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Task

The modelling of a Water Rocket's flight profile for a set of predefined variables.

TL;DR: I am wondering if my assumptions and mathematics are correct. If they are, this is merely an informative post.

Schematic

rocket

Concerns/Things to take into account:

As the rocket starts to lift off, (assuming that with the specs I have, it can), water is ejected through an opening at the bottom of the rocket. This causes a pressure decrease of the air in the tank and an increase in volume of the air. This in turn causes the force applied by the air to lessen over time and as a result, the acceleration of the rocket decreases.

Modelling

Subscript 1 and 2 represent the air and water, respectively. The physical dimensions of the rocket are described in the schematic.

Starting off with conservation of mass on the water over the control volume

$$ \frac{dm_2}{dt} = -\rho_2 v_2 A_2 \tag{1}$$

Applying Bernoulli's equation to the water:

$$ v_2^2 = 2gh_2 + \frac{2(P_1-P_{atm})}{\rho_2} \tag{2}$$

The height of the water in the rocket is given by

$$ h_2 = V_2/A_1 \tag{3}$$

It follows from Eq (1) that

$$ \frac{dh_2}{dt} = -v_2\frac{A_2}{A_1} \tag{4}$$

Differentiating Eq (2) with respect to time:

$$ 2v_2\frac{dv_2}{dt} = 2g\frac{dh_2}{dt} + \frac{2}{\rho_2}\frac{dP_1}{dt} \tag{5} $$

I will return to Eq (5) once it can be simplified.

Because the gas expands as the water leaves the rocket, the gas volume can be described by

$$ V_1 = V_3-V_2 \tag{6}$$

Differentiating with respect to time $$ \frac{dV_1}{dt} = \frac{dV_3}{dt} - \frac{dV_2}{dt} \tag{7}$$

Where $ V_3 $ is constant and $ V_2 = m_2/\rho_2 $. Substituting into Eq (7):

$$ \frac{dV_1}{dt} = - \frac{1}{\rho_2}\frac{dm_2}{dt} \tag{8}$$

Substituting Eq (1) into Eq (8) yields

$$ \frac{dV_1}{dt} = v_2 A_2 \tag{9}$$

Assuming that the gas undergoes adiabatic expansion,

$$ du = dW \tag{10}$$

where

$$ \frac{du_1}{dt} = C_{p_{1}}\frac{dT_1}{dt} \tag{11}$$

and $$ \frac{dW_1}{dt} = -P_1\frac{dV_1}{dt} \tag{12}$$

Substituting Eq (11) and (12) into (10) yields

$$ \frac{dT_1}{dt} = \frac{-P_1}{C_{p_{2}}}\frac{dV_1}{dt} \tag{13}$$

Then, substituting Eq (9) into (13) :

$$ \frac{dT_1}{dt} = \frac{-P_1}{C_{p_{2}}} v_2 A_2 \tag{14}$$

Now, I am able to find an expression for the $dP_1/dt$ term in Eq (5).

Assuming ideal gas law and differentiating it with respect to time with constant moles delivers:

$$ P_1\frac{dV_1}{dt} + V_1\frac{dP_1}{dt} = n_1 R \frac{dT_1}{dt} \tag{15}$$

$$\frac{dP_1}{dt} = \frac{n_1 R}{V_1} \frac{dT_1}{dt} - \frac{P_1}{V_1}\frac{dV_1}{dt} \tag{16}$$

Now, substituting Eq (9) and (14) into (16):

$$\frac{dP_1}{dt} = \frac{n_1 R}{V_1} (\frac{-P_1}{C_{p_{2}}} v_2 A_2) - \frac{P_1}{V_1}(v_2 A_2) \tag{17}$$

Returning to Eq (5) and substituting in Eq (4) and Eq (17):

$$ 2v_2\frac{dv_2}{dt} = 2g(-v_2\frac{A_2}{A_1}) + \frac{2}{\rho_2}(\frac{n_1 R}{V_1} (\frac{-P_1}{C_{p_{2}}} v_2 A_2) - \frac{P_1}{V_1}(v_2 A_2)) \tag{18} $$

Simplifying

$$ \frac{dv_2}{dt} = -g\frac{A_2}{A_1} - \frac{ n_1 R P_1}{V_1 C_{p_{2}} \rho_2} A_2 - \frac{P_1 A_2}{V_1 \rho_2} \tag{19}$$

Now, for the physical rocket, a force balance:

$$ F_r = F_g + F_D + F_T \tag{20}$$

Where $ F_g, F_D, F_T $ is the gravitational force, drag force, and thrust.

