Nitrogen pneumatic gas ram -- force to compress the piston in a cylinder

  • Thread starter Ted Farkas
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  • #1
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Hi,

Can some one point me to where I can workout the amount of force required to compress a niitrogen filled ram.

Ram has an 80 mm bore and 172 mm stroke. We have filled it with nitrogen to both 1 and 1.5 bar.

thanks
Brett
 

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  • #2
JBA
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The force will depend on how far you want to compress it.
 
  • #3
anorlunda
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What are you assuming about temperature? When you compress a gas it gets hot. But if you push the piston slowly enough for the heat to leak out and for the gas to stay cool, you get a different answer.
 
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  • #4
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Hi,

The ram is used as a holding force for a plow or digger arm. When the arm hits an obstruction it allows the a digger arm to bend out of the way. (see attached)

Once the obstruction is passed the ram pushes the digger back into the ground.

The normal 'breakout force' is approximate 1.8 tonnes.

It works well in problem 2.

However Problem 1 does not have strength to hold into ground.

It appears to be a leverage problem.

SO I am trying to work out all my knowns and calculate from there.

So the ram would work in what ever the ambient temp is. Probably from 10-30 degrees celsius. And I need to the know the INITIAL force needed to start to compress, as it would by minimum force. (oil is the daampener.

It should be a slow movement - tractor travels at approx 5 km/h and shouldn't be moving unless it hits a sizeable obstruction.

I was using a formula in the 'engineering toolbox' but it was only giving me 785 newtons which I know its more that that.

Thanks

Brett
 

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  • #5
JBA
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Hello again, I decided I couldn't do a good job on your lever ratio post; but, maybe I can be of more assistance this time around. As I stated above, the amount of resisting force is dependent upon how far you compress the cylinder. To start the initial amount of force is dependent upon how high you pressurize the cylinder and will increase as you reduce the volume in the cylinder by compressing it.
For a start, the force at any given point is dependent on the area of the cylinder bore x the pressure in the cylinder at that point.
For example, you state that you have an 80 mm bore which results in a piston area of 80^2 x pi() / 4 = 5027 mm^2 of area, for your first case of pressurizing the cylinder to 1 Bar then the resulting initial resisting force of the cylinder will = 503 N; and, for 1.5 Bar that force = 1.5/1 x 503 = 754 N.

Now that we that using a simple P1V1 = P2V2 relationship, which can be stated as: P1 x A x L1 = P2 x A x L2 where P1 is the initial pressure from above; A is the Area of the Piston = 5027 mm^2; L1 is initial distance at full rod extension = 172 mm; and P2 is the resulting pressure at L2 = L1 - S (Stroke distance)
By multiplying P1 x A = F1 (force at full rod extension) and P2 x A = F2 ( force at L1 - S), which leaves F1 x L1 = F2 X (L1- S) or better for your purpose:
F2 = (P1 x A) x L1 / (L1 - S).

Now for any initial cylinder Pressure, you can determine F2 at each selected Stroke length.

From a scientific point there are more sophisticated formulas based upon the basic gas laws; but, this calculation will give you a reasonably accurate resisting force of the cylinder at each initial cylinder pressure and stoke length combination. PS The force increase is linear with the stroke increase, if the stroke is slow enough to allow the cylinder wall to cool the gas during the stroke; if not, the above formula becomes F2 = [(P x A) / (L1- S)] x (T2 / T1) where T2 and T1 are given in °R and T2 must be calculated using the classic gas equation for compression heating based upon T2 = T1 x P2/P1 and the properties of the gas.
 
  • #6
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Hi Thanks for your reply.

It has been a great help. I was confused with the numbers initially than realised we used 100 and 150 bar. Bring it up to approx 5 ton and 7.5 ton.

This now makes sense. I will now move onto the 2nd part of the problem which is the leverage. I might post some more simple diagrams to make sure I have the leverage concept right e.g. with an L shaped lever do you measure along each arm for effort or does it act perpendicluar.

Regards
Brett
 
  • #7
JBA
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I am glad I could help.

As to your leverage analysis, one of the problems I had in attempting to analyze your leverage was that by posting the full drawing of the two units with all of the bolt heads etc, I simply could not sort on the basic lever configurations of the assembly. For me, it would help considerably if you would make simple dimensioned line diagrams with the correct locations of the pivots connected by lines.
 
  • #8
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Now that we that using a simple P1V1 = P2V2 relationship, which can be stated as: P1 x A x L1 = P2 x A x L2 where P1 is the initial pressure from above; A is the Area of the Piston = 5027 mm^2; L1 is initial distance at full rod extension = 172 mm; and P2 is the resulting pressure at L2 = L1 - S (Stroke distance)
By multiplying P1 x A = F1 (force at full rod extension) and P2 x A = F2 ( force at L1 - S), which leaves F1 x L1 = F2 X (L1- S) or better for your purpose:
F2 = (P1 x A) x L1 / (L1 - S).
Hi JBA,

now that I have worked leverage problem out I am now checking forces along the arc - but this gives me massive numbers.

The ram has been using 150 bar so using your formula

I use the initial 1 bar pressure 42729.5 Newtons=(100000 x .005027m2) x (172/172 -170)

This equates to 4.3 ton at almost full closure at 1 bar - multiply by 150 for 150 bar + 654 ton

This can't be right.

