Damping due to porous plate travelling through hydraulic oil

[Mech King]

Evening All,

I am looking at a problem where there is essentially a hydraulic cylinder containing a double acting type rod (with an annulus). There will be some holes in the annulus, which will allow fluid to pass through once there is a load on the end of the rod causing it to extend. The fluid in the cylinder is totally closed in by the bounds of the cylinder (diameter and ends). There is a shear pin holding the rod in a close position until it is loaded.

Given the force applied i am trying to get my head around how to determine how much damping there will be. i.e, if a force of 10kN acts to pull the rod out, what will be the force seen in the ram. I think that the influencing factors will be the viscosity of the fluid, load applied, size of holes in annulus, diameter of cylinder, diameter of rod. the stroke of the cylinder will determine the duration of damping.

I'm not asking for anyone to solve the problem for me, as i haven't supplied enough information, but i wanted to get pointed in the right direction, maybe an example problem, or some principles and formula. I've been a mechanical designer all of my career and haven't looked at much other than structural statics and dynamics - unfortunately no fluid dynamics since university days.

Might be a simple problem that I'm over complicating, but any advice would be greatly appreciated.

Cheers

 

[FredGarvin]

One of the best pieces of advice in fluids I think I was ever given was, if you are dealing with moving fluids and forces, you need to look at momentum.

Draw your control volume and examine the flows coming and going. That should give you your forces.

 

[dsol]

My first thoughts are:
Dampers of this design work by the loss that occurs when a viscous fluid is forced through a small orifice (most of the losses occur because the jet that is induced through the holes creates turbulence and drag).
A first shot would be to find the velocity of those jets (assume incompressible fluid). I seem to remember a formula for head loss and/or drag for a fluid passing through an orifice, with the factors being velocity and diameter of the orifice.
Thats my guess.

 

[Mech King]

Cheers guys, will have another stab at it and let you know how i get on.

Thanks

 

[Mech King]

Ok guys, this is where i am at the moment, please refer to the attached pdfs for basic free body diagram.

In this hypothetical situation, a 100kg block falls from 3m onto a shock absorber. The shock absorber has six 4 mm Diameter holes in the annulus as seen on the attached pdfs.

Now, the PE of the block is:

PE=mgh
PE=100kg * 9.81ms-2 * 3m
PE = 2943 N.m

Now, the shock abosorber needs to transfer this 2943 N.m of energy into KE in order to absorb the shock.

KE=PE:

0.5mv^2=2943N.m

Velocity at impact = 7.7ms-1.

The absorber will transfer this energy by squeezing the fluid in the cylinder through the orifices on the annulus.

I have looked at several ways of calculating the energy absorbing force, but I'm confused as to which route to go down,

I have struggled to get my head around the momentum of the fluids as mentioned by Fred Gavin, so have tried to determine the drag force on the anulus due to the impact of the block on the rod of the absorber, where, drage force is:

Fd=Cd*0.5*(density of fluid)*(flow velocity)^2*Area

I guess I am stuggling to transfer the mechanics of the impact energy over to the fluid dynamics.

Is the energy approach at the beggining the correct way to approach this problem? I am really at a loss at the moment, any pointers would be greatly appreciated.

Many Thanks

 

[nucleus]

I don’t think the PE is converted into KE. It is converted into heat and wasted.

Dampers often have air or nitrogen in them. In chapter 6 of the following book it shows a cutaway of a motorcycle damper (shock absorber).
http://books.google.ca/books?id=84h...esult&ct=result&resnum=5#v=onepage&q=&f=false

 

[Mech King]

The energy is absorbed by the the piston traveling through the hydraulic oil. The oil is squeezed through the orifices in the anulus of the piston, and this squeezing of fluid through the orifice is what abosrbes the energy. So in essence it converts the impact energy into another form of kinetic energy by the piston traveling throught he fluid.

The difficult part is how to calculate how much damping the system does and how much stroke would be needed in the cylinder to absorb this energy. The bottom of the falling load will see a force, which will be less than the force it would see if it simply fell onto the ground.

Just need a few pointers as to where to go from here really.

Cheers,

 

[Mech King]

Back again, right...

I have worked with some arbitrary numbers: 100kg falling 1.5m onto the piston rod and I am trying to figure out the required size and number of holes required in the piston annulus in order to absorb the hypothetical load of impact:

Please let me know what you reckon?

Using the conservation of Energy, we have:

PE = m*g*h = 100kg*9.81m/s*1.5m = 1471.5 Pa r N.m

KE = 0.5*100*v^2

If PE = KE, then:

{2*(1471.5 N.m)}/100 = v^2

v = 5.43 ms-1


Work Done = F*d. And if work done is equal to the energy that needs to be absorbed, then:

KE = PE = F*d

d = the cylinder stroke = 0.37m

1471.5 N.m= F *0.37

F = 3977 N ……………………this is the damping force that the cylinder does per unit length over 0.37m in order to absorb all of the energy from the impact.

The velocity of the piston on impact will be as calculated above as 5.43 ms-1. The bore diameter of the cylinder is 40 mm. Assuming that 6mm diameter holes are used in the orifice:

A1*V1 = A2*V2

A1 = bore diameter
A2 = orifice diameter
V1 = Bore fluid velocity
V2 = orifice fluid velocity

(1.256E-3)m^2 * (5.43 ms-1) = (2.826E-5)m^2 * V2

V2 = 241 ms-1

Fluid dynamic pressure through orifice is:

PD = (Rho*v^2)/2

PD = [(900 kg.m-3) * (241 ms-1)^2]/2

PD = 26,136,450 Pa

Force damping through one 6mmm diameter orifice:

P = F/A:

(26,136,450 Pa)*(2.826E-5 m^2) = F

F = 739 N.

We need to absorb 3977 N, so the number of 6 mm orifices :

(3977/739)N = 5.4 holes



Now I know this doesn’t take account of the viscosity of the fluid, which I’m guessing it should do, or is that taken into account via the density of the fluid which I have plugged in as 900 kg.m-3?

Please let me know if you agree with my method and calculation?

cheers

 

[Ranger Mike]

One huge factor you must consider is HEAT. we run Penske gas shocks on the race car. compression and rebound can be adjusted by dialing the applicable valve knob. we can change the characteristic of the damper by adding or subtracting various discs attached to the piston. We use a shock dynometer that cycles the shock and plots out the actual curve so we have consistency for each wheel. and these shocks do get HOT. The oil carries away the heat and these shocks even have a remote reservoir for the nitrogen gas. so a dampers performance is impacted by HEAT.

 

[Mech King]

Yep that's a good point, i was going to take into account this effect after i had sorted the main approach to the sums as shown above.

But for a rough estimate before buiding and testing a prototype, would you all agree with my method for the sums etc?

cheers

 

[Mech King]

ANyone have any thoughts on my calcualtions in post 8?

cheers

 

[Mech King]

No thoughts? :(