12.7 Power in a Balanced System

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12.7 Power in a Balanced System

Let us now consider the power in a balanced three-phase system. We begin by examining the instantaneous power absorbed by the load. This requires that the analysis be done in the time domain. For a Y-connected load, the phase voltages are

$$v_{AN} = \sqrt{2} V_p \cos \omega t, \qquad v_{BN} = \sqrt{2} V_p \cos(\omega t - 120^\circ)$$ $$v_{CN} = \sqrt{2} V_p \cos(\omega t + 120^\circ)$$

(12.41)

where the factor √ 2is necessary because Vp has been defined as the rms value of the phase voltage. If ZY = Z⧸θ, the phase currents lag behind their corresponding phase voltages by θ. Thus,

$$i_a = \sqrt{2} I_p \cos(\omega t - \theta), \qquad i_b = \sqrt{2} I_p \cos(\omega t - \theta - 120^\circ) \quad (12.42)$$ $$i_c = \sqrt{2} I_p \cos(\omega t - \theta + 120^\circ)$$

Example 12.5

where Ip is the rms value of the phase current. The total instantaneous power in the load is the sum of the instantaneous powers in the three phases; that is,

$$p = p_a + p_b + p_c = v_{AN}i_a + v_{BN}i_b + v_{CN}i_c$$

= $2V_pI_p[\cos \omega t \cos(\omega t - \theta)$

• $\cos(\omega t - 120^\circ) \cos(\omega t - \theta - 120^\circ)$
• $\cos(\omega t + 120^\circ) \cos(\omega t - \theta + 120^\circ)$ (12.43)

Applying the trigonometric identity

$$\cos A \cos B = \frac{1}{2} [\cos(A+B) + \cos(A-B)] \quad (12.44)$$

gives

$$p = V_p I_p [3 \cos \theta + \cos(2\omega t - \theta) + \cos(2\omega t - \theta - 240^\circ) + \cos(2\omega t - \theta + 240^\circ)]$$

= $V_p I_p [3 \cos \theta + \cos \alpha + \cos \alpha \cos 240^\circ + \sin \alpha \sin 240^\circ + \cos \alpha \cos 240^\circ - \sin \alpha \sin 240^\circ]$ (12.45)
where $\alpha = 2\omega t - \theta$
= $V_p I_p [3 \cos \theta + \cos \alpha + 2(-\frac{1}{2}) \cos \alpha] = 3V_p I_p \cos \theta$

Thus the total instantaneous power in a balanced three-phase system is constant—it does not change with time as the instantaneous power of each phase does. This result is true whether the load is Y- or ∆- connected. This is one important reason for using a three-phase system to generate and distribute power. We will look into another reason a little later.

Since the total instantaneous po wer is independent of time, the average po wer per phase Pp for either the ∆-connected load or the Y-connected load is p∕3, or

$$P_p = V_p I_p \cos \theta \tag{12.46}$$

and the reactive power per phase is

$$Q_p = V_p I_p \sin \theta \tag{12.47}$$

The apparent power per phase is

$$S_p = V_p I_p \tag{12.48}$$

The complex power per phase is

$$S_p = P_p + jQ_p = V_p I_p^* \tag{12.49}$$

__

where Vp and Ip are the phase voltage and phase current with magnitudes Vp and Ip, respectively. The total average power is the sum of the average powers in the phases:

$$P = P_a + P_b + P_c = 3P_p = 3V_p I_p \cos \theta = \sqrt{3} V_L I_L \cos \theta \qquad (12.50)$$

For a Y-connected load, IL = Ip but VL = √ 3 Vp, whereas for a ∆-connected load, IL = √ 3 Ip but VL = Vp. Thus, Eq. (12.50) applies for both Y-connected and ∆-connected loads. Similarly, the total reactive power is

$$Q = 3V_p I_p \sin \theta = 3Q_p = \sqrt{3} V_L I_L \sin \theta \qquad (12.51)$$

and the total complex power is

$$\mathbf{S} = 3\mathbf{S}_p = 3\mathbf{V}_p \mathbf{I}_p^* = 3I_p^2 \mathbf{Z}_p = \frac{3V_p^2}{\mathbf{Z}_p^*}$$

