Force required to lift a water propelled rocket

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Discussion Overview

The discussion revolves around calculating the force required to lift a water-propelled rocket, focusing on the rocket's weight, the pressure of the water, and the dynamics of thrust generation. Participants explore various formulas and concepts related to thrust, mass, and nozzle efficiency, with an emphasis on practical application and experimentation.

Discussion Character

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

Main Points Raised

  • One participant seeks to determine the force needed for lift-off, questioning whether to use the weight of the rocket multiplied by gravity or another method.
  • Another participant suggests selecting a desired acceleration and using the formula F=ma, while also considering the mass of the water in the rocket.
  • Concerns are raised about the changing mass of the rocket as water is expelled, complicating accurate force calculations.
  • A formula is provided by a participant to calculate the pressure needed for lift-off, incorporating various parameters such as exhaust velocity and ambient pressure.
  • One participant calculates thrust force based on nozzle area and pressure, predicting a thrust of 146.4456 N.
  • Another participant calculates the force needed to lift the rocket without water, arriving at 2.1805 Newtons, but acknowledges the need to adjust for the water's mass.
  • Questions are raised about the efficiency of the nozzle and how to determine it, as well as the optimal nozzle size for performance.

Areas of Agreement / Disagreement

Participants express varying viewpoints on the calculations and methods for determining the necessary force for lift-off. There is no consensus on the exact approach or values, and multiple competing views remain regarding the influence of water mass and nozzle efficiency.

Contextual Notes

Participants note the complexity of accurately calculating lift-off force due to the variable mass of the water and the challenges in predicting nozzle efficiency. The discussion reflects a reliance on experimental testing to find optimal parameters.

Who May Find This Useful

This discussion may be useful for hobbyists and students interested in rocketry, fluid dynamics, and experimental physics, particularly those looking to understand the principles of thrust and lift in water-propelled rockets.

beamer09
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Hello, I need to figure out the force I require to lift the water rocket that I'm building. It weighs about three ounces or about .0850 kilograms to possibly make it easier. How much force will I need to provide lift-off. Would I take the acceleration due to gravity (9.8) times the weight of the object (which i think is correct), divide the weight by gravity, or neither.
 
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Follow up- When the water is coming out of the rocket at a force of 100 psi, is it a sufficient amount to provide lift-off?
 
What you might do is select a desired acceleration (say, 3 g's), then use f=ma to calculate the force, remembering to add the 1g it has while sitting still. Also, does your mass include the mass of the water?

For the force available, it is just the pressure times the exit area of the nozzle tines the efficiency of the nozzle (try 50%).
 
no, not the water that is why the answer wouldn't be one hundred percent accurate, i did not take into consideration the weight of the water. But, the water is constantly leaving the bottle and the weight will go down slowly causing it to be almost impossible to calculate a very accurate answer to the question because of this variable, correct?
 
Oh, wait nevermind i forgot that the question was how much force needed for lift, not how high it would go.
 
Now I understand your answer. Thank you very much. This will be a very big help.
 
How do I find the efficiency of the nozzle? By the effeciency of the nozzle, what do you mean?
 
One last quick question, how can I find the amount of pressure leaving the nozzle? In order to have the pressure escaping equal to the ambient pressure, I need to find out what the amount of pressure would be, in order to know if my nozzle is a sufficient one. The ambient pressure would be 100 psi in my case right?
 
beamer09 said:
Hello, I need to figure out the force I require to lift the water rocket that I'm building. It weighs about three ounces or about .0850 kilograms to possibly make it easier. How much force will I need to provide lift-off. Would I take the acceleration due to gravity (9.8) times the weight of the object (which i think is correct), divide the weight by gravity, or neither.

Here is a formula that can help you figure out how much pressure you need in order for the rocket to lift of the pad.
Fnet = m * Vexhaust = m * Ve - act + Ae(Pe - Pamb)

m = exhaust gas mass flow
Vexhaust = effective exhaust velocity
Ve - act = actual jet velocity at nozzle exit plane
Ae = flow area at nozzle exit plane
Pe = static pressure at nozzle exit plane
Pamb = ambient (or atmospheric) pressure

Subtract the Fnet from the weight of the rocket with the water as well which will look like this:
Fnet - the weight of the rocket = the thrust that you will need to lift the rocket.

You can then figure out the acceleration using Newton's Second Law which is:
Fnet = mass * acceleration
or
acceleration = Fnet/mass
 
  • #10
Well the mass of the rocket now is .2225 kg. I found the area of my nozzle which would be .2124 cm^2. Then i used the formula F= P*A to find the thrust force of my rocket which was 689.48 kpa that I am using, and the area of the nozzle which was .2124 cm^2, to get the answer of 146.4456 N. That is what I predict will be the thrust force of my rocket. Is this correct?
 
  • #11
And as far as the Newtons needed to lift the bottle, I used the formula F=M*A(or G). I got a Force of 2.1805 Newtons needed to lift the bottle without water. I can not use this answer yet because I need to find what my mass would be with water. But, am i on the right track and have the right formula's?
 
  • #12
Would a 2 cm nozzle work better?
 
  • #13
In terms of area.
 
  • #14
Looks fine - nozzle efficiency is tough to predict, but 50% is probably a reasonable estimate.

Mass with water is going to be arbitrary. Not enough and it'll empty quickly, too much and you'll be lifting too much weight. You'll need to do some testing to find the right balance.
 

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