Electric Circuits, 9th Edition P40 docx

366 Sinusoidal Steady-State Power Calculations Example 10.1 Calculating Average and Reactive Power a) Calculate the average power and the reactive power at the terminals of the network shown in Fig. 10.6 if v = 100 cos (tat + 15") V, i = 4sin(W - 15°) A. b) State whether the network inside the box is absorbing or delivering average power. c) State whether the network inside the box is absorbing or supplying magnetizing vars. '-*• + V Figure 10.6 A A pair of terminals used for calculating power. Solution a) Because i is expressed in terms of the sine func- tion, the first step in the calculation for P and Q is to rewrite i as a cosine function: i = 4cos(o>f - 105°) A. We now calculate P and Q directly from Eqs. 10.10 and 10.11. Thus P = -(100)(4) cos [15 - (-105)] = -100 W, Q = -100(4) sin [15 - (-105)] = 173.21 VAR. b) Note from Fig. 10.6 the use of the passive sign convention. Because of this, the negative value of -100 W means that the network inside the box is delivering average power to the terminals. c) The passive sign convention means that, because Q is positive, the network inside the box is absorbing magnetizing vars at its terminals. ^ASSESSMENT PR0BLE Objective 1—Understand ac power concepts, their relationships to one another, and how to calcuate them in a circuit 10.1 For each of the following sets of voltage and current, calculate the real and reactive power in the line between networks A and B in the circuit shown. In each case, state whether the power flow is from A to B or vice versa. Also state whether magnetizing vars are being trans- ferred from A to B or vice versa. a) v = 100 cos (at - 45°) V; i = 20cos(wr + 15°) A. b) v = 100 cos (cot - 45°) V; i = 20cos(&rf + 165°) A. c) v = 100 cos (at - 45°) V; i = 20 cos (w* - 105°) A. d) V = 100 cos at V; i = 20 cos (art + 120°) A. A i —^- + V B Answer: (a)P Q 500 W -866.03 VAR (b) P = -866.03 W Q = 500 VAR (c) P = 500 W Q = 866.03 VAR (d) P = -500 W Q = -866.03 VAR A to B), B to A); B to A), A to B); A to B), AtoB); BtoA), BtoA). 10.2 Compute the power factor and the reactive fac- tor for the network inside the box in Fig. 10.6, whose voltage and current are described in Example 10.1. Hint: Use -i to calculate the power and reac- tive factors. Answer: pf = 0.5 leading; rf = -0.866. NOTE: Also try Chapter Problem 10.1, 10.2 Average and Reactive Power 367 Appliance Ratings Average power is used to quantify the power needs of household appliances. The average power rating and estimated annual kilowatt-hour consumption of some common appliances are presented in Table 10.1. The energy con- sumption values are obtained by estimating the number of hours annually that the appliances are in use. For example, a coffeemaker has an estimated annual consumption of 140 kWh and an average power consumption during operation of 1.2 kW. Therefore a coffeemaker is assumed to be in operation 140/1.2, or 116.67, hours per year, or approximately 19 minutes per day. Example 10.2 uses Table 10.1 to determine whether four common appliances can all be in operation without exceeding the current-carrying capacity of the household. | Making Power The branch circuit supplying the Calculations Involving Household Appliances : outlets in a typical Solution home kitchen is wired with #12 conductor and is protected by either a 20 A fuse or a 20 A circuit breaker. Assume that the following 120 V appli- ances are in operation at the same time: a cof- feemaker, egg cooker, frying pan, and toaster. Will the circuit be interrupted by the protective device? From Table 10.1, the total average power demanded by the four appliances is P = 1200 + 516 + 1196 4- 146 The total current in the pro 4058 efr " 120 ' = 4058 W. ective device is 33.82 A. Yes, the protective device will interrupt the circuit. TABLE 10.1 Annual Energy Requirements of Electric Household Appliances Appliance Food preparation Coffeemaker Dishwasher Egg cooker Frying pan Mixer Oven, microwave (only) Range, with oven Toaster Laundry Clothes dryer Washing machine, automatic Water heater Quick recovery type Comfort conditioning Air conditioner (room) Dehumidifier Fan (circulating) Heater (portable) Average Wattage 1,200 1,201 516 1,196 127 1,450 12,200 1,146 4,856 512 2,475 4,474 860 257 88 1,322 NOTE: Assess your understanding of this Est. kWh Consumed Annually 3 140 165 14 100 2 190 596 39 993 103 4,219 4,811 860 b 377 43 176 Appliance Health and beauty Hair dryer Shaver Sunlamp Home entertainment Radio Television, color, tube type Solid-state type Housewares Clock Vacuum cleaner a) Based on normal usage. When using such factors as the size of the specific Average Wattage 600 15 279 71 240 145 2 630 hese figure appliance, I Est. kWh Consumed Annually 3 25 0.5 16 86 528 320 17 46 ; for projections. he geographical area of use, and individual usage should be taken into considera- tion. Note that the wattages are not additive, since all units are normally not in operation at the same time. b) Based on 1000 hours of operation pe r year. This "igure will vary widely depending on the area and the specific size of the unit. See EEI-Pub #76-2, "Air Conditioning Usage Study," for an estimate for your location. Source: Edison Electric Institute. material by trying Chapter Problem J0.2. 