Hydraulic project,how does a numpty calculate requirements??

[David Edgar]

When I built my press, I just bought whatever was available, welded the frame up, mounted the cylinder and valves, got the largest motor with a pump attached and plumbed it up and by sheer good luck it worked. I did get a professional to check hoses,connections etc. My point is I didn't do any calcs.

My son has asked me what would be necessary to move 1000kg (1 ton) through 300mm (1 foot) and back 30 times a minute. He is ready to buy the equipment necessary, I do not know if this is feasible with hydraulics or whether an electric motor and linear drive is better.

 If anyone is clever enough and can be bothered with yet another request for help, it would be greatly appreciated. Even if it is just to say it is not possible. Thank you for reading this.

[arftist]

Really need a lot more info to offer any help at all. 

 

Specific questions tend to get answered. 

[David Edgar]

Yes sorry,the hydraulic ram has to create a vertical, reciprocating movement of 300mm @ 30 cycles/ min. The load of 1000 kg is a bit over the top, 500 kg might do it. it has to be 240 volt single phase as well. the limit switches and speed control have to be  programmeable (sp) with a buffer at each end of the travel. This is all I know so far. If no one can help, do not worry about replying. It was a shot in the dark that someone could help.Even just knowing if it is possible in principle would be an important step forward.

Thank you Arftist for taking the time.

[HIGHSIDER]

I would like to help here and give my opinion that I believe will assist with a few guide lines. First off I’m putting on the kettle and putting on some tea as this is going take a little setting up to get my point over. Ah that’s better, oh that’s nice, I also found a bici.. lol.

 

Our friends across the Atlantic are going to love this as its all in metric….sorry guys!

 

Let’s get a little background set up for the force calculations.

 

Force is equal to Pressure times Area or **F = P [A]

Where F is measured in Newton’s (N)

Where P is measured in Pascal’s (Pa)

Where A is measured in m^2 and uses [(π.d^2)/4] for cylinder calculations

 

Now let’s try and make a bit of sense of the above

 

Take your 1Ton load. Well we’ve got to change that into standard notation where

 

1Ton = 1,000kg. Now we’ve got to use that science in the air called gravity and change the 1,000kg into N’s. To do this we call in another multiplying factor called g or gravity or 9.81m/s^2. We just simply multiply the two of these to get the force valve required for the above formula **

 

1,000kg x 9.81m/s^2 = 9810N.

 

Next factor required in the formula is P where we need to work out the Pascal value. To do this we’ll take the more familiar term used in hydraulics and convert Bar pressure to Pascal’s. 1 Bar = 100,000 Pascal’s or 1 x 10^5 Pa

 

Let’s look at a pressure on a power pack and use a realistic figure, say 200Bar pressure which I would say is a fair assumption where most mid range power packs can achieve this pressure. We have to take a figure so we’ll go with 200Bar for now. So..

 

200 Bar = 200,000,00Pa or 20 MPa. That’s the pressure sorted, Now lets look at what cylinder we need and to do this we will look at [(π.d^2)/4] where d in this formula represents the diameter of the cylinder.

 

Now let’s start some number crunching and to do this lets list what we’ve got and take it in stages.

 

We have:-

 

F = 9810N

P = 20MPa (or 20 x 10^6)

A = we don’t have d (dia of cylinder) so now we can calculate.

 

First off we need to reconfigure the formula so we can compute the value of d.

 

F = P [(π.d^2)/4]

4[F] = P [(π.d^2)]

(4[F])/P[π] = d^2

 

√ {4[F])/P[π]} = d (or turn this around)

 

d = √ {4[F])/P[π]}

 

Let’s fill out what we have...

d = √ {4[9810])/20 x 10^6[π]}

d = .024m or 24mm (or lets round this off, no cylinder manufacturer will make you a cylinder with a 24mm barrel diameter so the nearest standard will be 25mm diameter)

 

So in summary, you will need a cylinder of 25mm diameter (barrel end) operating at 200Bar pressure that will move a load of 1Ton for the project listed above. Altering the pressure value will alter the d value. Put in a pressure of 100Bar (10 x 10^6) and see what d value you get?

 

Now I’ve had a bit of fun putting that together but lets bring this study into a realistic situation. It’s very unlikely that you’re going to go out and buy a cylinder of 25mm Dia. That’s not likely to happen. You will probably end up going for a cylinder that has a decent body (say 80 or 100mm barrel diameter) that would generate a lot more force and do more work with a good power source say we use 200Bar again. So, let’s see what force is generated by using 200Bar pressure and using a cylinder with a 100mm barrel diameter. So again..

