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The Hydraulic Jack #4720860
04/08/18 06:53 PM
04/08/18 06:53 PM
Joined: Jun 2002
Posts: 19,235
Iowegia - USA
MolaKule Offline OP
MolaKule  Offline OP
Joined: Jun 2002
Posts: 19,235
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.




Last edited by MolaKule; 04/08/18 07:04 PM.

Reading is fundamental; understanding is not a given ability.
Re: The Hydraulic Jack [Re: MolaKule] #4720883
04/08/18 07:27 PM
04/08/18 07:27 PM
Joined: Dec 2017
Posts: 1,046
US
JLTD Online shocked
JLTD  Online Shocked
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Posts: 1,046
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.

Last edited by JLTD; 04/08/18 07:41 PM.

I'm a thickie; but a 20 has its place.

I use AMSOIL

Hers: 2008 Jeep Liberty 138k, SS 5w30/Amsoil

His: 2015 4Runner 41k, OE 5w20/Wix

Re: The Hydraulic Jack [Re: JLTD] #4720892
04/08/18 07:47 PM
04/08/18 07:47 PM
Joined: Jun 2002
Posts: 19,235
Iowegia - USA
MolaKule Offline OP
MolaKule  Offline OP
Joined: Jun 2002
Posts: 19,235
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.

Last edited by MolaKule; 04/08/18 07:48 PM.

Reading is fundamental; understanding is not a given ability.
Re: The Hydraulic Jack [Re: MolaKule] #4720906
04/08/18 08:05 PM
04/08/18 08:05 PM
Joined: Dec 2017
Posts: 1,046
US
JLTD Online shocked
JLTD  Online Shocked
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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)


I'm a thickie; but a 20 has its place.

I use AMSOIL

Hers: 2008 Jeep Liberty 138k, SS 5w30/Amsoil

His: 2015 4Runner 41k, OE 5w20/Wix

Re: The Hydraulic Jack [Re: MolaKule] #4720917
04/08/18 08:20 PM
04/08/18 08:20 PM
Joined: Jun 2002
Posts: 19,235
Iowegia - USA
MolaKule Offline OP
MolaKule  Offline OP
Joined: Jun 2002
Posts: 19,235
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.


Reading is fundamental; understanding is not a given ability.
Re: The Hydraulic Jack [Re: MolaKule] #4721007
04/08/18 10:07 PM
04/08/18 10:07 PM
Joined: Dec 2016
Posts: 6,247
Waco, TX
Linctex Online content
Linctex  Online Content
Joined: Dec 2016
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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






Last edited by Linctex; 04/08/18 10:10 PM.

"The evidence demands a verdict".
(Re:VOA)"it's nearly impossible to actually know the particular additives that are in there at what concentrations."
Re: The Hydraulic Jack [Re: MolaKule] #4721069
04/08/18 11:25 PM
04/08/18 11:25 PM
Joined: Jun 2002
Posts: 19,235
Iowegia - USA
MolaKule Offline OP
MolaKule  Offline OP
Joined: Jun 2002
Posts: 19,235
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

Last edited by MolaKule; 04/08/18 11:35 PM.

Reading is fundamental; understanding is not a given ability.
Re: The Hydraulic Jack [Re: MolaKule] #4722023
04/09/18 06:57 PM
04/09/18 06:57 PM
Joined: Jun 2002
Posts: 19,235
Iowegia - USA
MolaKule Offline OP
MolaKule  Offline OP
Joined: Jun 2002
Posts: 19,235
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.


Reading is fundamental; understanding is not a given ability.
Re: The Hydraulic Jack [Re: MolaKule] #4723074
04/10/18 08:10 PM
04/10/18 08:10 PM
Joined: Jun 2017
Posts: 133
Tennessee
ChevyMan93 Offline
ChevyMan93  Offline
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Tennessee
42


2016 Chevrolet Silverado Z71 5.3
2017 Honda CR-V 2.4
Re: The Hydraulic Jack [Re: MolaKule] #4723500
04/11/18 10:56 AM
04/11/18 10:56 AM
Joined: Nov 2003
Posts: 1,114
Middle of Iowa
DriveHard Offline
DriveHard  Offline
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Posts: 1,114
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


Smile, it increases your face value
2011 Silverado Crew Cab LT 6.2L/6spd
2017 Jeep Renegade Trailhawk (wife's)
2007 Nissan Xterra (trail toy)
2013 Fiat 500 Abarth (my DD)
Elio...?
Re: The Hydraulic Jack [Re: DriveHard] #4723536
04/11/18 11:39 AM
04/11/18 11:39 AM
Joined: Jun 2002
Posts: 19,235
Iowegia - USA
MolaKule Offline OP
MolaKule  Offline OP
Joined: Jun 2002
Posts: 19,235
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




Last edited by MolaKule; 04/11/18 12:09 PM.

Reading is fundamental; understanding is not a given ability.
Re: The Hydraulic Jack [Re: MolaKule] #4723575
04/11/18 12:10 PM
04/11/18 12:10 PM
Joined: Jun 2002
Posts: 19,235
Iowegia - USA
MolaKule Offline OP
MolaKule  Offline OP
Joined: Jun 2002
Posts: 19,235
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.

Last edited by MolaKule; 04/11/18 12:14 PM.

Reading is fundamental; understanding is not a given ability.
Re: The Hydraulic Jack [Re: MolaKule] #4723641
04/11/18 01:06 PM
04/11/18 01:06 PM
Joined: Nov 2003
Posts: 1,114
Middle of Iowa
DriveHard Offline
DriveHard  Offline
Joined: Nov 2003
Posts: 1,114
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.


Smile, it increases your face value
2011 Silverado Crew Cab LT 6.2L/6spd
2017 Jeep Renegade Trailhawk (wife's)
2007 Nissan Xterra (trail toy)
2013 Fiat 500 Abarth (my DD)
Elio...?
Re: The Hydraulic Jack [Re: MolaKule] #4723967
04/11/18 06:54 PM
04/11/18 06:54 PM
Joined: Jun 2002
Posts: 19,235
Iowegia - USA
MolaKule Offline OP
MolaKule  Offline OP
Joined: Jun 2002
Posts: 19,235
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.








Last edited by MolaKule; 04/11/18 06:54 PM.

Reading is fundamental; understanding is not a given ability.
Re: The Hydraulic Jack [Re: MolaKule] #4724025
04/11/18 07:55 PM
04/11/18 07:55 PM
Joined: Jun 2002
Posts: 19,235
Iowegia - USA
MolaKule Offline OP
MolaKule  Offline OP
Joined: Jun 2002
Posts: 19,235
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.


Reading is fundamental; understanding is not a given ability.
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