368 Sinusoidal Steady-State Power Calculations K, „cos (tot + 6 r ) | R Figure 10.7 • A sinusoidal voltage applied to the terminals of a resistor. 10.3 The rms Value and Power Calculations In introducing the rms value of a sinusoidal voltage (or current) in Section 9.1, we mentioned that it would play an important role in power calculations. We can now discuss this role. Assume that a sinusoidal voltage is applied to the terminals of a resis- tor, as shown in Fig. 10.7, and that we want to determine the average power delivered to the resistor. From Eq. 10.12, P = TL -R dt h+T Vl 1 cos 2 (cot + $ v )dt (10.18) Comparing Eq. 10.18 with Eq. 9.5 reveals that the average power deliv- ered to R is simply the rms value of the voltage squared divided by R, or R (10.19) If the resistor is carrying a sinusoidal current, say, I,,, cos (cot + <£,-), the average power delivered to the resistor is P = IlnJt. (10.20) The rms value is also referred to as the effective value of the sinu- soidal voltage (or current). The rms value has an interesting property: Given an equivalent resistive load, R, and an equivalent time period, T, the rms value of a sinusoidal source delivers the same energy to R as does a dc source of the same value. For example, a dc source of 100 V delivers the same energy in T seconds that a sinusoidal source of 100 V nns delivers, assuming equivalent load resistances (see Problem 10.12). Figure 10.8 demonstrates this equivalence. Energywise, the effect of the two sources is identical. This has led to the term effective value being used inter- changeably with rms value. The average power given by Eq. 10.10 and the reactive power given by Eq. 10.11 can be written in terms of effective values: V I P = cos (0 V - 0,-) V I = —•=—7= cos (0,, - Bs) V2 V2 K ' = K-ff'eff cos (fl„ -0,.); (10.21) lOOV(rms) R V s = 100V(dc) R Figure 10.8 A The effective value of v, (100 V rms) delivers the same power to R as the dc voltage V s (100 V dc). 10.3 The rms Value and Power Calculations 369 and, by similar manipulation, Q = V cff I cff sm($ v -ed. (10.22) The effective value of the sinusoidal signal in power calculations is so widely used that voltage and current ratings of circuits and equipment involved in power utilization are given in terms of rms values. For exam- ple, the voltage rating of residential electric wiring is often 240 V/120 V service. These voltage levels are the rms values of the sinusoidal voltages supplied by the utility company, which provides power at two voltage lev- els to accommodate low-voltage appliances (such as televisions) and higher voltage appliances (such as electric ranges). Appliances such as electric lamps, irons, and toasters all carry rms ratings on their nameplates. For example, a 120 V, 100 W lamp has a resistance of 120 2 /100, or 144 ft, and draws an rms current of 120/144, or 0.833 A. The peak value of the lamp current is 0.833 V2~, or 1.18 A. The phasor transform of a sinusoidal function may also be expressed in terms of the rms value. The magnitude of the rms phasor is equal to the rms value of the sinusoidal function. If a phasor is based on the rms value, we indicate this by either an explicit statement, a parenthetical "rms" adja- cent to the phasor quantity, or the subscript "eff," as in Eq. 10.21. In Example 10.3, we illustrate the use of rms values for calculating power. Example 10.3 Determining Average Power Delivered to a Resistor by Sinusoidal Voltage a) A sinusoidal voltage having a maximum ampli- tude of 625 V is applied to the terminals of a 50 fl resistor. Find the average power delivered to the resistor. b) Repeat (a) by first finding the current in the resistor. Solution a) The rms value of the sinusoidal voltage is 625/V2, or approximately 441.94 V. From Eq. 10.19, the average power delivered to the 50 Cl resistor is P = (441.94)2 50 = 3906.25 W. b) The maximum amplitude of the current in the resistor is 625/50, or 12.5 A. The rms value of the current is 12.5/V2, or approximately 8.84 A. Hence the average power delivered to the resistor is P = (8.84) 2 50 = 3906.25 W. i/ASSESSMENT PROBLEM Objective 1—Understand ac power concepts, their relationships to one another, and how to calculate them in a drcuit 10.3 The periodic triangular current in Example 9.4, repeated here, has a peak value of 180 mA. Find the average power that this current deliv- ers to a 5 kQ, resistor. Answer: 54 W. NOTE: Also try Chapter Problem 10.15. 