[sol]

So a leaky piston moves faster?
I am definitely missing something.
The fluid still has to fill the volume below the piston for the piston to move...

[skippy]

To get your mind around it, pretend the piston was removed altogether. Whether I have one hole or 50 holes in the piston, the end result is the same. (providing of course that in the case of one hole, it is not too small and becomes a restriction) It's as if the piston wasn't there so all you have is a cylinder and a rod (minus piston). The pressure then acts upon nothing more than the diameter (area measured in sq/ins) of the cylinder rod which in this case is 0.5" which equals 0.1964sq/ins. The end result of this imaginary (B) cylinder is a unit that is fast but is a lot less powerful.
The original cylinder is your everyday two-way unit (push/pull) but by drilling a hole in it it became a .......????? (Garg69, help me here with a name) displacement (??) cylinder of the type that you see in a forklift (the main lift cylinder). Not exactly but almost.
Hope this helps

[sol]

Here are some options.

The flow rate in the left cylinder is 4.2 cubic inches per second.
To lift the 50 lb. weight the pressure is 7 psi. on the piston bottom.
Atmospheric pressure is 14+ psi.
With the port closed the trapped air cannot exit.
The piston is not moving because fluid cannot even get into the cylinder.

Or maybe the pump is putting out much more pressure than needed but the flow rate is still the same 4.2 cu in/sec to lift the "A" load in 10 seconds.
The entering fluid has to compress the trapped air.
At first the fluid will raise the piston a bit, as it fills the cylinder, the air will compress. This pressure will build the further the piston raises.
The pressure needed to raise the 50lbs on the 0.5" rod is 250 psi. With the cylinder outlet closed that air will eventually compress to 250 psi but its compression slows the speed of raising the piston. The exact answer requires more math than I want to deal with.

Or there is nothing in the cylinder, a vacuum. The fluid enters at 4.2 cu.in./sec and fills both sides of the piston before it begins to raise the load. Filling The entire volume will take appx 10 seconds, now that the fluid has something to push against it will begin to lift the load with a pressure of 250psi. So it takes 10sec + 0.28 sec to raise the load.

Or there is fluid on both sides of the piston to begin with. The flow rate is 4.2. cu.in/sec.
The rod displaces 1.2 cubic inches. It will take 0 .28 sec to raise the load as the fluid above the cylinder leaks into the lower section through the drilled hole.

And that is why I despise these questions on standardized tests. Too many assumptions.

Now where is my barometer? I have to find the height of a building

[cp8071]

The difference between the two systems is called "double-ended" and "single-ended". In a double ended system the ram can be pushed either direction by controlling which side of the piston has pressure on it. In a single ended system the ram can only be extended (or retracted depending on the connections). The forklift cylinder is a great example of this type. It will always be pressing up on the load, to lower the load the cylinder is allowed bleed down slowly. But normally double ended cylinders are used in single ended applications so that the bore diameter gives greater force then that of the shaft diameter. In this type configuration the unused end will be vented back into the holding tank so that any bypassed fluid will return to the system.

For those of you who don't think cylinder "B" will move; How do you brush your teeth in the morning without toothpaste? Because squeezing that tube and having a small cylinder of paste come out is the same system as this.

CP8071

[skippy]

CP thanks for the excel table. That makes life easy. Here I sit in my office with one of those evironmently friendly light bulbs attempting to use my solar powered calculator and every time I do a calculation I have to stand up and put the calculator close to the light to be able to see the answer. Arghhh... you make me sick! hehehe Hey, brushing my teeth will never be the same again. :cheers:

Hi Sol.
I was meaning that the cylinder was purged of all air BUT it doesn't matter if it is or not as it won't affect the equation to any great extent.

"To lift the 50 lb. weight the pressure is 7 psi"
Yes but zero pressure on a pressure guage (i.e. 0 psi) is actually one atmosphere which is 14.7psi. So the 7psi would actually be 21.7psi absolute but very few industries use absolute measurements.

"Or maybe the pump is putting out much more pressure than needed but the flow rate is still the same 4.2 cu in/sec to lift the "A" load in 10 seconds."
Not this one because in my example, pressure is only a function of flow rate and the resistance (a 50lb load) applied against that flow rate.

"Or there is nothing in the cylinder, a vacuum."
You won't find a vaccum in a hydraulic cylinder unless someone put it there in the first place in which case I would have mentioned it"

If the cylinder (B) had no hydraulic oil in it but only air (at atmospheric pressure) and I introduced hyd. fluid at the flow rate mentioned, the pressure to shift the load would still be the same except it would take slightly longer to achieve full travel because of the extra oil required to occupy the reduced space occupied by the compressed air. It's now 2:30am here so I'll work it out tomorrow but I can tell you it won't be a big difference.

Or there is fluid on both sides of the piston to begin with. The flow rate is 4.2. cu.in/sec.
The rod displaces 1.2 cubic inches. It will take 0 .28 sec to raise the load as the fluid above the cylinder leaks into the lower section through the drilled hole.

Yes this one is what I was talking about.

[trubleshtr]

A good example of Boyles law.

[sol]

Hey, Thanks Skippy!

