Bottle Mechanics


The basic free body diagram of the launching bottle rocket using Newton’s Second Law (F = ma) includes the thrust force up from water expelling (Ft), mass weight down (W), and drag force from the plastic bottle (Cd).

\sum_{}^{}{F =}F_{t} - W - C_{d} = ma = m\frac{\text{dv}}{\text{dt}}


Weight assumes all mass comes from the water and the plastic mass from the rocket shell, neglecting air mass.


The thrust force (Ft) derives from the compressed air inside the rocket pushing out the water through the small area of the nozzle (An). This is counteracted by the weight of the water.

F = {\dot{m}}_{e}v_{e}


The exit force is written in terms of the momentum equation using the exit velocity of water (v_{e}) and mass flow rate of water (\dot{m_{e}}).


In order to relate velocity with force, we use the momentum p\ = \ mv and F = \frac{\text{dp}}{\text{dt}} = m\frac{\text{dv}}{\text{dt}}

Since we’re working with incremental values, we can manipulate and plug in actual values, working with Δ.

F = \frac{\mathrm{\Delta}m}{\mathrm{\Delta}t}v = \dot{m}v


The definition of mass, m = \rho V\ , adjusted to account for change over time can be written as

\frac{\mathrm{\Delta}m}{\mathrm{\Delta}t} = \rho\frac{\mathrm{\Delta}V}{\mathrm{\Delta}t}\ \Rightarrow \dot{m} = \ \rho\dot{V}

Substituting \dot{m} into the thrust force equation [2] you get

F = {(\rho\dot{V})v}_{e}

The volume flow rate can also be written as the water exiting in the cylindrical stream from the nozzle, with the length of water outside the bottle represented by x.

V = A_{n}x\ \Longrightarrow \ \dot{V} = A_{n}\dot{x} = \ A_{n}v_{e}


Substituting the volume flow rate, you get

F = \rho_{w}A_{n}v_{e}^{2}



From the thrust force equation, the exit velocity of water is determined by the pressurized 2 inside the bottle. Using Bernoulli’s Equation, we can relate the internal (1) and external (2) parts of the constant streamline coming out of the bottle. Since it is constant, we can equate both parts.

p_{1\ } + \frac{1}{2}\rho_{w}v_{1}^{2} + \ \gamma z_{1} = \ p_{2\ } + \frac{1}{2}\rho_{w}v_{2}^{2} + \ \gamma z_{2}

Dividing by the specific weight (\gamma = \rho g),

\frac{p_{1}}{\rho{g}} + \frac{v_{1}^{2}}{2g} + \ z_{1} = \ \frac{p_{2}}{\rho{g}} + \frac{v_{2}^{2}}{2g} + \ z_{2}

The three parts on each side each respectively represent the energy due to air pressure compressed inside (1) and in the atmosphere (2), kinetic energy from the exit velocity of water, and potential energy from the height of starting water (z). We assume the heights of water in negligible because its minuscule in comparison to the pressure and kinetic energy. We also assume the inner velocity of water is zero due to the exit velocity being much greater, making its kinetic energy negligible (v_{1} \ll v_{2})

Moving pressures to the left and cancelling out gravity, we are left with the equation:

\frac{p_{1 - \ }p_{2\ }}{\rho} = \ \frac{v_{2}^{2}}{2}

Writing this in terms of the exit velocity (v_{2}),

v_{2\ } = v_{e} = \sqrt{2\ \frac{p_{1 - \ }p_{\text{atm\ }}}{\rho_{w}}}


Substituting this into the thrust force equation [4], we have force in terms of the incremental air pressure.

F = 2A_{n}(p_{1 - \ }p_{\text{atm\ }})



Since the pressure inside the rocket varies with time unpredictably, we replace it with a function of volume still remaining in the bottle. Pressure can be related to volume using the ideal gas law, based under the assumption that in the 0.5 seconds, air expands too rapidly with little time for heat to exchange with the environment. In this adiabatic process, PV1.4 is constant (derived in 2.5). Therefore,

p = p_{0}\left( \frac{V_{0}}{V} \right)^{1.4}


Plugging this into the exit velocity [5],

v_{e} = \sqrt{2\ \frac{p_{0}\left( \frac{V_{0}}{V} \right)^{1.4} - p_{\text{atm\ }}}{\rho_{w}}}

