Electrical wireman, Electrical Supervisor malayalam Note കള്‍ക്ക് താഴോട്ട് scroll ചെയ്ത് പോകുക.PAGES എന്നതില്‍ നിന്ന് നിങ്ങള്‍ക്ക് ആവശ്യമുള്ളത് തിരഞ്ഞെടുക്കുക. നിങ്ങളുടെ വിലയേറിയ അഭിപ്പ്രായങ്ങള്‍ അറിയീക്കുക. ഇത് നിങ്ങള്‍ക്ക് ഉപകാരപ്പെട്ടാല്‍ മറ്റുള്ള WIRING തൊഴിലാളികള്‍ക്ക് ഇത് എത്തിച്ച് കൊടുക്കുക.പുതിയ Address www.sselectricgroup.com.

Production of Rotating Magnetic Field

Production of Rotating Magnetic Field

The production of Rotating magnetic field in 3 phase supply is very interesting.
When a 3-phase winding is energized from a 3-phase supply, a rotating magnetic field is produced. This field is such that its poles do no remain in a fixed position on the stator but go on shifting their positions around the stator. For this reason, it is called a rotating field.
It can be shown that the magnitude of this rotating field is constant and is equal to 1.5 fm where fm is the maximum flux due to any phase.
A three-phase induction motor consists of three phases winding as its stationary part called stator. The three-phase stator winding is connected in star or delta.
The three-phase windings are displaced from each other by 120°. The windings are supplied by a balanced three phase ac supply. 
 
The three-phase currents flow simultaneously through the windings and are displaced from each other by 120° electrical. Each alternating phase current produces its own flux which is sinusoidal.
So all three fluxes are sinusoidal and are separated from each other by 120°.
If the phase sequence of the windings is R-Y-B, then mathematical equations for the instantaneous values of the three fluxes ΦR , ΦY ,ΦB can be written as,
   Φ=  Φmsin(ωt)                
Φ=  Φmsin(ωt – 120) 
ΦB  =  Φmsin(ωt – 240)
As windings are identical and supply is balanced, the magnitude of each flux is Φm .  
 

Case 1 : ωt = 0

Φ=  Φmsin(0) =  0
Φ=  Φmsin(0 – 120) =  -0.866 Φm
ΦB  =  Φmsin(0 – 240) = +0.866 Φm

Case 2 : ωt = 60

Φ=  Φmsin(60) =  +0.866 Φm
Φ=  Φmsin(- 60) =  -0.866 Φm
ΦB  =  Φmsin(- 180) = 0

Case 3 : ωt = 120

Φ=  Φmsin(120) =  +0.866 Φm
Φ=  Φmsin(0) =  0
ΦB  =  Φmsin(- 120) = -0.866 Φm

Case 4 : ωt = 180

Φ=  Φmsin(180) =  0
Φ=  Φmsin(60) =  +.866 Φm
ΦB  =  Φmsin(- 60) = -0.866 Φm
By comparing the electrical and phasor diagrams we can find the flux rotates one complete 360 degrees on the 180-degree displacement of flux. 

No comments:

Post a Comment

Theme images by Jason Morrow. Powered by Blogger.