(12.52)

where Zp = Zp⧸θ is the load impedance per phase. (Zp could be ZY or Z∆.) Alternatively, we may write Eq. (12.52) as

$$\mathbf{S} = P + jQ = \sqrt{3} V_L I_L \underline{\theta} \tag{12.53}$$

Remember that Vp, Ip, VL, and IL are all rms values and that θ is the angle of the load impedance or the angle between the phase voltage and the phase current.

A second major adv antage of three-phase systems for po wer distribution is that the three-phase system uses a lesser amount of wire than the single-phase system for the same line voltage VL and the same absorbed power PL. We will compare these cases and assume in both that the wires are of the same material (e.g., copper with resisti vity ρ), of the same length ℓ, and that the loads are resisti ve (i.e., unity power factor). For the tw o-wire single-phase system in Fig. 12.21(a), IL = PL∕VL, so the power loss in the two wires is

$$P_{\text{loss}} = 2I_L^2 R = 2R \frac{P_L^2}{V_L^2}$$

(12.54)

Figure 12.21

Comparing the power loss in (a) a single-phase system, and (b) a three-phase system.

For the three-wire three-phase system in Fig. 12.21(b), IL′ = |Ia| = |Ib| = |Ic| = PL∕ √ __ 3 VL from Eq. (12.50). The power loss in the three wires is

$$P'_{\text{loss}} = 3(I'_L)^2 R' = 3R' \frac{P_L^2}{3V_L^2} = R' \frac{P_L^2}{V_L^2}$$

(12.55)

Equations (12.54) and (12.55) show that for the same total power delivered PL and same line voltage VL,

$$\frac{P_{\text{loss}}}{P'_{\text{loss}}} = \frac{2R}{R'}$$

\n(12.56)

But from Chapter 2, R = ρℓ∕πr 2 and R′ = ρℓ∕πr′ 2 , where r and r′ are the radii of the wires. Thus,

$$\frac{P_{\text{loss}}}{P'_{\text{loss}}} = \frac{2r'^2}{r^2}$$

\n(12.57)

If the same power loss is tolerated in both systems, then r 2 = 2r′ 2 . The ratio of material required is determined by the number of wires and their volumes, so _____________________ Material for three-phase = 2(π*^r*

$$\frac{\text{Material for single-phase}}{\text{Material for three-phase}} = \frac{2(\pi r^2 \ell)}{3(\pi r^2 \ell)} = \frac{2r^2}{3r^2}$$

\n

$$= \frac{2}{3}(2) = 1.333$$

\n(12.58)

since r 2 = 2r′ 2 . Equation (12.58) shows that the single-phase system uses 33 percent more material than the three-phase system or that the threephase system uses only 75 percent of the material used in the equivalent single-phase system. In other words, considerably less material is needed to deliver the same power with a three-phase system than is required for a single-phase system.

Refer to the circuit in Fig. 12.13 (in Example 12.2). Determine the total average power, reactive power, and complex power at the source and at the load.

Solution:

It is sufficient to consider one phase, as the system is balanced. For phase a,

$$V_p = 110\frac{0}{0} \text{ V}$$

and $I_p = 6.81\frac{1}{21.8^{\circ}} \text{ A}$

Thus, at the source, the complex power absorbed is

$$\mathbf{S}_s = -3\mathbf{V}_p \mathbf{I}_p^* = -3(110/0^\circ)(6.81/21.8^\circ)$$

= -2247/21.8° = -(2087 + j834.6) VA

The real or average power absorbed is −2087 W and the reactive power is −834.6 VAR.

