Dimensioning Example @ 690 V - gefran BUy-1020 Manuel D'utilisation

Motor rated torque:
M = P / ω = (10400) / ( 2Π * 1785 / 60) = 557 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 = 678 - 55.7 = 622 Nm
bM b S
The average power of the braking process is given by:
P = (M * ∆ω) / 2 = 622 * 178 * 0.5 = 55300 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 = 120 kW/ 965 = 125A and
PBR PBR BR
R ≤ V / I = 965 / 125= 7.7 Ω
BR BR PBR
It is therefore clear that the requirements are satisfied with 2 BUy-1075-5 units.
Resistor choice
The resistor rated power must be:
P = (P / 2 ) * (t
NBR PBR
For this reason the final choice is the model BR T8K0-7R7 (rated P
320kWsec) which has E

4.3 Dimensioning example @ 690 V

Data:
- Mains voltage
- Motor efficiency
- Motor rated power
M = 0.1 M = 55.7 Nm
S M
* (n -n ) = (38.1 / 2) * (0.10472)
2 2 2
1 2
= 38.1 * 178 / 10 = 678 Nm
= 55300 * 10 = 553000 Joules or Ws
* 60) = 120 kW
BR
/ T) = (60000 / 2 ) * (10 / 120) = 2500 W (2 res. x 2 BUy = 5000 W)
BR
≥ (553/2)kWsec for a t
BR
—————— Braking Unit ——————
* 1700 = 603000 Joules or Wsec
2 2
therefore
NBR
= 10 sec.
br
3 x 690 V
(%) 95.6
(P ) 132 kW
M
= 8000 W, rated E =
56-GB