Filter
Top Products
6KBU300
—————— DIMENSIONING... AND CORRESPONDING... ——————
4
3
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 E
B
as follows:
The required braking torque is
M
b
= (J
TOT
* ∆ω) / t
BR
= 0.983 * 314 / 10 = 30.9 Nm
In reality the friction torque “helps” the motor, so we obtain
M
bM
= M
b
- M
S
= 30.9 - 3.04 = 27.86 Nm
The brake process average power is given by
P
AV E
= (M
bM
* ∆ω) / 2 = 27.86 * 314 * 0.5 = 4374 W
And the new value of braking energy that we obtain in this way is
New E
BR
= P
AV E
* t
BR
= 4374 * 10 = 43740 Joules or Ws
which is obviously lower than the previous one.
The peak braking power is given by
P
PBR
= (J
TOT
* n
1
* ∆ω * 2Π) / (t
BR
* 60) = 9.7 kW then we continue with
I
PBR
= P
PBR
/ V
BR
= 9700 / 745 = 13A and
R
BR
≤ V
BR
/ I
PBR
= 745 / 13 = 57 Ω
Being I
PBR
= 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
P
NBR
= (P
PBR
* t
BR
) / 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 E
BR
is too low (21kWsec).
For this reason our final choice is the type BRR 1K3T 31R that has E
BR
= 44kWsec.