At the load, the complex power absorbed is

$$\mathbf{S}_{L} = 3|\mathbf{I}_{p}|^{2}\mathbf{Z}_{p}$$

where $\mathbf{Z}_{p} = 10 + j8 = 12.81 \underline{ / 38.66^{\circ}}$ and $\mathbf{I}_{p} = \mathbf{I}_{a} = 6.81 \underline{ / -21.8^{\circ}}$ . Hence,

$$\mathbf{S}_{L} = 3(6.81)^{2} 12.81 \underline{ / 38.66^{\circ}} = 1782 \underline{ / 38.66^{\circ}}$$ $$= (1392 + j1113) \text{ VA}$$

The real power absorbed is 1391.7 W and the reactive power absorbed is 1113.3 VAR. The difference between the two complex powers is ab sorbed by the line impedance (5 − j2) Ω. To show that this is the case, we find the complex power absorbed by the line as

$$\mathbf{S}_{\ell} = 3|\mathbf{I}_{p}|^{2}\mathbf{Z}_{\ell} = 3(6.81)^{2}(5 - j2) = 695.6 - j278.3 \text{ VA}$$

which is the difference between Ss and SL; that is, Ss + Sℓ + SL = 0, as expected.

Example 12.6

For the Y-Y circuit in Practice Prob. 12.2, calculate the complex power at the source and at the load.

Answer: −(1054.2 + j843.3) VA, (1012 + j801.6) VA.

A three-phase motor can be regarded as a balanced Y-load. A three-phase motor draws 5.6 kW when the line voltage is 220 V and the line current is 18.2 A. Determine the power factor of the motor.

Solution:

The apparent power is

$$S = \sqrt{3}V_L I_L = \sqrt{3}(220)(18.2) = 6935.13 \text{ VA}$$

Since the real power is

$$P = S \cos \theta = 5600 \text{ W}$$

the power factor is

pf =

$$\cos \theta = \frac{P}{S} = \frac{5600}{6935.13} = 0.8075$$

Calculate the line current required for a 30-kW three-phase motor having a power factor of 0.85 lagging if it is connected to a balanced source with a line voltage of 550 V.

Answer: 37.05 A.

Two balanced loads are connected to a 240-kV rms 60-Hz line, as shown in Fig. 12.22(a). Load 1 dra ws 30 kW at a po wer factor of 0.6 lagging, while load 2 draws 45 kVAR at a power factor of 0.8 lagging. Assuming the abc sequence, determine: (a) the complex, real, and reactive powers absorbed by the combined load, (b) the line currents, and (c) the kVAR rating of the three capacitors ∆-connected in parallel with the load that will raise the power factor to 0.9 lagging and the capacitance of each capacitor.

Solution:

(a) For load 1, given that P1 = 30 kW and cos θ1 = 0.6, then sin θ1 = 0.8. Hence,

$$S_1 = \frac{P_1}{\cos \theta_1} = \frac{30 \text{ kW}}{0.6} = 50 \text{ kVA}$$

and Q1 = S1 sin θ1 = 50(0.8) = 40 kVAR. Thus, the complex power due to load 1 is

$$S_1 = P_1 + jQ_1 = 30 + j40 \text{ kVA}$$

(12.8.1)

Practice Problem 12.7

Example 12.8

Practice Problem 12.6

Example 12.7

Figure 12.22

For Example 12.8: (a) The original balanced loads, (b) the combined load with improved power factor.

For load 2, if

$$Q_2 = 45$$

kVAR and $\cos \theta_2 = 0.8$ , then $\sin \theta_2 = 0.6$ . We find

$$S_2 = \frac{Q_2}{\sin \theta_2} = \frac{45 \text{ kVA}}{0.6} = 75 \text{ kVA}$$

and P2 = S2 cos θ2 75(0.8) = 60 kW. Therefore the complex power due to load 2 is

$$S_2 = P_2 + jQ_2 = 60 + j45 \text{ kVA}$$

(12.8.2)

From Eqs. (12.8.1) and (12.8.2), the total complex power absorbed by the load is

$$S = S_1 + S_2 = 90 + j85 \text{ kVA} = 123.8 \underline{43.36^{\circ}} \text{ kVA} \quad (12.8.3)$$

which has a power factor of cos 43.36° = 0.727 lagging. The real power is then 90 kW, while the reactive power is 85 kVAR.

