Percent Excess Air (Combustion)

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  • Опубліковано 17 гру 2024

КОМЕНТАРІ • 18

  • @BaldurKhr
    @BaldurKhr 5 років тому +9

    I think the answer for the 100% excess air has to be 3334 moles of air, because 100%= (n-1667)/1667 => n= 1667 + 1667 =3334. Isn't it so? Nice explanation by the way, thanks!

  • @sulaiman1630
    @sulaiman1630 6 років тому +7

    Thank you. I hope I dont fail material and energy balance tomorrow

  • @adrianpalencia5272
    @adrianpalencia5272 6 років тому

    A furnace is fire with coal with 72.65% C, 13% ash, 1.6% N, and 3.2% S.
    The Residue in the furnace contains negligible combustible matter. The air enter the furnace at 38 degress celsius and 5 psig and 75% RH.
    The stack gas leaves at 300 degress celsius with 8.4% CO2; 2.5% CO; 0.8% H2 and 10% O2. Calculate:
    a. % Excess Air
    b. Ultimate analysis of the coal
    c. m3 of stack gas/kg of coal

    • @kelumo7981
      @kelumo7981 5 років тому +1

      interesting, i will attempt it

  • @siirinsahibi
    @siirinsahibi 4 роки тому +2

    Hi, I am planning to heat ethanol under vacuum. Will try to produce carbon nanotubes.
    Is there an explosion risk? Air will be all vacuumed. In case of a small air leakage to vacuum chamber will it immediately explode?
    Second question After separating carbon molecules, water will also appear in the chamber. Does water vapor and Ethanol vapor create any explosion on hot surfaces above 700C degrees?
    What is the maximum air pressure I must have to prevent an explosion?
    I can go down to 0.01 Bar. If there are some little amount of air remaining, can this reaction still create a big explosion?
    Amount of ethanol 1 ml
    Size of the tank 0,05m3
    Tank material steel 2mm thick.
    There is already such a method actually but I will do it alone for the first time and have some worries that I have overseen something.
    Thanks!

  • @ואתהחשבתשאנייהודי

    which JRE episode is this

  • @AsimAli-dm3gy
    @AsimAli-dm3gy 6 років тому +1

    how did you get 3333 for air fed I don't understand how you arrived at that number

    • @akshatthakre1511
      @akshatthakre1511 6 років тому +1

      Ok I know another method.
      1.First calculate theoretically required air
      2.as %excess of air=% excess of O2 use this formula.
      O2 supplied=O2 theoretically required x(1+percent excess of O2 /100)
      3.After this , you will get O2 supplied then, by material balancing i.e 100 kmol air= 21 kmol O2 then x kmol air = the amount u got after the above step. After this u will get the amount of air fed.

    • @mustafaburakzeybek3619
      @mustafaburakzeybek3619 4 роки тому

      you need 3333 mol air for obtain 100% excess rate

  • @akashrajpoot2125
    @akashrajpoot2125 6 років тому +1

    Thank you very much u clear my doubt ver well

  • @dir2310
    @dir2310 3 роки тому

    is it possible if theres 0 percent of air?

  • @Ariniarum04
    @Ariniarum04 7 років тому

    A separator is designed to remove 75% of the CO2 in flue gas (assume N2) that is fed to adsorption unit, not the fresh feed. The CO2 is removed from the bottom of the separator. The exit composition of the stream leaving the separator contains 10% CO2 and the balance is N2. a recycle stream is used, where part of the exit of the separator is recycled and unites with the fresh feed. Draw and label a process flow chart. Calculate the flow rate and compositions of all unknown streams??

  • @rheadelmundo8793
    @rheadelmundo8793 5 років тому

    How to get that 0.2 mol and 0.79 mol?

    • @LearnChemE
      @LearnChemE  5 років тому +1

      Are you referring to the amounts of oxygen (0.21) and nitrogen (0.79) in air? This is a simplified assumption of the composition of air.

  • @adrianpalencia5272
    @adrianpalencia5272 6 років тому

    1. A bituminous coal has 27.15% VCM; 62.55% FC; 7.06% Ash; 0.95% S and .28% N. It has a calorific value of 32.3 MJ/kg.
    The coal is fired with excess air supplied at 30 degrees celsius and 1 atm. Partial Orsat analysis reports 9.78% CO2, and 2.45% CO. Calculate:
    a. % Excess Air
    b. Ultimate Analysis of the coal
    c. Complete analysis of the stack gas.

  • @Bilal626339
    @Bilal626339 8 років тому +2

    For percent excess, it's easier to simply multiply 2024 by 1.25 and you will get the answer.

    • @LearnChemE
      @LearnChemE  8 років тому +1

      Yes. That is a great shortcut.