# calculation of voc in coatings

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appendix a CALCULATION OF VOC IN COATINGS

This appendix provides an overview of the most important calculations, which environmental engineers and paint facility staff need to perform on a regular basis.

The calculations are actually quite easy to perform, and for each new concept one example is provided.

CALCULATING THE VOC OF A SIMPLE COATING MIXTURE You wish to calculate the VOC content for a mixture comprising 1 gal of red alkyd enamel and 1/4 gal (0.25 gal) thinners. The respective VOC levels are 2.5 lb/gal for the enamel and 6.8Ib/gal for the thinners.

To perform the calculation, set up a table where the first column lists the names of the components that you will be mixing together and the second contains the number of gallons to be added. In the third, you enter the VOC content as given to you by the paint manufacturer on the Material Safety Data Sheet (MSDS) and the last column is the total amount of solvent (or VOC) for each of the components. In this case, the table would appear as shown in Table 1.

Because the EPA assumes that all of the VOC (mostly the solvents) evaporates to cause emissions, the last column is headed "Emissions," instead of "VOC."

Step 1: Multiply the number of gallons by the VOC content to arrive at the to

tal amount ofVOC (or Emissions) for each component. Step 2: After completing the last column, total the number of gallons in Col

umn 2 as well as the lb ofVOC (Emissions) in the mixture, Column 4. In this case the total number of gallons is 1.25 and the total Emissions 4.2 lb. Note that you do not need to total Column 3 as the result is meaningless.

Step 3: The final step requires you to divide the number of gallons in the mixture into the total Emissions.

Therefore, the VOC of one gallon of the mixed coating is: VOC = 4.2Ib/1.25 gal = 3.36Ib/gal

In other words, if someone were to sample the coating you have just mixed, the VOC content would be 3.36 lb for one gallon of the mixture.

If your state coating rule specifies a maximum VOC content of 3.5 lb/gal, your mixture is in compliance.

CALCULATING THE VOC OF A TWO-COMPONENT COATING MIXTURE You wish to mix a two-component coating in the ratio four parts Component A

liable I. Calculating VOC fora Simple Coating

Component

Red Alkyd Enamel Thinner

TOIaI

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Volume (gal)

1.0 0.25

1.25

voc COnTenT Ilb/llal)

2.5 6.8

Emiss/r)ns (fb)

2.5 1.7

4.2

Table II. Calculating VOC for a Two.Component Coating Without Thinners

Volume VOC Coment Component (gal) lib/gal) Emission, (lb)

Component A 4 3.6 14.4 Component B I 1.5 1.5

Total 15.9

and one part Component B. For the time being, you will not add any thinners. From the MSDS you get the following information:

Component A ~ 3.6Ib/gal Component B ~ 2.2 lb/gal

Set up the same table as before and insert the given information into columns 1,2 and 3 as shown in Table II. Following the same steps given in the previous example, you simply divide the total number of gallons of the mixture into the total Emissions.

Therefore, the VOC of one gallon of the mixture is: VOC ~ 15.9/5 ~ 3.18Ib/gal

Once again, if someone were to sample the coating, the VOC content would be 3.181b for 1 gal of the mixture. Because the VOC content of the mixture is below 3.5 lb/gal, you are allowed to add thinners, if necessary, but under no circumstances are you allowed to exceed the 3.5 lb/gallimit.

CALCULATING THE VOC OF A TWO-COMPONENT COATING MIXTURE PLUS THINNERS Suppose you do want to add thinners having a VOC of 6.4lb/ gal to the coating mixture in Table II. How much can you add before exceeding the regulatory limit?

There are two methods for making this determination. The first is quite simple and does not require any knowledge of algebra. The second is slightly more difficult, but a whiz for those who still remember their high school math.

Here is the simpler of the two methods. Suppose you were to add only one quart (0.25 gal) of thinner, would you exceed the VOC limit?

Solution Using Trial and Error The results of the first trial are shown in Table III. Following the same steps as given in the previous examples,

VOC ~ 17.5/5.25 ~ 3.33 lb/gal Because the VOC content of the mixture is still below 3.5 lb/gal, you can afford

to add slightly more, if necessary, but remember that under no circumstances may you exceed the 3.5 lb/gallimit.

