LOAD SHARING BY TWO TRANSFORMERS

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=V₂+I_AZ_A     (1.71)

and          E_B=V₂+I_BZ_B     (1.72)

∴      I_AZ_A=I_BZ_B

∴    

Equation (1.73) suggests that if two transformers with different kVA ratings are connected in parallel, the total load will be divided in proportion to their kVA ratings if their equivalent impedances are inversely proportional to their respective ratings.

Since    

i.e.,    

i.e.,    

Similarly,    

Similarly, load shared by transformer A,

Similarly,    

Total    S=S_A+S_B=V₂I×10^-3 kVA

∴    

2 Unequal Voltage Ratios

For unequal voltage turns ratio, if the primary is connected to the supply, a circulating current will flow in the primary even at no load. The circulating current will be superimposed on the currents drawn by the load when the transformers share a load.

Let V₁ be the primary supply voltage, a₁ be the turns ratio of transformer A, a₂ be the turns ratio of transformer B, Z_A be the equivalent impedance of transformer A (= R_A + jX_A) referred to as secondary, Z_B be the equivalent impedance of transformer B (= R_B + jX_B) referred to as secondary, I_A be the output current of transformer A and I_B be the output current of transformer B.

The induced emf in the secondary of transformer A is

The induced emf in the secondary of transformer B is

Again, V₂ = IZ_L where Z_L is the impedance of the load

∴    V₂=(I_A+I_B)ZL    (1.80)

From Equations (1.78), (1.79) and (1.80), we have

E_A=I_AZ_A+(I_A+I_B)Z_L    (1.81)

and    E_A=I_BZ_B+(I_A+I_B)Z_L    (1.82)

∴ E_A − E_B = I_AZ_A − I_BZ_B

i.e.,    

Substituting I_A from Equation (1.83) in Equation (1.82), we have

i.e.,    

i.e.,    

Similarly,    

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PARALLEL OPERATION OF SINGLE-PHASE TRANSFORMER

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 magnitude and the X/R ratio must be the same to avoid circulating current and operation at different power factors.
• The equivalent impedances must be inversely proportional to the respective kVA ratings.

The above conditions must be satisfied by paralleling transformers of identical ratings of the same make/model. With different kVA ratings of even the same make/model, the effects in steps 1, 2 and 3 may appear in undesirable amounts. Step 2 must be carried out satisfactorily even if steps 1, 3 and 4 are slightly modified.

Figure 1.46 Parallel Operation of Transformers

Figure 1.47 Parallel Operation of Transformers with Incorrect Polarities

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SUMPNER’S TEST

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 W₁ 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 V₂ becomes zero. With the help of voltage regulator fed from source, a voltage is injected to the secondary of the transformers, which is adjusted until the rated secondary current flows. The voltmeter reads a voltage, which is the leakage impedance drop of the two transformers. The reading of W₁ remains unaltered. The wattmeter W₂ reads the total copper (Cu) loss of the two transformers. Although the transformers are not supplying any load current, this test measures the full iron loss as well as copper loss of the transformers. The net input during this test is W₁ + W₂. To measure the temperature rise, the two transformers are kept under rated loss conditions for several hours.

Figure 1.45 Sumpner’s Test

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POLARITY TEST OF A SINGLE-PHASE TRANSFORMER

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 A₁ and A₂, respectively, as shown in Figures 1.44(a) and 1.45(b), respectively. The secondary winding terminals are also marked as a₁ and a₂, shown in Figures 1.44(a) and 1.45(b), respectively. Now a voltmeter is connected across A₂ and a₂. if it measures the difference of E₁ and E₂, A₂ and a₂ are of the same polarity. If it measures the addition of E₁ and E₂, A₂ and a₂ are of opposite polarity.

Figure 1.44 Polarity Test of a single-phase Two Winding Transformer

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ALL-DAY EFFICIENCY

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. Calculate the all-day efficiency of the transformer.

Solution

Full-load output = 200 kVA, Full-load Cu loss = 4 kW, Iron loss = 1.8 kW.

