scottmurphy
Member
- Join Date
- Oct 2004
- Posts
- 363
This not a PLC direct related question, but I know there are plenty out there with a lot of theoretical knowledge.
I have 2 drums that I need to match surface speed on, they do not have to be exact, but within tolerance.
My approach has been to work backwards to get the desired RPM to the motor. If I take the surface speed required, and divide by the drum size, I get the drum RPM req, then multiply by the gearing ratio between drum & motor gives me the motor RPM req. If I then calculate the % of the motors rated speed, I can work out the required frequency, from this I can calculate the value of the signal I need to send from the PLC.
Surface Speed = 10m/min #multiply by 1000 to give mm/min
Drum Size = 640mm
Gearing Ration = 6.25:1
Motor RPM = 1455
Drum RPM = (10 x 1000) / 640 = 15.63
Motor RPM = 15.63 x 6.25 = 97.66
Speed % = (97.66 / 1455) x 100 = 6.71
My maximum value on Analogue OP is 4095, so the value I need to give to VSD is (4095 x (6.71 / 100 ) = 275.
One of the drums is within acceptable tolerance, but the 2nd does not seem to follow the same formula.
Both motors are running in open loop control, the accuracy is not that critical they both need to be exact, I would just like them to match each other.
Am I on the correct track with this theory, what else do I need to take into consideration, motor slip, quality, model VSD type, etc?
I have 2 drums that I need to match surface speed on, they do not have to be exact, but within tolerance.
My approach has been to work backwards to get the desired RPM to the motor. If I take the surface speed required, and divide by the drum size, I get the drum RPM req, then multiply by the gearing ratio between drum & motor gives me the motor RPM req. If I then calculate the % of the motors rated speed, I can work out the required frequency, from this I can calculate the value of the signal I need to send from the PLC.
Surface Speed = 10m/min #multiply by 1000 to give mm/min
Drum Size = 640mm
Gearing Ration = 6.25:1
Motor RPM = 1455
Drum RPM = (10 x 1000) / 640 = 15.63
Motor RPM = 15.63 x 6.25 = 97.66
Speed % = (97.66 / 1455) x 100 = 6.71
My maximum value on Analogue OP is 4095, so the value I need to give to VSD is (4095 x (6.71 / 100 ) = 275.
One of the drums is within acceptable tolerance, but the 2nd does not seem to follow the same formula.
Both motors are running in open loop control, the accuracy is not that critical they both need to be exact, I would just like them to match each other.
Am I on the correct track with this theory, what else do I need to take into consideration, motor slip, quality, model VSD type, etc?