I am obviously missing something.

thanks
Ted
 
  • #9
JBA
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Ted Farkas, post: 6053604, member: 649202 "I use the initial 1 bar pressure 42729.5 Newtons=(100000 x .005027m2) x (172/172 -170)"

If the maximum stroke length of the cylinder is 172 mm then for a closing stroke of 170 mm the compression ratio = 172 / (172-170) = 86 to1

As a result, for a 72 mm maximum stroke and 150 Bar pre-pressure, the cylinder pressure at a 70 mm stroke = 86 x 150 Bar = 12,900 Bar.

So, while I cannot decipher your above SI units calculation, it is no surprise that you have resulted in a very high piston load.

As an note regarding your above 172 mm apparent maximum stroke of the cylinder. If that is the correct maximum stroke for your cylinder, I have an error in my prior for the 25° plow tip rotation force results; because, I used the 350 mm overall cylinder dimension as the actual maximum rod travel in my force calculations. If the maximum cylinder travel (cylinder compressing stroke) is actually 172 mm then all of my force results will be higher and for:

Case 2 - D to C to Tip @ 25° Rotation (38.5 mm stroke) the resulting tip breakaway force = 1.33 tonnes
Case 3 - D to A to tip at 0°Rotation w/ Tip Fully Rotated 25° (38.5 mm stroke) the resulting tip breakaway force = 2.14 tonnes
Case 4 - D to A to tip at 25° Rotation w/ no D-C-Tip Rotation (38.5 mm stroke) the resulting tip breakaway force = 1.41 tonnes
(Since there is no effect on the pressure at zero stroke regardless of the maximum stroke length neither Case 1 or Case 5 results are affected.
 
  • #10
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So, while I cannot decipher your above SI units calculation, it is no surprise that you have resulted in a very high piston load.


I used bar initially , I assume that you can just multiply answer by 150 to get result for 150 bar???



F2 = (P1 x A) x L1 / (L1 - S).
42729.5 Newtons= (100000 pascals x .005027m2) x (172/172 -170)

The first set of brackets I thought are basic SI units e.g pascals, m², & newtons.

Or am I missing something .Is it as simple as this: if cylinder is 1 bar or what ever pressure I just multiply by L1/L1-S?

e.g for above its 86 Bar therefore 86000000 pascals x .005027 m² = 432322 Newtons

432322N/9.81=44069 Kgs

Seems like the cylinder would explode.

I know I out of wack somewhere.

Thanks
Ted
 
  • #11
JBA
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Ted, as I stated in my #5 post the basic gas rule is P1 x V1 = P2 x V2 which converts to P2 = P1 x V2 / V1 and this covers all cases of compression of gases and in inviolable, except for temperature effects as also mentioned in that post. You are correct that if you were to apply enough force to compress your cylinder to a 70 mm stroke something on the cylinder will most likely fail catastrophically at some point.

Any manufactured gas filled shock absorbing cylinder unit comes with a pre-charged pressure and maximum stroke length safe for its design and without any port that would allow a user to charge the unit to any other pressure.

On the other hand, if you are taking a standard ram cylinder unit and simply pressuring it and plugging its inlet port to create your own shock absorbing type of cylinder then you are definitely creating a very dangerous situation and a clear risk of a serious potential failure because the cylinder is not designed for that type of service and potential internal overpressure. A standard ram cylinder has a maximum safe pressure rating stated in the manufacturer's specification sheet for the unit and generally stamped on the cylinder nameplate; and, the user of that unit is expected to equip the nitrogen supply system for operating the cylinder with a safety pressure relief valve to prevent exceeding the manufacturer's specified maximum operating pressure.
 
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  • #12
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HI,

I understand your concern, this has been manufactured for this purpose with a max 150 bar pressure - it compresses on a ploughing machine if it hits an obstacle. These things work and are built for that exact purpose. My point is that I want to understand the workings, its a self contained unit and can be charged to allow movement between machines. They compress to the full stroke length 172 mm. This won't happen with the speed of a factory type line.

Thanks
Ted
 

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  • #13
JBA
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Hi,

Thanks for the drawing and explanation. I do notice that the manufacturers drawing describes this as being a "Nitrogen Operating Cylinder" and I accept that it has been utilized in your application for some time as a compression type unit; but, doing so still results a potential level of safety risk if applied in a new type of design without realizing that its combined N2 charging pressure, load and stroke must be such that the cylinder does not exceed its 250 Bar safe operating pressure at its intended operating full stroke, as you have now identified by your own calculations. For example: My calculations indicate that for your unit with a 150 Bar charge pressure, the maximum stroke without exceeding the 250 Bar cylinder limit is 68 mm with the maximum cylinder stroke of 172 mm (a 31° rotation of the plow tip around the "A" pivot), assuming that there is no retained volume in the cylinder at its maximum 172 mm stroke; .

I am not intending to be critical of what you are doing, I actually applaud your efforts to understand the basics of the mechanics on your unit. I am just trying to assist you by explaining some of the elements you need to consider based upon my over 20 years of equipment design and development for high pressure operating systems.

PS Thank you for taking the time to restate your calculations with the applicable SI units because that allows me to understand what type of unit conversions are being performed. One of the frustrating things for me, being accustomed to US units, is multiple zeros required for some of the basic SI unit conversions.
 
  • #14
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Thanks for your help,

No problems, once I understand the rudimentary design i.e whats happening , we then moved forward an get it signed off before manufacture.

regards
Ted
 

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