How Many Pores and What Sizes for 20 ppi Porosity in a Metal Foam?

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Discussion Overview

The discussion revolves around the simulation of a 3D model of a metal foam with a specified porosity of 20 pores per inch (ppi) and a volume ratio of 25% aluminum. Participants are exploring how to determine the number and size of pores in the foam, considering factors such as pore shape, packing density, and potential overlaps.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions the interpretation of "20 ppi," suggesting it means a 1 inch cube should contain 8000 pores, leading to a specific volume for each pore.
  • Another participant suggests calculating the radius of each pore based on the assumption of spherical pores and discusses the implications of arranging them in a grid.
  • Participants reference packing theories, with one mentioning that Cubic Close packing achieves a 74% packing density, but express uncertainty about calculating PPI in that context.
  • There is a discussion about whether the pores can overlap and how that affects the expected volume occupied by the pores.
  • One participant calculates the volume of a single pore considering overlaps and compares it to a calculation that ignores overlaps, raising concerns about the accuracy of the volume ratio.

Areas of Agreement / Disagreement

Participants express differing views on the implications of pore overlap and its effect on the volume ratio. There is no consensus on how to handle overlaps in the calculations or the impact on the prescribed composition of aluminum and voids.

Contextual Notes

Participants note limitations in their understanding of packing theories and the implications of pore overlap, which may affect the accuracy of their calculations. The discussion does not resolve these uncertainties.

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Homework Statement


This should be a pretty simple question but I can't find a straight answer in the literature. I want to simulate a 3D model of a metal foam by starting with an aluminum block and filling it with randomly placed spherical holes of constant volume. The foam should have a porosity of 20 ppi (pores per inch) and a volume ratio of 25% aluminum. How many pores should there be, and what are their sizes?

Homework Equations


20 pores per inch
25% aluminum, 75% void by volume

The Attempt at a Solution


I'm not clear on precisely what "ppi" means. Does 20 ppi imply that a 1 inch cube of aluminum should contain 20^3 = 8000 pores? And that each pore should have a volume of 0.75/8000 = 0.00009375 in^3? This seems pretty obvious but I want to make sure before I proceed. Thank you for your help!
 
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Assume the pores are spherical. Work out what the radius of each pore would be. Can you put them on a 1/20" pitch/grid without them intersecting each other :-)

I'm afraid I'm not familiar with packing theories.
 
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The pores in a foam are allowed to overlap with each other. This isn't a sphere packing problem though. This is a question of the expected value of the volume these pores will occupy.
 
Does 20 ppi imply that a 1 inch cube of aluminum should contain 20^3 = 8000 pores? And that each pore should have a volume of 0.75/8000 = 0.00009375 in^3?
Yes, that is how I see it. As far as averages go, it should make no difference whether the pores are neatly arrayed in equi-spaced rows and columns, or randomly positioned. So, take the easy route, consider them all neatly arrayed in 3D ranks. :smile:

Work out the radius each would have if it were spherical, and compare this dimension with their centre-to-centre spacing to see whether the model seems realistic for your application.
 
If the pores overlap each other won't the volume ratio be incorrect. (eg some of the volume is double counted).

PS I believe they do overlap.
 
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Will the expected overlap be likely to alter the prescribed composition of 25% by more than ±1 sig fig?
 
If I did the sums right the volume of a single pore after deducting the 6 overlaps is

8.86 * 10^-5 cubic inches compared to
9.38 * 10^-5 calculated by ignoring the overlap

8.86 * 10^-5 * 8000 = 0.71

eg 71% rather than the 75% requested.
 

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