It will help with the calculations to assume that the loads are wye connected and then to work with the phase voltages, i.e. the magnitude of VAN = (240∕√ __ 3 ) kV.

(b) Since

$$S = 3((240 \text{ kV}/\sqrt{3})I_L) = \sqrt{3}(240 \text{ kV})I_L$$

the magnitude of the line current is

$$I_L = \frac{S}{\sqrt{3}(240,000)}$$

(12.8.4)

We apply this to each load keeping in mind that the magnitude of the phase voltages is equal to (240∕√ __ 3 ) kV. For load 1,

$$I_{L1} = \frac{50,000}{\sqrt{3} \, 240,000} = 120.28 \, \text{mA}$$

Since the power factor is lagging, the line current lags the line voltage by θ1 = cos−1 0.6 = 53.13°. Thus,

$$I_{a1} = 120.28 \sqrt{-53.13^{\circ}}$$

For load 2,

$$I_{L2} = \frac{75,000}{\sqrt{3} \, 240,000} = 180.42 \, \text{mA}$$

and the line current lags the line voltage by θ2 = cos−1 0.8 = 36.87°. Hence,

$$I_{a2} = 180.42 \underline{/-36.87^{\circ}}$$

The total line current is

$$\mathbf{I}_a = \mathbf{I}_{a1} + \mathbf{I}_{a2} = 120.28 \underline{/ -53.13^\circ} + 180.42 \underline{/ -36.87^\circ}$$

= (72.168 - j96.224) + (144.336 - j108.252)
= 216.5 - j204.472 = 297.8 \underline{/ -43.36^\circ} mA

Alternatively, we could obtain the current from the total complex power using Eq. (12.8.4) as

$$I_L = \frac{123,800}{\sqrt{3} \, 240,000} = 297.82 \, \text{mA}$$

and

$$I_a = 297.82 \div 43.36^\circ \text{ mA}$$

which is the same as before. The other line currents, Ib2 and Ica, can be obtained according to the abc sequence (i.e., Ib = 297.82⧸−163.36° mA and Ic = 297.82⧸76.64° mA).

(c) We can find the reactive power needed to bring the power factor to 0.9 lagging using Eq. (11.59),

$$QC = P(\tan \theta_{\text{old}} - \tan \theta_{\text{new}})$$

where P = 90 kW, θold = 43.36°, and θnew = cos−1 0.9 = 25.84°. Hence,

$$Q_C = 90,000 \text{(tan } 43.36^\circ - \text{tan } 25.84^\circ) = 41.4 \text{ kVAR}$$

This reactive power is for the three capacitors. For each capacitor, the rating QC′ = 13.8 kVAR. From Eq. (11.60), the required capacitance is

$$C = \frac{Q'_C}{\omega V_{\text{rms}}^2}$$

Since the capacitors are ∆-connected as shown in Fig. 12.22(b), Vrms in the above formula is the line-to-line or line voltage, which is 240 kV. Thus,

is the line-to-ine of the voltage,

$$C = \frac{13,800}{(2\pi60)(240,000)^2} = 635.5 \text{ pF}$$

Assume that the two balanced loads in Fig. 12.22(a) are supplied by an 840-V rms 60-Hz line. Load 1 is Y-connected with 30 + j40 Ω per phase, while load 2 is a balanced three-phase motor drawing 48 kW at a power factor of 0.8 lagging. Assuming the abc sequence, calculate: (a) the complex power absorbed by the combined load, (b) the kVAR rating of each of the three capacitors ∆-connected in parallel with the load to raise the power factor to unity, and (c) the current drawn from the supply at unity power factor condition.

Answer: (a) 56.47 + j47.29 kVA, (b) 15.76 kVAR, (c) 38.81 A.