# Compressed Air Bottle Rocket (Variable Mass)

1. Oct 15, 2008

### Aftermarth

1. The problem statement, all variables and given/known data
I need to find the maximum distance a rocket (which is actually a bottle filled with compressed air to 654kPa) can travel when the exhuast nozzle is 22cm (and can be changed to any smaller size if needed). Neglect air resistance. The bottles original mass is 50g

2. Relevant equations
Ideal Gas Law: PV = nRT
Bernoullis equation: $$\rho$$gh + (1/2)$$\rho$$V^2 + P = constant
Thrust Equation: T = mv
$$\rho_{1}$$$$A_{1}$$$$V_{1}$$=$$rho_{2}A_{2}V_{2}$$=m
R = $$\dot{m}$$v
= $$\rho$$AV(v-$$v_{0}$$

Density of Air = 1.2062kg/m^3
Pressure of Air = 101.3kPa

3. The attempt at a solution
Well i dont know.
I cant seem to figure out anything here. Bernoulli's equation wont work cause i end up with unknowns on both sides.
m doesnt make sense. I dont know if i just just sub in a number
veloctity of the gases - i dont know how to find them

im just confused how to do this :x

(ps - i dont want the thing solved. can i just get some help in formulating expressions which can be integrated between 1. Angles of launch and 2. Area of the nozzle)

2. Oct 16, 2008

### Andrew Mason

This is a complicated question, even if you ignore air resistance. The mass of the rocket is continually changing as the gas is expelled. If you ignore this change in mass, the problem becomes more manageable but is still complicated because the thrust force is not constant - it decreases as the gas is expelled.

When the rocket reaches maximum height, it has kinetic and potential energy. The kinetic energy due to its horizontal speed (vertical speed = 0) and its potential energy at maximum height a function of that height. This energy has to come from the work done by rocket propulsion. Determine the expression for that work (ie. thrust x distance) and that will give you the total energy imparted to the rocket. It is not a trivial calculation.

AM

3. Oct 16, 2008

### Aftermarth

ok. here is an update on my progress....

Thrust = $$\dot{m}$$v
where v = (v - $$v_{0}$$)
and $$\dot{m}$$ = $$\rho$$AV

using the gas laws (PV = nRT) and Boyle's Law ($$\rho$$ = (MP)/(RT)
along with Pressure/ Temperature = constant

T(inside bottle) = 654/T = 101.325/293.15 (NOTE - i have assumed the atmosphere outside the bottle is STP P = 101.325kPa, T = 293.15K)

T = 1892K (this seems very high can someone please verify this!!)

Subbing average molar mass of air = 28.97g/mol and T into boyle's law gives
$$\rho$$ = 1.2062
(so density of the air does not change inside the bottle??)

Now i have for INSIDE the bottle: P = 654kPa, T = 1892K $$\rho$$=1.2062kg/m^3
for OUTSDIE the bottle P = 101.325kPa T = 293.15K $$\rho$$= 1.2062kg/m^3

subbing into Bernoulli's Equation and rearranging gives
($$V^{2}_{1}$$ - $$V^{2}_{2}$$ = 916.39 (u haven't worked out units for this value yet)

Also i now have the equation
mv + pA -mg - R = m$$v^{.}$$
ignoring R as the resistance force gives
T + pA - mg = m$$v^{.}$$ (as thrust is m`v)

and i dont know how to work out p which is supposed to be nozzle exit pressure or something like that :s

4. Oct 22, 2008

### Paul McHugh

See me tomorrow
- paul

5. Oct 22, 2008

### Paul McHugh

lol jks

6. Oct 22, 2008