Magnitization Current 4

[oranjeep]

How does magnitization current work?
I have a motor that has a magnitization current of 2.9 Amps at 460Volts.

 

[electricpete]

If you take a coil of wire wrapped around iron core, now apply ac voltage, you will get a magnetiziation current approx I = V / (2*Pi*f*L).

If you wrap a second set of coils around that coil, you have a transformer. Leave that second set of coils open, and it is an open-circuited transformer. Magnetization curent same as before.

Equivalent circuit of a motor is a transformer. At no load it has it's secondary set of coils open-circuited. You still have to supply that same magnetiziation current I = V / (2*Pi*f*L)

 

[jbartos]

Explanation: The magnetomotive force produced by a set of balanced polyphase currents in a symmetrical set of sinusoidally distributed winding is:
Eq. 5.15
Ft=(Ns x Is,max x ms / 2) x cos(ws x t + alfa,s - theta), in Amperes.
The vector addition of the stator magnetomotive force Fst and the rotor magnetomotive force Frt gives the net or magnetizing magnetomotive force:
Eq. 5.26
Fmt=Fmax x cos(ws x t + alfa,m - theta), in Amperes
From Eq. 5.15, this magnetizing force can be considered as the result of a magnetizing component of the stator current. In phase a, this current component would be:
Eq. 5.27
Ima=Ims,max x sin(ws x t + alfa,m)
where:
Eq. 5.28
Ims,max = 2 x Fmax / (Ns x ms), in Amperes.
With the rotor open circuited, all of the stator current Is (notice that this is rms value) can be considered as magnetizing current Ims (again, rms value). From Eq. 5.27, the magnetizing current in phase a is:
Eq. 5.69
Ima=Ims,max x sin(ws x t + alfa,m) in Amperes.
Tying this result to the original posting 2.9 A (assumed rms value)
Ima,rms=2.9A=Ims,max/sqrt2.
See Reference:
Gordon R. Slemon "Magnetoelectric Devices, Transducers, Transformers, and Machines," John Wiley and Sons, Inc., 1966, Chapter 5 Polyphase Machines (or a similar textbook).

 

[rodar]

Hi
In AC machinery (Transformers and Motors) the resistance
of the copper windings are very low. The inductance of the
windings contribute the majority of the impedance to current flow through the coil. So when no load is applied
and neglecting losses, the self inductance limits the current drawn by the machine. This current is called the
magnetizing current. It is the current necessary to flow
in the load inductance to cancel the line voltage.
rodar

 

[gsimson]

Let me try to explain without any equation
- In any coil when you pass current magnetic flux is produced. Lenz law
- When you pass 3 phase current in a winding of motor ( the phases of the winding are at 120 degrees electrical) magnetic flux produced will be the vector sum of all the three phases.
- If you give three phase current to a 3 phase motor winding, the vector sum will be always same, as the sum of the total voltage at any point in a 3 phase sine wave is same.
- As the voltage is alternating, the total vector also starts moving from 0 degree to 360 degree.
- When we insert the rotor in the magnetic flux, Voltage is induced in the rotor. This voltage generates rotor current. As per Fleming's rule torque is generated due to Stator flux & rotor current.

 

[oranjeep]

Thanks for the responses.

 

[PUMPDESIGNER]

jabartos,
Do you agree with rodar completely?
I know nothing, but rodar's explanation is simple and useful to me therefore.

PUMPDESIGNER

 

[jbartos]

