I am still working on calculating the energy that is required to disassociate the Water in the Water path of Stan's injector drawing posted by Russ. First I had to calculate the Volume of the water path in the injector cell using the dementions on the drawings posted by Russ.

The volume can be calculated using the Volume of a Frustrum of a Cone formula. First you calculate the Volume of the open area using the Conical inner surface of .100 dia exit port and .176 dia inlet hole. Next calculate the Volume of the Electrode cone given .080 diameter tip and .110 diameter at the start of the cone.
Both have a length of .993 inches. Then the Volme of the Water path = Volume of Inner surface - Volume of the Electrode Cone.

Given that Volume of a Frustrum =
(1/3) pi * h ( R sqr + R * r + r sqr )
 reference - http://www.roymech.co.uk/Useful_Tables/Maths/Mathematics.html

The Volume of the inner surface conical hole:
 where r = .050, R = .088, and h = .993
Volume = (1/3) Pi * .993 * (.088 sqr + (.088 * .050) + .050 sqr)
= 1.039867168 * .014644
= .152278148 cubic inches

The volume of the tip surface of the probe:
where r = .040, R = .055, and h = .993
Volume = (1/3) Pi * .993 * (.055 sqr + ( .055 * .040 ) + .040 sqr)
=1.039867168 * .006825
= .007097093 cubic inches

The water path Volume:
Path Volume = Inner Surface Volume - Tip Surfce Volume
= .152278148 -  .007097093
= .14518105457 cubic inches

Converting to cubic cm = .14518105457 * 16.387064
= 2.379091232 cubic cm
ref: https://www.google.com/#hl=en&gs_nf=1&tok=p69iBiIwpIPZCmXy-OE2qQ&cp=19&gs_id=2s&xhr=t&q=cubic+in+to+cubic+cm&pf=p&sclient=psy-ab&oq=cubic+in+to+cubic+c&aq=0&aqi=g2g-K1g-m1&aql=&gs_l=&pbx=1&bav=on.2,or.r_gc.r_pw.r_qf.,cf.osb&fp=190edb001faa8126&biw=1193&bih=736

Now, how much Water does this Water Path hold?

The average atomic mass of natural hydrogen is 1.00794 u and that of natural oxygen is 15.9994 u; therefore, the molecular mass of natural water with formula H2O is (2 × 1.00794 u) + 15.9994 u = 18.01528 u. Therefore, one mole of water has a mass of 18.01528 grams. However, the exact mass of hydrogen-1 (the most common hydrogen isotope) is 1.00783, and the exact mass of oxygen-16 (the most common oxygen isotope) is 15.9949, so the mass of the most common molecule of water is 18.01056 u.
ref : http://en.wikipedia.org/wiki/Molecular_mass

The Density oif Water at 25 deg C = .997044
ref :http://www2.volstate.edu/CHEM/Density_of_Water.htm

We can convert to grams of Water with the formula
Mass (grams) = Volume (cubic cm) * Density (gram/cubic cm)
ref: http://wiki.answers.com/Q/How_do_you_convert_ml_into_mole

Therefore Mass = 2.379091232  * .997044 = 2.372058638 grams of Water
or  2.372058638 ml of Water

We can turn this into moles by using the formula:
mm = m / n
molar mass = mass / moles
Therefore:
2.372058638 ml  /  18.01056 u. = 0.131703769 moles of Water in the Water Path

ref: http://wiki.answers.com/Q/How_do_you_convert_ml_into_mole#ixzz1x2lpCZxB
ref : http://calculator-converter.com/converter_g_to_ml_grams_to_milliliters_calculator.php

Terry Dixon

Quote from terry.dixon on June 6th, 2012, 02:42 PM

I am still working on calculating the energy that is required to disassociate the Water in the Water path of Stan's injector drawing posted by Russ. First I had to calculate the Volume of the water path in the injector cell using the dementions on the drawings posted by Russ.

The volume can be calculated using the Volume of a Frustrum of a Cone formula. First you calculate the Volume of the open area using the Conical inner surface of .100 dia exit port and .176 dia inlet hole. Next calculate the Volume of the Electrode cone given .080 diameter tip and .110 diameter at the start of the cone.
Both have a length of .993 inches. Then the Volme of the Water path = Volume of Inner surface - Volume of the Electrode Cone.

Given that Volume of a Frustrum =
(1/3) pi * h ( R sqr + R * r + r sqr )
 reference - http://www.roymech.co.uk/Useful_Tables/Maths/Mathematics.html

The Volume of the inner surface conical hole:
 where r = .050, R = .088, and h = .993
Volume = (1/3) Pi * .993 * (.088 sqr + (.088 * .050) + .050 sqr)
= 1.039867168 * .014644
= .152278148 cubic inches

The volume of the tip surface of the probe:
where r = .040, R = .055, and h = .993
Volume = (1/3) Pi * .993 * (.055 sqr + ( .055 * .040 ) + .040 sqr)
=1.039867168 * .006825
= .007097093 cubic inches

The water path Volume:
Path Volume = Inner Surface Volume - Tip Surfce Volume
= .152278148 -  .007097093
= .14518105457 cubic inches

Converting to cubic cm = .14518105457 * 16.387064
= 2.379091232 cubic cm
ref: https://www.google.com/#hl=en&gs_nf=1&tok=p69iBiIwpIPZCmXy-OE2qQ&cp=19&gs_id=2s&xhr=t&q=cubic+in+to+cubic+cm&pf=p&sclient=psy-ab&oq=cubic+in+to+cubic+c&aq=0&aqi=g2g-K1g-m1&aql=&gs_l=&pbx=1&bav=on.2,or.r_gc.r_pw.r_qf.,cf.osb&fp=190edb001faa8126&biw=1193&bih=736

Now, how much Water does this Water Path hold?

The average atomic mass of natural hydrogen is 1.00794 u and that of natural oxygen is 15.9994 u; therefore, the molecular mass of natural water with formula H2O is (2 × 1.00794 u) + 15.9994 u = 18.01528 u. Therefore, one mole of water has a mass of 18.01528 grams. However, the exact mass of hydrogen-1 (the most common hydrogen isotope) is 1.00783, and the exact mass of oxygen-16 (the most common oxygen isotope) is 15.9949, so the mass of the most common molecule of water is 18.01056 u.
ref : http://en.wikipedia.org/wiki/Molecular_mass

The Density oif Water at 25 deg C = .997044
ref :http://www2.volstate.edu/CHEM/Density_of_Water.htm

We can convert to grams of Water with the formula
Mass (grams) = Volume (cubic cm) * Density (gram/cubic cm)
ref: http://wiki.answers.com/Q/How_do_you_convert_ml_into_mole

Therefore Mass = 2.379091232  * .997044 = 2.372058638 grams of Water
or  2.372058638 ml of Water

We can turn this into moles by using the formula:
mm = m / n
molar mass = mass / moles
Therefore:
2.372058638 ml  /  18.01056 u. = 0.131703769 moles of Water in the Water Path

ref: http://wiki.answers.com/Q/How_do_you_convert_ml_into_mole#ixzz1x2lpCZxB
ref : http://calculator-converter.com/converter_g_to_ml_grams_to_milliliters_calculator.php

Terry Dixon

Hi Terry, question, will the electrical pulse change the volume or flow of the water? :D

Quote from Jeff Nading on June 6th, 2012, 04:56 PM

Quote from terry.dixon on June 6th, 2012, 02:42 PM

I am still working on calculating the energy that is required to disassociate the Water in the Water path of Stan's injector drawing posted by Russ. First I had to calculate the Volume of the water path in the injector cell using the dementions on the drawings posted by Russ.

The volume can be calculated using the Volume of a Frustrum of a Cone formula. First you calculate the Volume of the open area using the Conical inner surface of .100 dia exit port and .176 dia inlet hole. Next calculate the Volume of the Electrode cone given .080 diameter tip and .110 diameter at the start of the cone.
Both have a length of .993 inches. Then the Volme of the Water path = Volume of Inner surface - Volume of the Electrode Cone.

Given that Volume of a Frustrum =
(1/3) pi * h ( R sqr + R * r + r sqr )
 reference - http://www.roymech.co.uk/Useful_Tables/Maths/Mathematics.html

The Volume of the inner surface conical hole:
 where r = .050, R = .088, and h = .993
Volume = (1/3) Pi * .993 * (.088 sqr + (.088 * .050) + .050 sqr)
= 1.039867168 * .014644
= .152278148 cubic inches

The volume of the tip surface of the probe:
where r = .040, R = .055, and h = .993
Volume = (1/3) Pi * .993 * (.055 sqr + ( .055 * .040 ) + .040 sqr)
=1.039867168 * .006825
= .007097093 cubic inches

The water path Volume:
Path Volume = Inner Surface Volume - Tip Surfce Volume
= .152278148 -  .007097093
= .14518105457 cubic inches

Converting to cubic cm = .14518105457 * 16.387064
= 2.379091232 cubic cm
ref: https://www.google.com/#hl=en&gs_nf=1&tok=p69iBiIwpIPZCmXy-OE2qQ&cp=19&gs_id=2s&xhr=t&q=cubic+in+to+cubic+cm&pf=p&sclient=psy-ab&oq=cubic+in+to+cubic+c&aq=0&aqi=g2g-K1g-m1&aql=&gs_l=&pbx=1&bav=on.2,or.r_gc.r_pw.r_qf.,cf.osb&fp=190edb001faa8126&biw=1193&bih=736

Now, how much Water does this Water Path hold?

The average atomic mass of natural hydrogen is 1.00794 u and that of natural oxygen is 15.9994 u; therefore, the molecular mass of natural water with formula H2O is (2 × 1.00794 u) + 15.9994 u = 18.01528 u. Therefore, one mole of water has a mass of 18.01528 grams. However, the exact mass of hydrogen-1 (the most common hydrogen isotope) is 1.00783, and the exact mass of oxygen-16 (the most common oxygen isotope) is 15.9949, so the mass of the most common molecule of water is 18.01056 u.
ref : http://en.wikipedia.org/wiki/Molecular_mass

The Density oif Water at 25 deg C = .997044
ref :http://www2.volstate.edu/CHEM/Density_of_Water.htm

We can convert to grams of Water with the formula
Mass (grams) = Volume (cubic cm) * Density (gram/cubic cm)
ref: http://wiki.answers.com/Q/How_do_you_convert_ml_into_mole

Therefore Mass = 2.379091232  * .997044 = 2.372058638 grams of Water
or  2.372058638 ml of Water

We can turn this into moles by using the formula:
mm = m / n
molar mass = mass / moles
Therefore:
2.372058638 ml  /  18.01056 u. = 0.131703769 moles of Water in the Water Path

ref: http://wiki.answers.com/Q/How_do_you_convert_ml_into_mole#ixzz1x2lpCZxB
ref : http://calculator-converter.com/converter_g_to_ml_grams_to_milliliters_calculator.php

Terry Dixon

Hi Terry, question, will the electrical pulse change the volume or flow of the water? :D

Hey Jeff, Good to hear from you. Yes it will, Thats what I am trying to work out now. The Molar Volume of the water will change to the Molar volume of the H2 and O2 Gas. I will get back to you when I do the calculations. Just about got it figured out.