370 Sinusoidal Steady-State Power Calculations 10.4 Complex Power Complex power • TABLE 10.2 Three Power Quantities and Their Units Quantity Complex power Average power Reactive power Units volt-amps watts var |.V| - apparent power reactive power P - average power Figure 10.9 • A power triangle. Before proceeding to the various methods of calculating real and reactive power in circuits operating in the sinusoidal steady state, we need to intro- duce and define complex power. Complex power is the complex sum of real power and reactive power, or S = P + jQ. (10.23) As you will see, we can compute the complex power directly from the volt- age and current phasors for a circuit. Equation 10.23 can then be used to compute the average power and the reactive power, because P = !R {S} and<2 = 3{S}. Dimensionally, complex power is the same as average or reactive power. However, to distinguish complex power from either average or reactive power, we use the units volt-amps (VA).Thus we use volt-amps for complex power, watts for average power, and vars for reactive power, as summarized in Table 10.2. Another advantage of using complex power is the geometric interpre- tation it provides. When working with Eq. 10.23, think of P, Q, and \S\ as the sides of a right triangle, as shown in Fig. 10.9. It is easy to show that the angle 6 in the power triangle is the power factor angle 0 I} — 0,. For the right triangle shown in Fig. 10.9, Q tan0 = |. (10.24) But from the definitions of P and Q (Eqs. [10.10] and [10.11 J, respectively), Q _ (V m IJ2)sm(d v -dd P (V m l m /2)cos(0 v -ed tan (0 V - 0,-). (10.25) Therefore, 0 = 0 V - 0, The geometric relations for a right triangle mean also that the four power triangle dimensions (the three sides and the power factor angle) can be determined if any two of the four are known. The magnitude of complex power is referred to as apparent power. Specifically, Apparent power • \S\ = 2 P 2 + Q 2 (10.26) Apparent power, like complex power, is measured in volt-amps. The apparent power, or volt-amp, requirement of a device designed to convert electric energy to a nonelectric form is more important than the average power requirement. Although the average power represents the useful output of the energy-converting device, the apparent power represents the volt-amp capacity required to supply the average power. As you can see from the power triangle in Fig. 10.9, unless the power factor angle is 0° (that is, the device is purely resistive, pf = 1, and Q = 0), the volt-amp capacity required by the device is larger than the average power used by the device. As we will see in Example 10.6, it makes sense to operate devices at a power factor close to 1. Many useful appliances (such as refrigerators, fans, air conditioners, fluorescent lighting fixtures, and washing machines) and most industrial loads operate at a lagging power factor. The power factor of these loads sometimes is corrected either by adding a capacitor to the device itself or 10.5 Power Calculations 371 by connecting capacitors across the line feeding the load; the latter method is often used for large industrial loads. Many of the Chapter Problems give you a chance to make some calculations that correct a lag- ging power factor load and improve the operation of a circuit. Example 10.4 uses a power triangle to calculate several quantities associated with an electrical load. Example 10.4 Calculating Complex Power An electrical load operates at 240 V rms. The load absorbs an average power of 8 kW at a lagging power factor of 0.8. a) Calculate the complex power of the load. b) Calculate the impedance of the load. Solution a) The power factor is described as lagging, so we know that the load is inductive and that the algebraic sign of the reactive power is positive. From the power triangle shown in Fig. 10.10, P = \S\ cos 0, Q = \S\ sin 0. cos 0 - 0.8, sin 0 = 0.6. Now, because Therefore Q = SkW cos0 10 sine 0.8 6 kVAR, = lOkVA, and 5 = 8 + /6 kVA. b) From the computation of the complex power of the load, we see that P = 8 kW. Using Eq. 10.21, = K>f f 4ffcos(0 t; = (240)/ eff (0.8) = 8000 W. 0i) Solving for / efr , / cff = 41.67 A. We already know the angle of the load imped- ance, because it is the power factor angle: 0 = cos -1 (0-8) = 36.87°. We also know that 0 is positive because the power factor is lagging, indicating an inductive load. We compute the magnitude of the load impedance from its definition as the ratio of the magnitude of the voltage to the magnitude of the current: \Z\ = lK-a-1 l/effl 240 41.67 5.76. Hence, Z = 5.76 /36.87° D, = 4.608 + y'3.456 O. Figure 10.10 • A power triangle. 