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Why does a motor tend to work on rated power always? (I have a series of doubts one by one I am planning to clear).

ustpower mentions

A low voltage forces a motor to draw extra current to deliver the power expected of it thus overheating the motor windings.

Why does it force to make a motor run in the rated power even at low voltage? How does a motor know what's it's rated power at low voltage?

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    \$\begingroup\$ It doesn't. Motors are not constant power devices. If you lower the voltage to a motor, it will spin slower. The current draw will stay approximately the same steady state, and will only vary as a function of load and efficiency. \$\endgroup\$ – Ocanath Jun 4 '20 at 17:40
  • \$\begingroup\$ Ratings are how a thing is designed to work at. They don't always work at the rated values. Your computer has a 500GB hard drive in it, but that doesn't mean you actually have 500GB of movies saved on it. It also has a keyboard which is designed for 5000 key presses per minute, but you still cannot type that fast, and you certainly don't have to type that fast. \$\endgroup\$ – user253751 Jun 4 '20 at 18:35
  • \$\begingroup\$ That quote is not strictly corrected as worded so you should not make an extrapolations regarding rated power from it. It says nothing about the motor running at its rated power even at a low voltage. It doesn't even mention it. At best, it could be interpreted as the human who installed it expects the motor to deliver its rated power, despite starving the motor of the resources it needs to actually deliver its rated power. \$\endgroup\$ – DKNguyen Jun 4 '20 at 18:46
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A low voltage forces a motor to draw extra current to deliver the power expected of it thus overheating the motor windings.

The quoted statement applies to AC induction or synchronous motors, It does not apply to universal motors or DC motors. It does not mention "rated power." It says "the power expected." The power expected is really the power required to drive the load at the speed at which the motor is trying to run. The speed of an AC induction or synchronous motor is mostly determine by the frequency. If the voltage is reduced without reducing the frequency, the motor still tries to run at about the same speed. That requires about the same torque and the same power.

There is a big exception to that rule. Fans and most centrifugal pumps require much less power when the speed is reduced. With many induction motors, the speed at which the motor is trying to run drops sufficiently with reduced voltage to allow the motor to run at reduced power when the voltage is reduced.

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  • \$\begingroup\$ I suspected it might have been about AC induction/synchronous motors. good to know \$\endgroup\$ – Ocanath Jun 7 '20 at 17:17
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The link in the question is to a company that make AC supply conditioning cabinets: -

enter image description here

So, you have to read what they say in context - they will be talking about AC motors (not DC motors) and, given that the main type of AC motor is the induction motor, if you run these at too low a voltage, it will certainly have problems with excessive slip and god-forbid it might stall and smoke.

See this Q and A entitled How does low voltage burn out induction motor - it nicely explains the problem.

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It depends...

First off, not all motors are the same; AC or DC makes a difference, then within AC, you have Induction, Slip Ring Induction, Synchronous, Universal Motors, Switched Reluctance, single phase, 3 phase (even 2 phase), etc etc. Reducing the voltage does NOT always make a motor run slower, it depends on which TYPE of motor is being discussed and what the load is at that moment.

But speaking of AC induction motors, the most common type, the motor produces torque that spins the rotor and thus the load. The torque is created by the motor drawing power from the line and the torque created by the motor has to equal the torque required by the load. The amount of torque the motor CAN produce is a function of the correct ratio of voltage and frequency in the design of the motor windings. Any change in that ratio results in a change in the torque it produces. Too low of a ratio and the torque drops off, too high of a ratio and the windings saturate, producing more heat instead of torque. That ratio must be +-10% of that the motor was designed for in order to operate within specification tolerances. But in addition, the amount of torque the motor ACTUALLY produces is a function of the LOAD on the motor.

If you lower the voltage alone and keep the same frequency, the amount of peak torque the motor can produce will drop at the SQUARE of the change in voltage, i.e. at 80% voltage, the motor can only product 64% of rated torque at its peak. IF the load requires more than 64% of peak rated torque of the motor to accelerate or re-accelerate after a step change in lood, then the motor slows down. By slowing down, the "slip" of the motor increases, which causes it to draw more current in an attempt to get back to normal speed. But in drawing more current, that causes an increase in the internal losses in the motor and an increase in the heat it has to get rid of. If it can't get rid of that heat effectively, the motor burns up.

BUT IF at the time the voltage was dropped to 80%, the load was only requiring 50% of the motor's peak torque capability, the 64% that the motor is delivering is still more than adequate, so in reality the motor will NOT be overloaded and heat up.

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  • \$\begingroup\$ An upvote for the answer you produced a while ago that I have linked in my answer above. \$\endgroup\$ – Andy aka Jun 4 '20 at 18:55
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A motor converts electrical power into mechanical power. Electrical power = voltage x current. So if the motor is run on lower voltage then to deliver "the power expected of it" it must draw more current.

But that doesn't mean it will always work at constant power output regardless of voltage. The mechanical power output must be absorbed by a load. If the load is constant then the motor will have to deliver constant power. But if the motor is a type that slows down at lower voltage (which many are), and the load absorbs less power at lower rpm (which many do) then the motor will not have to deliver the same amount of power. So the operative word in your quote is 'expected'.

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The operating point of a motor is generally dependent on the load. The ouput power is not necessarily the rated power of the motor and i don't think the referenced text is saying this. I think the text tries to convey, that a specific mechanical load (speed and torque) should rather be handled at higher voltage and lower currents for lower losses in the windings. To achieve this, you might have to use a gearbox together with the motor depending on the load. For a better understanding it might help to think of the voltage being approximately proportional to the speed and current being approximately proportional to the torque, at least for DC motors.

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