12.4 Balanced Wye-Delta Connection

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12.4 Balanced Wye-Delta Connection

A balanced Y-∆ system consists of a balanced Y-connected source feeding a balanced ∆-connected load.

This is perhaps the most practical three-phase system, as the three-phase sources are usually Y-connected while the three-phase loads are usually ∆-connected.

The balanced Y-delta system is sho wn in Fig. 12.14, where the source is Y-connected and the load is ∆-connected. There is, of course, no neutral connection from source to load for this case. Assuming the positive sequence, the phase voltages are again

$$\mathbf{V}_{an} = V_p \underline{/0^{\circ}} \n\mathbf{V}_{bn} = V_p \underline{/ -120^{\circ}}, \qquad \mathbf{V}_{cn} = V_p \underline{/ +120^{\circ}}.$$

\n(12.19)

As shown in Section 12.3, the line voltages are

$$\mathbf{V}_{ab} = \sqrt{3} V_p \frac{\Delta 30^\circ}{\mathbf{V}_{ca}} = \mathbf{V}_{AB}, \qquad \mathbf{V}_{bc} = \sqrt{3} V_p \frac{\Delta 90^\circ}{\Delta 10^\circ} = \mathbf{V}_{BC}$$

\n

$$\mathbf{V}_{ca} = \sqrt{3} V_p \frac{\Delta 150^\circ}{\Delta 10^\circ} = \mathbf{V}_{CA}$$

\n(12.20)

showing that the line voltages are equal to the voltages across the load impedances for this system configuration. From these voltages, we can obtain the phase currents as

$$\mathbf{I}_{AB} = \frac{\mathbf{V}_{AB}}{\mathbf{Z}_{\Delta}}, \qquad \mathbf{I}_{BC} = \frac{\mathbf{V}_{BC}}{\mathbf{Z}_{\Delta}}, \qquad \mathbf{I}_{CA} = \frac{\mathbf{V}_{CA}}{\mathbf{Z}_{\Delta}}$$

(12.21)

These currents have the same magnitude but are out of phase with each other by 120°.

Figure 12.14 Balanced Y-∆ connection.

Another way to get these phase currents is to apply KVL. For example, applying KVL around loop aABbna gives

$$-\mathbf{V}_{an} + \mathbf{Z}_{\Delta} \mathbf{I}_{AB} + \mathbf{V}_{bn} = 0$$

or

$$\mathbf{I}_{AB} = \frac{\mathbf{V}_{an} - \mathbf{V}_{bn}}{\mathbf{Z}_{\Delta}} = \frac{\mathbf{V}_{ab}}{\mathbf{Z}_{\Delta}} = \frac{\mathbf{V}_{AB}}{\mathbf{Z}_{\Delta}}$$

(12.22)

which is the same as Eq. (12.21). This is the more general way of finding the phase currents.

The line currents are obtained from the phase currents by applying KCL at nodes A, B, and C. Thus,

$$\mathbf{I}_a = \mathbf{I}_{AB} - \mathbf{I}_{CA}, \qquad \mathbf{I}_b = \mathbf{I}_{BC} - \mathbf{I}_{AB}, \qquad \mathbf{I}_c = \mathbf{I}_{CA} - \mathbf{I}_{BC} \quad (12.23)$$

Since ICA = IAB^⧸−240°,

$$\mathbf{I}_a = \mathbf{I}_{AB} - \mathbf{I}_{CA} = \mathbf{I}_{AB} (1 - 1/240^\circ)$$

= $\mathbf{I}_{AB} (1 + 0.5 - j0.866) = \mathbf{I}_{AB} \sqrt{3}/-30^\circ$ (12.24)

showing that the magnitude IL of the line current is √ __ 3 times the magnitude Ip of the phase current, or

$$I^L = \sqrt{3}I_p \tag{12.25}$$

where

$$I_L = |\mathbf{I}_a| = |\mathbf{I}_b| = |\mathbf{I}_c| \tag{12.26}$$

and

$$I_p = |\mathbf{I}_{AB}| = |\mathbf{I}_{BC}| = |\mathbf{I}_{CA}| \tag{12.27}$$

Also, the line currents lag the corresponding phase currents by 30°, assuming the positive sequence. Figure 12.15 is a phasor diagram illustrating the relationship between the phase and line currents.

