I am trying to either explain or derive where the power savings come from with a variable frequency drive (VFD). I am well aware of all the literature and generally accepted assumption that energy savings resulting from a VFD are proportional to the cube of the ratio of the reduced speed to the original speed of the motor. This is from the affinity laws.

However, if I consider the rotational power equation where power is equal to torque multiplied by rotational velocity OR the fluid power equation where power is equal to pressure multiplied by flow rate, it would appear to me that power should be directly proportional to speed.

Or, if I consider that VFDs maintain a constant volts to hertz ratio and assume that motor current is proportional to load and then use the electrical power equation where power equals current multiplied by voltage multiplied by power factor - I would assume that power is proportional to the square of the ratio of the reduced speed to the original speed.

Can anyone clarify things? Where is the discrepancy or what am I missing here? And are the affinity laws just stated as fact? Are they fundamental principles or do they have a derivation relating to fluid flow.

I will update this question with equations to make it easier to read.

Any thoughts or explanations are greatly appreciated!


  • \$\begingroup\$ This is way outside my area. But when you run a pump faster, the back pressure and flow rate both go up, so that is not linear. Maybe square law? Concerning the rotational power equation, it may be a mistake to assume that torque remains constant. That is up to the mechanical load. With an agricultural irrigation pump, the torque required to spin the pump will be higher at higher rotational speeds, not constant. So as you speed up your VFD, the current will rise along with voltage. So all of your assumptions seem to be invalid. \$\endgroup\$
    – user57037
    Commented Jun 1, 2016 at 4:48

1 Answer 1


You are looking at the wrong end of the system. Instead of looking at the electrical end, which only supplies the power+losses to make the motor do what you ask it, look at what you are asking the motor to do.

If the motor is driving a fan, or driving an impeller churning any Newtonian fluid, where the torque varies as the square of the speed, then the power rises as the cube of the speed.

If the motor is driving a winch, say in a crane, then when lifting a weight, the torque required is independent of speed, the power will vary as the speed.

Two different types of load, two very different scaling laws.

In any real case, the detail will be more complicated than a pure 'scale speed raised to a power' law, but this covers the gross behaviour. You have to measure your load, and see what it does.

  • \$\begingroup\$ are you referring to constant power vs constant torque electric motors? The second paragraph is still a vague explanation - the Darcy–Weisbach equation states that pressure drop is proportional to the speed (or flow) squared. If fluid power is defined as the product of pressure and flow, and pressure is proportional to the square of flow, then power would be proportional to the flow cubed - this explanation isn't really great either - the motor just has to deliver that same power and I guess not concerned right now how the VFD delivers it because Power In = Power Out. \$\endgroup\$
    – Prevost
    Commented Jun 1, 2016 at 21:04
  • \$\begingroup\$ I think you're agreeing with me, but I'm not totally sure. The motor delivers the power that the load requires. \$\endgroup\$
    – Neil_UK
    Commented Jun 2, 2016 at 7:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.