gefran BUy-1020 Manuel D'utilisation page 57

- 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 = 35 + 2,3 = 37,3 kgm
TOT M L
∆ω = [2Π * (n - n )] / 60 sec/min = 2Π * 1486 / 60 = 155 sec
1 2
Motor rated torque:
M = P / ω = (123000) / ( 2Π * 1486 / 60) = 795 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 = 580 - 79,5 = 500 Nm
bM b S
The average power of the braking process is given by:
P = (M * ∆ω) / 2 = 500 * 1558 * 0.5 = 39000 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.
The peak braking power is given by
P = (J * n * ∆ω * 2Π) / (t
PBR TOT 1
I = P / V = 90 kW/ 1150 = 125A
PBR PBR BR
R ≤ V / I = 1150 / 125= 9,2 Ω
BR BR PBR
It is therefore clear that the requirements are satisfied with 1 BUy-1065-6 unit.
57-GB
2
M = 0.1 M = 79,5 Nm
S M
* (n -n ) = (37,3 / 2) * (0.10472)
2 2 2
1 2
= 37,3 * 155 / 10 = 580 Nm
= 39000 * 10 = 390000 Joules o Wsec
* 60) = 90 kW
BR
—————— Braking Unit ——————
(n )
n
(J )
M
(J )
L
(M )
S
(n )
1
(n )
2
(t )
BR
(T)
and
-1
* 1486
2
therefore
and
1486 rpm
2.3 kgm
2
35 kgm
2
10% of the rated torque
1486 rpm
0 rpm
10 sec
120 sec
= 451620 Joules or Wsec
2