Terry Dixon

Quote from terry.dixon on June 6th, 2012, 06:32 PM

Quote from Jeff Nading on June 6th, 2012, 04:56 PM

Quote from terry.dixon on June 6th, 2012, 02:42 PM

I am still working on calculating the energy that is required to disassociate the Water in the Water path of Stan's injector drawing posted by Russ. First I had to calculate the Volume of the water path in the injector cell using the dementions on the drawings posted by Russ.

The volume can be calculated using the Volume of a Frustrum of a Cone formula. First you calculate the Volume of the open area using the Conical inner surface of .100 dia exit port and .176 dia inlet hole. Next calculate the Volume of the Electrode cone given .080 diameter tip and .110 diameter at the start of the cone.
Both have a length of .993 inches. Then the Volme of the Water path = Volume of Inner surface - Volume of the Electrode Cone.

Given that Volume of a Frustrum =
(1/3) pi * h ( R sqr + R * r + r sqr )
 reference - http://www.roymech.co.uk/Useful_Tables/Maths/Mathematics.html

The Volume of the inner surface conical hole:
 where r = .050, R = .088, and h = .993
Volume = (1/3) Pi * .993 * (.088 sqr + (.088 * .050) + .050 sqr)
= 1.039867168 * .014644
= .152278148 cubic inches

The volume of the tip surface of the probe:
where r = .040, R = .055, and h = .993
Volume = (1/3) Pi * .993 * (.055 sqr + ( .055 * .040 ) + .040 sqr)
=1.039867168 * .006825
= .007097093 cubic inches

The water path Volume:
Path Volume = Inner Surface Volume - Tip Surfce Volume
= .152278148 -  .007097093
= .14518105457 cubic inches

Converting to cubic cm = .14518105457 * 16.387064
= 2.379091232 cubic cm
ref: https://www.google.com/#hl=en&gs_nf=1&tok=p69iBiIwpIPZCmXy-OE2qQ&cp=19&gs_id=2s&xhr=t&q=cubic+in+to+cubic+cm&pf=p&sclient=psy-ab&oq=cubic+in+to+cubic+c&aq=0&aqi=g2g-K1g-m1&aql=&gs_l=&pbx=1&bav=on.2,or.r_gc.r_pw.r_qf.,cf.osb&fp=190edb001faa8126&biw=1193&bih=736

Now, how much Water does this Water Path hold?

The average atomic mass of natural hydrogen is 1.00794 u and that of natural oxygen is 15.9994 u; therefore, the molecular mass of natural water with formula H2O is (2 × 1.00794 u) + 15.9994 u = 18.01528 u. Therefore, one mole of water has a mass of 18.01528 grams. However, the exact mass of hydrogen-1 (the most common hydrogen isotope) is 1.00783, and the exact mass of oxygen-16 (the most common oxygen isotope) is 15.9949, so the mass of the most common molecule of water is 18.01056 u.
ref : http://en.wikipedia.org/wiki/Molecular_mass

The Density oif Water at 25 deg C = .997044
ref :http://www2.volstate.edu/CHEM/Density_of_Water.htm

We can convert to grams of Water with the formula
Mass (grams) = Volume (cubic cm) * Density (gram/cubic cm)
ref: http://wiki.answers.com/Q/How_do_you_convert_ml_into_mole

Therefore Mass = 2.379091232  * .997044 = 2.372058638 grams of Water
or  2.372058638 ml of Water

We can turn this into moles by using the formula:
mm = m / n
molar mass = mass / moles
Therefore:
2.372058638 ml  /  18.01056 u. = 0.131703769 moles of Water in the Water Path

ref: http://wiki.answers.com/Q/How_do_you_convert_ml_into_mole#ixzz1x2lpCZxB
ref : http://calculator-converter.com/converter_g_to_ml_grams_to_milliliters_calculator.php

Terry Dixon

Hi Terry, question, will the electrical pulse change the volume or flow of the water? :D

Hey Jeff, Good to hear from you. Yes it will, Thats what I am trying to work out now. The Molar Volume of the water will change to the Molar volume of the H2 and O2 Gas. I will get back to you when I do the calculations. Just about got it figured out.

Terry Dixon

Thanks Terry great job, Jeff.:D:cool:

Now that we have the volume of Water in the Water Path of Stan's injector,
0.131703769 moles of Water ,
What is the Volume of the Hydrogen and Oxygen after the disassociation or decomposition of that volume of Water?

When we expose the Water Capacitor to the High Voltage produced by the  Coil in the Resonant Circuit, the Water will break down into its constituents: hydrogen and oxygen. Both hydrogen and oxygen gas exist as diatomic molecules so the equation for this chemical reaction is:
 2H2O -> 2H2 + O2
The prefixes tell us that it takes 2 molecules of water to produce one molecule of oxygen and two molecules of hydrogen. If we decompose 36 grams of water (2 moles), we produce 2 moles of hydrogen gas (4 g) and one mole of oxygen gas (32 g).

2H20 (l) + Energy --> 2H2 (g) + O2 (g)

If electrolysis occurs at standard temp and pressure (s.t.p. = 298.15 K ; 1 atm), then:

PV = nRT

For Hydrogen:
(1 atm) * (Xhydrogen) = 2 moles * (0.08206 L * Atm / K * mol) * (298.15 K)
Xhydrogen = 48.9 L of Hydrogen

since there is half as many moles of Oxygen, then there will be half as much gas or:
Xoxygen = 24.45 L of oxygen (the volume of one mole of gas at s.t.p.)

Remember, this is from 2 moles of water, so if you simplify the balanced chemical equation to one mole of water and a fraction of a mole of oxygen, then you need to decrease the volumes of gasses produced by 0.5.

Applying this to our Volume in the Water Path of Stan's Cell, We have:

0.131703769 moles of Water in the Water Path,
of which 66.66% will be the Hydrogen Volume and 33.33 will be the Oxygen Volume.

Sinse Std Molar Volume of a gas = 24.45 L at S.T.P.
ref: http://wiki.answers.com/Q/What_is_the_volume_hydrogen_to_oxygen_after_electrolisis

Volume of Hydrogen = .6666 * .131703769 *  24.45 L of Hydrogen per mole
= 2.14655675 Liters of Hydrogen

Volume of Oxygen = .3333 * .131703769 * 24.45 L of Oxygen per mole
= 1.073278378 Liters of Oxygen

Total Volume of the H-H-O Gas after disassociation = 3.219835128 Liters of Gas

ref: http://wiki.answers.com/Q/What_is_the_volume_hydrogen_to_oxygen_after_electrolisis

ref: http://wiki.answers.com/Q/What_is_the_volume_hydrogen_to_oxygen_after_electrolisis#ixzz1x8nL6WDT

ref: http://www2.volstate.edu/CHEM/Density_of_Water.htm

Now, How much energy does it take to convert the Water to its Hydrogen and Oxygen constituents ?

 For any particular chemical bond, say the covalent bond between hydrogen and oxygen, the amount of energy it takes to break that bond is exactly the same as the amount of energy released when the bond is formed. This value is called the bond energy.
 There are many forms of energy:
•electrical
•mechanical
•chemical
 but all forms are ultimately converted into heat. So it is convenient for biologists to measure energy in units of heat. The unit we shall use most often is the kilocalorie (kcal): the amount of heat needed to warm 1 liter of water 1 degree Celsius.
 
The kilocalorie is also the unit used to describe the energy content of foods. It is the "Calorie" used on food labels.
 It takes a net of 118 kcal to decompose 2 moles of H2O into its elements. Actually it takes more than 118 kcal to decompose the water into its atoms, but some of the energy is given back as the atoms immediately bond together to form molecules of hydrogen and oxygen.
Let's look at the numbers.
•The bond energy of the H-O bond is 110 kcal.
•The bond energy of H-H bonds is 103 kcal.
•The bond energy of the O=O bonds is 116 kcal.
•The decomposition of 2 molecules of water requires breaking 4 H-O bonds and thus the input of 440 kcal.
 •The formation of 2 moles of hydrogen yields 206 kcal (2 x 103).
•The formation of 1 mole of oxygen yields 116 kcal.
•The difference between ◦the energy released (206 + 116 = 322 kcal) and
◦the energy consumed (4 x 110 = 440 kcal)
 
•gives the net energy consumed = 118 kcal.

If we calculate the energy required for the Volume of Water in the Water path of Stan's injector to break the 4 HO Bonds We have to divide by two to get the Energy per mole so 440 kcal / 2 = 220 kcal per mole
So:  220 kcal (per mole) * (0.131703769 moles of Water in the Water Path)
= 28.97482918 kcal

ref: http://wiki.answers.com/Q/What_is_the_volume_hydrogen_to_oxygen_after_electrolisis
ref: http://users.rcn.com/jkimball.ma.ultranet/BiologyPages/B/BondEnergy.html

Terry Dixon

Now that we have the volume of Water in the Water Path of Stan's injector,
0.131703769 moles of Water ,
What is the Volume of the Hydrogen and Oxygen after the disassociation or decomposition of that volume of Water?

When we expose the Water Capacitor to the High Voltage produced by the  Coil in the Resonant Circuit, the Water will break down into its constituents: hydrogen and oxygen. Both hydrogen and oxygen gas exist as diatomic molecules so the equation for this chemical reaction is:
 2H2O -> 2H2 + O2
The prefixes tell us that it takes 2 molecules of water to produce one molecule of oxygen and two molecules of hydrogen. If we decompose 36 grams of water (2 moles), we produce 2 moles of hydrogen gas (4 g) and one mole of oxygen gas (32 g).

2H20 (l) + Energy --> 2H2 (g) + O2 (g)

If electrolysis occurs at standard temp and pressure (s.t.p. = 298.15 K ; 1 atm), then:

PV = nRT

For Hydrogen:
(1 atm) * (Xhydrogen) = 2 moles * (0.08206 L * Atm / K * mol) * (298.15 K)
Xhydrogen = 48.9 L of Hydrogen

since there is half as many moles of Oxygen, then there will be half as much gas or:
Xoxygen = 24.45 L of oxygen (the volume of one mole of gas at s.t.p.)

Remember, this is from 2 moles of water, so if you simplify the balanced chemical equation to one mole of water and a fraction of a mole of oxygen, then you need to decrease the volumes of gasses produced by 0.5.

Applying this to our Volume in the Water Path of Stan's Cell, We have:

0.131703769 moles of Water in the Water Path,
of which 66.66% will be the Hydrogen Volume and 33.33 will be the Oxygen Volume.

Sinse Std Molar Volume of a gas = 24.45 L at S.T.P.
ref: http://wiki.answers.com/Q/What_is_the_volume_hydrogen_to_oxygen_after_electrolisis

Volume of Hydrogen = .6666 * .131703769 *  24.45 L of Hydrogen per mole
= 2.14655675 Liters of Hydrogen

Volume of Oxygen = .3333 * .131703769 * 24.45 L of Oxygen per mole
= 1.073278378 Liters of Oxygen

Total Volume of the H-H-O Gas after disassociation = 3.219835128 Liters of Gas

ref: http://wiki.answers.com/Q/What_is_the_volume_hydrogen_to_oxygen_after_electrolisis

ref: http://wiki.answers.com/Q/What_is_the_volume_hydrogen_to_oxygen_after_electrolisis#ixzz1x8nL6WDT

ref: http://www2.volstate.edu/CHEM/Density_of_Water.htm

Now, How much energy does it take to convert the Water to its Hydrogen and Oxygen constituents ?

