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fluidpower1

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Join Date
Jan 2005
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Newburgh, Indiana
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Here are some problems I use to reinforce the teaching in my Fluid Power classes that should prove interesting to the posters on this Forum. The Teasers are in Chapter 20. Take a look first at Problem 13 since it is one that seems be hardest for most mechanical types to figure out.

Unfortuanately the answers are posted also, But, decide your answer before looking at the posted answers.

http://www.hydraulicspneumatics.com/200/eBooks/Article/True/75038/

The difference in the answers from Electrical forum posters as related to the Mechanical posters should be interesting.

I hope I have not put this up before. If I did I apologize in advance. I can blame it on Old Age and the forgetfullness that accomapnes it if I did.
 
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The electrical guys don't always do that well - the ones who do I think are the ones who take it upon themselves to learn it, whether proactively or out of necessity.

I used this one on one of our EEs and MEs to illustrate a principle a couple of weeks ago. The ME got it right away. The EE made a couple of wild wrong guesses, and then didn't believe the correct answer when it was provided.

750382015jpg_00000048055.jpg
 
My guess on the first one would be that A had 0 or close to zero and B would have 1000 psi. My math is 10000/10.
Not sure what the pressure would be in the second condition, but I believe the weight would push the pistion down to the bottom of the cylinder. I assume you would have some kind of pressure reading until the pistion bottomed out.I would this A and B would be equal.

So how close am I?
 
I've no formal hydraulic training but it appears to me that the "condition 2" piston travels to the bottom and the pressure on those gauges is zero (or, some other equal pressure top and bottom).

What say you?

Bill
 
10,000 pds over 10" would be 1000 pds per square inch, or 1000psi the pressure above where there is no weight on would remain 0 unless the pressure was put on the cap side then the pressure would change but in this problem rod area is not needed because all of the pressure is going to the 10", Rod must be made of pretty good materials to not buckle with that pressure on it. I hope this is right someone correct me if I am wrong
 
Clay B. said:
My guess on the first one would be that A had 0 or close to zero and B would have 1000 psi. My math is 10000/10.
Click to expand...
Correct.

Not sure what the pressure would be in the second condition, but I believe the weight would push the pistion down to the bottom of the cylinder.
Click to expand...
In order for the piston to go to the bottom of the cylinder more rod must enter the cylinder, rod displaces volume, and since there is no where for the volume to of oil displaced to go, then the piston cannot sink to the bottom, so something must change. What?

I assume you would have some kind of pressure reading until the pistion bottomed out.I would this A and B would be equal.

So how close am I?
Click to expand...
A = B. The pressure in the cylinder is dictated by Pascal's law. Hint: What is the effective working area?
 
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Problem #6 from Bud's brain teasers is related to principles in another recent discussion we have had here, does flow make it go, or does force make it go?
 
Alaric said:
Correct.

In order for the piston to go to the bottom of the cylinder more rod must enter the cylinder, rod displaces volume, and since there is no where for the volume to of oil displaced to go, then the piston cannot sink to the bottom, so something must change. What?


A = B. The pressure in the cylinder is dictated by Pascal's law. Hint: What is the effective working area?
Click to expand...

It would have to be the 1 square inch of the piston, so 10,000PSI on A and B.
 
Alaric said:
For your prize you get to get off work in 30 mintues and then you can buy yourself a beer. 🍺
Click to expand...

Well, heck, that's an hour later than I usually get off (er, I mean, leave work).

But really, I can't take credit. I missed the whole piston displacing fluid issue. Once you gave that hint, it was clear. Fluids, other than alcohol based, are not my forte ;)
 
That is a neat problem that I heard in 1955 from a Maintenance Man who had been through the Vickers Hydraulic course. I would have bet the Farm on the Cylinder dropping when the seals blew and never really understood the fact that a 1 Sq.In. Rod could hold up 10,000# until I started selling Fluid Power in 1966 and had some intensive training over a 4 week period as a new Fluid Power salesman.

That's when I found out most Fluid Power circuits are designed by the salesman and SALES was the least important part of the job.

Te other Brain Teasers came from classes and from situations on the job and were used in classes in Adult Evening and in plant situations.
 

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