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DAPPH
as dyslexik as I'm daft
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This might help ...
It very much depends on the type of motor and on the mechanical loading and how that varies with motor speed.
If the variac was supplying a pure resistive load, that dissipated 2000W at 240V, then at 60V, the power dissipated would be one sixteenth, or 125W (since power dissipated by a resistor is related to voltage squared).
For an induction motor rated to deliver 3hp, a mechanical load utilising the rated power would result in the motor drawing around 2400W from the supply. Reducing the supply voltage would result in an increase in the phase lag, eventually to the point that the motor would not be able to sustain that 3hp output, and it would stall. Up to that stall point, if the load remained constant, the motor would draw more and more current. On the other hand, the same induction motor rated at 3hp that was free running, with no mechanical load applied, would draw considerably less than 2400W from a 240V supply, maybe 100W. Decreasing the supply voltage would reduce the power drawn, although the motor would remain synchronously locked to the mains supply, until it was unable to overcome the mechanical losses, and it would stall.
For a universal type motor i.e. one using brushes and a commutator, the relationship is somewhat more complicated, and will be influenced by whether the motor's rotor and field coils are series, or parallel wound, and also by the load.
If the variac was supplying a pure resistive load, that dissipated 2000W at 240V, then at 60V, the power dissipated would be one sixteenth, or 125W (since power dissipated by a resistor is related to voltage squared).
For an induction motor rated to deliver 3hp, a mechanical load utilising the rated power would result in the motor drawing around 2400W from the supply. Reducing the supply voltage would result in an increase in the phase lag, eventually to the point that the motor would not be able to sustain that 3hp output, and it would stall. Up to that stall point, if the load remained constant, the motor would draw more and more current. On the other hand, the same induction motor rated at 3hp that was free running, with no mechanical load applied, would draw considerably less than 2400W from a 240V supply, maybe 100W. Decreasing the supply voltage would reduce the power drawn, although the motor would remain synchronously locked to the mains supply, until it was unable to overcome the mechanical losses, and it would stall.
For a universal type motor i.e. one using brushes and a commutator, the relationship is somewhat more complicated, and will be influenced by whether the motor's rotor and field coils are series, or parallel wound, and also by the load.
Indeed, picture a squirrel cage ,motor which will turn at a fixed rpm according to the mains frequency and number of poles. If this is connected to a centrifugal pump, and you turn down the volts, the power available to turn the pump isn't there any more. This will increase slip, current goes up, disaster results.
This.
If it's a typical induction/squirrel cage AC motor, then the speed will remain the same, as the speed is directly related to the power frequency.
If it's a brushed motor, then generally it will control speed, but will depend on the exact motor design/control circuit.
Kram
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What is it for and what are you trying to do?
A VFD will let you vary the frequency of an induction motor and vary the speed.
A induction motor, and generally most motors, will only draw their rated power when they are loaded. At full speed and unloaded, my 1.5kw motor draws about 300w due to friction in the drive and belt and lathe spindle its attached to.
A VFD will let you vary the frequency of an induction motor and vary the speed.
A induction motor, and generally most motors, will only draw their rated power when they are loaded. At full speed and unloaded, my 1.5kw motor draws about 300w due to friction in the drive and belt and lathe spindle its attached to.
