Practical Transformer on Load - HA.incmyjiit.weebly.com/uploads/8/0/3/2/8032658/transformer... ·...

Practical Transformer on Load We now consider the deviations from the last two

ideality conditions :

1. The resistance of its windings is zero.

2. There is no leakage flux.

The effects of these deviations become more prominent when a practical transformer is put on load.

Monday, August 01, 2011 Transformers 1Next

(1) Effect of Winding Resistance Current flow through the windings causes a power loss

called I2R loss or copper loss.

This effect is accounted for by including a resistance R1

in the primary and resistance R2 in the secondary


(2) Effect of Flux Leakage The difference between the total flux linking with the

primary and the useful mutual flux Φu linking with both the windings is called the primary leakage flux, ΦL1.

Similarly, ΦL2 represents the secondary leakage flux.

Flux leakage results in energy being alternately stored in and discharged from the magnetic fields with each cycle of the power supply.

It is not directly a power loss, but causes the secondary voltage to fail to be directly proportional to the primary voltage, particularly under heavy loads.


Leakage flux in a transformer

Monday, August 01, 2011 Transformers 4

(a) Its definition. (b) Its effect accounted for.

• The useful mutual flux Φu is responsible for the transformer action.• The leakage flux ΦL1 induces an emf EL1 in the primary winding.

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Similarly, flux ΦL2 induces an emf EL2 in the secondary.

Hence, we include reactances X1 and X2 in the primary and secondary windings, in the equivalent circuit.

The paths of leakage fluxes ΦL1 and ΦL2 are almost entirely due to the long air paths and are therefore practically constant.

The reluctance of the paths being very high, X1 and X2

are relatively small even on full load.

However, the useful flux Φu remains almost independent of the load.


Equivalent Circuit of a Transformer


It is merely a representation of the following KVL equations :

1 1 1 1 1 1 1 1 1 1( )I R jI X I R jX V E E

2 2 2 2 2 2 2 2 2 2( )I R jI X I R jX E V V

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Points to Draw Phasor Diagram

1. Resistive voltage drop in phase with current phasor.

2. Inductive voltage drop in quadrature with current.

3. To get V1, add I1Z1 to –E1.

4. Add V2 and I2Z2 , to get E2.

5. Current I1 is vector sum of I0 and I2’.

6. Angle between V1 and I1 gives the power factor angle of the transformer.



m

E2

E1

-E1

I1R1

I1X1V1

I1

I0

I1'

I2

V2

I2R2

I2X2

0

1

Ph

aso

r D

iag

ram

fo

r

Pra

ctic

al T

ran

sfo

rmer

on

R

esi

stiv

e L

oa

d

1 1 1 1 1

2 2 2 2

( )

( )

I R jX

V I R jX

2

V E

E

O

I1Z1

I2Z2

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Pra

ctic

al T

ran

sfo

rmer

on

In

du

ctiv

e L

oa

d

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Pra

ctic

al T

ran

sfo

rmer

on

C

ap

aci

tive

Lo

ad

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Simplified Equivalent Circuit The no-load current I0 is only about 3-5 % percent of the

full-load current.

The exciting circuit R0-X0 in is shifted to the left of impedance R1-X1.


Transforming the impedances from the secondary to the primary side.

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Equivalent resistance and reactance referred to the primary side

2 2

e1 1 2 e1 1 2( / ) and ( / )R R R K X X X K

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Approximate Equivalent Circuit


Example 5 A single-phase, 50-kVA, 4400-V/220-V, 50-Hz transformer

has R1 = 3.45 Ω, R2 = 0.009 Ω, X1 = 5.2 Ω and X2 = 0.015 Ω. Calculate

(a) the Re as referred to the primary,

(b) the Re as referred to the secondary,

(c) the Xe as referred to the primary,

(d) the Xe as referred to the secondary,

(e) the Ze as referred to the primary,

(f) the Ze as referred to the secondary, and

(g) the total copper loss.



