GE 6KBU300 Braking Unit

6KBU300

—————— DIMENSIONING... AND CORRESPONDING... —————— 4

But, if we want to take into account also the friction of the system, the braking energy that the braking unit will

need to dissipate is lower. To do this we can calculate EB as follows:

The required braking torque is

Mb = (JTOT * ∆ω) / tBR = 0.983 * 314 / 10 = 30.9 Nm

In reality the friction torque “helps” the motor, so we obtain

MbM = Mb - MS = 30.9 - 3.04 = 27.86 Nm

The brake process average power is given by

PAVE = (MbM * ∆ω) / 2 = 27.86 * 314 * 0.5 = 4374 W

And the new value of braking energy that we obtain in this way is

New EBR = PAVE * tBR = 4374 * 10 = 43740 Joules or Ws

which is obviously lower than the previous one.

The peak braking power is given by

PPBR = (JTOT * n1 * ∆ω * 2Π) / (tBR * 60) = 9.7 kW then we continue with

IPBR = PPBR / VBR = 9700 / 745 = 13A and

RBR ≤ VBR / IPBR = 745 / 13 = 57 Ω

Being IPBR = 13A, here we can already see that the unit 6KBU300-20 covers our needs. Now we have to

choose the resistor:

The nominal power of the resistor has to be

PNBR = (PPBR * tBR) / 2T = (9700 * 10) / 240 = 404 W

As we can see, the nominal power of the resistor is relative low due to a low duty-cycle (10 / 120) but the resistor

must be able to withstand the energy that is applied to it during the 10 seconds of braking. This energy is 43740

Joules. If we go on the table of normalized resistors, the type BRR 1K0T 49R has a nominal power that would

be sufficient but the value of EBR is too low (21kWsec).

For this reason our final choice is the type BRR 1K3T 31R that has EBR = 44kWsec.