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Hydrogen/Oxygen reaction underwater burning -

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Joined: 10 Mar 2010
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Location: Van Nuys, CA

PostPosted: Wed Mar 10, 2010 11:23 pm    Post subject: Hydrogen/Oxygen reaction underwater burning - Reply with quote

I am working on a show for NatGeo and looking to prove a statement in our narration.
Essentially the show is about deep-sea commercial divers and one of the tasks they do is called "burning" - it is basically underwater cutting with an oxygen-arc tool.
It releases oxygen and creates a flame hot enough to burn through steel. When the oxygen meets with hydrogen underwater it creates what the divers call an O2pop. Sometimes the hydrogen will get trapped in an underwater well casing. If this happens a bubble is created and the larger it gets the more explosive it is. We are trying to prove that if the bubble got as large as a grapefruit that it could explode with a force equivalent to a stick of dynamite. Remember we are 300 feet underwater and under 150psi.
Does this sound reasonable? How can I "measure" the force of the bubble and compare it to a stick of dynamite exploding?
ANY and all help is appreciated!
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Joined: 23 Jun 2007
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PostPosted: Sat Mar 13, 2010 1:54 pm    Post subject: Reply with quote

This is more of a chemistry question so any chemists, feel free to weigh in if i am wrong but I guess you will need to find the Enthalpy of combustion which is the difference between the enthalpy of the products and the reactants.

The stoichiometry of the gases is also important so that all the hydrogen and oxygen are used in the explosion.

The ideal gas equation relates pressure, volume and temperature.
PV =n R T where P is the pressure, V is the volume of gas, T the temperature, n the number of mols of gas and R the molar gas constant.

At 300ft below the water the pressure is about 10 times atmospheric pressure and the temperature is around 307 K.
A grapefruit sized volume would be approximately, (4/3)*3*(0.05)^3 = 5.2 x 10^-4 m^3. We will use this information and the ideal gas equation to find the size of the gas bubble at Standard Temperature and Pressure (STP).

Since the ideal gas equation has the same constant on the right-hand side of the equation we can calculate the volume of the gas at STP

(P1 V1)/T1 = (P2 V2)/T2
with P1, V1 and T1 the pressure, volume and temperate at a depth of 100m and P2, V2 and T2 the pressure, volume and temperature at the surface.

V2 = (P1 V1 T2)/(T1 P2)
= (10 x 5.2e-4 x 330)/(307 x 1)
= 5.6e -3 m^3 or roughly ten times the volume.

Next we need to find the number of moles of gas in this volume. There are 1000 litres in a cubic metre of gas and 1 mole approximately 22.4 litres at STP. Therefore number of moles, so
5.6 /22.4 = 0.25 moles.

The enthalpy for the reaction
2 H2(g) + O(g) -> 2 H2 O produces -483.6 kJ per mol of energy. So you will get around 121 kJ from the bubble as it combusts.

If you compare this with the energy released from a stick of dynamite which is around 2.1 MJ (2100 kJ) you can see it falls way short.
However, if you are the diver under water the number may be academic.

The explosion generates a shock wave that travels much more efficiently in water than in air because it is incompressible.
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