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#4720860 - 04/08/18 06:53 PM The Hydraulic Jack
MolaKule Offline


Registered: 06/05/02
Posts: 18951
Loc: Iowegia - USA
Using a 6.5 [email protected] fluid, we are are going to determine some hydraulic jack results.

The jack has two pistons, two "check" valves, a bleed valve, and a small reservoir.

The area of the Drive piston is 1 square inch,

The area of the Lift piston is 20 square inches.

Question 1: If the Drive piston is depressed with 1 inch of travel at a Force of 100 lb., what is the Force created by the Lift piston AND what is its travel distance (total upward movement)?

Question 2: Assuming no frictional losses and an incompressible fluid, what are the Input and Output Energies?

Show how you derived the answers.





Edited by MolaKule (04/08/18 07:04 PM)
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#4720883 - 04/08/18 07:27 PM Re: The Hydraulic Jack [Re: MolaKule]
JLTD Online   content


Registered: 12/15/17
Posts: 396
Loc: US
I'm likely wrong but...

1: 2000# and 1/20 of an inch, simple ratio.

2: With no friction the input and output energies are always equivalent, there is simply leverage and a mechanical advantage to apply force to move the fluid into the lift piston.

With no specified weight on the lift piston and no frictional loss there is zero initial cost, but there is a cost to move the weight of fluid 1/40 of an inch upward (halved from 1/20 assuming the fluid also moves downward from the drive piston). The cost to move the fluid ... 1 cubic inch of oil. 7.6 pounds/quart converted to weight per cubic inch.


Edited by JLTD (04/08/18 07:41 PM)
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#4720892 - 04/08/18 07:47 PM Re: The Hydraulic Jack [Re: JLTD]
MolaKule Offline


Registered: 06/05/02
Posts: 18951
Loc: Iowegia - USA
Originally Posted By: JLTD
I'm likely wrong but...

1: 2000# and 1/20 of an inch, simple ratio.

2: With no friction the input and output energies are always equivalent, there is simply leverage and a mechanical advantage to apply force to move the fluid into the lift piston.

With no specified weight on the lift piston and no frictional loss there is zero initial cost, but there is a cost to move the weight of fluid 1/40 of an inch upward (halved from 1/20 assuming the fluid also moves downward from the drive piston). The cost to move the fluid ... 1 cubic inch of oil. 7.6 pounds/quart converted to weight per cubic inch.


Try again.

Were assuming a 100% efficient system hence,
Quote:
"Question 2: Assuming no frictional losses and an incompressible fluid, what are the Input and Output Energies?

Show how you derived the answers.


Edited by MolaKule (04/08/18 07:48 PM)
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#4720906 - 04/08/18 08:05 PM Re: The Hydraulic Jack [Re: MolaKule]
JLTD Online   content


Registered: 12/15/17
Posts: 396
Loc: US
Lack of friction doesn't remove the weight of the fluid being moved...wait...equivalent volume of fluid is equivalent weight.

However, there's the difference in piston weights, so without weight on the lift piston:
Output energy = input energy + (Lift piston weight - Drive piston weight)
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#4720917 - 04/08/18 08:20 PM Re: The Hydraulic Jack [Re: MolaKule]
MolaKule Offline


Registered: 06/05/02
Posts: 18951
Loc: Iowegia - USA
The force vector F2 of the Lift Piston is pointing Upward; no weight is needed to determine it.

The force vector F1 of the drive piston is pointing downward and is given as F1 = 100 lbs. with a d1 of 1 inch.

Determine F2 and d2.
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#4721007 - 04/08/18 10:07 PM Re: The Hydraulic Jack [Re: MolaKule]
Linctex Offline


Registered: 12/31/16
Posts: 5718
Loc: Waco, TX
Originally Posted By: MolaKule

The area of the Drive piston is 1 square inch,......... the Drive piston is depressed with 1 inch of travel at a Force of 100 lb.,


1sq" x 1" movement = 1 cubic inch

OK, so exactly 1 cubic inch of oil is being moved by the drive piston at a force of 100 pounds per square inch (PSI)

x 20 square inches of surface MUST mean (20sq" x 100 p.s.i.) = 2000 pounds of upforce on the driven

the round piston of 20 sq in has a diameter of 5.046" (2.5231" radius - just making a mental picture)

I say travel is .050 inches







Edited by Linctex (04/08/18 10:10 PM)
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#4721069 - 04/08/18 11:25 PM Re: The Hydraulic Jack [Re: MolaKule]
MolaKule Offline


Registered: 06/05/02
Posts: 18951
Loc: Iowegia - USA
OK, I'll give you the Equations from Pascal's Law:

F2 = (A2/A1)*F1

d2 = (A1/A2)*d1

Energy (Force X Distance) Input = F1*d1
Energy (Force X Distance) Output = F2*d2


Edited by MolaKule (04/08/18 11:35 PM)
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#4722023 - 04/09/18 06:57 PM Re: The Hydraulic Jack [Re: MolaKule]
MolaKule Offline


Registered: 06/05/02
Posts: 18951
Loc: Iowegia - USA
Anyone, Anyone?

