TM 5-3895-374-24-1
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Resistance calculation example
A motor with the following characteristics: - power
4 kW,
-rated speed 1450 rpm,
- moment of inertia 0,0135 kg m2
drives a machine with: - inertia 10 times that of the motor,
- resistive torque one tenth of the rated motor torque
The requirement is to stop in 5 seconds from rated speed at a rate of 2 cycles per minute
Rated angular speed
:wn = 2pN = 2p1450= 152 rad/s
60
60
Rated motor torque
Cn
Pn = 4000 = 26,3 N m
wn
152
*/
Resistive torqueCr = 26,3 = 2,63 N m
10
Total inertia
J = 0,0135 + 10 x 0,0135 = 0,1485 kg m2
Braking torque
C= J Dw = 0,1485 x 152 =4,52 N m
Dt
5
Motor braking torque
Cb = C - Cr = 4,52 - 2,63 = 1,89 N m
Instantaneous
braking power
Pb = Cb w = 1,89 x 152 = 287 W
Average braking
power during
deceleration
Pbd = 0,5 Cb Dw = 0,5 x 1,89 x 152 = 144 W
Cycle time
T =
60 = 30 s
2
Average braking power
during one cycle
Pbm = Pbd t = 144 x 5 = 24 W
T
30
Standardized resistance VY1-ADR100W072 is suitable
- rated power Pn = 72 W, donc> Pfm,
- maximum power possible for 5 seconds (curve p 65)
Pmax = 10 x 72 = 720 W, thus > Pf
CAUTION
A Precise calculation of the resistance, as shown in the example above, is absolutely essential
for severe applications requiring high braking powers hoisting (vertical movements), machines
with very high inertia, driving loads,....
If the required braking torque is high, choose a resistance with an ohmic value equal to, or slight’
greater than the minimum value given in the table on table page 64
(page 3-218)
