Contents
Manual for K-Notes ............................................................................. ................................................................................... ...... 2 Transformers ............................................................................................. 3 DC Machines ........................................................................................... ................................................ ........................................... 11 Synchronous Machines ........................................................................... ............................................. .............................. 16 Induction Machines ............................................................................. ................................................................................. .... 27 27 Single Phase Induction Motor ................................................................. ................................... .............................. 34
© 2014 Kreatryx. All Rights Reserved. 1
Manual for K-Notes Why K-Notes? Towards the end of preparation, a student has lost the time to revise all the chapters from his / her class notes / standard text books. This is the reason why K-Notes is specifically intended for Quick Revision and should not be considered as comprehensive study material. What are K-Notes? A 40 page or less notebook for each subject which contains all concepts covered in GATE Curriculum in a concise manner to aid a student in final stages of his/her preparation. It is highly useful for both the students as well as working professionals who are preparing for GATE as it comes handy while traveling long distances. When do I start using K-Notes? It is highly recommended to use K-Notes in the last 2 months before GATE Exam (November end onwards). How do I use K-Notes? Once you finish the entire K-Notes for a particular subject, you should practice the respective Subject Test / Mixed Question Bag containing questions from all the Chapters to make best use of it.
© 2014 Kreatryx. All Rights Reserved.
2
Transformers Impact of dimensions on various parameters of Transformer KVA Rating
(Core Dimension)4
Voltage Rating
(Core Dimension)2
Current Rating
(Core Dimension)2
No-Load Current Core Loss
Core Dimension
Core Volume
Induced EMF in a Transformer
E1 N1 E2 N2
d dt d
dt E1 (rms) 4.44fN1 m E2 (rms) 4.44fN2 m
Where E1 and E2 are emf in primary and secondary windings of Transformer respectively.
Φ is the flux in the transformer and Φm is maximum value of flux.
The polarity of emf is decided on basis of Lenz Law as currents in primary and secondary should be such that primary and secondary flux should oppose each other.
Also, primary current enters the positive terminal of primary winding as primary absorbs power and secondary current leaves the positive terminal of secondary winding as secondary delivers power and this way we can mark emf polarities.
Exact equivalent circuit
3
Exact equivalent circuit w.r.t. primary
2
2
2
N N N R 2 = R 2 1 ; X 2 = X 2 1 ; Z L = Z L 1 ; N2 N2 N2
Approximately Equivalent Circuit
R 01 = R1 R 2 X 01 = X1 X 2
Tests Conducted on a Transformer (i) Open Circuit Test o
Conducted on LV side keeping HV side open circuited
o
Equivalent Circuit
4
o
Power reading = P = V1 I 0 cos 0 =
o
Ammeter reading
o
cos 0 =
o
o
Calculate
V1
2
Rc
-------- (i)
I = I0
P V1 I0
2 Calculate sin 0 = 1 - cos 0
Q = V1I 0 sin 0 =
R c from
(i) &
V1
2
Xm
Xm from
------- (ii)
(ii)
(ii) Short Circuit Test o
Conducted on HV side keeping LV side short circuited
o
Equivalent Circuit
o
R 01 & X 01 are
equivalent winding resistance & equivalent leakage reactor referred to
HV side. 2
o
o
o
Wattmeter reading = P = Isc R01 from this equation, we can calculate R 01
Z 01 =
Vsc Isc
& X01 =
Z01
2
R012
We obtain R 01 , X 01 & full load copper losses from this test.
