Institute of Engineering Studies (IES,Bangalore) Electrical Machines Formula Sheet
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DC MACHINES : -
Lap Winding Wave Winding
(1) Coil Span : =
=
(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 A = P A = 2
(6) Conductor Current =
=
(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) =
*
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AT(Inter pole) = A +
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 =
* 100 % =
* 100 %
Voltage regulation , VR =
* 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:
I f = V / R f
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The armature induced voltage Ea and torque T with magnetic flux at angular
speed are:
Ea = k f = km
T = k fIa = 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 fIa2 = kmIa
2
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
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= +
- (
)
Where = No of series field turns
= No of shunt field turns
Differentially compounded DC generator : - ( long shunt)
(a) = +
(b) = - ( + )
(c) =
= shunt field current
(d) 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: I f = V / R f
The armature induced voltage Ea and torque T with magnetic flux at angular
speed are:
Ea = k f = km
T = k fIa = 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 , =
-
T
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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 fIa2 = kmIa
2
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 = , Where = Brush voltage drop
The total machine losses , = + +
Efficiency
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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
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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 = (VR
2 + 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 = 2fCRS
RS = XCtan = tan / 2fC
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°
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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
Transformer emf equations :-
= 4.44 (1)
= 4.44 (2)
Instantaneous value
of emf in primary
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Emf per turn in =
= 4.44
Emf per turn in
= 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
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During operation of transformer :-
⟹
= constant
Equivalent ckt of t/f under N.L condition :-
No load current =
No load power = Iron losses.
⟹
Transferring from to :-
=
∴
From to :-
No load /shunt branch.
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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 :-
1. Cu losses in t/f:
Total Cu loss =
=
=
Rated current on
Similarly current on =
Cu losses or
. Hence there are called as variable losses.
1. Copper losses
2. Iron losses
3. Stray load losses
4. Dielectric losses
major losses
minor losses
t/f windings
t/f core cu parts
Iron parts
insulating materials.
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P.U Full load Cu loss =
=
If VA rating of t/f is taken as base then P.U Cu loss as remaining terms are constant.
P.U Cu loss at x of FL = 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 :-
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.
Constant Supply freq
thickness of laminations.
(it is a function of )
Area under one hysteresis loop. . f . v
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During operation of transformer :-
Case (i) :-
= constant, = const.
Case (ii) :-
constant, 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 :-
= Loss in t/f under S.C condition
Const.
∴
When = const.
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= 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 :-
=
=
O.C test S.C test
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Transformer efficiency =
Efficiency = =
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 =
*100 %=
*100 %
KVA corresponding to = F.L KVA
Voltage drop in t/f at a Specific load p.f =
% Voltage regulation =
100
=
P.U resistance P.U reactance
% Regulation = × 100
Condition for max. regulation :-
% regulation = (% R) cos
= 0
Tan
lagging
At maximum regulation
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=
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
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Scott Connection:
= 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
86.6%
0.289
2
:
1
M
N
0.866 0.577
A
B
C
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Let
Capacity of Scott Connection :-
↙ ↓
Vol. rating of 1 – t/f Current rating of 1 – 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
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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 =
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Pitch factor or coil span factor or chording factor :- ( )
=
=
Pitch factor for harmonic i.e,
chording angle to eliminate harmonics (α)=
coil spam to eliminate harmonics ,() = 180
Distribution factor | spread factor | belt factor | breadth factor(kd) :-
=
=
Kd =
γ
γ
The distribution factor for uniformly distributed winding is
For harmonic, =
γ
γ
To eliminate harmonics ,phase spread (mγ) =
Generally, KVA rating, power output kd and (induce emf) .
∴
=
= 1.15
= cos /2
=
=
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= 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,
Is
γ
γ
i.e., same that of fundamental
Pith factor for slot harmonics,
=
The synchronous speed Ns and synchronous angular speed s of a machine with p pole pairs running on a supply of frequency fs are:
s = 2fs / p
Slip S =
Where
= synchronous speed
The magnitude of voltage induced in a given stator phase is =
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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/ 2fs = 120PM / 2Ns
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 =
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=
100
∴ 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
∴ machine size SCR.
Cost SCR
Power =
P
SCR
Air gap length SCR
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∴ 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 SCR Air gap length
When the stator mmf is aligned with the d – axis of field poles then flux perpole is set up
and the effective reactance offered by the alternator is .
= Direct axis reactance
When the stator mmf is aligned with the q – axis of field poles then flux per pole is set up
and the effective reactance offered by the alternator is .
=
= 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 =
δ
δ
Power SCR
∴ Stability SCR
∴ Stability Air gap length
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Condition for max. power:-
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 =
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If = 0; = δ = 90 ; then max power is given by
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
:
:
If = 0 ;
Condition for max power :-
= 0 0 +
Sin ( + δ) = 0 = sin 180
=
cos θ
=
Q =
8 0
P =
Q =
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Expression for mechanical power developed :-
Mechanical power developed = = active component +
Condition for max. mechanical power developed :-
δ
δ = 0
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 =
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 :-
Tg =
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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
The slip of induction machine is (S) =
Mechanical power
developed,
Rotor i/p power
= airgap power
Power i/p to stator
from mains
Power of
rotor shaft
Windage
loss
Friction loss
at bearings
and sliprings
of (if any)
Rotor core loss
(negligible for
small slips)
Rotor
R
loss
Stator
core
loss
Stator
R
loss
(input)
Stator
copper loss
3
Mechanical power
developed in armature
= 3 . cos
( ± )
+ ve lead
–ve for lag
=
2 π Ns
60.
output
Friction
and Iron
losses
=
2π Ns
60
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=
Where 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
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Rotor leakage reactance = 2 (Rotor frequency)
(Rotor leakage Inductance)
∴ Rotor leakage reactance at stand still = 2
=
Rotor leakage reactance at any slips = 2
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
=
= s
=
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∴
`
=
=
is the power transferred from stator to rotor across the air gap. There fore is called air
gap power
=
= (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
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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 is given by
stator power input – stator loss – stator core loss.
Ratio of Rotor input power, rotor copper losses and gross mechanical output is
:
:
∴ 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
1 : S : (1 – S)
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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
. In order to get a simple and approximate
expression, stator resistance , or the stator equivalent resistance , is neglected.
∴
Since r1 or Re is neglected
The slip at which maximum torque occurs is
∴
∴
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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
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∴
Case (iii) : If frequency is constant and voltage is variable then
=
Short circuit current with normal voltage applied to stator is
I = short circuit current with normal voltage
= short circuit current with voltage .
Power factor on short circuit is found from
As 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
I =
Cos =
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∴ Rotor Cu loss =
∴
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
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∴
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.
=
=
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