 

F = P [A]

 

F = (unknown, we’re going to calculate this)

P = 20MPa (or 20 x 10^6)

A = [(π. (.1) ^2)/4] where .1 is 100mm converted to m) dia of cyl

F = 20 x 10^6 [(π. (.1) ^2)/4]

 

F = 157079.63N or 16012.19kg or 16Ton force (divide 157079.63 by 9.81 to get kg)

 

So, again in summary, a 100mm dia cylinder will give you 16Ton force using a power pack at 200Bar pressure. That’s more like it, more scope in the tonnage that can do more variety of work.

 

Two figures I like to use on a press is the pressure that the power pack is delivering (200Bar in the above study) and the diameter of the cylinder used. Again in this case, 100mm Dia. Straight away I know by the above calculations, I’m getting 16T of force available at the press. Use a pressure gauge fitted to the system or a hydraulic hose and note the pressure before the crack off point on the relief valve so you know the force available at the press. If you alter the pressure, you automatically alter the force. It takes a little getting used too but if you use the above calculations, it will calculate what force is being generated by what pressure is available and what cylinder you are using. The strokes per/minute is one of those questions that can be answered by a good hydraulic part supplier. It’s one of these set ups that can be achieved by using limit switches relayed back to solenoid/spool valves. It’s not that difficult to set up. The flow rate of a power pack along with the hydraulic power it can deliver is key information here. I hope this has helped…..time for another brew of tea!!

[arftist]

Yes sorry,the hydraulic ram has to create a vertical, reciprocating movement of 300mm @ 30 cycles/ min. The load of 1000 kg is a bit over the top, 500 kg might do it. it has to be 240 volt single phase as well. the limit switches and speed control have to be  programmeable (sp) with a buffer at each end of the travel. This is all I know so far. If no one can help, do not worry about replying. It was a shot in the dark that someone could help.Even just knowing if it is possible in principle would be an important step forward.

Thank you Arftist for taking the time.

 

Yes, it is totally doable. 

[ianinsa]

David, 1000kg in terms of hydraulics is very light, think small motor car jack, 2t you can wrap your hand around easily!

Sadly this usually means slow, if you speed it up it heats up(speed and heat are easy to deal with but pricey) sorry but most smiths are by nature 'cheap'(er. Frugal) look at air as a possibility. 500 to 1000kg, 300mm at 30/min easy even with a dinky compressor.

Just food for thought!

[Alan Evans]

Basic but...Don't forget when making your calculations that if the specified two-way application is only using a single, double acting cylinder it should be based on the return stroke power which will be about half that of the outward stroke. :)

Alan

[Alan Evans]

Excellent post Highsider, I have pasted that into my useful things to remember folder, thank you.

Alan

[David Edgar]

Thank you so so much for going to all that trouble Highsider, I really didn't know where to start. My son will be over the moon to know this idea of his is viable. I know I will get the job of welding up the frame and ordering the bits. When it is finished I will give you the full background and details and of course photos if my wife does it for me. I hope the info Highsider has posted will be of interest to many more other than Alan and myself. Thanks again to all who took an interest.

David

[HIGHSIDER]

You're both very welcome gentlemen. It is my pleasure. I enjoyed putting that post together.

 

Hydraulics is a very interesting topic. It's always a very worth while exercise to sit down with a note pad & calculator and work out the forces exerted by a hydraulic system if you know the operating pressure of a power pack and the diameters of the associated cylinders. Likewise it is also very beneficial to spec out; Cylinder Dia & Pressure values, by rearranging the formula:-

 

F = P [A]     or   P =F/A     or    A = F/P 

[David Edgar]

Highsider,while you are still here, so to speak.Before you disappear back into the ether.The press I built without doing any sums, can I reverse the formula above to get an idea of the force it exerts, (F?)   where my cylinder diameter (A?) is 125 mm (5") and the maximum pressure reading I can get on the gauge is 1250 psi from a power pack driven by a 2.2 kw motor (3hp?) I do not know the size of the pump. The cylinder stroke is only 200mm (8") and ram diameter is 60mm. would that be ?

   F=pressure (P)(A)  = tons??

    Force = 1250  x  5=  6250lbs   Or Have I got it wrong?  Don't forget I am a Numpty.

 

Thanks David

[HIGHSIDER]

Hi David,

 

Rather than say reverse the formula we just reconfigure it. Let’s run over it again…

 

Force is equal to Pressure multiplied by Area or

 

F = P[A]

 

Now reconfigure the formula to get Pressure or [P]

 

Where Pressure is Force divided by Area or

 

P = F/A

 

Now reconfigure the formula to get A or [A]

 

Where Force is divided by Pressure or

 

A = F/P

 

Now let’s put this into practice and work with your system and see what turns up. First off, let’s define what exactly is being asked in the question. Okay, we’re looking for a Force. You want to find out what force is being generated in your press.

