Lust CDA3000 - Example 2: Lifting Drive

2-12
Engineering Guide CDA3000
2 Drive definition
2.3.2 Example 2:
Lifting drive
1. Determine power requirement to move the application
2. Select motor
The selected motor must have a power rating higher than P
Lift
. Select the
motor from the list.
3. Calculate gross output
P
Gross
= P
a
+ P
F
+ P
H
+ P
aR
= 43W + 1W + 42W + 164W= 250W
For more details on “Selection of inverter modules” refer to sections 3.3 to
3.6.
Example: Z-axis of a manipulator
m = 2.5 kg
a = 10 m/s
2
v = 1.5 m/s
η = 0.88
ta = 0.15
µ = 0.01
Selected motor:
Type 71S/4, 250W, I
M
= 0.00056 kgm
2
The motor is to be run at max. 2000 rpm
(70 Hz characteristic).
P
a
mav⋅⋅
η
------------------
25kg, 10m s
2
⁄ 15ms⁄,⋅⋅
088,
---------------------------------------------------------------
43W===
P
F
mgµ v⋅⋅⋅
η
--------------------------
25kg, 98ms
s
⁄, 001, 15ms⁄,⋅⋅⋅
088,
----------------------------------------------------------------------------------
1W===
P
H
mgv⋅⋅
η
------------------
25kg, 98ms
s
⁄, 15ms⁄,⋅⋅
088,
-----------------------------------------------------------------
42W===
P
Lift
P
a
P
F
P
H
86W=++=
P
aR
J
M
n
2
M
⋅
91 2, t
a
⋅
--------------------
0 00056kgm
2
, 2000
2
min
1–
⋅
91 2, 015,⋅
------------------------------------------------------------------ 164W== =