12.2 Balanced Three-Phase Voltages

← Back to Fundamentals of Electric Circuits Overview

12.2 Balanced Three-Phase Voltages

Three-phase voltages are often produced with a three-phase ac genera tor (or alternator) whose cross-sectional view is shown in Fig. 12.4. The generator basically consists of a rotating magnet (called the rotor) surrounded by a stationary winding (called the stator). Three separate windings or coils with terminals a-a′, b-b′, and c-c′ are physically placed 120° apart around the stator. Terminals a and a′, for example, stand for one of the ends of coils going into and the other end coming out of the

Figure 12.3 Three-phase four-wire system.

A three-phase generator.

Figure 12.5 The generated voltages are 120° apart from each other.

page. As the rotor rotates, its magnetic field “cuts” the flux from the three coils and induces voltages in the coils. Because the coils are placed 120° apart, the induced voltages in the coils are equal in magnitude but out of phase by 120° (Fig. 12.5). Since each coil can be regarded as a singlephase generator by itself, the three-phase generator can supply power to both single-phase and three-phase loads.

A typical three-phase system consists of three voltage sources connected to loads by three or four wires (or transmission lines). (Threephase current sources are very scarce.) A three-phase system is equivalent to three single-phase circuits. The voltage sources can be either wyeconnected as shown in Fig. 12.6(a) or delta-connected as in Fig. 12.6(b).

Figure 12.6

Three-phase voltage sources: (a) Y-connected source, (b) ∆-connected source.

Let us consider the wye-connected voltages in Fig. 12.6(a) for now. The voltages Van, Vbn, and Vcn are respectively between lines a, b, and c, and the neutral line n. These voltages are called phase voltages. If the voltage sources have the same amplitude and frequency ω and are out of phase with each other by 120°, the voltages are said to be balanced. This implies that

$$\mathbf{V}_{an} + \mathbf{V}_{bn} + \mathbf{V}_{cn} = 0 \tag{12.1}$$ $$|\mathbf{V}_{an}| = |\mathbf{V}_{bn}| = |\mathbf{V}_{cn}| \tag{12.2}$$

Thus,

As a common tradition in power systems, voltage and current in this chapter are in rms values unless otherwise stated.

Balanced phase voltages are equal in magnitude and are out of phase with each other by 120°.

Because the three-phase voltages are 120° out of phase with each other, there are tw o possible combinations. One possibility is sho wn in Fig. 12.7(a) and expressed mathematically as

$$\mathbf{V}_{an} = V_p / \mathbf{0}^{\circ}$$

\n

$$\mathbf{V}_{bn} = V_p / -120^{\circ}$$

\n

$$\mathbf{V}_{cn} = V_p / -240^{\circ} = V_p / +120^{\circ}$$

\n(12.3)

where Vp is the effective or rms value of the phase voltages. This is known as the abc sequence or positive sequence. In this phase sequence, Van leads Vbn, which in turn leads Vcn. This sequence is produced when the rotor in Fig. 12.4 rotates counterclockwise. The other possibility is shown in Fig. 12.7(b) and is given by

$$\mathbf{V}_{an} = V_p \underline{/0^{\circ}}$$

\n

$$\mathbf{V}_{cn} = V_p \underline{/ -120^{\circ}}$$

\n

$$\mathbf{V}_{bn} = V_p \underline{/ -240^{\circ}} = V_p \underline{/ +120^{\circ}}$$

(12.4)

This is called the acb sequence or negative sequence. For this phase sequence, Van leads Vcn, which in turn leads Vbn. The acb sequence is produced when the rotor in Fig. 12.4 rotates in the clockwise direction. It is easy to show that the voltages in Eqs. (12.3) or (12.4) satisfy Eqs. (12.1) and (12.2). For example, from Eq. (12.3),

$$\mathbf{V}_{an} + \mathbf{V}_{bn} + \mathbf{V}_{cn} = V_p \underline{\hspace{0.3cm}} \left( 0^{\circ} + V_p \underline{\hspace{0.3cm}} \right) + V_p \underline{\hspace{0.3cm}} \left( 120^{\circ} + V_p \underline{\hspace{0.3cm}} \right)$$

\n

$$= V_p (1.0 - 0.5 - j0.866 - 0.5 + j0.866) \quad (12.5)$$

\n

$$= 0$$

The phase sequence is the time order in which the voltages pass through their respective maximum values.

