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Synchronous Generator or Alternator
Transcript of Synchronous Generator or Alternator
Alternator is a machine which generates an alternating emf
Alternator runs at constant speed called synchronous speed, hence they have
given the name synchronous Generator
Alternator consists of two parts
Stator
Rotor
It is a static part and outer part
Stator consists of steel frame which encloses
hollow cylindrical core
Core is made up of laminated silicon steel to
reduce Eddy current loss and hysteresis loss
Core has slots on its inner periphery
3 phase star connected winding is placed in
the slots and it is called armature winding.
It is a rotating part it is placed inside the stator
Rotor is mounted on the shaft which is connected to the prime mover
It has field windings which are excited by the pilot exciter which is mounted
on the shaft
There are two types of rotors
Salient pole type rotor
Non salient pole type rotor
Poles are mounted on the steel frame called magnetic
wheel
Field windings are connected in a series such a way
that when they are excited by DC exciter alternative
poles behaves as NORTH-SOUTH poles
Poles are projected
Used for low speed
Large number of poles N=120f/P
Water turbine is prime mover
Rotor is made of smooth solid cylinder having
numerous slots on outer periphery
Field windings are placed in these slots
Regions which are left un slotted form the poles
Poles are not projected
Used for high speed
Less number of poles N=120f/P
Steam turbines are prime movers
It works on the principle of electromagnetic
induction
Rotor field windings are excited by 110V Or
220 V DC voltage generated by pilot exciter
Rotor is rotated by prime mover
Flux produced by the rotor conductor sweeps
across the stator conductor and hence EMF is
induced.
Ns = Synchronous Speed = Rotar Speed in r.p.m
P = number of poles
F = frequency of e.m.f in Hz
No of cycles/revolution = P/2 ……………..(1)
No of revolution / second = Ns/60………….(2)
(1)*(2)
No of cycles / Second = No of cycles/revolution X No of revolution/Second
f = P/2 X Ns/60
f = Ns P/120
OR
Ns = 120 f/P
Winding Terminology
Conductor
Turn
Coil
Coil Side
Pole pitch
n = 180 electrical = slots / pole
Slot angle(𝛽)
𝛽 = 1 slot angle = 180
ƞ
Coil Span
Full pitch
Short pitch
Angle of short pitch (𝛼)
𝛼 = 𝛽 𝑋 𝑁𝑜 𝑜𝑓 𝑠𝑙𝑜𝑡𝑠 𝑏𝑦 𝑤ℎ𝑖𝑐ℎ 𝑐𝑜𝑖𝑙𝑠 𝑎𝑟𝑒 𝑠ℎ𝑜𝑟𝑡 𝑝𝑖𝑡𝑐ℎ Concentrated windings :-
Distributed Windings
Pitch Factor (Kp)
Kp = Cos(𝛼
2)
Distribution Factor (Kd)
Kd = 𝑆𝑖𝑛(
𝑚𝛽
2)
𝑚 𝑆𝑖𝑛(𝛽
2)
Winding factor [Kw]:
Kw = Kp X Kd
EMF equation of an alternator
Let Zph = No of stator conductors/Phase
Zph = 2Tph
Or Tph = Zph/2 = No of turns / Phase
P = No of Poles
f = Frequency of the induced e.m.f in Hz
∅ = flux/pole in weber
The flux cut by the conductor in one revolution is
d∅ = 𝑃∅
The time taken for one revolution is
dt =60
𝑁𝑠
Hence Average e.m.f induced in one conductor
= 𝑑∅
𝑑𝑡
= 𝑃∅ 60
𝑁𝑠
= 𝑁𝑠 𝑃∅
60
Since there are Zph conductors/phase
Hence Average e.m.f / phase = 𝑁𝑠 𝑃∅
60 XZph
Eavg = 𝑍𝑝 𝑃∅
60 X Ns
Eavg = 𝑍𝑝 𝑃∅
60 X
120𝑓
𝑃
Eavg/phase = 2 Zphf∅………………………(1)
W . K . T form factor , Kf = 𝐸𝑟𝑚𝑠
𝐸𝑎𝑣𝑔
Erms = Kf Eavg
Erms/phase = Kf Eavg/phase
Erms/phase = 1.11 Eavg/phase ……………….(2)
(1) In (2) we get
Erms/phase = 1.11 X 2 Zph f ∅
Eph = 2.22 Zph f ∅
Due to Kp and Kd the induced e.m.f gets reduced by a small quantity
Eph = 2.22 Kp Kd Zph f ∅
Voltage Regulation: Change in terminal voltage from no load to full load
% Regulation = 𝐸 −𝑉
𝑉 X 100%
E = No load voltage/phase
V = Full load voltage/phase
Important formulae
Ns = 120𝑓
𝑃 OR P =
120𝑓
𝑁
Were , Ns = Synchronous speed or rotor speed in r.p.m
f = frequency of e.m.f in Hz
P = No of poles
Tph = Zph / 2
were , Tph = No of turns conductor /phase
Zph = No of stator conductors / phase
n = 𝑇𝑜𝑡𝑎𝑙 𝑁𝑜 𝑜𝑓 𝑠𝑙𝑜𝑡
𝑝𝑜𝑙𝑒
were , n = Slots per pole
𝛼 = 180
𝑛
were 𝛼 = Slot angle
n = Slots per pole
𝛽 = 𝑛𝑜 𝑜𝑓 𝑠ℎ𝑜𝑟𝑡 𝑝𝑖𝑡𝑐ℎ𝑒𝑑 𝑋 𝛼
Were, 𝛽 =
Kp = cos (𝛽
2)
were , Kp =
𝛽 =
Kd = =sin (𝑞
𝛼
2)
𝑞 sin (𝛼
2) =
q = no of slots / pole per phase
Eph = 4.44 f ∗ ∅ ∗ 𝑇𝑝ℎ ∗ 𝐾𝑝 ∗ 𝐾𝑑
were , Eph =
f =
Tph =
Kp =
Kd =
1) A six pole dc generator is running and producing a voltage at a frequency of 60 Hz . Calculate the revolutions
per minute of the generator . If the frequency of the generated voltage is required to be decreased to 20 Hz , how
many poles would be needed on the generator ,if it still runs at same speed ?
