19 Mar '15 14:421 edit

Hello All,

Out of shear curiosity I'm looking for some help on validating a model equation for the interaction of the fuel(compressed air)/propellant(water) for the rocket in question. I'm trying to obtain the thrust force as a function of time, not the rockets position in time. I am also unsure of certain assumptions that I made, and am more importantly looking for clarification on how to properly apply the mechanics.

If I model the soda bottle as a cylinder with constant cross-sectional area; ignoring the bottle neck (truncated nozzle) with respect to fuel expansion (technically this would be a two stage rocket, but I'm only concerned with the first stage), and that the nozzle has appreciable volume when compared to the volume of the propellant (water).

Coordinate system:

Picture a rocket in launch position (nozzle capped) partially filled with water. Directly above the water is a cavity of compressed air.

The single coordinate will be

I'm only concerned with the forces acting internally (not the external holding force). I'm not sure if this is/isn't valid if I neglect that external "holding force" in the subsequent analysis.

Newtons Second Law:

Σ

Σ

Where "F_p" is the force from pressure, and "W" is the weight of the propellant. They are both functions of

F_p = P*A

Assuming the compressed air behaves as an Ideal Gas, expands adiabatically, and k = 1.4

Po*Vo^1.4 = P*V^1.4

P = Po*(Vo/V)^1.4

Assume:

Vo = A*Lo

V = A*(Lo+

Substitute into "P"

P = Po*(Lo/(Lo+

F_p = P*A = A*Po*(Lo/(Lo+

Moving on to "W"

m(

W = (mo - µ*A*

g = acceleration from gravity

µ = mass density of water

That takes care of the right side.

LHS:

d(m*

The first term on the right represents the "Force" associated with the accelerated mass (water) inside the bottle (neglecting mass from air), and the second term is the "Thrust force" from ejecting mass.

Looking at the first term on the right:

m*d(

The instantaneous mass, m(x) =mo - µ*A*

Acceleration of that mass (from gas expansion) d(

m*d(

The "Thrust" (second term on the right)

with v_e = the velocity at the exit of the nozzel, from continuity

A_e*

Then dm/dt is equal to the displaced mass from compressed air expansion

dm/dt = µ*A*

dm/dt = µ*A*d

Then If everything I did was correct? Solve:

A*Po*(Lo/(Lo+

Like I said, mostly interested in the correct application of the mechanics at this point?

Out of shear curiosity I'm looking for some help on validating a model equation for the interaction of the fuel(compressed air)/propellant(water) for the rocket in question. I'm trying to obtain the thrust force as a function of time, not the rockets position in time. I am also unsure of certain assumptions that I made, and am more importantly looking for clarification on how to properly apply the mechanics.

If I model the soda bottle as a cylinder with constant cross-sectional area; ignoring the bottle neck (truncated nozzle) with respect to fuel expansion (technically this would be a two stage rocket, but I'm only concerned with the first stage), and that the nozzle has appreciable volume when compared to the volume of the propellant (water).

Coordinate system:

Picture a rocket in launch position (nozzle capped) partially filled with water. Directly above the water is a cavity of compressed air.

The single coordinate will be

**x**↓+, and it will be zero at the gas/water interface.**The rocket is held in place, and will be kept from accelerating (keeping the frame inertial)**I'm only concerned with the forces acting internally (not the external holding force). I'm not sure if this is/isn't valid if I neglect that external "holding force" in the subsequent analysis.

Newtons Second Law:

Σ

**F**= d(m**v**)/dtΣ

**F**= F_p + WWhere "F_p" is the force from pressure, and "W" is the weight of the propellant. They are both functions of

**x**( and time)F_p = P*A

Assuming the compressed air behaves as an Ideal Gas, expands adiabatically, and k = 1.4

Po*Vo^1.4 = P*V^1.4

P = Po*(Vo/V)^1.4

Assume:

Vo = A*Lo

V = A*(Lo+

**x**)Substitute into "P"

P = Po*(Lo/(Lo+

**x**))^1.4F_p = P*A = A*Po*(Lo/(Lo+

**x**))^1.4Moving on to "W"

m(

**x**) = mo - µ*A***x**W = (mo - µ*A*

**x**)*gg = acceleration from gravity

µ = mass density of water

That takes care of the right side.

LHS:

d(m*

**v**)/dt = m*d(**v**)/dt +**v_e***dm/dtThe first term on the right represents the "Force" associated with the accelerated mass (water) inside the bottle (neglecting mass from air), and the second term is the "Thrust force" from ejecting mass.

Looking at the first term on the right:

m*d(

**v**)/dtThe instantaneous mass, m(x) =mo - µ*A*

**x**Acceleration of that mass (from gas expansion) d(

**v**)/dt = d²**x**/dt²m*d(

**v**)/dt = (mo - µ*A***x**)*d²**x**/dt²The "Thrust" (second term on the right)

with v_e = the velocity at the exit of the nozzel, from continuity

A_e*

**v_e**= A***v****v_e**= A/A_e***v****v_e**= A/A_e*d**x**/dtThen dm/dt is equal to the displaced mass from compressed air expansion

dm/dt = µ*A*

**v**dm/dt = µ*A*d

**x**/dt**v_e***dm/dt = µ*A²/A_e*(d**x**/dt)²Then If everything I did was correct? Solve:

A*Po*(Lo/(Lo+

**x**))^1.4 + (mo - µ*A***x**)*g = (mo - µ*A***x**)*d²**x**/dt² + µ*A²/A_e*(d**x**/dt)²Like I said, mostly interested in the correct application of the mechanics at this point?