Air and water pollution control., HTML programming, GNU Linux.

Tuesday, 5 August 2008

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.