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Posted by on August 21, 2007, 10:18 pm
You typed all that, just to spam your site?
LMAO
>
>
> Calculating the BTU output of a torch, or single hole orifice.
> BTU output is the number of cubic feet of gas coming through an
> orifice, times the number of BTU's per cubic foot over time.
>
> Cubic feet of gas through the orifice is:
> Pressure differential, Hole size (area), Hole coefficient (how a thin
> stream flows through a hole. This will be different for a hole under a
> minimum size or over a maximum size, specific gravity (and maybe
> viscosity of the gas).
>
> We are calculating the cubic feet (moles, since a mole of any gas
> occupies the same volume under standard pressure and temperature. ) of
> gas that comes through a hole at each PSI (atmosphere being 0 for the
> differential).
>
> We are assuming that we get enough air (20% oxygen, Nitrogen mix) to
> fully combust the fuel.
>
> Starting with the diameter we need the area of the hole in inches.
>
> A .04 inch hole has a surface area of PI * R Squared. Radius of .04
> inch hole is: .02 PI * .02 = 0.0628318 squared = 0.00394783509124 sq
> inches.
>
> Now we have the square inches of area for the hole.
>
> At pressure for the density of the gas how much goes through in a
> given timespan?
>
> ( ((3.14159 * (Dia/2) Squared) * Coeff/144)
> Therefore:
> ((PI * Diameter / 2) Squared) * (Coefficient / 144) * (21407 * (sqroot
> of PSI)) * BTUperCubicFoot * 60
>
> After entering the new formula into the calculator, guess what!
> It does not match the chart in the Berquist book
> (or any other standard table for orifice capacity that we can find.)!
>
> After fooling around with the Coefficient / 144 figure,
> "430" makes it match the chart almost exactly for all hole sizes and
> pressures.
>
> How much air do we need?
> The combustion reaction always has these these components:
> fuel molecules + O2 --> CO2 + H2O.
>
> A Mole is a Mole.
>
> Knowing that a mole of any gas at standard pressure and temperature
> occupies the same amount of space tells us that we can use the
> calculated mole number of oxygen for each fuel to determine how many
> cubic feet of oxygen we need for combustion.
>
> For example: One mole of Methane needs two moles of Oxygen
> for complete combustion (see the chart).
>
> So one Cubic Foot of Methane needs two cubic feet of oxygen.
>
> One Cubic foot of Propane needs Five Cubic feet of oxygen.
>
> The only Wrinkle here is that Hydrogen, for example is 2 molecules
> using 1 molecule of O2,
> so the actual figure is .5 for One molecule.
>
> Woodgas, Acetylene, Butane, Diesel, are all molecules that don't work
> out unless you use two molecules instead of one for the calculations.
>
> These end up being actually half of the stated figure for oxygen when
> thinking of one molecule.
>
> Air has 20% oxygen per volume. So 5 / .2 = 25.
>
> Propane needs 25 cubic feet of air per cubic foot of Propane at
> standard pressure and temperature.
>
> The other thing that you will notice is that vs BTU the fuels need for
> air doesn't seem to make sense unless you take into account the
> elemental make-up of each fuel.
>
> If your fuel is predominently Carbon then it needs only one Oxygen,
> but if your fuel is predominently Hydrogen then it needs TWO Oxygen
> atoms to combust as the products of combustion are either CO2 or H2O
> (or both in varying qtys).
>
> http://www.frostic.com/software/BTUcalcu/index.htm
>
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