Calculating Water Rocket Performance: Thrust, Velocity & More

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Discussion Overview

The discussion revolves around calculating the performance of a water rocket, focusing on parameters such as thrust, velocity, acceleration, momentum, and maximum height. Participants explore various factors that may influence these calculations, including gravity, wind velocity, and the physical properties of the rocket and water. The conversation includes theoretical modeling and practical considerations for a physics class project.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant outlines initial calculations for thrust and height, questioning the influence of air volume and density on these parameters.
  • Another participant suggests a revised expression for height, indicating a potential improvement in the model.
  • Discussion includes the need to account for changing thrust and mass as air expands and water is ejected.
  • Several participants propose using conservation of momentum to approximate flow rate and thrust, while others challenge the assumptions made regarding viscosity and drag.
  • Numerical methods are suggested for integrating the equations governing rocket motion, with emphasis on the complexities involved in calculating altitude.
  • Participants express uncertainty about the best approach to model the dynamics of the rocket, particularly regarding the treatment of air pressure and water ejection.

Areas of Agreement / Disagreement

There is no consensus on the best approach to model the water rocket's performance. Participants present multiple competing views on how to account for various factors, including thrust, drag, and the effects of air pressure and volume. The discussion remains unresolved with differing opinions on the significance of certain variables.

Contextual Notes

Participants acknowledge limitations in their models, such as the neglect of viscosity and the complexities of integrating equations analytically. The discussion highlights the dependence on assumptions regarding the behavior of air and water under varying conditions.

Who May Find This Useful

This discussion may be useful for students and enthusiasts interested in experimental physics, fluid dynamics, and engineering principles related to rocket design and performance analysis.

  • #31
mfb said:
Your initial V_air is V0, and it increases to V0+Vf (assuming the pressure is enough to eject all water). Therefore, the total volume is V0+Vf. I fixed this, which gives the formula V0=Vtotal-Vf with the total volume Vtotal.

No, it increases to Vf. Have you found something in the spreadsheet that says otherwise?
 
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  • #32
Oh... sorry, I got confused by the decimal places.
You are right.
 
  • #33
P(t) or P0 is the pressure of the air inside the bottle? If so, that is obtained by the amount of psi used in the experiment, correct, which in my case, will vary from 50 psi to 65 psi (but in pascals)?
And how is the volume of the air, V0 obtained?

I apologize for all the questions, I just want to be completely clear with all the steps used.
 
  • #34
P(t) or P0 is the pressure of the air inside the bottle? If so, that is obtained by the amount of psi used in the experiment, correct, which in my case, will vary from 50 psi to 65 psi (but in pascals)?
Right

And how is the volume of the air, V0 obtained?
The volume of the bottle minus the volume of the water inside.
 
  • #35
Okay, so haruspex's spreadsheet is fine except that I need to fix the time and change the variables to the ones I need (mass, etc).

Edit: 2 L is 0.002 cubic meters and the volume of the water is 0.001 cubic meters and that gives me 0.001 cubic meters. When I change V0 to that, all the other variables because the same.
 
Last edited:
  • #36
Help. Anyone? This is due in a few hours and I don't know how you got the volume of air so low when I get 0.001 cubic meters and that ruins the whole spreadsheet as it must be, apparently, at least 0.0001 cubic meters to take any effect.
 
  • #37
Alcubierre said:
Help. Anyone? This is due in a few hours and I don't know how you got the volume of air so low when I get 0.001 cubic meters and that ruins the whole spreadsheet as it must be, apparently, at least 0.0001 cubic meters to take any effect.
This is probably too late. I plugged in Vf = .002, P0 = 4*atmospheric ("=4*I2"), M = 0.5, A = 0.002 (pi*.025^2). I played around with V0 to maximise height. I got nearly 5m at V0 = 0.00113. But go just a fraction over that and you lose big time.
Not sure about P0. If your gauge says 45 psi, is that 3 atmospheres, or 3 atmospheres in addition to the background 1 atmosphere? I'm assuming it's the second.
 

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