Losses on Transformers o
Copper Loss 2
2
PCu = I1 R1 I2 R 2 2
= I1 R01
Where
I1 =
2
I2 R02
primary current
I2 = secondary current
5
R 1 = primary winding resistance R 2 = secondary winding
resistance
2
R 01
o
2
N N = R 1 1 R2 ; R02 = R2 2 R1 N2 N1
Core Loss (i) Hysteresis Loss x
Pn = KnBmf
X = 1.6 Bm =
maximum value of flux density 1.6
Pn = KnBm
Bm
f
V f
V = applied voltage f = frequency 1.6
V Pn = Kh f = KhV1.6 f 0.6 f If V is constant & f is increased, Ph decreases (ii) Eddy Current Loss 2 2
Pe = Ke Bm f
Bm
V f 2
V Pe = K e f 2 = K eV 2 f Core loss = Pc = Pe
Pn
6
Efficiency
=
x KVA cos x KVA cos Pi x 2PCu,FL
X = % loading of Transformer
cos = power factor Pi = iron loss
PCu,FL = Full load copper losses KVA = Power rating of Transformer For maximum efficiency,
Pi
x=
PCu,FL
Voltage Regulation of Transformer Regulation down
Regulation up
VNL
VFL
VNL VNL
VFL
VFL
100
100
Equivalent circuit with respect to secondary K = Transformation Ratio
No-load voltage
N2 N1
V2
Full-load voltage V2 Approximate Voltage Regulation VR =
I 2 R 02 cos 2
X 02 sin 2
V2
7
cos 2 = power factor of load ZL + sign is used for lagging pf load - sign is used for leading pf load Condition for zero voltage regulation
-1
= tan 2
R X 02
02
The power factor is leading, Voltage Regulation can never be zero for lagging pf load. Condition for maximum voltage regulation
-1
= tan 2
X R 02
02
The power factor is leading, Voltage Regulation can never be negative for lagging pf loads Three – Phase Transformers In a 3-Phase transformers; the windings placed parallel to each other at as primary & secondary of single phase transformer. Rules to draw Phasor diagram 1) Always draw phasors from A to B, B to C & C to A for line voltages. 2) The end points should have same naming as the input or output terminals. 3) If we draw primary phasor from dotted to undotted terminal and if secondary voltage is also from dotted to undotted, then secondary voltage is in same phase else in opposite phase. Some examples
8
Phasor
o
If you observe carefully, we traverse from dotted to undotted terminal in primary while going from a2 to b2 , b2 to
c2 & c2 to a2 .
Same is the case when we traverse the secondary winding, so secondary voltage are inphase to primary. o
Then, we draw reference phasors from neutral to terminal and mark it with phase with same name as terminal it is pointed to.
Then we plot it on clock & we observe it is like 12 0 clock so name is Dd12 connection. Another example
Phasor
9
o
Here, we traversed primary from dotted to undotted terminal & in secondary from undotted to dotted so all secondary phasor are out of phase wrt primary. Parallel operation of Transformer Necessary Conditions 1) Voltage ratings of both transformers should be same. 2) Transformers should have same polarity. 3) Phase sequence of both transformers must be same in case of 3- phase transformers. 4) Phase displacement between secondary’s of both transformers must be If there are 2 transformers A & B supplying a load power S A = SL
ZB ZA
ZB
; S B = SL
0
.
SL .
ZB ZA
ZB
ZB =
impedance of transformer B (in ohms)
Z A =
impedance of transformer A (in ohms)
Auto Transformer o
Generally, auto transformer is created from 2- winding transformer.
o
If rating of auto – transformer is LV/HV or HV/LV LV = low voltage HV = high voltage Transformation Ratio =
K =
LV HV
1 (KVA rating of 2- winding Transformer) 1 - R
o
KVA rating of auto transfer =
o
In auto- transformer, power is transferred from primary to secondary by 2 methods induction & conduction.
o
KVAinduction = 1 - K Input KVA
o
KVAconduction = K Input KVA
o
% Full load losses = 1 - K %FL losses in2 winding Transformer
o
If copper & core losses are not given separately, then we consider losses as constant, same as that of two winding transformer while calculating efficiency 10
DC Machines Induced emf equation Ea =
NZ P 60 A
= flux per pole wb
N = speed of machine rpm P = number of poles A = number of paralled path Z = number of conductors
A = 2 for wave winding A = P for lap winding If speed is given in rad/sec
Ea =
Z P 2 A
PZ 2A Km =
PZ 2A
Developed Torque
T = KmIa Km =
PZ 2A
= machine constant
= flux per pole Ia = armature current
11
where ω = speed (rad/s)
= Km
= machine constant
Classification of DC Machine (i) Separately excited
(ii) Shunt excited
(iii) Series excited
(iv) Compound Excited
12
Terminologies R a : Armature Resistance
R se : Series Field winding Resistance R sh : Shunt Field winding Resistance o
The only difference between Generator & Motor will be that the direction of armature current is coming out of positive terminal of emf Ea. In case of motor, armature current flows into Ea. Performance Equations of DC Machines For shunt & separately excited machine Generator:
Ea = Vt
I aR a
Motor:
Ea = Vt
IaR a
For series & compound excited machine Generator:
Ea = Vt
Motor:
Ea = Vt
Ia
Ia R a
R
a
R se
R se
Power Flow Shaft Power
Armature Power
Electrical Power
Pa EaIa Rotational loss
Copper loss
o
This power flow diagram is for a dc generator.
