ELECTRICAL MACHINES (FORMULA NOTES)
Features of Sing]
‘Transformer.
phase Transformers
Transformer is a static device which performs:
It works on the principle of electromagnetic induction
Transfers electrical energy from one electrical circuit to another without changing of
frequency.
The energy transfer usually takes place with change in voltage, although this is not always
necessary.
Secondary winding
Primary winding Laminated
steel core
Basic Diagram of Transformer
When the transformer raises the voltage, i.e. when output voltage of a transformer is higher
than its input voltage, it is called step-up transformer and when output voltage is lowered, it
is called step-down transformer.
‘Transformer doesn’t change the frequency of the system. Hence, it can be treated as
constant frequency device, it transfers almost same amount of power from one circuit to
another so it can be treated as constant power device and the amount of flux in its core
remains constant, so it is also known as constant flux device.
Ideal Transformer
Ideal transformer does not have any loss, so resistances are also assumed to be zero.
Magnetic leakage flux in transformer is completely zero. (K=1, coefficient of coupling)
Magnetic permeability of transformer core 11 = 99. So, the flux can be established without any
excitation current.
Magnetization curve of transformer core is assumed to be linear.
For an ideal two-winding transformer with primary voltage Vi applied across Ni primary turns
and secondary voltage V2 appearing across Nz secondary turns: Ve Ny
©The primary current I: and secondary current Iz are related by:
LN
bh Nn M
© Fora single-phase transformer with rated primary voltage Vi, rated primary current Ii, rated
secondary voltage V2 and rated secondary current Iz, the volt ampere rating S is:
S=VjI, = VJ,
‘+ During operation of transformer:
BOM
B, «Sha Mt
mn EES
+ Baar = Constant =: constant
Note:
+ Gross cross-sectional area = Area occupied by magnetic material + Insulation material.
‘+ Net cross-sectional area = Area occupied by only magnetic material excluding area of
insulation material.
‘+ Hence for all calculations, net cross-sectional area is taken since majority 4(flux) flows in
magnetic material
$=BA,
E.M.F, EQUATION OF TRANSFORMER
Let, Flux © = ®m sin(wt)
As the flux is assumed to be sinusoidal,
In primary winding sid:
According to faraday’s law,
Ban
dt dt
Instantaneous value of emf in primary
(4, Sinot)
N, {ot-=)
e, =N6,0sin{ ot 2
: (2)
Ea = Vast,
In secondary winding side
Similarly, [Enns = V2nf6,.N|
BLN
BON From the above equation,
EMF per turn in primary = EMF per turn in secondary
Note:
EB
+ Transformation ratio = K = =
+ Turn ratio= 2=N:N,
AIH
Equivalent circuit of Transformer under No-Load condition:
if
No load/shunt branch
+ No load current/ exciting current = 1, =I, +1, = 1,26
1, =I, cose,
1 -i,sing,
In = core loss component
1, = magnetizing component
+ No load power = VJ, cos, = V\L, = Iron losses
No load power
M
EQUIVALENT CIRCUIT OF A TRANSFORMER
Equivalent circuit
The equivalent circuit of a transformer having voltages transformation ratio, k
in figure below:
£ is shown
& Ideal Transformer
Exact Equivalent Circuit
Practical representation of simple two-winding transformer:
Ideal Transformer
Referred to Primary side
x, = Equivalent reactance referred to primary = (x, 2)
ale
= Equivalent resistance referred to primary = (s
When all the parameters are referred to secondary side
ry ik ix, "
N, No
i
it
Ideal Transformer
Referred to Secondary side Equivalent resistance to secondary side = (Fr, + k*r,)
X,, = Equivalent reactance referred to secondary side = (x, + k*x,)
Core loss resistance
X,, = Magnetising reactance
Ny
Transformation Ratio= 7°
Where, k
TESTS ON TRANSFORMER:
1, Open Circuit Test (0.C.) or No-load Test:
+ This test is performed to determine core or iron loss Pi and no-load current Io.
+ This test is helpful in determining the magnetizing current I, and the core loss
components Iw and hence no-load shunt branch parameters Re and Xo.
w
Dh Ppa 3
vy Vie
‘ HV. side open
ut HT.
Open Circuit configuration
Iron loss (P\) = Input no-load Power = We (wattmeter reading)
No-load current = To
w,
Hence no load p-f., cos a
P. * vp
Now from no-load phasor diagram
YM
L
At no load efficiency of the transformer is zero because output power i
zero.
