12.8 Unbalanced Three-Phase Systems

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12.8 Unbalanced Three-Phase Systems

This chapter would be incomplete without mentioning unbalanced threephase systems. An unbalanced system is caused by two possible situa tions: (1) The source voltages are not equal in magnitude and/or differ in phase by angles that are unequal, or (2) load impedances are unequal. Thus,

An unbalanced system is due to unbalanced voltage sources or an unbalanced load.

To simplify analysis, we will assume balanced source voltages, but an unbalanced load.

Unbalanced three-phase systems are solved by direct application of mesh and nodal analysis. Figure 12.23 sho ws an e xample of an unbal anced three-phase system that consists of balanced source v oltages (not shown in the figure) and an unbalanced Y-connected load (shown in the figure). Since the load is unbalanced, ZA, ZB, and ZC are not equal. The line currents are determined by Ohm’s law as

$$\mathbf{I}_a = \frac{\mathbf{V}_{AN}}{\mathbf{Z}_A}, \qquad \mathbf{I}_b = \frac{\mathbf{V}_{BN}}{\mathbf{Z}_B}, \qquad \mathbf{I}_c = \frac{\mathbf{V}_{CN}}{\mathbf{Z}_C}$$

(12.59)

Figure 12.23 Unbalanced three-phase Y-connected load.

A special technique for handling unbalanced three-phase systems is the method of symmetrical components, which is beyond the scope of this text.

Ia

Practice Problem 12.8

This set of unbalanced line currents produces current in the neutral line, which is not zero as in a balanced system. Applying KCL at node N gives the neutral line current as

$$\mathbf{I}_n = -(\mathbf{I}_a + \mathbf{I}_b + \mathbf{I}_c) \tag{12.60}$$

In a three-wire system where the neutral line is absent, we can still find the line currents Ia, Ib, and Ic using mesh analysis. At node N, KCL must be satisfied so that Ia + Ib + Ic = 0 in this case. The same could be done for an unbalanced ∆-Y, Y-∆, or ∆-∆ three-wire system. As mentioned earlier, in long distance po wer transmission, conductors in mul tiples of three (multiple three-wire systems) are used, with the earth itself acting as the neutral conductor.

To calculate po wer in an unbalanced three-phase system requires that we find the power in each phase using Eqs. (12.46) to (12.49). The total power is not simply three times the power in one phase but the sum of the powers in the three phases.

The unbalanced Y-load of Fig. 12.23 has balanced voltages of 100 V and the acb sequence. Calculate the line currents and the neutral current. Take ZA = 15 Ω, ZB = 10 + j5 Ω, ZC = 6 − j8 Ω.

Solution:

Using Eq. (12.59), the line currents are

$$\mathbf{I}_a = \frac{100/0^{\circ}}{15} = 6.67/0^{\circ} \text{ A}$$ $$\mathbf{I}_b = \frac{100/120^{\circ}}{10 + j5} = \frac{100/120^{\circ}}{11.18/26.56^{\circ}} = 8.94/93.44^{\circ} \text{ A}$$ $$\mathbf{I}_c = \frac{100/-120^{\circ}}{6 - j8} = \frac{100/-120^{\circ}}{10/-53.13^{\circ}} = 10/-66.87^{\circ} \text{ A}$$

Using Eq. (12.60), the current in the neutral line is

$$\mathbf{I}_n = -(\mathbf{I}_a + \mathbf{I}_b + \mathbf{I}_c) = -(6.67 - 0.54 + j8.92 + 3.93 - j9.2)$$

= -10.06 + j0.28 = 10.06/178.4° A

The unbalanced ∆-load of Fig. 12.24 is supplied by balanced line-to-line voltages of 440 V in the positive sequence. Find the line currents. Take Vab as reference.

Answer:

$$39.71 \underline{/-41.06^{\circ}}$$

A, $64.12 \underline{/-139.8^{\circ}}$ A, $70.13 \underline{/74.27^{\circ}}$ A.

Example 12.9

Practice Problem 12.9

Figure 12.24 Unbalanced ∆-load; for Practice Prob. 12.9. For the unbalanced circuit in Fig. 12.25, find: (a) the line currents, (b) the total complex power absorbed by the load, and (c) the total complex power absorbed by the source.

For Example 12.10.

