P q. Power (ac) 19.1 INTRODUCTION

Transcription

1 19 P q Power (ac) 19.1 INTRODUCTION The dicuion of power in Chapter 14 included only the average power delivered to an ac network. We will now examine the total power equation in a lightly different form and will introduce two additional type of power: apparent and reactive. For any ytem uch a in Fig. 19.1, the power delivered to a load at any intant i defined by the product of the applied voltage and the reulting current; that i, p vi In thi cae, ince v and i are inuoidal quantitie, let u etablih a general cae where v V m in(qt v) and i I m in qt The choen v and i include all poibilitie becaue, if the load i purely reitive, v 0. If the load i purely inductive or capacitive, v 90 or v 90, repectively. For a network that i primarily inductive, v i poitive (v lead i), and for a network that i primarily capacitive, v i negative (i lead v). i p v Load FIG Defining the power delivered to a load.

2 850 POWER (ac) P q Subtituting the above equation for v and i into the power equation will reult in p V m I m in qt in(qt v) If we now apply a number of trigonometric identitie, the following form for the power equation will reult: p VI co v(1 co 2qt) VI in v(in 2qt) (19.1) where V and I are the rm value. The converion from peak value V m and I m to rm value reulted from the operation performed uing the trigonometric identitie. It would appear initially that nothing ha been gained by putting the equation in thi form. However, the uefulne of the form of Eq. (19.1) will be demontrated in the following ection. The derivation of Eq. (19.1) from the initial form will appear a an aignment at the end of the chapter. If Equation (19.1) i expanded to the form p VI co v VI co v co 2 t VI in v in 2 t Average Peak 2x Peak 2x there are two obviou point that can be made. Firt, the average power till appear a an iolated term that i time independent. Second, both term that follow vary at a frequency twice that of the applied voltage or current, with peak value having a very imilar format. In an effort to enure completene and order in preentation, each baic element (R, L, and C) will be treated eparately RESISTIVE CIRCUIT i p R v R For a purely reitive circuit (uch a that in Fig. 19.2), v and i are in phae, and v 0, a appearing in Fig Subtituting v 0 into Eq. (19.1), we obtain FIG Determining the power delivered to a purely reitive load. p R VI co(0 )(1 co 2qt) VI in(0 ) in 2qt VI(1 co 2qt) 0 or p R VI VI co 2qt (19.2) where VI i the average or dc term and VI co 2qt i a negative coine wave with twice the frequency of either input quantity (v or i) and a peak value of VI. Plotting the waveform for p R (Fig. 19.3), we ee that T 1 period of input quantitie T 2 period of power curve p R Note that in Fig the power curve pae through two cycle about it average value of VI for each cycle of either v or i (T 1 2T 2 or f 2 2f 1 ). Conider alo that ince the peak and average value of the power curve are the ame, the curve i alway above the horizontal axi. Thi indicate that the total power delivered to a reitor will be diipated in the form of heat.

3 P q APPARENT POWER 851 T 2 Power delivered to element by ource p R Energy diipated p 1 Energy diipated VI VI (Average) 0 t 1 i t Power returned to ource by element T 1 v FIG Power veru time for a purely reitive load. The power returned to the ource i repreented by the portion of the curve below the axi, which i zero in thi cae. The power diipated by the reitor at any intant of time t 1 can be found by imply ubtituting the time t 1 into Eq. (19.2) to find p 1,aindicated in Fig The average (real) power from Eq. (19.2), or Fig. 19.3, i VI; or, a a ummary, P VI V m m I 2I 2 R V R 2 (watt, W) (19.3) a derived in Chapter 14. The energy diipated by the reitor (W R ) over one full cycle of the applied voltage (Fig. 19.3) can be found uing the following equation: W Pt where P i the average value and t i the period of the applied voltage; that i, W R VIT 1 (joule, J) (19.4) or, ince T 1 1/f 1, W R V f 1 I (joule, J) (19.5) 19.3 APPARENT POWER From our analyi of dc network (and reitive element above), it would eem apparent that the power delivered to the load of Fig i imply determined by the product of the applied voltage and current, with no concern for the component of the load; that i, P VI. However, we found in Chapter 14 that the power factor (co v) of the load will have a pronounced effect on the power diipated, le pronounced for more reactive load. Although the product of the voltage and current i not alway the power delivered, it i a power rating of ignificant ue- V I FIG Defining the apparent power to a load. Z

4 852 POWER (ac) P q fulne in the decription and analyi of inuoidal ac network and in the maximum rating of a number of electrical component and ytem. It i called the apparent power and i repreented ymbolically by S. * Since it i imply the product of voltage and current, it unit are voltampere, for which the abbreviation i VA. It magnitude i determined by S VI (volt-ampere, VA) (19.6) or, ince V IZ and I V Z then S I 2 Z (VA) (19.7) 2 and S V (VA) (19.8) Z The average power to the load of Fig i P VI co v However, S VI Therefore, P S co v (W) (19.9) and the power factor of a ytem F p i F p co v P S (unitle) (19.10) The power factor of a circuit, therefore, i the ratio of the average power to the apparent power. For a purely reitive circuit, we have P VI S and F p co v P 1 S In general, power equipment i rated in volt-ampere (VA) or in kilovolt-ampere (kva) and not in watt. By knowing the volt-ampere rating and the rated voltage of a device, we can readily determine the maximum current rating. For example, a device rated at 10 kva at 200 V ha a maximum current rating of I 10,000 VA/200 V 50 A when operated under rated condition. The volt-ampere rating of a piece of equipment i equal to the wattage rating only when the F p i 1. It i therefore a maximum power diipation rating. Thi condition exit only when the total impedance of a ytem Z v i uch that v 0. The exact current demand of a device, when ued under normal operating condition, could be determined if the wattage rating and power factor were given intead of the volt-ampere rating. However, the power factor i ometime not available, or it may vary with the load. *Prior to 1968, the ymbol for apparent power wa the more decriptive P a.

