Standard AC Motors
Guide for AC Motor Selection
In this section, we will discuss the fundamental criteria
involved in selecting small-size, standard AC motors such as
2. Calculating the Required Torque
induction motors and reversible motors. On a belt conveyor, the greatest torque is needed when
starting the belt. To calculate the torque needed for start-up,
Selection Procedure the friction coefficient (F) of the sliding surface is first
1. Required Specifications determined:
First, determine the basic required specifications such as
F m1.g 0.3159.8 ≒ 45 [N]
Product Line of
operating speed, load torque, power supply voltage and
Features and
AC Motors
frequency.
2. Calculate the Operating Speed Load torque (TL ) is then calculated by:
Induction and reversible motor speeds cannot be adjusted.
F · D 45100
Motor speed must be reduced with gearheads to match the TL 2550 [mN・m]
required machine speed. It is therefore necessary to 2· 20.9
determine the correct gear ratio.
The load torque obtained is actually the load torque at the
3. Calculate the Required Torque gearhead drive shaft, so this value must be converted into
Standard AC
Motor Types
Calculate the required torque for motor by the load torque. load torque at the motor output shaft. If the required torque at
List of
4. Select a Motor and Gearhead the motor output shaft is TM, then:
Use the required torque and speed to select a motor and
gearhead. TL 2500
TM 75.8 [mN・m]
5. Confirm the speed i · G 500.66
In a single-phase induction motor, starting torque is always
lower than the rated torque. Therefore, to drive a frictional (Gearhead transmission efficiency G = 0.66)
Use a safety margin of two for the required torque, taking into
Motor Selection
load, select the speed on the basis of starting torque. This will
consideration commercial power voltage fluctuation etc.
Guide for
cause the actual speed to exceed the rated speed. Also, the
and Use
motors are designed so that increases in motor temperature
are at their lowest when operating close to the rated speed of 75.82 ≒ 152 [mN・m]
rotation.
The suitable motor is one with a starting torque of 152 [mN・m]
EXAMPLE 1 or more. Therefore, motor 5IK40GN-CWE is the best choice.
Since a gear ratio of 50 is required, select the gearhead
Here is an example of how to select an induction motor to
5GN50K which may be connected to the 5IK40GN-CWE
drive a belt conveyor.
Information
motor.
General
In this case, a motor must be selected that meets the following
basic specifications.
Required specifications and structural specifications 3. Inertia load check
V Roller moment of inertia
1 1
J1 m2D 22 10.1222510-4 [kgm2]
8 8
Belt and work moment of inertia
Q&A
D Belt Conveyor
(D)2 0.12
Motor J2 m1 15 37510-4 [kgm2]
Gearhead
42 4
Total mass of belt and work .................... m1 ＝15kg Gear head shaft load inertia
Friction coefficient of sliding surface.......... μ＝0.3
Roller diameter ..................................... D＝100mm JJ1
J240010-4 [kgm2]
Mass of roller ........................................... m2 ＝1kg
Here, the 5GN50K permitted load inertia is:
Glossary
Belt roller efficiency.................................... η＝0.9
Belt speed .............................. V＝140mm/s±10％ J G0.7510-4502
Motor power supply ........... Single-Phase230V50Hz 187510-4 [kgm2]
Movement time........................ Eight hours per day
Refer to page A-21 to confirm this calculated value.
1. Determining the Gearhead Reduction Therefore, J < JG, the load inertia is less than the permitted
inertia, so there is no problem.
Ratio
Speed at the gearhead output shaft: Since the motor selected has a rated torque of 300 [mN・m],
which is somewhat larger than the actual load torque, the
V · 60 (14014)60 motor will run at a higher speed than the rated speed.
NG 26.72.7 [r/min]
·D 100 Therefore the speed is used under no-load conditions
(approximately 1470r/min) to calculate belt speed, and thus
Because the rated speed for a 4-pole motor at 50Hz is determine whether the product selected meets the required
12001300 r/min, the gear ratio (i ) is calculated as follows: specifications.
i
12001300 12001300
40.854.2 NM · · D 1470100
NG 26.72.7 V 154 [mm/s]
60 · i 6050
From within this range a gear ratio of i =50 is selected. (Where NM is the motor speed)
The motor meets the specifications.
