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VOC abatement control devices and calculating VOC from MSDS
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VOC Abatement Control Devices and Calculating VOC from M S D S by Ron Joseph
he first two questions in this month's column concern abatement control devices, and I have asked my fried and colleague, Dr. Robert Kenson, an expert on this topic, to please answer them.
T
EXCEEOING SOLVENT CONCENTlllATIONS IN CATALYTIC OXIDIZER
Our facility added a catalytic oxidizer to our coating line in order to control solvent air emissions. Every time we start up the line we exceed the high temperature limit and experience an automatic shutdown of the catalytic oxidizer about one hour later. The supplier says that we experience higher than design value for solvent emission concentration because we removed our original process exhaust f a n without telling them. They say that the f a n they supplied with their system can't maintain the coating process design exhaust flow rate without our fan still operating. They also say that this reduced exhaust flow rate raises the solvent emission concentration even though the solvent emission rate in p o u n d s / h o u r VOCs remains the same. The supplier won't raise the automatic s h u t d o w n temperature of the oxidizer because that will damage the catalyst which he supplied. Is their explanation correct and, i f so, how can we fix the problem? The catalytic oxidizer may be correct but you should also ask for the fan performance curve to make sure that they are telling you the truth. You need to determine if the design exhaust flow rate of the air emissions can be maintained at the design pressure drop with the oxidizer system fan below its maximum electric motor horsepower. If so, then you probably need a larger fan motor, or worse, a second fan in parallel with the one supplied with the catalytic oxidizer. Replacing the fan motor is not a high cost solution but will require down time of your coating line to accomplish. It will, however, be difficult to operate two fans in parallel without changes in the electrical controls for the oxidizer as well as automatic control dampers and potentially adding fan variable frequency drives. SOLVENT RECOVERY USING A D S O R P T I O N A B A T E M E N T CONTROLS
We are investigating use of ~ o r p t i o n systems to recovRon Joseph is an independent coating consultant in San Jose, Calif. E-mail, drrojo~aol.com. 58
er solvent air emissions from our coating lines and some suppliers of noncarbon adsorption systems tell us that their adsorbents can be regenerated at a higher temperature than carbon adsorbents. They have also told us that the noncarbon media won't catch fire like carbon-based systems. Can you tell us more about these noncarbon adsorbents and whether we should still consider carbon-based adsorption systems? Both zeolites and synthetic polymers have been used in adsorption systems in place of carbon with satisfactory results. What you are being told is true, but you need to determine the most important factors in your specific solvent recovery application. Ask yourself the following five questions.
1. Is n-hexane or ketones a major component of your coating solvents? 2. Are the technical, cost and safety problems of preventing potential carbon bed fires worrying you and your management excessively? 3. Is there a contaminant in your coating air emissions that cannot be removed from a carbon based adsorbent at the recommended regeneration temperature? 4. It does not concern you that the replacement cost of the noncarbon adsorbent is more expensive than carbon? 5. It does not concern you that the industry has less experience with noncarbon adsorbents in solvent recovery than carbon-based adsorbents? If you can answer yes to the above questions, noncarbon adsorption systems may be more appropriate for your solvent recovery application. CALCULATIONS OF r o e FROM M S D S
From B.A.S: Previously I sent you an M S D S for a product and I asked how to calculate the VOC content based from the data given, which was the following: * Under physical data it says the compound is 100% volatile by volume * It also stated on the M S D S that the compound is an aliphatic carbon with a concentration of >98% by weight, which means that some other trace solvents might also be in the mixture. • S.G of the mixture is 0.746 You told me my assumption must be, regardless of what the other traces are, the entire mixture is VOC. Metal Finishing
Your calculation was as follows:
Then you said, hence the VOC content of this comp o u n d is 6.21 lb /gal. I'm not sure if I a m assuming this correctly, but in essence you are multiplying the above equation by 100% right? The 100% that is volatile? Yes, since all of the material is volatile, regardless of its composition, the VOC content is the same as the density expressed in lb/gal. If my above assumption is correct then I am really confused because in your recent calculations class we learned that you calculate VOC content you multiply density of coating x % weight of VOC, not density of coating x % by volume. In this specific case you know the S.G. of the entire 1-gallon can of VOC. It is 0.746. When you express it in terms oflb/gal, you multiply S.G by the density of water 8.33 lb/gal. Your confusion arises because in this problem 100% by volume or by weight (it doesn't m a t t e r which) is VOC. The MSDS could equally have stated that the volatiles by weight are 100%.
Okay. Here then is another question. How do I handle an M S D S for the following product? • S . G . = 1.35 • WPG = 11.2 lb/gal • Volatile by volume is 57% • Volatile by weight is 35% First, the weight per gallon (WPG) is calculated by S.G. x density of water (8.33 lb/gal) = 11.2 lb/gal. As you can see, the MSDS gives you both numbers. You should know that in this case the volatile by volume of 57% does not help you to determine the VOC content. It is only useful in determining the volume solids of the coating, which is of no interest to you right now. (It is of interest when we perform calculations for the MMPP MACT in lb HAP/gal solids). For your immediate purposes the MSDS gives the volatile by weight as 35%. Since one gallon of the coating weighs 11.2 lb/gal, 35% of its weight is VOC. Therefore the VOC content is 0.35 x 11.2 = 3.92 lb/gal. The first two questions w e r e a n s w e r e d by Dr. Robert Kenson, [email protected], an associate of Ron Joseph & Associates Inc., www.ronjoseph.com MF
Circle 048 on reader service card
Circle 061 on reader service card
Density of the mixture = 0.746 x 8.33 lb / gal [Density of water] = 6.21 Ib / gal
April 2003
59
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VOC Abatement Control Devices and Calculating VOC from M S D S by Ron Joseph
he first two questions in this month's column concern abatement control devices, and I have asked my fried and colleague, Dr. Robert Kenson, an expert on this topic, to please answer them.
