Ethanol Rocket

From wikiluntti

Introduction

Theory

Ethanol 3d model

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

C2H6O+3O2→2CO2+3H2O

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 3×32/46=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=0.209×500=104.5 ml =0.1045×1.314=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 ρV/9=1.2041×0.5/9=0.066 gram. That amount equals to V/ρEthanol=0.066g/(789g/l)=0.084 ml =84mm3. Almost the same result using the methods. See the attached spreadsheet for detailed calculations.

The energy released by burning ethanol is 17.9 kJ/ml. Thus, 0.084 ml of ethanol releases 1,50 kJ of energy. This energy is converted into heat, sound and projectile motion (plus others).

Densities
Density ρ 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=πr2h⟺h=Vπr2=84mm3π12.52mm2=0.17 mm.

Exothermic Reaction and Energy release

Specific Heat

The energy is transferred into pressure, sound, etc. The isochoric specific heat Cv of air is Cv=0.7171 kJ/(kgK) at 18 centigrade. At 180 degrees Celsius Cv=0.7352 kJ/(kgK). Thus, the energy released heats

ΔE=CvΔTmΔT=ΔECvm=1.5kJ0.7kJ/(kgK)×0.5×10−3kg=2100K

Ideal gas law

The simplest idea is to use ideal gas law for isochoric process p/T= constant gives p2=T2T1p1 which gives force F=pA where A is the diameter of the nozzle.

Thus, the propulsive force is F=pA=T2T1p1A=T1+ΔTT1p1A=(1+ΔTT1)p1A=(1+ΔE/(Cvm)T1)p1A

Detonation velocity

Burn rate, detonation velocity. https://en.wikipedia.org/wiki/Table_of_explosive_detonation_velocities

http://www.explosionsolutions.co.uk/110411016.pdf

https://link.springer.com/article/10.1007/s00193-015-0554-7 v=1500 m/s???


Horizontal(?) accelaration due to rapidly expanding air.

Benjamin Robins:

F(x)=RPAcx

where x distance in barrel, R is the initial ratio of hot gas pressure to atmospheric pressure, P is the atmospheric pressure, A is the cross-sectional area of the ball or bore, c is the length of the barrel occupied by the powder charge before ignition.

Ek=∫0LF(x)dx⟺v02=2RPmπd2c4ln⁡(L/c)

where d is the barrel diameter (the bore)

The powder change p is given by p=πd2c4η where η is the density of gunpowder.


--

https://www.arc.id.au/RobinsOnBallistics.html

The pressure falls as 1/x.

--

Bernoulli?

--

∑F=0


F=ma=pA⟺F=md2xdt2

F=mdvdt=mvdvdx=pA⟺mvdv=pAdx

∫mvdv=∫0LpAdx

where

1L∫0LPdx=averagepressure=p→

If A is constant

Failed to parse (syntax error): {\displaystyle \frac12 mv^2 = A \int_0^L p dx = A \vec p L }

then

v=2p→ALmcf and p→=cfmv22AL

Projectile friction, rotational energy, heat transfer: correction factor cf.

The average pressure is 25% of peak pressure

--


http://closefocusresearch.com/calculating-barrel-pressure-and-projectile-velocity-gun-systems

https://www.arc.id.au/CannonBallistics.html

Maximum Flying Distance

By the conservation of energy, all explosive energy is transferred into the kinetic energy.

Eexplosion=Ekinetic

The distance covered by a projectile with initial velocity $v_0$ is

s=v02gsin⁡(2α)=

The drag coefficient cd needs to be found. For the circular disc (a coin) the drag coefficient is almost constant for all velocities (Reynold numbers). The coefficient of drag for a cylinder in this orientation is about 0.81 so long as the length to diameter ratio is greater than 2, see http://www.aerospaceweb.org/question/aerodynamics/q0231.shtml. The cone in either end gives some complications.

The drag equation for the drag force is FD=12ρv2cdA.

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

http://www.users.miamioh.edu/sommerad/NSF%20Files/drag_coefficient_calculation.pdf

DRAG COEFFICIENTS FOR FLAT PLATES , SPHERES, AND CYLINDERS MOVING AT LOW REYNOLDS NUMBERS IN A VISCOUS FLUID by ALVA MERLE JONES

http://www.aerospaceweb.org/question/aerodynamics/q0231.shtml

Ethanol rockets

https://www.youtube.com/watch?v=zTwz6FGobCA Ethanol Rocket - Cool Science Experiment


https://www.youtube.com/watch?v=4s-SZypWxeg Ethanol Explosion - Cool Science Experiment