Solution:

(a) We use mesh analysis to find the required currents. For mesh 1,

$$120 \underline{/- 120^{\circ}} - 120 \underline{/0^{\circ}} + (10 + j5)I_1 - 10I_2 = 0$$

or

$$(10+j5)\mathbf{I}_1 - 10\mathbf{I}_2 = 120\sqrt{3}/30^{\circ}$$

(12.10.1)

For mesh 2,

$$120\underline{\bigg/120^{\circ}} - 120\underline{\bigg/}-120^{\circ} + (10 - j10)\mathbf{I}_2 - 10\mathbf{I}_1 = 0$$

or

$$-10I1 + (10 - j10)I2 = 120\sqrt{3} / -90^{\circ}$$

(12.10.2)

Equations (12.10.1) and (12.10.2) form a matrix equation:

2.10.1) and (12.10.2) form a matrix equation:
\n

$$\begin{bmatrix}\n10 + j5 & -10 \\ -10 & 10 - j10\n\end{bmatrix}\n\begin{bmatrix}\n\mathbf{I}_1 \\ \mathbf{I}_2\n\end{bmatrix} = \n\begin{bmatrix}\n120\sqrt{3}/30^\circ \\ 120\sqrt{3}/-90^\circ\n\end{bmatrix}$$

The determinants are

erminants are
\n

$$\Delta = \begin{vmatrix} 10 + j5 & -10 \\ -10 & 10 - j10 \end{vmatrix} = 50 - j50 = 70.71 \underline{/-45^{\circ}}$$

\n

$$\Delta_1 = \begin{vmatrix} 120\sqrt{3}/30^{\circ} & -10 \\ 120\sqrt{3}/-90^{\circ} & 10 - j10 \end{vmatrix} = 207.85(13.66 - j13.66)$$

\n

$$= 4015 \underline{/-45^{\circ}}$$

\n

$$\Delta_2 = \begin{vmatrix} 10 + j5 & 120\sqrt{3}/30^{\circ} \\ -10 & 120\sqrt{3}/-90^{\circ} \end{vmatrix} = 207.85(13.66 - j5)$$

\n

$$= 3023.4 \underline{/-20.1^{\circ}}$$

The mesh currents are

rents are
\n

$$\mathbf{I}_1 = \frac{\Delta_1}{\Delta} = \frac{4015.23/-45^{\circ}}{70.71/-45^{\circ}} = 56.78 \text{ A}$$

\n

$$\mathbf{I}_2 = \frac{\Delta_2}{\Delta} = \frac{3023.4/-20.1^{\circ}}{70.71/-45^{\circ}} = 42.75/24.9^{\circ} \text{ A}$$

The line currents are

$$\mathbf{I}_a = \mathbf{I}_1 = 56.78 \text{ A}, \qquad \mathbf{I}_c = -\mathbf{I}_2 = 42.75 \underline{\smash{\big)}\, - 155.1^\circ} \text{ A}$$

\n

$$\mathbf{I}_b = \mathbf{I}_2 - \mathbf{I}_1 = 38.78 + j18 - 56.78 = 25.46 \underline{\smash{\big)}\, 135^\circ} \text{ A}$$

(b) We can now calculate the complex power absorbed by the load. For phase A,

$$\mathbf{S}_A = \mathbf{I} \mathbf{I}_a \mathbf{I}^2 \mathbf{Z}_A = (56.78)^2 (j5) = j16,120 \text{ VA}$$

For phase B,

$$\mathbf{S}_B = |\mathbf{I}_b|^2 \mathbf{Z}_B = (25.46)^2 (10) = 6480 \text{ VA}$$

For phase C,

$$S_C = I I_c^2 Z_C = (42.75)^2(-j10) = -j18,276 \text{ VA}$$

The total complex power absorbed by the load is

$$S_L = S_A + S_B + S_C = 6480 - j2156$$

VA

(c) We check the result above by finding the power absorbed by the source. For the voltage source in phase a,

$$S_a = -V_{an}I_a^* = -(120/0^\circ)(56.78) = -6813.6$$

VA

For the source in phase b,

$$\mathbf{S}_b = -\mathbf{V}_{bn}\mathbf{I}_b^* = -(120/-120°)(25.46/-135°)$$

= -3055.2/105° = 790 - j2951.1 VA

For the source in phase c,

$$\mathbf{S}_c = -\mathbf{V}_{bn}\mathbf{I}_c^* = -(120/120^\circ)(42.75/155.1^\circ)$$

= -5130/275.1° = -456.03 + j5109.7 VA

The total complex power absorbed by the three-phase source is

$$S_s = S_a + S_b + S_c = -6480 + j2156$$

VA

showing that Ss + SL = 0 and confirming the conservation principle of ac power.

Answer: 128.01⧸ 80.1° A, 76.21⧸−60° A, 85⧸−135° A, 19.36 kW.