This can be used in the volume flow rate equation [3]

\dot{V} = \ A_{n}\sqrt{2\ \frac{p_{0}\left( \frac{V_{0}}{V(t)} \right)^{1.4} - p_{\text{atm\ }}}{\rho_{w}}}

Since we will be working with theoretical data, we can use calculated numbers from known initial volume values to find the volume of air left in the bottle at each time step. This can then be plugged into the thrust force equation. To calculate volume at each time step, we use

\boxed{V(t + \mathrm{\Delta}t) = V(t) + \mathrm{\Delta}t\ \left\lbrack \text{\ A}{n}\sqrt{2\ \frac{p{0}\left( \frac{V_{0}}{V\left( t \right)} \right)^{1.4} - p_{\text{atm\ }}}{\rho_{w}}} \right\rbrack}

And plug in the next volume into

\boxed{F = 2A_{n}\left\lbrack p_{0}\left( \frac{V_{0}}{V(t)} \right)^{1.4} - p_{\text{atm\ }} \right\rbrack}



Since we’re assuming no temperature, no energy is added as heat in the First Law of Thermodynamics equation. This allows us to use the constant \text{PV}^{\gamma}, where gamma is the heat ratio capacity, or adiabatic gas constant, for air. This is valued at \gamma = 1.4

\gamma = \ \frac{f + 2}{f} This representation of gamma is necessary because the molecular structure of gases vary, with α denoting the degrees of freedom. Since air is comprised of primarily diatomic gases (nitrogen, oxygen, etc), f = 5.

3.0 — DRAG

The drag equation with respect to time is defined as the drag constant (C_{d} = 0.82) multiplied by the pressure of the bottle rocket moving through air.

D\left( t \right) = C_{d}\left( \frac{1}{2}\rho_{\text{air}}A_{c}v^{2} \right)


We chose to model the bottle rocket as a cylinder moving through the fluid with cross sectional area (A_{c}) which we defined as the average area of the large end of the bottle.


Taking the time-varying equations of thrust force [8] and drag [9], we can plug them back into the original equation from Newton’s Second Law [1]. Dividing each side by mass, we get

a\left( t \right) = \frac{F\left( t \right)}{m\left( t \right)} - g - \frac{D\left( t \right)}{m\left( t \right)}

We can model the mass as varying with time by the volume of water plus the bottle mass. The mass of the air is negligible.

m\left( t \right) = V_{w}\rho_{w} = \left( V_{\text{total}} - V_{\text{air}} \right)\rho_{w} + m_{b}

Plugging in each component,

\boxed{a\left( t \right) = \ \frac{2A_{n}\left\lbrack p_{0}\left( \frac{V_{0}}{V\left( t \right)} \right)^{1.4} - p_{\text{atm\ }} \right\rbrack}{\left( V_{\text{total}} - V_{\text{air}} \right)\rho_{w} + m_{b}} - g - \frac{C_{d}\left( \frac{1}{2}\rho_{\text{air}}A_{c}{v(t)}^{2} \right)\ }{\left( V_{\text{total}} - V_{\text{air}} \right)\rho_{w} + m_{b}}}

This final acceleration equation can be used to model the path of the rocket in the first 50 seconds. It utilizes the time-step function of the volume as well as velocity. The velocity can be modeled for this purpose by the previous velocity plus the acceleration multiplied by the time step.

v\left( t + \mathrm{\Delta}t \right) = v\left( t \right) + a(t)\mathrm{\Delta}t

This is initialized with the starting velocity at zero (v_{0} = 0), initializing drag at zero.


Though there is no heat exchange with the environment, there is still a temperature change when work is being done inside the bottle throughout the change in pressure. From the First Law of Thermodynamics, Q = 0 and the energy is

U = Q - W\ \Rightarrow dU = - dW

Work is done by the pressure due to the volume change

dW = p({V}_{2} - {V}_{1}),

and energy change can be modeled as the specific heat at a constant pressure multiplied by the temperature change, with an initial T0 = 300K

dU = \ c_{p}(T_{2} - T_{1})

Together, the equation can be written as a function of the new temperature

\boxed{T(t + \mathrm{\Delta}t) = T(t) - \frac{p\lbrack V\left( t + \mathrm{\Delta}t \right) - V\left( t \right)\rbrack}{c_{p}m_{\text{air}}}}

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