Scrap my first answer since I read the question as Rod End Vented and Cap End Full of Oil. Don't ask me why cause I have no idea why/how I read it that way. Maybe, Senility????After a re-read I see it is a completely different problem. On top of that, Others need to learn, I've had my time.
I know you are fishing for something but I don't believe you have provided enough info for anyone to bite-i.e. weight of the rod, type and fit of piston/ring assembly, rod gland etc. These factors could very well make Bud's first answer correct.
No rod and reel needed here. I believe there is a definite correct answer.
Open ended question-with the info provided the rod could remain stationary.
The actuator will move until the some of forces acting on it is zero. There is 1 atm x the area on the cap. The pressure on the rod side will decrease below atmospheric until the sum of forces is zero. If there is a lot of frictional force then the actuator will not move much at all. So now, if the cap side is at atmospheric pressure, say 14.7 psi, what must the pressure be on the rod side for the sum of forces to be 0? Assuming the bulk modulus of oil is 200,000 psi, how far does the actuator move before equilibrium is reached? All this can be calculated. The unknown is the friction, but we can assume that is 0 for these calculations, and the weight of the piston and rod because the initial pressure on the rod side will not be exactly atmospheric. So how far does the actuator move?The exact solution requires calculus.Bud, this question is similar to my under pressure thread on www.patchn.com Since the mass is moved by a force, how fast does it move to the new position? Does the actuator oscillate around the final position?Solving this requires solving a system of non-linear differential equations.Force makes it go.
I will stick to the posted question.Situation 1: There is enough friction from the piston and rod seals that the weight of the Piston, Rod, and Load cannot overcome the starting friction. In this case, the cylinder doesn't move.Situation 2: The seals provide negligible friction and the rod side of the cylinder is all air-no oil.Assume the piston is 4" thick and made of steel. Wght = approx 88 lbsAssume the Rod is 16" lgWght = approx 56 lbsAssume the piston and rod are supported when the plug is installed and then let go.The volume of air at atmospheric pressure when the plug is installed is approx 1/4 of the stroke or 165 cu in=V1At a constant TemperaturePV1=PV2; P =PSI(absolute) V=cu inP2=14.7-[144(wght pis+rod)/66(rod area)]=14.7-2.2 PSI=12.5 PSI abs14.7(165)=12.5(V2); V2=194 cu inwith 500 lb loadP2=14.7-[644/66]=14.7-9.8=4.9 PSIa14.7(165)=4.9(V2); V2= 495 cu inV fully retracted= 660 cu in495/660=75%Piston and Rod supported 25% retCapped no supp=30% retCapped no supp w/load 75% retMy final answer: Piston and Rod retract somewhere between 25% and 75% approximately depending on friction and amount of oil in rod end
I think this addresses 2 principles:1. Max vacuum is 14.7 PSI at sea level2. You can't generate any more force (exclude friction)than 14.7 X area when working in vacuum.What do you think?
JDK wrote:"I think this addresses 2 principles:1. Max vacuum is 14.7 PSI at sea level2. You can't generate any more force (exclude friction)than 14.7 X area when working in vacuum.What do you think?"That is the way I see it also.AnyMouse
Post a Comment