Selection criteria
5.
224
Ref.1912
DDS
HARDWARE
· 210 ·
Limited acceleration and choke. Choke = (acceleration / t).
The capability demanded from the motor is determined by the following
formulas:
We will now give several examples of calculations using a mechanical
specifications and for a standard motor. The results could vary from real
ones through mechanical losses, fluctuations in mains voltage, or
inaccuracies of mechanical data.
F. H5/10
Acceleration and choke limit.
T. H5/4 Acceleration and choke limit.
Method
Acceleration and choke limit.
Control
Progressive linear acceleration, avoiding abrupt variations of
transmitted torque.
Comment
Approach square sine function (bell shape) for the speed.
Capacity required by the motor
in the constant torque area:
(0 < N
M
< N
B
)
Capacity required by the motor
in the constant torque and
constant power area:
(0 < N
M
< N
max
)
J
M
Inertia of the load in kg·m² as viewed from the motor shaft
P
N
Rated power at base speed kW
Nmax
Maximum motor speed in rpm.
N
B
Motor base speed in rev/min.
N
M
Motor speed reached after a time period t in rpm
t
Acceleration time until N
M
(in seconds) is reached
Example
Data:
Acceleration time:
Between 0 and 1 500 rpm in 0.5 s. (1)
Between 0 and 6 000 rpm in 2.5 s. (2)
Motor inertia:
J
motor
= 0.13 kg·m²
Motor base speed:
N
b
= 1 500 rpm
Calculations:
1.
With speed between 0 and 1 500 rpm.
P
N
2
60
------
2
J
M
N
M
2
1000 t
--------------------kW=
P
N
2
60
------
2
J
M
N
M
2
N
B
2
+
2000 t
---------------------------------------kW=
2. With speed between 0 and 6 000 rpm.
P
N
2
60
-------
2
J
M
N
M
2
1000t
---------------------- k W
2
60
-------
2
0.13 1500
2
1000 0.5
--------------------------------
6.41 kW 1===
P
N
2
60
-------
2
J
M
N
M
2
N
B
2
+
2000t
--------------------------------------kW
2
60
-------
2
0.13 6000
2
1500
2
+
2000 2.5
-------------------------------------------------------
10.89 kW2===
To Next Page
Loading...
Need help?
General
