11/29/2012 12:54 PM | |
Posts: 24 Rating: (1) |
Hi! You have to calculate first the torque produced by your actual DC motor: M=9554*P[KW]/rated speed => M=9554*68/2500=259,86Nm so you can say your DC motor can deliver 260Nm of torque at 2500rpm rated speed. I used SIZER to configure a suitable motor to replace your DC motor. The results of the SIZER configuration is attached. You can use 1PH8186-1HF00-1DA1 servomotor and a S120 AC/AC drive system. This servomotor has forced ventilation(like your DC motor has) and can be used at constant torque application. It cand deliver 260Nm constantly between 0-2500rpm altghouh it is 1500rpm rated. You can see in the torque characteristic in the attachment. Good luck! AttachmentProject2.pdf (1334 Downloads) |
This contribution was helpful to2 thankful Users |
12/1/2012 5:23 AM | |
Joined: 6/16/2012 Last visit: 8/13/2024 Posts: 28 Rating: (0) |
Thank's Ni_ant |
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