Fuji Electric FRENIC-Ace series, Руководство пользователя

Страница: 592 / 788

10.3 Equations for Selections
10-10
(3) For a load running horizontally
Assume a carrier table driven by a motor as shown in Figure 10.3-1. If the table speed is υ (m/s) when the motor
speed is N
M
(r/min), then an equivalent distance from the shaft is equal to 60· υ / (2 π ·N
M
) (m). The moment of inertia
of the table and load to the shaft is calculated as follows:
) m kg ( ) W W ( )
N
2
60
( J
2
0
2
M
• •
•
•
+
π
υ
=
(Equation 10.3-13)
(4) For a vertical or inclined lift load
The moment of inertia J (kg·m
2
) of the loads connected with a rope as shown in Figure 10.3-2 and Figure 10.3-3 is
calculated with the following equation using the mass of all moving objects, although the motion directions of those
loads are different.
) m (kg ) W W (W )
N
2
60
( J
2
B 0
2
M
• •
•
•
+ +
π
υ
=
(Equation 10.3-14)
[ 2 ] Calculation of the acceleration time
Figure 10.3-5 shows a general load model. Assume that a motor drives a load via a reduction-gear with efficiency
η
G
. The time required to accelerate this load in stop state to a speed of N
M
(r/min) is calculated with the following
equation:
) s (
60
) 0
N
( 2
J J
t
M
G
L M
G
2 1
ACC
−
−
•
•
π
η
τ τ
η
+
=
(Equation 10.3-15)
where,
J
1
: Motor shaft moment of inertia (kg·m
2
)
J
2
: Load shaft moment of inertia converted to motor shaft (kg·m
2
)
τ
M
: Minimum motor output torque in driving motor (N·m)
τ
L
: Maximum load torque converted to motor shaft (N·m)
η
G
: Reduction-gear efficiency.
As clarified in the above equation, the equivalent moment of inertia becomes (J
1
+J
2
/ η
G
) by considering the
reduction-gear efficiency.
Figure 10.3-5 Load Model Including Reduction-gear