 For any particular chemical bond, say the covalent bond between hydrogen and oxygen, the amount of energy it takes to break that bond is exactly the same as the amount of energy released when the bond is formed. This value is called the bond energy.
 There are many forms of energy:
•electrical
•mechanical
•chemical
 but all forms are ultimately converted into heat. So it is convenient for biologists to measure energy in units of heat. The unit we shall use most often is the kilocalorie (kcal): the amount of heat needed to warm 1 liter of water 1 degree Celsius.
 
The kilocalorie is also the unit used to describe the energy content of foods. It is the "Calorie" used on food labels.
 It takes a net of 118 kcal to decompose 2 moles of H2O into its elements. Actually it takes more than 118 kcal to decompose the water into its atoms, but some of the energy is given back as the atoms immediately bond together to form molecules of hydrogen and oxygen.
Let's look at the numbers.
•The bond energy of the H-O bond is 110 kcal.
•The bond energy of H-H bonds is 103 kcal.
•The bond energy of the O=O bonds is 116 kcal.
•The decomposition of 2 molecules of water requires breaking 4 H-O bonds and thus the input of 440 kcal.
 •The formation of 2 moles of hydrogen yields 206 kcal (2 x 103).
•The formation of 1 mole of oxygen yields 116 kcal.
•The difference between ◦the energy released (206 + 116 = 322 kcal) and
◦the energy consumed (4 x 110 = 440 kcal)
 
•gives the net energy consumed = 118 kcal.

If we calculate the energy required for the Volume of Water in the Water path of Stan's injector to break the 4 HO Bonds We have to divide by two to get the Energy per mole so 440 kcal / 2 = 220 kcal per mole
So:  220 kcal (per mole) * (0.131703769 moles of Water in the Water Path)
= 28.97482918 kcal

ref: http://wiki.answers.com/Q/What_is_the_volume_hydrogen_to_oxygen_after_electrolisis
ref: http://users.rcn.com/jkimball.ma.ultranet/BiologyPages/B/BondEnergy.html

Terry Dixon

Quote from terry.dixon on June 7th, 2012, 03:37 PM

Now that we have the volume of Water in the Water Path of Stan's injector,
0.131703769 moles of Water ,
What is the Volume of the Hydrogen and Oxygen after the disassociation or decomposition of that volume of Water?

When we expose the Water Capacitor to the High Voltage produced by the  Coil in the Resonant Circuit, the Water will break down into its constituents: hydrogen and oxygen. Both hydrogen and oxygen gas exist as diatomic molecules so the equation for this chemical reaction is:
 2H2O -> 2H2 + O2
The prefixes tell us that it takes 2 molecules of water to produce one molecule of oxygen and two molecules of hydrogen. If we decompose 36 grams of water (2 moles), we produce 2 moles of hydrogen gas (4 g) and one mole of oxygen gas (32 g).

2H20 (l) + Energy --> 2H2 (g) + O2 (g)

If electrolysis occurs at standard temp and pressure (s.t.p. = 298.15 K ; 1 atm), then:

PV = nRT

For Hydrogen:
(1 atm) * (Xhydrogen) = 2 moles * (0.08206 L * Atm / K * mol) * (298.15 K)
Xhydrogen = 48.9 L of Hydrogen

since there is half as many moles of Oxygen, then there will be half as much gas or:
Xoxygen = 24.45 L of oxygen (the volume of one mole of gas at s.t.p.)

Remember, this is from 2 moles of water, so if you simplify the balanced chemical equation to one mole of water and a fraction of a mole of oxygen, then you need to decrease the volumes of gasses produced by 0.5.

Applying this to our Volume in the Water Path of Stan's Cell, We have:

0.131703769 moles of Water in the Water Path,
of which 66.66% will be the Hydrogen Volume and 33.33 will be the Oxygen Volume.

Sinse Std Molar Volume of a gas = 24.45 L at S.T.P.
ref: http://wiki.answers.com/Q/What_is_the_volume_hydrogen_to_oxygen_after_electrolisis

Volume of Hydrogen = .6666 * .131703769 *  24.45 L of Hydrogen per mole
= 2.14655675 Liters of Hydrogen

Volume of Oxygen = .3333 * .131703769 * 24.45 L of Oxygen per mole
= 1.073278378 Liters of Oxygen

Total Volume of the H-H-O Gas after disassociation = 3.219835128 Liters of Gas

ref: http://wiki.answers.com/Q/What_is_the_volume_hydrogen_to_oxygen_after_electrolisis

ref: http://wiki.answers.com/Q/What_is_the_volume_hydrogen_to_oxygen_after_electrolisis#ixzz1x8nL6WDT

ref: http://www2.volstate.edu/CHEM/Density_of_Water.htm

Now, How much energy does it take to convert the Water to its Hydrogen and Oxygen constituents ?

 For any particular chemical bond, say the covalent bond between hydrogen and oxygen, the amount of energy it takes to break that bond is exactly the same as the amount of energy released when the bond is formed. This value is called the bond energy.
 There are many forms of energy:
•electrical
•mechanical
•chemical
 but all forms are ultimately converted into heat. So it is convenient for biologists to measure energy in units of heat. The unit we shall use most often is the kilocalorie (kcal): the amount of heat needed to warm 1 liter of water 1 degree Celsius.
 
The kilocalorie is also the unit used to describe the energy content of foods. It is the "Calorie" used on food labels.
 It takes a net of 118 kcal to decompose 2 moles of H2O into its elements. Actually it takes more than 118 kcal to decompose the water into its atoms, but some of the energy is given back as the atoms immediately bond together to form molecules of hydrogen and oxygen.
Let's look at the numbers.
•The bond energy of the H-O bond is 110 kcal.
•The bond energy of H-H bonds is 103 kcal.
•The bond energy of the O=O bonds is 116 kcal.
•The decomposition of 2 molecules of water requires breaking 4 H-O bonds and thus the input of 440 kcal.
 •The formation of 2 moles of hydrogen yields 206 kcal (2 x 103).
•The formation of 1 mole of oxygen yields 116 kcal.
•The difference between ◦the energy released (206 + 116 = 322 kcal) and
◦the energy consumed (4 x 110 = 440 kcal)
 
•gives the net energy consumed = 118 kcal.

If we calculate the energy required for the Volume of Water in the Water path of Stan's injector to break the 4 HO Bonds We have to divide by two to get the Energy per mole so 440 kcal / 2 = 220 kcal per mole
So:  220 kcal (per mole) * (0.131703769 moles of Water in the Water Path)
= 28.97482918 kcal

ref: http://wiki.answers.com/Q/What_is_the_volume_hydrogen_to_oxygen_after_electrolisis
ref: http://users.rcn.com/jkimball.ma.ultranet/BiologyPages/B/BondEnergy.html

Terry Dixon

Thanks Terry, now another question,  these were all standard measurements of known science for the calculations preformed, right ? "kilocalorie (kcal): the amount of heat needed to warm 1 liter of water 1 degree Celsius" . The reason I ask is that Stan claimed he could drive his dune buggy from coast to coast on 22 gallons of water, knowing this would this change standard measurements of known science and or change the end result of the calculations, or can we go with the end results of the calculations you have just preformed ? Thanks again Terry for your hard work, Jeff.:cool::D:P

Jeff's Question :
Thanks Terry, now another question, these were all standard measurements of known science for the calculations preformed, right ? "kilocalorie (kcal): the amount of heat needed to warm 1 liter of water 1 degree Celsius" . The reason I ask is that Stan claimed he could drive his dune buggy from coast to coast on 22 gallons of water, knowing this would this change standard measurements of known science and or change the end result of the calculations, or can we go with the end results of the calculations you have just preformed ? Thanks again Terry for your hard work, Jeff.

Thanks for the question Jeff. I came up with some rough calculations to trry and answer your question.
First, The number I came up with is for all of the Water in the cell to be decomposed. I don't think that happens every time there is an intake stroke on the piston. I think it takes time to decompose this Volume of Water just like in a regular HHO cell that produces say 3.0 Liters a minute.
The Piston in the Volkswagon 1500 engine has a displacement of 1,493 cc (91.1 cu in) Volume. This is for all four of the Pistons and is running at appx 2000 revs per minute going down the road at 60 mph.
ref: http://en.wikipedia.org/wiki/Volkswagen_air-cooled_engine
That means that each piston has a displacement of 373.25 cc. Sinse it is a four cycle engine, intake only happens at 500 cycles per minute. That yeilds a Total intake of Gas Volume for each Piston of 186625 cc  per minute or 3110.41 cc per second.
Also, the Ratio of HHO gas to the non explosive Air we mix with the HHO just has to be slightly above the Lower Explosive Level or LEL of the Hydrogen per Volume. Hydrogen LEL and UEL at normal temperature and pressure is given as:
Oxyhydrogen can burn when it is between about 4% and 94% hydrogen by volume.
ref: http://hhofuel.wordpress.com/glossary/
So lets say you kept the mixture at 10% Hydrogen by Volume to Air then only 10% of the Piston Volume would need to filled on each intake cycle. The HHO Gas is 2/3 Hydrogen by Volume so contains 2.14655675 Liters of Hydrogen Gas in the water path of the cell.
H2 consumption is appx 10% of the piston Volume of (186625 cc per minute), which = 18662.5 cc per minute. That is 18.6625 liters per minute.  
If we kept the mixture at 5% Hydrogen by Volume, just above the 4% LEL for Hydrogen in air then the H2 consumption would be 9.33125 liters per minute.
Sinse 9.33125 / 2.14655675 = 4.34707. The cell would need to decompose its contents appx 4.4 times a minute or a little over appx 15 seconds for each Volume of 2.379091232 cubic cm of water for each piston. or 9.5164928 cc for all four pistons.
These calculatios are very rough estimates but consuming 9.5cc of water 4.4 times a minute is appx two times a normal consumption rate of a modern Gasoline car.
9.5164928 cc * 4.34707 = 41.36830448 cc per minute.
41.36830448 = 0.0413683044  Liters per minute. That is 2.48209826 Liters per hour water consumption. or for every 60 miles at 60 mph.
Miles per Gallon translates at this consumption rate to:
Sinse 1 liter = 0.264172052 US gallons, Then  2.482098264  * .6336 = 1.572657 Gallons every 60 miles or  38.15199372 miles per Gallon of water.
22 Gallons of water would give you 839.348 miles at this rate.
Again these calculations are very rough estimates but these estimates show that the consumption is about half of the rate of a very good modern Gasoline engine. I remember from Stan's videos,  he said, Water had about two times the energy that Gasoline had. So these calculations even thopugh they are rough compare with what Stan had said about the energy contained in water.
Another thing - Sinse we can reuse the water again if we collect the waste product (Water) at the tail pipe, consumption can be very low indeed.
Remember the only thing we are doing is breaking the HO bonds to use the HHO gas then reuniteing the HO bonds in the explosion of the HHO in the piston.
The end product is what we started out with. The perfect fuel contained in a non explosive form untill we need it.

