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Finding particulate matter emissions

  1. Feb 22, 2015 #1
    1. The problem statement, all variables and given/known data

    A power plant burns oil that is 4% ash and 3% Sulfur. At 50% excess air, what particulate (mg/m^3) emissions would you expect?

    2. Relevant equations
    PM in grams = 2.8*.04=.112 grams of PM per 1000 kg of oil
    Start with 1000 g of oil
    oil = 4% ash, 3%sulfur, and 93% carbon
    oxygen is 20% of air
    PM will be found in units of mg/m^3
    PV = nrt
    where p = 1 atm, r= .08206 l*atm/mole*k, t= 298K

    3. The attempt at a solution
    I started with the simple combustion equation provided by the professor:

    C + O2 ---> CO2

    The mole ratio of each part is 1.
    I then found the amount in grams or carbon. Since we started with 1000g of oil, we would have 930g of Carbon.

    930g C x (1 mole carbon/12g of carbon)x(1 moleO2/1mole C) = 77.5 moles of carbon.

    Since the mole ratio was one to one. Then there would be 77.5 moles of oxygen as well.

    77.5 moles of carbon x (1 mole of O2/ 1 mole of C) = 77.5 moles of O2. Then found the amount of O2 burned intially.

    77.5 moles of O2 X (32g of O2/1 mole of O2) = 2480 g of O2. Now to find the moles of air used.

    77.5 moles of O2 = .20 x (moles of air)
    x = 387.5 moles of air

    Now that the moles have been found, you can find the volume of air using
    v= ((387.5 moles of air)*(.08206 L*atm/mol*K)*(273K))/ 1atm
    v= 8680.92 L

    air was in 50% excess in the reaction so there needs to be a conversion.

    8680.92 L x ( 1.5)= 13021.4 L of air
    volume needs to be in m^3

    13021.4 L x (1 m^3/1000 L) = 13.021 m^3

    The amount of PM is equal to: 2.8 x .04 = .112 grams of PM
    This needs to be in mg

    .112 grams of PM x (1000 mg/ 1 gram) = 112 grams of PM

    Final solution of PM: 112 grams of PM / 13.021 m^3 = 8.60 mg/m^3 of PM

    ***** I am not sure if i correctly interpreted the excess air part of the formula. Is my method correct for finding moles of O2 to then find moles of air and using the ideal gas law? Thanks.
  2. jcsd
  3. Feb 22, 2015 #2
    I actually went back and calculated the grams of O2 and used the molar mass of air which was:

    77.5 moles of Carbon x (1 mole of O2 / 1 mole of C) x ( 32 grams of 02 / 1 mole of 02) = 2480 g of O2

    2480g of O2 = .20 x ( grams of Air)
    grams of air = 12400

    12400 grams of air x (1 mole of air / 28.967 g of Air ) = 428.073 moles of air

    v=((428.073 moles of air)*(.08206 L*atm/mol*K)*(273K))/ 1atm
    v=9589.85 L of air

    air is in excess of 50%...so 9589.85 X 1.5 = 14384.5 L
    converting 14384.5 L to m^3 is 14.3845 m^3

    Final solution of PM: 112 grams of PM / 14.385 m^3 = 7.79 mg/m^3 of PM

    **** I am not sure which route is correct in solving the problem****
  4. Feb 22, 2015 #3


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    Well, excess air should an amount of air beyond that which is required for stoichiometric combustion.

    Since this is oil rather than coal, any thoughts about the amount of hydrogen in the oil? Combustion of oil (some heavy alkane?) would produce CO2 and H2O.

    See this example - http://www.ohio.edu/mechanical/thermo/Applied/Chapt.7_11/Chapter11.html

    Are there similar examples in one's text?
  5. Feb 22, 2015 #4
    Our professor just told us to use the simple combustion formula.... C+ O2 ---> CO2
    He gave us some hints saying that we needed to solve for the mass of carbon and oxygen initally burned in order to find the amount of air.
    He just told us to use the fact that oxygen makes up 20% of air.
    The breakdown of the oil was: 4% ash, 3% sulfur, and 93% carbon.

    That link was helpful, but the only example in the link I could find was similar was 11.1, but still did not really help me.
  6. Feb 22, 2015 #5
    There are no other problems like this one in the text.
  7. Nov 5, 2015 #6
    "PM in grams = 2.8*.04=.112 grams of PM per 1000 kg of oil"
    where did 2.8 come from?
  8. Nov 6, 2015 #7


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    Homework Helper

    chemman218 has dropped out of sight from PF, unfortunately. You may not receive a reply from him.
  9. Nov 6, 2015 #8


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    One would have to know if the 0.04 = 4% by mass of the PM in the oil, or some other basis, e.g., by volume.

    If 0.04 is dimensionless, then the units of 2.8 would be the same as the units of 0.112, but it's not clear what the 0.04 represents. I would think 4 kg ash/100 kg would be quite high. The result, 0.112 gm PM in 1000 kg of oil seems to represent a very low ash content. So, something seems amiss. I wonder if the results is gms of PM per kg (1000 g) of oil.
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