Electrical Machines 2
Winding Design Sample of Synchronous
Generator (Alternator) for Armature Winding
1. Wind a 12 slots, 4-poles, 60 Hz, alternator armature winding connected in Wye and Delta.
Solution:
*Speed (N) = (120)(f) / P = (120)(60) / 4 = 1800 rpm
*Coil Span = slots / pole = 12/4 = 3 slots/pole
*Pole Pitch = slots /pole = 12/4 = 3 slots/pole
*Angular displacement between slots (α) = (180° )/ (slots per pole) = 180° / (12/4) = 60°
*Angular displacement between phase winding = 120°
*Winding pitch or coil pitch (P) = (coil span in terms of number of slots) / (slots per pole) = 3 / 3 = 1
(Note: If the coil span is equal to the pole pitch, the coil pitch is called Full Pitch. If the coil
span is less than, then it is called Fractional Pitch or Short Pitch)
*Pitch Factor (kp) =
𝑣𝑒𝑐𝑡𝑜𝑟 𝑜𝑓 𝑝ℎ𝑎𝑠𝑜𝑟 𝑠𝑢𝑚 𝑜𝑓 𝑖𝑛𝑑𝑢𝑐𝑒𝑑 𝑒𝑚𝑓𝑠 𝑝𝑒𝑟 𝑐𝑜𝑖𝑙
𝑎𝑟𝑖𝑡ℎ𝑚𝑒𝑡𝑖𝑐 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑖𝑛𝑑𝑢𝑐𝑒𝑑 𝑒𝑚𝑓𝑠 𝑝𝑒𝑟 𝑐𝑜𝑖𝑙
Or *Pitch Factor (kp) = sin (90 P)
*(kp) = sin (90 P) = sin (90 x 1) = 1
*Distribution factor or breadth factor or spread factor or winding factor (kd):
kd=
𝑒𝑚𝑓 𝑖𝑛𝑑𝑢𝑐𝑒𝑑 𝑖𝑛 𝑎 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑒𝑑 𝑤𝑖𝑛𝑑𝑖𝑛𝑔
𝑒𝑚𝑓 𝑖𝑛𝑑𝑢𝑐𝑒𝑑 𝑖𝑓 𝑡ℎ𝑒 𝑤𝑖𝑛𝑑𝑖𝑛𝑔 𝑤𝑜𝑢𝑙𝑑 ℎ𝑎𝑣𝑒 𝑏𝑒𝑒𝑛 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑
or to simplify
kd =
𝑛α
)
2
𝑛α
n sin( 2 )
sin(
, where n= slots per pole per phase
*Distribution factor or breadth factor or spread factor or winding factor (kd): (n = 12 / 4 / 3 = 1)
kd =
𝑛α
)
2
𝑛α
n sin( 2 )
sin(
=
(1)(60)
)
2
(1)(60)
(1) sin( 2 )
sin(
=1
Front View of Synchronous Generator
(Alternator)
Rotor (Field Winding)
Stator (Armature Winding)
Alternator Armature
Winding in Wye
Connection
Alternator Armature
Winding in Delta
Connection
Project 1:
Wind a 24 slots, 4-poles, 60 Hz, alternator armature winding connected in Wye and Delta.
Winding Design Sample of Synchronous
Generator (Alternator)
Design a 100-kW alternator with a line voltage of 220 V, 3-phase, unity pf, 60 Hz, 100 armature
winding turns. Determine the following:
1. Size of Armature winding Conductor
2. Size of field pole winding conductor
3. Number of turns of field pole winding
4. Dimension of field pole with L = 3W.
Solution:
1.) P= 3 (VL)(IL) (pf)
100 x 1000 = 3 (220) (IL) (1)
IL = 262.43 A
(In determining the size of magnetic conductor, current capacity =
Therefore, 262.43 A x
Use 2 sets AWG # 0.
800 CM
Ampere
800 CM
)
Ampere
= 209,944 CM (Based on the table for Magnet Wire)
Table for the
Magnetic Wire
2.) (Note: For Good Design, Field winding current capacity is 9% of the Armature winding
capacity, thus, If = 9% of Ia)
If = 9% (262.43)
If = 23.62 A x
800 CM
Ampere
= 18,896 CM (Based on the table)
Use 1 set AWG #7
3.) (Note: For good design, Ampere turn per pole should not be more than 12,000 AT in the field)
Therefore, AT = Ampere x turns
12,000 AT = (23.62 A)( N)
N = 508 turns
4.) For the formula of induced voltage, Eind =
4.44 f N ɸ
10⁸
,B=
ɸ
A
Where: Eind = voltage induced, N= number of turns in the windings, f=frequency, ɸ=flux in the core,
B= flux density, A=cross sectional area of the core
(Note: For Good design at 60 Hz, the flux density (B)=24,000 to 35,000 lines of force per square inch.
4.44 f N BA
10⁸
4.44 (60)(100)(24000)A
10⁸
Eind =
220 =
A = 34.41 in²
Since A = L x W, and L=3W
A = 3W(W) = 3W²
W=
A
3
=
34.41
3
= 3.39 in.
L = 3W = 3(3.39) = 10.17 in.
Project 2:
Design a 300-kW alternator with a line voltage of 400 V, 3-phase, unity pf, 60 Hz, 200 armature
winding turns, flux density is 30,000 lines of force per sq. inch. Determine the following:
1. Size of Armature winding Conductor
2. Size of field pole winding conductor
3. Number of turns of field pole winding
4. Dimension of field pole with L = 3W.