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)
Therefo
((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

Holy Mackrel, I've been doing it all wrong.

Thanks for that



wrote:
>
> 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)
> Therefo
> ((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
>

wrote:
>
> 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)
> Therefo
> ((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
>
I knew that.

> wrote in message
oups.com...
>
>
> 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)
> Therefo
> ((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
>

Can you figure it out for wood? How about lump charcoal? Next, try
Kingsford.
;-)

BOB