Stoichiometry - Revision history

Benstern: added links

2008-06-01T14:25:27Z

added links

← Older revision		Revision as of 09:25, 1 June 2008
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	[http://en.wikipedia.org/wiki/Stoichiometric Stoichiometry] is an important concept in [[combustion cannon\|combustion]] and [[hybrid cannon\|hybrid]] launchers, since they use the energy from combustible ~~fuels~~. A stoichiometric mixture is one where the ratio of fuel and oxygen molecules is balanced, so all the fuel and oxygen in the chamber is (in optimal conditions) consumed in the reaction. A [[fuel meter]] system is usually designed to achieve this mixture, which will yield the most power out of the combustion reaction.		[http://en.wikipedia.org/wiki/Stoichiometric Stoichiometry] is an important concept in [[combustion cannon\|combustion]] and [[hybrid cannon\|hybrid]] launchers, since they use the energy from combustible [[fuel]]s. A stoichiometric mixture is one where the ratio of fuel and oxygen molecules is balanced, so all the fuel and oxygen in the [[chamber]] is (in optimal conditions) consumed in the reaction. A [[fuel meter]] system is usually designed to achieve this mixture, which will yield the most power out of the [[combustion reaction]].


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	fuel volume in a 3000 cc chamber = 3000*0.0402 = 120.6 cc if air is displaced		fuel volume in a 3000 cc chamber = 3000*0.0402 = 120.6 cc if air is displaced

	Most [[fuel ~~meter]]s~~ do not displace air in the chamber with fuel (the pressure in the chamber rises slightly when the fuel is injected), so for a meter the fuel volume is 4.2% of the chamber volume. If the fuel is injected with a syringe, and the fuel displaces some of the air in the chamber, then the fuel volume should be 4.02% of the chamber volume.		Most fuel meters do not displace air in the chamber with fuel (the [[pressure]] in the chamber rises slightly when the fuel is injected), so for a meter the fuel volume is 4.2% of the chamber volume. If the fuel is injected with a syringe, and the fuel displaces some of the air in the chamber, then the fuel volume should be 4.02% of the chamber volume.

	Single formula for the fractional fuel to air ratio for hydrocarbon fuels: (1/(C+(H/4)+(O/2)))*0.21		Single formula for the fractional fuel to air ratio for hydrocarbon fuels: (1/(C+(H/4)+(O/2)))*0.21

	[[category:Concepts]]		[[category:Concepts]]

Biopyro: /* Concrete example */

2008-06-01T13:54:14Z

Concrete example

← Older revision		Revision as of 08:54, 1 June 2008
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	*number of oxygen atoms: 0		*number of oxygen atoms: 0

	Number of oxygen molecules needed = 32 + 80.5 = 5 molecules<br/>		Number of oxygen molecules needed = 31 + 80.25 = 5 molecules<br/>
	The stoichiometric combustion equation is:<br/>		The stoichiometric combustion equation is:<br/>
	C<sub>3</sub>H<sub>8</sub> + 5O<sub>2</sub> --> 3CO<sub>2</sub> + 4H<sub>2</sub>O		C<sub>3</sub>H<sub>8</sub> + 5O<sub>2</sub> --> 3CO<sub>2</sub> + 4H<sub>2</sub>O

Biopyro: Corrected stoichemetric ratio example. carbon needs 2 oxygens and hydrogen needs 0.5!

2008-06-01T13:50:22Z

Corrected stoichemetric ratio example. carbon needs 2 oxygens and hydrogen needs 0.5!

← Older revision		Revision as of 08:50, 1 June 2008
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	*number of oxygen atoms: 0		*number of oxygen atoms: 0

	Number of oxygen molecules needed = 31 + 80.25 = 5 molecules<br/>		Number of oxygen molecules needed = 32 + 80.5 = 5 molecules<br/>
	The stoichiometric combustion equation is:<br/>		The stoichiometric combustion equation is:<br/>
	C<sub>3</sub>H<sub>8</sub> + 5O<sub>2</sub> --> 3CO<sub>2</sub> + 4H<sub>2</sub>O		C<sub>3</sub>H<sub>8</sub> + 5O<sub>2</sub> --> 3CO<sub>2</sub> + 4H<sub>2</sub>O

Jimmy101: corrected formating, additive versus displacive volume %, word choice

2007-11-13T16:57:21Z

corrected formating, additive versus displacive volume %, word choice

← Older revision		Revision as of 11:57, 13 November 2007
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	===Concrete example===		===Concrete example===

