For Practice Prob. 11.10. 11.6 Complex Power

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For Practice Prob. 11.10. 11.6 Complex Power

Considerable effort has been expended over the years to express power relations as simply as possible. Po wer engineers have coined the term complex power, which they use to find the total effect of parallel loads. Complex power is important in po wer analysis because it contains all the information pertaining to the power absorbed by a given load.

Consider the ac load in Fig. 11.20. Gi ven the phasor form V = Vm^⧸θv and I = Im⧸θi of voltage v(t) and current i(t), the complex power S absorbed by the ac load is the product of the v oltage and the comple x conjugate of the current, or

$$S = \frac{1}{2}VI^*$$

\n(11.40)

assuming the passi ve sign con vention (see Fig. 11.20). In terms of the rms values,

$$S = V_{\rm rms}I_{\rm rms}^* \tag{11.41}$$

where

$$\mathbf{V}_{\rm rms} = \frac{\mathbf{V}}{\sqrt{2}} = V_{\rm rms} \underline{\theta_v} \tag{11.42}$$

and

$$\mathbf{I}_{\rm rms} = \frac{\mathbf{I}}{\sqrt{2}} = I_{\rm rms} / \theta_i \tag{11.43}$$

10 Ω 8 Ω 165 0° V rms ^{+ j}4 ^Ω ‒j6 ^Ω ‒ Practice Problem 11.10

Figure 11.19

Figure 11.20

The voltage and current phasors associated with a load.

When working with the rms values of currents or voltages, we may drop the subscript rms if no confusion will be caused by doing so.

Thus, we may write Eq. (11.41) as

$$\mathbf{S} = V_{\text{rms}} I_{\text{rms}} \underline{\beta_v - \theta_i}$$

= $V_{\text{rms}} I_{\text{rms}} \cos(\theta_v - \theta_i) + j V_{\text{rms}} I_{\text{rms}} \sin(\theta_v - \theta_i)$ (11.44)

This equation can also be obtained from Eq. (11.9). We notice from Eq. (11.44) that the magnitude of the complex power is the apparent power; hence, the comple x power is measured in v olt-amperes (VA). Also, we notice that the angle of the complex power is the power factor angle.

The complex power may be expressed in terms of the load impedance Z. From Eq. (11.37), the load impedance Z may be written as

$$Z = \frac{V}{I} = \frac{V_{\text{rms}}}{I_{\text{rms}}} = \frac{V_{\text{rms}}}{I_{\text{rms}}} \frac{\beta_v - \theta_i}{\beta}$$

(11.45)

Thus, Vrms = ZIrms. Substituting this into Eq. (11.41) gives

$$S = I_{\rm rms}^2 Z = \frac{V_{\rm rms}^2}{Z^*} = V_{\rm rms}I_{\rm rms}^*$$

(11.46)

Since Z = R + jX, Eq. (11.46) becomes

$$S = I_{\text{rms}}^2(R + jX) = P + jQ \tag{11.47}$$

where P and Q are the real and imaginary parts of the complex power; that is,

$$P = \text{Re}(\mathbf{S}) = I_{\text{rms}}^2 R \tag{11.48}$$ $$Q = \text{Im}(\mathbf{S}) = I_{\text{rms}}^2 X \tag{11.49}$$

P is the a verage or real po wer and it depends on the load’ s resistance R. Q depends on the load’ s reactance X and is called the reactive (or quadrature) power.

Comparing Eq. (11.44) with Eq. (11.47), we notice that

$$P = V_{\text{rms}} I_{\text{rms}} \cos(\theta_{\nu} - \theta_{i}), \qquad Q = V_{\text{rms}} I_{\text{rms}} \sin(\theta_{\nu} - \theta_{i}) \tag{11.50}$$

The real power P is the a verage power in watts delivered to a load; it is the only useful po wer. It is the actual po wer dissipated by the load. The reactive power Q is a measure of the energy exchange between the source and the reactive part of the load. The unit of Q is the volt-ampere reactive (VAR) to distinguish it from the real po wer, whose unit is the w att. We know from Chapter 6 that ener gy storage elements neither dissipate nor supply power, but exchange power back and forth with the rest of the network. In the same w ay, the reactive power is being transferred back and forth between the load and the source. It represents a lossless interchange between the load and the source. Notice that:

• 1. Q = 0 for resistive loads (unity pf).
• 2. Q < 0 for capacitive loads (leading pf).
• 3. Q > 0 for inductive loads (lagging pf).

Thus,

Complex power (in VA) is the product of the rms voltage phasor and the complex conjugate of the rms current phasor. As a complex quantity, its real part is real power P and its imaginary part is reactive power Q.

Introducing the complex power enables us to obtain the real and reactive powers directly from voltage and current phasors.

Complex Power =

$$\mathbf{S} = P + jQ = \mathbf{V}_{\text{rms}}(\mathbf{I}_{\text{rms}})^*$$

\n= $|\mathbf{V}_{\text{rms}}| |\mathbf{I}_{\text{rms}}| / \theta_v - \theta_i$
\nApparent Power = $S = |\mathbf{S}| = |\mathbf{V}_{\text{rms}}| |\mathbf{I}_{\text{rms}}| = \sqrt{P^2 + Q^2}$
\nReal Power = $P = \text{Re}(\mathbf{S}) = S \cos(\theta_v - \theta_i)$
\nReactive Power = $Q = \text{Im}(\mathbf{S}) = S \sin(\theta_v - \theta_i)$
\nPower Factor = $\frac{P}{S} = \cos(\theta_v - \theta_i)$

This sho ws ho w the comple x po wer contains all the rele vant po wer information in a given load.