The phase sequence is determined by the order in which the phasors pass through a fixed point in the phase diagram.

In Fig. 12.7(a), as the phasors rotate in the counterclockwise direction with frequenc y ω, the y pass through the horizontal axis in a se quence abcabca … . Thus, the sequence is abc or bca or cab. Similarly, for the phasors in Fig. 12.7(b), as the y rotate in the counterclockwise direction, they pass the horizontal axis in a sequence acbacba … . This describes the acb sequence. The phase sequence is important in threephase power distribution. It determines the direction of the rotation of a motor connected to the power source, for example.

Like the generator connections, a three-phase load can be either wye-connected or delta-connected, depending on the end application. Figure 12.8(a) sho ws a wye-connected load, and Fig. 12.8(b) sho ws a delta-connected load. The neutral line in Fig. 12.8(a) may or may not be there, depending on whether the system is four - or three-wire. (And, of course, a neutral connection is topologically impossible for a delta con nection.) A wye- or delta-connected load is said to be unbalanced if the phase impedances are not equal in magnitude or phase.

Figure 12.7 Phase sequences: (a) abc or positive sequence, (b) acb or negative sequence.

The phase sequence may also be regarded as the order in which the phase voltages reach their peak (or maximum) values with respect to time.

Reminder: As time increases, each phasor (or sinor) rotates at an angular velocity ω.

Figure 12.8

Two possible three-phase load configurations: (a) a Y-connected load, (b) a ∆-connected load.

A balanced load is one in which the phase impedances are equal in magnitude and in phase.

For a balanced wye-connected load,

$$\mathbf{Z}_1 = \mathbf{Z}_2 = \mathbf{Z}_3 = \mathbf{Z}_Y \tag{12.6}$$

where ZY is the load impedance per phase. For a balanced delta-connected

$$\mathbf{Z}_a = \mathbf{Z}_b = \mathbf{Z}_c = \mathbf{Z}_{\Delta} \tag{12.7}$$ $$\mathbf{Z}_{\Delta} = 3\mathbf{Z}_{Y} \qquad \text{or} \qquad \mathbf{Z}_{Y} = \frac{1}{3}\mathbf{Z}_{\Delta} \qquad (12.8)$$

Because both the three-phase source and the three-phase load can be either wye- or delta-connected, we have four possible connections:

• Y-Y connection (i.e., Y-connected source with a Y-connected load).
• Y-∆ connection.

load,

Eq. (9.69) that

• ∆-∆ connection.
• ∆-Y connection.

In subsequent sections, we will consider each of these possible con figurations.

It is appropriate to mention here that a balanced delta-connected load is more common than a balanced wye-connected load. This is due to the ease with which loads may be added or remo ved from each phase of a deltaconnected load. This is very difficult with a wye-connected load because the neutral may not be accessible. On the other hand, delta-connected sources are not common in practice because of the circulating current that will result in the delta-mesh if the three-phase voltages are slightly unbalanced.

Example 12.1 Determine the phase sequence of the set of voltages

$$v_{an} = 200 \cos(\omega t + 10^{\circ})$$

vbn = 200 cos(ωt − 230°), vcn = 200 cos(ωt − 110°)

Solution:

The voltages can be expressed in phasor form as

$$\mathbf{V}_{an} = 200 \underline{10^{\circ}} \text{ V}, \qquad \mathbf{V}_{bn} = 200 \underline{100^{\circ}} \text{ V}, \qquad \mathbf{V}_{cn} = 200 \underline{110^{\circ}} \text{ V}$$

We notice that Van leads Vcn by 120° and Vcn in turn leads Vbn by 120°. Hence, we have an acb sequence.