5 P q INDUCTIVE CIRCUIT AND REACTIVE POWER 853 The reaon for rating ome electrical equipment in kilovolt-ampere rather than in kilowatt can be decribed uing the configuration of Fig The load ha an apparent power rating of 10 kva and a current rating of 50 A at the applied voltage, 200 V. A indicated, the current demand of 70 A i above the rated value and could damage the load element, yet the reading on the wattmeter i relatively low ince the load i highly reactive. In other word, the wattmeter reading i an indication of the watt diipated and may not reflect the magnitude of the current drawn. Theoretically, if the load were purely reactive, the wattmeter reading would be zero even if the load wa being damaged by a high current level. P = VI co θ 0 Wattmeter (kw) 10 [10 kva = (200 V)(50 A)] S = VI Load I = 70 A > 50 A 200 V ± I ± V R X L (X L >> R) FIG Demontrating the reaon for rating a load in kva rather than kw INDUCTIVE CIRCUIT AND REACTIVE POWER For a purely inductive circuit (uch a that in Fig. 19.6), v lead i by 90, a hown in Fig Therefore, in Eq. (19.1), v 90. Subtituting v 90 into Eq. (19.1) yield p L VI co(90 )(1 co 2qt) VI in(90 )(in 2qt) 0 VI in 2qt i p L v FIG Defining the power level for a purely inductive load. Power delivered to element by ource v Energy aborbed T 2 p L i Energy aborbed VI θ = 90 Power returned to ource by element VI Energy returned Energy returned t T 1 FIG The power curve for a purely inductive load.

6 854 POWER (ac) P q or p L VI in 2qt (19.11) where VI in 2qt i a ine wave with twice the frequency of either input quantity (v or i) and a peak value of VI. Note the abence of an average or contant term in the equation. Plotting the waveform for p L (Fig. 19.7), we obtain T 1 period of either input quantity T 2 period of p L curve Note that over one full cycle of p L (T 2 ), the area above the horizontal axi in Fig i exactly equal to that below the axi. Thi indicate that over a full cycle of p L, the power delivered by the ource to the inductor i exactly equal to that returned to the ource by the inductor. The net flow of power to the pure (ideal) inductor i zero over a full cycle, and no energy i lot in the tranaction. The power aborbed or returned by the inductor at any intant of time t 1 can be found imply by ubtituting t 1 into Eq. (19.11). The peak value of the curve VI i defined a the reactive power aociated with a pure inductor. In general, the reactive power aociated with any circuit i defined to be VI in v, a factor appearing in the econd term of Eq. (19.1). Note that it i the peak value of that term of the total power equation that produce no net tranfer of energy. The ymbol for reactive power i Q, and it unit of meaure i the volt-ampere reactive (VAR). * The Q i derived from the quadrature (90 ) relationhip between the variou power, to be dicued in detail in a later ection. Therefore, Q VI in v (volt-ampere reactive, VAR) (19.12) where v i the phae angle between V and I. For the inductor, Q L VI (VAR) (19.13) or, ince V IX L or I V/X L, Q L I 2 X L (VAR) (19.14) 2 V or Q L (VAR) (19.15) X L The apparent power aociated with an inductor i S VI, and the average power i P 0, a noted in Fig The power factor i therefore F p co v P S 0 0 VI *Prior to 1968, the ymbol for reactive power wa the more decriptive P q.

7 P q INDUCTIVE CIRCUIT AND REACTIVE POWER 855 If the average power i zero, and the energy upplied i returned within one cycle, why i reactive power of any ignificance? The reaon i not obviou but can be explained uing the curve of Fig At every intant of time along the power curve that the curve i above the axi (poitive), energy mut be upplied to the inductor, even though it will be returned during the negative portion of the cycle. Thi power requirement during the poitive portion of the cycle require that the generating plant provide thi energy during that interval. Therefore, the effect of reactive element uch a the inductor can be to raie the power requirement of the generating plant, even though the reactive power i not diipated but imply borrowed. The increaed power demand during thee interval i a cot factor that mut be paed on to the indutrial conumer. In fact, mot larger uer of electrical energy pay for the apparent power demand rather than the watt diipated ince the volt-ampere ued are enitive to the reactive power requirement (ee Section 19.6). In other word, the cloer the power factor of an indutrial outfit i to 1, the more efficient i the plant operation ince it i limiting it ue of borrowed power. The energy tored by the inductor during the poitive portion of the cycle (Fig. 19.7) i equal to that returned during the negative portion and can be determined uing the following equation: W Pt where P i the average value for the interval and t i the aociated interval of time. Recall from Chapter 14 that the average value of the poitive portion of a inuoid equal 2(peak value/p) and t T 2 /2. Therefore, W L 2VI T 2 p 2 and W L VI (J) (19.16) p or, ince T 2 1/f 2, where f 2 i the frequency of the p L curve, we have T 2 VI W L p f 2 (J) (19.17) Since the frequency f 2 of the power curve i twice that of the input quantity, if we ubtitute the frequency f 1 of the input voltage or current, Equation (19.17) become However, o that VI VI W L p( 2f 1 ) q1 V IX L Iq 1 L W L (Iq 1L)I q 1 and W L LI 2 (J) (19.18) providing an equation for the energy tored or releaed by the inductor in one half-cycle of the applied voltage in term of the inductance and rm value of the current quared.

