GEI-100350A
—————— DIMENSIONING... AND CORRESPONDING... ——————
42
Being normally n2 = 0 (stop), we will have that:
E=
BR
P*t
PBR BR
1
2
f006
Braking unit features:
II
PBU PBR
≥
f007
This means that the peak current admissible by the 6KBU300-... must be equal or higher than the effective one.
Then for the average current we will have:
I=
AVBR
EBR
t*V
BR BR
II
AVBU AVBR
≥
f008
Sample calculation
Data:
- AC Input voltage 3 x 460 V
- Drive model 6KAV3015
- Rated motor power (PM) 15 HP
- Rated motor speed (n n) 3515 rpm
- Moment of inertia of the motor (JM) 0.033 kgm2
- Moment of inertia loading the motor shaft (JL) 0.95 kgm2
- Friction of the system (MS) 10% of motor nominal torque
- Initial braking speed (n1) 3000 rpm
- Final braking speed (n2) 0 rpm
- Braking time (tBR) 10 sec
- Cycle time (T) 120 sec
We will have:
JTOT= JM + JL = 0.033 + 0.95 = 0.983 kgm2and
∆ω = [2Π * (n1 - n2)] / 60 sec/min = 2Π * 3000 / 60 = 314 sec-1
Rated motor torque:
MM = PM / ωn = (15 * 745.7) / ( 2Π * 3515 / 60) = 30.4 Nm it follows that
MS = 0.1 MM = 3.04 Nm
The braking energy is given by:
EBR = (JTOT / 2) * (2Π / 60)2 * (n12 -n22) = (0.983 / 2) * (0.10472)2 * 30002 = 48509 Joules or Wsec