o
If you traverse the diagram from right to left then it is a power flow diagram for a motor.
13
Losses
Rotational loss
Copper loss
I 2R I 2R I 2R V I a a se se f f BD a Ohmic loss Brush contact loss Hystersis N &
Friction & Windage loss
P
Friction
windage
Bearing
Brush
N
Eddy current
f w
2
N
2
N
Stray load 2
N
P
L L
Efficiency
=
VaIa ; for generator VaIa Ia2Ra VBD Ia Pk
Pk = sum of all constant loss
For maximum efficiency
For shunt & separately excited machine
Ia =
For series & compound excited machine
Ia =
14
Pk ra Pk ra
r
se
2
i
Characteristics of DC Generator External characteristics If no-load voltage is same for all types of generators:
There are two categories of compound generators/motors 1. Cumulative Compound
=> If series field flux aids the shunt fields flux.
2. Differentially Compound => If series field flux opposes the shunt field flux. If full – load voltage of all generators is kept same 1 series excited
5 separately excited
2 over compound
6 shunt excited
3 level compound
7 differentially compound
4 under compound Conditions for voltage build-up in Shunt Generator 1) There must be residual flux. 2) Correct polarity of field winding with respect to armature winding so that field flux aids residual flux for a given direction of rotation. 3) Field Resistance must be less than critical value R f < R
f cr
Critical resistance is equal to the slop of air-gap line. 4) Speed of rotation should be more than critical value for a given field r esistanceR f .
N > Ncr 15
Braking of DC Motor Plugging o
Supply to armature terminals is reversed whole field is left undisturbed.
o
The current reverses resulting into negative torque & that brings rotor quickly to r est.
V
'
Ia =
o
Plugging Torque
EaIa
Before plugging,
Load Torque
R
Ia
,
E
a
R
a
ex
= speed of rotor
V - E a
Ra
Ea Ia
Breaking Torque = (Load Torque + Plugging Torque)
Synchronous Machine Induced emf Phase voltage 4.44 Nphf
Nph : number of turns per phase : flux per pole
f : frequency This phase voltage is rms value Armature Winding o
Usually, coil span is
o
If coil span = 180 (electrical), coil is called as full pitch coil.
o
If coil span =
180
(electrical)
180 (electrical), coil is called as Chorded coil or short pitched winding. 16
o
Pitch Factor, KP = cos
o
Induced emf
o
For nth harmonic
2
4.44 Nphf KP 4.44 Nphf KP
Induced emf
n 2
KP = cos
To eliminate nth harmonic n 2
=
=
2
180 electrical n
Distributed Winding
m=
number of slots number of poles no. of phase
Coil Span =
=
number of slots number of poles
180 electrical ; coil span
m 2 m sin sin
Distribution Factor, K d
2
th For n harmonic, is replaced by
Kd
n
mn 2 n sin 2
sin m
17
o
n by 2
For uniform distribution replace sin
n 2
Winding Factor, K w = KPKd Induced emf = 4.44 Nphf Kw Armature Resistance Generally winding resistance is measured using voltmeter ammeter –method. For star connection
Rm =
voltmeter reading V = I ammeter reading
Rm = 2R
R=
Rm 2
For Delta Connection
Rm =
voltmeter reading ammeter reading
Rm =
2 R 3
R=
3 R 2 m
This resistance is dc resistance but ac resistance is higher due to skin effect. R
a ac
= 1.2 to 1.3 R
18
Armature Reaction Power factor Unity
Generator
Motor
Zero pf lagging
Zero pf leading
Lagging pf
cos
Leading pf cos
19
Leakage Flux Leakage flux links only one winding but not both so if it is present in stator, it won’t link to rotor & vise versa. Equivalent Circuit
