A Technical Question

The power requirement of any liquid pump is the pressure rise multiplied by the flow through it divided by the overall efficiency. The rest is arithmetic.

If the flow through the pump were zero, the power requirement would be zero too.

Yes. Pump suppliers can be found either with a Google search or by using the Products & Suppliers tab at the top of this page. The rest is telephone calls. Good luck.

SOORY i had anthor question ??

i had a pump pressure =420 bar carries a pistone mass=20 ton and its area = 3500c.m2 this pistone moves with velocity =10 cm/s so the motor needed how many horses should be and it r.p.m ??

Simply look it up on the label attached to the motor on the pump.

If it doesn't exist yet then a figure in excess of 200kW is suggested.

Apologies come across as abstruse.

A 6in pipe at that pressure is going to have a thick sidewall, and may not be readily available off the shelf.

The purpose of this pump is a curiosity, being too high a pressure for seawater reverse osmosis by a factor of 7.

dear,

IS it logic to get out 3500 liter of hydrolic oil from a nozzel 1 cm2 area in 1 sec ??

and if not i need to know the nozzel area if i had 3500 liter and time 1 sec ??

That's nonsense. The velocity would be hypersonic. You didn't mention that in the original message. Why are you bringing it up now?

My homework alarm just went off. This can't be a real-world matter. Lyn is right.

Hi Del,

I like your analogy of a 1 sq.cm area bottle.

Doesn't that mean the 1 litre bottle would be 1 meter long and the whole thing would be 3500 meters long and would need to pass through the orifice in 1 second. I think that equates to 210,000 meters/minute or 12,600,000 meters/hour which is 12,600 Kilometers/hour.

I tried picturing it and things got a bit blurred.

I would imagine any liquid you tried to do that to would "flash off" due to the heat but that's a whole new ball game.

Best regards,

John.

Hi moataz,

I must agree with Crabtree, the figures you are quoting are astronomical.

I might be "miles out" but figures of 2100 litres /minute (460 Gallons / Minute) at 420 Bar (6174 psi) is in the region of 1900 kilowatt (2500 Horse Power).

Just looking at 3500 square centimeters piston area at 10 centimeters / second is 35 litres per second (hence my 2100 litres / minute.

As I said I might be miles out and for your sake I hope I am, otherwise you have one hell of a project ahead.

Best of luck

John

Insufficient information again. The rpm will depend on the displacement of the hydraulic pump, which has not been given.

https://www.youtube.com/watch?v=8HqyEHqEYho

Do your own homework. CR4 is not a homework cheat site; however, if you have questions about understanding concepts or how a portion of a problem is derived, these types of questions will be accepted.

it is not a homework i am graduted from mechatronics department and i am working in a project now and i have forgotten the hydrolic laws and i have some questions and i thought u could help me

You have indeed. Well, you ought to be aware that this site doesn't do education catch-ups, so I would suggest you continue your quest elsewhere. We're clearly into Ex-Lax territory here, and I have other things to do.

"I am graduted from mechatronics department"

I believe that Walt Disney coined the term "mechatronics" to describe the animated mechanical devices he used in Disneyland.

So, this must be some kind of Mickey Mouse project you are working on?

And just exactly where was this ( mechatronics department ) located ?...

Power = pressure x flowrate.

PowerOut (ft lb/sec) = pressure (lb/sq in) x flow rate (cu ft/sec) x 144 (sq in/sq ft). Divide by 550 to get PowerOut (hp).

PowerIn = PowerOut/efficiency.

Insufficient information given; we need also to know the flow rate.

Theoretically HP = (delta P)(GPM)/1714, but to account for inefficiency you might estimate (delta P)(GPM)/1500.

Hydraulic, not hydrolic.

Hmm... Farmer math says,

420 Bar = ~6000 PSI

3500 cm x 10 x 60 = 2100 l/min

2100 l/min = ~555 GPM

rule of 1500 says 1 GPM @ 1500 PSI or 1500 GPM @ 1 PSI = 1 Horsepower

So (6000 x 555) / 1500 = 2220 Horsepower

General rules of hydraulic fluid flow suggest no less than 4 inch dia piping and or port/valve sizing for 555 GPM flow rates and recommends 5 - 6 inch for continuous flow applications.

Well that wasn't so hard.

Personal opinion also suggests buying books like this whenever you can find them.

Pocket Reference Book.

They can even make dummies like me look like real educated, experienced and widely diversified engineers.

The OP said he wanted to get 3500 liters in one second, so the volumetric flow rate is 3500 l/sec, not 2100.

To the OP: the formula you have forgotten is simple:

Q = A x V where Q = volume flow rate, A = area of the nozzle or pipe, V is the flow velocity.

The numbers you give are not credible. To get 3500 l/sec through a 1 cm2 opening would require a flow velocity of 114,444.4 ft/sec or approximately Mach 104. Hypersonic indeed. Converted to miles per hour it exceeds escape velocity of the earth's gravity by a factor of three.

If you look up typical flow velocities for hydraulic oil online, you will find a high pressure line might have 25-30 ft/sec. Put that value into the formula I gave you above and you will find you need about 4000 times as much area as you propose. The proof is left as an exercise for the student.

May I also suggest you bookmark www.engineeringtoolbox.com so you can find all the formulae you need there?

"i had a pump pressure =420 bar carries a pistone mass=20 ton and its area = 3500c.m2 this pistone moves with velocity =10 cm/s so the motor needed how many horses should be and it r.p.m ??"

I take 3500c.m2 to be 3500 square centimeters which if I was wrong and it's 3500 square meters (to lift 20 tons?)then everything I said would just need to be multiplied by 1000.

You can know that a 2" Viking positive displacement pump, without a bypass valve, driven by a 5 hp motor, will continue to develop head pressure against a closed valve, to the point at which the head bolts stretch, fluid sprays out, the motor trips out on overload, then runs backwards until the pressure bleeds off. The guys that put this together were dead heads.

I don't know if they are still available, but Womack had a great series of handbooks on fluid power. If you can locate them, they are a great source of information and formulae.

You don't need any if there is no flow. -- JHF

I can give you the answer for Camels. If you are interested PM me.

Lots of flow for Mechatronics assignment. There is not a single pump that can do this.

I disagree. The huge plunger pumps used for oil formation fracturing can easily do those pressures and flow rates.

Note that you said pumps! I have not come across a single pump that can do this flow rate and pressure. And even then 1320 barrel per minute is a big stretch.

Pumped at 6100 psi this is 200,000 HP. You need your own power plant or some very big diesel tanks plus assorted over hundred 2000 HP pumps and CAT engines.

I disagree that you disagree!

Okay where are you getting your numbers from?

We've already figured out that the OP's flow rate was ~555 GPM at ~6100 PSI which as I know is well within the working HP limits of a single FMC 5 piston high pressure plunger pump being driven by a 2500 HP Caterpillar diesel engine.

From the horse mouth! See here.

Clearly says 3500 liter per second.

We haven't figured a thing in this thread.

I took that as he was asking if it was logical not what the power requirements were like in the posts I was referring to.