Motor basics


Thread Starter


hi everyone,

i am new to motors,:).. so..,

if i don't couple a pump to the motor, then the motor has no load.

now if i couple a pump and there is no fluid in the pump line, is the motor on load?

i think yes, there should be some load. one more thing, load is specified in hp. so can i use maximum power instead of the term maximum load. i mean load term is same as power.

George bailey

Power and load are not the same property. Power is the amount of energy required to drive the load. Your motor when running nothing attached theoretically has no load. With no load there would be no energy required to turn it. Therefore no power required. Lets use watts for power. Watts = voltage times current or P=EI. You motor when running unattached to anything does have a load. The bearings cause drag, the air resistance inside the motor both require power to overcome. This power comes from the current the motor draws. If you attach the pump with no water you increase the resistance to the motors rotation, therefore you require more power to turn the motor. The motor then will then draw more current. If you add water to the pump then you again have more resistance to the motor and it will draw more current increasing the amount of power in the form of current drawn from your power source. So power is the amount of energy needed to perform the work and the load is the work being done.
I guess the way I look at it... load is torque or current (amps)... voltage is speed... HP or kW is derived from the combination of speed and torque...

So in my book...load is not the same as power.
thank you world,
what i believe now, the hp specified on the motor is the maximum power the motor can deliver at any point of time (considering the fact we can use vfd to control speed), and so the current drawn by the motor varies. the current drawn by the motor at any point of time is an indication of the then load on the motor. the current specified on the motor is the current at maximum load. do share your opinions.

i think we can keep this thread going.

also if somebody could refer some literature..
An <b>electric</b> motor is capable of producing more power than the nameplate rating. The nameplate rating is the optimum power output that results in the longest life of the motor (windings; bearings; frame; etc.).

It is up to the designer of an <b>electric</b> motor-operated device to choose a motor that can supply the anticipated load. Choosing an <b>electric</b> motor that has a lower rating than the anticipated load will likely result in tripping of the <b>electric</b> motor, at a minimum, or damage to the <b>electric</b>motor, at the worst. Choosing an <b>electric</b> motor that is rated higher than the anticipated load results in, at a minimum, a higher cost of the device due to the over-sized <b>electric</b> motor.

The anticipated load of a device should include the maximum anticipated load that will typically be experienced by the device being powered by the <b>electric</b> motor.

And, when you are posting to a forum such as this you should provide all the information about your application, such as the use of an adjustable speed/frequency drive. Also, there are many types of motors; <b>electric</b> is just one type of motor.

In general, the load driven by an <b>electric</b> motor is directly proportional to the amount of current being drawn by the <b>electric</b> motor.

If you want documentation, see any text or reference on <b>electric</b> motors. Use your preferred Internet search engine to search for how electric motors work; and are both good entry-level explanations for many topics.
actually i am looking for something on AC induction motors and VFDs. from my current level it seems a long journey though...

any good movie suggestion..:)..
I am not sure what you are really getting at with your question(s).

An electric motor is normally rated at the mechanical power (horsepower) supplied at the motor shaft (commonly called brake or shaft horsepower) if supplied with its nameplate voltage and current, which it can supply continuously.

Motors available horse power can also be modified by service factor (nominally 1.0 for standard electric motors). Service factor being defined as a measure of the periodic overload capacity at which a motor can operate without overload or damage. The relationship between motor horsepower and voltage and current is normally defined as:

Hp = (Voltage x Current x Efficiency)/746

Efficiency = supplied power/(supplied power + motor losses sustained
providing the "supplied power")

Motor losses are mechanical losses (windage, friction, etc) plus electrical losses (I2R, eddy current, magnetizing current, etc). So Brake or Shaft Horsepower (Hp) at rated current and voltage = Total power supplied to the motor at its terminals minus the motor losses (electrical + mechanical) when it is supplying its rated horsepower.

Power and Load are terms which are sometimes used interchangeably in industry and load is also commonly used refer to the equipment attached to the motor. The required motor load is essentially the power (Hp)required to perform the task that the equipment connected to the motor is required to do. If, for example, the connected load is a pump, it is the power to continuous supply so many feet of liquid head or psi. Motor no load condition is commonly referred as the motor load when the motor is uncoupled from its equipment (in this case the pump).

In the example of a pump, the required horsepower is given by the equation:

Pump Required HP = (GPM x Total Head in Feet x S.G.)/ (3960 x Mechanical Efficiency
of Pump)

As you can see, efficiency raises it ugly head again since the pump will have mechanical losses to account for.