Okay, my answer #1 was just me being a pedant to make a point, psig vs. psia.

Your answer to my Answer #2 "Not this one because in my example, pressure is only a function of flow rate and the resistance applied against that flow rate."
That's not how I read it. "...identical pumps and then I apply X pressure and Y flow ..."

No vac? Fair enough.

"If the cylinder (B) had no hydraulic oil in it but only air (at atmospheric pressure) and I introduced hyd. fluid at the flow rate mentioned, the pressure to shift the load would still be the same except it would take slightly longer to achieve full travel because of the extra oil required to occupy the reduced space occupied by the compressed air. It's now 2:30am so I'll work it out tomorrow but I can tell you it won't be a big difference."

It'll just be like brakes that need to be bled....

Thanks for the puzzle, took way too much of my time.

Still looking for that barometer.

[sbrpollock]

Now for a real example of this situation:

A few months ago I was working with a couple of millwrights to change an expanding mandrel in a "Decoiling" machine. This is not exactly the kind of mandrel familiar around here. This mandrel is twenty something feet long, twenty four inches in diameter, and weighs five or six tons. It fits into a taper and is expanded and collapsed by a double action hydraulic cylinder mounted on the back of the machine. This cylinder has a ten inch stroke, twelve inch piston and a six inch rod. It is fed by two 2" hydraulic lines (One on each end) at 700 pounds of pressure. I'm not sure what the available flow is, but trust me when I say it's plenty! (About 700 hosepower worth of pumps pumping from a resevoir of about 6000 gallons of fluid) The cylinder is rated for 1000 psi.

The standard procedure when changing this mandel is to remove all the lock nuts from the back end and use this cylinder to push the mandrel out of the taper. This particular time though, this cylinder didn't have the balls to break the mandrel loose with just 700 psi. Since we couldn't turn up the pressure because this system feeds many other parts of the plant, we decided to bring in a portable electric pump.

We took off the two 2" lines and hooked up a small high pressure pump to the back end. We left the rod end opened to drain into the containment pit below. When we started pumping with this portable pump we realised that everything we pumped in the back end just ran right out the rod end. The seals on the piston were bad. Our portable pump could build lots of pressure but it was a low volume pump and with the piston seals bad, everthing we pumped in just leaked right past the piston!

This combination of leaking piston combined with the low volume high pressure pump simulates your piston with the hole drilled in it.

After scratching our heads for a while, one of the millwrights decided we needed to PLUG the rod end of the cylinder! I could see where this was going and started to look for things to hide behind.

They plugged the rod end and started pumping into the back end. Of course with the leaking piston seals we were building equal amounts of pressure on both sides of the piston and the mandrel wasn't budging. The small amound of thrust we were getting on the six inch rod sure wasn't going to accomplish anything.

Remember I said this cylinder was rated for 1000 psi. Well, when we hit about 5200 psi on our portable pump, the seals, retaining rings and everything else just EXPLODED out the rod end of the cylinder! (Even part of the casting itself had blown out)

Now picture this: We have built 5200 psi on both sides of the piston and the rod end of the cylinder blows. This realeased the pressure on the rod end almost instantaneously, leaving the back side with 5200 psi in it. Since this couldn't leak past the piston seals as fast as the rod end seals had blown out, it created a tremendous amount of thrust momentarily.

This breif burst of thrust was enough to not only break the mandrel loose from its taper, but it also shot this five or six ton mandrel about five feet out of the front of the machine. Luckily there was a forklift sitting in the front of the machine that stopped it from flying all the way out! It did bend the roll cage when it hit!

Even though I was hidding behind another part of the machine when this cylinder blew, I was still literally soaked in hydraulic oil as was every one else in the vicinity. It was pretty spectacular.

The worst part of the whole thing was that now, in addition to putting in the new mandrel, we had to change this hydraulic cylinder.

[ghyman]

Piston on the right may wiggle a little initially, but nothing else.

[FranH]

Since there is no mention of a pressure relief valve in the system and it appears the 50# is a greater distance above the cylinder than the piston is to the bottom of the cylinder, we will have the piston bursting thru the bottom or the top of the cylinder bursting into the weight. Cylinder 'A' is a moot isue.

[sbrpollock]

ghyman & Franh:

The correct answer is explained in post #19.

[skippy]

Sol: To further answer your questions in post #23. Being a hydraulic cylinder I had meant that it was full of oil however, even if cylinder (B) had no oil in it, just air at atmospheric pressure, it would still generate 250psi in the lifting process. So that’s 17.86 times atmospheric pressure which we’ll call 18 times. Excluding bringing relative humidity into the equation, the amount of air (= internal volume of the cylinder) will be compressed to 1/18th of the size so the equation we need is: 17/18ths of cylinder volume plus rod displacement = the amount of oil needed for full travel = 17/18 x (42.4305cu/ins + 1.178625cu/ins + 0.03143cu/ins) (The last figure is the volume of the small hole in piston) = 17/18 x 43.64cu/ins = 41.22cu/ins of oil required to achieve full travel. So if it takes 10secs to move 42.4305cu/ins (achieve full travel with cyl (A)), then it should take cylinder (B) (with no oil in it which needs 41.22cu/ins of oil required to achieve full travel lift) a total of 9.7seconds to do its job. It may (possibly) initially get a bounce up but will do its job. Someone please correct me if I´m wrong when I say that for every atmosphere gained the volume occupied by the air will be halved (excluding relative humidity corrections). I’m not sure of this point.

