# Pascal's Principle with Hydraulic Pumps

• Myung
In summary: F2 = 2x10^6 NA2 = 8m^2... you can calculate the pressure at the large piston. This pressure will be the same as the pressure at the fluid at the small piston so you will have...P2 = P1 = ...Now, how does the height difference come into play? What will be the pressure difference between the large and small pistons?Do you know how to calculate the pressure due to a height difference in a fluid? Do you know what equation to use?In summary, the problem involves a hydraulic can crusher with a large piston exerting a force of 2x10^6 N on the cans through a pressure difference created
Myung

## Homework Statement

A hydraulic can crusher is shown in the figure. The large piston has an area of 8m2 and exerts a force F2 of magnitude 2x106 N on the cans. Calculate the magnitude of the force F1 exerted by the small piston (area 10cm2) on the fluid. Do not ignore that the large pistons is 1 m higher than the small piston.

## Homework Equations

P1 = P2
Thus
F1/A1 = F2/A2

Figure:
http://bit.ly/yuDaZ7

## The Attempt at a Solution

What is the significance of the height of the piston ? I cannot solve it without knowing that information is it very important?

Last edited by a moderator:
Myung said:

## Homework Statement

A hydraulic can crusher is shown in the figure. The large piston has an area of 8m2 and exerts a force F2 of magnitude 2x106 N on the cans. Calculate the magnitude of the force F1 exerted by the small piston (area 10cm2) on the fluid. Do not ignore that the large pistons is 1 m higher than the small piston.

## Homework Equations

P1 = P2
Thus
F1/A1 = F2/A2

Figure:
http://bit.ly/yuDaZ7

## The Attempt at a Solution

What is the significance of the height of the piston ? I cannot solve it without knowing that information is it very important?

The height difference will create a pressure difference due to the action of gravity and the weight of the working fluid. See Pascal's Principle.

The problem doesn't state what the hydraulic working fluid is. You'll need to know its density in order to calculate the pressure difference due to height.

Last edited by a moderator:
I don't know what you mean. Our professor never really elaborated in Pascal's Principle all she said was P1 = P2 :/

ρgh = F2 / A2

ρ(9.8m/s^2)(1m) = 2x10^6 N / 8m^2

Unknown Density? Can we assume that it is water?

Myung said:
I don't know what you mean. Our professor never really elaborated in Pascal's Principle all she said was P1 = P2 :/

Pascal's Principle implies that any pressure induced into a (confined) fluid at one location will be felt equally at every other location in the fluid. This is independent of any pressure differences that may exist due to other reasons like gravitational potential differences (height differences). When the hydraulic system is assumed to be all at the same height then the pressure is the same everywhere, hence your P1 = P2.

In this problem you are told that there is a height difference, so the pressure at the large piston will be less than the pressure at the small piston by an amount fixed by the height difference and the density of the working fluid: ρgh.

You should be able to write an equation that incorporates this fixed pressure difference into a relation between the forces at the piston faces.

Myung said:
ρgh = F2 / A2

ρ(9.8m/s^2)(1m) = 2x10^6 N / 8m^2

Unknown Density? Can we assume that it is water?

If you are not told what the working fluid is then you can either make the assumption that it is water (state this assumption in the solution that you hand in!), or you can leave the density in symbolic form and present a symbolic result.

gneill said:
Pascal's Principle implies that any pressure induced into a (confined) fluid at one location will be felt equally at every other location in the fluid. This is independent of any pressure differences that may exist due to other reasons like gravitational potential differences (height differences). When the hydraulic system is assumed to be all at the same height then the pressure is the same everywhere, hence your P1 = P2.

In this problem you are told that there is a height difference, so the pressure at the large piston will be less than the pressure at the small piston by an amount fixed by the height difference and the density of the working fluid: ρgh.

You should be able to write an equation that incorporates this fixed pressure difference into a relation between the forces at the piston faces.

So what you are saying is that P1<P2 when there is height? So must I change the equation to inequalities then? or its still the same as P1 = P2 but this time it will be --> ρgh = F/A?

Myung said:
So what you are saying is that P1<P2 when there is height? So must I change the equation to inequalities then? or its still the same as P1 = P2 but this time it will be --> ρgh = F/A?

First let's make it clear which piston is which. The problem places piston 2 higher than piston 1. That means that the pressure P2 at piston 2 will be less than the pressure P1 at piston 1 by a fixed amount.

The equation will be an equality, not an inequality.

gneill said:
If you are not told what the working fluid is then you can either make the assumption that it is water (state this assumption in the solution that you hand in!), or you can leave the density in symbolic form and present a symbolic result.
I don't need to consider the pressure of the Larger Piston because it is in a different elevation.

