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TRANSCRIPT
Jordan University of Science & Technology
Department of Electrical Engineering
ELECTRICAL MACHINES I LAB
EE 336
Experiment # 1
SINGLE PHASE TRANSFORMER TESTS
Name : الجّراح ماضي مروان Univ. no : 20010024014 Section : 3 Group : 4
Objectives
To measure the equivalent circuit and efficiency of the single phase transformer from no-load and short circuit tests .
To investigate how the secondary voltage of the transformer changes with different loads and to measure 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 = N1/N2 = 138/60 = 2.3
LOAD increases FULL LOAD
I1 (A) 0.18 0.51 1.00 1.35 1.51 2.29 2.89I2 (A) 0.414 1.173 2.3 3.105 3.47 5.267 6.65
V1 ( V ) 220 220 220 220 220 220 220V2 ( V ) 94.2 93.4 92.8 92.3 91.8 90.1 89.9P1 ( W ) 88 116 224 296 336 500 620P2 ( W ) 39 109.56 213.44 286.6 318.55 474.56 597.84η ( % ) 44.3 94.4 95.29 96.82 94.8 94.9 96.43
2
Hysteresis loop
The graphs are plotted on the graph paper .
Graphs: Graphs are shown in Fig1, Fig2, Fig3 .
Questions :
1. KVArated = 5 * 220 = 1.1 KVA , p.f = 1 Pcu ( f.l ) = Ps.c = 68 W Pc ( f.l ) = Po.c = 7.7 W η ( f.l ) = 1.1 / (1.1 + 0.068 + 0.0077) = 93.56 %
2. V.R = { [ VN.L – VF.L ] / VF.L } * 100% = [( 95.65 – 89.9 ) / 89.9 ] * 100% = 6.4 %
3. At No-Load the current I2 = 0 A , and at short circuit test I2 = rated value .
4. Because at low frequency the transformer will be like a D.C, and at high frequency the impedance ( jwL ) will become very high .
5. Because by varying the load, then I2 changes so the Pcu changes
Pcu α I22
6. O.C TESTRc1 = (220)2 / 7.7 = 6.29 KΩIc1 = 220 / 6.29 K = 35 mAIm
2 = Io.c2 – 0.0352
Im = 41.65 mAXm = 220 / Im = 5.28 KΩ
S.C TESTReq = Ps.c / Is.c
2 = 68 / (11.5)2 = 0.514 ΩZeq = Vs.c / Is.c = 17 / 11.5 = 1.48 ΩXeq2 = Zeq
2 – Req2
Xeq = 1.387 Ω
7. By evaluating its area .
3
Conclusions
The open circuit test is conducted by applying rated voltage at rated frequency .
The primary current is equal to No-load current and it is so small, about 5% of the rated value .
The short circuit test is used to determine the equivalent series resistance and reactance .
The short circuit test is done on the high voltage side usually .
The core loss is equal to open circuit power .
The copper loss at full load is equal to short circuit test power .
4
5
0 0.5 1 1.5 2 2.5 340
50
60
70
80
90
100
I1(A)
Efficiency(%)
Fig1
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)
V2(V)
Fig2
6
7