No load test conducted on rated voltage and rated frequency on LV side of transformer.
‘cuit Test (S.C.):
S.C. test is carried out at rated current to determine the Cu loss at full load condition.
Full load Cu-loss, P. = Fr, =W i IV. side
HT LT
Short circuit of transformer
Equivalent resistance referred to HV side, r,
Equivalent impedance referred to HV side, Z, it
Thus, equivalent reactance, X,, =
Note: As the applied voltage required to circulate full load current at short-circuit would be
limited to 5% to 10% of the rated voltage, the core loss at such reduced voltage is ignored.
Also, the exciting current at such low voltage may be neglected.
VOLTAGE REGULATION OF A TRANSFORMER
It is defined as the percentage change in secondary terminal voltage from no load to full load,
expressed as a percentage (or pu) of the secondary rated voltage.
If,
Vo = secondary terminal voltage at rated load, and
E2 = secondary terminal voltage at no load,
Then,
Voltage regulation (in pu)
Voltage regulation Mi. <100 (in %)
The magnitude of change in secondary voltage depends on the load power factor, load
current, total resistances and total leakage reactance of the transformer.
Therefore, voltage regulation equation can be given as
Voltage drop in the secondary terminal voltage,
E,-V,=1y,, cosa,
x, SiN,
Hence for any load current Iz, Per unit voltage regulation is given by: where,
Lt
= PU Equivalent resistance or PU resistance drop = Rou
1
&
“7 = PU Equivalent reactance or PU reactance drops = Xow
At rated load voltage regulation in PU = (R,., cos 4, + X,,, Sind)
Percentage voltage drop =(R,, cos#, + X,, sin@,) x 100
Condition for Zero Voltage Regulation:
\
Zero voltage regulation takes place at leading pf and when load pf is (=).
Condition for Maximum Voltage Regulation:
Maximum voltage regulation occurs at lagging power factor and when pf is (=
Voltage,
R
oes 7 os 6
—+ Load (lagging PF)
Load <—|
(leading Pf)
Plot between voltage regulation and power factor
Voltage regulation is a figure of merit of a transformer and its low value is always desired.
n may be reduced by reducing the per unit impedance of the
+ The voltage regulat
transformer.
‘© Per unit impedance, Zpu can be reduced for reducing Rou or Xou
© Per unit resistance, Rou is directly kept at an optimally low value,
+ Xpu can be reduced by reducing leakage flux. Per Unit System
A per-unit system is the expression of system quantities as fractions of a defined base unit quantity.
Calculations are simplified because quantities expressed as per-unit do not change when they are
referred from one side of a transformer to the other.
Per Unit Value
It is usual to assume the base values as:-
Base voltage = rated voltage of the machine
Base current = rated current of the machine
Base impedance = base voltage /base current
vvvy
Base power = base voltage x base current
Hence it can be concluded that out of 4 parameters, any two can be selected as base value
and according to that the remaining values are determined.
Values which we will choose: 1) Base Apparent Power = Sp; 2) Base Voltage
Values which we calculate: 1) Base Current I, 2) Base Impedance Z, = 2
For an ideal Transformer Eieu = Ez.
Also, Zip = Zaeu
Direct relation for changing the base of an impedance:
Ss_vew . Vo.
- sow
Zou_new = Zru_o *"G ve
s_o1o Ve_new
Losses present in transformer
Copper losses: Transformer windings major losses
Iron losses: Transformer core
Stray load losses: Cu parts and Iron parts minor losses
ayer
Dielectric losses: Insulating materials.
Cu losses in transformer:
Total Cu loss = TR, +ER, =ER,, = ER,
VA rating of transformer
* Rated current on primary side = =
VA rating of transformer
E
* Similarly, current on secondary side * Cu losses « F or 12. Hence these are called as variable losses.
FLCulossin watts _ ERyy
* PLU, full load Cu loss = TL CU loss in watts _
Varating of transformer EI,
+ If VA rating of transformer is taken as base then P.U Cu loss «I? as remaining terms are
constant
+ P.U. Cu loss at x (Loading factor) of full load = x? x PU full load Cu loss
+ PU.resistance drop ref to primary = BR. ER
& hE,
.U. Full Load Cu loss
P.U, resistance drop =
2. Iron (or) Core losses in Transformer:
2.1 Hysteresis loss:
Steinmetz formula
Wh = 7) Bast £V
‘Area under one hysteresis loop.
where,
1 = Stienmetz coefficient
Bmax = Maximum flux density in transformer core.
f = Frequency of magnetic reversal = supply frequency.
V = Volume of core material
x = Hysteresis coefficient (or) stienmetz exponent = 1.6 (Si or CRGO steel)
2.2 Eddy current loss:
Eddy current loss, W, © R,. x12
As area decreases in laminated core resistance as a result conductivity decrease.
W, = K (Cons tant)B%,.,f#(Supply freq.) x t?(Thickness of lamination)
(It is a function of o)
During operation of transformer:
y
B, «Mb
mee
Case(i): = Constant, Bmax = Constant
Wo w,
The transformers must have the same polarities.
> The transformers should have equal turn ratios.
Desirable conditions:
> The voltages at full load across transformers internal impedance should be equal.
> The ratios of their winding resistances to reactance should be equal for both
transformers. This condition ensures that both transformers operate at the same
power factor, thus sharing active power and reactive volt amperes according to
their ratings. Conditions for parallel operation of Three-Phase transformers:
The condition for the parallel operation of single phase and three phase transformers is the
same but with the following additions:
> The phase sequence of the transformers must be identical
> The primary and secondary voltages of all the transformers connected in parallel
must have the same phase shift.
1
ise, Leakage impedance * RyA rating
Let two transformers A and B having Zes and Ze» equivalent impedances are in parallel and Sar
and Sw: are the kVA rating respectively
Zn
Hence, kVA shared by A=| 5—S>
25 + Ze)
Z,
o-(572,-|s
and kVA shared b}
YO \Ze4 Ze
where, S = Total kVA
AUTO - TRANSFORMER
‘A transformer whose winding is common to both primary and secondary circuits is called
autotransformer. The primary and secondary are connected electrically as well as coupled
magnetically.
Auto-transformer construction uv
+ K of autotransformer = LY.
0 HV
(VA) induction = (Vi ~ Va)Is
Input kVA = Vil,
(VA)paucson _ (
input KVA|
s. (KVA)incueton = (1 = K) input kVA
(RVA)concucton = Input kVA ~ (KVA)nd
(KVA)concucton = K x input kVA
Weight of conductor in section AB of auto t/f « (N,-N,)I;
Weight of conductor in section BC of auto t/f « (I, -I,)N,
Total Weight of conductor in auto t/f is
oT, (Ny — Na) Ny + (No — Ni) No
« 2(N, -N,)I,
Total Weight of conductor in two winding transformer
ee hN, +1N, « 2hN,
We. of conductor in an auto t/f _ 2(N,-N)I, _
-eas -1-k
Wt. of conductor in 2 wda t/F 2NI,
Weight of conductor in auto-transformer = (1 - K) (Weight of conductor in in two winding
transformer)
Thus, saving of conductor material if auto-transformer is used = K x (Conductor weight in
two winding transformer)
Some Important Relations:
Weight of conductor in auto transformer __ (4 _ gy
5 Weight of conductor in 2-winding transformer
kVA rating as an auto transformer _{_1 |
g KVA rating as 2-winding transformer (1k)
.u, full load losses as an auto transformer
Pw full load losses as an auto transformer _ (1 -k)
(5 Bu. fll load losses as a 2-winding transformer
au. impedance drop in an auto transformer _ 4 y
Prt Impedance drop in 2-winding transformer ~ )
Voltage regulation as i
Voltage regulation in 2-1
auto transformer
inding transformer
@-%) TERTIARY WINDING
Tertiary winding of transformer
MMF balance:
Nui, - NGL NL, = NE,
Application of Different 3-phase Connections:
1, B/ > This connection is used where we have simple 3-g load, specially at low voltage
level. No mix loading possible.
2. ¥/A > This connection is used for step down application.
3. A/Y > This connection is used for step up application except in distribution system where,
A/Y is used for step down application for mix loading.
4. Y/Y — This connection is quite attractive for HV applications. It is not generally used
without a tertiary A.
Impact of dimensions on varius parameters
Voltage rating of transformer depends on the level of insulation of the conductor.
Voltage rating is directly proportional to the area of the transformer.
Voltage rating is directly proportional to square of the dimensions of the transformer.
Current rating of the transformer is depends on the cross sectional area of the conductor.
Current rating is directly proportional to the area of the transformer. Current rating is directly proportional to square of the dimensions of the transformer.
KVA rating of the transformer is quartic of the dimensions of the transformer.
DC Machines
EMF EQUATION AND DEVELOPED TORQUE
Notations:
9 + Flux/pole (Wb)
P — No. of poles
Z -+ Total no of conductors
N — Rotor speed (rpm)
‘A — Number of parallel paths
A = 2, for wave windings
A= P, for lap winding
PONZ
60A
As we know that,
Volts
on ~ 2N rad / sec
60
600
n- 800
Qn
Put this value in the above formula-
Where k= £2 = Machine constant
Developed Torque
Developed power Ps = Es Is and Pa = T.tdm
2 Pa = T.tom
T= Eh gl, = kot
On
IT = kl, Induced emf > koo,,
Developed Torque = [Ty = kdl, COMMUTATION PROCESS
The reversal of current in the armature coil by means of brush and commutator bars, is called
commutation process. Good commutation means no sparking at the brushes and with
commutator surface remaining unaffected during continuous operation of the DC machine.
+ Under commutation: Tesmmutaton > Te
+ Over commutation: Teommuttion < Te
The time required by the coil current to change from +Ic to -Ic is called the commutation
period Te.
Brush _width Where, Tc is commutation period
-ommutator peripheral spees
ARMATURE REACTION
The effect of armature flux due to armature current over main field flux is known as
Armature reaction effect.
ona
faa (ontoad)
Resultant thx
ietibution
Armature Reaction Effect
Effects of Armature Reaction
(i) If strengthening effect = weakening effect, the average flux under the pole remains same
and therefore no demagnetizing effect of AR.
But due to saturation, strengthening effect is less than the weakening effect so average flux
under each pole reduces and hence Armature reaction is “Demagnetizing”.
(ii) The MNA shifts in the direction of rotation for generator action and in a direction opposite
to rotation for motor.
(iii) Armature flux is perpendicular to field flux. So, the nature of armature reaction is cross
magnetizing. INTERPOLES
+ In large machine, interpoles also called commutating poles are used to overcome
‘commutation problems.
+ The interpolar windi
is designed to neutralize armature MMF in interpolar region
‘+The interpoles winding carries the armature current as it is connected in series with the
armature winding. The presence of interpoles ensures sparkless linear commutation
COMPENSATING WINDING
compensating winding consists of conductors embedded in pole faces and carry armature
current in a direction opposite to armature conductor current under one pole arc. The
compensating winding may be designed to completely neutralize the armature MMF of the
conductors that lie under the pole are resulting into restoration of main field flux.
DC machine compensating winding mmf per pole is:
2/2 { Pole arc) L
P “(Pole Pitch } “A
© Foomp(MMF) |AT/pole
2/2, (1 _ Pole arc) 1, , Brresute
Pp “\*” Pole Pitch } Hy
xL, |AT/pole
interpetar
ta
CLASSIFICATION OF DC MACHINE:
DC Machines have been classified into multiple categories based on the connection of field
winding and the armature winding,
Gsolation b/w field (a) Series Excited
winding and armature (b) shunt excited
winding) (©) Compound Excited
(ji) Cumulative compound
(ii) Differential compound
Shunt Generator:
+ Fora shunt generator with armature induced voltage Es, armature current Is and armature
resistance Re, the terminal voltage V is:
V =. DRe
+ The field current Ir fora field resistance Reis: 1, =
+ The armature induced voltage Es and torque T with magnetic flux @ at angular speed
are E, =k,o0,, =k,o, T=k ol, =k,I,
Where, k:and km are design coefficients of the machine.
Note that for a shunt generator:
- Induced voltage is proportional to speed.
= Torque is proportional to armature current.
+ The airgap power Pe for shunt generator
P, = OgT = Ely =kyOm Ty
Series Generator:
+ For a series generator with armature induced voltage Es, armature current Iz, armature
resistance Re and field resistance Rr, the terminal voltage V is:
V = Es (IaRe + LR) = Eo Ia(Ro + Re)
The field current is equal to the armature current.
+ The armature induced voltage Es and torque T with magnetic flux @ at angular speed wm
are
E, =k,o,1, = kyOnle
T-kor
k,E
Where kr and km are design coefficient of the machine,
Note that for a series generator:
- Induced voltage is proportional to both speed and armature current,
- Torque is proportional to the square of armature current.
- Armature current is inversely proportional to speed for a constant Es
+ The airgaps power Pe for a series generator is
P,=0,T =E,], = kon
Cumulatively compounded DC generator: (long shunt):
(2) b=k+h
(b) Vi = Ee ~ Ia (Ra + Rs)
(©) 1, = =Shunt field current
(d) The equivalent effective shunt field current for this machine is given by
{ Armature reaction MMF
v N
Where, Nee = No. of series field turns
Nr = No of shunt field turns
Differentially compounded DC generato:
(a) b=r+h
long shunt): (b) Ve = Ea - Ia (Ra + Re)
© Shunt field current
(4) The equivalent effective shunt field current for this machine is given by
{ Armature reaction we)
L N
Where, Nese = No. of series field turns
Nr = No of shunt field turns
Shunt Motor:
+ For a shunt generator with armature induced voltage Es, armature current Is and armature
resistance Re, the terminal voltage V is:
V = Es + aR
The field current Ir for a field resistance Rr is: 1, =
R
+ The armature induced voltage Es and torque T with magnetic flux @ at angular speed w
are:
E, =k,@o =k,
T =k, =k,I,
Where, ke and km are design coefficients of the machine.
Note that for a shunt motor:
- Induced voltage is proportional to speed.
= Torque is proportional to armature current.
+ The airgap power Pe for shunt generator is:
P, R,
=0
# If the rotor resistance is made equal to its reactance at stand still than the
motor starts with maximum starting torque.
# Under such conditions the rotor power factor will be 0.707 lag.
Motor Torque in terms of Tem:
The torque expression of an induction motor can also be expressed in terms of maximum
torque Ton and dimension less ratio <>. In order to get a simple and approximate
expression, stator resistance r: or the stator equivalent resistance Re is neglected.
te 2x
* The slip at which maximum torque occurs is TORQUE SLIP CHARACTERISTIC
Torque equation is given by:
sR,
T.-K
Re + (SX)
Low slip region:
SR,
R24 (5X,)
le
For low slip, (sX2)? <<< Ro?
Tye oa >Texs
For high slip region:
(sX2)? >>> R?
Low slip (operating region)
soa
N>N, S—> negative
slip
+-Braking-Region-+|+—Motoring-Region—>|
«Generating Region—>
‘ic on induction machine
Torque-slip characteri
In an operating region, the torque-slip characteristic is essentially a straight line, It has good
speed regulation
Speed Regulation: It is the change in speed when the full load across shaft is disconnected.
It is expressed in % of full load speed. SR% = 100
N
Note: As the rotor resistance increases under operating region, to maintain constant torque,
slip also increases correspondingly.
Constant torque, 7, «2S
POWER FLOW IN INDUCTION MOTOR
The Power Flow Diagram of an Induction Motor is shown below.
Rotor i/p to
stator = airgay
Power i/pto stator power
from mainse
Mechanical power
developed, Poe
Power
> J+ rotor
sehatt
Rotor Stator { | windage
FR coreloss Rotor Rotor core Friction 95°
loss rR 108s oibte sat
‘= (negligible bearings
ae and sliprings
small slips) of (if any)
Power Flow Diagram of Induction Motor
* The power flow diagram of 3- induction motor is:
* Per phase power input to rotor is
Py = Eal2 cos 0,
Per phase rotor resis tance /s
Per phase rotorimpedance — Te 75)". (x,)
=f hx B/S
Ve/sy toy yin / sy +0)
+ Pa is the power transferred from stator to rotor across the air gap. Therefore, Ps is called air
cos@, =
aap power
p, - EE -Br + tA |
$ 3
P, = (Rotor ohmic loss) + Internal mechanical power developed in rotor (Pm) = sPs +(1 —
s)Po
Py = (1-S)P, =
Rotor ohmic loss = Pe =s?,
T-5
+ Internal (or gross) torque developed per phase is given by Internal mechanical power developed in rotor
Rotor speed in mechanical radian per sec.
Br, loss
_ Rotor ohmic loss
(esi
t
+ Power available at the shaft can be obtained from Ps as follows. Output or shaft power,
Poy = Pm — Mechanical losses
+ Mechanical losses imply frication and windage losses
Pen = Pp — Rotor ohmic loss ~ Friction and windage losses = Net mechanical power output or
net power output output or shaft power
Ty = Pat Pa,
“Rotor speed (1-s)o,
+ If the stator input is known. Then air gap power Py is given by Py = stator power input -
stator PR loss - stator core loss,
* Ratio of Rotor input power, rotor copper losses and gross mechanical output is
>1:s:(1-s)
Rotor copper losses = s x Rotor input
Gross Mechanical output = (1-s) x Rotor input.
Rotor copper losses = (Gross mechanical output) x
Efficiency of the rotor is approximately equal to
Gross mechanical power output
Nrotor =
Rotor input
1.-s)Rotor
(2=s)Rotor input
Rotor input
N
Maar
DETERMINATION OF EQUIVALENT CIRCUIT PARAMETERS
It requires two tests to be conducted on a given induction motor to find out its parameter,
(1) No-load test
(2) Block Rotor test
1, No-Load Test + This test is used to determine the core losses
+ The motor is made to run at rated voltage and frequency on no load and stator voltage
Vn, input current Int and input power Pai are recorded.
+ Rotor core loss and rotor copper loss are very less and can be neglected because current
in rotor at no load is very less.
+ Wy -3ER, = Constant losses
2. Blocked Rotor Test:
This test is done by blocking the rotor and applying small voltage around (5 to 10%) across
stator winding while ensuring rated current drawn by the motor.
Note: As voltage is low, core loss is neglected on stator and rotor.
Wax = stator copper loss + Rotor copper loss
fae = ERoa = Tagan
Equivalent circuit at Blocked rotor test
Vox / 13
Teg
Xan = \Zin Ree
On calculating
X, =X = Xoq /2
RY =a R SP]
Hence, all the parameters of the equivalent circuit can be calculated by conducting the these
Zea
tests.
STARTING OF THREE PHASE INDUCTION MOTORS
1. Direct-on line (across the line) starting:
+ The relation between starting torque and full load torque is:
1-4.
0, ES ‘The above equation valid for rotor resistance remains constant.
Tex [lay cg
Tow Ln)
Where,
(Effective rotor to stator turns ratio), «
1, _ (Effective rotor to stator turns ratio)I,
+ Per phase short ~ circuit current at stand still (or at starting) is,
z,
Where, Zse = (ri + r2) + j(X1 + X2)
Here shunt branch parameters of equivalent circuit are neglected,
+ Therefore, for direct switching.
2. Stator resistor (or reactor) starting:
Since per phase voltage is reduced to xV, the per phase starting current Iet is given by
xv,
Ze
1,
As before
Starting torque with reactor starting
Starting torque with direct switching
3. Auto transformer starting:
* Per phase starting current from the supply mains is Ist = x"Ise
Per phase starting current in motor winding , «
Per phase motor full load current ~y
v,
T..,With an auto transformer
“Ty awith direct switching —
4, Star — Delta method of Starting:
Starting torque with star delta starter
Starting torque with direct switching in delta [v7
Star delta starter also reduces the starting torque to one-third of that produce by direct
switching in delta.
‘+ With star-delta starter, a motor behaves as if it were started by an auto transformer starter
with x = = 0.577 i.e, with 57.7% tapping.
B
Ju pe
Starting torque with star delta starter, Ty.
Starting torque with direct switching in delta, Ty, 16,
SPEED CONTROL OF 3 - © INDUCTION MOTORS
Running speed of an induction motor is given as:
N Fas)
The method involved in speed control of induction motor is not simple and efficient.
Tt does not offer wide range of speed control.
1. Line Voltage Control:
As we known,
Tx sv?
+ If voltage is reduced by keeping constant torque, the slip had to be increased.
Hence, the speed decreases.
+ Itdraws more current and does not offer wide range of speed control
2. V/f control:
It is basically a frequency control
To keep V/f ratio constant, it requires variable voltage also otherwise stator and rotor core
gets saturated
120f
P
= Nf
= 5,0 £ (a5 reactance depends on frequency) v
Faq > constant
(as ¥ is constant)
3. Rotor resistance control:
* Speed control by this method is used when speed is to be reduced for a short period.
* Because increase in rotor resistance efficiency becomes less and poor speed
regulations.
+ Tt is possible in slip wound rotor only as rotor resistance cannot be changed in squirrel
cage winding
4. Speed control by Cascade Arrangement:
The two motors may be mechanically coupled together to drive a common load
Generally, the rotor output of first machine is connected to the stator of second
machine in such a way that the revolving field of both the machines are in the same
direction.
Under this condition, the resulting speed of field will be
w, (226
A+R,
| (When the machines are cumulatively coupled)
(120 F)
R-B)
N= (When the machines are differentially coupled)
Where, f = supply frequency
Pi and P2 = number of poles of machines I and II, respectively.
5. Slip power Recovery:
In this method, external voltage is injected into the rotor through the slip ring at slip
frequency.
As,
T, «svi
(i) Injecting the voltage to add the existing rotor voltage which will reduce slip to keep
torque constant
(ii) Injecting the voltage to oppose the e
keep torque constant.
1g rotor voltage which will increase slip to Synchronous Machines
A three-phase synchronous machine is a doubly excited ac machine because its field winding is
energized from a DC source and its armature winding is connected to an AC source. It rotates with
speed of revolving field i.e, synchronous speed.
Synchronous machines are divided in two parts based on their rotor construction
1, Cylindrical Rotor Machines
2. Salient pole Rotor Machines
‘+ The synchronous speed Ns and synchronous angular speed of a machine with p-pole pairs
running on a supply of frequency fs are:
Where Ns= 120f
P
synchronous speed
EMF EQUATION OF SYNCHRONOUS GENERATOR OR ALTERNATOR
The emf equation of Synchronous Generator or Alternator is given as
Let,
© = Flux per pole, in Wb
P = Number of poles
Non = Turns per phase
f
Z = Total number of conductors
Zp = Conductors per phase connected in series = Z/3 as number of phases
Consider a single conductor placed in a slot.
Ey =NIALON py, Volts,
This is the general emf equation for an induced emf per phase for full pitch, concentrated
type of winding.
Frequency of induced emf in Hz
1. Pitch Factor or Coil Span Factor (Ke)
It is defined as the ratio of resultant emf when the coil is short pitch to the result emf when
the coil is full pitched. It is always less than one.
2
Ex when coil is short pitched _
E, when coil is full pitched
K, -on( 4)
180°
+ Chording angle to eliminate n® harmonics (x) = #8¢ + Coil span to eliminate nt harmonics, (f) = 180[
2. Distribution Factor (Ke)
‘The distribution factor is defined as the ratio of the resultant emf when coils are distributed
to the resultant emf when coils are concentrated. It is always less than one.
2esin( 22)
E,_ when coils are distributed 2
E, when coils are concentrated
2m
Ky=
m = Slots per pole per phase
Slot angle = 180°/n
Slots per pole
When 8 is very small and m is large then the total phase spread is (mB). So, generalised
expression for the derivation of emf equation of Synchronous generator or Alternator can be
written as
Epn = V2nKe Ka f ®Npn Volts
For full pitch coil, Ke = 1
For concentrated winding Ka= 1
Note:
+ Coil span : It is the distance between two sides of the coil. It is expressed in terms of
degrees, pole pitch, no. of slots/pole etc.
+ Pole pitch: It is the distance between two identical points on two adjacent poles.
Pole pitch is always 180° electrical = slots / pole.
+ Slot pitch or slot angle: (B) Slot angle is the angle for each slot.
1
(a
‘+ Speed of space harmonics of order (6k + 1)is,
Where, Ns = Synchronous speed = 220f
+ The order of slot harmonics is
where s = No, of slots, P = No. of poles
‘+ Slot harmonics can be eliminated by skewing the armature slots and fractional slot winding.
The angle of skew = @; = (slot angle) = 2(harmonic pole pitches) = 1 slot pitch. ‘+ Distribution factor for slot harmonics,
i.e., same that of fundamental one.
(25 ..\ «
‘+ Pitch factor for slot harmonics, k, | Bat} k, =cos=
iP nT COST
Synchronous Generator:
For a synchronous generator with stator induced voltage Es, stator current Is and synchronous
impedance Zs, the terminal voltage V is
Equivalent circuit of synchronous generator
Vy =E,Z5- ER, + 3X)
Where, Rs is the stator resistance and Xs is the synchronous reactance.
oR
x,
(Vcosé+1,R,) +(Vsing +
+= lag pf.
— = leading pf.
Synchronous Motor:
For a synchronous motor with stator induced voltage Es, stator current Is and synchronous
impedance Zs, the terminal voltage V is
Xu x R, I,
ann,
EO E vi
1
Equivalent circi
of synchronous motor
VY =E4-8+ LR, + 3%) Where, Xs is synchronous reactance.
E- \(vcose-1J
-= lag pf.
+= leading pf.
where Re is the stator resistance and Xs is the synchronous reactance
Voltage Regulation:
+ % regulation = I-M 100
M
E
V = hiZs
= % regulatior
regulation « Zs
As Zs increases, voltages regulation increases.
1Z,
2
+ Condition for zero/minimum voltage regulation is, cos(® + g) = -
* Condition for max. Voltage regulation is, » = 8
Short
cuit ratio (SCR)
It is defined as the ratio of field current required to produce rated voltage on open circuit to
the field current required to produce rated armature current.
‘As SCC is linear,
(V, (rated) / V3)
scr -—is X, (saturated) Zoose
“Ifrated) ‘I,(rated)——‘X, (Saturated)
1
SCR =
X,(Pu)
SCR « 1 1
‘Armature reaction
Voltage regulation « Armature reaction
1
SCR « ——__
Voltage regulation
« Small value of SCR represents poor regulation.
_ Armature mmf
Reluctance
But reluctance « Air gap
Armature mmf
“0: Airgap % © Rirgap length
1
Armature reaction & dy. ——+—___
Airgap length
1
SCR x ——__>__
‘Armature reaction
x Airgap length
Air gap length « SCR
Machine size «SCR, Cost «SCR
Power = sing Px + «SCR
x x
Power x SCR
Large value of SCR represents more power output.
POWER FLOW IN CYLINDRICAL ROTOR SYNCHRONOUS MACHINE
Power flow in generator:
Sox = Vib *
Sour = Pour + JQoue
= hh cose a) - ME cose
m2 Zz,
EN, ve
= > sin(@-8)- sino
Qu Z, (0 - 8) Z
Pow will be maximum when 8 = 5
Prrim Xt -¥ cos
outinas)
Z,
# Neglecting armature resistance, Re
Then, Zs = Xs and 0 = 90°
By putting the value in power equation,
p= ME sing|
Reactive power:
ve
aut = 5° (E, Coss - Vy)
x,
Power Flow in motor:
Shut = Proput + JQinour = Vela ve
P, == cose
Lcos(0 +
=F cos( +8)
ve VE,
. stsing -+sin(0 +3}
Oy = FE sind - Fe sin(o~8)
# Neglecting armature resistance, Ra
Then, Z: = Xs and @ 10°
By putting the value in power equation,
VEL
= “Etsina|
xX
Reactive power
a... = Mv, ~£, cosa)
x,
Conclusion:
For generator:
0 and therefore operating at unity
Case 1: When E, cosé = \V, i.e. normally excited, then Qou
factor.
Case 2: When E, cosé > V, i.e. overexcited, then Qo: = +ve, that is supplying lagging VARS and
therefore operating at lagging p-f.
Case 3: When E, cosé < V, i.e. under excited, then Qo = -ve, that is supplying leading VARs and
therefore operating at leading p.f.
For motor:
Case 1: When E, cosé = \, i.e. normally excited motor than Qin = 0 and therefore operating at
unity power factor.
Case 2: When E, cosé > V i.e. overexcited motor than Qn = -ve i.e. taking leading VARs and
therefore operating at leading pf.
Case 3: When E, cosé < V i.e. under excited motor than Qi = +ve i.e. taking lagging VARs and
therefore operating at lagging p.f.
EFFECT OF CHANGE IN EXCITATION AT CONSTANT (KW) OUTPUT
Power can be expressed as:
VE; ina
P= Sesina P= VI, cos}
* Pel,
P cE, sind 2cos¢ A) Generator:
Under excited over-exciting
(lagging p.f) “— [—* (lagging load)
Normally
excited
(unity p.f.)
Effect of change in excitation on alternator
Conclusion:
1. For unity power factor load (Is!), generator is normally excited (E)
2. For lagging power factor load (Is?), generator is over-excited (E)
3. For leading power factor load (Is), generator is under-excited (E?)
B) Motor:
—+1, cos ¢-rconstant
rere
wl Lycontant
Under excited over excited
(lagging p.f) *— |—* (loading pf)
Normally
excited
: (anity pt
Effect of change in excitation on synchronous motor
Conclusion:
1. For unity power factor load (Is'), motor is normally excited (E+)
2. For leading power factor load (122), motor is over-excited (E2)
3. For lagging power factor load (1.*), motor is under-excited (Er) Vcurve:
It is the plot of armature current (Is) versus field current (Ir) at constant load.
Generator (lead p.f) Unity p.f
Motor(lag p-f)
Generator (lag p-f)
Motor (lead p.f)
V-curves
Note: Armature current is minimum at normally excited machine which is at unity p.f load.
Inverted ‘V’ curve:
‘These are the curves plotted between power factor (p.f) and field current (Ir)
Unity p.f
Af _—__, Generator (Ia
pf T"(ead) Motor (lead)
Inverted V-curves
Note: p-f is unity at normally excited machine at unity p-f. load.
Synchronizi
9 power coefficient or stability factor
Synchronizing power coefficient or stability factor Psy is given as:
dp_dfev.) ev
= 2-8) cing] - cose
ds aX),
P.y is a measure of stability
Stability
