How to convert ppm to mg/m3

According to BIPM(Bureau International des Poids et Mesures) “ppm” is “The term "ppm", meaning 10–6 relative value, or 1 in 106, or parts per million, is also used. This is analogous to the meaning of percent as parts per hundred”

According to Wikipedia “ppm” is “"Parts-per" notation is used, especially in science and engineering, to denote relative proportions in measured quantities; particularly in low-value (high-ratio) proportions at the parts-per-million (ppm), parts-per-billion (ppb), and parts-per-trillion (ppt) level.”

I think one example will be more clear. These days when everyone talks for ecology we in EU know that there is a limits of the sulfur contains in the liquid fuels. In the EU since 2007 the contain of sulfur shall not exceed 50 ppm(Euro IV). And from 2009 it must be 10 ppm.

What this digits mean actually – if you burn one litter of diesel then you will burn and 50 ppm of sulfur or 50 parts from all million. This means that you will burn 50 milliliters of sulfur! Of course the sulfur is not liquid – the stuff sulfur is a yellow crystal and it melt on 115.21 degrees. So it is impossible to be liquid in the diesel. This shows us that the ppm measurement is not so correct in when we talk about stuffs with different states – liquid and crystal.

Let see how the ppm is used in gases from chimneys and how it is used. Years ago when I have started to learn about measuring of pollutions in chimney gases everything was in mg/m3 or milligrams per cubic meter. This mean mass in volume which is also strange because we talk about gases and their masses differ and depends from temperatures and pressure. Many of the limits are given in these measurement units and if you saw the limits in EU, UK or even Bulgaria every limits are in mg/m3. For example in UK the limits of NOx in chimney gases on appliances which are new and work with liquid fuels is 450 mg/m3.

So how we should work to transfer ppm to mg/m3 at first let look for example when we have concentration of 35 ppm SO2. This mean that in one cubic meter we have 1 000 000 cubic centimeters. Then if we have 35 ppm this means that in one cubic meter there is 35 cubic centimeters of sulfur dioxide – 35 cm3/m3. Now we must find how the weight of these 35 cm3 is. And we have 25 degrees and 1 atm. pressure. From equation of Klapeyron

ρ = (M/22.4)*(ToP/PoT) – this give us a density of the gases in different temperature and pressure.
Where:
M – molecular mass of the stuff
To – normal temperature – 273K
Po – normal pressure 1.03 kg/cm2
T – measured temperature
P – measured pressure

Then in 25 Celsius degrees and 1 atm. the equitation will be

ρ = (64/22.4)*(273*1.03/1.03*298)
ρ = 2.85*0.916 = 2.617 kg/m3
So we have found the density of the 35 cubic centimeters of sulfur dioxide and it is 2.617 kg/m3. Now we could find how is the mass of sulfur dioxide in this 35 cm3:

1.35 cm3/m3 can be present as 35 x 10-6 m3/m3 = 0.000035 m3/m3
2.2.617 kg/m3 = 2 617 000 mg/m3
3.Then 2 617 000 mg/m3 * 0.000035 m3/m3= 91.56 mg/m3 sulfur dioxide

There is a simple way to calculate this here is the formula:


Xppm=(Ymg/m3*24.45)/(molecular weight)
Xmg/m3=(Xppm*molecular weight)/(24.45)You should change the needed values.

And if you are so lazy try online converter here.

Air pollution calculations part III

At this article we will see how to calculate the emissions from chimney stack by the CORINAIR methodology.
As we saw before the methodology is prepared by the European Topic Centre on Air Emissions under contract to the European Environment Agency and it describe the methods of emission calculation.
At firs we must choose what calculation to use. I am working in oil refinery and the fuels that we use are a gases and liquids. So at first we must define what emissions to calculate – as example this could be Sulfur Dioxide emissions. From CORINAIR we took the data for small combustion plants and refineries which SNAP Code is: 030103. The emission factor for SO2 is 490. Of course we should know how many tons of fuel is burned. Lets imagine that our combustion were used 250 000 tones of naphtha. We must convert this of MegaGrams or Mg so there is 250 000 Mg of naphtha. Another important think that we should to know is the concentration of sulfur in the naphtha in percents. I will take it 4% it is not normal but it could be. Actually the biggest sulfur contains can damage the pipes and machinery in every appliance and most of the refineries tried to remove it. So to generalize we have 250 000 Mg with 4% sulfur contain which is burned for one year. And to calculate the emission of one pollution compound we must multiplies the emission factor and total thermal productivity of the combustion. Total thermal productivity is product of volume of burned fuel and Lower heating value (J/mol or GJ/MG) which for naphtha vary from 35 GJ/Mg to 45 GJ/Mg:

250 000 Mg x 40 GJ/Mg =1 000 000 GJ

Emission factor of SO2 will be

EF=490 x 4 = 1960 g/GJ

The total volume of emissions should be found when:

1 000 000 GJ x 1960 g/GJ = 1 960 000 000 grams or 1960 tons per year.

If we know how many hours are worked the furnace then we could calculate what the emissions per hour are. Lets take a 8000 hours per year then

1 960 000 000 g / 8000 = 245 000 g/h

Now let divide the fuel consumption to 8000

250 000Mg / 8000 = 31.25 Mg/h

As we say before one ton of burned naphtha gave to us 12 500 cubic meters smoke gases then:

31.25 Mg/h x 12 500 = 390 625 cubic meters of smoke gases per hour

And the concentration in it must be

245 000 g/h / 390 625 m3/h = 0.627 g/m3 or 627 mg/m3

Not so bad. But of course it depends from limits
Now I will not prepare calculator for this but in the next article I promise to do this and I will put it together with another calculation unit.