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(vi) Calculation of actual torque (Trms)
Actual torque can be calculated by the following formula.
2
×
×
((0.86
0.05 2) + (0.03
T
=
rms
0 5
.
Since the rated torque of 0.2 [kW] motor is 0.637 [N・m], a relation of "actual torque < rated torque " is satisfied. Therefore, the temporarily
selected motor is actually applicable.
(vii) Examination of braking resistor
In the above torque pattern, the hatched area corresponds to the regenerative braking mode. Braking power is as follow.
P
= 0.105 × T
× N × (1/2) = 0.105 × 0.86 × 3000 × (1/2) = 135 [W]
1
AC
From the above result, mean braking power per cycle is :
×
135
0.05
= 14 [W]
P =
0.5
(4) Determination of external braking resistor
Because of the braking resistor is not built-in the 0.2 [kW] motor, requiring of external braking resistor provision is reviewed as follows :
(a) Energy (E
) on the mechanical equipment system during deceleration
G
1
=
+
E
(
J
J
) (
2
N
/
60
・ π
G
M
L
2
1
−
=
×
3
+
( .
0 0216 10
0 11 10
.
2
(b) Consumption energy (E
) due to load torque
L
E
= ( 2π / 60 )× T
× Ｎ × t
L
L
= ( 2π / 60 )× 0.03 × 3000 × 0.05 × ( 1/2 ) = 0.24 [ J ]
(c) Consumption energy (E
) in motor winding
M
2
E
= 3 × ( R × I
)× t
= 3 × R × (( T
M
DC
= 3 × 2 × (( 0.86 / 0.637 × 1.5 )
(d) Absorbing energy (E
) of motor
S
1
1
−
=
2
=
×
6
E
CV
(
660 10
S
2
2
Where,
R : Phase resistance of stator winding of GYC201 type motor is 2 [Ω], at room temperature
V : Power supply voltage 200 [ V ]
C : DC intermediate capacity of RYS201 amplifier 660 [μF]
(e) Consumption and absorbing energies ( E
system ( E
= 6.5 [ J ]). Therefore, external braking resistor provision is not required in this example.
G
2
×
×
0 05
.
))
1
=
0 38
.
[N・m]
2
)
2
×
2
3000
π
−
×
3
×
=
)
6 5
. [ ]
60
× ( 1/2 )
DC
/ T
× I
DC
R
2
) × 0.05 = 1.2 [ J ]
1
×
2
−
×
2
=
)
(
370
(
200
2
) )
18 7
2
+ E
+ E
) is 20 [ J ], which is larger than the energy on the mechanical equipment
L
M
S
J
2
)
) × t
R
DC
. [ ]
J
11-9
MHT260a (Engl.)
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