# coating weight calculations

by neptune
Tags: calculations, coating, weight
 P: 9 Hello, This is not homework etc. I am in the work place (automotive industry) and have a question. It has been some time since I have taken any math courses. I am trying to calculate the theoretical weight of a known coating on the surface of a hollow cylinder and set this up in an excell program. I already have the surface area of the cylinder, but how will I get the surface weight of the coating? Will I need the density and/or approximate thickness of the coating. (I have many coatings to put in the program, but for this example I am using a Zinc Phosphate coating.) Thanks for any help in advance.
 Admin P: 21,595 For a thin coating, all you need is 1) surface area, 2) coating thinkness and 3) density. Mass = Area * thickness * density, since Area * thickness = volume, and Mass = Density (mass/unit volume) * total volume.
 Math Emeritus Sci Advisor Thanks PF Gold P: 38,706 Calculating the volume as "surface area times thickness" is exact for a flat surface but not a curved surface such as the lateral surface of a cylinder- though it might well be a good enough approximation. For an exact volume, look at the pipe from the end and imagine that you can see the "edge" of the coating. It looks like a "washer"- the area between two concentric circles. If the radius of the inner circle is r (it's not clear to me whether you are coating the outside of the cylinder in which case this would be the outer radius of the cylinder, or the inside of the cylinder in which case this would be the inner radius of the cylinder minus the thickness of the coating) and the thickness of the coating is d (so the radius of the outer circle is r+d) then the two concentric circles has area $\pi r^2$ and $\pi(r+d)^2= \pir r^2+ 2\pi rd+ \pi d^2$ and the area between them (the side area of the coating) is [itex]\pi(2rd+ d^2)[/tex]. If the cylinder has length L, then the volume of the coating would be $$\pi(2rd+ d^2)L$$ rather than the $$2\pi rdL$$ of Astronuc's formula. Of course, the difference is only $$\pi d^2L$$.
P: 9

## coating weight calculations

I have set up calculations for both the inner and outer surface area as well as the total surface area. The coating will be very thin, I have a range of 150-1350mg/ft2. The coating thickness won't actually be measured. I will have the surface area and hopefully I will be able to find the densities of all the coatings that I need. Would I need to set up an integral for the thickness range? TIA
Math
Emeritus
Thanks
PF Gold
P: 38,706
 Quote by neptune I have set up calculations for both the inner and outer surface area as well as the total surface area. The coating will be very thin, I have a range of 150-1350mg/ft2. The coating thickness won't actually be measured. I will have the surface area and hopefully I will be able to find the densities of all the coatings that I need. Would I need to set up an integral for the thickness range? TIA
No, as long as the density is uniform, just multiply density times volume.

And if the coating is very thin, Astronuc's approximation will work fine.
 P: 9 for the coating weight that I am trying to calculate... I know that Coating Weight=Thickness x Surface Area x Density It seems that my units are not working out right. I have a Thickness=0.02032mm Surface Area=252.898mm2 Density of 0.0081mg/mm2 I end up with units of mg mm? What is wrong with this?? Thanks in advance!
 P: 9 Anyone??
 Math Emeritus Sci Advisor Thanks PF Gold P: 38,706 This is impossible: Density of 0.0081mg/mm2 Density is in unites of "mass over volume": mg/mm3
 Sci Advisor HW Helper P: 2,535 If you're getting this from a spec sheet, you may be looking at 'mass per area' rather than 'mass per volume' as the specific mass. In that case you can just take: $$\rm{area} \times \frac{\rm{mass}}{\rm{area}}=\rm{mass}$$
 P: 9 Thanks Nate! Now if I only have the surface area and an average thickness, how will that work. Since, Coating Weight=Thickness x Surface Area x Density I have an avg Thickness of 0.0076mm Surface Area of 252.898mm2 How are the units going to work out for that? I don't know why this is seeming so difficult to me to see this. Thanks in advance.
 P: 9 any thoughts?
HW Helper
P: 2,535
 Quote by neptune any thoughts?
Let me try it this way:
If you're looking at information that has been provided by the manufacturer of the coating, the number that you have may be the proporitionality constant so that you can simply take
constat * surface area
to get the mass. That would explain the unit disparity.

However, you should really double-check that to make sure.
 P: 9 The coating average thicknesses that I have are from a range that our facility has for min and max coating thickness. I have simply averaged the min and max. I don't think that I can just leave out the units in mm of the average thickness. It would seem that I would need the density of the coating to calculate this correctly. The problem being that I do not have all the densities of the coatings. It doesn't seem that I will arrive at a correct number with only having... Avg Thickness=0.0076mm Surface Area=252.898mm2 I will end up with mm3 and not be able to convert to mg??? Advise is possible. Thanks in advance.
HW Helper
P: 2,535
 Quote by neptune It doesn't seem that I will arrive at a correct number with only having... Avg Thickness=0.0076mm Surface Area=252.898mm2 I will end up with mm3 and not be able to convert to mg???
Correct. You've got 1.9 cubic mm of material. If you're coating with gold (the most dense material that comes to mind) that's about 38 grams. If you're coating with aerogels, it's next to nothing.

N.B. It's not necessarily accurate to assume that the average thickness will be the mean of the minimum and maximum thicknesses.

Depending on your situation, and the degree of precision you need it may be easier to simply weigh a bunch of these parts before and after the coating process.
 P: 9 Nate, i have located a density for one particular coating... so now I will have... SA=252.898mm2 Thickness=0.0076mm Density= 8.01lb/gal Also, for all intensive purposes, the avg of the max and min will suffice. This will give me units of mm3lb/gal How can I convert this to only mg??
 P: 9 Any thoughts??
$$\frac{1 \rm{gal}}{3785.4 \rm{mm}^3}=1$$