An alternative way of analyzing the Y-∆ circuit is to transform the ∆-connected load to an equi valent Y-connected load. Using the ∆-Y transformation formula in Eq. (12.8),

$$Z_Y = \frac{Z_{\Delta}}{3}$$

(12.28)

After this transformation, we now have a Y-Y system as in Fig. 12.10. The three-phase Y-∆ system in Fig. 12.14 can be replaced by the singlephase equivalent circuit in Fig. 12.16. This allows us to calculate only the line currents. The phase currents are obtained using Eq. (12.25) and utilizing the fact that each of the phase currents leads the corresponding line current by 30°.

A balanced abc-sequence Y-connected source with Van = 100⧸ 10° V is Example 12.3 connected to a ∆-connected balanced load (8 + j4) Ω per phase. Calculate the phase and line currents.

Phasor diagram illustrating the relationship between phase and line currents.

Figure 12.16 A single-phase equivalent circuit of a balanced Y-∆ circuit.

Solution:

This can be solved in two ways.

■ METHOD 1 The load impedance is

$$\mathbf{Z}_{\Delta} = 8 + j4 = 8.944 / 26.57^{\circ} \,\Omega$$

If the phase voltage Van = 100⧸ 10°, then the line voltage is

$$\mathbf{V}_{ab} = \mathbf{V}_{an} \sqrt{3} / 30^{\circ} = 100 \sqrt{3} / 10^{\circ} + 30^{\circ} = \mathbf{V}_{AB}$$

or

$$V_{AB} = 173.2 \angle 40^{\circ}$$

V

The phase currents are

$$\begin{aligned}\n\text{I}_{AB} &= \frac{\mathbf{V}_{AB}}{\mathbf{Z}_{\Delta}} = \frac{173.2/40^{\circ}}{8.944/26.57^{\circ}} = 19.36/13.43^{\circ} \text{ A} \\ \text{I}_{BC} &= \mathbf{I}_{AB} / -120^{\circ} = 19.36 / -106.57^{\circ} \text{ A} \\ \text{I}_{CA} &= \mathbf{I}_{AB} / +120^{\circ} = 19.36 / 133.43^{\circ} \text{ A}\n\end{aligned}$$

The line currents are

$$\mathbf{I}_a = \mathbf{I}_{AB} \sqrt{3} \underline{/-30^\circ} = \sqrt{3} (19.36) \underline{/13.43^\circ - 30^\circ}$$

= 33.53} \underline{/-16.57^\circ} A

$$\mathbf{I}_b = \mathbf{I}_a \underline{/-120^\circ} = 33.53 \underline{/-136.57^\circ} A$$ $$\mathbf{I}_c = \mathbf{I}_a \underline{/+120^\circ} = 33.53 \underline{/-103.43^\circ} A$$

■ METHOD 2 Alternatively, using single-phase analysis,

$$I_a = \frac{V_{an}}{Z_{\Delta}/3} = \frac{100/10^{\circ}}{2.981/26.57^{\circ}} = 33.54/-16.57^{\circ}$$

A

as above. Other line currents are obtained using the abc phase sequence.

Practice Problem 12.3 One line voltage of a balanced Y-connected source is VAB = 120⧸−20° V. If the source is connected to a ∆-connected load of 20 ^⧸40° Ω, find the phase and line currents. Assume the abc sequence.

Answer: 6⧸−60° A, 6⧸−180° A, 6⧸60° A, 10.392⧸−90° A, 10.392⧸ 150° A, 10.392⧸30° A.

12.5 Balanced Delta-Delta Connection

A balanced ∆-∆ system is one in which both the balanced source and balanced load are ∆-connected.

The source as well as the load may be delta-connected as sho wn in Fig. 12.17. Our goal is to obtain the phase and line currents as usual.

