Speed vs Amperes

Nothing beats actually observing the result. While driving a fan, we reduced the frequency and I saw the current drop (as well as the voltage due to V/F). However, there comes a point when the frequency drops too low and the current starts to rise.

So yes, the current drops but only to a certain point.

regards,

Vulcan

It's the watts that tell the tale, if the load is decreased then the watts will decrease, this then depends on the design of the motor and vfd as the overall efficiency also plays a role...If on the other hand you are increasing the load as you lower the RPM, then the same amount of work is being done...

The amperes will drop, but not as fast as the work is being reduced. As Solar Eagle notes, the watts tell the tale, less work, less watts. Strictly speaking, the amps include both real current and the reactive current, the amperes required to maintain the magnetic field in the stator. As the speed decreases, the reactive load becomes a larger part of the total, until at no load, just before zero speed, the real is just windage, winding I squared R heat & bearing friction, simply considered. A VFD reduces the voltage with the frequency, but at base speed and below for typical VFD designs, the air gap still develops motor rated torque. The amperes will drop with less work being done, but the magnetic field still needs to be maintained, until you get to the end point where no work is being done, but the air gap is still charged. If a variable torque excitation curve is selected, you still have the endpoint of zero speed, no work, but some lesser reactive ampere draw.

The definition I take of the proposed constant load is a machine such as a fan or a pump coupled to the motor, as opposed to some driven device that consumes the same energy at any speed, unlikely in mechanics.

Depends on your definition of a "constant load".

Whose, mine or the OP's?

The OP's.

I gave your answer a GA, and, I just tried to make the problem with the OP's question succinct.

OK thanks. Be nice if AP#1 came back to confirm!

If constant load means constant required torque, amps should remain the same. Torque equates to amps, speed equates to volts. For a motor, input power = volts x amps, output power = speed x torque.

https://www.pumpsandsystems.com/pumps/motor-horsepower-torque-versus-vfd-frequency

Odd that your Pumps&System link only deals with constant-torque applications. There must be far more centrifugal piumps out there than PD.

And it doesn't inspire confidence when you see statements like

"Power is the rate at which torque (work) is performed" and

"...torque produced per unit of time..."

Odd that your Pumps&System link only deals with constant-torque applications. There must be far more centrifugal piumps out there than PD.

Well, they had the graph I was looking for. I hope they don't mind, I did link the source.

The OP's question:

If a constant load is applied to a 3 phase electric motor, and the rpm is decreased by a VFD, will the amps drop accordingly?

My interpretation of "constant load" is "constant required torque". The torque required by the load is equal to the torque supplied by the motor, which is proportional to the motor current. (An example of a "constant load" might be a motor in a crane lifting a specified weight. The torque required is independent of the speed. Most loads require more torque at greater speed, e.g., when friction is involved.)

So my vote is "Equal Load implies Equal Current."

If a constant load is applied to a 3 phase electric motor, and the rpm is decreased by a VFD, will the amps drop accordingly?

As I said in #4, that doesn't specify the type of load. You give another example of a constant torque load.

But then you say "Most loads require more torque at greater speed, e.g., when friction is involved.)"! Squared-torque loads also require more torque at greater speed, but not because of friction. My guess is squared-torque is more common than constant.

Until we hear more from the OP we won't know.

It all depends on what you mean by "constant load". If you mean constant torque, then Amps should remain about the same as Volts are almost invariably linear with frequency. If you mean "attached to the same load device", it depends on the load/speed characteristics of that device.

One person referenced an experiment with a fan, which did see a drop in Amps as its torque requirement can be expected to be quadratic with speed. A hoist likely would not see Amps significantly decline.

BTW, please always CAPITALIZE Amps, Volts, and Ohms, as they are derived from PERSONAL NAMES (André-Marie Ampère, Allessandro Volta, and George Ohm). Capitalization is basic respect.

That was nearly all covered in earlier posts

I had only scanned a couple of the responses at the time, but still maintain my response was useful, especially in that I pointed out that most VFD units are programmed for linear dependency of Voltage to frequency. My point about capitalization is, in my opinion, important. I question the value of your criticism.

If the <...load...> is <...constant...>, then there is no scope for <...the rpm...> to be <...decreased...>.

All rotating mechanical loads have a torque and speed characteristic curve. It is the job of the individual selecting the <...electric motor...> to match its torque and speed characteristic curve to that of the load; the operating point is where these two curves intersect.

Torque and current are almost proportional. Therefore, is the speed <....is decreased by...> any form of speed control, the change will be to a different operating point. Which is why the question is inherently nonsensical.