19 - PVD3627-August 2011
3.1.4.3. Numerical example:
The MGV motor taken to illustrate the calculations is the type
MGV840CAD
Corresponding data are as follows:
P
S1
=
P
S6
=
63
kW
N
base
= 10300 rpm
N
max
= 24000 rpm
M
S1
=
58.4 Nm
M
S6
=
93.5 Nm
J
motor
=
0.01455 kgm²
We will consider:
M
resistant
=
6 Nm
J
load
=
J
motor
0.01455 kgm²
Method #1
rad/s
1079
60
10300
*
*
2
60
*
*
2
=
=
Ω
π
π
base
base
N
rad/s
2513
60
24000
*
*
2
60
*
*
2
max
max
=
=
Ω
π
π
N
kgm²
0.02910
001455
.
0
01455
.
0
=
+
+
=
Σ
load
motor
J
J
J
W
10776
2
)
1079
2513
(
*
6
2
)
(
*
max
tan
tan
=
+
=
Ω
+
Ω
≈
base
t
resis
t
resis
M
P
Nm
52.4
6
4
.
58
tan
=
−
=
−
=
t
resis
motor
M
M
M
Acceleration duration t
1
“at constant Torque” from 0 to the
Base speed
:
s
0.599
4
.
52
02910
.
0
*
1079
*
1
=
=
Σ
Ω
=
M
J
t
base
Nm
64.4
6
4
.
58
tan
=
+
=
+
=
t
resis
motor
M
M
M
Deceleration time t
1
“at constant Torque” from the
Base speed
to 0:
s
0.487
4
.
64
02910
.
0
*
1079
*
1
=
=
Σ
Ω
=
M
J
t
base
The power available for the acceleration is equal to:
Acceleration time t
2
“at constant Power” from
Base speed
to
Maximal speed
:
W
52224
10776
63000
tan
=
−
=
−
=
t
resis
mot
P
P
P
So the estimated
s
1.436
52224
²)
1079
²
2513
(
*
02910
.
0
*
5
.
0
²)
²
(
2
1
max
2
=
−
=
Ω
−
Ω
Σ
=
P
J
t
base
duration
t
2
is given by the formula:
s
t
t
t
2.035
436
.
1
599
.
0
2
1
=
+
+
=
Λ
Total acceleration time
∆
t
from 0 to
Maximal speed
is given by the sum of t
1
and t
2
:
