1445016948Krash Sample Machines

Krash (Sample)

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Manual for Krash

Why Krash?

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.

It’s very overwhelming for a student to even think about finishing 300-400 questions per

subject when the clock is ticking at the last moment. This is the reason why Kuestion serves

the purpose of being the bare minimum set of questions to be solved from each chapter

during revision.

What is Krash?

Krash is a combination of K-Notes and Kuestion which effectively becomes a great tool for

revision. A 50 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.

A set of 100 questions or less for each subject covering almost every type which has been

previously asked in GATE. Along with the Solved examples to refer from, a student can try

similar unsolved questions to improve his/her problem solving skills.

When do I start using Krash?

It is highly recommended to use Krash in the last 2-3 months before GATE Exam.

How do I use Krash?

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. Kuestion should be used as a tool to improve your speed and accuracy subject-

wise. It should be treated as a supplement to our K-Notes and should be attempted once

you are comfortable with the concepts and basic problem solving ability of the subject. You

should refer K-Notes before solving any “Type” problems from Kuestion.

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Krash (Sample)


Krash - Sample

Electrical Machines

(Transformers)

Krash (Sample)


Krash (Sample)


Transformers

MMF and Flux

C

C

C C

C

C

C

Ni

A l

lNi

A

F R

c

c

lR

A

Reluctance of transformer core

Flux in core

F=MMF

Construction

In a Transformer the primary and secondary windings are wound around the core of a

transformer. Core is a magnetic material which allows the flow of magnetic flux lines to link

both primary and secondary windings.

1. Core should have low reluctance and high permeability to the flow of magnetic flux.

2. Core is generally made of Silicon steel

3. Generally, this steel is cold rolled grain oriented CRGO steel, to increase permeability

along the direction of magnetization and reduce core losses.

4. KVA Rating (Core Dimension)4

5. Voltage Rating (Core Dimension)2

6. Current Rating (Core Dimension)2

7. No-Load Current Core Dimension

8. Core Loss Core Volume

Properties of Ideal Transformer

Permeability of Transformer is infinity.

Iron loss in the Transformer core are zero

Resistance of transformer winding is zero

No magnetic leakage flux, so coefficient of coupling is 1.

Magnetization curve of transformer is linear.

Induced EMF in a Transformer

For a practical transformer with finite permeability the induced emf can be calculated as,

Krash (Sample)


The instantaneous values of induced emf in primary and secondary are given by,

1 1

2 2

de N

dtd

e Ndt

The rms values of induced emf is given as,

1 1 m

2 2 m

E (rms) 4.44fN

E (rms) 4.44fN

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.

m m nB A where Bm is the maximum value of magnetic flux density and An is the cross

sectional area of the core.

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.

Induced emf always lags the flux by 900.

If the frequency of operation of a transformer is reduced then KVA rating of transformer

reduces proportionately as induced emf varies linearly with frequency.

For a given KVA rating and for given maximum value of B of core, more the designed

frequency lesser is size and weight of transformers.

Exact equivalent circuit

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Exact equivalent circuit w.r.t. primary

2 2 2

1 1 1

2 2 2 2 L L

2 2 2

N N NR = R ; X = X ; Z = Z

N N N

;

Approximately Equivalent Circuit

01 1 2R = R R

01 1 2X = X X

In this approximation, we are neglecting the no-load copper losses and we are

overestimating the core losses.

The phasor diagram for the exact equivalent circuit is shown below,

I

: Magnetizing Current

wI : Core-loss Component of No-Load Current

0I : No-Load Current

o70 to 75

0cos 0.2lag (No-Load pf)

Here, o is the angle between E1 and I0

Transformer has poor No load PF because w

I I

11 2 2NI ' N I

Load component of primary MMF = Secondary MMF

2 1 2

1 2 1

N I ' EK

N I E = Transformation ratio

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Tests Conducted on a Transformer

(i) Open Circuit Test

o Conducted on LV side keeping HV side open circuited

o Equivalent Circuit

o Power reading =

2

11 0 0

c

VP = V I cos =

R -------- (i)

o Ammeter reading 0I = I

o 0

1 0

Pcos =

V I

o Calculate 2

0 0sin = 1 - cos

o

2

11 0 0

m

VQ = V I sin =

X ------- (ii)

Calculate cR from (i) & mX from (ii)

if OC Test is conducted at rated voltage but at less than required frequency

o Due to increase in flux, the magnetizing current increases.

o With decrease in frequency, hysteresis loss increases and thus core loss increase.

o Hence, core loss component of current also increases.

o Due to increase in magnetizing and core loss current the no-load current also increases.

o Due to increase in No-Load Current the No-Load Cu Loss increases.

o Due to increase in core loss, wattmeter reading will increase.

o The no-load power factor reduces as magnetizing current dominates core loss component

of current.

(ii) Short Circuit Test

o Conducted on HV side keeping LV side short circuited

o Equivalent Circuit

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o 01R & 01X are equivalent winding resistance & equivalent leakage reactor referred to HV

side.

o Wattmeter reading = 2

sc 01P = I R from this equation, we can calculate 01R

o sc01

sc

VZ =

I &

2 2

01 01 01X = Z R

o We obtain 01R , 01X & full load copper losses from this test.

Deviations if SC test is conducted at rated voltage but less than rated frequency,

o Since, reactance is proportional to frequency so 01

X decreases but 01

R remains same.

o Short-circuit Impedance decreases so Short Circuit Current increases.

o So, Full Load Copper Loss increases.

o Short Circuit Power Factor, 01

sc

01

Rcos

Z increases as Short Circuit Impedance decreases.

Losses on Transformers

Copper Loss

2 2

Cu 1 1 2 2P = I R I R

2 2

1 01 2 02= I R I R

Where 1I = primary current

2I = secondary current

1R = primary winding resistance

2R = secondary winding resistance

2 2

1 201 1 2 02 2 1

2 1

N NR = R R ; R = R R

N N

Core Loss

(i) Hysteresis Loss

xmh h

P = K B f

x = 1.6

mB = maximum value of flux density

1.6mh h

P = K B f

m

VB

f

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V = applied voltage

f = frequency

V

f = constant then

maxB = constant; then

hP f

If V is constant & f is increased, h

P decreases

(ii) Eddy Current Loss

2 2

e e mP = K B f

m

VB

f

If m

Vconst B const

f

2eP f

So, eddy current losses are proportional to square of frequency.

Otherwise,

2

2 2

e e e

VP = K f = K V

f

Core loss = c e h

P = P P

To further reducee

W , we use laminations

2 2 2

max

e

k.B f tW

f= frequency of eddy current e

(I )

t= thickness of lamination

= Resistivity of the core

High frequency Transformer should have thinner laminations to reduce e

W

Separation of iron losses

To separate iron losses into Hysteresis and eddy current loss, Open Circuit test is conducted

at variable f and voltage such that 1V

f constant

1V

f Constant

maxB Constant

1.6

1.6 0.6h h h

VP = K f = K V f

f

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2

i

i

W Af Bf

WA Bf

f

The constants A and B can thus be determined

by the slope and intercept of the curve whose

equation is given above.

Frequency can be changed by Cyclo-converters.

Efficiency

2

i Cu,FL

100%x KVA cos

= x KVA cos P x P

FL

Ix

I =% loading of Transformer

cos = power factor

iP = iron loss

Cu,FLP = Full load copper losses

KVA = Power rating of Transformer

For maximum efficiency,

i

Cu,FL

Px =

P

If loading is kept fixed then maximum efficiency occurs at unity power factor.

Voltage Regulation of Transformer

Regulation down NL FL

NL

V V 100

V

Regulation up NL FL

FL

V V 100

V

K = Transformation Ratio 2

1

N N

No-load voltage 2 V

Full-load voltage 2 V

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Approximate Voltage Regulation

2 02 2 02 2

2

I R cos X sinVR =

V

2 Lcos = power factor of load Z

+ sign is used for lagging pf load

- sign is used for leading pf load

If 2 2

V and I are chosen as base values, then base impedance is given by

2

base

2

VZ

I

Then, 02

02

base

RR pu

Z and 02

02

base

XX pu

Z

So, Voltage Regulation can be expressed as,

Lagging pf load, VR = 02 02R pu cos X pu sin 100%

Leading pf load, VR = 02 02R pu cos X pu sin 100%

Condition for zero voltage regulation

-1 022

02

R = tan

X

The power factor is leading, Voltage Regulation can never be zero for lagging pf load.

Condition for maximum voltage regulation

-1 022

02

X = tan

R

The power factor is leading, Voltage Regulation can never be negative for lagging pf loads

Non-Linear Core

If applied voltage is sinusoidal then flux is also sinusoidal but magnetizing current is

peaky wave due to 3rd harmonic.

For sinusoidal magnetizing current, flux is flap topped and emf is peaky wave containing a

strong 3rd harmonic.

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Magnetizing Inrush Current

When Transformer is switched ON it draws high magnetizing current which is 15-20 times

the full load current. This current exists only for a short duration.

This current is rich in second harmonics and can be eliminated if Transformer is switched ON

at zero flux or maximum voltage condition and this is most severe if Transformer is switched

on at zero voltage point.

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) The secondary phasors are drawn parallel to primary phasors i.e. a1a2 will always be

parallel to A1A2.

Some examples

The connection shown below is a delta-delta transformer,

Phasor

o If you observe carefully, we traverse from A2 to A1(or B2) in primary and same way we

traverse in secondary. So, both phasors are parallel.

o Then, we draw reference phasors from neutral to terminal and mark it with phase with

same name as terminal it is pointed to.

Krash (Sample)


Then we plot it on clock & we observe it is like 12 0 clock so name is Dd12 connection.

Another example

Phasor

o Here, primary phasor is drawn from A2 to A1 but secondary phasor is drawn from a1 or n

to a2. So, secondary phasor is anti-parallel to primary phasor.

Important Points

Delta-Star connection has the highest transformation ratio so it is used as Step-up

Transformer but Star-Delta connection has least Transformation Ratio so it is used as Step-

Down Transformer.

Star-Star connection has triplen harmonics in the phase voltages but they are absent in

line voltages so single phase load must not be driven by Star-Star Transformer. To remove

this, we can ground the neutral of Star on both sides or we can use a delta connected

Tertiary winding.

Star-Star Connection has higher voltage rating than delta-delta connection but lesser

current rating.

For same rating, Star-Star connection requires lesser insulation as compared to Delta –

Delta Connection but has higher cross-sectional area of the conductor.

If one-unit of delta-delta connection is open-circuited then it operates at reduced

capacity of 57.7% of original capacity. If it supplies same power as Delta-Delta connection

then each phase is 73.2% over-loaded.

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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 0 .

If there are 2 transformers A & B supplying a load power LS .

B BA L B L

A B A B

Z ZS = S ; S = S

Z Z Z Z

BZ = impedance of transformer B (in ohms)

AZ = impedance of transformer A (in ohms)

Note: If impedances are in pu then they must be on the same base.

If A B

Z & Z are given as complex values, then

* *

B B

A L B L

A B A B

Z ZS = S ; S = S

Z Z Z Z

Auto Transformer

Generally, auto transformer is created from 2- winding transformer.

If rating of auto – transformer is LV/HV or HV/LV

LV = low voltage

HV = high voltage

Transformation Ratio = LV

K = HV

KVA rating of auto transformer = 1

1 - K

(KVA rating of 2- winding Transformer)

In auto- transformer, power is transferred from primary to secondary by 2 methods

induction & conduction.

inductionKVA = 1 - K Input KVA

conductionKVA = K Input KVA

% Full load losses = 1 - K %FL losses in2 winding Transformer

If copper & core losses are not given separately, then we consider losses as constant,

same as that of two winding transformer while calculating efficiency.

Krash (Sample)


Type 1: Dimensions of a Transformer

For Concept, refer to Electrical Machines K-Notes, Transformers

Sample Problem 1:

A single phase 10 kVA, 50 Hz transformer with 1 kV primary winding draws 0.5 A and 55W, at

rated voltage and frequency, on no load. A second transformer has a core with all its linear

dimensions √2 times the corresponding dimensions of the first transformer. The core

material and lamination thickness are the same in both transformer. The primary winding of

both the transformers have the save number of turns. If a rate voltage of 2 kV at 50 Hz is

applied to the primary of the second transformer, then the no load current and power,

respectively, are

(A) 0.7 A, 77.8W (B) 0.7 A, 155.6W

(C) 1A, 110W (D) 1A, 220W

Solution: Option (B) is correct.

No-load current of second transformer becomes 2 times.

2 2 .5 .707mI A

Core Loss becomes2 2 times

2 2 2 55 155.6 P W

Problems:

01. Two transformers of the same type, using the same grade of iron and conductor

materials, are designed to work at the same flux and current densities, but the linear

dimensions of one are three times those of the other in all respects. The ratio of KVA of the

two transformers closely equals

(A) 9 (B) 81

(C) 27 (D) 3

02. A 11000V, 50 Hz transformer has a flux density of 1.2T and a core loss of 3000 watts at

rated voltage and frequency. Now all the linear dimensions of the core are doubled, primarily

and secondary turns are halved and the new transformer is energized from 22000 V, 50 Hz

supply. Both the transformers have the same core material and the same lamination

thickness. Core losses of the transformer are

(A) 6000 W (B) 9000 W

(C) 12,000 W (D) 24,000 W

Krash (Sample)


Type 2: Induced EMF in a Transformer

For Concept, refer to Electrical Machines K-Notes, Transformers Common Mistake: Generally we mark the polarities correct and then use Lenz Law again, so we will negative of actual emf so once you know polarities of EMF in primary and secondary just use Faraday’s Law instead of Lenz Law. Sample Problem 2:

The circuit diagram shows a two-winding, lossless transformer with no leakage flux, excited

from a current source, i(t), whose waveform is also shown. The transformer has a

magnetizing inductance of 400/π mH.

The peak voltage across A and B, with S open is

(A) 400/π V (B) 800 V

(C) 4000/π V (D) 800/π V

Solution: (D) is correct option

Peak voltage across A and B with S open is

diV m m (slope of I-t)

dt

3

3

400 10 800V 10

5 10

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Problems:

01. The core of a two winding transformer is subjected to a magnetic flux variation as shown

in the figure

The induced emf (Epq) in the primary winding will be the form

(A) (B)

(C) (D)

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02. In previous question the induced emf (Ers) in the secondary winding will be of the form

(A) (B)

(C) (D)

03. An Ideal transformer has Linear B – H curve with finite slope and a turns ratio of 1:1.The

primary of transformer is energized with an ideal current source, producing the signal ‘i’ as

shown in the figure. Then shape of the secondary terminal voltage v2(t) is

(A) (B)

(C) (D)

i

0 1 2 3

4 5

6

1 2 3 4 5 6 1 2 3 4 5 6

1 2 3 4 5 6 1 2 3 4 5 6

Krash (Sample)


Type 3: Equivalent Circuit of Transformer

For Concept, refer to Electrical Machines K-Notes, Transformers Common Mistake: Sometimes we add primary current to no-load current as scalar though they must be added as phasors. Sample Problem 3:

A Single-phase transformer has a turns ratio 1:2, and is connected to a purely resistive load

as shown in the figure. The magnetizing current drawn is 1 A, and the secondary current is 1

A. If core losses and leakage reactance’s are neglected, the primary current is

(A) 1.41 A

(B) 2 A

(C) 2.24 A

(D) 3 A

Solution: (C) is correct option

I0 = 1 amp (magnetizing current)

Primary current IP = ?

I2 = 1 A

I2p = secondary current referred to Primary

2 2

p

p

2

2

o p

21 2 amp

1

I i i 1 4 2.24 amp

I

Problems:

01. Across the HV side of a single phase 200 V / 400 V, 50 Hz transformer, an impedance of

32 + j24Ω is connected, with LV side supplied with rated voltage & frequency. The supply

current and the impedance seen by the supply are respectively:

(A) 20 A & (128 + j96)Ω (B) 20 A & (8 + j6)Ω

(C) 5 A & (8 + j6)Ω (D) 20 A & (16 + j12)Ω

Krash (Sample)


02. Consider the circuit shown below

If the ideal source supplies 1000W, half of which is delivered to the 100 Ω load, then the

value of b is____

(A) 1.5 (B) 0.89

(C) 0.56 (D) 0.67

03. A 1200/300, turns transformer when loaded, current on the secondary is 100A at 0.8

power factor lagging and primary current is 50A at 0.707 power factor lagging. Determine

the no-load current of the transformer with respect to the voltage?

(A) 25.5A (B) 26.5A

(C) 27.5A (D) 28.5A

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Solutions

Type 1: Dimensions of a Transformer

01. Ans: (B)

Solution: B & J are same

Also V

V B.A

B is constant in both transformers.

So V A................. 1

And I= J.A

J is same in both transformers

I A.................... 2

KVA rating 2A .................. 3

Since linear dimensions are 3 times

2

l 3l

A l ................. 4

From equation (3) & (4)

4KVA l

So KVA of other transformer is 4

KVA l

4

4

KVA 3l

KVA l

KVA 81 KVA

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02. Ans: (D)

Solution: V 2f N

Where V=Voltage

f=Supply frequency

Flux

N= No. of turns

1 111KV 2 f N

1 1 111KV 2 fB A N ............... 1

Also 2 2 222KV 2 fB A N ................. 2

1

2

NN

2 and

2 1l 2l =>

2 1A 4A

So from equation (2)

1

2 1

N22KV 2 fB 4A

2

Solving this 2 1 12 fB N A 11KV................ 3

Comparing equation (1) & (3)

1 2B B i.e flux densities are same

So core loss Core volume

Volume 3 3

l Volume 2l 8 Volume

New core loss=8 original core loss=83000=24000Watt

Krash (Sample)


Type 2: Induced EMF in a Transformer

01. Ans: (C)

Solution: pq 1

dE N

dt

From 0 to 0.06

pq

.009 0E 200 30Volts

.06 0

From 0.06 to 0.1

pq

d0

dt

E 0Volts

From 0.1 to 0.12

pq

0 0.009E 200 90Volts

0.12 0.1

02. Ans: (D)

Solution: rs 2

dE N

dt

From 0 to 0.06

rs

0.009 0E 500 75Volts

0.06 0

From 0.06 to 0.1

rs

d0

dt

E 0Volts

From 0.1 to 0.12

rs

0 0.009E 500 225Volts

0.12 0.1

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03. Ans: (B)

Solution: Since transformer is an ideal transformer with linear B-H curve. So wave shape of

flux will be same as the waveform of current

So flux in the core will have similar wave shape.

2

dV

dt

From given curve of current

From t=0 to t=1 slope is positive

So 2

V ve

From t=2 to t=3 slope is negative

So 2

V ve

From t=3 to t=5 slope is positive

So 2

V ve

And so on ……..

Type 3: Equivalent Circuit of Transformer

01. Ans: (B)

Solution: 2

Z 32 j24

T

400a 2

200

So load impedance seen by primary side 2

2 2

T

ZZ 8 j6

a

Current at supply side200

8 j6

020 36.86

02. Ans: (B)

Solution:

1 2 3 4 5 6

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Since source is supplying 1000W

So current from voltage source 1

1000I 10A

100

Since 500W is delivered to 100Ω resistor 2

3 3I 100 500 I 5A

Power consumed in 4Ω resistor= 2

10 4 400W

Total power consumed=Power consumed is 4Ω+power consumed in 25Ω+Power consumed

in 100Ω

Power consumed in 25Ω=1000-500-400=100W 2

2I 25 100

2I 2A

Since 2

3

I b

I 1

2

3

I 2b 0.89433

I 5

03. Ans: (A)

Solution: T

1200a 4

300

Taking voltages at reference phasor

Then 1

2I 100 cos 0.8 0100 36.86

0

2

2

I 100 36.86I

a 4

025 36.86 A

1

1I 50 cos 0.707 050 45 A No load current

0 1 2I I I

0 0 050 45 25 36.86 25.5 52.97

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