SINAMICS G130
Engineering Information
SINAMICS Engineering Manual – November 2015
Ó Siemens AG
299/528
1. The result of the mean braking power calculation is as follows:
P
mean
= [ (150 kW • 20 s) + (0 kW • 70 s) ] / 90 s
= 33.33 kW
The braking unit must have a continuous power capability of more than 1.125 • P
mean
. The following thus applies:
P
DB
≥ 1.125 • 33.33 kW
≥ 37.5 kW
2a. Checking the required peak power for a factory-set upper response threshold of V
DClink
= 774 V according to k = 1
P
peak
> 5 • k • P
DB
?
150 kW > 5 • 1 • 37.5 kW ?
> 187.5 kW ?
The condition is not fulfilled, i.e. the required peak power of 150 kW is not higher than the peak power of 187.5 kW
which can be supplied by a braking unit with a continuous power rating of 37.5 kW. The mean braking power is thus
the decisive criterion for selecting the Braking Module and braking resistor
A braking unit with a continuous power rating of
P
DB
≥ 1.125 • P
mean
≥ 37.5 kW
is therefore needed. The braking unit with P
DB
= 50 kW or P
20
= 200 kW which can be selected for the Power Module
is therefore suitable for this application.
2b. Checking the required peak power for a reduced lower response threshold of V
DClink
= 673 V according to
k = 0.756:
P
peak
> 5 • 0.756 • P
DB
?
150 kW > 5 • 0.756 • 37.5 kW ?
> 141.75 kW ?
The condition is fulfilled, i.e. the required peak power of 150 kW is higher than the peak power of 141.75 kW which
can be supplied by the braking unit with a continuous power rating of 37.5 kW. The peak power of the braking unit is
thus the decisive criterion for selecting the Braking Module and braking resistor.
A braking unit with a continuous power rating of
P
DB
≥ [1 / (5 • k)] • P
peak
≥ [1 / (5 • 0.756)] • 150 kW
≥ 39.68 kW
is therefore needed. The braking unit with P
DB
= 50 kW or P
20
= 200 kW which can be selected for the Power Module
is therefore suitable for this application.
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