How viscous forces affect force transmission in hydraulic systems.

[DiMbex]

Hi everyone,

I have a question that I am struggling and need your help. So I am working on a project which is essentially two syringes, A & B, with different cross-sectional areas, A1 & A2, connected via a tube filled with water. If we assume loss-less system, Pascal's law, the input force ,F1 ,will be transmitted to the second syringe and the output force, F2, will be a multiplication of the two cross-sectional areas [F2=F1*(A2/A1)].

So my question is if that is the case because in real life?

Hydraulic systems are affected by viscous forces (minor and major losses) which cause a pressure gradient in the system. In simple words the efficiency is not 100% but something else. Is there a way to calculate that efficiency and what would be the actual output force (F2) in such a system. Is Bernoulli's equation including losses a suitable equation?

Thank you very much.

 

[Andy Resnick]

Is there a way to calculate that efficiency and what would be the actual output force (F2) in such a system. Is Bernoulli's equation including losses a suitable equation?

Thank you very much.

From what you describe, non-zero fluid viscosity will introduce time-dependence to the process- if you compress one syringe (with rigid walls), the other syringe (also with rigid walls) will respond over time and asymptotically approach the invisicd case. If you apply a time-dependent compression to the first syringe (say, an oscillatory driving force), then the dynamics will follow that of a damped driven oscillator.

I don't think you can tweak Bernoulli's equation, since that holds for inviscid fluids (lossless processes). You may be able to adapt Brinkman's equation or Darcy's law, tho.

Does that help?

 

[jrmichler]

Is there a way to calculate that efficiency and what would be the actual output force (F2) in such a system.

Yes. You have two cases:

Case 1: Force and pressure without motion. In that case, the forces are proportional to the areas as you stated.

Case 2: Force and pressure with motion. The syringe pistons are moving and the fluid has a velocity through the tube. Each piston has friction, so one piston has pressure slightly less than predicted from force and area, while the other piston has force slightly less than predicted from pressure and area. There is a pressure loss due to fluid viscosity in the tube. That pressure loss can be calculated using a Moody chart (search the term). The sum of these losses means that the actual output force will be less than the force with zero motion. The ratio of force with motion to the force without motion is the efficiency.

The end result is that hydraulic efficiency is variable.

 

[DiMbex]

Just to clarify then.

If using the Moody Chart and the relevant equation we calculate that Pressure drop from syringe A to B, let's say the ΔP= 6 KPa. So if F1/A1 = 10 KPa, then the output force (F2) will be 10-6KPa = 4KPa or will it be 4 KPa less than the F2= F1*A2/A1 result?