 

Straight away we are going to use: F = P[A] The first formula listed above, the remaining two as said are manipulations of this one.

 

So, let’s list out what we have and have not;

 

F =? (We don’t have F and will now calculate this)

 

P = 1250 psi which is 86 Bar which is 8,600,000 Pa (86 x 10^5 Pa) or 8.6MPa (M for Mega (10^6))

 

[To get Bar divide psi by 14.5]

 

A = 125mm (again, we must change mm to m (standard notation) so 125mm is .125m)

 

Area:-  Formula for cylinders = (π.d^2)/4 where d = .125m

 

(π.(.125^2))/4 = .012271846 m^2

 

 

In summary

 

F = ?

 

P = 86 x 10^5 Pa

 

A = .012271846 m^2

 

Again: F = P[A]

 

So

 

F = (86 x 10^5) x (.012271846)

 

F = 105537.87N (divide this by 9.81 to bring N’s to kg) 105537.87/9.81 = 10758.19kg (and finally divide 10758kg by 1000 to bring kg’s to Ton’s) 10758.19/1000 = 10.7Ton’s. Almost 11Ton.

 

So in summary; your press operating at 1250psi, using a cylinder with a 125mm barrel bore will generate a Force of 10.7Tons.

 

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

 

Now let’s look at the rod end and work out what’s the returning force on that cylinder seeing you supplied the rod diameter (60mm). Now have a think about this, the barrel end is always going to generate the largest force for a cylinder because of the surface area available for the hydraulic oil. The rod end of the barrel will not have the same surface area and the force generated will be less. Where you have a reduced surface area, you will have a reduction in force, allow me to demonstrate;

 

The barrel end diameter of your cylinder: 125mm (or .125m)

 

The rod diameter of your cylinder: 60mm (or .06m)

 

We use a slightly different formula for the area calculation for a rod end of a cylinder

 

A = (π. [D – d] ^2)/4,

 

Where D = Barrel end Diameter of cylinder

 

Where d = rod diameter of cylinder

 

D = .125

 

d = .06

 

A = (π. [.125 – .06] ^2)/4,

 

A = .003318307m^2

 

So let’s pop that into the formula and see what the rod end force of the cylinder is:-

 

F = P[A]

 

F =? (We don’t have F and will now calculate this)

 

P = 86 x 10^5 Pa

 

A = .003318307m^2

 

F = (86 x 10^5) x (.003318307)

 

F =28537.44N or 2909.01kg or 2.9Ton (say 3Ton nice round figure)

 

So, in summary

 

Your press is generating 10.7Ton when your power pack is supplying 1250psi and using a 125mm cylinder.

 

Also the return or rod end force is 3Ton while using the same pressure, (1250psi) where the rod diameter is 60mm with the main barrel dia again 125mm

 

In the words of the late great Fred Dibnah….did yer like that?? lol

 

Take your time and run through the above figures and have a go with a calculator. Nothing numpty about the questions you have asked but it does help if you get a little practice in with the math’s where it can be a big help understanding a hydraulic system. Good luck.

[David Edgar]

Once again thank you for giving me all the answers, I am pleased that the press will give me nearly 11 tons on the down stroke as it is not too slow either. I will now do as you say and work backwards from the answer towards the questions .

Thanks Highsider. I feel quite guilty that you have done all this work. If we ever meet up I owe you several drinks of your choice.

David

[HIGHSIDER]

You're very welcome David, I'm glad you've found my input helpful.

 

If you're ever over on my island, do drop by for those drinks. I'll have the kettle boiling.

[Badger1875]

Hi folks, I have set an excel-sheet for hydraulic calculation, but I don´t know how to upload it here, any hints? If someone want´s it for testing please mail, I will send it to you. There is only one to say: all units are metrical using german specs. greets from Germany Freddie

[Badger1875]

Hi folks,

I finally found out how to upload something, sometimes you don´t see the woods cause of the trees....
Well, the motor calculation in the excel-sheet is done for german/ european standards, you might change the power-calculation to 110/ 220 volts for american specs.

The calculated power of the electrical motor might seem a little high, but if a quality moter is used, you can split the power to half, because the press is working in a pulsing mode and the high power is not needed all the time forging. I run my press at 25 tons (calculated) with a 7,5 kW motor.

The motor will get warm/ hot during running, specially if you use lower power motors, the temperature might get up to 200 to 220° Fahrenheit, a good motor can handle this. I never reached a higher temp than handwarm.

 

Freddie

hydraulic pump english.xlsx