2 - Fagor Automation

Selection criteria
5.
SELECTING CRITERIA
Asynchronous spindle motor and servo drive selection
232
DDS
HARDWARE
Ref.1209
216
P N
P N
Calculations:
1. With speed between 0 and 1500 rpm.
2
------
60
2 J M N 2
M
2
---------------------- kW ------
1000t
60
2 0.13 1500 2
= = -------------------------------- = 6.41 kW 1 1000 0.5
2. With speed between 0 and 6000 rpm.
2
------
60
2 J M N 2 2
M
+ N
B
2 2 0.13 6000
------------------------------------- kW ------
2000t
60
2 1500 2
+
= = ------------------------------------------------------ = 10.89 kW 2 2000 2.5
Calculation of acceleration and braking time
After selecting the mechanical characteristics and the power of the drive,
the acceleration and braking time is calculated as follows:
Constant torque
area:
(0 < N M < N B )
Constant power
area:
(N B < N M < N max )
Constant torque &
power area:
(N B < N M < N max )
JM Inertia of the load in kg·m² as viewed from the motor shaft
TM Rated torque at base speed in N·m
Nmax Maximum motor speed in rpm.
N B
N M
Motor base speed in rpm.
Motor speed reached after a time period t in rev/min
Calculation of power with intermittent load
Forming the drive to the right dimensions has to be done with the greatest
care when the application involves a periodical starting and stopping
operation, frequently repeated as in the case of threading with a miller.
0
0
torque
speed
F. H5/11
Tr Ts Tf
Tc
Periodic start-stop operation
t 1
t 2
2 JM
NM = ------------------------------ s 60 TM 2 2
NB
2 JM
NM–
= ---------------------------------------------------- s 120 TMNB 2 2
NB
2 JMNM+
t 3 = t1+ t 2
= -------------------------------------------------- s 120 TMNB T R
=
For a cycle like the one shown in the figure F. H5/11 which includes acceleration
and stopping, the equivalent effective torque T R of equation must
be within the S1 dimension given for the drive torque.
2
TP -Tp
tr+ tf + TL
------------------------------------------------------
t c
2 ts
TL
Nm
Tp
time
time
Nm