How Do You Calculate Coating Weight for a Cylinder in the Automotive Industry?

[neptune]

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.

 

[Astronuc]

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.

 

[HallsofIvy]

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 $$\pi d^2L$$.

 

[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

 

[HallsofIvy]

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.

 

[neptune]

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!

 

[neptune]

Anyone??

 

[HallsofIvy]

This is impossible: Density of 0.0081mg/mm2

Density is in unites of "mass over volume": mg/mm³

 

[NateTG]

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:
$$\rm{area} \times \frac{\rm{mass}}{\rm{area}}=\rm{mass}$$

 

[neptune]

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.

 

[neptune]

any thoughts?

 

[NateTG]

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.

 

[neptune]

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.

 

[NateTG]

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.

 

[neptune]

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??

 

[neptune]

Any thoughts??

 

[NateTG]

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??

I happen to know that:
1 gallon = 3785.4 cubic millimeters
So, to do the the conversion, you can multiply by
$$\frac{1 \rm{gal}}{3785.4 \rm{mm}^3}=1$$

NB: The density per gallon may well be the wet rather than dry density.

 

[sparrow]

Surface Area/Ton

hello
my math is a little rusty.. can you please help me?

Im looking for AREA/TON (m^2/ton) for painting purposes of a 219dia X 6(wall) X 1m tube. The weight of the tube is 31.5 kg.
1)area of the tube
2)area per ton
It is just for the outer surface
The answer should be 21.92m^2/ton, but my formulas don't give me that answer.
Can you please help me with a formula?

Thank you
Sparrow

 

[HallsofIvy]

hello
my math is a little rusty.. can you please help me?

Im looking for AREA/TON (m^2/ton) for painting purposes of a 219dia X 6(wall) X 1m tube. The weight of the tube is 31.5 kg.
1)area of the tube
2)area per ton
It is just for the outer surface
The answer should be 21.92m^2/ton, but my formulas don't give me that answer.
Can you please help me with a formula?

Thank you
Sparrow

First, please do not hijack someone else's thread to post a question of your own. It is a rude and less likely to get a response then if you post your own thread.

Now: you say the tube is "219dia X 6(wall) X 1m tube." that's meaningless if you don't say what units all the measurements are in. Since you mention "m", I might assume that all are in meters but I simply don't believe a 6 meter think wall 1 m high!

If d x w x L are the (inner diameter) of the tube, in m, the thickness of the wall, in m, and the height of the tube, in m, then the tube has outer diameter of 2w+ d. The circumference of a circle of diameter 2w+ d is just that times $\pi$: $(2w+d)\pi$ meters and so the surface area of the outer wall is $\pi(2w+d)h$.

I am a bit confused about what you mean by "AREA/TON". I would presume mean "area divided by the weight of the paint" but that would require knowing the density of the paint and thickness of the paint on the wall, neither of which you mention.

You do say "The weight of the tube is 31.5 kg", but what does the weight of the tube have to do with painting it?