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- 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
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