8 856 POWER (ac) P q 19.5 CAPACITIVE CIRCUIT i p C v C FIG Defining the power level for a purely capacitive load. For a purely capacitive circuit (uch a that in Fig. 19.8), i lead v by 90, a hown in Fig Therefore, in Eq. (19.1), v 90. Subtituting v 90 into Eq. (19.1), we obtain p C VI co( 90 )(1 co 2qt) VI in( 90 )(in 2qt) 0 VI in 2qt or p C VI in 2qt (19.19) where VI in 2qt i a negative ine wave with twice the frequency of either input (v or i) and a peak value of VI. Again, note the abence of an average or contant term. Power delivered to element by ource θ = 90 i VI T 2 Energy aborbed p C v Energy aborbed t Power returned to ource by element Energy returned VI Energy returned T 1 FIG The power curve for a purely capacitive load. Plotting the waveform for p C (Fig. 19.9) give u T 1 period of either input quantity T 2 period of p C curve Note that the ame ituation exit here for the p C curve a exited for the p L curve. The power delivered by the ource to the capacitor i exactly equal to that returned to the ource by the capacitor over one full cycle. The net flow of power to the pure (ideal) capacitor i zero over a full cycle, and no energy i lot in the tranaction. The power aborbed or returned by the capacitor at any intant of time t 1 can be found by ubtituting t 1 into Eq. (19.19). The reactive power aociated with the capacitor i equal to the peak value of the p C curve, a follow: Q C VI (VAR) (19.20) But, ince V IX C and I V/X C, the reactive power to the capacitor can alo be written Q C I 2 X C (VAR) (19.21)

9 P q THE POWER TRIANGLE 857 and 2 V Q C (VAR) (19.22) X C The apparent power aociated with the capacitor i S VI (VA) (19.23) and the average power i P 0, a noted from Eq. (19.19) or Fig The power factor i, therefore, F p co v P S 0 0 VI The energy tored by the capacitor during the poitive portion of the cycle (Fig. 19.9) i equal to that returned during the negative portion and can be determined uing the equation W Pt. Proceeding in a manner imilar to that ued for the inductor, we can how that W C VI T 2 p (J) (19.24) or, ince T 2 1/f 2, where f 2 i the frequency of the p C curve, VI W C p f 2 (J) (19.25) In term of the frequency f 1 of the input quantitie v and i, VI VI V(Vq 1 C) W C p(2f1 ) q1 q1 and W C CV 2 (J) (19.26) providing an equation for the energy tored or releaed by the capacitor in one half-cycle of the applied voltage in term of the capacitance and rm value of the voltage quared THE POWER TRIANGLE The three quantitie average power, apparent power, and reactive power can be related in the vector domain by S P Q (19.27) with P P 0 Q L Q L 90 Q C Q C 90 For an inductive load, the phaor power S, a it i often called, i defined by S P j Q L

10 858 POWER (ac) P q v S P Q L FIG Power diagram for inductive load. v P S Q C FIG Power diagram for capacitive load. a hown in Fig The 90 hift in Q L from P i the ource of another term for reactive power: quadrature power. For a capacitive load, the phaor power S i defined by S P j Q C a hown in Fig If a network ha both capacitive and inductive element, the reactive component of the power triangle will be determined by the difference between the reactive power delivered to each. If Q L > Q C, the reultant power triangle will be imilar to Fig If Q C > Q L, the reultant power triangle will be imilar to Fig That the total reactive power i the difference between the reactive power of the inductive and capacitive element can be demontrated by conidering Eq. (19.11) and (19.19). Uing thee equation, the reactive power delivered to each reactive element ha been plotted for a erie L-C circuit on the ame et of axe in Fig The reactive element were choen uch that X L > X C. Note that the power curve for each i exactly 180 out of phae. The curve for the reultant reactive power i therefore determined by the algebraic reultant of the two at each intant of time. Since the reactive power i defined a the peak value, the reactive component of the power triangle i a indicated in the figure: I 2 (X L X C ). P L = V L I in 2qt V L I Q T = Q L Q C = V L I V C I = I(V L V C ) = I(IX L IX C ) = I 2 X L I 2 X C = I 2 (X L X C ) V C I qt p C = V C I in 2qt Q T FIG Demontrating why the net reactive power i the difference between that delivered to inductive and capacitive element. j X L X C Z R X L X C An additional verification can be derived by firt conidering the impedance diagram of a erie R-L-C circuit (Fig ). If we multiply each radiu vector by the current quared (I 2 ), we obtain the reult hown in Fig , which i the power triangle for a predominantly inductive circuit. Since the reactive power and average power are alway angled 90 to each other, the three power are related by the Pythagorean theorem; that i, FIG Impedance diagram for a erie R-L-C circuit. S 2 P 2 Q 2 (19.28)

11 P q THE POWER TRIANGLE 859 j I 2 X L = Q L S = I 2 Z Q (reultant) = Q L Q C = I 2 (X L X C ) I 2 X C = Q C P R = I 2 R FIG The reult of multiplying each vector of Fig by I 2 for a erie R-L-C circuit. Therefore, the third power can alway be found if the other two are known. It i particularly intereting that the equation S VI * (19.29) will provide the vector form of the apparent power of a ytem. Here, V i the voltage acro the ytem, and I * i the complex conjugate of the current. Conider, for example, the imple R-L circuit of Fig , where V 10 V 0 10 V 0 I 2 A ZT 3 j The real power (the term real being derived from the poitive real axi of the complex plane) i and the reactive power i P I 2 R (2 A) 2 (3 ) 12 W I V = 10 V 0 R 3 X L 4 FIG Demontrating the validity of Eq. (19.29). Q L I 2 X L (2 A) 2 (4 ) 16 VAR (L) with S P j Q L 12 W j 16 VAR (L) 20 VA a hown in Fig Applying Eq. (19.29) yield S VI * (10 V 0 )(2 A ) 20 VA a obtained above. The angle v aociated with S and appearing in Fig , 19.11, and i the power-factor angle of the network. Since or P VI co v P S co v S = 20 VA v = P = 12 W Q L = 16 VAR FIG The power triangle for the circuit of Fig then F p co v P (19.30) S