A-8 ORIENTAL MOTOR GENERAL CATALOGUE
STANDARD
Standard AC Motors
EXAMPLE 2
This example demonstrates how to select a motor with an
electromagnetic brake for use on a tabletop moving vertically
2. Calculating the Required Torque
MOTORS
on a ball screw. F, the load weight in the direction of the ball screw shaft, is
In this case, a motor must be selected that meets the following obtained as follows:
basic specifications.
F FA
m1.g (sin
cos )
0
309.8 (sin 90°
0.05 cos 90°)
Required and structural specifications 300 [N]
Preload F0:
Motor
Product Line of
Features and
AC Motors
F
Gearhead F0 100 [N]
Coupling 3
Load torque TL:
Ball Screw FPB 0F0PB 3005 0.31005
TL
FA 2 2 20.9 2
Slide Guide
289 [mN·m]
v
Standard AC
Motor Types
List of
w1
This value is the load torque at the gearhead drive shaft, and
must be converted into load torque at the motor output shaft.
The required torque at the motor output shaft (TM ) is given by:
TL 289
TM 39.6 [mN·m]
i · G 90.81
Total mass of table and work ..................... m1 = 30kg
Table speed .................................... V = 122[mm/s] (Gearhead transmission efficiency G = 0.81)
Motor Selection
Ball screw pitch....................................... PB = 5[mm] Use a safety margin of two for the required torque, taking into
Guide for
and Use
Ball screw efficiency....................................... = 0.9 consideration commercial power voltage fluctuation etc.
Ball screw friction coefficient........................ 0 = 0.3 39.62 = 79.2 [mN·m]
Friction coefficient of sliding surface
(Slide guide)............................................... = 0.05 To find a motor with a start-up torque of 79.2 [mN·m] or more,
select motor 4RK25GN-CWME. This motor is equipped with
Motor power supply ............ Single phase 230V 50Hz
an electromagnetic brake to hold a load. The gearhead with a
Ball screw total length...........................LB = 800[mm]
reduction ratio of 9 that can be connected to motor model
Ball screw shaft diameter .......................DB = 20[mm]
4RK25GN-CWME is 4GN9K.
Information
Ball screw material
General
.................................Iron (density =7.9103kg/m3)
Distance moved for one rotation of ball screw .........A = 5[mm]
External force................................................FA =0[kg] 3. Load inertia check
Ball screw tilt angle ........................................... =90
LBDB4
Movement time........................................................... Ball screw moment of inertia J1
32
Brake must provide holding torque Intermittent
7.91030.8(0.02)4
operation, five hours per day. Load with repeated
32
starts and stops must be held.
Q&A
0.9910-4 [kgm2]
m1A2
Table and work moment of inertia J2
42
1. Determining the Gear Ratio
300.0052
42
Speed at the gearhead output shaft:
0.1910-4 [kgm2]
Glossary
V · 60 (122) 60
NG 14424 [r/min] Gear head shaft total load inertia J1.1810-4 [kgm2]
PB 5
Here, the 4GN9K permitted load inertia is:
Because the rated speed for a 4-pole motor at 50Hz is
J G0.310-492
12001300r/min, the gear ratio (i ) is calculated as follows:
25.110-4 [kgm2]
12001300 12001300
i 7.110.8
NG 14424 Refer to page A-21 to confirm this calculated value.
Therefore, J < JG, the load inertia is less than the permitted
From within this range a gear ratio of (i ) = 9 is selected. inertia, so there is no problem.The same as for selected
Example 1, there is margin for the torque, so the rotation rate
is checked with the no-load rotation rate (about 1470 r/min).
NM · P
V 13.6 [mm/s]
60 · i
(where NM is the motor speed).
This confirms that the motor meets the specifications.
ORIENTAL MOTOR GENERAL CATALOGUE A-9