	Formula of [[propane]]: ~~C3H8~~		Formula of [[propane]]: C<sub>3</sub>H<sub>8</sub>

	*number of carbon ~~molecules~~: 3		*number of carbon atoms: 3
	*number of hydrogen ~~molecules~~: 8		*number of hydrogen atoms: 8
	*number of oxygen ~~molecules~~: 0		*number of oxygen atoms: 0

	~~number~~ of oxygen molecules needed = 31 + 80.25 = 5 molecules<br/>		Number of oxygen molecules needed = 31 + 80.25 = 5 molecules<br/>
	~~the actual~~ combustion ~~reaction~~ is ~~as follows~~: ~~C3H8~~ + 5O<sub>2</sub> --> 3CO<sub>2</sub> + 4H<sub>2</sub>O		The stoichiometric combustion equation is:<br/>
			C<sub>3</sub>H<sub>8</sub> + 5O<sub>2</sub> --> 3CO<sub>2</sub> + 4H<sub>2</sub>O

	fuel/oxygen ratio = 1/5 = 0.2<br/>		fuel/oxygen ratio = 1/5 = 0.2<br/>
	fuel/air ratio = 0.2*0.21 = 0.042, or 4.2%<br/>		fuel/air ratio = 0.2*0.21 = 0.042, or 4.2%<br/>
	~~fuel volume in a 3000 cc chamber = 3000*0.042 = 126 cc~~

	Single formula for fractional fuel to air ratio: (1/(C+(H/4)+(O/2)))*0.21		The actual percentage of fuel for a stoichiometric mixture will depend on whether the fuel is '''added''' to the air in the chamber or the fuel '''displaces''' some of the air in the chamber. If the fuel is added to the air in the chamber then the fuel volume is '''4.2%''' of the chamber volume. If the fuel displaces some of the air in the chamber then the fuel volume is '''4.02%''' of the chamber volume.

			fuel volume in a 3000 cc chamber = 3000*0.042 = 126 cc if no air is displaced<br/>
			fuel volume in a 3000 cc chamber = 3000*0.0402 = 120.6 cc if air is displaced

			Most [[fuel meter]]s do not displace air in the chamber with fuel (the pressure in the chamber rises slightly when the fuel is injected), so for a meter the fuel volume is 4.2% of the chamber volume. If the fuel is injected with a syringe, and the fuel displaces some of the air in the chamber, then the fuel volume should be 4.02% of the chamber volume.

			Single formula for the fractional fuel to air ratio for hydrocarbon fuels: (1/(C+(H/4)+(O/2)))*0.21

	[[category:Concepts]]		[[category:Concepts]]

Boilingleadbath: categorized

2007-04-15T12:33:57Z

categorized

← Older revision		Revision as of 07:33, 15 April 2007
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	Single formula for fractional fuel to air ratio: (1/(C+(H/4)+(O/2)))*0.21		Single formula for fractional fuel to air ratio: (1/(C+(H/4)+(O/2)))*0.21

			[[category:Concepts]]

GalFisk: /* Calculations */ legibility

2007-03-18T14:35:24Z

Calculations: legibility

← Older revision		Revision as of 09:35, 18 March 2007
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	*for each carbon atom, add 1		*for each carbon atom, add 1
	*for each oxygen atom, subtract 0.5		*for each oxygen atom, subtract 0.5
	The resulting number is the oxygen to fuel ratio. To find the fuel to oxygen ratio, divide 1 by ~~the oxygen to fuel ratio~~. The amount of oxygen in air is 21%, so to find the fractional fuel to air ratio, the fuel to oxygen ratio is multiplied with 0.21. To find the ratio in percent, multiply this value by 100.		The resulting number is the oxygen to fuel ratio. To find the fuel to oxygen ratio, divide 1 by this number. The amount of oxygen in air is 21%, so to find the fractional fuel to air ratio, the fuel to oxygen ratio is multiplied with 0.21. To find the ratio in percent, multiply this value by 100.

	To find the volume of fuel needed for a particular chamber, the fractional ratio is multiplied by the chamber volume.		To find the volume of fuel needed for a particular chamber, the fractional ratio is multiplied by the chamber volume.

