We use about 50 containers a day like that and the variability in size of the spheres is beyond control right now.

We have been trying to measure what is the true density of the spheres in the container. As it depends upon how the spheres are packed inside the container(random or random closely packed) , whenever we measure the weight of the container, it varies.

In order to measure the true density of the spheres, we must know the weight of the beads in the container and how much space is left between the spheres in that container so that we could get the true volume occupied by those spheres. I know that there are some mathematical formulas to calculate this when the spheres are of equal size. But when size is a variable, volume occupied by the spheres is a variable since it depends upon how they are packed in that container, the weight will be a variable too .. in this scenario, what form on statistical sampling or techniques I could use to measure the true density accurately? Also, are there any photo-analyzers, laser tools, etc that exist to do this job accurately? (50, 1 m^3 containers filled with either 60 mm, 40mm, 10 mm or 4 mm diameter spheres a day)Is there any technology that exists to do this job at this scale?

Thank you in advance. Please let me know if you need any clarification.
 
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dgriffith

Quite Involved in Discussions
#2
Re: True density of spheres

Thank you in advance. Please let me know if you need any clarification.
Yes, by all means please do.
As it depends upon how the spheres are packed inside the container(random or random closely packed) , whenever we measure the weight of the container, it varies.
Are the balls homogenous? Seems the weight would vary because sphere size varies, therefor the number of spheres in the container would also vary. Not sure the packing distribution matters--unless there is extra packing materiel inside also.

In order to measure the true density of the spheres, we must know the weight of the beads in the container and how much space is left between the spheres in that container so that we could get the true volume occupied by those spheres. I know that there are some mathematical formulas to calculate this when the spheres are of equal size. But when size is a variable, volume occupied by the spheres is a variable since it depends upon how they are packed in that container, the weight will be a variable too .. in this scenario, what form on statistical sampling or techniques I could use to measure the true density accurately? Also, are there any photo-analyzers, laser tools, etc that exist to do this job accurately? (50, 1 m^3 containers filled with either 60 mm, 40mm, 10 mm or 4 mm diameter spheres a day)Is there any technology that exists to do this job at this scale?
And when you say either 60, 40, 10 or 4, you means the package contains only one nominal size, not a mixture of sizes, yes?
And certainly you don't produce 10mm spheres that are ?4mm do you?:mg:
Are you looking to put only a specific number in the container (count by photo), or keep them all the same size (weigh-in-motion and kick), etc.?
 
Last edited:

Quality27

Involved In Discussions
#4
Re: True density of spheres

Yes, by all means please do.Are the balls homogenous? Seems the weight would vary because sphere size varies, therefor the number of spheres in the container would also vary. Not sure the packing distribution matters--unless there is extra packing materiel inside also..?
Homogeneous in the sense of nominal size. Packing distribution on the spheres varies depending on the way they are packed but since the quantities are very high, they are just going to be randomly packed. There is a way to come up with the volume occupied by equal sized spheres in a container but given the tolerence on the diameter, it becomes more complex. I could measure the variability of diameter in one container but its enormous amount of work to measure in for every container we use. So that throws me off.

And when you say either 60, 40, 10 or 4, you means the package contains only one nominal size, not a mixture of sizes, yes?
And certainly you don't produce 10mm spheres that are ?4mm do you?:mg:
Are you looking to put only a specific number in the container (count by photo), or keep them all the same size (weigh-in-motion and kick), etc
Yes, one container will have only one nominal size, not a mixture of those sizes. Ofcourse, tolerances are different for different diameters. Neither I am trying to put a certain number in the container or put a certain size in the container, I am trying to calculate the true volume occupied by the spheres in the container when the size is within tolerance. But I do not know how many spheres would be above the nominal and hoow many of them will be below the nominal value. For the application these spheres are being used, it is important for me to know the true density of spheres we are going to use. For an accurate solution, capex would not be a hurdle.

Please let me know if there are any questions.

Thank you.
 

Bev D

Heretical Statistician
Leader
Super Moderator
#5
what is the material type?
you can perform a particle size distribution measurement and then integrate to get the volume but you'll need an instrument to quantify the particle counts. they are not too expensive adn relatively easy to use....
 