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EFFICIENCY OF A 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 = V₂I_2ftcosθ, where cosθ is the power factor of the load, I_2ft is the secondary current at full load and V₂ is the rated secondary voltage of the transformer.

Full-load copper loss of the transformer = I_2ftR₀₂.

∴     Efficiency of the transformer at full load is given by

Now V₂I_2ft = 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 voltage and frequency, its efficiency varies with the load. Its efficiency increases as the load increases. At a certain load, its efficiency becomes maximum. If the transformer is further loaded, its efficiency starts decreasing. Figure 1.43 shows the plot of efficiency versus load current.

Figure 1.43 Comparison Efficiency and Current

To determine the condition of maximum efficiency, let us assume that the power factor of the load remains constant and the secondary terminal voltage (V₂) is constant. Therefore, efficiency becomes only a function of load current (I₂).

For maximum efficiency

Now,    

∴    

i.e.,  

i.e.,    V₂I₂cosθ+P_i+I₂²R₀₂–V₂I₂cosθ–2I₂²R₀₂=0

i.e.,    P_i=I₂²R₀₂    (1.66)

To achieve maximum efficiency, Iron loss = Cu loss

i.e.,     Constant loss = Variable loss

1.34.1 Load Current at Maximum Efficiency

Let I_2M be the load current at maximum efficiency.

∴    I_2M²R₀₂=P_i

i.e.,    

Let I_2ft be the full-load current.

∴    

i.e.,    

Equation (1.67) shows the load current in terms of full-load current at maximum efficiency.

1.34.2 kVA Supplied at Maximum Efficiency

For constant V₂ the kVA supplied is the function of load current only.

∴    

In general,

where    

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LOSSES IN A TRANSFORMER

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 loss (P_h), which is expressed by

P_h=K_hB_m^1.6fV 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_eB_m²f²t² W per model W       (1.60)

where K_e is the eddy current constant and t is the thickness of the core.

Since the supply voltage V₁ at rated frequency f is always constant, the flux in the core is almost constant. Therefore, flux density in the core remains constant. Hence, hysteresis and eddy current losses are constant at all loads. Thus, the core loss or iron loss is also known as constant loss. The iron loss is denoted by P_i.

Iron loss is reduced using high-grade core material such as silicon steel having very low hysteresis loop for reducing hysteresis loss and laminated core for reducing the eddy current loss.

1.32.2 Copper Loss

The loss of power in the form I²R due to the resistances of the primary and secondary windings is known as copper losses. The copper loss also depends on the magnitude of currents flowing through the windings. The total Cu loss is given by

Copper losses are determined on the basis of R₀₁ or R₀₂ which is determined from short circuit test. Since the standard operating temperature of electrical machine is taken as 75 °C, it is then corrected to 75 °C.

The copper loss due to full-load current is known as full-load Cu loss. If the load on the transformer is half, the Cu loss is known as half-load Cu loss, which is less than the full-load Cu loss. The Cu loss is also known as variable loss.

There are two other losses known as stray loss and dielectric loss. Since leakage field is present in a transformer, eddy currents are induced in the conductors, tanks walls and bolts etc. Stray losses occur due to this eddy currents. Dielectric loss occurs in insulating materials coil and solid insulation. These two losses are small and hence neglected.

Therefore, the total loss of the transformer = Iron loss + Cu loss = P_i + P_Cu

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Kapp’s Regulation

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 (I₂) is taken as a reference phasor. OA representing V₂ is drawn at angle θ₂ with I₂. AB represents I₂R₀₂ drawn parallel to I₂, whereas BC represents I₂X₀₂ drawn perpendicular to AB, i.e., I₂. Here OC represents secondary emf (₀V₂ = E₂) 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 I₂Z₀₂. 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 regulation, which is drawn through O and drawn parallel to AC. The regulation at any power factor angle θ is obtained by extending OA to meet the outer circle at H. The regulation at the required power factor cosθ is represented by AH, which is the intercept between the two circles.

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