Repeat the calculations by adding another quart (0.25 gal) of thinners. In re-

Table III. Calculating VOC rOr a Two-Co_mponent Coating With Thinners (First Trial)

Volume VOC Conlenl Emission, Component (gal) (lbfgal) (Ib)

Component A 4.0 3.6 14.4 Componenl B 1.0 1.5 1.5 Thinners 0.25 6.4 1.6

Total 5.25 17.5

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fable IV. Calculating VOC for a Two.component Coating With Thinners (Second TrIal)

Volume vac Conttnt Emissions Compo/W1l (gill) (lblgal) (lb)

CI,mponcnt A 4,0 3,6 14.4 Component B 1.0 l.S 1.S Thlnner~ D,S 6.4 3,2

Total 5,5 19,1

calculating the problem, a total of 1/2 gal of thinners has been added. (See Table IV).

VOC ~ 19.1/5.5 ~ 3.47\ Ib/gal This is as far as you should go. By adding any more thinner to the coating, you

will overshoot the 3.51b/gallimit. It is important to understand that you should only add thinners if you can

not properly atomize the coating. By adding thinners you might be able to overcome a problem such as orange peel, or excessive film build (dry film thickness), but the trade off is that you are adding to air pollution.

Note These regulations have been written for the sole purpose of reducing air pollution; therefore, you should avoid adding thinners unless it is really required.

There is one other critical point that must be borne in mind. When an EPA or state inspector takes a sample, it is sent to an analytical laboratory where the VOC content is determined. In the analytical test, a few drops (approximately 0.3-0.5 g) of the mixed coating are weighed into a small aluminum dish and heated to 230 0 F for 1 hr, after which the sample is weighed again. The loss of weight between the first and second weighing is due to the loss ofVOC from the mixture. At this relatively high temperature, it is possible for some of the coating resin and other ingredients to evaporate.

According to the EPA definition ofVOC, everything that evaporates, with the exception of water, inorganic compounds, and a few select number of exempt organic compounds, is considered to be Voc. Therefore, even though you calculated the VOC content of the mixture to be 3.471b/gal, it is possible that if you subjected a sample of the mixture to the laboratory procedure, the VOC content could be higher than 3.471b/gal. In fact, it could exceed the regulatory 3.51b/gal limit, which would cause you to be in violation of the regulation. For this reason you should always be cautious when adding thinners to ensure that you stay well below the regulatory limit. In the case of this problem, add less than 1/2 gal of thinners, thereby playing it safe.

Solution Using Algebra This method assumes that the amount of thinners added is the unknown "y." Set up the table as before. The result is shown in Table V.

VOC ~ (15.9 + 6.4y)/(5.0+y) Because you are constrained by the regulatory limit of 3.5 Ib/gal, you can

solve fory.

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3.5 ~ (15.9 + 6.4y)/(5.0+y) Y ~ 0.55 gal

Table V. Calculating VOC for a Two-Component Coating With Thinners (Algebraic Solution)

Volume VOC C()nlenr Emissions Component (gal) (lblgal) (Ib)

Component A 4.0 3.6 14.4 Component B 1.0 1.5 1.5 Thinner! y 6.4 My

Total 5.0 ... y 15.9 ... 6.4y

If you wanted to add thinners right up to the limit, you could add 0.55 gal, which is slightly higher than the 0.5 gal that was calculated by trial and error, but it is strongly advised not to go so close to the limit as you can easily overshoot the mark and find yourself with a costly violation. More importantly, since pollution prevention is now the name of the game, it is good practice to keep your addition of thinners to a minimum.

PROBLEMS THAT INCLUDE WATER The following calculations will demonstrate how the EPA and states differentiate between solvent-borne and waterborne coatings.

If the coating is solely based on organic solvents, there is only one voe content to report, such as 3.5 Ibj gal. If you are dealing with a waterborne coating, such as a latex or water reducible, most EPA or state regulations require that the coating contain less than, e.g., 3.5 Ibj gal, less water. What this means is that if you were able to remove the water from the coating, then the new "hypothetical" coating would need to have a voe content less than 3.51bjgal.

To clarifY this, imagine that you have a I-gal can of waterborne coating, which contains a small amount ofVOe, and a large amount of water. The coating is represented by Figure la, for which the volume solids is, say 25%. Assume that the ratio ofVOe to solids is 1: 10 by volume.

Now suppose you had another I-gal can with identically the same solid ingredients (resins, pigments, extenders, and additives), and maintained the voe to solid ratio at 1:10, such as in Figure lb. Imagine now that you wanted to paint two identical large walls so that you would deposit exactly 1 mil (0.001 in.) of solid coating onto each.