Thus:

P1 = ρgh

F/A = ρgh

F = A(ρgh)

F = 1x10^-3 m^3 (ρ)(9.8 m/s^2)(1m)

F = ρ( 9.8 x 10 ^-3 m/s^2 )

Correct?

*EDIT* I think this is highly wrong.. I am really confused.

Can you just walk me through ?

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Myung said:
I don't need to consider the pressure of the Larger Piston because it is in a different elevation.
How do you justify that?

Of course you must consider the pressure at the large piston! It's the pressure at the large piston that is creating the force to crush the cans. You are given a value for that force, and you are given the area of the piston...
Thus:

P1 = ρgh

F/A = ρgh

F = A(ρgh)

F = 1x10^-3 m^3 (ρ)(9.8 m/s^2)(1m)

F = ρ( 9.8 x 10 ^-3 m/s^2 )

Correct?

*EDIT* I think this is highly wrong.. I am really confused.

Can you just walk me through ?

I cannot just present the solution to you. I can continue to give hints and suggestions.

Since you have the can-crushing force applied by piston 2, and you have the area of piston 2, what is the pressure of the hydraulic fluid at piston 2?

gneill said:
How do you justify that?

Of course you must consider the pressure at the large piston! It's the pressure at the large piston that is creating the force to crush the cans. You are given a value for that force, and you are given the area of the piston...

I cannot just present the solution to you. I can continue to give hints and suggestions.

Since you have the can-crushing force applied by piston 2, and you have the area of piston 2, what is the pressure of the hydraulic fluid at piston 2?

Pressure = 2x10^6 N / 8 m^2

Pressure = 250000 Pa

P1 = P2 + ρgh ? Can this be? so that two pressures will be equal?

F/A = 250,000Pa + ρ(9.8m/s^2)(1m)
F = 1x10^-3 m^2 [ 250,000Pa + ρ(9.8m^2/s^2) ]

F = 250 N + (9.8 x10^-3 m^4/s^2) ρ

ASSUMING FLUID IS WATER

F = 250 N + ((9.8 x10^-3 m^2/s^2) 1000 kg/m^3

F = 250 N + 9.8 N
F = 259.8 N

Correct?

What if the machine has the efficiency of 90%? what does this mean?

gneill said:
How do you justify that?

Of course you must consider the pressure at the large piston! It's the pressure at the large piston that is creating the force to crush the cans. You are given a value for that force, and you are given the area of the piston...

I cannot just present the solution to you. I can continue to give hints and suggestions.

Since you have the can-crushing force applied by piston 2, and you have the area of piston 2, what is the pressure of the hydraulic fluid at piston 2?

Pressure = 2x10^6 N / 8 m^2

Pressure = 250000 Pa

P1 = P2 + ρgh ? Can this be? so that two pressures will be equal?

F/A = 250,000Pa + ρ(9.8m/s^2)(1m)
F = 1x10^-3 m^2 [ 250,000Pa + ρ(9.8m^2/s^2) ]

F = 250 N + (9.8 x10^-3 m^4/s^2) ρ

ASSUMING FLUID IS WATER

F = 250 N + ((9.8 x10^-3 m^2/s^2) 1000 kg/m^3

F = 250 N + 9.8 N
F = 259.8 N

Correct?

What if the machine has the efficiency of 90%? what does this mean?

Myung said:
Pressure = 2x10^6 N / 8 m^2

Pressure = 250000 Pa

P1 = P2 + ρgh ? Can this be? so that two pressures will be equal?

F/A = 250,000Pa + ρ(9.8m/s^2)(1m)
F = 1x10^-3 m^2 [ 250,000Pa + ρ(9.8m^2/s^2) ]

F = 250 N + (9.8 x10^-3 m^4/s^2) ρ

ASSUMING FLUID IS WATER

F = 250 N + ((9.8 x10^-3 m^2/s^2) 1000 kg/m^3

F = 250 N + 9.8 N
F = 259.8 N

Correct?
Yes! That result looks good.
What if the machine has the efficiency of 90%? what does this mean?

Generally such a statement means that only 90% of the work done at the input end of a machine gets delivered to the output. Typically this is due to frictional or other losses in the mechanism.

The result will be that a force higher than you calculated without taking such losses into consideration will be required at piston 1 in order to deliver the desired can-crushing force at piston 2.

gneill said:
Yes! That result looks good.Generally such a statement means that only 90% of the work done at the input end of a machine gets delivered to the output. Typically this is due to frictional or other losses in the mechanism.

The result will be that a force higher than you calculated without taking such losses into consideration will be required at piston 1 in order to deliver the desired can-crushing force at piston 2.

