Electrical World

Let us consider the following two cases: Equal voltage ratios. Unequal voltage ratios. 1.39.1 Equal Voltage Ratios Assume no-load voltages E A and E B are identical and in phase. Under these conditions if the primary and secondary are connected in parallel, there will be no circulating current between them on no load. Figure 1.48 Equal Voltage Ratios Figure 1.48 shows two impedances in parallel. Let R A , X A and Z A be the total equivalent resistance, reactance and impedance of transformer A and R B , X B and Z B be the total equivalent resistance, reactance and impedance of transformer B . From Figure 1.48, we have E A = V 2 + I A Z A      (1.71) and           E B = V 2 + I B Z B      (1.72) ∴       I A Z A = I B Z B ∴     Equation (1.73) suggests that if two transformers with different kVA ratings...

It is required to connect a second transformer in parallel with the first transformer if the load exceeds the rating of the transformer shown in Figure 1.46. The primary windings are connected to the supply bus bars while the secondary windings are connected to the load bus bars. During paralleling of the transformer, similar polarities of the transformers should be connected to the same bus bars shown in Figure 1.46. Otherwise, the two emfs induced in the secondary windings with incorrect polarities will produce the equivalent of a dead short circuit shown in Figure 1.47. The following conditions are important for parallel opera-tion of transformers: The voltage ratings of both the primary and the secondary of the transformers should be identical. Small differences are permissible if the resultant circulating currents can be tolerated. The connections of the transformers should be proper with respect to their polarities. The percentage impedances should be equal in...

To determine the rise of maximum temperature of a transformer, its load test is of utmost importance. Using suitable load impedance, small transformers can be put on full load. The full-load test of large transformers is not possible because considerable wastage of energy occurs and it is difficult to get a suitable load for absorbing full-load power. Sumpner’s test is used to put large transformer on full load. This test can also be used to determine the efficiency of a transformer. Figure 1.45 shows the schematic diagram of Sumpner’s test. This test is also known as back to back test or load test . This test requires two identical transformers. The two primaries are connected in parallel and are energized at rated voltage and rated frequency. The wattmeter W 1 records the reading of core loss of both the transformers. Next the two secondaries are connected in series in such a way that their polarities are in phase opposition and the reading of the voltmeter V 2 becomes ...

The ratio of output in watts to input in watts is called commercial efficiency of a transformer. Distribution transformers are used for supplying lighting and general networks. Distribution transformers are energized throughout the day. Their secondaries are at no load most of the time in a day except during the hours of lighting period. Core loss occurs throughout the day. Copper loss occurs only when they are loaded and hence is less important. To judge their performance, all-day efficiency or operational efficiency is calculated. The all-day efficiency is defined by The all-day efficiency is less than the commercial efficiency of a transformer. Example 1.16 A 200 kVA single-phase transformer is in circuit throughout 24 hours. For 8 hours in a day, the load is 150 kW at 0.8 power factor lagging and for 7 hours, the load is 90 kW at 0.9 power factor. Remaining time or the rest period, it is at no-load condition. Full-load Cu loss is 4 kW and the iron loss is 1.8 kW. Calcu...

Polarity testing of transformers is vital before connecting them in parallel. Otherwise, with incorrect polarity, it is not possible to connect them in parallel. The rated voltage is applied to the primary and its two terminals are marked as A 1 and A 2 , respectively, as shown in Figures 1.44(a) and 1.45(b), respectively. The secondary winding terminals are also marked as a 1 and a 2 , shown in Figures 1.44(a) and 1.45(b), respectively. Now a voltmeter is connected across A 2 and a 2 . if it measures the difference of E 1 and E 2 , A 2 and a 2 are of the same polarity. If it measures the addition of E 1 and E 2 , A 2 and a 2 are of opposite polarity. Figure 1.44 Polarity Test of a single-phase Two Winding Transformer