In post #27 you said “That's not how I read it. "…identical pumps and then I apply X pressure and Y flow” It’s true I said that but in post # 6 I said “we'll forget about the pressure and just say that the hydraulic pump driving this thing has the same flow rate (gpm) for both cylinders.” When I said that I should have also edited the original post to say the same thing. Sorry!

Re the cylinder with no oil you said “It'll just be like brakes that need to be bled....” Well sort of in that it would become spongy but not really in that: The brakes won’t work properly only because the master cylinder runs out of travel. If you had air in the disc callipers or wheel cylinders and the master cylinder had double or triple the travel, full pressure would be obtained and the car would stop normally.

2/ Patrick: Yeouch!!!

3/ FranH I purposely let the pressure relief valve out of the equation not to confuse things but in this instance if I had one and had it set to say 500psi, the rod would extend as normal and when it reached full travel and had no further to go, the pressure would increase and blow off the pressure relief valve. In the real world with this cylinder and no relief valve it would first extend, reach full travel then explode. I have the feeling you don’t believe it would first extend in a normal manner? Bring me your cylinder and I’ll gladly attack it with my drill.
Skippy

[skippy]

Hey Sol you said "thanks for the quiz which took way too much of my time".
If you want me to waste some more of your time (and anyone else's time) Go back to the original cylinder (A) and forget about cylinder (B) and hydraulics altogether. The cylinder is empty (just air) and the rod/piston is sitting at the half travel point (i.e. in the middle). I have a compressor with a pressure regulator valve set to 40psi and from there a shut off valve connected to a "T" junction going to each (top and bottom) port of the cylinder. What will happen to the cylinder? EDIT When I open the shut off valve of course.
(A) It will sit stationary.
(B) It will retract.
(C) It will extend.

Remember, NO WHYs, just a or b or c. 24hours.
Geez, I should go and get a job and stop wasting other people's time!
Sorry guys!

[FranH]

I'm slightly confused.

1. We have 2 identical cylinders (except one has a hole thru the piston).

2. We have identical pumps (plural) which states 1 pump for each piston.

3. You state "(the top port flows back to the reservoir)." This is a statement to assume the top of the piston to the top of the cylinder is filled with fluid and moves fluid to the reservoir. Why have a reservoir full of air?

4. "and I plug the top port" states we have a cylinder that is full of fluid above and below the piston...or a vessel full of fluid. If that is so with any pressure exerted that pressure is equal in all directions, how can a piston with equal pressure above and below move anywhere???

[sbrpollock]

Fran:

With regard to:
"
4. "and I plug the top port" states we have a cylinder that is full of fluid above and below the piston...or a vessel full of fluid. If that is so with any pressure exerted that pressure is equal in all directions, how can a piston with equal pressure above and below move anywhere???
"

Yes, there is "equal pressure on both sides of the piston". But the piston does not have the same surface area on both sides. Part of one side is occupied by the rod.

Before I try to explain further, try picturing the cylinder with no piston, just the rod. Then try picturing it with the piston with the hole in it. Picture what the difference would be.

[cp8071]

Skippy Post#33

It will extend with 7.85 pounds of force.

CP8071

[sol]

It goes up...barely.

[skippy]

Only two attempts to answer to post #33 and CP8071 got it. :cheers:

The bottom side of the piston of cylinder (A) has a 3.0” diameter piston which gives an effective area of 7.07175 sq/ins. The effective area of the top side of the piston is 6.8753125 sq/ins (the area of a 3.0” piston minus the area of a 0.5” cylinder rod). The end result is equivalent to having a 0.1964375 sq/ins piston area being pushed from the bottom side making the cylinder extend. We have 40psi acting on 0.1964375 sq/ins so divide 40psi by 5.0906776 = cylinder will extend, top out and have a total force of 7.85lbs.

[sol]

Wait! I want to change my answer yet again!

The top of the piston is experiencing 6.88 sq.in. times 40 lb/sq in or 275 lbs of downward force.
The bottom is being pushed up with 7.07 sq.in times 40 lb/sq.in. for 282.8 lbs of force.
The piston is being pushed down with a 50 lb weight. So the downward force is 275lbs + 50lbs for a total of 325 lbs. Which is more than the 283 lbs pushing up so the piston will fall.

Also what is the difference between the tee valve and a hole in the piston?

[skippy]

Now you got me Sol! I was meaning without the weight, just the cylinder BUT I didn't write it so I guess that makes you the winner! Hmm.....those fine details..... I haven't done the maths yet, probably won't, but one thing is for sure it's going to retract. In fact now that I think about it I would have said 7.85lbs upward force minus 50lbs downwards force equals a downward force of 42.15lbs HOWEVER, I did only ask if it would extend, retract or not move so you're the winner if we can allow the fact that your answer came in after the horn blew.
cheers
skippy