M1 3rd With Report

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Electrical Power & Machines Dept. 3rd

year Lab.

M1 Three Phase Transformer Back to Back Test

Objective

To determine the parameters of large transformers and obtain their performance

characteristics.

Background

Two very simple tests are used to determine the constants of equivalent circuit and the powerlosses in the transformer. These consist in measuring the input voltage, current, and power tothe primary, first with the secondary short circuited and then with secondary open circuited. Thecore losses are determined from the open circuit test. The copper losses are determined fromthe short circuit test.Stray load loss consists of the losses arising from the non-unifrom current distribution in the

copper and the additional core losses produced in the iron by distortion of the magnetic flux bythe load current. It is difficult to determine such losses accurately by conventional no-load andshort circuit load tests.To obtain exact equivalent circuit and losses, the input and output parameters are directly

measured under different loading conditions. This is easy for small rating transformers.However for large transformers, it is difficult and expensive to take direct measurements.A Back to BackTest is used in this case. This test requires two identical transformers having

some tapping in the windings.Why Back-to-back test is used in case of large transformers ?First, the short circuit test is difficult to be applied, since applying a reduced voltage is very

difficult and unpractical. Second, this test can simulate the loading conditions on thetransformer without using real loads. Third, a large transformer supplying large essential loadshas usually a second identical transformer installed in the same location for back-up, so usingback-to-back transformer in this case is very practical.

Procedure

This test requires two identical three phase transformers. Take complete particulars of either ofthe two transformers.Connect the two transformers as shown in figure 1 and check the correct polarity using a

voltmeter connected across the switch. The reading of the voltmeter should be zero for correctpolarity connection. Then carry out the following procedures:

1. With the switch open, apply the rated voltage and read the various instruments. The inputpower in this case covers the iron losses of the two transformers, thus the parameters R c

and X of the transformer equivalent circuit can be determined. The turns ratio of thetransformer can be calculated from the reading of the primary and secondary voltages.

2. It is to be noted that if no taps are used and the corresponding terminals of the twotransformers are connected together, by closing the switch no circulating current will passand the transformers behave as if they were open circuited.

3. With the terminals of either of the two transformers remain unaltered and changing the

taps of the other, a small voltage will appear across the switch when it is open. This smallvoltage results in a circulating current in the transformer windings if the switch is closed. It


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year Lab.

is to be pointed out here that the voltage which appears across the switch (when it isopen) is the open circuit voltage calculated when Thevenin's theory is applied.Correspondingly the circulating current which passes after closing the switch is due to theimpedance seen from the switch terminals. This impedance is nothing but the short circuitimpedance of the two transformers as if they were connected in series. This input power in

this case covers the iron losses of the two transformers as well as their correspondingcopper losses.

4. Repeat step 3 using other tapping points.

5. From the readings of steps 2 to 4, the resistance and leakage reactance of the transformerreferred to the secondary side can be determined. Hence along with the results of step 1the complete equivalent circuit referred either to the primary or the secondary side can beobtained.

Report

1- From the results obtained determine:a- The parameters of the equivalent circuit of the transformer under test referred to the

primary and secondary sides.b- Draw the loss curve of either of the two transformers (power losses vs. load factor).c- The efficiency curves at unity power factor, 0.8 lagging, and 0.8 leading.d- The regulation curves at different power factors at x=0.8 and x=1.

2- Comment on the shape of the curves obtained.

3- Comment on the method used to determine the equivalent resistance and reactance andcompare with that used in the normal short circuit test.

Comment on your results whenever necessary.


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Induction RegulatorA stationary 3-phase induction machine can also be used as a source of variable 3-phasevoltage if it is connected as an induction regulator as shown in Figure (2). The phasor diagramshown illustrates the principle. As the rotor rotated through 360o, the output voltage Vo follows acircular locus of variable magnitude. If the induced voltages E1 and E2 are of the same

magnitude (i.e. identical stator and rotor windings) the output voltage may be adjusted fromzero to twice the supply voltage.The induction regulator has the following advantages over a variable auto transformer:

A continuous step-less variation of the output voltage is possible.

No sliding electrical connections are necessary.

Figure 2


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M1 Back-to-back transformer

Lab results

V1o = V2o =

turns ratio (a) = =

V (V) I1 (A) I2 (A) W1 (W) W2 (W)

Tap 1(no-load)

W10 = W20 =

Tap 2

Tap 3

Tap 4

At No-load (Tap 1)

=

At loading conditions (other taps)

[(W1 + W2) - (W10 + W20 ) ] = 2 * 3 I22 Req Req =

2Zeq = = Zeq = Xeq =

PcuFL= 3 I22 Req * ( )

2 I2FL = =

PcuFL= ReqLV= XeqLV=

ReqHV= XeqHV=

Transformer Ratings

S =

V1/V2 =

f =

Sb = Srated= S/3V1b = V1rated V2b = V2rated

2

1

VV

o

oI

V cos2

31

10

o

o

c

II cos

3

2=

o

o

m

II sin

3

2=

==

c

o

cI

VR 1 ==

m

o

mI

VX 1 =ironP

b

b

bSVZ

2

1

1=

b

b

b

S

VZ

2

2

2=

22

2

2 eqeq XRI

V+=

32

1

o

o

V

V

2

21 ooWW +


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Loss Curve

Plosses = Piron + Pcu

Pcu = x2PcuFL where x = load factor =

=

Piron = PcuFL =Plosses = Piron +x

2PcuFL

x 0.1 0.3 0.4 0.6 0.7 0.8 1

Plosses (w)

Efficiency Curves

= ++

= +

+

x 0.1 0.3 0.4 0.6 0.8 1

(0.8 lag)

(unity)

Regulation Curves

= ( ) - lagging pf / + leading pf

=

=

lag unity lead

(degrees) 80 30 10 0 10 20 30

Reg (x=0.8)

Reg (x=1)


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M1 3rd With Report

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