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**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 following proportionality applies:

pressure p ~ height h.

The force that a fluid exerts on a surface is calculated using Pascal's law:

**Law of Pascal: **F = p * A **Units: **1 bar = 10^{5} Pa (Pascal) = 100 kPa; 1 Pa = 1 N/m^{2}

**Example: **A cylinder with a diameter of d_{1} = 15 cm has to hold a mass of 1500 kg on height. How much rises the pressure within the system? *Solution: p = 8,327 bar*

Work machines, which can be described by the law of Pascal:

**Pressure intensifier:**

Principle: Two differently sized pistons are mechanically connected by a piston rod. If the piston surface A_{1} is pressurized, so the force F_{1} on the piston rod acts directly on the small piston area A_{2}:

F_{1} = F_{2 }

p_{e1} * A_{1 }* η = p_{e2} * A_{2}

p_{e} operating pressure in bar

A piston area in cm^{2}

F piston force in N

**Example: **By using a pressure intensifier the pressure from P_{E1} = 30 bar has to be increased to P_{E2} = 250bar. How big must be the diameter d_{2} when d_{1} = 300 mm? The efficiency is 0.9. (*Solution: d2 = 98,6 mm)*

**Hydraulic press**

Principle: A force F_{1} 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 A_{2.}

**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:

_{ } __F___{1}__ __ = __F___{2}__ __ respectively: __F____ _{1}__ =

__A__

_{1}. A

_{1 }A

_{2}F

_{2}A

_{2}

**Oil quantity:** The by the fist piston displaced volume V_{1} supplies the volume V_{2}. 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: ** V_{1} = V_{2 }**approach: **W_{1} = W_{2}

. => A_{1} * s_{1} = A_{2} * s_{2} F_{1} * s_{1} = F_{2} * s_{2} ** **

**2A3:** With the help of a hydraulic press, a load of 80 kN is to be lifted. A force of F_{1} = 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 s
_{2}, when the fist piston has a stroke of s_{1}= 400 mm.

- d2 = 600 mm
- s2 = 1 mm