# Motors and VFD's

SevenToFive
How does one calculate the amount of HP a motor makes when it is not running at a full 60Hz?

The reason for my asking is a customer wants to use a 1/4HP motor and a gearbox to vary the speed of a conveyor. At a full 60Hz we are figuring 1750rpm for the motor, and through the 7:1 gearbox the output speed would be 250rpm, but they also want to slow it down to 3.3rpm. This can be done using a VFD and having the motor output speed at 27.25rpm. However most gearboxes are rated for HP, so as speed goes down torque goes up and we do not want to exceed the torque rating by exceeding the tooth strength of the gearbox. Other applications will be using a 1/2HP motor, 1HP, and 2HP motor with larger gearboxes but in the same speed ranges.

Is there a way to calculate the motor HP when the frequency and speed decrease?

Thanks to all of hose who reply.

Staff Emeritus
not enough information. The steady state will be where the motor speed-power curve intersects the conveyor speed-power curve.

You need conveyor data or at least an assumption such as conveyor power varies with the square of speed. With that assumption, half speed needs 1/4 power and that's what the motor will deliver.

Gold Member
half speed needs 1/4 power and that's what the motor will deliver.

That's what the motor will try to deliver. If the power required at the final motor speed is still 1/4HP, then the needed torque will go up in the same ratio as the VFD slows the motor. (That ratio is either 1/76 or 1/64, the numbers given in the OP are conflicting.) Since VFD's adjust their output voltage for a constant Volts/Hertz (to limit motor current), there will be less than 2V applied to the motor at about 1Hertz. Motor current and motor torque are directly proportional to each other. So at 2V, the motor winding resistance will be the limiting factor on available torque; quite low.

Overall, a VFD does not seem a practical option, I recommend a much higher ratio gearbox.

jim hardy
Asymptotic
Has the 1/4 HP, 1750 RPM motor and 7:1 gearbox combo been used successfully?

Does a similar conveyor exist at their facility?
If so, determine what power is required "no load" with nothing on the conveyor, and fully loaded power with the 'worst case' load that must be conveyed.

Is there a way to calculate the motor HP when the frequency and speed decrease?
A VFD-driven motor operates with variable horsepower and constant torque from zero to full rated speed, and constant horsepower with variable torque if driven above base speed. Study these V/f curves (Google search on "constant torque vfd curves"). The amount of power required at any given speed depends upon the nature of the load.

Is it a roller conveyor, belt conveyor, or something else? This can make a big difference. Generally speaking, conveyors are constant torque loads.

EC&M has a good write-up on VFD selection you may find useful.

dlgoff
Staff Emeritus
That's what the motor will try to deliver

Good point. It might not succeed. I should have added "if a steady state is achieved." That's simply conservation of energy.

Dr.D
It would be good to get @jim hardy to respond on this question. He know more about motors than most of the rest.

dlgoff
Gold Member
Dearly Missed
You didnt say whether this is an induction motor or a brushed universal motor.
I'll assume it's induction.

However most gearboxes are rated for HP, so as speed goes down torque goes up and we do not want to exceed the torque rating by exceeding the tooth strength of the gearbox.

As @Tom.G mentioned there's a volts per hertz ratio that must be observed. That means at reduced frequency you must reduce applied voltage. Else the motor will draw excessive current and burn up.

Torque falls off as the square of applied voltage,
and at any given value of slip,,,
at 70.7% voltage you'll get half the torque.
Now most induction motors are capable of 200% torque but at greatly increased amps (Starting Current)
so at 70.7% frequency and 70.7% voltage you will likely be able to get full torque.
But the motor might be unable to accelerate the load and reach its pull-in speed. It would then draw starting current and quickly burn up.

For that reason VFD's typically limit current to rated running current . That limits torque to the motor's rated value and is called "Constant Torque" operation. Smart VFDs allow brief overcurrent to start a sticky load like a mixer or to accelerate a high inertia load like an elevator or conveyor..

So i think any VFD you are likely to encounter will limit the motor's torque to just about its rated value. That ought to protect your gear teeth. But you'll have to see by trial whether it'll start and keep your conveyor moving.

Is there a way to calculate the motor HP when the frequency and speed decrease?
Sure. The available horsepower, that is.
A motor's horsepower is its RPM X its torque(ft-pounds) divided by 5252 .
And within its operating range torque and current are roughly proportional.
So a VFD that protects the motor from overcurrent will cause the maximum available torque to be very nearly constant (except for a brief starting interval . )
That means the power available from the motor will be linear with speed : half speed = half power, quarter speed = quarter power etc.
Actual power will be whatever the load takes , if the motor can deliver it at that speed .

I would worry less about the gear teeth than the motor windings. At reduced speed the motor's internal cooling fan, those vanes you see when looking in the end, no longer move enough air through to to cool it.

Sorry if the above sound vague -
while i do understand your point about torque and RPM vs power,
you should think of your motor when combined with a VFD as a nearly constant torque entity not a constant power one. I think that's the main confusion factor .

Any help ?

old jim

Tom.G, Asymptotic and dlgoff