I am a mechanical engineer and going to commission a small grinding mill in couple of weeks. My knowledge on electrical stuff is limited so I appreciate if someone help me out on this. We measure the motor power draw by measuring the current drawn by motor. But how can I calculate the motor power draw using information on the motor name plate and current reading?
I just want to make sure the motor is not drawing more power than it should knowing that the PF varies depending on motor load.
Power formula is KW/VxIxPfx1.732, Current drawn by motor and voltage applied to motor and power factor. these values putting in formula you can get the power drawn by the motor. A suitable energy analyzer like TES 3600 3P4W power analyzer, manual can be searched through google can measure without calculation all these value easily.
For a three-phase AC motor the formula for power consumed by the motor is:
Watts = V * I * 3^0.5 * pf.
(The 3^0.5 is the only way I can express "the square root of three" on this forum, which is approximately 1.732.)
Most AC motor nameplates list the current drawn at rated voltage when producing rated power in watts, or KW, (all values shown on the nameplate) and also list the power factor of the motor when producing rated power.
I have not heard nor can I really find any sources that relate load and power factor, or that say power factor changes with load. Electric motors can produce more than rated power as long they aren't overloaded to the point of stalling or reducing speed much below nameplate rated (also listed on the motor nameplate). So, for the most part when determining if a motor is overloaded by measuring current using an ammeter (in-circuit or clamp-on), the power factor is presumed to be "constant" (at or near rated power output) and if the voltage is at or near rated, then the current drawn by the motor is the real indicator of load. (This is on a per-phase basis, also, for a three-phase motor.)
As long as the current drawn by the motor doesn't exceed nameplate (presuming the voltage is at or near rated), then the motor is presumed to be operating at or below rated power output. If the current drawn by the motor (presuming the voltage is at or near rated) is greater than nameplate, then the motor is considered to be producing more than rated power.
The service factor of the motor (also usually listed on the nameplate) will basically indicate how long the motor can be operated continuously at an overload condition, or how much overload can be tolerated by the motor continuously. But, good practice is to size the motor not to exceed rated current under normal continuous loaded operation. Occasional overloads, even for "extended" periods of time, are not necessarily injurious, but over time can be.
Hope this helps!
By the way, the above presumes the motor is operating at a constant speed, and is not being operated using a VFD (variable frequency drive) of some sort to reduce speed.
CSA, Here's a link.
Thanks for the link. I'm not accustomed to motors being lightly loaded.... The site explains it very well; reactive load is constant, and therefore power factor will vary with real load!
Thanks again; makes perfect sense.
One more thing. For most AC motors, when operating at or near rated power they are very efficient at converting electrical power (watts) to mechanical power (torque; horsepower; etc.)--on the order of 95+%, if not higher. Again, that's when at or near rated current/voltage (which should be at/near rated power output).
Thanks for your response
So can we say if the motor current is at the nameplate value assuming the voltage is also at rated number then the motor is drawing full power? or in another word the PF is already built in the current value on the name plate so I don't need to use the formula kw=V x I x PF x 1.7 ?
Try this link:
(remove any spaces inserted by the forum software)
It's been a very long time since I've had to calculate motor performance to such a fine degree, but I believe there are standards which most motor manufacturers adhere to, and one of them is that the rating (output) of the motor accurately reflects losses due to expected power factor.
The (real--not reactive) current drawn by an electric motor IS the basis for determining how much power the motor is producing. If the current drawn equals nameplate the presumption is the motor is producing rated output power (again, presuming voltage is at or very near rated, and frequency is at rated, and speed is at rated).
I think if you read this, and do the maths with the motor nameplate data from the motor in question you will be able to see how all of the numbers work.
In my experience, when an AC induction motor is operating at or near rated voltage, at rated frequency, and at rated speed, when the measured current drawn by the motor is at or near rated then the motor is producing nameplate rated power (specifically, torque, from which HP and/or watts can be calculated). Without some way to measure torque (a dynamometer or strain gauge or some other method) it's not really possible to say much more about the power output than it's meeting the required parameters (air flow; pressure and flow-rate; conveyor load/speed; etc.). Most of my experience is with pumps and fans, and with few exceptions (mostly because the motors were running backwards on certain types of fans (reverse tangential fans)) the current being drawn is slightly less than nameplate, or much more than nameplate--in which case the thermal overload relay is operating which is why motor current is being measured.
Please write back to let us know how your start-up goes.
Thank you Kam for your efforts. I understand the implication of using a VFD to reduce speed. how about if i use a v-belt instead of VFD, will the result be the same?
Unless it's a Very small mill it will have a wound rotor and be running as a synchronous motor. It won't be running lightly loaded because of the grinding media, balls or rods most likely.
Because the load a grinding mill draws is a significant portion I'd be surprised if it doesn't have a KW readout.
The motor nameplate should have FLA (full load amps). There should be an approximately linear relationship between amps and power:
0 amps = 0 HP (or 0 KW)
FLA = rated HP (or rated KW)
> The motor nameplate should have FLA (full load amps). There should be an
> approximately linear relationship between amps and power:
> 0 amps = 0 HP (or 0 KW)
> FLA = rated HP (or rated KW)
That is just not true. A completely unloaded motor will often draw 50% or more of its FLA.
Shaheed's formula is incorrect because he omitted Efficiency, that is, Nin, which varies with load, but not linearly!
CSA's formula covers kWin, that is, it equals:
Sqrt(3) x Vin x Iin x PFin, for any load, but that presents a problem. It is not an accurate indication of kWout or HPout!
Asumming Vin is the rated value, and is constant, then, Iin, PFin, and Nin all vary with kWout or HPout!
Thus, for a given load with Amps known, then, PF, and Nin vs load-curves must be used to determine kWin. Typically, these parameters are available from the manufacturer.
For an approximate value, one can assume that Iin is proportional to nameplate kW or HP, but only if load is more than, say 60%, but increases non-linearly below 60%.
The relationship between PFin, Nin and load are non-linear. For example PFin and Nin could be constant, say between 70 and 110% of load, decrease drastically for smaller loads.
If accuracy is a concern, then there are two very reliable field-test methods: a)measurement of Temperature-Rise over Ambient; and b) measurement of slip!
I suggest you search the Control.com Archives. There are many, many threads covering this topic.
Of course, most of above is moot if Power Analyser measurements are made as alluded to by Connoly!
Regards, Phil Corso