Terry Dixon
 

Jeff's Question :
Thanks Terry, now another question, these were all standard measurements of known science for the calculations preformed, right ? "kilocalorie (kcal): the amount of heat needed to warm 1 liter of water 1 degree Celsius" . The reason I ask is that Stan claimed he could drive his dune buggy from coast to coast on 22 gallons of water, knowing this would this change standard measurements of known science and or change the end result of the calculations, or can we go with the end results of the calculations you have just preformed ? Thanks again Terry for your hard work, Jeff.

Thanks for the question Jeff. I came up with some rough calculations to trry and answer your question.
First, The number I came up with is for the Total Volume of the Water in the cell path. I don't think that Total Volume is decomposed every time there is an intake stroke on the piston. I think it takes time to decompose this Volume of Water just like in a regular HHO cell that produces say 3.0 Liters a minute.
The Piston in the Volkswagon 1500 engine has a displacement of 1,493 cc (91.1 cu in) Volume. This is for all four of the Pistons and is running at appx 2000 revs per minute going down the road at 60 mph.
ref: http://en.wikipedia.org/wiki/Volkswagen_air-cooled_engine
That means that each piston has a displacement of 373.25 cc. Sinse it is a four cycle engine, intake only happens at 500 cycles per minute. That yeilds a Total intake of Gas Volume for each Piston of 186625 cc  per minute or 3110.41 cc per second.
Also, the Ratio of HHO gas to the non explosive Air we mix with the HHO just has to be slightly above the Lower Explosive Level or LEL of the Hydrogen per Volume. Hydrogen LEL and UEL at normal temperature and pressure is given as:
Oxyhydrogen can burn when it is between about 4% and 94% hydrogen by volume.
ref: http://hhofuel.wordpress.com/glossary/
So lets say you kept the mixture at 10% Hydrogen by Volume to Air then only 10% of the Piston Volume would need to filled on each intake cycle. The HHO Gas is 2/3 Hydrogen by Volume so contains 2.14655675 Liters of Hydrogen Gas in the water path of the cell.
H2 consumption is appx 10% of the piston Volume of (186625 cc per minute), which = 18662.5 cc per minute. That is 18.6625 liters per minute.  
If we kept the mixture at 5% Hydrogen by Volume, just above the 4% LEL for Hydrogen in air then the H2 consumption would be 9.33125 liters per minute.
Sinse 9.33125 / 2.14655675 = 4.34707. The cell would need to decompose its contents appx 4.4 times a minute or a little over appx 15 seconds for each Volume of 2.379091232 cubic cm of water for each piston. or 9.5164928 cc for all four pistons.
These calculatios are very rough estimates but consuming 9.5cc of water 4.4 times a minute is appx two times a normal consumption rate of a modern Gasoline car.
9.5164928 cc * 4.34707 = 41.36830448 cc per minute.
41.36830448 = 0.0413683044  Liters per minute. That is 2.48209826 Liters per hour water consumption. or for every 60 miles at 60 mph.
Miles per Gallon translates at this consumption rate to:
Sinse 1 liter = 0.264172052 US gallons, Then  2.482098264  * .6336 = 1.572657 Gallons every 60 miles or  38.15199372 miles per Gallon of water.
22 Gallons of water would give you 839.348 miles at this rate.
Again these calculations are very rough estimates but these estimates show that the consumption is about half of the rate of a very good modern Gasoline engine. I remember from Stan's videos,  he said, Water had about two times the energy that Gasoline had. So these calculations even though they are rough compare with what Stan had said about the energy contained in water.
Another thing - Sinse we can reuse the water again if we collect the waste product (Water) at the tail pipe, consumption can be very low indeed.
Remember the only thing we are doing is breaking the HO bonds to use the HHO gas then reuniteing the HO bonds in the explosion of the HHO in the piston.
The end product is what we started out with. The perfect fuel contained in a non explosive form untill we need it.

Terry Dixon
 

Jeff's Question :
Thanks Terry, now another question, these were all standard measurements of known science for the calculations preformed, right ? "kilocalorie (kcal): the amount of heat needed to warm 1 liter of water 1 degree Celsius" . The reason I ask is that Stan claimed he could drive his dune buggy from coast to coast on 22 gallons of water, knowing this would this change standard measurements of known science and or change the end result of the calculations, or can we go with the end results of the calculations you have just preformed ? Thanks again Terry for your hard work, Jeff.

Thanks for the question Jeff. I came up with some rough calculations to try and answer your question.
First, The number I came up with is for the Total Volume of the Water in the cell path. I don't think that Total Volume is decomposed every time there is an intake stroke on the piston. I think it takes time to decompose this Volume of Water just like in a regular HHO cell that produces say 3.0 Liters a minute.
The Piston in the Volkswagon 1500 engine has a displacement of 1,493 cc (91.1 cu in) Volume. This is for all four of the Pistons and is running at appx 2000 revs per minute going down the road at 60 mph.
ref: http://en.wikipedia.org/wiki/Volkswagen_air-cooled_engine
That means that each piston has a displacement of 373.25 cc. Sinse it is a four cycle engine, intake only happens at 500 cycles per minute. That yeilds a Total intake of Gas Volume for each Piston of 186625 cc  per minute or 3110.41 cc per second.
Also, the Ratio of HHO gas to the non explosive Air we mix with the HHO just has to be slightly above the Lower Explosive Level or LEL of the Hydrogen per Volume. Hydrogen LEL and UEL at normal temperature and pressure is given as:
Oxyhydrogen can burn when it is between about 4% and 94% hydrogen by volume.
ref: http://hhofuel.wordpress.com/glossary/
So lets say you kept the mixture at 10% Hydrogen by Volume to Air then only 10% of the Piston Volume would need to filled on each intake cycle. The HHO Gas is 2/3 Hydrogen by Volume so contains 2.14655675 Liters of Hydrogen Gas in the water path of the cell.
H2 consumption is appx 10% of the piston Volume of (186625 cc per minute), which = 18662.5 cc per minute. That is 18.6625 liters per minute.  
If we kept the mixture at 5% Hydrogen by Volume, just above the 4% LEL for Hydrogen in air then the H2 consumption would be 9.33125 liters per minute.
Sinse 9.33125 / 2.14655675 = 4.34707. The cell would need to decompose its contents appx 4.4 times a minute or a little over appx 15 seconds for each Volume of 2.379091232 cubic cm of water for each piston. or 9.5164928 cc for all four pistons.
These calculatios are very rough estimates but consuming 9.5cc of water 4.4 times a minute is appx two times a normal consumption rate of a modern Gasoline car.
9.5164928 cc * 4.34707 = 41.36830448 cc per minute.
41.36830448 = 0.0413683044  Liters per minute. That is 2.48209826 Liters per hour water consumption. or for every 60 miles at 60 mph.
Miles per Gallon translates at this consumption rate to:
Sinse 1 liter = 0.264172052 US gallons, Then  2.482098264  * .6336 = 1.572657 Gallons every 60 miles or  38.15199372 miles per Gallon of water.
22 Gallons of water would give you 839.348 miles at this rate.
Again these calculations are very rough estimates but these estimates show that the consumption is about half of the rate of a very good modern Gasoline engine. I remember from Stan's videos,  he said, Water had about two times the energy that Gasoline had. So these calculations even though they are rough compare with what Stan had said about the energy contained in water.
Another thing - Sinse we can reuse the water again if we collect the waste product (Water) at the tail pipe, consumption can be very low indeed.
Remember the only thing we are doing is breaking the HO bonds to use the HHO gas then reuniteing the HO bonds in the explosion of the HHO in the piston.
The end product is what we started out with. The perfect fuel contained in a non explosive form untill we need it.

Terry Dixon
 

Jeff's Question :
Thanks Terry, now another question, these were all standard measurements of known science for the calculations preformed, right ? "kilocalorie (kcal): the amount of heat needed to warm 1 liter of water 1 degree Celsius" . The reason I ask is that Stan claimed he could drive his dune buggy from coast to coast on 22 gallons of water, knowing this would this change standard measurements of known science and or change the end result of the calculations, or can we go with the end results of the calculations you have just preformed ? Thanks again Terry for your hard work, Jeff.

Thanks for the question Jeff. I came up with some rough calculations to try and answer your question.
First, The number I came up with is for the Total Volume of the Water in the cell path. I don't think that Total Volume is decomposed every time there is an intake stroke on the piston. I think it takes time to decompose this Volume of Water just like in a regular HHO cell that produces say 3.0 Liters a minute.
The Piston in the Volkswagon 1500 engine has a displacement of 1,493 cc (91.1 cu in) Volume. This is for all four of the Pistons and is running at appx 2000 revs per minute going down the road at 60 mph.
ref: http://en.wikipedia.org/wiki/Volkswagen_air-cooled_engine
That means that each piston has a displacement of 373.25 cc. Sinse it is a four cycle engine, intake only happens at 500 cycles per minute. That yeilds a Total intake of Gas Volume for each Piston of 186625 cc  per minute or 3110.41 cc per second.
Also, the Ratio of HHO gas to the non explosive Air we mix with the HHO just has to be slightly above the Lower Explosive Level or LEL of the Hydrogen per Volume. Hydrogen LEL and UEL at normal temperature and pressure is given as:
Oxyhydrogen can burn when it is between about 4% and 94% hydrogen by volume.
ref: http://hhofuel.wordpress.com/glossary/
So lets say you kept the mixture at 10% Hydrogen by Volume to Air then only 10% of the Piston Volume would need to filled on each intake cycle with HHO gas. The HHO Gas is 2/3 Hydrogen by Volume so contains 2.14655675 Liters of Hydrogen Gas in the water path of the cell in the form of water.
H2 consumption is appx 10% of the piston Volume of (186625 cc per minute), which = 18662.5 cc per minute. That is 18.6625 liters per minute of H2.  
If we kept the mixture at 5% Hydrogen by Volume which is just above the 4% LEL for Hydrogen in air then the H2 consumption would be 9.33125 liters per minute.
Sinse 9.33125 / 2.14655675 = 4.34707. The cell would need to decompose its contents appx 4.4 times a minute or a little over appx 15 seconds for each Volume of 2.379091232 cubic cm of water for each piston. or 9.5164928 cc for all four pistons.
These calculatios are very rough estimates but consuming 9.5cc of water 4.4 times a minute is appx two times a normal consumption rate of a modern Gasoline car.
9.5164928 cc * 4.34707 = 41.36830448 cc per minute.
41.36830448 = 0.0413683044  Liters per minute. That is 2.48209826 Liters per hour water consumption. or for every 60 miles at 60 mph.
Miles per Gallon translates at this consumption rate to:
Sinse 1 liter = 0.264172052 US gallons, Then  2.482098264  * .6336 = 1.572657 Gallons every 60 miles or  38.15199372 miles per Gallon of water.
22 Gallons of water would give you 839.348 miles at this rate.
Again these calculations are very rough estimates but these estimates show that the consumption is about half of the rate of a very good modern Gasoline engine. I remember from Stan's videos,  he said, Water had about two times the energy that Gasoline had. So these calculations even though they are rough compare with what Stan had said about the energy contained in water.
Another thing - Sinse we can reuse the water again if we collect the waste product (Water) at the tail pipe, consumption can be very low indeed.
Remember the only thing we are doing is breaking the HO bonds to use the HHO gas then reuniteing the HO bonds in the explosion of the HHO in the piston.
The end product is what we started out with. The perfect fuel contained in a non explosive form untill we need it.