T
EXCEEOING SOLVENT CONCENTlllATIONS IN CATALYTIC OXIDIZER
Our facility added a catalytic oxidizer to our coating line in order to control solvent air emissions. Every time we start up the line we exceed the high temperature limit and experience an automatic shutdown of the catalytic oxidizer about one hour later. The supplier says that we experience higher than design value for solvent emission concentration because we removed our original process exhaust f a n without telling them. They say that the f a n they supplied with their system can't maintain the coating process design exhaust flow rate without our fan still operating. They also say that this reduced exhaust flow rate raises the solvent emission concentration even though the solvent emission rate in p o u n d s / h o u r VOCs remains the same. The supplier won't raise the automatic s h u t d o w n temperature of the oxidizer because that will damage the catalyst which he supplied. Is their explanation correct and, i f so, how can we fix the problem? The catalytic oxidizer may be correct but you should also ask for the fan performance curve to make sure that they are telling you the truth. You need to determine if the design exhaust flow rate of the air emissions can be maintained at the design pressure drop with the oxidizer system fan below its maximum electric motor horsepower. If so, then you probably need a larger fan motor, or worse, a second fan in parallel with the one supplied with the catalytic oxidizer. Replacing the fan motor is not a high cost solution but will require down time of your coating line to accomplish. It will, however, be difficult to operate two fans in parallel without changes in the electrical controls for the oxidizer as well as automatic control dampers and potentially adding fan variable frequency drives. SOLVENT RECOVERY USING A D S O R P T I O N A B A T E M E N T CONTROLS
We are investigating use of ~ o r p t i o n systems to recovRon Joseph is an independent coating consultant in San Jose, Calif. E-mail, drrojo~aol.com. 58
er solvent air emissions from our coating lines and some suppliers of noncarbon adsorption systems tell us that their adsorbents can be regenerated at a higher temperature than carbon adsorbents. They have also told us that the noncarbon media won't catch fire like carbon-based systems. Can you tell us more about these noncarbon adsorbents and whether we should still consider carbon-based adsorption systems? Both zeolites and synthetic polymers have been used in adsorption systems in place of carbon with satisfactory results. What you are being told is true, but you need to determine the most important factors in your specific solvent recovery application. Ask yourself the following five questions.
1. Is n-hexane or ketones a major component of your coating solvents? 2. Are the technical, cost and safety problems of preventing potential carbon bed fires worrying you and your management excessively? 3. Is there a contaminant in your coating air emissions that cannot be removed from a carbon based adsorbent at the recommended regeneration temperature? 4. It does not concern you that the replacement cost of the noncarbon adsorbent is more expensive than carbon? 5. It does not concern you that the industry has less experience with noncarbon adsorbents in solvent recovery than carbon-based adsorbents? If you can answer yes to the above questions, noncarbon adsorption systems may be more appropriate for your solvent recovery application. CALCULATIONS OF r o e FROM M S D S
From B.A.S: Previously I sent you an M S D S for a product and I asked how to calculate the VOC content based from the data given, which was the following: * Under physical data it says the compound is 100% volatile by volume * It also stated on the M S D S that the compound is an aliphatic carbon with a concentration of >98% by weight, which means that some other trace solvents might also be in the mixture. • S.G of the mixture is 0.746 You told me my assumption must be, regardless of what the other traces are, the entire mixture is VOC. Metal Finishing
Your calculation was as follows:
Then you said, hence the VOC content of this comp o u n d is 6.21 lb /gal. I'm not sure if I a m assuming this correctly, but in essence you are multiplying the above equation by 100% right? The 100% that is volatile? Yes, since all of the material is volatile, regardless of its composition, the VOC content is the same as the density expressed in lb/gal. If my above assumption is correct then I am really confused because in your recent calculations class we learned that you calculate VOC content you multiply density of coating x % weight of VOC, not density of coating x % by volume. In this specific case you know the S.G. of the entire 1-gallon can of VOC. It is 0.746. When you express it in terms oflb/gal, you multiply S.G by the density of water 8.33 lb/gal. Your confusion arises because in this problem 100% by volume or by weight (it doesn't m a t t e r which) is VOC. The MSDS could equally have stated that the volatiles by weight are 100%.
Okay. Here then is another question. How do I handle an M S D S for the following product? • S . G . = 1.35 • WPG = 11.2 lb/gal • Volatile by volume is 57% • Volatile by weight is 35% First, the weight per gallon (WPG) is calculated by S.G. x density of water (8.33 lb/gal) = 11.2 lb/gal. As you can see, the MSDS gives you both numbers. You should know that in this case the volatile by volume of 57% does not help you to determine the VOC content. It is only useful in determining the volume solids of the coating, which is of no interest to you right now. (It is of interest when we perform calculations for the MMPP MACT in lb HAP/gal solids). For your immediate purposes the MSDS gives the volatile by weight as 35%. Since one gallon of the coating weighs 11.2 lb/gal, 35% of its weight is VOC. Therefore the VOC content is 0.35 x 11.2 = 3.92 lb/gal. The first two questions w e r e a n s w e r e d by Dr. Robert Kenson, [email protected], an associate of Ron Joseph & Associates Inc., www.ronjoseph.com MF
Circle 048 on reader service card
Circle 061 on reader service card
Density of the mixture = 0.746 x 8.33 lb / gal [Density of water] = 6.21 Ib / gal
April 2003
59