Motor is inductive load so how does our electricity residential meters calculate the power consume by motor?
Motor needs current to produce magnetic field so that current and voltage would be reactive so how does there is power in KW of motor and where does the KW volatage and current comes ? Like in transformer it is easy to understand as there are core losses and copper losses and magnetization current is responsible of making magnetic field
Hello Syed Taha Ahmed,
A motor surely consumes active / real power. In a motor, you also have core losses and copper losses. It is an alternating magnetization that is responsible for the varying air gap magnetic field, used for torque production.
If you are talking about power factor, there is a nice on-site report that you could read to get a better understanding: http://ecmweb.com/motors/motor-efficiency-power-factor-and-load
If not, could you describe your question a bit more?
Hope I could help,
Regards
No inductor(L) or capacitor(C) are ideal, it consumes active power which could be metered and billed....
A motor is not very different from a transformer. Even the motor has iron and copper losses. Rotating magnetic field created by the stator. Magnetic field is induced by the stator. By electro magnetic effect the rotor tries to catch up with rotating field and hence rotates. This consumes some real power. Multiplication of Voltage ,current and power factor gives the value of the Power (watt) consumed. There are few methods to calculate the power. Example two watt meter method.
Induction motor or transformer working on same induction principle, for keeping working flux(density) in core constant the increased secondary current(I2) reflect to primary drawing more primary current(power), only the secondary loading in transformer is electrical(I2, with constant E2), while in induction motor is mechanical(SE2), where slip(S) increased (with more mechanical load(T=F(liver arm)).
The 'ideal' motor under no-load conditions would be an inductive load - consuming no 'real' power.
As soon as the motor is loaded (thus delivering mechanical 'work'), the rotor starts to lag in phase vs. the stator field. As a result, voltage and current no longer are in a phase of 90° - and the motor starts to morph into a non-reactive load.
At no load the no load current could split in two component eddy current(Ie) and magnetizing(Im), where Ie, is in phase with voltage, and is called loss component, Io = Ie + Im....
Hi Ahmed,
To answer your question briefly, there are three primary considerations when setting tariff rules for induction motor loads (includes pumps, mill motors, drilling machines, industrial tools, and whatever one can think of !!):
We find that in most cases, the tariff rules that apply to motor loads are variations of the three above.
With best wishes.
-Sanjay
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