Terry Dixon
 

Jeff's Question :
Thanks Terry, now another question, these were all standard measurements of known science for the calculations preformed, right ? "kilocalorie (kcal): the amount of heat needed to warm 1 liter of water 1 degree Celsius" . The reason I ask is that Stan claimed he could drive his dune buggy from coast to coast on 22 gallons of water, knowing this would this change standard measurements of known science and or change the end result of the calculations, or can we go with the end results of the calculations you have just preformed ? Thanks again Terry for your hard work, Jeff.

Thanks for the question Jeff. I came up with some rough calculations to try and answer your question.
First, The number I came up with is for the Total Volume of the Water in the cell path. I don't think that Total Volume is decomposed every time there is an intake stroke on the piston. I think it takes time to decompose this Volume of Water just like in a regular HHO cell that produces say 3.0 Liters a minute.
The Piston in the Volkswagon 1500 engine has a displacement of 1,493 cc (91.1 cu in) Volume. This is for all four of the Pistons and is running at appx 2000 revs per minute going down the road at 60 mph.
ref: http://en.wikipedia.org/wiki/Volkswagen_air-cooled_engine
That means that each piston has a displacement of 373.25 cc. Sinse it is a four cycle engine, intake only happens at 500 cycles per minute. That yeilds a Total intake of Gas Volume for each Piston of 186625 cc  per minute or 3110.41 cc per second.
Also, the Ratio of HHO gas to the non explosive Air we mix with the HHO just has to be slightly above the Lower Explosive Level or LEL of the Hydrogen per Volume. Hydrogen LEL and UEL at normal temperature and pressure is given as:
Oxyhydrogen can burn when it is between about 4% and 94% hydrogen by volume.
ref: http://hhofuel.wordpress.com/glossary/
So lets say you kept the mixture at 10% Hydrogen by Volume to Air then only 10% of the Piston Volume would need to filled on each intake cycle with HHO gas. The HHO Gas is 2/3 Hydrogen by Volume so contains 2.14655675 Liters of Hydrogen Gas in the water path of the cell in the form of water.
H2 consumption is appx 10% of the piston Volume of (186625 cc per minute), which = 18662.5 cc per minute. That is 18.6625 liters per minute of H2.  
If we kept the mixture at 5% Hydrogen by Volume which is just above the 4% LEL for Hydrogen in air then the H2 consumption would be 9.33125 liters per minute.
Sinse 9.33125 / 2.14655675 = 4.34707. The cell would need to decompose its contents appx 4.4 times a minute or a little over appx 15 seconds for each Volume of 2.379091232 cubic cm of water for each piston. or 9.5164928 cc for all four injectors on all four pistons.
These calculatios are very rough estimates but consuming 9.5cc of water 4.4 times a minute is appx two times a normal consumption rate of a modern Gasoline car.
9.5164928 cc * 4.34707 = 41.36830448 cc per minute.
41.36830448 = 0.0413683044  Liters per minute. That is 2.48209826 Liters per hour water consumption. or for every 60 miles at 60 mph.
Miles per Gallon translates at this consumption rate to:
Sinse 1 liter = 0.264172052 US gallons, Then  2.482098264  * .6336 = 1.572657 Gallons every 60 miles or  38.15199372 miles per Gallon of water.
22 Gallons of water would give you 839.348 miles at this rate.
Again these calculations are very rough estimates but these estimates show that the consumption is about half of the rate of a very good modern Gasoline engine. I remember from Stan's videos,  he said, Water had about two times the energy that Gasoline had. So these calculations even though they are rough compare with what Stan had said about the energy contained in water.
Another thing - Sinse we can reuse the water again if we collect the waste product (Water) at the tail pipe, consumption can be very low indeed.
Remember the only thing we are doing is breaking the HO bonds to use the HHO gas then reuniteing the HO bonds in the explosion of the HHO in the piston.
The end product is what we started out with. The perfect fuel contained in a non explosive form untill we need it.

Terry Dixon
 

Jeff's Question :
Thanks Terry, now another question, these were all standard measurements of known science for the calculations preformed, right ? "kilocalorie (kcal): the amount of heat needed to warm 1 liter of water 1 degree Celsius" . The reason I ask is that Stan claimed he could drive his dune buggy from coast to coast on 22 gallons of water, knowing this would this change standard measurements of known science and or change the end result of the calculations, or can we go with the end results of the calculations you have just preformed ? Thanks again Terry for your hard work, Jeff.

Thanks for the question Jeff. I came up with some rough calculations to try and answer your question.
First, The number I came up with is for the Total Volume of the Water in the cell path. I don't think that Total Volume is decomposed every time there is an intake stroke on the piston. I think it takes time to decompose this Volume of Water just like in a regular HHO cell that produces say 3.0 Liters a minute.
The Piston in the Volkswagon 1500 engine has a displacement of 1,493 cc (91.1 cu in) Volume. This is for all four of the Pistons and is running at appx 2000 revs per minute going down the road at 60 mph.
ref: http://en.wikipedia.org/wiki/Volkswagen_air-cooled_engine
That means that each piston has a displacement of 373.25 cc. Sinse it is a four cycle engine, intake only happens at 500 cycles per minute. That yeilds a Total intake of Gas Volume for each Piston of 186625 cc  per minute or 3110.41 cc per second.
Also, the Ratio of HHO gas to the non explosive Air we mix with the HHO just has to be slightly above the Lower Explosive Level or LEL of the Hydrogen per Volume. Hydrogen LEL and UEL at normal temperature and pressure is given as:
Oxyhydrogen can burn when it is between about 4% and 94% hydrogen by volume.
ref: http://hhofuel.wordpress.com/glossary/
So lets say you kept the mixture at 10% Hydrogen by Volume to Air then only 10% of the Piston Volume would need to filled on each intake cycle with HHO gas. The HHO Gas is 2/3 Hydrogen by Volume so contains 2.14655675 Liters of Hydrogen Gas in the water path of the cell in the form of water.
H2 consumption is appx 10% of the piston Volume of (186625 cc per minute), which = 18662.5 cc per minute. That is 18.6625 liters per minute of H2.  
If we kept the mixture at 5% Hydrogen by Volume which is just above the 4% LEL for Hydrogen in air then the H2 consumption would be 9.33125 liters per minute.
Sinse 9.33125 / 2.14655675 = 4.34707. The cell would need to decompose its contents appx 4.4 times a minute or a little over appx 15 seconds for each Volume of 2.379091232 cubic cm of water for each piston. or 9.5164928 cc for all four injectors on all four pistons.
These calculatios are very rough estimates but consuming 9.5cc of water 4.4 times a minute is appx two times a normal consumption rate of a modern Gasoline car.
9.5164928 cc * 4.34707 = 41.36830448 cc per minute.
41.36830448 = 0.0413683044  Liters per minute. That is 2.48209826 Liters per hour water consumption. or for every 60 miles at 60 mph.
Miles per Gallon translates at this consumption rate to:
Sinse 1 liter = 0.264172052 US gallons, Then  2.482098264  * .6336 = 1.572657 Gallons every 60 miles or  38.15199372 miles per Gallon of water.
22 Gallons of water would give you 839.348 miles at this rate.
Again these calculations are very rough estimates but these estimates show that the consumption is about half of the rate of a very good modern Gasoline engine. I remember from Stan's videos,  he said, Water had about two times the energy that Gasoline had. So these calculations even though they are rough compare with what Stan had said about the energy contained in water.
Another thing to consider is:
Sinse we can reuse the water again if we collect the waste product (Water) at the tail pipe, consumption could be very low indeed.
Remember the only thing we are doing is breaking the HO bonds to use the HHO gas then reuniteing the HO bonds in the explosion of the HHO in the piston.
The end product is what we started out with. The perfect fuel contained in a non explosive form untill we need it.

Terry Dixon
 

Jeff's Question :
Thanks Terry, now another question, these were all standard measurements of known science for the calculations preformed, right ? "kilocalorie (kcal): the amount of heat needed to warm 1 liter of water 1 degree Celsius" . The reason I ask is that Stan claimed he could drive his dune buggy from coast to coast on 22 gallons of water, knowing this would this change standard measurements of known science and or change the end result of the calculations, or can we go with the end results of the calculations you have just preformed ? Thanks again Terry for your hard work, Jeff.

Thanks for the question Jeff. I came up with some rough calculations to try and answer your question.
First, The number I came up with is for the Total Volume of the Water in the cell path. I don't think that Total Volume is decomposed every time there is an intake stroke on the piston. I think it takes time to decompose this Volume of Water just like in a regular HHO cell that produces say 3.0 Liters a minute.
The Piston in the Volkswagon 1500 engine has a displacement of 1,493 cc (91.1 cu in) Volume. This is for all four of the Pistons and is running at appx 2000 revs per minute going down the road at 60 mph.
ref: http://en.wikipedia.org/wiki/Volkswagen_air-cooled_engine
That means that each piston has a displacement of 373.25 cc. Sinse it is a four cycle engine, intake only happens at 500 cycles per minute. That yeilds a Total intake of Gas Volume for each Piston of 186625 cc  per minute or 3110.41 cc per second.
Also, the Ratio of HHO gas to the non explosive Air we mix with the HHO just has to be slightly above the Lower Explosive Level or LEL of the Hydrogen per Volume. Hydrogen LEL and UEL at normal temperature and pressure is given as:
Oxyhydrogen can burn when it is between about 4% and 94% hydrogen by volume.
ref: http://hhofuel.wordpress.com/glossary/
So lets say you kept the mixture at 10% Hydrogen by Volume to Air then only 10% of the Piston Volume would need to filled on each intake cycle with HHO gas. The HHO Gas is 2/3 Hydrogen by Volume so contains 2.14655675 Liters of Hydrogen Gas in the water path of the cell in the form of water.
H2 consumption is appx 10% of the piston Volume of (186625 cc per minute), which = 18662.5 cc per minute. That is 18.6625 liters per minute of H2.  
If we kept the mixture at 5% Hydrogen by Volume which is just above the 4% LEL for Hydrogen in air then the H2 consumption would be 9.33125 liters per minute.
Sinse 9.33125 / 2.14655675 = 4.34707. The cell would need to decompose its contents appx 4.4 times a minute or a little over appx 15 seconds for each Volume of 2.379091232 cubic cm of water for each piston. or 9.5164928 cc for all four injectors on all four pistons.
These calculatios are very rough estimates but consuming 9.5cc of water 4.4 times a minute is appx half of the normal consumption rate of Gasoline in a modern Gasoline engine.
9.5164928 cc * 4.34707 = 41.36830448 cc per minute.
41.36830448 = 0.0413683044  Liters per minute. That is 2.48209826 Liters per hour water consumption. or for every 60 miles at 60 mph.
Miles per Gallon translates at this consumption rate to:
Sinse 1 liter = 0.264172052 US gallons, Then  2.482098264  * .6336 = 1.572657 Gallons every 60 miles or  38.15199372 miles per Gallon of water.
22 Gallons of water would give you 839.348 miles at this rate.
Again these calculations are very rough estimates but these estimates show that the consumption is about half of the rate of a very good modern Gasoline engine. I remember from Stan's videos,  he said, Water had about two times the energy that Gasoline had. So these calculations even though they are rough compare with what Stan had said about the energy contained in water.
Another thing to consider is:
Sinse we can reuse the water again if we collect the waste product (Water) at the tail pipe, consumption could be very low indeed.
Remember the only thing we are doing is breaking the HO bonds to use the HHO gas then reuniteing the HO bonds in the explosion of the HHO in the piston.
The end product is what we started out with. The perfect fuel contained in a non explosive form untill we need it.