Practice Problem 12.1 Given that Vbn = 220⧸ 30° V, find Van and Vcn, assuming a positive (abc) sequence.

Answer: 220⧸150° V, 220⧸−90° V.

12.3 Balanced Wye-Wye Connection

We begin with the Y-Y system, because any balanced three-phase sys tem can be reduced to an equivalent Y-Y system. Therefore, analysis of this system should be regarded as the key to solving all balanced threephase systems.

A balanced Y-Y system is a three-phase system with a balanced Y-connected source and a balanced Y-connected load.

Consider the balanced four-wire Y-Y system of Fig. 12.9, where a Y-connected load is connected to a Y-connected source. We assume a balanced load so that load impedances are equal. Although the impedance ZY is the total load impedance per phase, it may also be re garded as the sum of the source impedance Zs, line impedance Zℓ, and load impedance ZL for each phase, since these impedances are in series. As illustrated in Fig. 12.9, Zs denotes the internal impedance of the phase winding of the generator; Zℓ is the impedance of the line join ing a phase of the source with a phase of the load; ZL is the impedance of each phase of the load; and Zn is the impedance of the neutral line. Thus, in general

$$\mathbf{Z}_{Y} = \mathbf{Z}_{s} + \mathbf{Z}_{\ell} + \mathbf{Z}_{L} \tag{12.9}$$

A balanced Y-Y system, showing the source, line, and load impedances.

Zs and Zℓ are often very small compared with ZL, so one can assume that ZY = ZL if no source or line impedance is given. In any event, by lump ing the impedances together, the Y-Y system in Fig. 12.9 can be simplified to that shown in Fig. 12.10.

Assuming the positi ve sequence, the phase v oltages (or line-toneutral voltages) are

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

(12.10)

Figure 12.10 Balanced Y-Y connection.

The line-to-line voltages or simply line voltages Vab, Vbc, and Vca are related to the phase voltages. For example,

$$\mathbf{V}_{ab} = \mathbf{V}_{an} + \mathbf{V}_{nb} = \mathbf{V}_{an} - \mathbf{V}_{bn} = V_p \left( \frac{0^\circ}{\rho} - V_p \right) - 120^\circ$$

= $V_p \left( 1 + \frac{1}{2} + j \frac{\sqrt{3}}{2} \right) = \sqrt{3} V_p / 30^\circ$ (12.11a)

Similarly, we can obtain

$$\mathbf{V}_{bc} = \mathbf{V}_{bn} - \mathbf{V}_{cn} = \sqrt{3} V_p \sqrt{-90^\circ}$$

(12.11b)

$$\mathbf{V}_{ca} = \mathbf{V}_{cn} - \mathbf{V}_{an} = \sqrt{3} V_p \underline{/-210^\circ}$$

(12.11c)

Thus, the magnitude of the line voltages VL is √ 3 times the magnitude of the phase voltages Vp, or

$$V_L = \sqrt{3} V_P \tag{12.12}$$

where

$$V_p = |\mathbf{V}_{an}| = |\mathbf{V}_{bn}| = |\mathbf{V}_{cn}| \tag{12.13}$$

and

$$V_L = |\mathbf{V}_{ab}| = |\mathbf{V}_{bc}| = |\mathbf{V}_{ca}| \tag{12.14}$$

Also the line voltages lead their corresponding phase voltages by 30°. Figure 12.11(a) illustrates this. Figure 12.11(a) also shows how to determine Vab from the phase voltages, while Fig. 12.11(b) shows the same for the three line voltages. Notice that Vab leads Vbc by 120°, and Vbc leads Vca by 120°, so that the line voltages sum up to zero as do the phase voltages.