Assuming a positive sequence, the phase voltages for a delta-connected source are

$$\mathbf{V}_{ab} = V_p / \underline{\mathbf{0}^{\circ}}$$ $$\mathbf{V}_{bc} = V_p / \underline{\mathbf{-120}^{\circ}}, \qquad \mathbf{V}_{ca} = V_p / \underline{\mathbf{+120}^{\circ}}$$

(12.29)

The line voltages are the same as the phase voltages. From Fig. 12.17, assuming there is no line impedances, the phase voltages of the de ltaconnected source are equal to the voltages across the impedances; that is,

$$\mathbf{V}_{ab} = \mathbf{V}_{AB}, \qquad \mathbf{V}_{bc} = \mathbf{V}_{BC}, \qquad \mathbf{V}_{ca} = \mathbf{V}_{CA} \tag{12.30}$$

Hence, the phase currents are

$$\mathbf{I}_{AB} = \frac{\mathbf{V}_{AB}}{Z_{\Delta}} = \frac{\mathbf{V}_{ab}}{Z_{\Delta}}, \qquad \mathbf{I}_{BC} = \frac{\mathbf{V}_{BC}}{Z_{\Delta}} = \frac{\mathbf{V}_{bc}}{Z_{\Delta}}$$

\n

$$\mathbf{I}_{CA} = \frac{\mathbf{V}_{CA}}{Z_{\Delta}} = \frac{\mathbf{V}_{ca}}{Z_{\Delta}}$$

\n(12.31)

Because the load is delta-connected just as in the previous section, some of the formulas derived there apply here. The line currents are obtained from the phase currents by applying KCL at nodes A, B, and C, as we did in the previous section:

$$\mathbf{I}_a = \mathbf{I}_{AB} - \mathbf{I}_{CA}, \qquad \mathbf{I}_b = \mathbf{I}_{BC} - \mathbf{I}_{AB}, \qquad \mathbf{I}_c = \mathbf{I}_{CA} - \mathbf{I}_{BC} \tag{12.32}$$

Also, as shown in the last section, each line current lags the corre sponding phase current by 30°; the magnitude IL of the line current is √ __ 3 times the magnitude Ip of the phase current,

$$I_L = \sqrt{3}I_p \tag{12.33}$$

An alternative way of analyzing the ∆-∆ circuit is to convert both the source and the load to their Y equivalents. We already kno w that ZY = Z∆∕3. To convert a ∆-connected source to a Y-connected source, see the next section.

A balanced ∆-connected load ha ving an impedance 20 ^{− j}15 Ω is Example 12.4 connected to a ∆-connected, positi ve-sequence generator ha ving Vab = 330⧸ 0° V. Calculate the phase currents of the load and the line currents.

Solution:

The load impedance per phase is

$$Z_{\Delta} = 20 - j15 = 25 \sqrt{-36.57^{\circ}} \,\Omega$$

Since VAB = Vab, the phase currents are

$$\mathbf{I}_{AB} = \frac{\mathbf{V}_{AB}}{\mathbf{Z}_{\Delta}} = \frac{330/0^{\circ}}{25/-36.87^{\circ}} = 13.2/36.87^{\circ} \text{ A}$$ $$\mathbf{I}_{BC} = \mathbf{I}_{AB}/-120^{\circ} = 13.2/-83.13^{\circ} \text{ A}$$ $$\mathbf{I}_{CA} = \mathbf{I}_{AB}/+120^{\circ} = 13.2/156.87^{\circ} \text{ A}$$

For a delta load, the line current always lags the corresponding phase current by 30° and has a magnitude √ __ 3 times that of the phase current. Hence, the line currents are

$$\mathbf{I}_a = \mathbf{I}_{AB}\sqrt{3}/-30^\circ = (13.2/36.87^\circ)(\sqrt{3}/-30^\circ)$$

= 22.86/6.87° A

$$\mathbf{I}_b = \mathbf{I}_a/-120^\circ = 22.86/-113.13^\circ$$

A

$$\mathbf{I}_c = \mathbf{I}_a/+120^\circ = 22.86/126.87^\circ$$

A

A positive-sequence, balanced ∆-connected source supplies a balanced ∆-connected load. If the impedance per phase of the load is 18 + j12 Ω and Ia = 9.609⧸35° A, find IAB and VAB. Practice Problem 12.4

Answer: 5.548⧸65° A, 120⧸98.69° V.