Terry Dixon
 

Jeff's Question :
Thanks Terry, now another question, these were all standard measurements of known science for the calculations preformed, right ? "kilocalorie (kcal): the amount of heat needed to warm 1 liter of water 1 degree Celsius" . The reason I ask is that Stan claimed he could drive his dune buggy from coast to coast on 22 gallons of water, knowing this would this change standard measurements of known science and or change the end result of the calculations, or can we go with the end results of the calculations you have just preformed ? Thanks again Terry for your hard work, Jeff.

Thanks for the question Jeff. I came up with some rough calculations to try and answer your question.
First, The number I came up with is for the Total Volume of the Water in the cell path. I don't think that Total Volume is decomposed every time there is an intake stroke on the piston. I think it takes time to decompose this Volume of Water just like in a regular HHO cell that produces say 3.0 Liters a minute.
The Piston in the Volkswagon 1500 engine has a displacement of 1,493 cc (91.1 cu in) Volume. This is for all four of the Pistons and is running at appx 2000 revs per minute going down the road at 60 mph.
ref: http://en.wikipedia.org/wiki/Volkswagen_air-cooled_engine
That means that each piston has a displacement of 373.25 cc. Sinse it is a four cycle engine, intake only happens at 500 cycles per minute. That yeilds a Total intake of Gas Volume for each Piston of 186625 cc  per minute or 3110.41 cc per second.
Also, the Ratio of HHO gas to the non explosive Air we mix with the HHO just has to be slightly above the Lower Explosive Level or LEL of the Hydrogen per Volume. Hydrogen LEL and UEL at normal temperature and pressure is given as:
Oxyhydrogen can burn when it is between about 4% and 94% hydrogen by volume.
ref: http://hhofuel.wordpress.com/glossary/
So lets say you kept the mixture at 10% Hydrogen by Volume to Air then only 10% of the Piston Volume would need to filled on each intake cycle with HHO gas. The HHO Gas is 2/3 Hydrogen by Volume so contains 2.14655675 Liters of Hydrogen Gas in the water path of the cell in the form of water.
H2 consumption is appx 10% of the piston Volume of (186625 cc per minute), which = 18662.5 cc per minute. That is 18.6625 liters per minute of H2.  
If we kept the mixture at 5% Hydrogen by Volume which is just above the 4% LEL for Hydrogen in air then the H2 consumption would be 9.33125 liters per minute.
Sinse 9.33125 / 2.14655675 = 4.34707. The cell would need to decompose its contents appx 4.4 times a minute or a little over appx 15 seconds for each Volume of 2.379091232 cubic cm of water for each piston. or 9.5164928 cc for all four injectors on all four pistons.
These calculatios are very rough estimates but consuming 9.5cc of water 4.4 times a minute is appx half of the normal consumption rate of Gasoline in a modern Gasoline engine.
9.5164928 cc * 4.34707 = 41.36830448 cc per minute.
41.36830448 = 0.0413683044  Liters per minute. That is 2.48209826 Liters per hour water consumption. or for every 60 miles at 60 mph.
Miles per Gallon translates at this consumption rate to:
Sinse 1 liter = 0.264172052 US gallons, Then  2.482098264  * .6336 = 1.572657 Gallons every 60 miles or  38.15199372 miles per Gallon of water.
22 Gallons of water would give you 839.348 miles at this rate.
Again these calculations are very rough estimates but these estimates show that the consumption is about half of the rate of what the old VW engine gets in Gas Mileage.
ref: http://answers.yahoo.com/question/index?qid=20061006153749AAWo13L
 I remember from Stan's videos,  he said, Water had about two times the energy that Gasoline had. So these calculations even though they are rough compare with what Stan had said about the energy contained in water.
Another thing to consider is:
Sinse we can reuse the water again if we collect the waste product (Water) at the tail pipe, consumption could be very low indeed.
Remember the only thing we are doing is breaking the HO bonds to use the HHO gas then reuniteing the HO bonds in the explosion of the HHO in the piston.
The end product is what we started out with. The perfect fuel contained in a non explosive form untill we need it.

Terry Dixon
 

Jeff's Question :
Thanks Terry, now another question, these were all standard measurements of known science for the calculations preformed, right ? "kilocalorie (kcal): the amount of heat needed to warm 1 liter of water 1 degree Celsius" . The reason I ask is that Stan claimed he could drive his dune buggy from coast to coast on 22 gallons of water, knowing this would this change standard measurements of known science and or change the end result of the calculations, or can we go with the end results of the calculations you have just preformed ? Thanks again Terry for your hard work, Jeff.

Thanks for the question Jeff. I came up with some rough calculations to try and answer your question.
First, The number I came up with is for the Total Volume of the Water in the cell path. I don't think that Total Volume is decomposed every time there is an intake stroke on the piston. I think it takes time to decompose this Volume of Water just like in a regular HHO cell that produces say 3.0 Liters a minute.
The Piston in the Volkswagon 1500 engine has a displacement of 1,493 cc (91.1 cu in) Volume. This is for all four of the Pistons and is running at appx 2000 revs per minute going down the road at 60 mph.
ref: http://en.wikipedia.org/wiki/Volkswagen_air-cooled_engine
That means that each piston has a displacement of 373.25 cc. Sinse it is a four cycle engine, intake only happens at 500 cycles per minute. That yeilds a Total intake of Gas Volume for each Piston of 186625 cc  per minute or 3110.41 cc per second.
Also, the Ratio of HHO gas to the non explosive Air we mix with the HHO just has to be slightly above the Lower Explosive Level or LEL of the Hydrogen per Volume. Hydrogen LEL and UEL at normal temperature and pressure is given as:
Oxyhydrogen can burn when it is between about 4% and 94% hydrogen by volume.
ref: http://hhofuel.wordpress.com/glossary/
So lets say you kept the mixture at 10% Hydrogen by Volume to Air then only 10% of the Piston Volume would need to filled on each intake cycle with HHO gas. The HHO Gas is 2/3 Hydrogen by Volume so contains 2.14655675 Liters of Hydrogen Gas in the water path of the cell in the form of water.
H2 consumption is appx 10% of the piston Volume of (186625 cc per minute), which = 18662.5 cc per minute. That is 18.6625 liters per minute of H2.  
If we kept the mixture at 5% Hydrogen by Volume which is just above the 4% LEL for Hydrogen in air then the H2 consumption would be 9.33125 liters per minute.
Sinse 9.33125 / 2.14655675 = 4.34707. The cell would need to decompose its contents appx 4.4 times a minute or a little over appx 15 seconds for each Volume of 2.379091232 cubic cm of water for each piston. or 9.5164928 cc for all four injectors on all four pistons.
These calculatios are very rough estimates but consuming 9.5cc of water 4.4 times a minute is appx half of the normal consumption rate of Gasoline in a modern Gasoline engine.
9.5164928 cc * 4.34707 = 41.36830448 cc per minute.
41.36830448 = 0.0413683044  Liters per minute. That is 2.48209826 Liters per hour water consumption. or for every 60 miles at 60 mph.
Miles per Gallon translates at this consumption rate to:
Sinse 1 liter = 0.264172052 US gallons, Then  2.482098264  * .6336 = 1.572657 Gallons every 60 miles or  38.15199372 miles per Gallon of water.
22 Gallons of water would give you 839.348 miles at this rate.
Again these calculations are very rough estimates but these estimates show that the MPG using water is about Two Times the MPG the old VW engine gets using Gasoline.
ref: http://answers.yahoo.com/question/index?qid=20061006153749AAWo13L
 I remember from Stan's videos,  he said, Water had about two times the energy that Gasoline had. So these calculations even though they are rough compare with what Stan had said about the energy contained in water.
Another thing to consider is:
Sinse we can reuse the water again if we collect the waste product (Water) at the tail pipe, consumption could be very low indeed.
Remember the only thing we are doing is breaking the HO bonds to use the HHO gas then reuniteing the HO bonds in the explosion of the HHO in the piston.
The end product is what we started out with. The perfect fuel contained in a non explosive form untill we need it.

Terry Dixon
 

Jeff's Question :
Thanks Terry, now another question, these were all standard measurements of known science for the calculations preformed, right ? "kilocalorie (kcal): the amount of heat needed to warm 1 liter of water 1 degree Celsius" . The reason I ask is that Stan claimed he could drive his dune buggy from coast to coast on 22 gallons of water, knowing this would this change standard measurements of known science and or change the end result of the calculations, or can we go with the end results of the calculations you have just preformed ? Thanks again Terry for your hard work, Jeff.