Quality27

Involved In Discussions
#6
what is the material type?
you can perform a particle size distribution measurement and then integrate to get the volume but you'll need an instrument to quantify the particle counts. they are not too expensive adn relatively easy to use....

They are expanded polystyrene spheres coated with resin and hardener.

Thanks
 

Bev D

Heretical Statistician
Leader
Super Moderator
#7
OK - well I just realized you stated that the size was in mm. I was thinking microns. these are fairly large partles. you could try sieving them in progressive sieves to separate out the size bins.
 

Ninja

Looking for Reality
Trusted Information Resource
#8
We have been trying to measure what is the true density of the spheres in the container.
What are you specifically looking for? The density of the spheres, or the density of the packed bed in the container?

For the density of the spheres themselves, Archimedes will only work if you use a fluid in which they will sink (unless you have an awesomely accurate method for measuring the exposed depth above the fluid surface)
...but then again if they are spheres you would only need the sphere weight and a measured diameter...(use a CMM)

For the second, simply knowing the weight of the filled container, the weight of the empty container, and the container volume is all you need.

Another permutation is to know the packed bed density only of the filled portion of the container...which would simply be the diameter of the container, the filled height, and the container weight filled and empty. This would change with simply vibrating the container to rearrange the packing.

Which permutation are you seeking?

Edit/Addition:
Depending on the accuracy needed, you could also measure the free space in the container by adding a liquid to the container with known specific gravity and measuring the weight gain to fill the air space.
Since this is polystyrene, this may be considered a destructive test since drying it back out might be a real pain...
 
Last edited:

David-D

Involved In Discussions
#9
I'll try this a third time as my previous attempts don't seem to have posted.

It sounds like you're trying to determine the free volume. In that case, I'd suggest you need some sort of pycnometer, although I've generally use them for small particle sizes (metal powders, etc.). Perhaps a take-off on a two chamber leak test would work as long as the spheres are impermeable and rigid.

Here's how I'd envision it. The spheres (volume = Vs) would go in the first chamber of known volume (V1) and it would be closed. The first chamber would be connected to the second chamber (known volume V2 and pressure gage) with a closed valve and another closed valve to a pressurized gas source (air or N2). Open the valve to the source, pressurize chamber 2 to a known, initial pressure (Pi) and close it again. Open the valve between the chambers and let them equilibrate to a final pressure (Pf). Based upon the ideal gas law and that the product of the initial pressure and volume which should equal the product of the final pressure and volume you should be able to determine the volume of the spheres. I think it should look something like:
V2*Pi = Pf*[V2 + (V1 - Vs)]

If your container is rigid, airtight, and known, you might even be able to use it as the 1st chamber (like a drum with port).

Good luck,

David
 

Quality27

Involved In Discussions
#10
I'll try this a third time as my previous attempts don't seem to have posted.

It sounds like you're trying to determine the free volume. In that case, I'd suggest you need some sort of pycnometer, although I've generally use them for small particle sizes (metal powders, etc.). Perhaps a take-off on a two chamber leak test would work as long as the spheres are impermeable and rigid.

Here's how I'd envision it. The spheres (volume = Vs) would go in the first chamber of known volume (V1) and it would be closed. The first chamber would be connected to the second chamber (known volume V2 and pressure gage) with a closed valve and another closed valve to a pressurized gas source (air or N2). Open the valve to the source, pressurize chamber 2 to a known, initial pressure (Pi) and close it again. Open the valve between the chambers and let them equilibrate to a final pressure (Pf). Based upon the ideal gas law and that the product of the initial pressure and volume which should equal the product of the final pressure and volume you should be able to determine the volume of the spheres. I think it should look something like:
V2*Pi = Pf*[V2 + (V1 - Vs)]

If your container is rigid, airtight, and known, you might even be able to use it as the 1st chamber (like a drum with port).

Good luck,

David

Thank you all for your inputs. After several hours of brain storming, I feel nothing else helps apart from the boyle's law. I agree with David's suggestion above. I am just in the process of figuring out how to get a pycnometer made for such industrial scale application with proper reliability and repeatability.

Thank you everyone once again.
 
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