Laws of hydrostatics and its application
Pressure by external forces
The pressure within a closed system increases by the impact of an external force. The pressure spreads within this system in all directions equally and has therefore at all points of the system the same value. Example of this would be the hydraulic cylinder
The opposite to the closed system forms the open system, such as a pipeline. Here, the proportionality applies_ pressure p ~ height h.
Law of Pascal: F = p * A Units: 1 bar = 105 Pa (Pascal) = 100 kPa; 1 Pa = 1 N/m2
2A1: A cylinder with a diameter of d1 = 15 cm has to hold a mass of 1500 kg on height. How high rises the pressure within the system? Solution: p = 8,327 bar
Work machines, which can be described by the law of Pascal:
Principle: Two differently sized pistons are mechanically connected by a piston rod. If the piston surface A1 is pressurized, so the force F1 on the piston rod acts directly on the small piston area A2:
F1 = F2
pe1 * A1 * η = pe2 * A2
pe operating pressure in bar
A piston area in cm2
F piston force in N
2A2: By using a pressure intensifier the pressure from PE1 = 30 bar has to be increased to PE2 = 250bar. How big must be the diameter d2 when d1 = 300 mm? The efficiency is 0.9.
Solution: d2 = 98,6 mm
Principle: A force F1 acts on a pressure piston with a small piston area A1 - a pressure within the closed system is generated. Since the pressure within a closed system is equal, this pressure acts also on the second piston with a large piston area A2.
Relation between the first and second piston:
first step: The pressure in both piston is the same => with the law of Pascal p = F/A you get:
F1 = F2 respectively: F1 = A1
. A1 A2 F2 A2
Oil quantity: The by the fist piston displaced volume V1 supplies the volume V2. Similarly, the work done on both pistons W = F * s is equal. The following applies:
the displaced volume V: the work carried out W:
approach: V1 = V2 approach: W1 = W2
. => A1 * s1 = A2 * s2 F1 * s1 = F2 * s2
2A3: With the help of a hydraulic press, a load of 80 kN is to be lifted. A force of F1 = 200 N is applied to the first piston with the diameter of 30 mm. Calculate:
- The required diameter of the load piston in mm
- The piston stroke s2, when the fist piston has a stroke of s1= 400 mm.
- d2 = 600 mm
- s2 = 1 mm