Converting to ppm

[op9]

I am involved with a particular asphalt plant burner (diesel fueled) installation which has a problem with the emissions of hydrocarbon smoke at low fire which translates into "smell" for the local residents. My analyser only measures O2, CO, NOx and not hydrocarbons. I am able to visually reduce the smoke with no material in the drum. In production it is a bit more difficult as there is the accompanying steam being emitted. On this particular burner which is unfortunately quite oversized for the job the CO (partly burnt) can be controlled and managed but we need to keep a check on the unburnts which I think is vaporised diesel.

A testing company came (at great expense) and did VOC readings as well as CO and NO2 in mg/nm^3. I can easily convert the NO2 and CO to ppm knowing molar volumes and they are about the same as mine. How do you convert the VOC to a similar volume relationship?
They came up with a figure for TVOC of 22mg/nm^3 at start up which reduces to 8.8 during normal running. What I have available to measure the emissions of "hydrocarbons" is a Drager test pump which uses chemical filled test tubes which "stain" to show a reading. They are available in only certain ranges. The two which measure Petroleum Hydrocarbons available are 100-2500ppm and 10-300ppm. So I need to know how to relate ppm to mg/ nm^3. Both of these also show the CO reading as a hydrocarbon. However I can subtract this (known value) from the total reading to achieve a representative value for comparison.

In this way I can get a feel for whether my burner adjustments have achieved a reduction in the TVOC's before the proper measurements are done.

Would appreciate any help.
Rod. Rod Nissen.
Combustion & Engineering Diagnostics
nissenr@iprimus.com.au

http://www.canded.com.au

 

[grimma]

Rod,

Usually the TOVC are expressed in equivalent sthg, as they have different molar weight. It comes from when they are analysed by chromatography, ie equivalent methane.
After that, I don't know if, to convert, you still use methane (in this case) or you use the major pollutant of the TVOC.
I don't know how it is written on the Draeger tubes, isn't it explained in the Instructions for Use (in the tubes box)?
I have one here, not at all for the TVOC but for ethylen oxide, where it is written (under the value of a norm): " 1 ppm ethylen oxide = 1,83 mg / m3 ". So maybe you can find it there too? Or ask Draeger...

I hope you'll find the answer

Delphine
d.ranslant@europe-environnement.com

http://www.esil.univ-mrs.fr/~dranslan

 

[op9]

Thanks for the response Delphine,
I have some more info. about the sampling method that was used for TVOC's. A portable Photo-Ionization Detector, referenced to benzene was used.

If I use the molar weight and vol of benzene I get:
1ppm= 3.48 mg/ Nm^3

The Draeger tubes I am considering use n-octane as the reference. Here I get 1ppm= 5.09 mg/ Nm^3.

So how do you equate their reading of 8.8 or 22mg/ Nm^3 to what I would expect on the Draeger tube? The highest readings I would expect on the tube if I used the benzene factor would be:
8.8= 2.5ppm and 22= 6.3ppm Either would be too low to read on either Draeger hydrocarbon tube.

Now one other complicating factor.....the tubes are also cross sensitive to the following substances:

50ppm n-hexane shows as 70ppm on tube
100ppm n-heptane shows as 150ppm on tube
10ppm iso-octane.........15ppm.........
100ppm iso-octane........150ppm.......
200ppm iso-octane........350ppm......
50ppm n-nonane............50ppm.......
50ppm perchloroethylene...50ppm......
30ppm CO..................20ppm.....

I know I have a CO ppm at present up to 300ppm which I would guess would show as say 200ppm on the tube. what other organic compounds are present in the diesel "smoke" or vapour I don't know. So I am not sure if these tubes will be of use. Even used as a comparitor for checking before and after burner adjustments, may not work because the ppm relating to the measured TVOC's could be such a small proportion. I really don't know. Unfortunately Draeger in Australia do not appear to know the answer themselves. They have made some approach to Germany which was only answered after some days but I am not confident they have asked the right questions. I tried to find an E-mail on the website but it gives no E-mail address, just provides a form back to your country of origin.

I would appreciate your comments,
Ta,
Rod. Rod Nissen.
Combustion & Engineering Diagnostics
nissenr@iprimus.com.au

http://www.canded.com.au

 

[grimma]

Rod,

That's a really interesting problem you've raised, unfortunately I can't help you much.
I will ask my teachers (yeah, still in internship) and the VOC specialist of my company, maybe they can give some tips.

We use the Draeger tubes too, but for other compounds. When it is possible, we try to compare the tube's results to other means of analysis, and it has shown that Draeger tubes are less accurate, first because of the graduations, the size of the cristals in it and what our eyes want to see, then I guess because the ranges are sometimes large - maybe there are other reasons.

I'll let you know if there's something new.

Delphine
d.ranslant@europe-environnement.com

http://www.esil.univ-mrs.fr/~dranslan

 

[grimma]

Rod,

It seems that, if you don't know the proportions of your different VOc, you won't be able to go from ppm to mg/m3 or the opposite.
Moreover, this forum isn't very active, maybe you should try to post it in another one to get more answers.

However, here is what I have found. It's long and translated directly from French, so some sentences may seem weird...

It is for FID and no PID, but I guess the theory for this case can be the same.

--------------------------------------

The response of the detector towards one particular molecule isn't directly proportional to its number of carbon atoms, because the response of an atom depends on the nature of the atoms next to it and of the kind of link in which each atom is engaged.

Therefore, the result isn't a "voc concentration" but a "relative indication to the organic compounds".

Here are the estimations, generally used, for the response coefficients of a carbon atom, function of the kind of link:
______________________________________
|NATURE OF THE LINK RESPONSE COEF. |
--------------------------------------
|aliphatic 1 (convention)|
|aromatic 1 |
|olefinic (?) 0.95 |
|acétylenic (?) 1.3 |
|carbonyl(ex. HCHO) 0 |
|nitril (?), amide 0.3 |
|ether 0.5 |
|alcohol 0.2 - 0.6 |
---------------------------------------

ex. Xylene : 6 aromatic + 2 aliphatic -> 6x1 + 2*1 = 8

UNITS CONVERSION

Calculations take into account:
- molar mass of the reference compounds (16g methane, 44g propane,...)
- mass of carbon atom (12g)
- molar volume (22.4L in normal conditions / for Nm3 24L)

You admit by convention that one propane molecule (3 carbon atoms) gives a three times higher signal than a methane molecule.
Thus, an effluent which concentration is X ppm in propane equivalent will have a 3X ppm concnetration in methane equivalent.
In massic (?) units, this effluent will have a concentration:
- X*44/22.4 = 1.96X mg/m3 propane equivalent
- 3X*16/22.4 = 2.15X mg/m3 methane equivalent
- 3X*12/22.4 = 1.61X mg/m3 total organic carbon.

EXAMPLE OF THE ESTIMATION OF THE CONCENTRATION OF VOC based on the FID indication

You have an effluent with a flow of 10,000 m3/h (normal conditions)
The FID indication is 200 mg/m3 equivalent propane
It contains ethyl acetate, MEK, isopropanol and ethanol at these proportions: 40/30/20/10 %
What is the value in VOC, in kg/h ?

Let's say X is the concentration of COV of the effluent.

- an ethyl acetate molecule (CH3COOC2H5, molar mass 88g) has 3 carbon atoms seen in FID (cf. response coefficient). Thus, it is detected as a propane molecule (molar mass 44g). The concentration of ethyl acetate (0.4X) will provide a signal corresponding to
0.4X*44/88 = 0.2X eq. propane

-For the MEK (CH3COOC2H5, molar mass 72g), 3 carbon atomes are seen in FID. Its concentration in the effluent (0.3X) will provide a signal corresponding to
0.3X*44/72 = 0.183X

- isopropanol ( (CH3)2CHOH, molar mass 60g): we admit that one molecule is seen as if it had 2.5 carbon atoms. The corresponding signal will be
0.2X*2.5/3*44/60 = 0.123X

- ethanol (C2H5OH, molar mass 46g) : one molecule is seen as 1.5 carbon atom. The corresponding signal will be
0.1X*1.5/3*44/66 = 0.05X

So, the total calculated signal is like a concentration of 0.556X eq. propane

The concentration of VOC is near 200/0.556, that is 360 mg/m3, which leads to an "exit flow of VOC" of 3,6 kg VOC/h

-------------------------------
Hope this can help, at least a bit...

Delphine
d.ranslant@europe-environnement.com

http://www.esil.univ-mrs.fr/~dranslan

PS: sorry, but French here, what's "Ta"???

 

[op9]

Dear Delphine,
Thanks for all your efforts and input. I'll have to sit down and "digest" all that.
"Ta" is an abbreviated Thank You. Probably more used in England than Australia.

Ta,
Rod.

Rod Nissen.
Combustion & Engineering Diagnostics
nissenr@iprimus.com.au

http://www.canded.com.au