Solution : Full-load primary current,

1

1

kVA 5000011.36 A

4400I

V

Full-load secondary current, 2

2

kVA 50000227.27 A

220I

V

2

1

220 10.05

4400 20

VK

V

(a) 2 2

e1 1 2( / ) 3.45 [0.009/(0.05) ]R R R K 7.05 Ω

(b)

(c)

2 2

e2 1 2 (0.05) 3.45 0.009R K R R 0.0176 Ω

2 2

e1 1 2( / ) 5.2 [0.015/(0.05) ]X X X K 11.2 Ω

(d) 2 2

e2 1 2 (0.05) 5.2 0.015X K X X 0.028 Ω

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(e) 2 2 2 2

e1 e1 e1 (7.05) (11.2)Z R X 13.23 Ω

(f)

(g) Total copper loss

2 2 2 2

e2 e2 e2 (0.0176) (0.028)Z R X 0.0331 Ω

2 2 2 2

1 1 2 2 (11.36) 3.45 (227) 0.009I R I R 909 W

Alternatively, by considering equivalent resistances, total copper loss

2 2

1 e1 (11.36) 7.05I R 909.8 W

2 2

2 e2 (227.27) 0.0176I R 909 W

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Voltage Regulation


V2(0) = secondary terminal voltage at no load,

and V2 = secondary terminal voltage at full load.

The voltage regulation of a transformer is defined as the change in its secondary terminal voltage from no load to full load, the primary voltage being assumed constant.

The voltage drop V2(0) - V2 is called the inherent regulation.

2(0) 2

2(0)

( ) Per unit V V

iV

regulation down

2(0) 2

2(0)

% 100V V

V

regulation down

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2(0) 2

2

( ) Per unit V V

iiV

regulation up

2(0) 2

2

% 100V V

V

regulation up

Normally, when nothing is specified, ‘regulation’ means ‘regulation down’.

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Exact voltage drop =

2(0) 2 OC OA OG OA AG AF+FGV V

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Approximate Voltage Drop

The secondary terminal voltage at no load,2(0) 2 1 1V E KE KV


In case of leading power factor,

2 e2 2 e2

Approximate voltage drop, AF AE EF AE BD

cos sinI R I X

2 e2 2 e2

Approximate voltage drop, AF AE EF AE BD

cos sinI R I X

In general,

2 e2 2 e2Approximate voltage drop cos sinI R I X

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2 2 2 2

2(0)

cos sin% Regulation 100

cos sin

e e

r x

I R I X

V

V V

Condition for Zero Regulation :Possible only if the load has leading power factor.

2 2 2 2cos sin 0e eI R I X 2

2

tan e

e

R

X

Use + sign for lagging power factor and – sign for leading power factor.

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Condition for Maximum Regulation


2 2 2 2( cos sin ) 0e e

dI R I X

d

2 2 2 2( sin cos ) 0e eI R I X

2

2

tan e

e

X

R

Maximum regulation can occur only for inductive load. The voltage drop is maximum when

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Example 6

Solution :

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Example 7

Solution :

2the load voltage, 240 6V 234V

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Example 8 A single-phase, 40-kVA, 6600-V/250-V,

transformer has primary and secondary resistances R1 = 10 Ω and R2 = 0.02 Ω, respectively. The equivalent leakage reactance as referred to the primary is 35 Ω. Find the full-load regulation for the load power factor of

(a) unity,

(b) 0.8 lagging, and

(c) 0.8 leading.



2 2

e2 1 2 (0.0379) 10 0.02 0.0343R K R R

2 2

e2 e1and (0.0379) 35 0.0502X K X

(a) For power factor, cos = 1; sin = 0. Hence,

2 2 2 2

2(0)


160 0.0343 1 0100

250

e eI R I X

V

2.195 %

Solution : Given : R1 = 10 Ω; R2 = 0.02 Ω; Xe1 = 35 Ω

2

250 the turns-ratio, 0.0379

6600

40000the full-load current, 160 A

250

K

I

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(b) For power factor, cos = 0.8 (lagging, positive);

(c) For power factor, cos = 0.8 (leading, negative);

sin 0.6

2 e2 2 e2

2(0)


160 0.0343 0.8 160 0.0502 0.6100

250

I R I X

V

0.172 %

2sin 1 cos 0.6

2 e2 2 e2

2(0)


160 0.0343 0.8 160 0.0502 0.6100

250

I R I X

V

3.68 %

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