It is a standard practice to represent S, P, and Q in the form of a triangle, known as the power triangle, shown in Fig. 11.21(a). This is similar to the impedance triangle sho wing the relationship between Z, R, and X, illustrated in Fig. 11.21(b). The power triangle has four items—the apparent/complex power, real power, reactive power, and the power factor angle. Given two of these items, the other two can easily be obtained from the triangle. As shown in Fig. 11.22, when S lies in the first quadrant, we have an inductive load and a lagging pf. When S lies in the fourth quadrant, the load is capacitive and the pf is leading. It is also possible for the comple x power to lie in the second or third quadrant. This requires that the load impedance ha ve a negative resistance, which is possible with active circuits.

S contains all power information of a load. The real part of S is the real power P; its imaginary part is the reactive power Q; its magnitude is the apparent power S; and the cosine of its phase angle is the power factor pf.

The voltage across a load is v(t) = 60 cos(ωt − 10°) V and the cur - Example 11.11 rent through the element in the direction of the v oltage drop is i(t) = 1.5 cos( ωt + 50°) A. Find: (a) the comple x and apparent po wers, (b) the real and reacti ve powers, and (c) the po wer factor and the load impedance.

Solution:

(a) For the rms values of the voltage and current, we write

$$\mathbf{V}_{\rm rms} = \frac{60}{\sqrt{2}} \angle 10^{\circ}, \qquad \mathbf{I}_{\rm rms} = \frac{1.5}{\sqrt{2}} \angle 50^{\circ}$$

The complex power is

$$\mathbf{S} = \mathbf{V}_{\rm rms} \mathbf{I}_{\rm rms}^* = \left(\frac{60}{\sqrt{2}} \angle 10^\circ \right) \left(\frac{1.5}{\sqrt{2}} \angle 50^\circ \right) = 45 \angle 60^\circ \text{ VA}$$

The apparent power is

$$S = |\mathbf{S}| = 45 \text{ VA}$$

(b) We can express the complex power in rectangular form as

$$S = 45 \underline{/-60^{\circ}} = 45 [\cos(-60^{\circ}) + j \sin(-60^{\circ})] = 22.5 - j38.97$$

Since S = P + jQ, the real power is

$$P = 22.5 \, \mathrm{W}$$

while the reactive power is

$$Q = -38.97$$

VAR

(c) The power factor is

$$pf = \cos(-60^\circ) = 0.5 \text{ (leading)}$$

It is leading, because the reactive power is negative. The load impedance is

$$Z = \frac{V}{I} = \frac{60/-10^{\circ}}{1.5/+50^{\circ}} = 40/-60^{\circ} \,\Omega$$

which is a capacitive impedance.

For a load, Vrms = 110⧸ 85° V, Irms = 3⧸ 15° A. Determine: (a) the complex and apparent powers, (b) the real and reactive powers, and (c) the power factor and the load impedance. Practice Problem 11.11

Answer: (a) 330 ^⧸ 70° VA, 44 VA, (b) 112.87 W, 310.1 VAR, (c) 0.342 lagging, (12.541 + j34.46) Ω.

Example 11.12 A load Z dra ws 12 kV A at a po wer f actor of 0.856 lagging from a 120-V rms sinusoidal source. Calculate: (a) the average and reactive powers delivered to the load, (b) the peak current, and (c) the load impedance.

Solution:

(a) Given that pf = cos θ = 0.856, we obtain the power angle as θ = cos−1 0.856 = 31.13°. If the apparent power is S = 12,000 VA, then the average or real power is

P = S cos θ = 12,000 × 0.856 = 10.272 kW

while the reactive power is

$$Q = S \sin \theta = 12,000 \times 0.517 = 6.204$$

kVA

(b) Since the pf is lagging, the complex power is

$$S = P + jQ = 10.272 + j6.204$$

kVA

From S = VrmsI*rms, we obtain

$$\text{Im } \mathbf{S} = \mathbf{V}_{\text{rms}} \mathbf{I}_{\text{rms}}^* \text{, we obtain}$$

\n

$$\mathbf{I}_{\text{rms}}^* = \frac{\mathbf{S}}{\mathbf{V}_{\text{rms}}} = \frac{10,272 + j6204}{120/0^{\circ}} = 85.6 + j51.7 \text{ A} = 100/31.13^{\circ} \text{ A}$$

Thus Irms = 100⧸ −31.13° and the peak current is

$$I_m = \sqrt{2}I_{\text{rms}} = \sqrt{2}(100) = 141.4 \text{ A}$$

(c) The load impedance

update

\n

$$\mathbf{Z} = \frac{\mathbf{V}_{\text{rms}}}{\mathbf{I}_{\text{rms}}} = \frac{120/0^{\circ}}{100/-31.13^{\circ}} = 1.2/31.13^{\circ} \ \Omega$$

which is an inductive impedance.

A sinusoidal source supplies 100 kVAR reactive power to load Z = 250⧸ −75° Ω. Determine: (a) the power factor, (b) the apparent power delivered to the load, and (c) the rms voltage.

Answer: (a) 0.2588 leading, (b) 103.53 kVA, (c) 5.087 kV.