UNIVERSITY OF BASRAH

ENGINEERING COLLEGE

ELECTRICAL ENGINEERING

LOAD TEST OF SINGLE PHASE

TRANSFORMER

NAME : AMAR ANMAR AHMEAD

CLASS :−2 ND¿

GROUP NAMES :−¿

اللطيف.1 عبد عتبةاسماعيل.2 علي3. حسون عماد

The transformer is one of the most widely used and useful electrical devices available. It can be used to transform an ac voltage or current in a circuit from one level to another and it can provide electrical isolation between circuits. By using transformers we can efficiently transmit electricity over a great distance at a high voltage and then reduce the voltage to safe level at the point of use (Q: Why is it more efficient to transmit electricity at a high voltage?). The voltage can then be further reduced and used to obtain the low voltage dc power supplies used in appliances such as computers, printers and CD players. The Ideal Transformer The transformer works by electromagnetic induction. If a sinusoidal ac voltage of is applied to a coil then it will produce a magnetic field where the flux will alternate at the same frequency as the voltage. Conversely if a coil is placed in an alternating magnetic field then a voltage will be induced in the coil that will alternate at the same frequency as the magnetic field. This is Faraday’s law of electromagnetic induction, which can be expressed as:

Where: e = voltage induced in the coil N = number of turns in the coil Φ = flux within the coil in Webbers (Wb) λ = flux linkage = NΦ (Wb) t = time in seconds If we now consider two coils that are closely coupled and we apply a sinusoidal voltage to one of the coils where the voltage is defined as: )cos(maxtVvω= Where: v = the instantaneous value of the voltage Vmax = the peak value of the voltage

= 2лf The magnetic field produced by the first coil will induce a voltage in the second coil. This is transformer action. To ensure that the coils are closely coupled the coils should be wound on an

iron core, which will provide a path for the mutual flux linking the coils (Q: Why will this be better than air?), to further improve the coupling the second coil should be wound on top of the first coil. For now the coupling between the coils will be assumed to be ideal. Apply Lenz’s law to the coils gives:

Where: L = the self inductance of the coil = N2/(reluctance) i = current producing the mutual flux C = a constant of integration that will be zero in this case The current that produces the mutual flux will be a sine wave that lags the voltage by 900, the flux will be in phase with the current. In the case of the ideal transformer the reluctance of the core will be zero (μr = ∞) and the current required to produce the mutual flux will be zero. With reference to figure 1and by applying Faraday’s and Lenz’s laws, we can say:

LOAD TEST OF SINGLE PHASE TRANSFORMER

AIM :−¿ perform load test for different power factors .

procedures :−¿

connection are mad asshown∈fig .3 supply is given¿ the hight voltage side

¿ load is applied on low voltage side .(a).UPF Load:-The switch S2 open, the variac is adjusted so that the primary voltage v1 is at its rated value.this value keep constant through the expirmebt for all load conditions. The secondary voltage v2 is noted . the resistive load is now put on and if necessary variac is adjusted to keep v1at its rated value . The readings of i1,i2,w1,w2 and v2 the secondary voltage under loading condition are noted .The resistive load is varied step by step ( and the above readings are taken) till full load.(b).Regulation characteristics :-The variation of regulation with PF , under constant output current i2, is determined as follows: resistive (R), inductive(L) ,capacitive(C) loads are connected to bring i2 to its rated value and the value of the secondary voltage v2 under this condition is noted. the inductive load only is put on and adjusted to bring i2 to its rated value with v1 still held constant by the variac.v2,i1,w1 and w2 are noted. The load PF now is lagging and low.

The PF can be increased now by increasing L and increasing R still keeping i2 at its rated value. The readings as above are taken .The PF is increased step by step till unity PF which is obtain only R in circuit . then PF is made leading by putting R and C in circuit. Finally the PF is made zero lead with only C in circuit.

practical Reasults (R only)

I 1( A ) I 2(A ) V 1(V ) V 2(V ) W 1(watt) W 2(watt)0.26 0 220 115 12 00.64 0.9 220 114.5 117 1041.58 2.75 220 112.2 328 3042.16 3.85 220 111 452 4222.63 4.71 220 110.3 553 515

Report :−¿

1.For UFPtest , calculate %regulation for each load calculate alsothe

% efficiency∧the input PF cos∅ 1

I 1( A ) I 2(A ) V 1(V ) V 2(V ) W 1(watt) W 2(watt) % regulation input PF % efficiency0.36 0 220 115.7 13 00.57 0.58 220 111.5 80 65 0.43 0.637 81.250.89 1.23 220 114.3 154 138 1.225 0.7865 89.61.19 1.81 220 113.9 221 202 1.58 0.844 91.41.39 2.27 220 113.6 275 254 1.849 0.899 92.361.93 3.45 220 113.6 275 254 2.57 0.96 93.382.45 4.5 220 111.6 526 493 3.67 0.965 93.73

2. plot % regulation, %efficiency∧input PF against output w2

¿∈th graphsheet

3. Determineoutput PF∧reguationfor eachset of readingof ( R∧C )

¿ RL

practical Reasults of RL∧the calculations of point 3

I 1( A ) I 2(A ) V 1(V ) V 2(V ) W 1(watt) W 2(watt) % regulation Out put PF2.66 4.5 220 114.1 89 54 1.402 0.15.65 4.5 220 1119 363 331 3.395 0.625.57 4.5 220 111.7 498 465 3.58 0.8782.51 4.5 220 111.6 526 491 3.67 0.97

practical Reasults of RC∧the calculations of point 3

I 1( A ) I 2(A ) V 1(V ) V 2(V ) W 1(watt) W 2(watt) % regulation Out put PF2.31 4.54 220 111.9 517 484 -0.268 0.8892.1 4.53 220 113.8 407 375 -1.933 0.7272.07 4.55 220 114.8 334 303 -2.787 0.582.16 4.85 220 118.6 39 3 -5.9 5.2∗10−3

4. plot %regulation against PF cos∅ (on x−axis ) expirment

¿∈th graphsheet

5.discuss your results∧give your comments .

We seen from the result of table 1( R only) that the regulation voltage decreased , the input PF increasing also the efficiency by the increasing of load current, while the load voltage (output transformer voltage) decreased from it's rated value ( at no load ) to its minimum value at rated output load current . For RL Load at rated output load current the load voltage decreased ,while output PF increased by increasing L For RC load and at rated output current we seen that the load voltage increased also the output PF by increasing C and the regulation voltage is negative .