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Bosch Rexroth AG Hydraulics
Calculation
Oil volume
According to the pressures p
0
V
... V
result.
0
2
In this connection, V
is also the accumulator's nominal volume.
0
The available oil volume V corresponds to the difference of
the gas volumes V
and V
:
1
2
(3)
V V
– V
1
2
The gas volume that can be changed within one pressure dif-
ference is determined by the following equations:
a) To isothermal changes of condition of gases, i.e. if the
gas cushion changes so slowly that there is enough time
for the complete heat exchange between the nitrogen and
its environment and the temperature therefore remains
constant, the following applies:
p
• V
= p
• V
= p
• V
0
0
1
1
2
Calculation diagram
For the graphical determination, the formulas (4.1) and (4.2)
in diagrams on pages 7 to 10 are implemented. Depending
on the task, the available oil volume, the accumulator size or
the pressures can be determined.
Correction factor K
and K
i
a
The equation (4.1) or (4.2) is only true for ideal gases. In
the behavior of real gases, considerable deviations result at
operating pressures of more than 200 bar, which have to be
considered by correction factors. They can be seen from the
following diagrams. The correction factors by which the ideal
sampling volume V is to be multiplied lie within the range
from 0.6 ... 1.
Isothermal
1,0
0,9
0,8
0,7
0,6
0,1
0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1,0
... p
, the gas volumes
2
(4.1)
2
V
= V
• K
real
ideal
i
p
/ p
→
1
2
b) To adiabatic changes of condition, i.e. if the gas cushion
changes quickly with the nitrogen temperature changing as
well, the following applies:
p
• V
= p
• V
χ
χ
0
0
1
1
χ = Ratio of the specific heats of the gases
(adiabatic exponent), for nitrogen = 1.4
In practice, the changes of condition rather follow adiabatic
laws. The charging is often isothermal, the discharge adiabatic.
Considering the equations (1) and (2), V lies at
70 % of the nominal accumulator volume. As reference
point, the following applies:
V
= 1.5 ... 3 x V
0
Application of the calculation diagrams
(see page 7 to 10)
V
2
Available oil
volume
V
1
Adiabatic
1,0
0,9
0,8
0,7
0,6
0,1
0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1,0
HAB RE 50170/12.10
(4.2)
= p
• V
χ
2
2
50 % to
(5)
Gas filling pressure
s
2
s
1
p in bar →
P
P
1
2
Working pressure range
V
= V
• K
real
ideal
p
/ p
→
1
2
a
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