Please check my thinking

I want to calibrate several pressure guages. These are low pressure gauges, mostly 15 PSI max but a few up to 200 PSI. So I want to make a chamber that the gauges screw into and is filled with water. To pressurize I will have a tube sticking straight up with an ID just a little over the size that would make an area of .1 square inch. Then I will use a piston with a cross sectional area of exactly .1 square inch to pressurize the chamber. The little amount of water that leaks out past the piston will provide the seal. I don't want to use a lip seal because it will have a little drag. My thinking is that the piston, not the tube, needs to be exactly .1 square inch. If I used a lip seal then the tube would need to be exactly .1 square inch. Is my thinking on this correct? I will be using a lever of some sort to apply pressure to the piston. I want to use the small diameter piston to reduce the weight I will need to apply to the piston. Thanks, Eric

Reply to
etpm
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Reply to
Ignoramus25526

I will be using a lever of some sort to

I do not see anything wrong with your thinking. But here are a couple of i deas you might consider.

Make the cylinder and piston somewhat bigger and make your lever so you can hang a weight on it. I would use a brake cylinder and make the lever so the numbers came out nice. Maybe have two holes for attaching weights and have a 1 lb weight on one hole create 1 psi. and a 1 lb weight in the othe r hole create 10 psi. So a 5 lb weight in one hole and a 1 lb weight in th e other would create 15 psi.

Anything is easy for the guy not doing the work.

Dan

Reply to
dcaster

That's a pretty good description of a dead weight tester.

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I agree with your reasoning re piston area. If you'd like to buy the piston & cylinder, consider these products:

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Reply to
Ned Simmons

What you are designing is, I guess, a "dead weight tester". The ones I have used had a set of weights that were calibrated with the area of the piston, i.e., the piston probably was not an exact 1 inch, or whatever, in area. Example, a 1/2" piston with a 1 pound weight...

The systems I used were pressurized with a separate piston pump and when the dead weight tester weight lifted, a valve between the pump and the system was closed, maintaining pressure in the system. The dead weight tester simply acts as the "gauge".

The tester was a cylinder with a ground I.D. finish fitted with a piston, actually a rod, with a ground O.D. The cylinder is vertical and the piston drops in and weight sit on the top of the rod. Obviously the weight of the rod is calculated into the equation.

By the way, pressure in a closed vessel is equal throughout the vessel, regardless of what it contains.

Reply to
John B.

Thanks for the links Ned. I will be making my own piston and cylinder. I can hold a tolerance of less than .0001 and that will be accurate enough for me. Eric

Reply to
etpm

On Thu, 13 Feb 2014 11:56:14 -0600, Ignoramus25526 wrote: Thanks.

Reply to
etpm

Greetings Dan, I am going to use a lever with weights. I have some nice accurate wieght sets and scales. By using a small piston I can keep the weights small. I have had gauges checked for my pressure cooker several times. I do so once a year. But to do so I need to make a two hour round trip. This is what prompted me to think about making the tester. Eric

Reply to
etpm

Greetings John, I know about the pressure being equal. I assume you said that because I mentioned filling the thing with water. I chose water mostly because it's cheap and safe. And the viscosity is such that it will leak slowly past the piston, providing the seal, without being so viscous that it takes a long time for air bubbles to escape from the gauges under test. Eric

Reply to
etpm

As others have said, right...and as others noted, static pressure is uniform in the vessel. I only note the accuracy is dependent on the incompressibility of the working fluid. Water, at these pressures, is a excellent approximation for the purpose albeit it isn't perfect....well, let's see, the tables are just over here on the shelf--for the difference between 200 and 15 psia at 70F, the change in density is ~0.05%. Probably close enough... :)

Reply to
dpb

I'm surprised it's even that much. Lessee, 1 lb divived by 100 equals .01 divided by 100 equals .0001 x 5 equals .0005. So I will have an error of .0075 lbs at 15 lbs pressure. I think I can live with that. Eric

Reply to
etpm

Well, I just said that as a reminder....:-)

I've no idea what tolerance the piston and cylinder was on the "dead weight testers" that we used. There was certainly no problems with inserting the piston into the cylinder as there is on some of the high pressure hydraulic valves that don't use seals and no exotic instructions about greasing things with rendered elephant fat or anything. We just took it out of the box, connected it to the system being tested and pumped until the weights floated and you could spin them freely.

Piping systems were usually pressure tested for a 24 hour period and there was no noticeable leakage at the tester during that period using plain water as the test medium.

The highest test pressure I have used was 4,000 PSI on a 10 Km x 4 inch gas pipeline for 24 hours with no problems with the dead weight tester.

I believe that lever systems were not used with dead weight testers as it would have been simply one more unknown in the equation; is the lever ratio exact... is there any friction in the pivots... Of course lever operated dead weight pressure relief valves were common.

Reply to
John B.

But if you are using an air over water system to avoid getting water into the gauge and you read, say 500 psi, of water pressure and glance over at the air pressure gauge I'll bet it will be reading 500 psi also :-)

Reply to
John B.

On 2/14/2014 7:26 PM, John B. wrote: ...

Indeed...it came to me later that his system doesn't depend on knowing displacement where is the place the compressibility comes into play.

Reply to
dpb

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's_law "Pascal's principle applies to all fluids, whether gases or liquids."

Reply to
Jim Wilkins

Well, yes, but that wasn't the point of the problem -- I was thinking about a displacement of the cylinder but in fact he doesn't need to know for his purpose what that is...I wasn't arguing a pressure change (albeit I didn't say that explicitly, granted, but it _is_ what I was thinking -- and that's my story and I'm sticking to it! :) ).

Reply to
dpb

Your thinking is sound.

The vertical force (including weight) on the piston must be matched by pressure * piston area. If this were not so, the piston would accelerate.

A minor amount of leakage will introduce a minor amount of error as the liquid passing upwards around the piston exerts a vertical force due to hydrodynamic drag, which will be additive to the force due to pressure * area. This leak rate will obviously be a function of both pressure and the area available for leakage -- and a whole bunch of other stuff too, but bottom line: with tolerances tight enough to make the drop rate slow, the velocity of the bypassing liquid will be low which will minimize the error force due to hydrodynamic drag.

Reply to
Don Foreman

Greetings Don, Haven't corresponded with you for quite a while. Thanks for the cogent reply. I sincerely hope all is well with you. Cheers, Eric

Reply to
etpm

And after all that, KISS kicks in - Just buy one Lab Calibrated 4" gauge and tee it in permanently - they're cheap if you aren't picky about looks, back or bottom, panel or free-standing, etc. Then you cross-check with the mystery gauge you're testing.

Reply to
Bruce L. Bergman (munged human readable)

This looks practical for tire gauges if you have a tall tree you can toss a line over:

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You could fill the hose from the ground through a wye valve until water spills out the top end, where you attached the tape measure. WalMart sells RV blowout plugs with a Schrader valve and hose thread.

Practice your explanation in advance, for when the neighbors congregate to watch and snicker. Mine just mumble about MacGyver.

jsw

Reply to
Jim Wilkins

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