Energy balance in hydraulic cylinder

[mounith]

hello everyone,

In a hydraulic cylinder high amount of load is lifted by applying lower loads in smaller diameter side..
for example, if the diameters of two cylinders is 2cm and 4 cm and i applied a force of 10 N on the 2cm cylinder for 4m downwards. so now the 4cm diameter cylinder could lift a load of 40 N through 1 m... so the work done on both the sides is balanced..

work done = force*displacement
10*4=40*1

But, what if i placed only 20 N load..
the work done on both sides is not balanced...

where does the remaining energy go..?

thanks..

 

[Delta2]

It will accelerate the load and the remaining energy will become kinetic energy of the load.

 

[gsal]

Don't forget to take into account the work done on the fluid itself inside the cylinder.

 

[Lsos]

In your example if you reduced the load from 40N to 20N, then likely the force you will apply to it wil also be reduced from ~10N to ~5N (if you were applying the force yourself).

In practice you will simply feel less resistance and adjust accordingly. If you attempted to still apply 20N, then the load will quickly accelerate away from you, forcing you to reduce how much force you apply.

 

[mounith]

thanks to all..
whats is the effeciency of a hydraulic cylinder?
output power/input power?

 

[gsal]

I am sure it depends on the design and quality of the cylinder...

do you have one in mind?
do you have access to the physical device?
is the efficiency written in the nameplate?
contact the manufacturer.