So to sum it up:

Simply put, the Motor Input Power at Motor Terminals for a Pump Service = (Motor Shaft Power + Motor Power Losses) + (Power Required to Provide the Pump Capacity + Pump Mechanical Power Losses)

Additional motor power information can be found at: Books/book-3/Chapter 3.2 Electical Motors.pdf

William (Bill) L. Mostia, Jr. PE
Sr. Consultant
SIS-TECH Solutions, LP
Any information is provided on Caveat Emptor basis.
Bill (if I may be so informal),

In my travels and toils I get asked similar questions a lot. I think the thing that amazes many people is that electric motors are kind of "dumb". They produce the amount of torque required by the device they are driving. If the device requires less torque (such as when flow is being throttled), the motor will produce less torque. Or, when the device requires more torque (viscous fluids or higher than normal flow rates) the motor will provide more torque.

The only "protection" or "smarts" is typically the thermal overload elements in the motor's starter, or, if a very large (high horsepower/high voltage) motor there might be a protective relay (such as a GE Multilin or equivalent).

It's something people don't give much thought to (like flipping the light switch or flushing the toilet) and are surprised when they do some thinking and/or research that they are really very simple, dumb devices (electric motors, that is).

This whole thing about VFDs is also very confusing to most people. It's just an efficient means of throttling flow for most applications.

If this isn't what the original poster is getting at, then I'm really lost, too. The choice of a particular electric motor for a particular application is all about the required torque at the required RPM, and it's the designer of the equipment that chooses the proper motor for the application. The motor has no "brains" and is just going to supply the torque being "requested" of it by the device being driven, even if that means an overload condition (if the motor can drive the overload condition for very long or at all).

I don't know how else to put it.

I think that when people do put some thought into the topic they tend to over-think it (as is customary of many things!).

I may have oversimplified to some extent, but I always try to think back to the first thoughts I had on a topic when putting more thought into it than just flipping a switch.
<i>what i believe now, the hp specified on the motor is the maximum power the motor >can deliver at any point of time (considering the fact we can use vfd to control speed)</i>

This is not necessarily true. from 0 RPM to Base RPM (using a VFD to control an induction motor) the HP produced is derived from torque being supplied AND speed of the motor... only at max torque and max speed does the motor deliver nameplate HP... so using "at any point in time" is an overstatement. Also if you overspeed the motor past base speed THEN you are in constant HP range... or what some call field weakening... your voltage increases, but torque starts to roll off... producing constant HP. again... all of this is with a VFD.

<i>and so the current drawn by the motor varies. the current drawn by the motor at any point of time is an indication of the then load on the motor. the current specified on the motor is the current at maximum load.</i>

and that last part I can pretty much agree with.
let me correct myself before someone else does. past/above base speed the voltage remains constant the freq will increase (to increase the speed of the motor), but HP is constant.. i.e. torque starts to roll off.
thank you all , that was really helpful. any motor issues again, and i will post on the same thread. do reply back.

I have a question about VFD specifications wrt to a motor. I have heard that it is often good practice to use a VFD with an output power larger than the motor that it is intended to drive. Other reading I have done has led me to believe that this would be very inefficient and the VFD should be chosen to properly drive the given motor.

For example a .5HP AC motor, was suggested to use a 1HP VFD
Such a big motor in the example: 0.5 HP.

How much difference (cost) do you think there is between a 0.5 HP drive vs. a 1.0 HP drive?

How much difference (cost) do you think there is between a 20 HP drive vs. a 30 HP drive, or a 25 HP drive?

You say you "heard" it's often a good practice to use a VFD with a higher output power that the motor it will be driving. What is the context of this statement? In what way, or ways, is having a larger drive a better practice?

And, then you say you "read" it would be inefficient to do so.

A lot of people say a lot of things. A lot of manufacturers write a lot of things. Everything needs to be considered in the context of how it's said or how it's written.

You need to examine each application in light of the needs of the application, and potential future needs, when making a decision.

And, there might be electrical reasons or economic reasons that would make the choice of one drive over another better or not. You have not provided any context for your question.

I am not sure if there is very much context in this question. I would just like to know in general, a coworker of mine is working on obtaining a VFD for that motor and had initially told me that he thinks it is good practice to size up the VFD and since the cost was only $20 difference between the .5 and 1HP VFD it is what would be recommended. Also, it would leave room if the motor needed to be sized up in the future.

We are now checking to see if this oversized VFD will not cause any detrimental effects on the motor whose cost is of magnitudes larger than that of the VFD.

As a result, I was just wondering if in general, looking at the compatibility between motors and their VFD's, and if the economics were a non issue, is it acceptable to have a VFD powering a motor with a lower power rating.
The simple answer to your question is yes, it is acceptable to drive a motor with a larger capacity VFD.
So, you've answered your own question. And put it in context. $20 difference between the drives; possible future upgrade potential.

Does this apply to every VFD/motor application? Likely not.

In my experience, impedance is the issue when matching/mating drives and motors. Some require "reactors" between the drive and the motor; some don't.

Again, every application needs to be reviewed in its own context, and with its own set of "filters".

Best of luck!