motor and reservoir size?

Hi bwire,

What is your question?

Mike

Mikefho,

Thanks for offering assistance of correctly sizing a motor and or fluid reservoir and quantity of fluid to adequately operate this small actuator.

That is a problem I do not understand since PSI is a pressure unit and not a force unit.

Do you need to work at the given pressure or do you want to have a force? In the second case please indicate the units so that the pressure can be computed.

Okay I need 1.875 tons force from described unit.

I worked on a similar system 6"id X 1 1/2" stroke 1800psi [hydraulic pressure] 20 strokes per minute.

20hp running a 20gpm pump & a 30 gallon reservoir. A air cooled heat exchanger was utilized since the cylinder was mounted in a fairly hot location, for the same reason 20 wt oil.

All the valves, flow controls & pump were Vickers. This set-up was very reliable, most of the problems were related to the heat & that the cylinder was also moving back & forth @ the same 20 cycles.

These units ran 24hrs, 6 1/2days/week, 360 day/yr.

Oh you're right about the need for cooling

This particular cooler set up was the worst part of the system, when the relief valve was by-passing the oil flowed through the cooler. @ 20 stroke/min this wasn't enough. [around 1/4-1/3 sec/cycle]

The operating conditions included a cylinder about 6 meters [of pipe] away from the valves.

The cylinder was mounted on top of [2] 1square meter platens @ about 350 degrees F

the oil in the area of the cylinder, didn't have enough flow, contributing to thermal breakdown. The cylinders had metal rings allowing a small amount of blow by. We tryed cheveron seals [rubber]on the pistons and the oil would overheat & boil in the vicinity of the cylinder.

the oil only lasted about 6 months.

The check valve in the bypass/cooler circuit would some times allow pressure spikes to the heat exchanger, the HE's would be good for a couple of years, before the welds would stress crack. A couple of rounds of repairs were about it, after that you were just junkies chasing crack!

Given the option I would have used a small auxillary pump to cool & filter the oil

Even with these limitations, still a very reliable set up, the failures were mostly heat or vibration related. There are 10 of these lines @ this production facility & hundrends corporate wide.

For clarification this unit will be a inside a cabinet and though cooling should be addresses there will be no other machinery adding to the heat.

Thanks

Cool! How to convert info into a 2-1/2" cylinder w/

4-1/2" stroke?

A cylinder of 2.5" ID will usually have a rod of 1.5" dia.

15 x Vol of cylinder extending = 5.43 Lt/min

15 x Vol of Cylinder retracting = 3.475 Lt/min (anulus vol)

Vol required = 8.905 Lt/min

Area of cylinder = 4.909 in²

1750/4.909 = 356.5psi = 24.59 Bar (THIS MIGHT BE A WRONG ASSUMPTION)

(24.59 x 8.905)/600 = 0.365kW

Mec Eff = 0.8 therefore = 0.456kW

Direct drive, Single Phase Motor of 2,900 RPM would require (gear) pump of 3.07 cc/rev conected to a reservoir of about 20Lts Vol (not including air space).

Anyway, what was the question?

Dear Beej50,

Your solution based on considering 1750 as a force (lb) not pressure (psi), why? may be there is a mistake in data given by our friend bwire.

I know your assumption, but what will happen if we considered it as a pressure, 1750 psi = 120.658 bar, it will be 5 times your figure 24.59 Bar. I interest to hear from bwire the exact unit psi or lb.

Dear Abdel

I have got tons of engineering programs on my computer and somewhere I have actually seen PSI Force. I've spent half the night looking but I'm buggered if I can find it!

Your perfectly correct in saying that my assumptions would be a factor of five out if incorrect and I did think of giving the answer with PSI as a pressure. It might have been the logical solution.

At least the volume of the reservoir should be 'good to go' dependant if this is a mobile or factory based system, with or without cooler etc, etc!

I think that, at least if we are 'transparent' with our workings, they can be accepted or disputed rationally. This question is a bit like using a calculator, sh*t in, sh*t out!

Many thanks for your comments though!

The original query doesn't specify over what area [how many square inches] this force is going to be applied

A 2-1/2" ID cylinder with 4-1/2" stroke needs to produce 1750 PSI force

my earlier solution solution is probably going to be oversized by a factor of 6-7, but there's no way to tell without a bit of clarification.

The cycle rate of every 4 seconds, the length/size of the piping & the method used to signal the direction change are going to be important also.

The original query doesn't specify over what area [how many square inches] this force is going to be applied

.250"

Oops!

Abdel Halim Galala,

I expect will be cylinder pressure about 800 psi and force of 1.875 ton. I think I think I can do all well with 1.250 ton force but want extra 4 testing.

Thank you for contributing.

Yep that about covers it except will probably be about 800 PSI and rod diameter of 1".

Thanks for the pointers.

Unless you get your cylinder built as a special size, you probably won't get a 1" rod on a 2.5" bore. The standard would be either:-

a) 2.5" Bore to 1.5" rod.

b) 2" Bore to 1.25" rod.

c) 1.5" Bore to 7/8" rod.

If you expect a working pressure of 800psi then you have the scope to reduce the size of the cylinder and increase the pressure of the pump and still maintain the working force required. A standard gear pump should be good for a little over 3000psi.

This would also reduce the amount of stored oil required in the reservoir although I would agree with a previous commentator that it is better to cool with convection (dependant on local ambient temperatures) so, probably keep reservoir volumes higher rather than lower.

If you do reduce the size of the cylinder, you may need to consider the 'slenderness ratio' although with a 4.5" stroke it is probably irrelevant!

Good luck with it!

I assume that pressure is 1750 PSI. Considering losses the system pressure has to be at least 2000. Based on 2.5"cylinder and in the assumption that the cylinder is NOT symmetrical with a rod 1.5" the pump has to deliver at least 9 L/min =2.4 gpm.

Motor power for a global efficiency of 0.8 is 2.6 kW.

If the reservoir has a capacity of more than 30 L it is possible to avoid a cooler since free convection over the reservoir lateral area can evacuate the 20% losses by an oil /air temp. difference of 40°C. A bigger reservoir has the advantage that oil is less stressed and will maintain a longer time its characteristics. Convection area has to be > 1.5m^2

In order to avoid not necessary losses it would be positive to have a small accumulator able to take the pump delivery during the commutations at each end of stroke. Care has to be taken for choice of valves and filtration of the oil. For a system running as mentioned it is most important to filter as low as possible( 1µ) since the very small particles of dirt in oil are the most aggressive and the most wear generating. Although pressure is not high it would be positive to use a ram with servoseals low friction and also long life. V-cups could present a too fast wear and require maintenance. The summ of strokes/year is over 1100 miles so that seals have to be able to accept it.

Thank you all I will keep you posted as to the result of the project...