VFD doubt
VFD doubt
(OP)
I have the following question, I have a conveyor feeder with a 10 HP motor drived by a VFD that was projected to work in a nominal speed (50Hz output for VFD) since that we have a 60Hz power supply line, I can also drive the motor to maximun speed (60Hz). But the client wants to drive the motor up to 70Hz without torque drop. This is possible?, it depends of the capacity of the VFD? and what are the motor requirements to work on this condition





RE: VFD doubt
hope this helps
Jeff
www.motors-direct.co.uk
RE: VFD doubt
Your 10 Hp gets it's rating based on this formula;
Horsepower = Torque x RPM / 5250.
So, ignoring other factors, a 10 Hp motor operated at 1800 RPM develops about 30 Lb/Ft of torque.
With a motor, as you reduce the speed from it's rated speed, you also reduce the Hp. But, the Torque tends to remain constant. This is called operating the motor in the "Constant Torque" range. As you increase the speed beyound it's rated, the Horsepower will remain constant. but, the Torque will be reduced. This is operating in the "Constant Horsepower" range.
What all that means is; if you want to go faster and maintain the same amount of torque, you got to get a bigger motor.
I hope that gives you a handle on your question.
Ed
RE: VFD doubt
HP= T*rpm/5250.
For instance, if the load rating is 10 HP at 50 HZ, the new demand at 70HZ will be 14 HP because the torque is constant. In my opinion a 15 HP motor is required.
The driver will see almost constant current so I expect it will handle the extra power demand.
RE: VFD doubt
The motor should also be derated for VFD use (current harmonics cause some extra heating) eg as per NEMA MG-1 though this may already have been allowed for by the manufacturer.
RE: VFD doubt
The performance of the motor when connected to the output of a VFD is independent of the frequency of the supply. The factors that affect performance are the design voltage and frequency of the motor and the voltage of the supply that you are using.
On a VFD, the motor is capable of delivering rated torque below it's rated speed, and rated HP above it's rated speed. If the supply voltage is higher than the rated voltage of the motor, then it will be able to supply higher than rated HP above rated speed until the output of the vfd becomes voltage limited by the supply.
To get rated torque above rated speed (motor), you need to increase the size of the motor.
If you drive a 50Hz motor to 70Hz, and the supply voltage equals rated voltage of the motor at 50Hz, then the output torque at 70Hz will drop to 71%.
If you drive a 60Hz motor to 70 Hz, and the supply voltage is equal to the rated voltage of the motor at 60Hz, then the torque will drop to 83%.
Best regards,
Mark Empson
http://www.lmphotonics.com
RE: VFD doubt
The fact you have a 10hp motor doesn't necessarily mean (and rarely does) that you actually need that amount of torque capability. You should go back a step and determine the torque you really need and then work out, using the information provided previously (83% x [approx] 30lb/ft=25lb/ft) if this is enough, before you start ripping out the motor etc. A simple step would be to look at the current being demanded at rated speed and if this is <the motor FLC then you could be fine increasing the speed without the motor stalling.
RE: VFD doubt
A very common seat-of-the-pants method of sizing motors is to determine the torque required at the desired motor speed, add 20%, calculte the HP, then buy the next motor size larger. So if for instance that formula came out to 8HP, they would size a 10HP motor rather than a 7.5HP. So the purchase overshoot combined with the 20% fudge factor usually means that your torque loss at 16% overfrequency might be insignificant. Another factor in this is that the torque loss is not linear, so at only 70Hz you have not yet lost too much. At 90Hz however you hay have a significant problem.
As skogsgurra said, it would pay you to run the numbers before swapping the motor.
"Venditori de oleum-vipera non vigere excordis populi"