Question 1: If the Drive piston is depressed with 1 inch of travel at a Force of 100 lb., what is the Force created by the Lift piston AND what is its travel distance (total upward movement)?

Question 2: Assuming no frictional losses and an incompressible fluid, what are the Input and Output Energies?

Show how you derived the answers.
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#4723074 - 04/10/18 08:10 PM Re: The Hydraulic Jack [Re: MolaKule]
ChevyMan93 Offline


Registered: 06/05/17
Posts: 131
Loc: Tennessee
42
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#4723500 - 04/11/18 10:56 AM Re: The Hydraulic Jack [Re: MolaKule]
DriveHard Offline


Registered: 11/24/03
Posts: 1033
Loc: Middle of Iowa
You got the answer above...what more are you looking for?

D2 = 0.05" (although if "depressed" is a positive value of 1", then technically the travel of cylinder 2 would be -0.05" if a traditional arrangement and parallel)

F2 = 2000 lbf

no need to calculate energies...if system is 100% efficient and no losses, then energy in = energy out
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#4723536 - 04/11/18 11:39 AM Re: The Hydraulic Jack [Re: DriveHard]
MolaKule Offline


Registered: 06/05/02
Posts: 18951
Loc: Iowegia - USA
Originally Posted By: DriveHard
You got the answer above...what more are you looking for?

D2 = 0.05" (although if "depressed" is a positive value of 1", then technically the travel of cylinder 2 would be -0.05" if a traditional arrangement and parallel)

F2 = 2000 lbf

no need to calculate energies...if system is 100% efficient and no losses, then energy in = energy out





Edited by MolaKule (04/11/18 12:09 PM)
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#4723575 - 04/11/18 12:10 PM Re: The Hydraulic Jack [Re: MolaKule]
MolaKule Offline


Registered: 06/05/02
Posts: 18951
Loc: Iowegia - USA
Edit got me.

Linctex did have the correct answer for the original problem. cool

Let's have one more try and make it simpler:

The area of the Drive piston is 2 square inches,

The area of the Lift piston is 20 square inches.

Question 1: If the Drive piston is depressed with 1 inch of travel at a Force of 100 lb., what is the Force created by the Lift piston AND what is its travel distance (total upward movement)?

Question 2: Assuming no frictional losses and an incompressible fluid, what are the Input and Output Energies?

Using,

F2 = (A2/A1)*F1

d2 = (A1/A2)*d1

Energy (Force X Distance) Input = F1*d1
Energy (Force X Distance) Output = F2*d2,

Show how you derived the answers.


Edited by MolaKule (04/11/18 12:14 PM)
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#4723641 - 04/11/18 01:06 PM Re: The Hydraulic Jack [Re: MolaKule]
DriveHard Offline


Registered: 11/24/03
Posts: 1033
Loc: Middle of Iowa
F2 = 1000 lbf
d2 = 0.1"

Not interested in showing work ;-)

The funny part, being a fluid power engineer, I would never think of a problem in the way you are presenting it. Rarely do we think of imposing a force on a fluid power system through the input. Instead, the pressure is always created by the resisting force. You start with the object that must be moved...or the pressure in the system that is resulting from trying to move the resisting force. Then back-calculate into the needed pressure, then into either pump torque required to create that pressure, or the input cylinder force required.

In short: The "input" to the fluid power system creates flow. The resisting output of the system creates pressure.
input = flow
load resistance = pressure

I come across far too many engineers that come into the industry with the mindset that a pump produces both flow and pressure...and it plagues their career until they can figure out that fundamental.

You don't ever create 100 lbf input force unless you have exactly 1000 lbf resisting it. It would be a much more plausible question if you posed a load and were looking for the input force on the cylinder required to lift it.
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#4723967 - 04/11/18 06:54 PM Re: The Hydraulic Jack [Re: MolaKule]
MolaKule Offline


Registered: 06/05/02
Posts: 18951
Loc: Iowegia - USA
I gave this problem and others in QOTD to review basic principles of mechanics, hydraulics, and other topics.

You are missing the point; in this problem we're reviewing Force Multiplication via Pascal's Law wrt the application of hydraulics.

I never asked for flow or pressure values, so that comment has no bearing on this problem.









Edited by MolaKule (04/11/18 06:54 PM)
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#4724025 - 04/11/18 07:55 PM Re: The Hydraulic Jack [Re: MolaKule]
MolaKule Offline


Registered: 06/05/02
Posts: 18951
Loc: Iowegia - USA
What the heck did you do 4WD? ??? This thing is upmteen miles long. smile

Dumping in a link does not constitute a valid answer.
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