machines lab exp01 just

8/13/2019 Machines Lab Exp01 JUST

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Jordan University of Science & Technology

Department of Electrical Engineering

ELECTRICAL MACHINES I LAB

EE 336

Experiment # 1

SINGLE PHASE TRANSFORMER

TESTS

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Objectives

To measure the equivalent circuit and efficiency of the single phase transformer from no-loadand short circuit tests .

To investigate how the secondary voltage of the transformer changes with different loads and tomeasure the efficiency and voltage regulation of the transformer .

To display the hysteresis loop .

Equipments

Power pack . 1 KVA, 240:120 V, Single-Phase Transformer . Load Resistor, Capacitor and Inductor . Wattmeter . 4* AVO . CRO .

Procedure

OPEN CIRCUIT TESTIo.c = 54.4 mAPo.c = 7.7 WVo.c = 220 VV2 = 95.65 V

SHORT CIRCUIT TESTIs.c = 5 APs.c = 68 WVs.c = 17 V

LOAD TEST

Turn ratio = N 1/N2 = 138/60 = 2.3

LOAD increases FULLLOAD

I1 (A) 0.18 0.51 1.00 1.35 1.51 2.29 2.89

I2 (A) 0.414 1.173 2.3 3.105 3.47 5.267 6.65V1 ( V ) 220 220 220 220 220 220 220V2 ( V ) 94.2 93.4 92.8 92.3 91.8 90.1 89.9P 1 ( W ) 88 116 224 296 336 500 620P 2 ( W ) 39 109.56 213.44 286.6 318.55 474.56 597.84 ( % ) 44.3 94.4 95.29 96.82 94.8 94.9 96.43


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Hysteresis loop

The graphs are plotted on the graph paper .

Graphs:Graphs are shown in Fig1, Fig2, Fig3 .

Questions :

1. KVA rated = 5 * 220 = 1.1 KVA , p.f = 1Pcu ( f.l ) = P s.c = 68 WPc ( f.l ) = P o.c = 7.7 W ( f.l ) = 1.1 / (1.1 + 0.068 + 0.0077) = 93.56 %

2. V.R = { [ V N.L V F.L ] / V F.L } * 100%= [( 95.65 89.9 ) / 89.9 ] * 100%= 6.4 %

3. At No-Load the current I 2 = 0 A , and at short circuit test I 2 = rated value .

4. Because at low frequency the transformer will be like a D.C, and at high frequency theimpedance ( jwL ) will become very high .

5. Because by varying the load, then I2 changes so the Pcu changes

Pcu I22

6.O.C TESTR c1 = (220) 2 / 7.7 = 6.29 K Ic1 = 220 / 6.29 K = 35 mAIm2 = Io.c 2 0.035 2 Im = 41.65 mAXm = 220 / Im = 5.28 K

S.C TESTR eq = P s.c / I s.c2 = 68 / (11.5) 2 = 0.514 Zeq = Vs.c / I s.c = 17 / 11.5 = 1.48 Xeq2 = Z eq 2 R eq 2 Xeq = 1.387

7. By evaluating its area .


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0 1 2 3 4 5 6 789.5

90

90.5

91

91.5

92

92.5

93

93.5

94

94.5

I2(A)

V 2 ( V )

Fig2

0 0.5 1 1.5 2 2.5 340

50

60

70

80

90

100

I1(A)

E f f i c i e n c y ( % )

Fig1


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machines lab exp01 just

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