12 860 POWER (ac) P q 19.7 THE TOTAL P, Q, AND S The total number of watt, volt-ampere reactive, and volt-ampere, and the power factor of any ytem can be found uing the following procedure: 1. Find the real power and reactive power for each branch of the circuit. 2. The total real power of the ytem (P T ) i then the um of the average power delivered to each branch. 3. The total reactive power (Q T ) i the difference between the reactive power of the inductive load and that of the capacitive load. 4. The total apparent power i S T P T 2 Q 2 T. 5. The total power factor i P T /S T. There are two important point in the above tabulation. Firt, the total apparent power mut be determined from the total average and reactive power and cannot be determined from the apparent power of each branch. Second, and more important, it i not neceary to conider the erie-parallel arrangement of branche. In other word, the total real, reactive, or apparent power i independent of whether the load are in erie, parallel, or erie-parallel. The following example will demontrate the relative eae with which all of the quantitie of interet can be found. EXAMPLE 19.1 Find the total number of watt, volt-ampere reactive, and volt-ampere, and the power factor F p of the network in Fig Draw the power triangle and find the current in phaor form. Load 1 I 0 VAR 100 W Load 2 Load 3 E = 100 V VAR (L) 200 W 1500 VAR (C) 300 W FIG Example Solution: Contruct a table uch a hown in Table TABLE 19.1 Load W VAR VA (L) (2 0 0 ) 2 ( ) (C) (3 0 0 ) 2 ( ) P T 600 Q T 800 (C) S T (6 0 0 ) 2 ( ) Total power diipated Reultant reactive power of network (Note that S T um of each branch: )

13 P q THE TOTAL P, Q, AND S 861 Thu, P F p T 600 W ST 0.6 leading (C) 1000 VA The power triangle i hown in Fig Since S T VI 1000 VA, I 1000 VA/100 V 10 A; and ince v of co v F p i the angle between the input voltage and current: I 10 A The plu ign i aociated with the phae angle ince the circuit i predominantly capacitive. P T = 600 W S T = 1000 VA = co Q T = 800 VAR (C) FIG Power triangle for Example EXAMPLE 19.2 a. Find the total number of watt, volt-ampere reactive, and voltampere, and the power factor F p for the network of Fig R X L E = 100 V 0 I 6 7 X C 15 FIG Example b. Sketch the power triangle. c. Find the energy diipated by the reitor over one full cycle of the input voltage if the frequency of the input quantitie i 60 Hz. d. Find the energy tored in, or returned by, the capacitor or inductor over one half-cycle of the power curve for each if the frequency of the input quantitie i 60 Hz. Solution: E 100 V V 0 a. I ZT 6 j 7 j A V R (10 A )(6 0 ) 60 V V L (10 A )(7 90 ) 70 V V C (10 A )(15 90 ) 150 V P T EI co v (100 V)(10 A) co W I 2 R (10 A) 2 (6 ) 600 W VR 2 (60 V) W R 6 S T EI (100 V)(10 A) 1000 VA I 2 Z T (10 A) 2 (10 ) 1000 VA E 2 (100 V) VA ZT 10 Q T EI in v (100 V)(10 A) in VAR Q C Q L I 2 (X C X L ) (10 A) 2 (15 7 ) 800 VAR

14 862 POWER (ac) P q V 2 C V 2 L (150 V) 2 (70 V) 2 Q T XC XL VAR 700 VAR 800 VAR P T 600 W F p 0.6 leading (C) ST 1000 VA P T = 600 W S T = 1000 VA Q T = 800 VAR (C) FIG Power triangle for Example b. The power triangle i a hown in Fig V R I (60 V)(10 A) c. W R 10 J f1 60 Hz V L I (70 V)(10 A) 700 J d. W L 1.86 J q1 (2p)(60 Hz) 377 V C I (150 V)(10 A) 1500 J W C 3.98 J q1 377 rad/ 377 EXAMPLE 19.3 For the ytem of Fig , E = 208 V W bulb Heating element Motor h = 82% 6.4 kw 5 Hp F p = 0.72 lagging X C R Capacitive load 9 12 FIG Example a. Find the average power, apparent power, reactive power, and F p for each branch. b. Find the total number of watt, volt-ampere reactive, and voltampere, and the power factor of the ytem. Sketch the power triangle. c. Find the ource current I. Solution: a. Bulb: Total diipation of applied power P 1 12(60 W) 720 W Q 1 0 VAR S 1 P VA F p1 1 Heating element: Total diipation of applied power P kw Q 2 0 VAR S 2 P kva F p2 1

