6 replies to this topic

[bob789]

Hi,

I am student in the UK and have a query regarding calculating ppm a of a particular waste stream. The process that I am looking at has a waste stream which contains 0.0666 Kg of Formaldehyde in 555.4 kg of water. I am unsure as to what I have done makes sense or not. I am sorry that is a really simple question but I am struggling to get my head around it.

As the process is at standard temperature and pressure I assumed the density to be taken as pure water as there is such as small amount of formaldehyde,


Therefore : mg/L = mg/kg = ppm.


Within the UK, the occupational exposure limit of formaldehyde after a duration of 8 hours is 2.5 mg/m3.


What I did was initially calculate concentration:


0.0666 kg into mg = 66600 mg



to get mg/kg = mg/L I did: 66600/555.4 = 119.91 mg/L = 119.991 ppm,


Is it wrong to convert this 119.91 mg/L into mg/m3? which would be 119910 mg/m3 which is definatley wrong!


I am unsure as to what this means with respect to the exposure limit








thanks in advance

[Zauberberg]

Bob,

1L = 1dm3 (not 1m3)

Therefore, mg/L = mg/dm3. If you want to convert mg/kg to mg/L, it is done by multiplying the result in mg/kg (ppmw) by the density of water stream (kg/L) at given conditions: mg/L = mg/kg * kg/L

As for the exposure, you need to confirm the basis of this value i.e. whether it refers to Formaldehyde concentration in Air, or water, or it has some other meaning. If it refers to the concentration of air, what you have to do is to calculate (or measure in the Lab) to how much the mg/kg of Formaldehyde in water corresponds to mg/m3 equilibrium concentration in air, at given pressure and temperature.

An initial estimate can be calculated by using Henry's law: K(a) = c(a) / P(a)

[kkala]

1. Your calculation seems basically right, but reported 8 hour exposure limit for HCHO must refer to air. See http://www.osha.gov/...table=STANDARDS, where OSHA specifies an 8 hour exposure limit for HCHO (in working environment) of 0.75 ppm, that is 0.75E-6 mol HCHO per mol air = 0.75E-6x30/22.414 g HCHO/l air = 0.75x30/22.414 mg HCHO/Nm3 air = 1.00 mg HCHO/Nm3 air. It seems that HCHO 8 hour exposure limit in UK is currently 2.5 times higher than in USA.
If the stream of HCHO + H2O is in an open channel or goes to an open tank, HCHO concentration in surrounding air should be estimated, following Zauberberg's advice.
2. Concerning the concentration of HCHO contained in the water stream: for liquids ppm means ppmw (e.g kg of HCHO per 1000000 kg of solution, practically of H2O), so ppm = mg/l is correct (precisely at 4 - 5 0C, where water density is 1000 kg/m3; but even at 25 oC, water density is 997 kg/m3 and resulting error is insignificant compared to uncertainty of concentration measurements).
3. Way of calculating HCHO concentration could be clarified as follows.
Assumed density of water solution 1000 kg/m3
Mass of solution 555.4 + 0.0666 = 555.4666 ~ 555.4 kg
Volume of solution 555.4 / 1000 = 0.5554 m3
HCHO content of solution 0.0666 kg
HCHO concentration 0.0666 / 0.5554 = 0.1199 ~ 0.120 kg/m3 = 120000 mg/m3 (mentioned 119910 mg/m3 is correct)
ppm of HCHO in the solution (0.0666/555.4)x1E6 = 119.9 ~ 120
4. Do not hesitate to ask "simple" questions after you have tried them. I (and many others) have fallen into pitfalls due to wrong application (or no consideration) of basics, better to clarify them in advance.

Edited by kkala, 28 December 2010 - 04:59 AM.

[bob789]

Hi guys,

Thanks for the response greatly appreciated. The issue now, however, is that I have a ppm of 120, and in Norway the limit of exposure is only 0.5 ppm.

Correct me if I am wrong, but this value is only going to be dangerous if staff are exposed to it in the surroudings? According to the HSE in the UK, exposure limits refer to both dermal contact and breathing it; the way have designed the plant is that the seaweed passes into the formalin tank then on to the remainder of the process and at no point is it open to the atmosphere. Therefore, the risk to exposure is minimal.

As a student where I lack knowledge is how where occupational safety risks may occur, I assume these may occur with making up the initial formalin preservative solution, and the issue of disposing of the waste formalin.

Formalin is biodegradable, but I am pretty sure that at 120 ppm we wont be able to dispose of it as normal. It may be necessary to install a dilution step before discharge. I have gotten in touch with the environmental agency in Norway for some insight into formaldehyde use and how we could dispose of it, but so far no reply. Do I need to know how much of the 120 ppm will actually be released in order to assess disposal? or is this wrong thinking? Are large dilutions normal in industry for hazardous chemical control?

My assumptions were that formalin is very small and the risk of evaporation will be limited due to the low volatility of the formaldehyde as there is so much water in the waste stream; is this wrong thinking? I was also assuming that there was going to be no formalin in the actual seaweed after this preservation step; I take it this is a wrong assumption as well.

The purpose of the formaldehyde is to remove the the phenols present within the seaweed and to prevent microbial growth which normal water preservation does not do. As there is between 0 to 2% phenol as part of seaweed composition, I can't remember what mass of this is but it's minimal that would have reacted, percentage conversion is unknown; we have assumed that the formalin is therefore all removed, or would you say that it has through process of osmosis gone into the? Surely the higher water concentration of the 555 kg of water would stop this

The reason I have assumed a lot of data is because the reaction kinetics are not available for seaweed and formaldehyde.

