GE Industrial Solutions 6KBU300 Braking Unit - Manual (Page 20)

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:
I I
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
I I
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
= [2
Π
* (n
1
- n
2
)] / 60 sec/min = 2
Π
* 3000 / 60 = 314 sec -1
Rated motor torque:
M
M
= P
M
/
ω
n
= (15 * 745.7) / ( 2
Π
* 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) * (2
Π
/ 60) 2 * (n
1
2 -n
2
2 ) = (0.983 / 2) * (0.10472) 2 * 3000 2 = 48509 Joules or Wsec