X s = synchronous reactance
X ar
Xl
= sum of armature reaction & leakage reactance
E V 0 + Ia (R a jX s ) , for Synchronous Generator
E V 0 - Ia (R a jX s ) , for Synchronous Motor Where Φ is power factor angle (leading) for lagging power factor we replace Φ by “– Φ” Voltage Regulation Voltage regulation
E
V
V
100%
For zero voltage regulation
Xs R a
= tan-1
= 180 cos = load pf leading
20
For maximum voltage regulation
=
cos = load pf lagging Characteristics of Alternator OCC & SCC Open circuit characteristics & short circuit characteristics
ZS =
open circuit voltage at same field current short circuit current at same field current
Generally, open circuit voltage is given as Line to Line value so, before calculating need to find phase voltage
ZS =
ZS =
Voc / 3 Isc
: For Star Connection If = constant
Voc Isc
: For Delta Connection If = constant
Short circuit ratio
SCR =
Field current required for rated open circuit voltage Field current required for rated short circuit current
1 X S pu
XS pu = synchronous reactance in pu
21
ZS ,
we
Finding Voltage Regulation There are usually 4 methods to find voltage regulation o
EMF Method
o
MMF Method
o
Potier Triangle Method
o
ASA Method
Order of voltage regulation:
EMF
ASA>ZPF>MMF
Power Angle Equation Output of generator VtEf
Pout =
ZS
VtEf
Qout =
ZS
cos
sin
Vt2 ZS Vt 2 ZS
cos
sin
Input of motor Pin =
Qin =
Vt
2
ZS
Vt
Vt Ef
cos S
2
ZS
sin
cos
ZS
VtEf ZS
sin
Synchronous Impedance = Z s = R a jX S = Z S
-1
tan
XS Ra
If R a = neglected, Z s = jX S = X S
Pout g
=
90
Ef Vt V sin ; Qout = t Ef cos Vt g XS XS
22
Developed power in synchronous motor Pdev =
Q dev = If
ra
Ef Vt ZS
Ef Vt ZS
is neglected, Pdev =
Q dev =
2
co s
sin
Ef
ZS
Ef 2 ZS
co s
sin
ZS = XS 90
Ef Vt ZS
sin
Ef Vt ZS
2
co s
Ef
ZS
o
Developed Power is the power available at armature of motor.
o
In all power expressions, all voltages are line voltages and if we want to use phase voltage, we must multiply all expressions by a factor of 3.
Parallel operation of Alternators Necessary Conditions 1) Terminal voltage of incoming alternator must be same as that of existing system. 2) Frequency should be same. 3) Phase sequence should be same.
23
Synchronization by Lamp Method
1) Observe if 3 lamps are bright & dark simultaneously, that means phase sequence of incoming alternator is same as that of existing system. Otherwise, phase sequence is opposite and stator terminals must be interchanged to reverse phase sequence of incoming generator. 2) The frequency of alternator is usually a bit higher than infinite bus. 3) To understand the concept better, refer Ques. 39 of GATE – 2014 EE-01 paper. o
If two alternators are supplying a load and we change either excitation or steam input of one machine is varied, then following effects will happen:
o
If excitation of machine 1 is increased Parameter Real Power Reactive Power Armature Current Power Factor
o
Machine 1 Same Increases Increases Decreases
Machine 2 Same Decreases Decreases Increases
If steam input of machine 1 is increased Parameter Real Power Reactive Power Armature Current Power Factor
Machine 1 Increases Constant Increases Increases
Machine 2 Decreases Constant Decreases Decreases 24
Droop Characteristics
droop of generator =
fNL
f FL
f FL
100%
Example: Refer Kuestions on Electrical Machines Type-8 Salient Pole Machine o
In case of salient pole machine, There are 2 reactances
Xd & Xq X d :
Direct axis reactance
Xq : quadrature axis reactance o
Id = Ia sin 90
Iq = Iacos
=
For synchronous generator
tan =
Vsin Ia Xq V cos IaRa
;
lagging pf
-
leading pf
leading pf
-
lagging pf
For synchronous motor
tan =
Vsin Ia Xq V cos IaRa
;
Power – Angle Characteristics
VtEf Vt2 1 1 sin2 P= sin Xd 2 X q Xd
Excitation power
Reluctance power
25
Slip Test If machine is run by prime mover at a speed other than synchronous speed & voltages & currents are observed
Xd =
Maximum Voltage Maximum Current
Xq =
Maximum Voltage Maximum Current
Power Flow Diagram 3 E Ia cos f
Input
Shaft Power
Pe
3Vt Ia cos Field
Rotational
SC load
Loss
loss
Circuit loss
3I r 2
a
a
Power Flow for Synchronous Generator
3 E Ia cos f
Input
Shaft Power
Pe
3V I
ta
Field Circuit loss
cos
SC load loss
Rotational
3I r 2
a
Loss
a
Power Flow Diagram for Synchronous Motor
26
Induction Machines Stator & Rotor Magnetic Fields o
When a 3-phase supply is connected to the stator, than a magnetic field is set up whose speed of rotation is NS =
120f P
f = frequency of supply o
If negative sequence currents are applied the rotating magnetic field rotates in opposite direction as compared to magnetic field produced by positive sequence currents.
o
The rotor rotates in same direction as the stator magnetic field with a speed, Nr .
slip s =
Ns
Nr
Ns
Nr = Ns 1
s
o
Speed of rotor magnetic field with respect to rotor
o
speed of rotor magnetic field with respect to stator
= sNs = Ns .
Hence, stator & rotor magnetic fields are at rest with respect to each other. o
Frequency of emf & current in rotor = sf
With respect to
Stator Stator Magnetic Field Rotor Rotor Magnetic Field
Relative Speed of Stator Stator Rotor Magnetic Field 0 Ns Ns(1-s) -Ns 0 -sNs
-Ns(1-s) -Ns
sNs 0
27
0 -sNs
Rotor Magnetic Field Ns 0
sNs 0
Inverted Induction Motor o
When a
3 supply
is connected to the rotor & stator terminals are shorted or are
connected to the resistive load. o
Then a rotor magnetic field is set up which rotates at speed Ns =
o
120f P
Ns with
respect to rotor ;
where f is frequency of supply.
If rotor rotates at speed s=
Ns
Nr ,
than slip
Nr
Ns
Here, the rotor rotates in a direction opposite to the direction of rotation of stator magnetic field. o
Speed of rotor magnetic field with respect to stator = Ns
Ns 1 s
= sNs
Speed of stator magnetic field = sNs o
Frequency of emf & current induced in stator = sf f = supply frequency on rotor.
With respect to
Stator Stator Magnetic Field Rotor Rotor Magnetic Field
Relative Speed of Stator Stator Rotor Magnetic Field 0 sNs Ns(1-s) -sNs 0 -Ns
-Ns(1-s) -sNs
Ns 0
Equivalent circuit of Induction Motor
28
0 -Ns
Rotor Magnetic Field sNs 0
Ns 0
If we refer all parameters on stator side
2
N N 1 r2 = r2 ; x2 = x2 1 N N 2 2
2
N1 = N1 k1
Where
N1 =
no. of turns per phase on stator
k1 = winding factor of stator winding N2 = N2 k2 N2 =
number of turns per phase on rotor
k2 = winding factor of rotor winding
Tests Conducted on Induction Motor (i) No-Load Test o
Conducted on Stator with no-load on rotor side.
o
It gives No-Load Losses ( Rotational Loss + Core Loss).
(ii) Blocked Rotor Test o
Conducted on stator side keeping rotor blocked
o
It gives full load Copper Losses and equivalent resistance and equivalent reactance referred to Stator Side.
