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4.1 Dimensioning example @ 460V
Data:
- Mains voltage
- Inverter
- Motor rated power
- Motor rated speed
- Moment of inertia of the motor
- Moment of inertia loading the motor shaft
- Friction of the system
- Initial braking speed
- Final braking speed
- Braking time
- Cycle time
Total moment of inertia:
J
= J
+ J
= 0.033 + 0.95 = 0.983 kgm
TOT
M
L
∆ω = [2Π * (n
- n
)] / 60 sec/min = 2Π * 3000 / 60 = 314 sec
1
2
Motor rated torque:
M
= P
/ ω
= (15 * 745.7) / ( 2Π * 3515 / 60) = 30.4 Nm
M
M
n
Machine friction:
The braking energy is given by:
E
= (J
/ 2) * (2Π / 60)
BR
TOT
Taking into account also the system friction, the braking energy to be dissipated by the
braking unit is lower.
The required braking torque is:
M
= (J
* ∆ω) / t
b
TOT
BR
The braking torque consists of two sections: the machine friction and the torque to be supplied
by the motor electric braking:
M
= M
- M
= 30.9 - 3.04 = 27.86 Nm
bM
b
S
The average power of the braking process is given by:
P
= (M
* ∆ω) / 2 = 27.86 * 314 * 0.5 = 4374 W
AVE
bM
The new value of the braking energy is therefore:
New E
= P
* t
BR
AVE
BR
it is obviously lower than the previous one.
M
= 0.1 M
= 3.04 Nm
S
M
* (n
-n
) = (0.983 / 2) * (0.10472)
2
2
2
1
2
= 0.983 * 314 / 10 = 30.9 Nm
= 4374 * 10 = 43740 Joules or Ws
—————— Braking Unit ——————
3 x 460 V
ADV3110
(P
)
M
(n
)
n
(J
)
M
(J
)
L
(M
)
S
(n
)
1
(n
)
2
(t
)
BR
(T)
and
2
-1
* 3000
2
15 HP
3515 rpm
0.033 kgm
2
0.95 kgm
2
10% of the rated torque
3000 rpm
0 rpm
10 sec
120 sec
= 48509 Joules or Wsec
2
54-GB
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