The speed of rotation is given by
Ns = 120𝑓
𝑃
If the frequency is decreased to 20 Hz , required number of poles ,
P = 120𝑓
𝑁
2) A 4-pole ,3 phase alternator has 36 slots . It has an armature winding which is short-pitched by one slot . Calculate
its coil span factor
Coil span factor as
Kp = cos (𝛽
2)
As the winding is short pitched by one slot , the angle of short pitch is given by
𝛽 = 1 𝑋 𝛼
Since one pole corresponds to 180 degree and there are n slots per pole , the angle corresponds to one slot or its slot
angle is
𝛼 = 180
𝑛
The number of slots per pole
n = 𝑇𝑜𝑡𝑎𝑙 𝑁𝑜 𝑜𝑓 𝑠𝑙𝑜𝑡
𝑝𝑜𝑙𝑒
3) An 8-pole , 3 phase , 120 slots alternator has distributed armature winding . Calculate its distribution
factor
Distribution factor as
Kd = sin (𝑞
𝛼
2)
𝑞 sin (𝛼
2)
Slot angle 𝛼 = 180
𝑛
Number of slots per pole ,
q = 𝑛
3
The number of slot per pole
n = 𝑇𝑜𝑡𝑎𝑙 𝑁𝑜 𝑜𝑓 𝑠𝑙𝑜𝑡
𝑝𝑜𝑙𝑒
4) Determine the distribution factor for a machine having 9 slots per pole for the following cases
a) A three phase winding with 120 degree phase group
b) A three phase winding with 60 degree phase group
The slot angle , 𝛼 = 180
𝑛
a) Since one phase occupies 120 degree , the number of slots in one phase group ,
q = 120
20
Kd = sin(𝑞
𝛼
2)
𝑞 sin(𝛼
2)
b) Since one phase occupies 60 degree , the number of slots in one phase group ,
q = 60
20
Kd = sin (𝑞
𝛼
2)
𝑞 sin (𝛼
2)
5) A 2-pole , 3-phase alternator running at 3000rpm has 42 slots with 2 conductors per slot . Claculate the flux per
pole required to generate line voltage of 2300 v . Assume Kd = 0.952, Kp = 0.956.
Ns = 120𝑓
𝑃
f =
Zph = Z/3
Z = slots X P
Zph =
Tph = Zph / 2
Assuming star connected armature ,
Eph = 𝐸𝐿
3
Eph =
Eph = 4.44 f ∗ ∅ ∗ 𝑇𝑝ℎ ∗ 𝐾𝑝 ∗ 𝐾𝑑
∅ =
6) A 3-phase , 16 pole , star –connected alternator has 144 slots on armature periphery . Each slot contains 10
conductors . It is driven at 375 r.p.m . The line value of e.m.f available across the terminals is observed
to be 2.657 KV . Find the frequency of the induced e.m.f and the flux per pole
Give ,P = 16 , Ns = 375 rpm ,EL = 2.657KV , slots = 144 , conductors/slot = 10 , f = ? Kp = ? Kd = ? ∅ =
Ns = 120𝑓
𝑃 = f = Ns P/120 = ?
Since nothing is mentioned about the pitch , assume full pitch , i.e. ., Kp = 1.
Now , the number of slots/pole , n = slots/pole = ?
The number of slots / pole /phase q = n/phases = ?
The slot angle . 𝛼 = 180
𝑛 = ?
Hence Kd = sin(𝑞
𝛼
2)
𝑞 sin(𝛼
2) = ?
Total number of conductors = Z = Slots X conductors/slot = ?
Conductors / phase , Zph = Z/3 = ?
Turns/phase , Tph = Zph/2
For star connected armature , Eph = 𝐸𝐿
3 = ?
Eph = 4.44 f ∗ ∅ ∗ 𝑇𝑝ℎ ∗ 𝐾𝑝 ∗ 𝐾𝑑
∅ = Eph
4.44 f 𝑇𝑝 𝐾𝑝 𝐾𝑑