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testing and models

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mmacomber

Mechanical
Oct 9, 2004
1
what effect does scaling up a miniature model, to test it, have on the test specifications, for example if something is supposed to be tested at 5 psi what would a 5x model have to be tested at? Is there a specific science that deals in this area?
 
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try searching on similitude and fluid dynamics
 
Similitude indeed. Buckingham Pi Theorum is another search term to use.

Best regards,

Matthew Ian Loew


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My fluid mechanics book from college covers the topic fairly well from a fluid mechanics standpoint. As the posts above indicate, you approach the problem by using nondimensional variables like Reynolds number and drag coefficient instead of velocity and force.

This approach doesn't necessarily clue you in to everything that might be going on in a model. For example, you might make a small scale model for fluid flow, and find that surface tension is important on the model but isn't on the real object. If you've ever seen a movie where they film model ships, you'll notice the drops become huge relative to the ship- you can't scale everything down!

In your case, pressure testing an object, I would be concerned with whether the actual mechanical properties of the model and the real object were identical.

If you have some form of approximate analysis for the item, the test would help to confirm the reliability of the analysis.

Seems like I remember reading long ago that the Wright brothers improperly scaled their wind tunnel tests and wrongly concluded that a "thin" wing was better than a "thick" wing.
 
Hi mmacomber

Dimensional analysis and similiarity are the subjects to look for in books or on the net.
However if your pressure testing somthing at 5psi and you want to scale the model up by 5x then what you need to do is ensure that say the hoop stress in the scaled up model is equal to that of the actual component.

example:- take a thin walled cylinder of 20mm i.d and a
wall thickness of say 0.5mm and an internal
pressure of 10N/mm^2.Then scale up the cylinder
by 5x which gives an i.d of 100mm and a wall
thickness of 2.5mm


now hoop stress in this cylinder is equal to:-

stress= P*r/t
where P= pressure

r= internal radius of cylinder

t = wall thickness

so using the above formula and the actual component cylinder I calculate a hoop stress of 200N/mm^2.

now in the model I require the same stress therefore


hoop stress = P*r/t so transpose the formula to find P


and P= hoop stress * t/r

therefore P = 200*2.5/50 = 10N/mm^2

so for a thin walled cylinder modelled at a scale of 5x there is no change to the internal pressure to that in the component. Note this is a very simple example and from your post I have no idea what geometrical shape your pressure testing. If you are testing a thick walled cylinder or a rectangular shaped vessel then the pressure to simulate the same stress within that vessel would probably be different when the component is scaled up.

regards desertfox
 
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