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Analysis: Although only a simple water bottle rocket, it still applies to basic rocket theory. The simplest equation which applies is Tsiolkovsky's equation which describes a device that can apply acceleration to itself (a thrust) by expelling part of its mass with high speed and thereby move due to the conservation of momentum. In short the momentum which the rocket gains is that momentum which the water loses as it is expelled. The equation is: ∆v=V_e ln(m_o/m_f ) Where: ∆v = change in velocity (m/s) V_e= velocity of exhaust leaving nozzle (m/s) m_o= initial mass of rocket, including water mass (kg) m_f = final mass of rocket, after water is expelled (kg) Velocity of exhaust can be calculated by: V_e=√(2(p_in-p_out ) )/ρ_w Where: p_in= internal*…show more content…*

The initial volume is (1-f)V, where f is the filling fraction of water of the rocket. The value of K can be determined from the initial conditions, where K=P(1-f)^γ V^γ. Substituting in we get: W=(P(1-f)^γ V^γ V^(-γ+1))/(-γ+1) [1-(1-f)^(-γ+1) ] =PV/(-γ+1) [(1-f)^γ-(1-f)] Using this final equation we studied how the work done expelling the water varies with filling fraction f. Taking: γ = 1.4 V = 0.002 (m^3) P = 4.0 10^5 (Pa) We drew up a graph of the amount of energy extracted by the water from the compressed air. It is clear to see that at high filling factors, that there is only a small volume of compressed air and so not much energy is*…show more content…*

When looking at these values it is important to think of real applications. By filling up a bottle to 57% you are creating quite a heavy rocket. This means that most of the energy used will be lifting water rather than the rocket itself. A simple estimation of optimal filling fraction can be found by dividing the Work Done by the mass of the rocket at launch: W/(rocket mass)=1/(m_o+ ρ_w*fV) (PV/(-γ+1) [(1-f)^γ-(1-f)]) Using this data we find that the ideal initial mass for our 2L bottle rocket is: m_o=m_r+f_c Vρ_w Where: m_r= mass of the empty rocket f_c= critical filling fraction of water Assuming the mass of air is negligible. So our initial mass is equal to: m_o=0.1+(0.21)(0.002)(1000) m_o=0.52 kg Plugging into Tsiolkovsky's equation: ∆v=(24.49 )ln(0.52/0.1) ∆v=40.37 m/s Given that the rocket is launching from rest we can expect the rocket to have a final velocity v_f=40.37 m/s. Next we must calculate the amount of work done by the gas itself after expelling the water. It is this energy that really adds acceleration to the

The initial volume is (1-f)V, where f is the filling fraction of water of the rocket. The value of K can be determined from the initial conditions, where K=P(1-f)^γ V^γ. Substituting in we get: W=(P(1-f)^γ V^γ V^(-γ+1))/(-γ+1) [1-(1-f)^(-γ+1) ] =PV/(-γ+1) [(1-f)^γ-(1-f)] Using this final equation we studied how the work done expelling the water varies with filling fraction f. Taking: γ = 1.4 V = 0.002 (m^3) P = 4.0 10^5 (Pa) We drew up a graph of the amount of energy extracted by the water from the compressed air. It is clear to see that at high filling factors, that there is only a small volume of compressed air and so not much energy is

When looking at these values it is important to think of real applications. By filling up a bottle to 57% you are creating quite a heavy rocket. This means that most of the energy used will be lifting water rather than the rocket itself. A simple estimation of optimal filling fraction can be found by dividing the Work Done by the mass of the rocket at launch: W/(rocket mass)=1/(m_o+ ρ_w*fV) (PV/(-γ+1) [(1-f)^γ-(1-f)]) Using this data we find that the ideal initial mass for our 2L bottle rocket is: m_o=m_r+f_c Vρ_w Where: m_r= mass of the empty rocket f_c= critical filling fraction of water Assuming the mass of air is negligible. So our initial mass is equal to: m_o=0.1+(0.21)(0.002)(1000) m_o=0.52 kg Plugging into Tsiolkovsky's equation: ∆v=(24.49 )ln(0.52/0.1) ∆v=40.37 m/s Given that the rocket is launching from rest we can expect the rocket to have a final velocity v_f=40.37 m/s. Next we must calculate the amount of work done by the gas itself after expelling the water. It is this energy that really adds acceleration to the

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