Dear academics,
I have a copper infiltrated carbon matrix composite material. I looked at the pore distribution of this sample with the imageJ program. How can I derive the volume fraction from this areal distribution?
I also looked at the densities of these materials with the geometric and archimedean principle. Based on these values, if I assume that the difference between the two densities is due to the penetration of water into the pores, this way I can determine the total volume ratio of carbon + copper. But I'm not sure how to specify these two separately.
Thank you for your help
selim burak Cantürk I refer you to the paper by F Chayes “The relation between area and volume in micrometric analysis” Mineralogical Magazine 30,147-149, (1953) and this webinar (registration required):
The mathematics of image analysis
https://www.malvernpanalytical.com/en/learn/events-and-training/webinars/W151119MathematicsImageAnalysis
The relevant extract (attached) is at 49:29 in the talk. The area and volume fractions are identical.
Very interesting reference from Alan F Rawle, I would just like to add one precision: This relation is only true if the orientation of the particles is random. It is not applicable in some cases like if you spread high aspect ratio particles onto a thin layer (lower than the large dimension of the particles). Off course the image you use for analysis should contain a statistically high number of particles.
Alan F Rawle thank you Sir. I'll look into this. One more thing, If I dissolve the copper by putting my samples into some chemical, I can easily measure the mass volume of graphite matrix. Do you think could it be possible to find a specific chemical that dissolve copper and not damage the graphite matrix at the same time ?
Rémi Rafael Thank you for your answer. As I asked Professor Rawle, Could be feasible if I dissolve the copper and then measure the mass volume of graphite matrix?
Thank you
selim burak Cantürk
Dilute nitric acid should dissolve the Cu and leave behind the C species.
Rémi Rafael You correctly state 'Of course the image you use for analysis should contain a statistically high number of particles' (I have corrected your typo). The math is simple. If you desire the mean to a standard error of a% then you'll need 1/a0.5 numbers of particles. Same applies to any point in the distribution. So, for example, if you require the x99 point of the distribution to 1% standard error then you'll need 10000 particles in the x99+ point of the distribution. Note that ISO uses x for particle size as d (or D) implies diameter and thus spherical particles. Conversely, if you measure 42 particles then you can only get the mean to a precision of 1/420.5 or ~ 15.4%. See this webinar (registration needed):
Keys for Successful Analysis – Representative Sampling & Estimation of Standard Error Calculation
https://www.malvernpanalytical.com/en/learn/events-and-training/webinars/W190613KeysLD
There are minimum mass requirements as a consequence and this is explained in the webinar.
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