10.5 Power Calculations We are now ready to develop additional equations that can be used to cal- culate real, reactive, and complex power. We begin by combining Eqs. 10.10, 10.11, and 10.23 to get VI VI S = ~Y~cos (0 V - 6,) + j—^—sm(e v - $i) V I r in 1 m [cos (0,,- 0i) +; sin (0,, - 0,)] iggefflrtt o \v m I m /{6 n - 9d. (10.27) 372 Sinusoidal Steady-State Power Calculations If we use the effective values of the sinusoidal voltage and current, Eq. 10.27 becomes S = KffW(0, ~ 0/). (10.28) Equations 10.27 and 10.28 are important relationships in power calcula- tions because they show that if the phasor current and voltage are known at a pair of terminals, the complex power associated with that pair of terminals is either one half the product of the voltage and the conjugate of the cur- rent, or the product of the rms phasor voltage and the conjugate of the rms phasor current. We can show this for the rms phasor voltage and current in Fig. 10.11 as follows: = V ea ei e >-I cii e-! 9 > Complex powers = \ cfi l*^ (10.29) Note that left = h& eJ$i follows from Euler's identity and the trigonomet- ric identities cos(—0) = cos(fl) and sin(-0) = — sin (#): I^e* - / cff cos (Si) + // eff sin (-0,-) = / eff cos (0/) - jl ei{ sin (6i) = Ifo The same derivation technique could be applied to Eq. 10.27 to yield S = -VI*. (10.30) Both Eqs. 10.29 and 10.30 are based on the passive sign convention. If the current reference is in the direction of the voltage rise across the termi- nals, we insert a minus sign on the right-hand side of each equation. To illustrate the use of Eq. 10.30 in a power calculation, let's use the same circuit that we used in Example 10.1. Expressed in terms of the pha- sor representation of the terminal voltage and current, V = 100 /15° V, I = 4/-105° A. Therefore S = -(100 /15°)(4 / + 105°) = 200 /120° •ell •*- 4- V t .ff . Circuit Figure 10.11 • The phasor voltage and current associ- ated with a pair of terminals. = -100 + /173.21 VA. 10.5 Power Calculations 373 Once we calculate the complex power, we can read off both the real and reactive powers, because S = P + jQ. Thus P = -100 W, Q = 173.21 VAR. The interpretations of the algebraic signs on P and Q are identical to those given in the solution of Example 10.1. let + v cff — z 1 Figure 10.12 A The general circuit of Fig. 10.11 replaced with an equivalent impedance. V cf f = Zltf. (10.31) Substituting Eq. 10.31 into Eq. 10.29 yields S - ZI efr Ie[f = Heff| 2 Z = |I eff | 2 (fl + IX) = HeffPi? + /|I cff | 2 X = P + jQ, (10.32) from which P = lleffl 2 ^ = \l\fr (10.33) Q = Ihd 2 * = \l\X- (10.34) In Eq. 10.34, X is the reactance of either the equivalent inductance or equivalent capacitance of the circuit. Recall from our earlier discussion of reactance that it is positive for inductive circuits and negative for capaci- tive circuits. A second useful variation of Eq. 10.29 comes from replacing the cur- rent with the voltage divided by the impedance: S = Veff(^y = ^ = P + jQ- (10.35) Alternate Forms for Complex Power Equations 10.29 and 10.30 have several useful variations. Here, we use the rms value form of the equations, because rms values are the most common type of representation for voltages and currents in power computations. The first variation of Eq. 10.29 is to replace the voltage with the prod- uct of the current times the impedance. In other words, we can always rep- resent the circuit inside the box of Fig. 10.11 by an equivalent impedance, as shown in Fig. 10.12. Then, 374 Sinusoidal Steady-State Power Calculations Note that if Z is a pure resistive element P = and if Z is a pure reactive element, Q = R ' |V eff | 2 (10.36) X (10.37) In Eq. 10.37, X is positive for an inductor and negative for a capacitor. The following examples demonstrate various power calculations in circuits operating in the sinusoidal steady state. Example 10.5 Calculating Average and Reactive Power In the circuit shown in Fig. 10.13, a load having an impedance of 39 + /26 O is fed from a voltage source through a line having an impedance of 1 + /4 O. The effective, or rms, value of the source voltage is 250 V. a) Calculate the load current I L and voltage V L . b) Calculate the average and reactive power deliv- ered to the load. c) Calculate the average and reactive power deliv- ered to the line. d) Calculate the average and reactive power sup- plied by the source. Thus the load is absorbing an average power of 975 W and a reactive power of 650 VAR. in /4 a JTYY\ »_ 6 D 250/0° V(rms) Line Source Figure 10.13 • The circuit for Example 10.5. 