[kkala]

"...The issue now however is that the I have a ppm of 120, in norway the limit of exposure 0.5ppm."
120 ppm concerns HCHO concentration (mg/l) in the water solution, 0.5 ppm concerns HCHO conentration (mol/E+6 mol, approx 0.67 mg/Nm3) in ambient air. From http://www.mpch-main.../res/henry.html (read clarifications) Henry's constant for HCHO can be about 3E+3 M/Atm. This results in HCHO partial pressure in air of (1/3E+3)*120E-3/30 = 1.33E-6 Atm, corresponding to 1.33 ppm HCHO vapor, higher than 0.5 ppm but not much, at least for ambient temperatures.

"Correct me if I am wrong, but this value is only going to be dangerous if staff are exposed to it in the surroudings? According to the HSE in the UK, exposure limits refer to both dermal contact and breathing it, the way we have designed the plant is that the seaweed passes into formalin tank then on to the remainder of the process at no point is it open to the atmosphere. Therefore the risk to exposure is minimal."
Agreed that the plant can be safe enough, complying with HSE on this matter. This is because all HCHO+H2O circuit is closed and leaks out (if any) are not expected to create 0.5 ppm average concentration in the working environment. An induced draft fan (if Authoroities require it) could practically nullify emissions into working environment, pushing them into atmosphere.

"As a student where I lack knowledge is how where occupational safety risks may occur, I assume these may occur with making up the initial formalin preservative solution, and the issue of disposing of the waste formalin. Formalin is biodegradable, but I am pretty sure that at 120ppm we wont be able to dispose of it as normal. It may be necessary to install a dilution step before discharge. I have got in touch will the environmental agency in Norway for some insight into formaldehyde use and how we could dispose of it but so far no reply. I need to know how much of the 120ppm will actually be released in order to assess disposal? or is this wrong thinking? Are large dilutions normal in industry for hazardous chemical control?"
Yes, you have to know max permissible HCHO concentration in wastewater out, the rest should be retained by some treatment (a specialist can clarify the treatment). Examples strongly indicate that dilution of liquid effluents (so that pollutants approach permissible limits) is not legally acceptable.
Safe preparation of formalin solution may require an induced draft fan. Medical schools are probably aware of formalin proper preparation and disposal, they may provide advise.

"My assumptions were that formalin is very small and the risk of evaporation will be limited due to the low volatility of the formaldehyde as there is so much water in the waste stream, is this wrong thinking?"
The thinking has been OK for the start, now verify it quantitatively through Henry's law.

"I was also assuming that there was going to be no formalin in the actual seaweed after this preservation step, I take it, this is a wrong assumption as well?"
If the process is usual, "absorbed" formalin by seaweed will be within acceptable limits; otherwise it has to be measured in the lab. Assumption of no "absorption" seems OK for the start only.

"The purpose of the formaldehyde is to remove the phenols present within the seaweed and to prevent microbial growth which normal water preservation does not do. As there is between 0 to 2% phenol as part of seaweed composition, I can't remember what mass of this is but its minimal that would have reacted, percentage conversion is unknown , we have assumed that the formalin is therefore all removed, or would you say that it has through process of osmosis gone into the seaweed."
Most probably a small part of formalin has remained in seaweed (this can be measured in lab), but I am not aware of the subject

"Surely the higher water concentration of the 555kg of water would stop this ? The reason I have assumed a lot of data is because the reaction kinetics are not available for seaweed and formaldehyde."
Not having experience, I can suppose that lower HCHO concentration (i.e. higher H2O concentration) reduces "absorbed" formalin, but also phenol reaction rate and microbe destruction. Lab measurements can be useful for clarifications.

[bob789]

I have a few queries from your last posts, I am sorry if these are simple I am making it harder than it is:

(1) This is a units issue, how have you gone from 0.75ppm to 0.75E-6 mol HCHO?

ppm is a concentration unit. It cannot be converted to amount of matter (moles) directly.

ppm = mg/litre. Work from there. If you have 0.75 ppm HCHO, then you have 0.75mg HCHO in 1 litre.

Atomic mass HCHO = 30g, therefore the molarity will be:
0.00075/30 = 2.5*10^-5 moles/litre

(2) In your post yesterday I am struggling to work out how you got this mol/E+6 mol approx 0.67 mg/Nm3

(3) With there being methanol in my formalin solution would that affect disposal of the formalin on its own?

Thank you in advance.

[kkala]

"(1) This is a units issue, how have you gone from 0.75ppm to 0.75E-6 mol HCHO?
ppm is a concentration unit. It cannot be converted to amount of matter (moles) directly.
ppm = mg/litre. Work from there. If you have 0.75 ppm HCHO, then you have 0.75mg HCHO in 1 litre.
Atomic mass HCHO = 30g, therefore the molarity will be: 0.00075/30 = 2.5*10^-5 moles/litre."
ppm=mg/l is for liquids (post 28 Dec); for gases ppm means mol/mol (or ppmv). 0.75 ppm HCHO in air means 0.75 molHCHO per E+6 mol air = 0.75E-6x30 g HCHO/mol air = 0.75E-6x30/22.414 g HCHO/ Nl air=0.75x30/22.414 mg HCHO / Nm3 air = 1.00 mg HCHO / Nm3 air.

"(2) In your post yesterday I am struggling to work out how you got this mol/E+6 mol approx 0.67 mg/Nm3"
Similarly 0.5 ppm in air means 0.67 mg HCHO / Nm3 air (you can also come to this figure from above result, by analogy).

"(3) With there being methanol in my formalin solution would that affect disposal of the formalin on its own?"
I suppose that methanol has also a concentration limit is waste water that should not be exceeded; I do not know whether formalin (assumed as mixture of methanol+formaldeyde) imposes stricter limits for its two mentioned components, yet this is considered unlikely - consult legislation.