Thanks for the question Jeff. I came up with some rough calculations to try and answer your question.
First, I don't know if the injector was the only source of HHO gas that Stan used on his final design of the Dune Buggy using the VW engine. I remember seeing an HHO cell along with the injectors. Lets assume the design used only the injectors for the source of HHO fuel.
Second, The number I came up with is for the Total Volume of the Water in the cell path. I don't think that Total Volume is decomposed every time there is an intake stroke on the piston. I think it takes time to decompose this Volume of Water just like in a regular HHO cell that produces say 3.0 Liters a minute.
The Piston in the Volkswagon 1500 engine has a displacement of 1,493 cc (91.1 cu in) Volume. This is for all four of the Pistons and is running at appx 2000 revs per minute going down the road at 60 mph.
ref: http://en.wikipedia.org/wiki/Volkswagen_air-cooled_engine
That means that each piston has a displacement of 373.25 cc. Sinse it is a four cycle engine, intake only happens at 500 cycles per minute. That yeilds a Total intake of Gas Volume for each Piston of 186625 cc  per minute or 3110.41 cc per second.
Also, the Ratio of HHO gas to the non explosive Air we mix with the HHO just has to be slightly above the Lower Explosive Level or LEL of the Hydrogen per Volume. Hydrogen LEL and UEL at normal temperature and pressure is given as:
Oxyhydrogen can burn when it is between about 4% and 94% hydrogen by volume.
ref: http://hhofuel.wordpress.com/glossary/
So lets say you kept the mixture at 10% Hydrogen by Volume to Air then only 10% of the Piston Volume would need to filled on each intake cycle with HHO gas. The HHO Gas is 2/3 Hydrogen by Volume so contains 2.14655675 Liters of Hydrogen Gas in the water path of the cell in the form of water.
H2 consumption is appx 10% of the piston Volume of (186625 cc per minute), which = 18662.5 cc per minute. That is 18.6625 liters per minute of H2.  
If we kept the mixture at 5% Hydrogen by Volume which is just above the 4% LEL for Hydrogen in air then the H2 consumption would be 9.33125 liters per minute.
Sinse 9.33125 / 2.14655675 = 4.34707. The cell would need to decompose its contents appx 4.4 times a minute or a little over appx 15 seconds for each Volume of 2.379091232 cubic cm of water for each piston. or 9.5164928 cc for all four injectors on all four pistons.
These calculatios are very rough estimates but consuming 9.5cc of water 4.4 times a minute is appx half of the normal consumption rate of Gasoline in a modern Gasoline engine.
9.5164928 cc * 4.34707 = 41.36830448 cc per minute.
41.36830448 = 0.0413683044  Liters per minute. That is 2.48209826 Liters per hour water consumption. or for every 60 miles at 60 mph.
Miles per Gallon translates at this consumption rate to:
Sinse 1 liter = 0.264172052 US gallons, Then  2.482098264  * .6336 = 1.572657 Gallons every 60 miles or  38.15199372 miles per Gallon of water.
22 Gallons of water would give you 839.348 miles at this rate.
Again these calculations are very rough estimates but these estimates show that the MPG using water is about Two Times the MPG the old VW engine gets using Gasoline.
ref: http://answers.yahoo.com/question/index?qid=20061006153749AAWo13L
 I remember from Stan's videos,  he said, Water had about two times the energy that Gasoline had. So these calculations even though they are rough compare with what Stan had said about the energy contained in water.
Another thing to consider is:
Sinse we can reuse the water again if we collect the waste product (Water) at the tail pipe, consumption could be very low indeed.
Remember the only thing we are doing is breaking the HO bonds to use the HHO gas then reuniteing the HO bonds in the explosion of the HHO in the piston.
The end product is what we started out with. The perfect fuel contained in a non explosive form untill we need it.

Terry Dixon
 

Jeff's Question :
Thanks Terry, now another question, these were all standard measurements of known science for the calculations preformed, right ? "kilocalorie (kcal): the amount of heat needed to warm 1 liter of water 1 degree Celsius" . The reason I ask is that Stan claimed he could drive his dune buggy from coast to coast on 22 gallons of water, knowing this would this change standard measurements of known science and or change the end result of the calculations, or can we go with the end results of the calculations you have just preformed ? Thanks again Terry for your hard work, Jeff.

Thanks for the question Jeff. I came up with some rough calculations to try and answer your question.
First, I don't know if the injector was the only source of HHO gas that Stan used on his final design of the Dune Buggy using the VW engine. I remember seeing an HHO cell along with the injectors. Lets assume the design used only the injectors for the source of HHO fuel. I also don't know which VW engine Stan used. I am going to use the 1500 for my example.
Second, The number I came up with is for the Total Volume of the Water in the cell path. I don't think that Total Volume is decomposed every time there is an intake stroke on the piston. I think it takes time to decompose this Volume of Water just like in a regular HHO cell that produces say 3.0 Liters a minute.
The Piston in the Volkswagon 1500 engine has a displacement of 1,493 cc (91.1 cu in) Volume. This is for all four of the Pistons and is running at appx 2000 revs per minute going down the road at 60 mph.
ref: http://en.wikipedia.org/wiki/Volkswagen_air-cooled_engine
That means that each piston has a displacement of 373.25 cc. Sinse it is a four cycle engine, intake only happens at 500 cycles per minute. That yeilds a Total intake of Gas Volume for each Piston of 186625 cc  per minute or 3110.41 cc per second.
Also, the Ratio of HHO gas to the non explosive Air we mix with the HHO just has to be slightly above the Lower Explosive Level or LEL of the Hydrogen per Volume. Hydrogen LEL and UEL at normal temperature and pressure is given as:
Oxyhydrogen can burn when it is between about 4% and 94% hydrogen by volume.
ref: http://hhofuel.wordpress.com/glossary/
So lets say you kept the mixture at 10% Hydrogen by Volume to Air then only 10% of the Piston Volume would need to filled on each intake cycle with HHO gas. The HHO Gas is 2/3 Hydrogen by Volume so contains 2.14655675 Liters of Hydrogen Gas in the water path of the cell in the form of water.
H2 consumption is appx 10% of the piston Volume of (186625 cc per minute), which = 18662.5 cc per minute. That is 18.6625 liters per minute of H2.  
If we kept the mixture at 5% Hydrogen by Volume which is just above the 4% LEL for Hydrogen in air then the H2 consumption would be 9.33125 liters per minute.
Sinse 9.33125 / 2.14655675 = 4.34707. The cell would need to decompose its contents appx 4.4 times a minute or a little over appx 15 seconds for each Volume of 2.379091232 cubic cm of water for each piston. or 9.5164928 cc for all four injectors on all four pistons.
These calculatios are very rough estimates but consuming 9.5cc of water 4.4 times a minute is appx half of the normal consumption rate of Gasoline in a modern Gasoline engine.
9.5164928 cc * 4.34707 = 41.36830448 cc per minute.
41.36830448 = 0.0413683044  Liters per minute. That is 2.48209826 Liters per hour water consumption. or for every 60 miles at 60 mph.
Miles per Gallon translates at this consumption rate to:
Sinse 1 liter = 0.264172052 US gallons, Then  2.482098264  * .6336 = 1.572657 Gallons every 60 miles or  38.15199372 miles per Gallon of water.
22 Gallons of water would give you 839.348 miles at this rate.
Again these calculations are very rough estimates but these estimates show that the MPG using water is about Two Times the MPG the old VW engine gets using Gasoline.
ref: http://answers.yahoo.com/question/index?qid=20061006153749AAWo13L
 I remember from Stan's videos,  he said, Water had about two times the energy that Gasoline had. So these calculations even though they are rough compare with what Stan had said about the energy contained in water.
Another thing to consider is:
Sinse we can reuse the water again if we collect the waste product (Water) at the tail pipe, consumption could be very low indeed.
Remember the only thing we are doing is breaking the HO bonds to use the HHO gas then reuniteing the HO bonds in the explosion of the HHO in the piston.
The end product is what we started out with. The perfect fuel contained in a non explosive form untill we need it.

Terry Dixon
 

Jeff's Question :
Thanks Terry, now another question, these were all standard measurements of known science for the calculations preformed, right ? "kilocalorie (kcal): the amount of heat needed to warm 1 liter of water 1 degree Celsius" . The reason I ask is that Stan claimed he could drive his dune buggy from coast to coast on 22 gallons of water, knowing this would this change standard measurements of known science and or change the end result of the calculations, or can we go with the end results of the calculations you have just preformed ? Thanks again Terry for your hard work, Jeff.

Thanks for the question Jeff. I came up with some rough calculations to try and answer your question.
First, I don't know if the injector was the only source of HHO gas that Stan used on his final design of the Dune Buggy using the VW engine. I remember seeing an HHO cell along with the injectors. Lets assume the design used only the injectors for the source of HHO fuel. I also don't know which VW engine Stan used. I am going to use the 1500 for my example.
Second, The number I came up with is for the Total Volume of the Water in the cell path. I don't think that Total Volume is decomposed every time there is an intake stroke on the piston. I think it takes time to decompose this Volume of Water just like in a regular HHO cell that produces say 3.0 Liters a minute.
The Volkswagon 1500 engine has a displacement of 1,493 cc (91.1 cu in) Volume. This is for all four of the Pistons and is running at appx 2000 revs per minute going down the road at 60 mph.
ref: http://en.wikipedia.org/wiki/Volkswagen_air-cooled_engine
That means that each piston has a displacement of 373.25 cc. Sinse it is a four cycle engine, intake only happens at 500 cycles per minute. That yeilds a Total intake of Gas Volume for each Piston of 186625 cc  per minute or 3110.41 cc per second.
Also, the Ratio of HHO gas to the non explosive Air we mix with the HHO just has to be slightly above the Lower Explosive Level or LEL of the Hydrogen per Volume. Hydrogen LEL and UEL at normal temperature and pressure is given as:
Oxyhydrogen can burn when it is between about 4% and 94% hydrogen by volume.
ref: http://hhofuel.wordpress.com/glossary/
So lets say you kept the mixture at 10% Hydrogen by Volume to Air then only 10% of the Piston Volume would need to filled on each intake cycle with HHO gas. The HHO Gas is 2/3 Hydrogen by Volume so contains 2.14655675 Liters of Hydrogen Gas in the water path of the cell in the form of water.
H2 consumption is appx 10% of the piston Volume of (186625 cc per minute), which = 18662.5 cc per minute. That is 18.6625 liters per minute of H2.  
If we kept the mixture at 5% Hydrogen by Volume which is just above the 4% LEL for Hydrogen in air then the H2 consumption would be 9.33125 liters per minute.
Sinse 9.33125 / 2.14655675 = 4.34707. The cell would need to decompose its contents appx 4.4 times a minute or a little over appx 15 seconds for each Volume of 2.379091232 cubic cm of water for each piston. or 9.5164928 cc for all four injectors on all four pistons.
These calculatios are very rough estimates but consuming 9.5cc of water 4.4 times a minute is appx half of the normal consumption rate of Gasoline in a modern Gasoline engine.
9.5164928 cc * 4.34707 = 41.36830448 cc per minute.
41.36830448 = 0.0413683044  Liters per minute. That is 2.48209826 Liters per hour water consumption. or for every 60 miles at 60 mph.
Miles per Gallon translates at this consumption rate to:
Sinse 1 liter = 0.264172052 US gallons, Then  2.482098264  * .6336 = 1.572657 Gallons every 60 miles or  38.15199372 miles per Gallon of water.
22 Gallons of water would give you 839.348 miles at this rate.
Again these calculations are very rough estimates but these estimates show that the MPG using water is about Two Times the MPG the old VW engine gets using Gasoline.
ref: http://answers.yahoo.com/question/index?qid=20061006153749AAWo13L
 I remember from Stan's videos,  he said, Water had about two times the energy that Gasoline had. So these calculations even though they are rough compare with what Stan had said about the energy contained in water.
Another thing to consider is:
Sinse we can reuse the water again if we collect the waste product (Water) at the tail pipe, consumption could be very low indeed.
Remember the only thing we are doing is breaking the HO bonds to use the HHO gas then reuniteing the HO bonds in the explosion of the HHO in the piston.
The end product is what we started out with. The perfect fuel contained in a non explosive form untill we need it.

Terry Dixon
 

Jeff's Question :
Thanks Terry, now another question, these were all standard measurements of known science for the calculations preformed, right ? "kilocalorie (kcal): the amount of heat needed to warm 1 liter of water 1 degree Celsius" . The reason I ask is that Stan claimed he could drive his dune buggy from coast to coast on 22 gallons of water, knowing this would this change standard measurements of known science and or change the end result of the calculations, or can we go with the end results of the calculations you have just preformed ? Thanks again Terry for your hard work, Jeff.