3912 II /26 ft; Load Solution a) The line and load impedances are in series across the voltage source, so the load current equals the voltage divided by the total impedance, or I, 250 /0° 40 + /30 = 4 - /3 = 5 /-36.87° A (rms). Because the voltage is given in terms of its rms value, the current also is rms. The load volt- age is the product of the load current and load impedance: V L = (39 + /26)1 L = 234 -/13 = 234.36 /-3.18° V (rms). b) The average and reactive power delivered to the load can be computed using Eq. 10.29. Therefore S = \Jl = (234 -/13)(4 +/3) = 975 + /650 VA. c) The average and reactive power delivered to the line are most easily calculated from Eqs. 10.33 and 10.34 because the line current is known. Thus P = (5) 2 (1) = 25 W, Q = (5)2(4) = 100 VAR. Note that the reactive power associated with the line is positive because the line reactance is inductive. d) One way to calculate the average and reactive power delivered by the source is to add the com- plex power delivered to the line to that delivered to the load, or S = 25 + /100 + 975 + /650 = 1000 +/750VA. The complex power at the source can also be cal- culated from Eq. 10.29: S s = -250IL- 10.5 Power Calculations 375 The minus sign is inserted in Eq. 10.29 whenever the current reference is in the direction of a volt- age rise. Thus 5, = -250(4 + /3) = -(1000 4- /750) VA. The minus sign implies that both average power and magnetizing reactive power are being deliv- ered by the source. Note that this result agrees with the previous calculation of 5, as it must, because the source must furnish all the average and reactive power absorbed by the line and load. Example 10.6 Calculating Power in Parallel Loads The two loads in the circuit shown in Fig. 10.14 can be described as follows: Load 1 absorbs an average power of 8 kW at a leading power factor of 0.8. Load 2 absorbs 20 kVA at a lagging power factor of 0.6. 0.05 n • VW- + /0.50 a v, 250/0° V (rms) I., L Figure 10.14 • The circuit for Example 10.6. a) Determine the power factor of the two loads in parallel. b) Determine the apparent power required to sup- ply the loads, the magnitude of the current, I v , and the average power loss in the transmission line. c) Given that the frequency of the source is 60 Hz, compute the value of the capacitor that would correct the power factor to 1 if placed in parallel with the two loads. Recompute the values in (b) for the load with the corrected power factor. Solution a) All voltage and current phasors in this problem are assumed to represent effective values. Note from the circuit diagram in Fig. 10.14 that I v = l { + I 2 . The total complex power absorbed by the two loads is S = (250)i; = (250)(1, + I 2 ) 8 = (250)1^ + (250)I 2 = s x + s 2 . We can sum the complex powers geometrically, using the power triangles for each load, as shown in Fig. 10.15. By hypothesis, 8000(.6) = 8000 - /6000 VA, S 2 = 20,000(.6) + /20,000(.8) = 12,000 +/16,000 VA. -36.87° 16kVAR 10 kVAR Figure 10.15 • (a) The power triangle for load 1. (b) The power triangle for load 2. (c) The sum of the power triangles. It follows that S = 20,000 + /10,000 VA, and , 20,000 + /10,000 Therefore I v = 80 - /40 = 89.44 /-26.57° A. Thus the power factor of the combined load is pf = cos(0 + 26.57°) = 0.8944 lagging. The power factor of the two loads in parallel is lagging because the net reactive power is positive. b) The apparent power which must be supplied to these loads is \S\ = |20 + /10| = 22.36 kVA. The magnitude of the current that supplies this apparent power is II.sl = 180 - j40| = 89.44 A. . low-voltage appliances (such as televisions) and higher voltage appliances (such as electric ranges). Appliances such as electric lamps, irons, and toasters all carry rms ratings on their nameplates volt-amps. The apparent power, or volt-amp, requirement of a device designed to convert electric energy to a nonelectric form is more important than the average power requirement. Although the average. power triangle to calculate several quantities associated with an electrical load. Example 10.4 Calculating Complex Power An electrical load operates at 240 V rms. The load absorbs an average