That would be my suggestion.

[ pi * radius2 * height ] - [ (4/3) * pi * radius3 * number of balls ] = porous volume

We know this only when we have known value of no. of lead balls. Is there any other method?

Gentlemen (Ladies): I hate to disillusion you about finding the void volume of a space packed with spheres. The liquid measurement method relies on the assumption that the liquid has a positive contact angle (i.e. does not wet the spheres). I would advise against using water as the fluid, since it wets nearly every surface it is in contact with (and therefore has a negative contact angle). If you want greater accuracy of measurement (and are concerned that all known sources of error in the measurement be reduced), you should not use decantation of the liquid as the means of measurement by difference, but you should prefer to measure the liquid by evaporation/condensation method to ensure that all of the liquid is removed from the void spaces, but then you must ensure that your experiment has the ability to recover all of the evaporate as condensate. That is all I have to offer on the subject.

I am confident that given certain assumptions, the mathematical models given certain boundary conditions (such as the finite volume and shape of the space), filling material limited to spheres of a related class (i.e. where the radius series is a geometric series of R, R/2, R/4,...etc.) is limited with respect to order (two to four sizes of balls for example), that a "clean" solution to the mathematics is possible.

Of course, the first order example is more likely to be the useful one in engineering, since usually engineers can come up with "uniform" particles (within the limits of engineering practices) when making a packed bed.

What does concern me about the original question is that the fellow wants to use "lead balls" as the packing material? Are you sure we are not discussing some sort of IED? Should I be discussing you with homeland security?

If you know the weight of the lead balls and there respective density you can calculate their volume and compare that to the volume of the cylinder.

Basic math and physics I think.

Simple way to figure out the volume of the voids and write the subsequent equation:

Install a water tight cap on the bottom of the tube, then place the cylinder it upright (plumb), then fill with water to the top. Stop. Drain the water into a bucket the while measuring it with a laboratory graduated cylinder or breaker, etc. Since you have calculated the internal volume of the cylinder, know the quantity of the balls and their individual volume it will be quite simple to verify the calculated void volume versus the measured void volume. The answer will have a very close correlation if you took care measuring the water as it's removed.

The empty cylinder volume is easy to calculate; no need to measure. It will be hard to determine in advance just how many lead balls will nestle into place for a good fit. But you can simply try. Then pour out the lead balls, weigh them, and divide by the density of lead to get their total volume. Then subtract.

You may have to create this equation yourself.

The porous volume will be the cylinder volume minus the occupied volume (sum of the ball volumes).

Some limitations:

The occupied volume will be dependent on the ball diameter and the cylinder to ball diameter ratio.

ball diameter must be <= cylinder diameter.

Ball diameter must be <= height of cylinder.

ball count must be an integer.

boundary cases:

cylinder diameter = ball diameter

cylinder height= ball diameter

Release the mathematicians!

For puls cryo coolers the packing density is generally specified by the manufacturer or supplier of the packings. As all experst have suggested packing density depends on cylinder dia and ball dia, still the cooling capacity depend on the density of packing. Total exact porous volume you have to find by filling liquid and measuring the same.

Mukesh I am doing project to make/manufacture pulse tube cryocooler which also include design and fabrication of regenerator which has porous material as matrix material. So, I need it.