Comment on rodar (Electrical) Jan 1, 2004 marked ///\\Hi
In AC machinery (Transformers and Motors) the resistance
of the copper windings are very low.
///True\\ The inductance of the windings contribute the majority of the impedance to current flow through the coil. So when no load is applied and neglecting losses, the self inductance limits the current drawn by the machine.
///True\\ This current is called the magnetizing current.
///This contradicts reference:
Donald G. Fink, H. Wayne Beaty "Standard Handbook for Electrical Engineers," 13th Edition, McGraw-Hill, Inc., 1993,
pages 20-26 to 20-27
Sections:
39. Analysis of Induction Motors
40. Circle Diagram
41. Equivalent Circuit
where magnetizing current Im flows in the parallel branch of the motor electrical equivalent circuit. This current is much smaller than the current flowing in the main branch. The magnetizing reactance Xm is usually eight or more times as great as X=X1+X2
where
X1 is primary leakage reactance in Ohms
X2 is secondary leakage reactance in Ohms
while R1 (primary resistance in Ohms) and R2 (secondary resistance in primary terms in Ohms) are usually much smaller than X, except in case of special motors designed for frequent starting service.\\ It is the current necessary to flow in the load inductance to cancel the line voltage.
///The cited reference refers to this current as to the primary current. Obviously, the primary current consists of resistive component and inductively reactive component.
The cited reference is somewhat condensed. The more detailed book will be needed to understand the equivalent circuit development. See for example:
M.G. Say "Alternating Current Machines," John Wiley and Sons, Inc., 1978.\\rodar

 

[electricpete]

The inductance of the windings contribute the majority of the impedance to current flow through the coil. So when no load is applied and neglecting losses, the self inductance limits the current drawn by the machine.
///True\\ This current is called the magnetizing current.
///This contradicts reference:
Donald G. Fink, H. Wayne Beaty "Standard Handbook for Electrical Engineers," 13th Edition, McGraw-Hill, Inc., 1993,
pages 20-26 to 20-27
Sections:
39. Analysis of Induction Motors
40. Circle Diagram
41. Equivalent Circuit
where magnetizing current Im flows in the parallel branch of the motor electrical equivalent circuit. This current is much smaller than the current flowing in the main branch"

There is no contradiction. The current flowing in the main branch is ~ 0 at no-load. The no-load current is (approximately) the same as the magnetizing current, as rodar has said.

 

[jbartos]

Clarification of the "main branch". The main branch is meant to be the primary current branch down to the point where the magnetizing branch starts.
The magnetizing current Im under no load condition is somewhat bigger than under the motor rated conditions since the voltage at the magnetizing branch will be higher. The Im under no load with the higher voltage across the parallel branch will be more nonlinear.

 

[aolalde]

The magnetizing current for a motor (or transformer) is the current required to produce the Ampere-turn or magneto-motive force ( Fmm) able to produce the magnetic flux ( Phi) which induces a voltage of the same magnitude as that applied voltage to the motor terminals. Saturation of the laminated core will change the required Ampere-turn. For motors, the air gap part of the magnetic circuit will add most of the magnetizing current requirements.

Fmm= Phi * Reluctance

Reluctance = 1/mu * L/A+ 1/muo* g/Ag

mu = steel permeability at working saturation
L = Magnetic circuit length
A = Magnetic circuit cross sectional area.
muo = air permeability
g = air gap effective length.
Ag = air gap effective cross sectional area.

I agree with electricpete and rodar, this does not contradict the equivalent circuit, which in my opinion has primary current path, magnetizing and core losses current paths and secondary current path. The primary current is the sum of secondary plus magnetizing and core loss currents. For no load the secondary is open.

Stator resistance (R1) and leakage reactance (X1) are very small as compared to the magnetizing reactance ( XM) and the core losses current is small. Therefore the primary NO LOAD current ( Io) of induction motors is practically the same as the magnetizing current (IM).

A Olalde.

 

[jbartos]

Suggestion to the previous posting marked ///\\Stator resistance (R1) and leakage reactance (X1) are very small as compared to the magnetizing reactance ( XM) and the core losses current is small.
///True\\ Therefore the primary NO LOAD current ( Io) of induction motors is practically the same as the magnetizing current (IM).
///Not quite. The parallel magnetizing branch has noticeably higher voltage than for the motor rated operating conditions. Now, depending on the motor operating point design of the motor magnetic iron, the motor unloaded conditions may cause the no load current to become noticeably nonlinear and somewhat higher than the rated Im. The losses on the resistive parts of the magnetizing parallel branch circuit will also increase somewhat under the motor no-load conditions.\\

 

[PUMPDESIGNER]

Thank you guys, now i is expert toooo.

PUMPDESIGNER

 

[jghrist]

Seems to me that the main difference between jbartos and electricpete and aolalde is that for motors, there is no "no-load" condition. There is always some mechanical load because of friction, so the slip is never zero and the R₂/s in the equivalent circuit is never infinite.