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VFD/Motor speed reduction

VFD/Motor speed reduction

VFD/Motor speed reduction

(OP)
Hello all! We have a situation where a 200HP, 480v motor controlled by VFD needs to have it's max speed limited to 54Hz. This is a fan motor (variable HP and Torque), so as I understand, if speed is reduced to 90%, the HP is reduced to (90%*90%*90%) = 72.9%, and Torque to 81%. But this is what the load (fan) will demand. The question is, would the motor have any problem meeting this demand? Would I have to look at motor curves to determine that? Thanks for the help!

RE: VFD/Motor speed reduction

If the motor is OK operating the fan at 60 Hz now, it will not have a problem at 54 Hz. The drive torque for the fan is a squared relationship with the speed, and nothing in a normal VFD+motor will make the available torque drop off that fast as you slow it down.

RE: VFD/Motor speed reduction

HP capacity of the MOTOR is reduced only by the speed change, so 90% speed = 90% HP capability. The motor will only draw current and produce torque at the level demanded by the load, in this case a centrifugal fan.

The demand upon the motor by the LOAD will drop by the cube of the speed change in a centrifugal machine, so the FAN will demand 73% less HP from the motor and therefore the motor now has even MORE capacity than the fan needs at that speed.


"You measure the size of the accomplishment by the obstacles you had to overcome to reach your goals" -- Booker T. Washington

RE: VFD/Motor speed reduction

(OP)
Thanks jraef...that makes sense!

RE: VFD/Motor speed reduction

Quoted /HP capacity of the MOTOR is reduced only by the speed change, so 90% speed = 90% HP capability /unquoted
Just for my learning...(considering you answered to OP already)

90% speed = 90 % capability (guess is meant for power?) assumes there is linear ratio; so is this because it is assumed scalar control which has V/F = constant ?
If correct, then what about VFD where vector control can be applied, I am guessing right to say that this would allow to change the ratio, thereby changing the picture of 90% speed = 90 % capability ?

Thanks

RE: VFD/Motor speed reduction

rotw, the horsepower output of the motor is given by torque times rpm. It is limited by the temperature limit of the motor. Heat in the motor is created by the current flowing through it. Torque and current are linearly proportional so, in effect, every motor has a max torque capability, which remains constant as speed changes. Therefore, if motor speed is reduced to 90%, then horsepower capability is reduced to 90%. At significantly lower speeds the cooling fan will become ineffective and power/torque capacity will be further reduced unless an external cooling fan is used.

RE: VFD/Motor speed reduction

make sense Compositepro. thanks

RE: VFD/Motor speed reduction

Quote:

Torque and current are linearly proportional so...

Oh, if only it was so simple. smile

That is reasonably true of the in-phase component of the line current, assuming losses are low, but totally neglects the quadrature magnetising current which can be 1/2 - 2/3 of the line current in some motors.

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