29
o
R 01 & X 01 are
equivalent winding resistance & equivalent leakage reactor referred to
Stator side. 2
o
o
Wattmeter reading = P = Isc R01 from this equation, we can calculate R 01 Z 01 =
Vsc I sc
& X01 =
2
Z 01 R 01
2
o
We obtain R 01 , X 01 & full load copper losses from this test.
o
R 01 =
R1+ R2’ ; X 01 = X1+ X2’
Power Flow Diagram
Rotor i/p = Pg (Airgap power)
Mechanical Power Developed
P
in
Stator
Stator
Rotor
I2R loss
core loss
2
I R
loss
2
Pg = I 2 =
3I 2 r2 s rotor current
s = slip r2 =
rotor resistance per phase 2
Rotor Cu Loss = 3I2 r2 = sPg
Mechanical power developed = Pg sPg = 1-s Pg
Developed Torque,
Te =
1-sPg Pm = wr 1-s ws 30
Pg ws
Rotor core loss
Friction & windage loss
Torque – Slip Characteristics If core loss is neglected then equivalent circuit looks like as shown
Ve =
Re =
V1 jXm r1 j X1 Xm r1Xm X1 X m
; Xe =
X1 X m
X1
Xm
2
Torque developed, Tc =
mVe
r ws Re 2 s
2 2 X Xe 2
For Approximate analysis,
V12 r2 3 Stator impedance is neglected; Tc = 2 ws s R2 X22 s 31
r2 s
o
At low slip, s 1 R 2
X 2 ,
s o
At high slip , s
Tc =
3 ws
2
sV1
R2
Tc
s
1 2
R 2 s
X 2
3 V1 R 2 1 Tc = ws X s s 2
,
For maximum torque
R2
Sm,T = Re
2
Xe X 2
2
It stator impedance is neglected
Sm,T =
And also,
T Tmax
R2
X2
=
and Tmax =
2
s s m,T s sm,T
3 V12 s
(2X 2 )
, where T is the torque at a slip ‘s’
For maximum power
Sm,P =
R2 2
2
R R X X R e e 2 2 2
Starting of Induction Motor (i) Direct on – line starting o
o
Directly motor is connected to supply. Te,st Te,FL
2
I = st SFL IFL 32
(ii) Auto Transformer Starting o
Instead of connecting the motor to direct supply we reduce the voltage from V1 to
o
xV 1
This is done with the help of auto – transformer. Te,st
o
1 Ist
=
Te,FL
SFL
X 2 IFL
Te,st auto X'mer
o
2
Te,FL direct
2
XV1 2 = =X V1
(iii)Star – Delta Starting o
At starting, stator winding is connected in star & in running state stator winding is connected in delta.
o
o
V1
Vph = IY =
3 1 3
TY
;
=
TD
3
V1
V1
2
=
2
1 3
ID 2
o
1 2 I st,d Ist,Y 3 S = S = I FL IFL ,d FL FL,d
Tst TFL
;
Speed Control of Induction Motor o
Constant V
f
Control T=
At low slip,
s=
180 2Ns
sV1
2
R 2
N N s
Ns
T =
180 2Ns
N
s
N
Ns
V 1 R 2 f V1
2
33
2
N
s
N
2
1 I st,Y = SFL TFL 3 IFL,d Tst
For constant torque, Ns N = constant So, by varying frequency we vary Ns & since Ns N = constant we vary N accordingly. Crawling o
Due to harmonies, the actual torque characteristics may look like
o
Due to this saddle region, the motor may become stable at a low speed & this is called as crawling.
Cogging o
If number of stator slots is equal to or integral multiple number of rotor slots, than at the time of start, the strong alignment forces between stator teeth & rotor teeth simultaneously at all rotor teeth may prevent movement of rotor. This is called cogging.
Single Phase Induction Motor o
According to Double field Revolving Theory, a single phase mmf can be resolved into two rotating fields one rotating clockwise called as Forward field & other rotating anti-clock wise called as Backward Field. Both fields rotate at synchronous speed
Ns = o
120f P
If rotor rotates at speed Nr , or a slips with respect to forward field. Than slip with respect to backward field is 2 s
34
o
Due to these two fields producing opposing torques on rotor single phase IM is not self starting.
o
To produce starting torque, we introduce an auxiliary winding which is used at the time of start & is disconnected during the run stage.
We generally design auxiliary winding such that phase difference is approximately 90 between main winding & auxiliary winding currents.
35
o
Capacitor Start Motor
o
Capacitor Run Motor
36
