Load Test Of Single Phase Transformer 29b1q

UNIVERSITY OF BASRAH ENGINEERING COLLEGE ELECTRICAL ENGINEERING

LOAD TEST OF SINGLE PHASE TRANSFORMER

𝑵𝑨𝑴𝑬: 𝑨𝑴𝑨𝑹 𝑨𝑵𝑴𝑨𝑹 𝑨𝑯𝑴𝑬𝑨𝑫 𝑪𝑳𝑨𝑺𝑺: −𝟐𝑵𝑫 _𝑩_𝟑𝟐

𝑮𝑹𝑶𝑼𝑷 𝑵𝑨𝑴𝑬𝑺: − 1. 2. 3. 4.

Single-Phase Transformers 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 V = the peak value of the voltage max

= 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:

2

Where: L = the self inductance of the coil = N /(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 0

voltage by 90 , the flux will be in phase with the current. In the case of the ideal transformer the reluctance of the core will be zero (μ = ∞) and the current r

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:

− −

(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.

0.26 0.64 1.58 2.16 2.63

0 0.9 2.75 3.85 4.71

220 220 220 220 220

115 114.5 112.2 111 110.3

12 117 328 452 553

0 104 304 422 515

𝑹

0.36 0.57 0.89 1.19 1.39 1.93 2.45

0 0.58 1.23 1.81 2.27 3.45 4.5

−

220 220 220 220 220 220 220

115.7 111.5 114.3 113.9 113.6 113.6 111.6

13 80 154 221 275 275 526

𝑹 𝑰 𝑨 2.66 5.65 5.57 2.51

𝑰𝟐 𝑨 4.5 4.5 4.5 4.5

0 65 138 202 254 254 493

0.43 1.225 1.58 1.849 2.57 3.67

81.25 89.6 91.4 92.36 93.38 93.73

𝑹𝑳

𝟑

𝟐

220 220 220 220

0.637 0.7865 0.844 0.899 0.96 0.965

114.1 1119 111.7 111.6

𝟐

89 363 498 526

54 331 465 491

1.402 3.395 3.58 3.67

𝑶 0.1 0.62 0.878 0.97

𝑹 𝑰 𝑨 2.31 2.1 2.07 2.16

4.54 4.53 4.55 4.85

220 220 220 220

𝟑 111.9 113.8 114.8 118.6

517 407 334 39

484 375 303 3

-0.268 -1.933 -2.787 -5.9

0.889 0.727 0.58

𝑷𝑭

−

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 .