Rsgpit at 17:55, 2 April 2006

2006-04-02T17:55:30Z

← Older revision		Revision as of 12:55, 2 April 2006
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	C + O<sub>2</sub> --> CO<sub>2</sub><br/>		C + O<sub>2</sub> --> CO<sub>2</sub><br/>
	4H + O<sub>2</sub> --> 2H<sub>2</sub>O<br/>		2H<sub>2</sub> + O<sub>2</sub> --> 2H<sub>2</sub>O<br/>
	C<sub>x</sub>H<sub>x</sub> + O<sub>2</sub> --> CO<sub>2</sub> + H<sub>2</sub>O		C<sub>x</sub>H<sub>x</sub> + O<sub>2</sub> --> CO<sub>2</sub> + H<sub>2</sub>O

70.31.208.207: spelling

2006-03-04T22:30:24Z

spelling

← Older revision		Revision as of 17:30, 4 March 2006
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	[http://en.wikipedia.org/wiki/Stoichiometric Stoichiometry] is an important concept in [[combustion cannon\|combustion]] and [[hybrid cannon\|hybrid]] launchers, since they use the energy from combustible fuels. A stoichiometric mixture is one where the ratio of fuel and oxygen molecules is balanced, so all the fuel and oxygen in the chamber is (in optimal conditions) consumed in the reaction. A [[fuel meter]] system is usually designed to ~~ahcieve~~ this mixture, which will yield the most power out of the combustion reaction.		[http://en.wikipedia.org/wiki/Stoichiometric Stoichiometry] is an important concept in [[combustion cannon\|combustion]] and [[hybrid cannon\|hybrid]] launchers, since they use the energy from combustible fuels. A stoichiometric mixture is one where the ratio of fuel and oxygen molecules is balanced, so all the fuel and oxygen in the chamber is (in optimal conditions) consumed in the reaction. A [[fuel meter]] system is usually designed to achieve this mixture, which will yield the most power out of the combustion reaction.


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	Carbon atoms combine with oxygen to form CO<sub>2</sub>, so each carbon atom will consume one entire O2 molecule.		Carbon atoms combine with oxygen to form CO<sub>2</sub>, so each carbon atom will consume one entire O2 molecule.

	Any oxygen in the fuel will lead to one less oxygen atom being consumed from the air. Two oxygen atoms will prevent one O<sub>2</sub> molecule from ~~beling~~ consumed, so each oxygen atom "saves" 0.5 O<sub>2</sub> molecules.		Any oxygen in the fuel will lead to one less oxygen atom being consumed from the air. Two oxygen atoms will prevent one O<sub>2</sub> molecule from being consumed, so each oxygen atom "saves" 0.5 O<sub>2</sub> molecules.

GalFisk: Stoichiometric moved to Stoichiometry

2005-10-23T10:05:18Z

Stoichiometric moved to Stoichiometry

← Older revision	Revision as of 05:05, 23 October 2005
(No difference)

GalFisk: +consistant chemical reaction notation

2005-10-20T11:32:23Z

+consistant chemical reaction notation

← Older revision		Revision as of 06:32, 20 October 2005
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	The stoichiometric ratio for gaseous fuels can easily be calculated. When the hydrogen and carbon in a fuel burns, it combines with oxygen as follows:		The stoichiometric ratio for gaseous fuels can easily be calculated. When the hydrogen and carbon in a fuel burns, it combines with oxygen as follows:

	C+O<sub>2</sub> --> CO<sub>2</sub><br/>		C + O<sub>2</sub> --> CO<sub>2</sub><br/>
	4H+O<sub>2</sub> --> 2H<sub>2</sub>O<br/>		4H + O<sub>2</sub> --> 2H<sub>2</sub>O<br/>
	C<sub>x</sub>H<sub>x</sub> + O<sub>2</sub> --> CO<sub>2</sub> + H<sub>2</sub>O		C<sub>x</sub>H<sub>x</sub> + O<sub>2</sub> --> CO<sub>2</sub> + H<sub>2</sub>O

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	number of oxygen molecules needed = 31 + 80.25 = 5 molecules<br/>		number of oxygen molecules needed = 31 + 80.25 = 5 molecules<br/>
	the actual combustion reaction is as follows: C3H8+5O<sub>2</sub> = 3CO<sub>2</sub>+4H<sub>2</sub>O		the actual combustion reaction is as follows: C3H8 + 5O<sub>2</sub> --> 3CO<sub>2</sub> + 4H<sub>2</sub>O

	fuel/oxygen ratio = 1/5 = 0.2<br/>		fuel/oxygen ratio = 1/5 = 0.2<br/>