Flow Equation

Discussion in 'General Science' started by rahulpsharma, May 2, 2013.

  1. rahulpsharma

    Thread Starter New Member

    Sep 5, 2010
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    Hello All,
    Request some help on the following:

    There's a Stack and we have to measure Flue Flow thru the Stack at a height of 20 meters....!!! We have following Measurements:

    1) Fuel Gas Flow to the Burner
    2) Fuel Gas Pressure in Fuel Gas Header
    3) Fuel Gas Temperature in Fuel Gas Header
    4) %Oxygen in the Final Flue thru the Stack

    Fuel Gas is Methane..

    Based on the above information, is it possible to derive the instantaneous Mass Flow Value of Flue Gases thru the Stack in Kilograms per unit Time

    Thanks and regards
    Rahul
     
  2. #12

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    Nov 30, 2010
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  3. SplitInfinity

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    Mar 3, 2013
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    Although you are right there are some other factors that must be taken into account.

    The geometry of the stack as it has to do with reducing or possibly increasing any air flow...ie...wind that will pass over the top of the stack and change the flow rate.

    In the past due to improper smoke stack construction wind would be caught as it passed over the stack or in some other cases...wind passing over the top would create overly increased flow rates, micro-vacuums, vortexes and these would lead to blow back, coal dust or other particulates being sucked up into the stack...igniting explosively...or perhaps adversly effecting burn rate.

    Stack design and materials used is very important as well as the use of stack wind damns or flowpaths.

    Split Infinity
     
  4. #12

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    I eagerly await you quantifying any of those factors.
     
  5. SplitInfinity

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    Mar 3, 2013
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    This is obvious stuff. Just think of how airflow behaves over different airfoils or airplane wings. The wrong design creates undesirable vortexes and poor lift and the right design creates lift were one wants it.

    The same thing can be applied to the top of a smoke stack. Depending upon the diameterof the opening...the hight of the stack...the material it's made of...etc...etc...air flow from wind will create different effects.

    If a stack is tapered too much the flow will slow as well if the stack opens up to greatly the wind will play havoc with blow back.

    Good enough for you?

    Split Infinity
     
  6. SplitInfinity

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    Mar 3, 2013
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    Well...you said you were eager. Anything you would like to say or ask?

    Split Infinity
     
  7. #12

    Expert

    Nov 30, 2010
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    Yes. The OP is trying to design a chimney. The equations are available. You say other things must be considered. Well, consider them. Quantify them. Give the OP what he needs instead of a bunch of nebulous talk.
     
    Papabravo and rahulpsharma like this.
  8. rahulpsharma

    Thread Starter New Member

    Sep 5, 2010
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    Hey thanks for the replies... Actually, I am not trying to design a chimney... I am too non-technical for that...!! I am just a regular maintenance guy.. No designing stuff...!!

    Our vendor has installed a Furnace Stack which will burn Methane... He was supposed to have installed a Mass Flow Meter at 20 meters height, which somehow for some reason he missed out...!! Now he's saying that he'll do some calculations based on Chemistry of Combustion and give Flue's Mass Flow Rate....

    His calculation block, as I said, takes the following inputs:
    1) Methane Flow to Burner (m3/hr)
    2) Pressure of Methane Gas Header (bar)
    3) Temperature of Methane Gas in Header (Deg C)
    4) %age O2 content in Flue Gas

    He claims that based on these inputs alone, he'll be able to derive the Mass Flow Rate of Flue Gas... Since, he's not measuring Air Flow Rate at any point, I am not sure, if Flue Gas Flow can actually be determined...!!

    The Stack is circular and Natural Draught Stack...!!
     
  9. #12

    Expert

    Nov 30, 2010
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    It is entirely possible to arrive at "The Required Chimney" by converting the fuel rate into BTU's per hour and looking it up in Standard Mechanical Code or Uniform Mechanical Code because those books are about building codes and they have it pretty well worked out for not very math conscious people.

    Now, because your Vendor has the numbers to find the BTU's per hour, it is very possible to arrive at the mass flow for a given chimney based on the equations I have already presented.

    Bottom line = yes.
    He can do that if he knows the math.
     
  10. SplitInfinity

    Member

    Mar 3, 2013
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    Actually...all he needs to do is look at the CODE.

    Depending upon the conditions he has set forth...he will probably be required to install a FAN to have the flow mandated by...probably both the Building and Plumbing Depts...to prevent exhaust backups and issues specific to the things I have posted about.

    We own many buildings that use a variety of Gas Burners for Heat and Hot Water and depending upon the PVC angles coming out of the building as well as the distance the exhaust travels and as 12 stated...the Heat...the current CODES call for the installation of exhaust fans which in a Gas Water Heater are fans installed at the very top of the tank prior to the rest of the exhaust tube.

    Split Infinity...p.s...the fan will do what is needed.
     
  11. SplitInfinity

    Member

    Mar 3, 2013
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    I would first contact the city or town hall Building Dept. where the property is located and ask the Building Inspector which steps have to be taken to have this project approved and which code books are applicable.

    You can either download the appropriate code books online or ask the Building Inspector where the physical code books can be purchased. You can also go online and run a search for the ICC...International Commercial Code...or the International Residential Code.

    Make certain that you get the most updated code books along with the endless number of amendments that are created constantly.

    Split Infinity...p.s...This info comes from my Legal Dept.

    These can be read online but make certain you are looking at the most recent amended code. I think 2012 is the most recent but it may have amendments.
     
  12. ramueller11

    New Member

    Dec 13, 2013
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    This problem can be solved using what is given, making a few assumption. Forget fluid flows; think in terms of mass balances. Considering your process, you have two streams coming in air and fuel and you have one stream leaving. At steady state, the mass entering has to be the same as the mass exiting.

    So:
    Ma + Mf = Ms (1)

    where a,f and s subscripts represent air, fuel and stack mass flow rates.

    CH4 reacts with O2 (burning in air) in the following reaction:

    CH4 + 2O2 --> CO2 + 2H2O

    MAJOR ASSUMPTION: This is the only reaction that takes place AND all of the methane is consumed.

    Therefore if methane is fed at molar flow rate F, oxygen is being consumed at 2F.

    Equate the amount of oxygen entering from air (21 mol% * the air flow rate, unknown) minus the amount consumed (2F) to the amount of oxygen leaving the stack (the mol % = vol % of O2) times the flow rate (the unknown)

    (0.21 * A - 2F ) = x * S (2)

    Again note that:
    S = A + F (3)

    Where x is the vol % of oxygen in the stack stream, A is the molar flow rate of air and F is the molar flow rate of fuel gas.

    Thus the air molar flow rate must be:
    (0.21 - x) * A = (2 + x)*F
    A = (2 + x) / (0.21 - x) * F (4)

    F can be calculated in terms of the known methane flow rate, Q ( m^3 / hr ), Pressure, P (bar) and the temperature, T in Deg. C.

    PF = QR(T + 273) ---> F = QR(T+273) / P, where R is the Gas constant = 0.08314

    F = 0.08314 * Q(T+273)/P in kmol / hr and
    Mf = 16 (kg/kmol) * F = 1.33024 * Q(T + 273) / P kg / hr

    A = (2 + x) / (0.21 - x) * F
    Ma = 28.84 (kg / kmol) * (2 + x) / (0.21 - x) * F

    Thus:
    Ms = Ma + Mf (eq. 1)

    Ms = (28.84 * (2 + x)/(0.21 - x) + 16) * 0.08314 * Q(T+273)/P (kg / hr)
     
    Last edited: Dec 13, 2013
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