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Rexroth IndraDrive
Mechanical power for servo
Mechanical power for main
DC bus continuous power for
servo drives
DC bus continuous power for
main drives
DOK-INDRV*-HMV-*******-PR01-EN-P
drives
P
:
continuous mech. power for servo drives [kW]
mSe
M
:
effective motor torque [Nm]
eff
n
:
average motor speed [min
av
Fig. 8-6:
Mechanical power for servo drives
Main drives are primarily used with constant power over a specific speed
range. This means that when planning power supply, nominal power is
drives
important. The mechanical nominal power of the main drives is illustrated
in the operating characteristics or it can be calculated using nominal
speed and torque.
P
:
mechanical nominal power for main drives (shaft output) [kW]
mHa
M
:
motor nominal torque [Nm]
n
n
:
motor nominal speed [min
n
Fig. 8-7:
Mechanical power for main drives
The power supply unit must make the DC bus continuous power available
to all servo drives. All drives are operated simultaneously in only a few
applications which means that only the simultaneously occurring output
needs to be considered. For the calculation of the required DC bus
continuous power for typical NC feed axes on tool machines, it has
proven in practice that a so-called coincidence factor is included:
Number of Axes
Coincidence factor (F
Fig. 8-8:
Coincidence factors
P
:
DC bus continuous power for servo drives [kW]
ZWSe
P
... P
mSe1
mSen
F
:
coincidence factor
G
1,25:
constant for motor and controller efficiency
Fig. 8-9:
DC bus continuous power for servo drives
If several main drives are operated on one DC bus, then add the
simultaneously required power:
P
:
DC bus continuous power for main drives [kW]
ZWHa
P
... P
mHa1
mHan
1,25:
constant for motor and controller efficiency
Fig. 8-10:
DC bus continuous power for main drives
Chokes and auxiliary capacitors are selected in terms of the actually
required DC bus continuous power. It is fixed by the nominal power of the
spindle drives.
Note:
When selecting the power supply unit make sure that the DC
bus continuous power does not limit the short-term power of
the main drives.
Determination of Appropriate Power Supply Units
M
=
eff
P
mSe
9550
-1
]
M
=
n
P
mHa
9550
-1
]
1
1
1,15
)
G
+
+
P (
P
=
mSe
1
mSe
2
P
F
:
cont. mech. power for servo drives [kW]
=
+
P
P (
P
ZWHa
mHa
1
mHa
2
: mech. cont. power for main drives [kW]
*
n
av
*
n
n
2
3
4
5
1,32
1,75
2,0
+
...
P
* )
, 1
25
mSen
G
+
+
...
P
* )
, 1
25
mHan
8-3
6
2,25
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