![if !IE]> <![endif]>
Design of the field System: Salient pole Alternator:
(i) Axial Length of the pole: Axial length of the pole may be assumed 1 to 1.5 cm less than that of the stator core.
(ii) Width of the pole: Leakage factor for the pole is assumed varying between 1.1 to 1.15.
Thus the flux in the pole body = 1.1 to 1.15
Area of the pole = Flux in the pole body/ Flux density in the pole body. Flux density in the pole body is assumed between 1.4 to 1.6 wb/m2.
Area of the pole = width of the pole x net axial length of the pole.
Net axial length of the pole = gross length x stacking factor Stacking factor may be assumed as 0.93 to 0.95.
Hence width of the pole = Area of the pole / net axial length of the pole.
(iii) Height of the pole:
Height of the pole is decided based on the mmf to be provided on the pole by the field winding at full load. Hence it is required to find out the mmf to be provided on the pole at full load before finding the height of the pole. Full load field ampere turns required for the pole can be calculated based on the armature ampere turns per pole.
Hence full load field ampere turns per pole can be assumed 1.7 to 2.0 times the armature ampere turns per pole.
Armature ampere turns per pole ATa = 1.35 Iph Tph Kw /p And
ATfl = (1.7 to 2.0) ATa
Height of the pole is calculated based on the height of the filed coil required and the insulation.
Height of the filed coil:
If = current in the field coil
af = area of the field conductor
Tf = number of turns in the field coil Rf = resistance of the field coil
lmt = length of the mean turn of the field coil
sf = copper space factor hf = height of the field coil df = depth of the field coil
pf = permissible loss per m2 of the cooling surface of the field coil ζ = specific resistance of copper
Watts radiated from the field coil = External surface in cm2 x watts/cm2
= External periphery of the field coil x Height of the field coil x watts/cm2
Total loss in the coil = (If2 x Rf) = ( If2 x ζ x lmt x Tf / af)
Total copper area in the field coil = af x Tf = sf hf df
Hence af = sf df hf / Tf
Thus watts lost per coil = ( If2 x ζ x lmt x Tf ) Tf / sf hf df
= (If Tf)2 ζ x lmt/ sf hf df
Loss dissipated form the field coil = qf x cooling surface of the field coil
Normally inner and outer surface of the coils are effective in dissipating the heat. The heat dissipated from the top and bottom surfaces are negligible.
Cooling surface of the field coil = 2 x lmt x hf
Hence loss dissipated from the field coil = 2 x lmt x hf x qf
For the temperature rise to be with in limitations
Watts lost per coil = watts radiated from the coil
(If Tf)2 ζ x lmt/ sf hf df = 2 x lmt x hf x qf
Hence hf = (If Tf) / [ 104 x √(sf df qf)]
= ATfl x 10-4/ √(sf df qf)
Depth of the field coil is assumed from 3 to 5 cm,
Copper space factor may be assumed as 0.6 to 0.8,
Loss per m2 may be assumed as 700 to 750 w/m2
Hence the height of the pole = hf + height of the pole shoe + height taken by insulation
Copyright © 2018-2023 BrainKart.com; All Rights Reserved. Developed by Therithal info, Chennai.