# Lesson 11 - Tests on Transformers

September 13, 2017 | Author: Kaelel Bilang | Category: Transformer, Voltage, Power Engineering, Quantity, Electromagnetism

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Electrical Machine...

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TESTS ON TRANSFORMER 1. Open Circuit Test or No Load Test 2. Short Circuit Test 3. Polarity Test

Open Circuit Test or No-load Test The low voltage side is supplied with rated voltage while the high voltage side is left open. Electrical measurements are done on the low voltage side.

Pcoreloss  Poc

S oc  Eoc I oc

2

2

Eoc Rm  Poc

Eoc Xm  Qoc

Qoc  S oc  Poc 2

2

where: Poc = reading of the wattmeter during the test (watt) Eoc = reading of the voltmeter during the test (volt) Ioc = reading of the ammeter during the test (ampere)

Example: The no-load current of a transformer is 4A at 0.25 pf when supplied at 250 V, 60 Hz. Determine the a. core loss b. resistance representing the core loss c. magnetizing reactance

A. Pcoreloss  Poc

Poc  Eoc Ioc pf  25040.25  250watts

B. 2

Eoc 2502 Rm    250 Poc 250

C. Soc  Eoc Ioc  2504  1000VA Qoc  Soc  Poc  10002  2502  968.24VAR 2

2

2

2 Eoc 250 Xm    64.55 Qoc 968.24

Short Circuit Test The low voltage side is short circuited while the high voltage side is supplied with voltage adjusted so that the high side will draw rated high side current. Electrical measurements are done on the high voltage side.

Re( high )

P sc  2 I sc

 Psc

Z e( high )

E sc  I sc

X e( high )  Z e( high )  Re( high ) 2

2

where: Psc = wattmeter reading during the test (watt) Isc = ammeter reading during the test (ampere) Esc = voltmeter reading during the test (volt) Re = equivalent resistance referred to the high side Xe = equivalent reactance referred to the high side Ze = equivalent impedance referred to the high side

Example: A short circuit test was performed upon a 10kVA, 2300/230 volt transformer with the following results: Esc = 137 volts; Psc = 192W; Isc = 4.34 A. Calculate in primary terms the equivalent resistance and reactance of the transformer.

E sc 137 Ze    31.567 I sc 4.34 Psc 192 Re  2   10.193 2 4.34  I sc X e  Z e  Re  2

X e 29.88

2

31.567  10.193 2

2

Polarity Test

where: Vp = test input voltage (volt) V = voltmeter reading (volt)

If V>Vp, the polarity is additive If V< Vp, the polarity is subtractive

Example: A 2300/230 volt distribution transformer is tested for polarity in accordance with the standard method. If 120 volts is impressed across the high voltage windings, determine the voltmeter reading if the transformer has a. additive polarity b. subtractive polarity

 N2  E1 N1    E2  E1  E2 N 2  N1   230  E2  120   12volts  2300  Reading = 120 + 12 = 132 volts Reading = 120 – 12 = 108 volts

TRANSFORMER LOSSES

Copper Loss - I2R loss in the primary and secondary windings.

Pcu  I R1  I 2 R2 2 1

Pcu  I Re1  I 2 Re 2

2

2 1

Core Loss or Iron Loss - eddy current loss + hysteresis loss

Peddy  ke ( f m )  ke ' E g 2

2

1.6

Physteresis  kh f

1.6 m

Eg  kh ' 0.6 f

2

where: Pe = eddy current loss (watt) Ph = hysteresis loss (watt) ke, kh = proportionality constant βm = maximum flux density (tesla) Eg = supply voltage (volt)

f = frequency of supply voltage (hertz)

Example 1:

In a 400V, 50 c/s transformer, the total iron loss is 2500 W. When the supplied voltage is 220V at 25 c/s, the corresponding loss is 850W. Calculate the eddy current loss at normal frequency and voltage.

 Eg  Ph  kh  0.6  f 

1.6

Pe  ke E g

2

   

 Eg  Pe  Ph  ke E g  kh  0.6  f 

1.6

Pcore

2

   

When supply is 400V, 50Hz: 1.6  400  2 2500  ke ( 400)  kh  0.6   50  2500  160000ke  1392.88kh

kh  1.795  114.87ke  Eq.1 When supply is 220V, 25Hz: 1.6   220 2 850  ke (220)  kh  0.6   25  850  48400ke  811.17kh  Eq.2

Substitute Eq. 1 in Eq. 2:

850  48400ke  811.17(1.795  114.87ke ) ke  0.01353 Pe  ke E g  0.01353(400) 2

Pe  2165watts

2

The efficiency of the transformer is the ratio of its output power (power drawn by the load) to the input power (power developed by the transformer).



Poutput Pinput

Poutput  V2 I 2 pf

Pinput  Poutput  Plosses

where: Poutput = output power or power delivered to the load Plosses = power losses

V2 = load voltage (volt) I2 = secondary current or load current (ampere) pf = power factor

TRANSFORMER BANKS FOR THREE-PHASE CIRCUITS

1. Delta-to-delta (Δ-Δ) connection - This connection is seldom used in threephase transformers.

2. Delta-to-wye (Δ-Y) connection - It is often used for distribution service where a four-wire secondary distribution circuit is desired.

3. Wye-to-delta (Y-Δ) connection - This is extensively used for power transmission and distribution.

4. Wye-to-wye (Y-Y) connection - It is used when tying together two high-voltage transmission system of unequal voltage.

Seatwork: 1. The following data were obtained when a short circuit test was performed upon a 100 kVA, 2400/240V distribution transformer: Esc = 72 volts; Isc = 41.6 A; Psc = 1180W. All instruments are on the high side during the short circuit test. Calculate the equivalent resistance and reactance of the transformer.

2. The no-load loss of a transformer at rated voltage is 100watts at 30 Hz and 300 watts at 60 Hz. What is the hysteresis loss at 60 Hz and rated voltage? (Note: Since not specified, the maximum flux density (βm) is assumed constant.)

Assignment:

When a 220V, 60Hz is impressed on a certain transformer at no load, the total core loss is 200W. When the frequency of the impressed voltage is changed to 25 Hz and the magnitude of the voltage is made such as to maintain the same maximum flux density as before, the core loss falls to 75W. Calculate the hysteresis losses at 60 Hz.