Force required to produce wave in tank

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SUMMARY

The discussion focuses on calculating the force required to move a flap in a tank to produce waves, utilizing two potential systems: a pneumatic system and a motor and cam system. The calculations provided indicate that the force required is approximately 4602.4 kN, derived from the mass of displaced water and the height of the water when the flap is extended. The participant expresses uncertainty about the application of the force and the incorporation of pressure equations. Feedback suggests considering the work-energy theorem for clarity in calculations.

PREREQUISITES
  • Understanding of fluid mechanics principles, specifically related to wave generation.
  • Knowledge of force calculations using Fxd = mgh.
  • Familiarity with pressure equations, particularly P = F/A.
  • Basic principles of the work-energy theorem in physics.
NEXT STEPS
  • Research the application of the work-energy theorem in mechanical systems.
  • Learn about pneumatic systems and their force calculations in fluid dynamics.
  • Explore motor and cam systems for mechanical wave generation.
  • Investigate advanced fluid dynamics concepts related to wave behavior in tanks.
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Engineers, physics students, and hobbyists involved in mechanical design and fluid dynamics, particularly those interested in wave generation systems.

Whatamiat
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Homework Statement


I am building a tank to produce waves.
I have two options to power the flap that will be produced to create the waves:

1. Pneumatic system
2. Motor and Cam system

Before I can order motors or pneumatic air cylinders I need to know what force will be required to move the flap the required distance to produce the waves.

Sketches of both systems are attached
Sketch of problem is attached

Homework Equations



Fxd = mgh

P= FxA ?

The Attempt at a Solution



From attached images:

Mass of Water = 7500kg

Finding Distance:
d = tan 18 x 1 = 32.5cm

Finding mass of displaced water:
d x h x w = .325 x 1 x .75 = 243.75kg
but we are dealing with triangle so
mass of displaced water =243.75/2 = 121.875kg of water displaced

Calculating Height of water when flap is extended:
Mass of water = length of tank x width of tank x height of tank
7.5 = (10-.1625) x .75 x h
h = 1.0165 cm

Calculating Force:
d = 32.5/2 = 16.25 (as we are dealing with triangle not rectangle)

F x 16.25 = 7500x9.81x1.0165

F = 4602.4kN

My Thoughts
Very sure I am missing something fundamental with this solution but not sure what!

I feel I have ignored the fact that the force is being applied at 1.25m not 1m but don't know how to incorporate this.

Im also unsure if I have used the wrong figures for my Fxd=mgh solution.

I feel P=F/A can be incorporated but not sure how?

Thanks for any feedback!
 

Attachments

  • wave generater pneumatic.JPG
    wave generater pneumatic.JPG
    16.1 KB · Views: 482
  • wave generater.JPG
    wave generater.JPG
    18.3 KB · Views: 431
  • detailed sketch.JPG
    detailed sketch.JPG
    15.6 KB · Views: 538
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Hi Whatamiat! :smile:

Are you using the work-energy theorem? (it would have been clearer if you'd said so)

If so, in the formula work done = force "dot" displacement, the displacement is the displacement of the point of application of the force. :wink:
 

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