15 P q THE TOTAL P, Q,AND S 863 Motor: P o P 5(746 W) h o P i W P 3 Pi h 0.82 F p 0.72 lagging P W P 3 S 3 co v S VA co v 0.72 Alo, v co , o that Q 3 S 3 in v ( VA)(in ) ( VA)(0.694) VAR (L) Capacitive load: E 208 V V 0 I A Z 9 j P 4 I 2 R (13.87 A) W Q 4 I 2 X C (13.87 A) VAR (C) S 4 P 2 4 Q 2 4 ( W ) 2 ( V A R ) VA P W F p 0.6 leading S VA b. P T P 1 P 2 P 3 P W 6400 W W W 13, W Q T Q 1 Q 2 Q 3 Q VAR (L) VAR (C) VAR (L) S T P T 2 Q T 2 (1 3, W ) 2 ( V A R ) 2 13, VA P T 13.4 kw F p lagging ST 13, VA v co Note Fig S T = 13, VA 8.89 P T = 13.4 kw Q T = VAR (L) FIG Power triangle for Example S T 13, VA c. S T EI I A E 208 V Lagging power factor: E lead I by 8.89, and I A 8.89

16 864 POWER (ac) P q Z T R 1.2 X L 1.6 FIG Example EXAMPLE 19.4 An electrical device i rated 5 kva, 100 V at a 0.6 power-factor lag. What i the impedance of the device in rectangular coordinate? Solution: S EI 5000 VA 5000 VA Therefore, I 50 A 100 V For F p 0.6, we have v co Since the power factor i lagging, the circuit i predominantly inductive, and I lag E. Or, for E 100 V 0, I 50 A However, E 100 V 0 Z T j 1.6 I 50 A which i the impedance of the circuit of Fig v v <θ S S <S Q T Q T < Q T FIG Demontrating the impact of power-factor correction on the power triangle of a network POWER-FACTOR CORRECTION The deign of any power tranmiion ytem i very enitive to the magnitude of the current in the line a determined by the applied load. Increaed current reult in increaed power loe (by a quared factor ince P I 2 R) in the tranmiion line due to the reitance of the line. Heavier current alo require larger conductor, increaing the amount of copper needed for the ytem, and, quite obviouly, they require increaed generating capacitie by the utility company. Every effort mut therefore be made to keep current level at a minimum. Since the line voltage of a tranmiion ytem i fixed, the apparent power i directly related to the current level. In turn, the maller the net apparent power, the maller the current drawn from the upply. Minimum current i therefore drawn from a upply when S P and Q T 0. Note the effect of decreaing level of Q T on the length (and magnitude) of S in Fig for the ame real power. Note alo that the power-factor angle approache zero degree and F p approache 1, revealing that the network i appearing more and more reitive at the input terminal. The proce of introducing reactive element to bring the power factor cloer to unity i called power-factor correction. Since mot load are inductive, the proce normally involve introducing element with capacitive terminal characteritic having the ole purpoe of improving the power factor. In Fig (a), for intance, an inductive load i drawing a current I L that ha a real and an imaginary component. In Fig (b), a capacitive load wa added in parallel with the original load to raie the power factor of the total ytem to the unity power-factor level. Note that by placing all the element in parallel, the load till receive the ame terminal voltage and draw the ame current I L. In other word, the load i unaware of and unconcerned about whether it i hooked up a hown in Fig (a) or Fig (b).

17 P q POWER-FACTOR CORRECTION 865 I L I E = E 0 Inductive load L X L > R F p < 1 R E F p = 1 I c X c I L Inductive load L X L > R F p < 1 R Z T = Z T 0 (a) (b) FIG Demontrating the impact of a capacitive element on the power factor of a network. Solving for the ource current in Fig (b): I I C I L j I C (I mag ) I L (R e ) j I L (I mag ) I L (R e ) j [I L (I mag ) I C (I mag )] If X C i choen uch that I C (I mag ) I L (I mag ), then I I L (R e ) j (0) I L (R e ) 0 The reult i a ource current whoe magnitude i imply equal to the real part of the load current, which can be coniderably le than the magnitude of the load current of Fig (a). In addition, ince the phae angle aociated with both the applied voltage and the ource current i the ame, the ytem appear reitive at the input terminal, and all of the power upplied i aborbed, creating maximum efficiency for a generating utility. EXAMPLE 19.5 A 5-hp motor with a 0.6 lagging power factor and an efficiency of 92% i connected to a 208-V, 60-Hz upply. a. Etablih the power triangle for the load. b. Determine the power-factor capacitor that mut be placed in parallel with the load to raie the power factor to unity. c. Determine the change in upply current from the uncompenated to the compenated ytem. d. Find the network equivalent of the above, and verify the concluion. Solution: a. Since 1 hp 746 W, P o 5 hp 5(746 W) 3730 W P o 3730 W and P i (drawn from the line) W h 0.92 Alo, F P co v 0.6 and v co Applying tan v Q L Pi we obtain Q L P i tan v ( W) tan VAR (L) and S P 2 i Q L 2 ( W ) 2 ( V A R ) VA

18 866 POWER (ac) P q S = VA v = P = W Q L = VAR (L) FIG Initial power triangle for the load of Example The power triangle appear in Fig b. A net unity power-factor level i etablihed by introducing a capacitive reactive power level of VAR to balance Q L. Since V 2 Q C XC V 2 (208 V) then X C 2 8 QC VAR (C) 1 1 and C mf 2pfXC (2p)(60 Hz)(8 ) c. At 0.6F p, S VI VA S VA and I A V 208 V At unity F p, S VI VA S VA and I A V 208 V producing a 40% reduction in upply current. d. For the motor, the angle by which the applied voltage lead the current i v co and P EI m co v W, from above, o that P W I m A (a above) E co v (208 V)(0.6) reulting in I m A Therefore, E 208 V 0 Z m Im A j 5.12 a hown in Fig (a). I = I m = A I = A I m = A E = 208 V 0 X L R E = 208 V 0 X C I C = 26 A 8 Z m R X L 8 (a) Motor (b) Motor FIG Demontrating the impact of power-factor correction on the ource current.