Gravitational force is given by

$$ F_g = -m_r g \tag{21}$$

where

$$ m_r = m_e + m_1 + m_2 \tag{22}$$

The empty mass of the rocket is given by $ m_e $ and is a constant. The mass of the gas inside the rocket is $ m_2 = n_2 MW_2 $ Drag is given by

$$ F_D = -\frac{1}{2}\rho_{atm} C_d A_c v_r^2 \tag{23}$$

The thrust force is a bit more complex:

$$ F_T = \frac{d (m_2 v_2)}{dt} \tag{24}$$

$$ F_T = m_2\frac{d v_2}{dt} + v_2\frac{dm_2}{dt} \tag{25}$$

Now, substituting Eq (1) and (19) into (25) yields:

$$ F_T = m_2 (-g\frac{A_2}{A_1} - \frac{n_1 R P_1}{V_1 C_{p_{2}} \rho_2} A_2 - \frac{P_1 A_2}{V_1 \rho_2}) + v_2 (-\rho_2 v_2 A_2) \tag{26}$$

$$ F_T = -g m_2\frac{A_2}{A_1} - \frac{ n_1 m_2 R P_1}{V_1 C_{p_{2}} \rho_2} A_2 - \frac{m_2 P_1 A_2}{V_1 \rho_2} -\rho_2 v_2^2 A_2 \tag{27}$$

Now, $F_r$ can be expanded to

$$ F_r = v_r \frac{dm_r}{dt} + m_r\frac{dv_r}{dt} \tag{28}$$

simplifying using Eq (22) where $ m_e $ and $ m_1$ are constant

$$ F_r = v_r \frac{dm_2}{dt} + m_r\frac{dv_r}{dt} \tag{29}$$

A new variable $F_{total}$ is used to describe the right hand side of Eq (20)

Substituting Eq (28) and $F_{total}$ into Eq (20) yields

$$ v_r \frac{dm_2}{dt} + m_r\frac{dv_r}{dt} = F_{total} \tag{30}$$

$$ \frac{dv_r}{dt} = (F_{total} + v_r \rho_2 v_2 A_2 m_r) / m_r \tag{31}$$

$$ \frac{d x_r}{dt} = v_r \tag{32}$$

The solution can be found by simultaneously solving Eq (1), (4), (9), (14), (17), (19), (21), (23), (27), (29), (30).

Questions and Concerns

Are the assumptions I made correct? Is the mathematics correct?

Future alterations to the model:

Non-adiabatic expansion

Joule-Thompson effect

Non-ideal gas

Non-ideal liquid

Non-ideal gas-liquid interface

Heat transfer through the rocket walls

Heating due to aerodynamic effects

Nozzle shape and design

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closed as too broad by tpg2114, sammy gerbil, Gert, user36790, John Rennie Aug 20 '16 at 9:55

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ This is an interesting topic and it's obvious that you've done your job of working the problem out yourself. Good job! However, on this site we generally consider "check my work" posts to be off-topic. It's certainly likely that this post will get closed for that reason. Can you find a specific question that you'd like to ask and narrow the post down to that? Asking "Is [sic] the mathematics correct?" is too broad and it's basically "check my work". $\endgroup$ – DanielSank Aug 20 '16 at 0:00
  • $\begingroup$ Would asking how I would implement my "future alteration" qualify as a proper question? $\endgroup$ – 22134484 Aug 20 '16 at 0:23
  • $\begingroup$ Well, that's still pretty vague. You asked the mathematics were correct: is there a specific step or concept that you're wondering about? $\endgroup$ – DanielSank Aug 20 '16 at 1:21
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    $\begingroup$ Every single one of your alterations is already a pretty big question on its own. Several of them are likely a good fit for this site on their own, though some others would likely be closed as engineering, and get a better answer here. $\endgroup$ – knzhou Aug 20 '16 at 3:44