13.7 Three-Phase Transformers

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13.7 Three-Phase Transformers

To meet the demand for three-phase po wer transmission, transformer connections compatible with three-phase operations are needed. We can achieve the transformer connections in tw o ways: by connecting three single-phase transformers, thereby forming a so-called transformer bank, or by using a special three-phase transformer . For the same kVA rating, a three-phase transformer is always smaller and cheaper than three single-phase transformers. When single-phase transformers are used, one must ensure that the y have the same turns ratio n to achie ve a balanced three-phase system. There are four standard w ays of con necting three single-phase transformers or a three-phase transformer for three-phase operations: Y-Y, Δ-Δ, Y-Δ, and Δ-Y.

For any of the four connections, the total apparent po wer ST, real power PT, and reactive power QT are obtained as

$$S_T = \sqrt{3} V_L I_L \tag{13.69a}$$ $$P_T = S_T \cos \theta = \sqrt{3} V_L I_L \cos \theta \tag{13.69b}$$ $$Q_T = S_T \sin \theta = \sqrt{3} V_L I_L \sin \theta \qquad (13.69c)$$

where VL and IL are, respecti vely, equal to the line v oltage VLp and the line current ILp for the primary side, or the line v oltage VLs and the line current ILs for the secondary side. Notice From Eq. (13.69) that for each of the four connections, VLs ILs = VLp ILp, since power must be conserved in an ideal transformer.

For the Y-Y connection (Fig. 13.46), the line v oltage VLp at the primary side, the line v oltage VLs on the secondary side, the line current ILp on the primary side, and the line current ILs on the secondary side are related to the transformer per phase turns ratio n according to Eqs. (13.52) and (13.55) as

$$V_{Ls} = nV_{Lp} \tag{13.70a}$$ $$I_{Ls} = \frac{I_{Lp}}{n} \tag{13.70b}$$

For the Δ-Δ connection (Fig. 13.47), Eq. (13.70) also applies for the line voltages and line currents. This connection is unique in the sense

Figure 13.46 Y-Y three-phase transformer connection.

Figure 13.47 Δ-Δ three-phase transformer connection.

that if one of the transformers is removed for repair or maintenance, the other two form an open delta, which can provide three-phase voltages at a reduced level of the original three-phase transformer. __

For the Y-Δ connection (Fig. 13.48), there is a factor of √ 3 arising from the line-phase values in addition to the transformer per phase turns ratio n. Thus,

__

n √ __ 3

$$V_{Ls} = \frac{nV_{Lp}}{\sqrt{3}}$$

(13.71a)

$$I_{Ls} = \frac{\sqrt{3}I_{Lp}}{n} \tag{13.71b}$$

Similarly, for the Δ-Y connection (Fig. 13.49),

$$V_{Ls} = n\sqrt{3} V_{Lp}$$

(13.72a)

$$I_{Ls} = \frac{I_{Lp}}{\sqrt{3}}$$

(13.72b)

Y-Δ three-phase transformer connection.

Figure 13.49 Δ-Y three-phase transformer connection.

Example 13.12

The 42-kVA balanced load depicted in Fig. 13.50 is supplied by a threephase transformer. (a) Determine the type of transformer connections. (b) Find the line voltage and current on the primary side. (c) Determine the kVA rating of each transformer used in the transformer bank. Assume that the transformers are ideal.

Solution:

(a) A careful observation of Fig. 13.50 shows that the primary side is Y-connected, while the secondary side is Δ-connected. Thus, the threephase transformer is Y-Δ, similar to the one shown in Fig. 13.48. (b) Given a load with total apparent power ST = 42 kVA, the turns ratio n  =  5, and the secondary line voltage VLs  =  240 V, we can find the secondary line current using Eq. (13.69a), by

$$I_{Ls} = \frac{S_T}{\sqrt{3} V_{Ls}} = \frac{42,000}{\sqrt{3}(240)} = 101 \text{ A}$$

From Eq. (13.71),

$$I_{Lp} = \frac{n}{\sqrt{3}} I_{Ls} = \frac{5 \times 101}{\sqrt{3}} = 292 \text{ A}$$ $$V_{Lp} = \frac{\sqrt{3}}{n} V_{Ls} = \frac{\sqrt{3} \times 240}{5} = 83.14 \text{ V}$$

(c) Because the load is balanced, each transformer equally shares the total load and since there are no losses (assuming ideal transformers), the kVA rating of each transformer is S = ST∕3 = 14 kVA. Alternatively, the transformer rating can be determined by the product of the phase current and phase voltage of the primary or secondary side. For the pri mary side, for example, we have a delta connection, so that the phase voltage is the same as the line voltage of 240 V, while the phase current is ILp∕√ __ 3  = 58.34 A. Hence, S = 240 × 58.34 = 14 kVA.

A three-phase Δ-Δ transformer is used to step down a line voltage of 625 kV, to supply a plant operating at a line voltage of 12.5 kV. The plant draws 40 MW with a lagging power factor of 85 percent. Find: (a) the current drawn by the plant, (b) the turns ratio, (c) the current on the primary side of the transformer, and (d) the load carried by each transformer. Practice Problem 13.12

Answer: (a) 2.174 kA, (b) 0.02, (c) 43.47 A, (d) 15.69 MVA.