9 - AHP Merkle GmbH

1.3 General calculations
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Conversions
N
1 Pa = 1 = 1
1 MPa = 1 = 10 bar
1 PSI = 6.8948 • 10 3 Pa = 6.8948 • 10 -2 bar
1 N =
1 W =
Force / piston diameter
A: effective piston surface [mm 2 ]
F: force [N]
p: pressure [bar]
D: piston diameter [mm]
d: rod diameter [mm]
η: effi ciency of the hydraulic cylinder
The effi ciency [η], which is for the most part the result of the frictional losses (seals, guides), can be approximated
as 0.8. The larger the cylinder, the smaller the effect of friction on the overall force. At speeds of less than 0.05 m/s
(0.164 feet/s), the friction is practically independent of the pressure.
For piston diameters of 100 mm (3.94 inches) and larger the percentage loss is not more than 2%, even in
the worst case. With even larger piston diameters it can even be regarded as insignifi cant.
Example:
For cylinders with a piston diameter of less than 20 mm (0.79 inches) and an operating pressure of approx.
140 bar (2030 PSI) the frictional losses can be about 20%. For a piston diameter of 100 mm (3.94 inches)
this value is reduced to 2%.
It has been noted in practice that new seals have relatively high frictional values, which however become lower
as the operating time increases, thus increasing the effi ciency of the hydraulic cylinder. This should be taken into
account above all when the cylinders are being operated at low speeds (stick-slip effect), or low operating
pressures are present.
For hydraulic cylinders, the interrelationship between the force [F], the system pressure [p] and the piston area
[A] is produced by the following formula:
F = 0.1 • A • p • η
The force resulting from the system pressure is lower at the rod end
than at the piston end. The effective surface is calculated as follows:
A = Apiston - Arod =
As a general rule, the circular area [A] is calculated from the diameter [D] using the following formula:
A =
Alternatively from the force to be applied [F] and the pressure [p]:
A =
Determination of the piston diameter as a function of the system pressure and the required force:
D =
kg • m
s 2
m 2
kg • m 2
D 2 • π
4
s 3
F
p • η
N
mm 2
4 • F
p • π • η
kg
m • s 2
(D
4
2 – d2 ) • π