19 P q POWER-FACTOR CORRECTION 867 Y The equivalent parallel load i determined from 1 1 Z S S j S a hown in Fig (b). It i now clear that the effect of the 8- inductive reactance can be compenated for by a parallel capacitive reactance of 8 uing a power-factor correction capacitor of 332 mf. Since Y T j X C R j X L R 1 1 j 8 I EY T E (208 V) R A a above In addition, the magnitude of the capacitive current can be determined a follow: E 208 V I C 26 A XC 8 1 EXAMPLE 19.6 a. A mall indutrial plant ha a 10-kW heating load and a 20-kVA inductive load due to a bank of induction motor. The heating element are conidered purely reitive (F p 1), and the induction motor have a lagging power factor of 0.7. If the upply i 1000 V at 60 Hz, determine the capacitive element required to raie the power factor to b. Compare the level of current drawn from the upply. Solution: a. For the induction motor, S VI 20 kva P S co v ( VA)(0.7) W v co and Q L VI in v ( VA)(0.714) VAR (L) The power triangle for the total ytem appear in Fig Note the addition of real power and the reulting S T : S T (2 4 k W ) 2 ( k V A R ) kva S kva with I T E T A 1000 V The deired power factor of 0.95 reult in an angle between S and P of v co P = 10 kw Heating S T S = 20 kva P = 14 kw Induction motor Q L = kvar (L) FIG Initial power triangle for the load of Example 19.6.

20 868 POWER (ac) P q v = P T = 24 kw Q L = 7.9 kvar (L) FIG Power triangle for the load of Example 19.6 after raiing the power factor to changing the power triangle to that of Fig : with tan v Q L Q L P T tan v ( W)(tan ) P T ( W)(0.329) 7.9 kvar (L) The inductive reactive power mut therefore be reduced by Q L Q L kvar (L) 7.9 kvar (L) 6.38 kvar (L) Therefore, Q C 6.38 kvar, and uing we obtain Q C E 2 X C E 2 (10 3 V) 2 X C Q C VAR 1 1 and C mf 2pfXC (2p)(60 Hz)( ) b. S T (2 4 k W ) 2 [ 7.9 k V A R ( L )] kva S kva I T E T A 1000 V The new I T i I T A A (original) FIG Digital wattmeter. (Courtey of Yokogawa Corporation of America) 19.9 WATTMETERS AND POWER-FACTOR METERS The electrodynamometer wattmeter wa introduced in Section 4.4 along with it movement and terminal connection. The ame meter can be ued to meaure the power in a dc or an ac network uing the ame connection trategy; in fact, it can be ued to meaure the wattage of any network with a periodic or a nonperiodic input. The digital diplay wattmeter of Fig employ a ophiticated electronic package to ene the voltage and current level and, through the ue of an analog-to-digital converion unit, diplay the proper digit on the diplay. It i capable of providing a digital readout for ditorted noninuoidal waveform, and it can provide the phae power, total power, apparent power, reactive power, and power factor. When uing a wattmeter, the operator mut take care not to exceed the current, voltage, or wattage rating. The product of the voltage and current rating may or may not equal the wattage rating. In the highpower-factor wattmeter, the product of the voltage and current rating i uually equal to the wattage rating, or at leat 80% of it. For a lowpower-factor wattmeter, the product of the current and voltage rating i much greater than the wattage rating. For obviou reaon, the lowpower-factor meter i ued only in circuit with low power factor (total impedance highly reactive). Typical rating for high-power-factor (HPF) and low-power-factor (LPF) meter are hown in Table Meter of both high and low power factor have an accuracy of 0.5% to 1% of full cale.

21 P q EFFECTIVE RESISTANCE 869 TABLE 19.2 Current Voltage Wattage Meter Rating Rating Rating HPF 2.5 A 150 V 1500/750/ A 300 V LPF 2.5 A 150 V 300/150/ A 300 V A the name implie, power-factor meter are deigned to read the power factor of a load under operating condition. Mot are deigned to be ued on ingle- or three-phae ytem. Both the voltage and the current are typically meaured uing nonintruive method; that i, connection are made directly to the terminal for the voltage meaurement, wherea clamp-on current tranformer are ued to ene the current level, a hown for the power-factor meter of Fig Once the power factor i known, mot power-factor meter come with a et of table that will help define the power-factor capacitor that hould be ued to improve the power factor. Power-factor capacitor are typically rated in kvar, with typical rating extending from 1 to 25 kvar at 240 V and 1 to 50 kvar at 480 V or 600 V EFFECTIVE RESISTANCE The reitance of a conductor a determined by the equation R r(l/a) i often called the dc, ohmic, or geometric reitance. It i a contant quantity determined only by the material ued and it phyical dimenion. In ac circuit, the actual reitance of a conductor (called the effective reitance) differ from the dc reitance becaue of the varying current and voltage that introduce effect not preent in dc circuit. Thee effect include radiation loe, kin effect, eddy current, and hyterei loe. The firt two effect apply to any network, while the latter two are concerned with the additional loe introduced by the preence of ferromagnetic material in a changing magnetic field. FIG Clamp-on power-factor meter. (Courtey of the AEMC Corporation.) Experimental Procedure The effective reitance of an ac circuit cannot be meaured by the ratio V/I ince thi ratio i now the impedance of a circuit that may have both reitance and reactance. The effective reitance can be found, however, by uing the power equation P I 2 R, where R eff I P 2 (19.31) A wattmeter and an ammeter are therefore neceary for meauring the effective reitance of an ac circuit. Radiation Loe Let u now examine the variou loe in greater detail. The radiation lo i the lo of energy in the form of electromagnetic wave during the tranfer of energy from one element to another. Thi lo in energy