The first wall will be painted using the waterborne coating, and the second with the solvent-borne. In each case, you would stop painting as soon as the entire surface had a coating film of exactly I-mil dry film thickness. Which of the two coatings will emit more voe into the air?

They will both emit exactly the same amount ofVOe, because in each case the amount of solid coating deposited is identical, and in each case the ratio ofVOe to solid is the same. For instance, if you were to apply 1 gal of solid of the waterborne coating you would emit 0.1 gal ofVOe (ratio ofVOe to solid is 1:10). In the case of the solvent-borne coating, you would also need to apply 1 gal of solid coating to achieve the 1 mil thickness, and for the same reason you would emit 0.1 galofVOe.

In other words, regardless of how much water is in the coating, the amount ofVOe emitted will solely depend on the ratio ofVOe to solid, in this case 1:10. For this reason, the regulations are written such that the coating may not contain more

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18ble VI. Calculating VOC for a Two.CompoDent Solvent-Borne Coating Plus Water (Example 2)

Componel# Volume (gal) VOC Content (Ib/gul) Emission" lib)

Componenl A 3 3.1 9.3 Componenl B 1 1.8 1.8 Waler 7 0.0 0.0

Total 4 11.1

VOC than "x" lb/gal, less water. In the case of a 3.5lb/gallimit, the regulation would specifically state that the coating may not contain more than 3.5Ib/gal, less water.

The following calculations will demonstrate this concept.

Example 1: Single-Component Solvent-Borne Coating Plus Water A state regulation limits you to a coating for which the VOC is less than 3.5 lb/gal, less water. You purchase 1 gal of solvent-borne coating, which contains no water, but has a VOC content of 5.0 lb/gal. To bring down the VOC content you decide to add 1 gal of water. What is the VOC content, less water, of the mixed coating? Would it now meet the regulatory requirement?

Regardless of how much water is added, the VOC content remains 5.0Ib/gal, less water, and continues to be in violation of the regulation.

Example 2: Two-Component Solvent-Borne Coating Plus Water What is the VOC content of the following two-component epoxy? Is it in compliance with a regulatory limit of2.8Ib/gal, less water? The coating consists of three parts of Component A at 3.1Ib/gal, one part of Component B at 1.8Ib/gal, and 7 parts of water. (See Table VI.)

Neither Component A nor B contains water. Notice that when you calculate the VOC of the coating mixture for compliance

purposes, you must ignore the water altogether. Hence, in Column 2 the total is 4 and not 7.

VOC = 11.1/4 = 2.75Ib/gal, less water The mixture is in compliance with the regulation.

Example 3: Two-Component Solvent-Borne Coating Plus Water You are provided with the following information concerning a waterborne coating. What is the VOC content, less water? The volume of water is 52%, the volume solids is 42%, and the density of the VOC only is approximately 7.36Ib/gal.

The % volume ofVOC in the coating = 100 - 42 - 52 = 6%. For 1 gal of coat-

18ble VI. Calculating VOC for a Two-CompoDent Solvent-Borne Coating Plus Water (Example 2)

Component

Componenl A Componenl B Waler

Total

396

Volume (gal)

3 1 7

4

VOC COn/ellt (Ib/gul)

3.1 1.8 0.0

Emission,' (/b I

9,3 1.8 0.0

11.1

Table VII. English to Metric Conversion

Ibfgul

1.0 2,3 2,8 3,0 3,5 5,7 6,5 6,8

~fL

119,8 275 340 360 420 680 780 815

ing, the volume ofVOe is 0.06 gal and the volume of solids is 0.52 gal. If you remove the water from the coating, the total volume of the voe plus solids is 0.58 gal.

Suppose you wanted to fill a I-gal can with only the voe plus solid, then the total volume ofVOe in the I-gal can would be:

0.06/0.58 ~ 0.103 gal If the approximate density of the voe is 7.36Ib/gal, then the VOC content of

the coating, less water, is given as 0.103 gal x 7.36 ~ 0.76Ib/gal

CONVERTING FROM ENGLISH TO METRIC UNITS Many of the regulations are not written in lb/gal, but rather in grams per liter (giL). The conversion is as follows:

lib/gal ~ 119.8 gil Table VII shows the voe limits most commonly found in regulations, bur be aware that the conversions are nor exact. With the exception of lib/gal, the values in the table have been rounded off

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