6. - Fagor Automation

Drive. CT series
3.
CT
Ref.1109
28
therefore:
M = J·ω/tb = J·π ·nN /30·tb = 1.479xπ x1500 / 30·tb = 1114.05 N·m
and therefore:
tb = 0.21 s is the minimum time the motor can be stopped and the time required by the
application for braking, 5 s, is therefore within the specs of the 100 kW drive.
3. Torque and power needed for a required motor braking time of 5 s
M = 1.479xπ x1500 / 30x5 = 46.46 N·m
and the power, therefore:
P = π x n x M / 30x10³ = π x n x M / 30x10³ with:
Symb. Description Units
P Power kW
n Motor Speed rpm
M Max. braking torque Nm
P = π x 1500 x 46.46 / 30x10³ = π x n x M / 30x10³ = 7.29 kW
4. Braking resistance
P = V²/R
7.29 x10³ = 780²/R
R = 83.45 Ω is the minimum braking resistance with a 110 kW drive.
The calculation is the result of assuming constant speed, but the speed gets lower as the
movement of the load slows down. Therefore, the avarage power needed to obtain the resistance
value is:
Paverage = 0.5 x J·ω²/t
Paverage = 0.5 x 1.479 x (2xπ x1500/60)²/5 = 3.64 kW. This braking power is needed for 4 seconds
every 30 seconds.
Assuming that the resistors admits (dissipates) this overload; then, for a continuous duty cycle:
PN = 3.64x5/30 = 0.6 kW that is the power that would be required for instantaneous braking.