22 870 POWER (ac) P q require that the input power be larger to etablih the ame current I, cauing R to increae a determined by Eq. (19.31). At a frequency of 60 Hz, the effect of radiation loe can be completely ignored. However, at radio frequencie, thi i an important effect and may in fact become the main effect in an electromagnetic device uch a an antenna. Φ FIG Demontrating the kin effect on the effective reitance of a conductor. I Skin Effect The explanation of kin effect require the ue of ome baic concept previouly decribed. Remember from Chapter 11 that a magnetic field exit around every current-carrying conductor (Fig ). Since the amount of charge flowing in ac circuit change with time, the magnetic field urrounding the moving charge (current) alo change. Recall alo that a wire placed in a changing magnetic field will have an induced voltage acro it terminal a determined by Faraday law, e N (df/dt). The higher the frequency of the changing flux a determined by an alternating current, the greater the induced voltage will be. For a conductor carrying alternating current, the changing magnetic field urrounding the wire link the wire itelf, thu developing within the wire an induced voltage that oppoe the original flow of charge or current. Thee effect are more pronounced at the center of the conductor than at the urface becaue the center i linked by the changing flux inide the wire a well a that outide the wire. A the frequency of the applied ignal increae, the flux linking the wire will change at a greater rate. An increae in frequency will therefore increae the counter-induced voltage at the center of the wire to the point where the current will, for all practical purpoe, flow on the urface of the conductor. At 60 Hz, the kin effect i almot noticeable. However, at radio frequencie the kin effect i o pronounced that conductor are frequently made hollow becaue the center part i relatively ineffective. The kin effect, therefore, reduce the effective area through which the current can flow, and it caue the reitance of the conductor, given by the equation R r(l/a ), to increae. I E Eddy current Coil Hyterei and Eddy Current Loe A mentioned earlier, hyterei and eddy current loe will appear when a ferromagnetic material i placed in the region of a changing magnetic field. To decribe eddy current loe in greater detail, we will conider the effect of an alternating current paing through a coil wrapped around a ferromagnetic core. A the alternating current pae through the coil, it will develop a changing magnetic flux linking both the coil and the core that will develop an induced voltage within the core a determined by Faraday law. Thi induced voltage and the geometric reitance of the core R C r(l/a) caue current to be developed within the core, i core (e ind /R C ), called eddy current. The current flow in circular path, a hown in Fig , changing direction with the applied ac potential. The eddy current loe are determined by Ferromagnetic core FIG Defining the eddy current loe of a ferromagnetic core. P eddy i 2 eddyr core The magnitude of thee loe i determined primarily by the type of core ued. If the core i nonferromagnetic and ha a high reitivity like wood or air the eddy current loe can be neglected. In term of the frequency of the applied ignal and the magnetic field trength pro-

23 P q EFFECTIVE RESISTANCE 871 duced, the eddy current lo i proportional to the quare of the frequency time the quare of the magnetic field trength: P eddy f 2 B 2 Eddy current loe can be reduced if the core i contructed of thin, laminated heet of ferromagnetic material inulated from one another and aligned parallel to the magnetic flux. Such contruction reduce the magnitude of the eddy current by placing more reitance in their path. Hyterei loe were decribed in Section You will recall that in term of the frequency of the applied ignal and the magnetic field trength produced, the hyterei lo i proportional to the frequency to the 1t power time the magnetic field trength to the nth power: P hy f 1 B n where n can vary from 1.4 to 2.6, depending on the material under conideration. Hyterei loe can be effectively reduced by the injection of mall amount of ilicon into the magnetic core, contituting ome 2% or 3% of the total compoition of the core. Thi mut be done carefully, however, becaue too much ilicon make the core brittle and difficult to machine into the hape deired. EXAMPLE 19.7 a. An air-core coil i connected to a 120-V, 60-Hz ource a hown in Fig The current i found to be 5 A, and a wattmeter reading of 75 W i oberved. Find the effective reitance and the inductance of the coil. Wattmeter E I 120 V 0 CC PC Coil f = 60 Hz FIG The baic component required to determine the effective reitance and inductance of the coil. b. A bra core i then inerted in the coil. The ammeter read 4 A, and the wattmeter 80 W. Calculate the effective reitance of the core. To what do you attribute the increae in value over that of part (a)? c. If a olid iron core i inerted in the coil, the current i found to be 2 A, and the wattmeter read 52 W. Calculate the reitance and the inductance of the coil. Compare thee value to thoe of part (a), and account for the change.

24 872 POWER (ac) P q Solution: P 75 W a. R 3 I 2 (5 A) 2 E 120 V Z T 24 I 5 A X L Z 2 T R 2 (2 4 ) 2 ( 3 ) and X L 2pfL X L or L mh 2pf 377 rad/ P 80 W 80 b. R 5 I 2 (4 A) 2 16 The bra core ha le reluctance than the air core. Therefore, a greater magnetic flux denity B will be created in it. Since P eddy f 2 B 2, and P hy f 1 B n, a the flux denity increae, the core loe and the effective reitance increae. P 52 W 52 c. R 13 I 2 (2 A) 2 4 E 120 V Z T 60 I 2 A X L Z T 2 R 2 (6 0 ) 2 ( 1 3 ) X L L mh 2pf 377 rad/ The iron core ha le reluctance than the air or bra core. Therefore, a greater magnetic flux denity B will be developed in the core. Again, ince P eddy f 2 B 2, and P hy f 1 B n, the increaed flux denity will caue the core loe and the effective reitance to increae. Since the inductance L i related to the change in flux by the equation L N (df/di), the inductance will be greater for the iron core becaue the changing flux linking the core will increae APPLICATIONS FIG Single-phae portable generator. (Courtey of Coleman Powermate, Inc.) Portable Power Generator Even though it may appear that 120 V ac are jut an extenion cord away, there are time uch a in a remote cabin, on a job ite, or while camping that we are reminded that not every corner of the globe i connected to an electric power ource. A you travel further away from large urban communitie, gaoline generator uch a hown in Fig appear in increaing number in hardware tore, lumber yard, and other retail etablihment to meet the need of the local community. Since ac generator are driven by a gaoline motor, they mut be properly ventilated and cannot be run indoor. Uually, becaue of the noie and fume that reult, they are placed a far away a poible and are connected by a long, heavy-duty, weather-reitant extenion cord. Any connection point mut be properly protected and placed to enure that the connection will not it in a puddle of water or be enitive to heavy rain or now. Although there i ome effort involved in etting up generator and contantly enuring that they have enough ga, mot uer will tell you that they are worth their weight in gold.