Thanks for the question Jeff. I came up with some rough calculations to try and answer your question.
First, I don't know if the injector was the only source of HHO gas that Stan used on his final design of the Dune Buggy using the VW engine. I remember seeing an HHO cell along with the injectors. Lets assume the design used only the injectors for the source of HHO fuel. I also don't know which VW engine Stan used. I am going to use the 1500 for my example.
Second, The number I came up with is for the Total Volume of the Water in the cell path. I don't think that Total Volume is decomposed every time there is an intake stroke on the piston. I think it takes time to decompose this Volume of Water just like in a regular HHO cell that produces say 3.0 Liters a minute.
The Volkswagon 1500 engine has a displacement of 1,493 cc (91.1 cu in) Volume. This is for all four of the Pistons and is running at appx 2000 revs per minute going down the road at 60 mph.
ref: http://en.wikipedia.org/wiki/Volkswagen_air-cooled_engine
That means that each piston has a displacement of 373.25 cc. Sinse it is a four cycle engine, intake only happens at 500 cycles per minute. That yeilds a Total intake of Gas Volume for each Piston of 186625 cc  per minute or 3110.41 cc per second.
Also, the Ratio of HHO gas to the non explosive Air we mix with the HHO just has to be slightly above the Lower Explosive Level or LEL of the Hydrogen per Volume. Hydrogen LEL and UEL at normal temperature and pressure is given as:
Oxyhydrogen can burn when it is between about 4% and 94% hydrogen by volume.
ref: http://hhofuel.wordpress.com/glossary/
So lets say you kept the mixture at 10% Hydrogen by Volume to Air then only 10% of the Piston Volume would need to filled on each intake cycle with HHO gas. The HHO Gas is 2/3 Hydrogen by Volume so contains 2.14655675 Liters of Hydrogen Gas in the water path of the cell in the form of water.
H2 consumption is appx 10% of the piston Volume of (186625 cc per minute), which = 18662.5 cc per minute. That is 18.6625 liters per minute of H2.  
If we kept the mixture at 5% Hydrogen by Volume which is just above the 4% LEL for Hydrogen in air then the H2 consumption would be 9.33125 liters per minute.
Sinse 9.33125 / 2.14655675 = 4.34707. The cell would need to decompose its contents appx 4.4 times a minute or a little over appx 15 seconds for each Volume of 2.379091232 cubic cm of water for each piston. or 9.5164928 cc for all four injectors on all four pistons.
These calculatios are very rough estimates but consuming 9.5cc of water 4.4 times a minute is appx half of the normal consumption rate of Gasoline in a modern Gasoline engine.
9.5164928 cc * 4.34707 = 41.36830448 cc per minute.
41.36830448 = 0.0413683044  Liters per minute. That is 2.48209826 Liters per hour water consumption. or for every 60 miles at 60 mph.
Miles per Gallon translates at this consumption rate to:
Sinse 1 liter = 0.264172052 US gallons, Then  2.482098264  * .6336 = 1.572657 Gallons every 60 miles or  38.15199372 miles per Gallon of water.
22 Gallons of water would give you 839.348 miles at this rate.
Again these calculations are very rough estimates but these estimates show that the MPG using water is about Two Times the MPG the old VW engine gets using Gasoline.
ref: http://answers.yahoo.com/question/index?qid=20061006153749AAWo13L
 I remember from Stan's videos,  he said, Water had about two times the energy that Gasoline had. So these calculations even though they are rough compare with what Stan had said about the energy contained in water.
Another thing to consider is:
Sinse we can reuse the water again if we collect the waste product (Water) at the tail pipe, consumption could be very low indeed.
Remember the only thing we are doing is breaking the HO bonds to use the HHO gas then reuniteing the HO bonds in the explosion of the HHO in the piston.
The end product is what we started out with. The perfect fuel contained in a non explosive form untill we need it.

Terry Dixon
 

Quote from terry.dixon on June 8th, 2012, 11:47 AM

Jeff's Question :
Thanks Terry, now another question, these were all standard measurements of known science for the calculations preformed, right ? "kilocalorie (kcal): the amount of heat needed to warm 1 liter of water 1 degree Celsius" . The reason I ask is that Stan claimed he could drive his dune buggy from coast to coast on 22 gallons of water, knowing this would this change standard measurements of known science and or change the end result of the calculations, or can we go with the end results of the calculations you have just preformed ? Thanks again Terry for your hard work, Jeff.

Thanks for the question Jeff. I came up with some rough calculations to try and answer your question.
First, I don't know if the injector was the only source of HHO gas that Stan used on his final design of the Dune Buggy using the VW engine. I remember seeing an HHO cell along with the injectors. Lets assume the design used only the injectors for the source of HHO fuel. I also don't know which VW engine Stan used. I am going to use the 1500 for my example.
Second, The number I came up with is for the Total Volume of the Water in the cell path. I don't think that Total Volume is decomposed every time there is an intake stroke on the piston. I think it takes time to decompose this Volume of Water just like in a regular HHO cell that produces say 3.0 Liters a minute.
The Volkswagon 1500 engine has a displacement of 1,493 cc (91.1 cu in) Volume. This is for all four of the Pistons and is running at appx 2000 revs per minute going down the road at 60 mph.
ref: http://en.wikipedia.org/wiki/Volkswagen_air-cooled_engine
That means that each piston has a displacement of 373.25 cc. Sinse it is a four cycle engine, intake only happens at 500 cycles per minute. That yeilds a Total intake of Gas Volume for each Piston of 186625 cc  per minute or 3110.41 cc per second.
Also, the Ratio of HHO gas to the non explosive Air we mix with the HHO just has to be slightly above the Lower Explosive Level or LEL of the Hydrogen per Volume. Hydrogen LEL and UEL at normal temperature and pressure is given as:
Oxyhydrogen can burn when it is between about 4% and 94% hydrogen by volume.
ref: http://hhofuel.wordpress.com/glossary/
So lets say you kept the mixture at 10% Hydrogen by Volume to Air then only 10% of the Piston Volume would need to filled on each intake cycle with HHO gas. The HHO Gas is 2/3 Hydrogen by Volume so contains 2.14655675 Liters of Hydrogen Gas in the water path of the cell in the form of water.
H2 consumption is appx 10% of the piston Volume of (186625 cc per minute), which = 18662.5 cc per minute. That is 18.6625 liters per minute of H2.  
If we kept the mixture at 5% Hydrogen by Volume which is just above the 4% LEL for Hydrogen in air then the H2 consumption would be 9.33125 liters per minute.
Sinse 9.33125 / 2.14655675 = 4.34707. The cell would need to decompose its contents appx 4.4 times a minute or a little over appx 15 seconds for each Volume of 2.379091232 cubic cm of water for each piston. or 9.5164928 cc for all four injectors on all four pistons.
These calculatios are very rough estimates but consuming 9.5cc of water 4.4 times a minute is appx half of the normal consumption rate of Gasoline in a modern Gasoline engine.
9.5164928 cc * 4.34707 = 41.36830448 cc per minute.
41.36830448 = 0.0413683044  Liters per minute. That is 2.48209826 Liters per hour water consumption. or for every 60 miles at 60 mph.
Miles per Gallon translates at this consumption rate to:
Sinse 1 liter = 0.264172052 US gallons, Then  2.482098264  * .6336 = 1.572657 Gallons every 60 miles or  38.15199372 miles per Gallon of water.
22 Gallons of water would give you 839.348 miles at this rate.
Again these calculations are very rough estimates but these estimates show that the MPG using water is about Two Times the MPG the old VW engine gets using Gasoline.
ref: http://answers.yahoo.com/question/index?qid=20061006153749AAWo13L
 I remember from Stan's videos,  he said, Water had about two times the energy that Gasoline had. So these calculations even though they are rough compare with what Stan had said about the energy contained in water.
Another thing to consider is:
Sinse we can reuse the water again if we collect the waste product (Water) at the tail pipe, consumption could be very low indeed.
Remember the only thing we are doing is breaking the HO bonds to use the HHO gas then reuniteing the HO bonds in the explosion of the HHO in the piston.
The end product is what we started out with. The perfect fuel contained in a non explosive form untill we need it.

Terry Dixon

I just started reading through this forum and I am collecting a lot of information in an attempt to get up to speed with all of the discussions so I apologize if I am restating some previous points.

I did not verify your calculations, but they do make good sense to me. I think it is cool that you came out to the same rough number that Stan used in his presentations (2.2 times the energy of gasoline). We must also remember that the buggy had a chamber that exposed the gases to light which (in theory) liberated some of the electrons which in turn increased the amount of energy expelled in the engine and provided an oportunity to use those electrons as electricity. Also if this is the case only some of the gasses used in the engine would come back together as water and others would go through some sort of exothermic reaction.

I have been thinking a lot about his claim of being able to drive a vehicle across the US on one tank and I think this takes into consideration two prerequisites.
-“Laser” energy input
- reclaiming/condensing exhaust

I have some equipment at my disposal that I think will be helpful in determining the amount of energy expelled with and with out laser energy.

Questions: Do you have any idea what kind of reaction would occur if the gasses did not form water? (I have a few theories)
I have background in nuclear technology, but I know nothing about circuts. Where can I get help with the VIC so I can build something?

Consider this: Lots of energy is used by the SWAT team to bust down the door, but the guy with the key uses much less…  I think Stan may have found the key to opening the covalent bond vice breaking it. Therefore, the amount of energy required (as we know it) may be less. He did claim to be able to “turn off” the bond.

Quote from shippy on November 16th, 2012, 10:24 AM

Where can I get help with the VIC so I can build something?

Here you can find information you need:

http://hereticalbuilders.com/showthread.php?t=174

Quote from bussi04 on November 16th, 2012, 01:49 PM

Quote from shippy on November 16th, 2012, 10:24 AM

Where can I get help with the VIC so I can build something?

Here you can find information you need:

http://hereticalbuilders.com/showthread.php?t=174

THAT IS GREAT! THANKS!

Dear Terry.Dixon,

In http://waterpoweredcar.com/pdf.files/Stan_Meyer_Full_Data.pdf at the end before the lawton circuit are Stan's calculations.
Stan mentioned that the energy content of Water is 2.5 times the content of fuel.  Hence consumption is 2.5 times lower.

I'm interested in your comments on this.

Rider

P.S. Russ, I couldn't find this data on your site RWGresearch. Need to add?

Thanks Terry, now another question,  these were all standard measurements of known science for the calculations preformed, right ? "kilocalorie (kcal): the amount of heat needed to warm 1 liter of water 1 degree Celsius" . The reason I ask is that Stan claimed he could drive his dune buggy from coast to coast on 22 gallons of water, knowing this would this change standard measurements of known science and or change the end result of the calculations, or can we go with the end results of the calculations you have just preformed ? Thanks again Terry for your hard work, Jeff.:cool::D:P[/quote]i think stan was wasting to much energy.
he should be able to go farther on a cup or 2 of water no?
 much may have been turned directly to steam?

need to find more results water per gallon. seems to high. the energy in the molecules should add up to more energy.
but perhaps the dune buggy wasted a lot through gearing.