Yay! Thanks! I also found out in the web that the formula is called "Pressure Vs Depth of Static Fluid."
P1 = P2 + ρgh.So in terms of computation of Efficiency, if the machine has only 90% efficiency it means it only delivers 0.9(259.8N) or 233.82 N to the output?

EDIT:

Or do we still need to take account that

Efficiency = AMA / TMA x100% ?

Last edited:
Myung said:
Yay! Thanks! I also found out in the web that the formula is called "Pressure Vs Depth of Static Fluid."
P1 = P2 + ρgh.

So in terms of computation of Efficiency, if the machine has only 90% efficiency it means it only delivers 0.9(259.8N) or 233.82 N to the output?

EDIT:

Or do we still need to take account that

Efficiency = AMA / TMA x100% ?

It is stated that the machine delivers a particular force at the output, so presumably you will want to calculate the required force at the input that will result in that force being delivered at the output despite the losses due to efficiency.

gneill said:
It is stated that the machine delivers a particular force at the output, so presumably you will want to calculate the required force at the input that will result in that force being delivered at the output despite the losses due to efficiency.

There is no given input force.. so is this even possible?

TMA = requires distance moved and is also not given

Myung said:
There is no given input force.. so is this even possible?

You calculated the force required at piston 1 given that the machine is 100% efficient. You need to modify this value to account for the stated efficiency.
TMA = requires distance moved and is also not given

I think you'll find that the work done (being force x distance) is proportional to the force. The distances involved can be assumed to remain the same despite efficiency. Thus you need to apply the efficiency consideration to the force at the input.

gneill said:
You calculated the force required at piston 1 given that the machine is 100% efficient. You need to modify this value to account for the stated efficiency.

I think you'll find that the work done (being force x distance) is proportional to the force. The distances involved can be assumed to remain the same despite efficiency. Thus you need to apply the efficiency consideration to the force at the input.

259.8N is the force of required of the piston 1 at 100% efficiency.
at 90% efficiency it only gives about 233.82 N of Force.

Correct?

Myung said:
259.8N is the force of required of the piston 1 at 100% efficiency.
at 90% efficiency it only gives about 233.82 N of Force.

Correct?

No, it should require MORE force at the input to achieve the same 2 x 106N at the output.

gneill said:
No, it should require MORE force at the input to achieve the same 2 x 106N at the output.

So, 90% of it is only 233.82

We need the force to be 110% so it can match the output force which will be

285.78 N of force in the input ? Correct? (

Bump!

Myung said:
So, 90% of it is only 233.82

We need the force to be 110% so it can match the output force which will be

285.78 N of force in the input ? Correct? (

Well, that's in the right ballpark, but not precise. For a more accurate value, first write out your equation relating F2 to F1 when the machine is 100% efficient:

$F2 = (F1 - \rho g h) \frac{A2}{A1}$

Now suppose that due to unspecified losses the actual value of F2 is only 90% of the value given by the above. So if you let eff = 0.9, then

$F2 = eff \;(F1 - \rho g h) \frac{A2}{A1}$

Rearranging the above to solve for F1 (which is then the actual force required to produce the given force F2):

$F1 = \frac{F2}{eff}\frac{A1}{A2} + \rho g h A1$

Interestingly, it appears that the efficiency factor effects the mechanical parts of the machine (pistons) but not the hydraulic pressure due to height difference! Now you can plug in your desired F2 and the efficiency and determine the required F1.

## What is Pascal's Principle?

Pascal's Principle, also known as the principle of transmission of fluid pressure, states that a change in pressure applied to an enclosed fluid at rest is transmitted equally to all parts of the fluid and the walls of the container.

## How does a hydraulic pump work?

A hydraulic pump uses Pascal's Principle to create movement and force by applying pressure to a confined fluid. As the pressure increases, the fluid is forced through a small opening which amplifies the force and allows for the movement of larger objects.

## What are the advantages of using hydraulic pumps?

Hydraulic pumps have the ability to generate a large amount of force with relatively small amounts of input force. They are also able to operate in a variety of environments and can be easily controlled and adjusted for different tasks.

## What are some common applications of hydraulic pumps?

Hydraulic pumps are commonly used in heavy machinery, such as construction equipment and cranes, to lift and move heavy objects. They are also used in hydraulic brakes in vehicles, as well as in elevators and aircraft control systems.

## Are there any limitations to using hydraulic pumps?

One limitation of hydraulic pumps is the possibility of leaks or failures in the system, which can result in loss of pressure and decreased efficiency. They also require regular maintenance to ensure proper functioning and to prevent potential hazards.

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