Applying KVL to each phase in Fig. 12.10, we obtain the line cur rents as

$$\mathbf{I}_{a} = \frac{\mathbf{V}_{an}}{\mathbf{Z}_{Y}}, \qquad \mathbf{I}_{b} = \frac{\mathbf{V}_{bn}}{\mathbf{Z}_{Y}} = \frac{\mathbf{V}_{an} / -120^{\circ}}{\mathbf{Z}_{Y}} = \mathbf{I}_{a} / -120^{\circ} \tag{12.15}$$

\n

$$\mathbf{I}_{c} = \frac{\mathbf{V}_{cn}}{\mathbf{Z}_{Y}} = \frac{\mathbf{V}_{an} / -240^{\circ}}{\mathbf{Z}_{Y}} = \mathbf{I}_{a} / -240^{\circ}$$

We can readily infer that the line currents add up to zero,

$$\mathbf{I}_a + \mathbf{I}_b + \mathbf{I}_c = 0 \tag{12.16}$$

so that

$$\mathbf{I}_n = -(\mathbf{I}_a + \mathbf{I}_b + \mathbf{I}_c) = 0 \tag{12.17a}$$

or

$$\mathbf{V}_{nN} = \mathbf{Z}_n \mathbf{I}_n = 0 \tag{12.17b}$$

that is, the voltage across the neutral wire is zero. The neutral line can thus be removed without affecting the system. In fact, in long distance power transmission, conductors in multiples of three are used with the earth itself acting as the neutral conductor. Power systems designed in this way are well grounded at all critical points to ensure safety.

While the line current is the current in each line, the phase current is the current in each phase of the source or load. In the Y-Y system, the line current is the same as the phase current. We will use single subscripts

Vnb Vab = Van + Vnb

Vcn

Vbn

Phasor diagrams illustrating the relationship between line voltages and phase voltages.

(b)

Vbc

From Ia, we use the phase sequence to obtain other line currents. Thus, as long as the system is balanced, we need only analyze one phase. We may do this even if the neutral line is absent, as in the three-wire system.

Calculate the line currents in the three-wire Y-Y system of Fig. 12.13. Example 12.2

Figure 12.13

Three-wire Y-Y system; for Example 12.2.

Solution:

The three-phase circuit in Fig. 12.13 is balanced; we may replace it with its single-phase equivalent circuit such as in Fig. 12.12. We obtain Ia from the single-phase analysis as

$$\mathbf{I}_a = \frac{\mathbf{V}_{an}}{\mathbf{Z}_Y}$$

where ZY = (5 − j2) + (10 + j8) = 15 + j6 = 16.155⧸21.8°. Hence,

$$\mathbf{I}_a = \frac{110/0^{\circ}}{16.155/21.8^{\circ}} = 6.81/-21.8^{\circ} \text{ A}$$

In as much as the source voltages in Fig. 12.13 are in positive sequence, the line currents are also in positive sequence:

$$\mathbf{I}_b = \mathbf{I}_a \underline{/-120^\circ} = 6.81 \underline{/-141.8^\circ} \text{ A}$$ $$\mathbf{I}_c = \mathbf{I}_a \underline{/-240^\circ} = 6.81 \underline{/-261.8^\circ} \text{ A} = 6.81 \underline{/-98.2^\circ} \text{ A}$$

A Y-connected balanced three-phase generator with an impedance of 0.4 + j0.3 Ω per phase is connected to a Y-connected balanced load with an impedance of 24 + j19 Ω per phase. The line joining the generator and the load has an impedance of 0.6 + j0.7 Ω per phase. Assuming a positive sequence for the source voltages and that Van = 120⧸30° V, find: (a) the line voltages, (b) the line currents. Practice Problem 12.2

Answer: (a) 207.8⧸60° V, 207.8⧸−60° V, 207.8⧸−180° V, (b) 3.75⧸−8.66° A, 3.75⧸−128.66° A, 3.75⧸ 111.34° A.