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GEI-100350A
—————— DIMENSIONING... AND CORRESPONDING... ——————
4
2
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
E
BR
t*V
BR BR
II
AVBU AVBR
≥
f008
Sample calculation
Data:
- AC Input voltage 3 x 460 V
- Drive model 6KAV3015
- Rated motor power (P
M
) 15 HP
- Rated motor speed (n
n
) 3515 rpm
- Moment of inertia of the motor (J
M
) 0.033 kgm
2
- Moment of inertia loading the motor shaft (J
L
) 0.95 kgm
2
- Friction of the system (M
S
) 10% of motor nominal torque
- Initial braking speed (n
1
) 3000 rpm
- Final braking speed (n
2
) 0 rpm
- Braking time (t
BR
) 10 sec
- Cycle time (T) 120 sec
We will have:
J
TOT
= J
M
+ J
L
= 0.033 + 0.95 = 0.983 kgm
2
and
∆ω = [2Π * (n
1
- n
2
)] / 60 sec/min = 2Π * 3000 / 60 = 314 sec
-1
Rated motor torque:
M
M
= P
M
/ ω
n
= (15 * 745.7) / ( 2Π * 3515 / 60) = 30.4 Nm it follows that
M
S
= 0.1 M
M
= 3.04 Nm
The braking energy is given by:
E
BR
= (J
TOT
/ 2) * (2Π / 60)
2
* (n
1
2
-n
2
2
) = (0.983 / 2) * (0.10472)
2
* 3000
2
= 48509 Joules or Wsec