Lust CDA3000 - Sensorless Flux Control (SFC)

2-10
Engineering Guide CDA3000
2 Drive definition
2.3.1 Example 1:
Traction drive
1. Determine power requirement to move the application
2. Select motor
The selected motor must have a power rating higher than P
Drive
. Select
the motor from the list.
3. Calculate gross output
P
Gross
= P
a
+ P
F
+ P
aR
= 264W + 9W + 65W = 338W
For more details on “Selection of inverter modules” refer to sections 3.3 to
3.6.
Example: Z-axis of a manipulator
m = 51.5 kg
a = 3 m/s
2
v = 1.5 m/s
η = 0.88
ta = 0.5
µ = 0.01
Selected motor: Type 71L/4, 370W, J
M
= 0.00073 kgm
The motor is to be run at max. 2000 rpm
(70 Hz characteristic).
P
a
mav⋅⋅
η
------------------
51 5kg, 3m s
2
⁄ 15ms⁄,⋅⋅
088,
---------------------------------------------------------------
264W===
P
F
mgµ v⋅⋅⋅
η
--------------------------
51 5kg, 98ms
s
⁄, 001, 15ms⁄,⋅⋅⋅
088,
------------------------------------------------------------------------------------
9W===
P
Fahr
P
a
P
F
273W=+=
P
aR
J
M
n
2
M
⋅
91 2, t
a
⋅
--------------------
0 00073kgm
2
, 2000
2
min
1–
⋅
91 2, 05,⋅
------------------------------------------------------------------ 65W== =