Due to the losses in a transformer, its output power is less than the input power. ∴ Power output = Power input – Total losses ∴ Power input = Power output + Total losses = Power output + P i + P Cu The ratio of power output to power input of any device is called its efficiency ( η ).        Output power of a transformer at full-load = V 2 I 2ft cos θ , where cos θ is the power factor of the load, I 2 ft is the secondary current at full load and V 2 is the rated secondary voltage of the transformer. Full-load copper loss of the transformer = I 2 ft R 02 . ∴     Efficiency of the transformer at full load is given by Now V 2 I 2 ft = VA rating of the transformer. ∴     i.e.,     Efficiency of the transformer at any load m is given by where m = and P Cuft is the Cu loss of the transformer at full load. 1.34 CONDITION FOR MAXIMUM EFFICIENCY During working of a transformer at constant vol...

Two types of losses occur in a transformer: Core loss or iron loss occurs in a transformer because it is subjected to an alternating flux. The windings carry current due to loading and hence copper losses occur. 1.32.1 Core or Iron Loss The separation of core losses has already been introduced. The alternating flux gets set up in the core and it undergoes a cycle of magnetization and demagnetization. Therefore, loss of energy occurs in this process due to hysteresis. This loss is called hysteresis los s ( P h ), which is expressed by P h = K h B m 1.6 f V W       (1.59) where K h is the hysteresis constant depending on the material, B m is the maximum flux density, f is the frequency and V is the volume of the core. The induced emf in the core sets up eddy current in the core, and hence eddy current loss ( P e ) occurs, which is given by P e = K e B m 2 f 2 t 2 W per model W       (1.60) where K e is the ed...

Kapp had designed a diagram shown in Figure 1.42 to determine the regulation at any power factor. The description of the construction of the diagram is shown below. Figure 1.42 Kapp’s Diagram Load current ( I 2 ) is taken as a reference phasor. OA representing V 2 is drawn at angle θ 2 with I 2 . AB represents I 2 R 02 drawn parallel to I 2 , whereas BC represents I 2 X 02 drawn perpendicular to AB, i.e., I 2 . Here OC represents secondary emf ( 0 V 2 = E 2 ) at no-load. The circle 1 known as often circuit EMF circle is drawn with O as centre and OC as radius. The line OO ′ is drawn parallel to AC representing I 2 Z 02 . With O ′ as centre and OA as radius, the circle 2 known as terminal voltage circle is drawn, which intersects with circle 1 at the points D and E. The region above and below the reference line represents the lagging and leading power factors region, respectively. The point D is the point corresponding to zero regulation. The intercept FG gives the maximum ...

The voltage regulation up is expressed mathematically by Positive sign is for lagging power factor and negative sign is for leading power factor. 1.31.1 Zero Voltage Regulation For lagging power factor and unity power factor, 0 V 2 > V 2 . Therefore, we get positive voltage regulation. For leading power factor, V 2 starts increasing. At a certain leading power factor, 0 V 2 = V 2 and hence regulation becomes zero. If the load power factor is further increased, 0 V 2 becomes less than V 2 and hence regulation becomes negative. For zero voltage regulation, we have 0 V 2 – V 2 = 0 i.e.,      I 2 ( R 02 cos θ – X 02 sin θ )=0             Equation (1.56) shows the leading power factor at which voltage regulation becomes zero. 1.31.2 Condition for Maximum Voltage Regulation Maximum voltage regulation can be obtained for lagging power factor. For maximum voltage regula-tion, we have i.e.,    ...

Full-load voltage of a transformer can be expressed as a fraction of the full-load terminal voltage. Let I 1 fl be the full-load primary current, I 2 fl be the full-load secondary current, V 1 be the rated pri-mary voltage and V 2 be the rated secondary voltage. Per unit resistance drop of a transformer ∴       Per unit reactance drop of a transformer Per unit reactance drop of a transformer is called per unit reactance and it is given by Per unit impedance drop of a transformer is called per unit impedance and it is given by