# Modeled Data

## Sample Calculations

### Calculations & Formulas

Excel Column Letter — Value description (Bottle Mechanics)
Equation
Excel formula taken at t = 0.001

H — Air volume (2.4) $V_{a}(t + \mathrm{\Delta}t) = V_{a}\left( t \right) + \mathrm{\Delta}t\ \left\lbrack A_{n}\sqrt{2\ \frac{p_{0}\left( \frac{V_{0}}{V\left( t \right)} \right)^{1.4} - p_{\text{atm\ }}}{\rho_{w}}} \right\rbrack$
=H2+$B$2*$B$6*SQRT((2*($B$8*($B$3/H2)^(1.4)-$B$5))/($B$9))

I — Thrust (2.4) $F_{t}(t) = 2A_{n}\left\lbrack p_{0}\left( \frac{V_{0}}{V\left( t \right)} \right)^{1.4} - p_{\text{atm\ }} \right\rbrack$
=2*($B$8*(($B$3/H3)^(1.4))-$B$5)*$B$6

J — Drag (3.0) $D\left( t \right) = C_{d}\left( \frac{1}{2}\rho_{\text{air}}A_{c}{v(t)}^{2} \right)$
=$B$12*0.5*$B$10*$B$7*K3^2

K — Velocity (4.0) $v\left( t + \mathrm{\Delta}t \right) = v\left( t \right) + a(t)\mathrm{\Delta}t$
=K2+$B$2*L2

L — Acceleration (4.0) $a\left( t \right) = \ \frac{2A_{n}\left\lbrack p_{0}\left( \frac{V_{0}}{V(t)} \right)^{1.4}{- p}_{\text{atm\ }} \right\rbrack}{\left( V_{\text{total}} - V_{\text{air}}(t) \right)\rho_{w} + m_{b}} - g - \frac{C_{d}\left( \frac{1}{2}\rho_{\text{air}}A_{c}{v(t)}^{2} \right)\ }{\left( V_{\text{total}} - V_{\text{air}}(t) \right)\rho_{w} + m_{b}}$
=(I3-(($B$9*(0.002-H3)+0.7)*$B$11)-J3)/(0.7+$B$9*(0.002-H3))

M — Weight $W\left( t \right) = \rho_{w}V_{w}\left( t \right)g$
=$B$9*R3*$B$11

N — Mass of water $m_{w}(t) = \rho_{w}(0.002 - V_{w}\left( t \right))$
=($B$9*(0.002-H3))

O — Air pressure (2.4) $p(t) = p_{0}\left( \frac{V_{0}}{V(t)} \right)^{1.4}$
=O2*(H2/H3)^(1.4)

P — Temperature of air (5.0) $T(t + \mathrm{\Delta}t) = T(t) - \frac{p\lbrack V\left( t + \mathrm{\Delta}t \right) - V\left( t \right)\rbrack}{c_{p}m_{\text{air}}}$
=P2-((O3*10^-3)*(H3-H2))/($B$13*$B$15)

Q — Height of bottle $h\left( t + \mathrm{\Delta}t \right) = v\left( t + \mathrm{\Delta}t \right)\mathrm{\Delta}t + h(t)$
=K3*$B$2+Q2

R — Volume of water $V_{w}(t) = 0.002 - V_{a}(t)$
=0.002-H3

S — Exit velocity $v_{e}(t) = \sqrt{2\ \frac{p_{0}\left( \frac{V_{0}}{V(t)} \right)^{1.4} - p_{\text{atm\ }}}{\rho_{w}}}$
=SQRT((2*(O3-$B$5))/$B$9)

T — Gauge pressure $p_{\text{in}}(t) = \frac{p_{\text{air}}(t) - p_{\text{atm}}}{1000}$
=(O3-101300)/1000

## Validation

Our model is able to give results that resemble reality closely, except the temperature drop inside the bottle. This may due to the assumption that the bottle is isolated and no heat transfer happens between the bottle and environment while in reality the effect of heat transferring is not negligible.

When the initial volume of water is 1.2 L and initial air gauge pressure is 172 kPa, the prediction from Hydrodynamics and thrust characteristics of a water-propelled rocket  gives:

The outputs of our model:

When the initial volume of water is 1.2 L and initial air gauge pressure is 275 kPa, the prediction from Hydrodynamics and thrust characteristics of a water-propelled rocket  gives:

The outputs of our model:

When the initial volume of water is 0.83 L and initial air gauge pressure is 172 kPa, the prediction from Hydrodynamics and thrust characteristics of a water-propelled rocket  gives:

The outputs of our model:

When the initial volume of water is 0.83 L and initial air gauge pressure is 172 kPa, the data from A more thorough analysis of water rockets: Moist adiabats, transient flows, and inertial forces in a soda bottle :

The output of our model: