Institute of Engineering Studies (IES,Bangalore)
Electrical Machines Formula Sheet
DC MACHINES : Lap Winding
(1) Coil Span :
Wave Winding
=
(2) Back Pitch
(3) Commutator Pitch
=
=
=1
for Progressive winding
= -1
for Retrogressive winding
=
=
for Progressive winding
=for Restrogressive winding
( Must be integer)
=
-
(4) Front Pitch
= +2
for Progressive winding
= -2
for Retrogressive winding
(5) Parallel Paths
(6) Conductor Current
A=P
=
A=2
=
(7) No of brushes
No of brushes = A = P
No of brushes = 2


S = No of commutator segments
P = No of poles

U = No of coil sides / No of poles =



C = No of coils on the rotor
A = No of armature parallel paths
= Armature current
 Distribution factor (
 Pitch factor (
)=
=
)=
=
*100%

 Armature mmf/Pole (Peak) , A
 AT (Compensating Winding) =
1
=
*
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 AT(Inter pole) = A
Electrical Machines Formula Sheet
+
Where = Flux density in inter pole airgap
= length of inter pole airgap ,
 No of turns in each interpole ,
=
 The no of compensating conductor per pole,
/pole =
 The Mechanical power that is converted is given by
Where T = Induced torque
(
)
=
= Angular speed of the machines rotor
 The resulting electric power produced
=
 The power balance equation of the DC Machine is
 The induced emf in the armature is
 Torque developed in Dc machine ,
Where
=
=
=
= Flux\pole , Z = No of armature conductors , P = No of poles , N = Speed in rpm ,
A = No of armature parallel paths,
rmature current
 The terminal voltage of the DC generator is given by
=  The terminal voltage of the DC motor is given by
= +
 Speed regulation of dc machine is given by ,SR =
 Voltage regulation , VR =
* 100 % =
* 100 %
* 100 %
Shunt Generator:
 For a shunt generator with armature induced voltage Ea, armature current Ia and
armature resistance Ra, the terminal voltage V is:
V = Ea - IaRa
 The field current I f for a field resistance R f is:
If = V / Rf
2
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Electrical Machines Formula Sheet
 The armature induced voltage Ea and torque T with magnetic flux  at angular
speed  are:
Ea = k f = km
T = k fIa = kmIa
where k f 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 a shunt generator is:
Pe = T = EaIa = km Ia
Series Generator:
 For a series generator with armature induced voltage Ea, armature current Ia,
armature resistance Ra and field resistance R f, the terminal voltage V is:
V = Ea - ( IaRa + IaR f )= Ea - Ia(Ra + R f)
The field current is equal to the armature current.
 The armature induced voltage Ea and torque T with magnetic flux  at angular
speed  are:
Ea = k f Ia = km Ia
T = k fIa2 = kmIa2
where k f and km are design coefficients 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 Ea
 The airgap power Pe for a series generator is:
Pe = T = EaIa = km Ia2
 Cumulatively compounded DC generator : - ( long shunt)
(a) = +
(b) = - ( +
)
(c)
=
= shunt field current
(d) The equivalent effective shunt field current for this machine is given by
3
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=
+
-(
Electrical Machines Formula Sheet
)
Where
= No of series field turns
= No of shunt field turns
 Differentially compounded DC generator : - ( long shunt)
(a)
= +
(b)
= - ( +
)
(c)
(d)
=
= shunt field current
The equivalent effective shunt field current for this machine is given by
=
Where
-
-(
)
= No of series field turns
= No of shunt field turns
Shunt Motor:
 For a shunt motor with armature induced voltage Ea, armature current Ia and
armature resistance Ra, the terminal voltage V is:
V = Ea + IaRa
The field current I f for a field resistance R f is:
If = V / Rf
 The armature induced voltage Ea and torque T with magnetic flux  at angular
speed  are:
Ea = k f = km
T = k fIa = kmIa
where k f 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 a shunt motor is:
Pe = T = EaIa = km Ia
 The speed of the shunt motor , = 
4
T
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Electrical Machines Formula Sheet
Where K =
Series Motor :
 For a series motor with armature induced voltage Ea, armature current Ia,
armature resistance Ra and field resistance R f, the terminal voltage V is:
V = Ea + IaRa + IaR f = Ea + Ia(Ra + R f)
The field current is equal to the armature current.
 The armature induced voltage Ea and torque T with magnetic flux  at angular
speed  are:
Ea = k f Ia = km Ia
T = k fIa2 = kmIa2
where k f and km are design coefficients of the machine.
Note that for a series motor:
- 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 Ea
 The airgap power Pe for a series motor is:
Pe = T = EaIa = km Ia2
Losses:

constant losses (P k) = Pw f + Pi o
Where,
= No of load core loss

= Windage & friction loss
 Variable losses ( ) = +
+
where
= Copper losses =
= Stray load loss = α
= Brush Contact drop =
 The total machine losses ,
, Where
=
+
= Brush voltage drop
+
Efficiency
5
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Electrical Machines Formula Sheet
 The per-unit efficiency  of an electrical machine with input power Pin, output
power Pout and power loss Ploss is:
 = Pout / Pin = Pout / (Pout + Ploss) = (Pin - Ploss) / Pin
 Rearranging the efficiency equations:
Pin = Pout + Ploss = Pout /  = Ploss / (1 - )
Pout = Pin - Ploss = Pin = Ploss / (1 - )
Ploss = Pin - Pout = (1 - )Pin = (1 - )Pout / 
Temperature Rise:
 The resistance of copper and aluminium windings increases with temperature,
and the relationship is quite linear over the normal range of operating
temperatures. For a linear relationship, if the winding resistance is R1 at
temperature 1 and R2 at temperature 2, then:
R1 / (1 - 0) = R2 / (2 - 0) = (R2 - R1) / (2 - 1)
where 0 is the extrapolated temperature for zero resistance.
 The ratio of resistances R2 and R1 is:
R2 / R1 = (2 - 0) / (1 - 0)
 The average temperature rise  of a winding under load may be estimated from
measured values of the cold winding resistance R1 at temperature 1 (usually
ambient temperature) and the hot winding resistance R2 at temperature 2, using:
 = 2 - 1 = (1 - 0) (R2 - R1) / R1
 Rearranging for per-unit change in resistance Rpu relative to R1:
Rpu = (R2 - R1) / R1 = (2 - 1) / (1 - 0) =  / (1 - 0)
.Copper Windings:
 The value of 0 for copper is - 234.5 °C, so that:
 = 2 - 1 = (1 + 234.5) (R2 - R1) / R1
6
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Electrical Machines Formula Sheet
 If 1 is 20 °C and  is 1 degC:
Rpu = (R2 - R1) / R1 =  / (1 - 0) = 1 / 254.5 = 0.00393
 The temperature coefficient of resistance of copper at 20 °C is 0.00393 per
degC.
Aluminium Windings:
 The value of 0 for aluminium is - 228 °C, so that:
 = 2 - 1 = (1 + 228) (R2 - R1) / R1
 If 1 is 20 °C and  is 1 degC:
Rpu = (R2 - R1) / R1 =  / (1 - 0) = 1 / 248 = 0.00403
 The temperature coefficient of resistance of aluminium at 20 °C is 0.00403 per
degC.
Dielectric Dissipation Factor:
 If an alternating voltage V of frequency f is applied across an insulation system
comprising capacitance C and equivalent series loss resistance RS, then the
voltage VR across RS and the voltage VC across C due to the resulting
current I are:
VR = IRS
VC = IXC
V = (VR2 + VC2)½
 The dielectric dissipation factor of the insulation system is the tangent of the
dielectric loss angle  between VC and V:
tan = VR / VC = RS / XC = 2fCRS
RS = XCtan = tan / 2fC
 The dielectric power loss P is related to the capacitive reactive power QC by:
P = I2RS = I2XCtan = QCtan
 The power factor of the insulation system is the cosine of the phase
angle  between VR and V:
cos = VR / V
so that  and  are related by:
 +  = 90°
7
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Electrical Machines Formula Sheet
 tan and cos are related by:
tan = 1 / tan = cos / sin = cos / (1 - cos2)½
so that when cos is close to zero, tan  cos
TRANSFORMERS:
 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 (flux) majorly
flows in magnetic material.
 Specific weight of t/f =
 Stacking/iron factor :- ( ) =

is always less than 1
 Gross c.s Area =
= length × breadth
 Net c.s Area =
=
 Utilization factor of transformer core =
U.F of cruciform core = 0.8 to
0.85
 Flux =
=
 According to faradays second law
Instantaneous value
of emf in primary
 Transformer emf equations := 4.44
= 4.44
8
 (1)
 (2)
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 Emf per turn in
 Emf per turn in
=
Electrical Machines Formula Sheet
= 4.44
= 4.44
⟹ Emf per turn on both sides of the transformer is same
⟹
Transformation ratio = K =
Turns ratio =
 For an ideal two-winding transformer with primary voltage V1 applied
across N1 primary turns and secondary voltage V2 appearing across N2 secondary
turns:
V1 / V2 = N1 / N2
 The primary current I1 and secondary current I2 are related by:
I1 / I2 = N2 / N1 = V2 / V1
 For an ideal step-down auto-transformer with primary voltage V1 applied
across (N1 + N2) primary turns and secondary voltage V2 appearing
across N2 secondary turns:
V1 / V2 = (N1 + N2) / N2
 The primary (input) current I1 and secondary (output) current I2 are related by:
I1 / I2 = N2 / (N1 + N2) = V2 / V1.
 For a single-phase transformer with rated primary voltage V1, rated primary
current I1, rated secondary voltage V2 and rated secondary current I2, the voltampere
rating S is:
S = V1I1 = V2I2
 For a balanced m-phase transformer with rated primary phase voltage V1, rated
primary current I1, rated secondary phase voltage V2 and rated secondary current I2,
the voltampere rating S is:
S = mV1I1 = mV2I2
 The primary circuit impedance Z1 referred to the secondary circuit for an ideal
transformer with N1 primary turns and N2 secondary turns is:
Z12 = Z1(N2 / N1)2
9
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Electrical Machines Formula Sheet
 During operation of transformer :-
⟹
= constant
Equivalent ckt of t/f under N.L condition :-
No load /shunt branch.
 No load current =
 No load power =
Iron losses.
⟹
Transferring from
to
:-
=
∴
From
10
to
:-
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Electrical Machines Formula Sheet
 Total resistance ref to primary =
/
 Total resistance ref to secondary =
 Total Cu loss =
Or
Per unit resistance drops : P.U primary resistance drop =
 P.U secondary resistance drop =
 Total P.U resistance drop ref to
 Total P.U resistance drop ref to
=
=
 The P.U resistance drops on both sides of the t/f is same
Losses present in transformer :t/f windings
major losses
t/f core
cu parts
Iron parts
minor losses
insulating materials.
1. Copper losses
2. Iron losses
3. Stray load losses
4. Dielectric losses
1. Cu losses in t/f:
Total Cu loss =
=
=
 Rated current on
Similarly current on
 Cu losses
11
or
=
. Hence there are called as variable losses.
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Electrical Machines Formula Sheet
 P.U Full load Cu loss =
=
 If VA rating of t/f is taken as base then P.U Cu loss

P.U Cu loss at x of FL =
as remaining terms are constant.
FL Cu loss

=
∴ P.U Resistance drop = P.U FL cu loss
% FL Cu loss = % R = % Resistance drop.
Iron (or) Core losses in t/f :1. Hysteresis loss :
Steinmetz formula :. f . v
Area under one hysteresis loop.
Where
= stienmetz coefficient
= max. flux density in transformer core.
f = frequency of magnetic reversal = supply freq.
v = volume of core material
x = Hysteresis coeff (or) stienmetz exponent
= 1.6 (Si or CRGo steel)
2. Eddycurrent loss:
Eddy current loss ,(
As area decreases in laminated core resistance increases as a result conductivity decreases.
thickness of laminations.
Supply freq
Constant
(it is a function of )
12
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During operation of transformer :-
Case (i) :-
= constant,
Electrical Machines Formula Sheet
= const.
Const.
∴
When
Case (ii) :-
constant,
= const.
const.
P.U iron loss : P.U iron loss =
 As VA rating is choosen as base then the P.U iron loss are also constant at all load conditions.
To find out constant losses :
= Losses in t/f under no load condition
= Iron losses + Dielectric loss + no load primary loss (
)
 Constant losses =
Where ,
= LV winding resistance.
To find out variable losses :
13
= Loss in t/f under S.C condition
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Electrical Machines Formula Sheet
= F.L Cu loss + stray load losses (Cu and Iron) + Iron losses in both wdgs
 Variable losses =
Iron losses corresponding to
O.C test :rated 
S.C test :
∴ Variable losses =
 Under the assumption that small amount of iron losses corresponds to
and stray load
losses are neglected the wattmeter reading in S.C test can be approximately taken as F.L
Cu losses in the transformer.

F.L Cu loss
Efficiency :
=
=
S.C test
14
O.C test
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 Transformer efficiency =
 Efficiency =
Electrical Machines Formula Sheet
=
 Total losses in transformer =
output
 Condition for maximum effieciency is, Cu losses = Iron losses
 Total losses at
=2
 %load at which maximum efficiency occurs % x =
 KVA corresponding to
*100 %=
*100 %
= F.L KVA

 Voltage drop in t/f at a Specific load p.f =
 % Voltage regulation =
100
=

P.U resistance
% Regulation =

P.U reactance
× 100
Condition for max. regulation :% regulation = (% R) cos
=0
Tan
lagging
At maximum regulation
15
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Electrical Machines Formula Sheet
=
Value of maximum regulation :% Regulation = (% R) cos
At max. regulation cos
Sin
max. % regulation = (% R)
=
=
max. % regn = % Z
= % of rated voltage required to produce rated short ckt current
.
Condition for zero regulation : If the voltage regulation in the t/f is zero, the t/f voltages are maintained at their nominal
values even under load condition
% Regn = (% R) cos
For zero regulation  occurs at leading p.f’s
(% R) cos
sin
=0
Tan
leading.
 At zero regulation condition :
 Regulation at x of FL = x [% R cos
X sin
]
= x × F.L regn
Regulation at U.P.F:Regulation at UPF = % R
= % F.L Cu loss
16
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Scott Connection:
Electrical Machines Formula Sheet
86.6%
A
0.577
0.866
N
2
:
1
0.289
M
B
C
= 0.866
= 0.577
= 0.866
0.577
= 0.289
= 0.577
0.289
=2:1
 If a neutral pt is located on 3 side, such that, voltage between any terminal to that neutral
point is 0.577
then such neutral point divides the primary of teaser transformer in the ratio
of 2 : 1
 Location of neutral point from top = 0.866
 Location of neutral point from bottom = 0.866
Operation of Scott Connection with 2 balanced load at UPF :Teaser t/f :-
Let
Main t/f
17
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Electrical Machines Formula Sheet
Let
 Capacity of Scott Connection :-
↙
 Vol. rating of 1 –
↓
Current rating of 1 –
t/f
t/f
 Utilization factor =
=
= 0.866
 Utilization factor of Scott connection with 2 identical 1 –
t/f’s is 86.6%
AUTO TRANSFORMER:
 Primary applied voltage,
= Secondary voltage referred to primary + primary leakage impedance
drop + secondary leakage impedance drop ref. to primary.
 K of auto transformer =
I/P KVA =
=1–
∴
=1–K
(KVA) induction = (1 – K) i/p KVA
(KVA) conduction = I/p KVA –
I/p KVA
18
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Electrical Machines Formula Sheet
 Wt. of conductor in section AB of auto t/f
 Wt of conductor in section BC of auto t/f
∴ Total wt. of conductor in auto t/f is
 Total wt. of conductor in 2 wdg transformer

=1–
=1–K
Wt. of conductor in auto t/f = (1 – K) (wt. of conductor in 2 wdg t/f)
 Thus saving of conductor material if auto – t/f is used} = K × {conductor wt in 2 wdg transformer.



.
SYNCHRONOUS MACHINES:
 Principle of operation :Whenever a conductor cuts the magnetic flux, an emf is induced in that conductor”
 Faraday’s law of electromagnetic induction.
 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° e = slots / pole.

 Slot pitch or slot angle :- (T)Slot angle is the angle for each slot.
 For a machine with ‘P’ poles and ‘s’ no. of slots, the slot angle = γ =
γ=
19
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 Pitch factor or coil span factor or chording factor :- (
Electrical Machines Formula Sheet
)
=
=
= cos /2
 Pitch factor for
harmonic i.e,
 chording angle to eliminate
 coil spam to eliminate
=
harmonics (α)=
harmonics ,() = 180
 Distribution factor | spread factor | belt factor | breadth factor(kd) :=
=
γ
Kd =
γ
 The distribution factor for uniformly distributed winding is
=
γ
For
harmonic,
 To eliminate
=
γ
harmonics ,phase spread (mγ) =
 Generally, KVA rating, power output
∴
20
kd and
(induce emf)
=
.
= 1.15
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Electrical Machines Formula Sheet
= 1.06
= 1.5
1.414
 Speed of space harmonics of order (6k ± 1) is
where
= synchronous speed =
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 =
= γ (slot angle)
= 2 harmonic pole pitches
= 1 slot pitch.
 Distribution factor for slot harmonics,
γ
Is
γ
i.e., same that of fundamental
 Pith factor for slot harmonics,
=
 The synchronous speed Ns and synchronous angular speed
pairs running on a supply of frequency fs are:
s
of a machine with p pole
s = 2fs / p
 Slip S =
Where
= synchronous speed
 The magnitude of voltage induced in a given stator phase is
21
=
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Electrical Machines Formula Sheet
Where K = constant
 The output power Pm for a load torque Tm is:
Pm = sTm
 The rated load torque TM for a rated output power PM is:
TM = PM / s = PM p/ 2fs = 120PM / 2Ns
Synchronous Generator:
 For a synchronous generator with stator induced voltage Es, stator current Is and
synchronous impedance Zs, the terminal voltage V is:
V = E - IsZs = Es - Is(Rs + jXs)
where Rs is the stator resistance and Xs is the synchronous reactance
E=
+  lag p.f
 leading p.f.
Synchronous Motor:
 For a synchronous motor with stator induced voltage Es, stator current Is and synchronous
impedance Zs, the terminal voltage V is:
V = Es + IsZs = Es + Is(Rs + jXs)
where Rs is the stator resistance and Xs is the synchronous reactance
Voltage regulation :
 % regulation =
100
E–V=
∴ % regulation =
22
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=
100
Electrical Machines Formula Sheet
∴ regulation
∴ As increases, voltages regulation increases.
 Condition for zero | min. voltage regulation is, Cos (θ + ) =
 Condition for max. Voltage regulation is,
=θ
 Short circuit ratio (SCR) =
SCR
Voltage regulation
Armature reaction
∴ SCR
∴ Small value of SCR represent poor regulation.
But reluctance
Air gap
∴
Armature reaction
∴ SCR
Airgap length
Air gap length
SCR
∴ machine size SCR.
Cost SCR
Power =
P
23
SCR
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Power
Electrical Machines Formula Sheet
SCR
∴ Large value of SCR represent more power output.
 Synchronizing power coefficient or stability factor
is given as
=
=
is a measure of stability
∴ stability
But
SCR
∴ Stability
Stability
SCR
∴ Stability
SCR
Air gap length
Air gap length
 When the stator mmf is aligned with the d – axis of field poles then flux
and the effective reactance offered by the alternator is .
perpole is set up
= Direct axis reactance
 When the stator mmf is aligned with the q – axis of field poles then flux
and the effective reactance offered by the alternator is .
=
per pole is set up
= Quadrature axis reactance
 Cylindrical rotor Synchronous machine ,
The per phase power delivered to the infinite bus is given by P =
sin δ
 Salient pole synchronous machine ,
The per phase power delivered to the infinite bus is given by
P=
24
δ
δ
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Condition for max. power:-
Electrical Machines Formula Sheet
 For cylindrical rotor machine :At constant
and
, the condition for max. power is obtained by putting
∴
=0
=0
Cos δ = 0
δ = 90
Hence maximum power occurs at δ = 90
 For salient – pole synchronous machine :
=0
=0
Cos δ =
The value of load angle is seed to be less than 90°.
∴ max. power occurs at δ < 90
 Synchronizing power =
. ∆ δ.
=
.
 Synchronizing torque =
.
Power flow in Alternator : Complex power = S = P + jQ = V
Where Active power flow (P) =
Reactive power flow (Q) =
;
;
 Condition for max. power output :P=
= 0 for max power condition
ie
–δ=0
θ=
25
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If
= 0; = δ = 90 ; then max power is given by
Electrical Machines Formula Sheet
cos θ
=
SYNCHRONOUS MOTORS:
 Speed regulation =
=
= 0%
 Slip S =
= 0%
 The speed can be controlled by varying the frequency

ratio control is preferred for rated torque operation
Power flow in synchronous motor is given by
complex power i/p s = p + jQ = V
where P = Resl power flow , Q = Reactive power flow
P=
:
Q=
:
 If
=0;
=
Q=
 Condition for max power :-
=00+
Sin ( + δ) = 0 = sin 180
8
26
0
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Electrical Machines Formula Sheet
Expression for mechanical power developed : Mechanical power developed =
= active component +


Condition for max. mechanical power developed :δ =0
δ
Sin ( – δ) = 0 = sin 0
δ=
This is the expression for the mechanical power developed interms of load angle and the
internal machine angle , for constant voltage
and constant E i.e., excitation
 Gross Torque =
=
synchronous speed in r.p.m
∴ Tg =
Tg =
 Condition for excitation when motor develops
For max power developed is
:-
=0
 Condition for excitation when motor develops
is
The corresponding value of max. power is
=
Power flow in synchronous motors :-
27
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Electrical Machines Formula Sheet
=
2π Ns
60
Stator
copper loss
3
(input)
Mechanical power
developed in armature
=3
.
cos
( ± )
=
+ ve lead
–ve for lag
2 π Ns
Friction
and Iron
losses
60.
output
 For leading p.f
tan δ =
 The mechanical power developed per phase is given by,
s
δ
δ
INDUCTION MACHINES:
 The power flow diagram of 3 –
induction motor is
Power i/p to stator
Rotor i/p power
from mains
= airgap power
Mechanical power
developed,
Power of
rotor shaft
Stator
R
loss
Stator
core
loss
Rotor Rotor core loss Friction loss Windage
R
(negligible for at bearings
loss
loss
small slips)
and sliprings
of (if any)
The slip of induction machine is (S) =
28
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=
Where


Electrical Machines Formula Sheet
is synchronous speed in rpm
is synchronous speed in rps
=S
∴ Rotor frequency,
For an induction machine with rotor resistance Rr and locked rotor leakage reactance Xr, the
rotor impedance Zr at slip s is:Zr = Rr + jsXr
The stator circuit equivalent impedance Zrf for a rotor / stator frequency ratio s is:
Zrf = Rrs / s + jXrs
For an induction motor with synchronous angular speed s running at angular speed m and
slip s, the airgap transfer power Pt, rotor copper loss Pr and gross output power Pm for a
gross output torque Tm are related by:
Pt = sTm = Pr / s = Pm / (1 - s)
Pr = sPt = sPm / (1 - s)
Pm = mTm = (1 - s)Pt
The power ratios are:
Pt : Pr : Pm = 1 : s : (1 - s)
The gross motor efficiency m (neglecting stator and mechanical losses) is:
m = Pm / Pt = 1 - s
Rotor emf, Current Power :At stand still, the relative speed between rotating magnetic field and rotor conductors is
synchronous speed ; under this condition let the per phase generated emf in rotor circuit
be .
∴
= 4.44
= 4.44
= Rotor winding factor
 But during running conditions the frequency of the rotor becomes, running with speed
S
∴
∴ Emf under running conditions is
E=
=S
29
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Electrical Machines Formula Sheet
 Rotor leakage reactance = 2 (Rotor frequency)
(Rotor leakage Inductance)
∴ Rotor leakage reactance at stand still = 2
=
 Rotor leakage reactance at any slips = 2
=s
 Rotor leakage impedance at stand still
=
 At any slip s, rotor
=
 Per phase rotor current at stand still
=
 Per phase rotor current at any slip s is given by
 The rotor current
lags the rotor voltage
by rotor power factor angle
given by
 Per phase power input to rotor is
=
30
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∴
Electrical Machines Formula Sheet
`
=
=

is the power transferred from stator to rotor across the air gap. There fore
gap power
is called air
=
= (Rotor ohmic loss) + Internal mechanical power developed in rotor (
=S
∴
)
=
Rotor ohmic loss =
=S
 Internal (or gross) torque developed per phase is given by
 Electromagnetic torque
can also be expressed as
∴
 Power available at the shaft can be obtained from as follows.
Output or shaft power,
Mechanical losses
 Mechanical losses implies frication and windage losses
31
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Electrical Machines Formula Sheet
Rotor ohmic loss – Friction and windage losses
= Net mechanical power output or net power output
Output or shaft torque
 If the stator input is known. Then air gap power
stator power input – stator
is given by
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 × Rotor input
Gross mechanical output =(1 – S) × Rotor input.
Rotor copper losses = (Gross Mechanical output) ×
Efficiency of the rotor is approximately
Equal to
=
=1–S
=1
=
Total torque is
m is the number of stator phases.
Torque equation can be written as
rotor input per phase.
Thus the slip
at which maximum torque occurs is given by
Substituting the value of maximum slip in the torque equation, gives maximum torque
32
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Electrical Machines Formula Sheet
If stator parameters are neglected then applying maximum transfer theorem to
then
=
Slip corresponding to maximum torque is
(Breakdown slip)

is the stalling speed at the maximum torque
Starting torque:At starting, slip S = 1.00, starting torque is given by
Test =
Motor torque in terms of
:
 The torque expression of an induction motor can also be expressed in terms of maximum
torque
and dimension less ratio
expression, stator resistance
. In order to get a simple and approximate
, or the stator equivalent resistance
, is neglected.
∴
 Since r1 or Re is neglected
 The slip at which maximum torque occurs is
∴
∴

33
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Electrical Machines Formula Sheet
Power slip characteristics : The total internal mechanical power developed is
Maximum power transfer theorem is invoked again to obtain maximum value of internal
mechanical power developed. Since
per phase is the power delivered to
, internal
mechanical power developed is maximum, when
In order to get maximum power
,substitute
, in place of
in power equation
In order to get maximum power output from an induction generator, the rotor must be deiven
at a speed given by
Losses and efficiency :There are three cases in iron losses.
Case (i) : If the ratio of voltage to frequency is constant and flux is also constant then
Iron loss = Hysteresis loss + eddy current loss
Given

∴
is constant. As
is constant
and
Case (ii) : If the ratio of voltage to frequency is not constant and flux is also not constant

≠ const
34
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Electrical Machines Formula Sheet
∴
Case (iii) : If frequency is constant and voltage is variable then
=
 Short circuit current with normal voltage applied to stator is
I=
I = short circuit current with normal voltage
= short circuit current with voltage
.
 Power factor on short circuit is found from

 As
Cos
=
is approximately equal to full load copper losses
The blocked rotor impedance is
∴ Blocked rotor reactance =
Efficiency of Induction machines :Generally efficiency =
∴ Efficiency of Induction motor =
∴ Efficiency of Induction generator =
Squirrel cage rotor:
Stator Cu loss = 3
35
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∴ Rotor Cu loss =
Electrical Machines Formula Sheet
∴
Wound rotor
Direct – on line (across the line) starting : The relation between starting torque and full load torque is
∴
=
The above equation valids of rotor resistance remains constant.
Where
 Per phase short – circuit current at stand still (or at starting) is,
Where
Here shunt branch parameters of equivalent circuit are neglected.
 Therefore, for direct switching,
∴
.
Stator resistor (or reactor) starting :Since per phase voltage is reduced to xv, the per phase starting current
is given by
As be fore
=
 In an induction motor, torque
36
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Electrical Machines Formula Sheet
∴
Auto transformer starting : Per phase starting current from the supply mains is
Star – delta method of starting :

=
∴ star delta starter also reduces the starting torque to one – third of that produced by direct switching
in delta.
 With star – delta starter, a motor behaves as if it were started by an auto transformer starter with x =
= 0.58 i.e with 58% tapping.

=
37
=
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