Ethanol Rocket

Theory

Ethanol 3d model

The Ethanol. The Oxygen is hiding at the back side.

Atom positions

-1.1712	0.2997	0.0000	O
-0.0463	-0.5665	0.0000	C
1.2175	0.2668	0.0000	C
-0.0958	-1.2120	0.8819	H
-0.0952	-1.1938	-0.8946	H
2.1050	-0.3720	-0.0177	H
1.2426	0.9307	-0.8704	H
1.2616	0.9052	0.8886	H
-1.1291	0.8364	0.8099	H

HCH bonds are assumed to be undistorded tetraherdal angle 109.5 degrees. Actually the electrons repeal each other. The HOC bond is 104.5 deg because. . .

Valence shell electron-pair repulsion theory (VSEPR theory). . .

Van der Waals radius

Atom	Radius	Relative radius
C	170 pm	1.42
H	120 pm	1.00
O	152 pm	1.27

Use CPK coloring convention, white (hydrogen), black (carbon) and red (oxygen).

Ethanol reaction with Oxygen and Air

Ethanol reaction with oxygen

${\textrm {C}}_{2}{\textrm {H}}_{6}{\textrm {O}}+3{\textrm {O}}_{2}\to 2{\textrm {CO}}_{2}+3{\textrm {H}}_{2}{\textrm {O}}$

The molecular weight of ethanol is $46.069$ g/mol, and the molar weight of oxygen is 32 g/mol. The oxygen--ethanol fuel ratio is $3\times 32/46=2.1$. We need $2.1$ kg of oxygen to $1$ kg of ethanol. The air consists of 23.2 mass-% of oxygen, thus the air--ethanol ratio is $2.1/0.232=9$.

The volume-% of oxygen in air is 20.9%. The volume of the bottle is $V=500$ ml which gives the amount of oxygen to be $0.209V=0.209\times 500=104.5$ ml $=0.1045\times 1.314=0.137$ g of oxygen, which gives the amount of ethanol $0.137/2.1=0.0659$ g $=0.0659/0.789=0.0834$ ml. OR directly using air--ethanol ratio we have $\rho V/9=1.2041\times 0.5/9=0.066$ gram. That amount equals to $V/\rho _{\textrm {Ethanol}}=0.066g/(789g/l)=0.084$ ml $=84$mm³. Almost the same result using the methods. See the attached spreadsheet for detailed calculations.

Densities

	Density $\rho$	At
Oxygen (g)	1.429 g/l	STP
Oxygen (g)	1.314 g/l	20 °C
Ethanol (l)	789.45 g/l	20 °C
Air (l)	1.2041 g/l	20 °C

File:Ethanol oxygen combustion.ods

The inner diameter of the rocket bottle is 25 mm. The height of the ethanol in the cap need to be $V=\pi r^{2}h\iff h={\frac {V}{\pi r^{2}}}={\frac {84mm^{3}}{\pi 12.5^{2}mm^{2}}}=0.17$ mm.

Piezo Crystal

To ignite the air--ethanol mixture, we use piezo crystal. When tjhe piezo crystal is compressed, it will generate an electric charge which creates a spark.

References

https://nptel.ac.in/content/storage2/courses/123106002/MODULE%20-%20I/Lecture%201.pdf

https://www.peacesoftware.de/einigewerte/o2_e.html