25 P q APPLICATIONS 873 The vat majority of generator are built to provide between 1750 W and 5000 W of power, although larger unit can provide up to 20,000 W. At firt encounter, you might aume that you can run the world on 5000 W. However, keep in mind that the unit purchaed hould be rated at leat 20% above your expected load becaue of urge current that reult when appliance, motor, tool, etc., are turned on. Remember that even a light bulb develop a large turn-on current due to the cold, low-reitance tate of the filament. If you work too cloely to the rated capacity, experience uch a a evere drop in lighting can reult when an electric aw i turned on almot to the point where it appear that the light will go out altogether. Generator are like any other piece of equipment: If you apply a load that i too heavy, they will hut down. Mot have protective fue or circuit breaker to enure that the excurion above rated condition are monitored and not exceeded beyond reaon. The 20% protective barrier drop the output power from a 5000-W unit to 4000 W, and already we begin to wonder about the load we can apply. Although 4000 W would be ufficient to run a number of 60-W bulb, a TV, an oil burner, and o on, trouble develop whenever a unit i hooked up for direct heating (uch a heater, hair dryer, and clothe dryer). Even microwave at 1200 W command quite a power drain. Pile on a mall electric heater at 1500 W with ix 60-W bulb (360 W), a 250-W TV, and a 250-W oil burner, and then turn on an electric hair dryer at 1500 W uddenly you are very cloe to your maximum of 4000 W. It doen t take long to puh the limit when it come to energy-conuming appliance. Table 19.3 provide a lit of pecification for the broad range of portable gaoline generator. Since the heaviet part of a generator i the gaoline motor, anything over 5 hp get pretty heavy, epecially when you add the weight of the gaoline. Mot good unit providing over 2400 W will have receptacle for 120 V and 220 V at variou current level, with an outlet for 12 V dc. They are alo built o that they tolerate outdoor condition of a reaonable nature and can run continuouly for long period of time. At 120 V, a 5000-W unit can provide a maximum current of about 42 A. TABLE 19.3 Specification for portable gaoline-driven ac generator. Continuou output power W W W Horepower of ga motor 411 hp 514 hp 516 hp Continuou At 120 V: 1525 A At 120 V: 1742 A At 120 V: 1963 A output current At 220 V(3f): 814 A At 220 V(3f): 923 A At 220 V(3f): 1034 A Output voltage 120 V or 120 V or 120 V or 3f: 120 V/220 V 3f: 120 V/220 V 3f: 120 V/220 V Receptacle Fuel tank 1 2 to 2 gallon 1 2 to 3 gallon 1 to 5 gallon gaoline gaoline gaoline Buine Sene Becaue of the cot involved, every large indutrial plant mut continuouly review it electric utility bill to enure it accuracy and to conider way that will keep it in check. A decribed in thi chapter, the

26 874 POWER (ac) P q power factor aociated with the plant a a whole can have a meaurable effect on the drain current and therefore the kva drain on the power line. Power companie are aware of thi problem and actually add a urcharge if the power factor fade below about 0.9. In other word, to enure that the load appear a reitive in nature a poible, the power company i aking every uer to try to enure that hi power factor i between 0.9 and 1 o that the kw demand i very cloe to the kva demand. Power companie do give ome leeway, but they don t want it to get out of hand. Conider the following monthly bill for a fairly large indutrial plant: kwh conumption peak kw demand kw demand kva demand MWh kw kw kva The rate chedule provided by the local power authority i the following: Energy Firt /kwh Next /kwh Additional 8.9 /kwh Power Firt 240 free Additional $12.05/kW Note that thi rate chedule ha an energy cot breakdown and a power breakdown. Thi econd fee i the one enitive to the overall power factor of the plant. The electric bill for the month i then calculated a follow: Cot (450 kwh)(22.3 /kwh) (12 MWh)(17.1 /kwh) [146.2 MWh (12 MWh 450 kwh)](8.9 /kwh) $ $ $11, $14, Before examining the effect of the power fee tructure, we can find the overall power factor of the load for the month with the following ratio taken from the monthly tatement: P 233 kw F p P a 2 50 kva Since the power factor i larger than 0.9, the chance are that there will not be a urcharge or that the urcharge will be minimal. When the power component of the bill i determined, the kva demand i multiplied by the magic number of 0.9 to determine a kw level at thi power factor. Thi kw level i compared to the metered level, and the conumer pay for the higher level. In thi cae, if we multiply the 250 kva by 0.9, we obtain 225 kw which i lightly le than the metered level of 233 kw. However, both level are le than the free level of 240 kw, o there i no additional charge for the power component. The total bill remain at $14, If the kva demand of the bill were 388 kva with the kw demand taying at 233 kw, the ituation would change becaue 0.9 time 388 kva would reult in kw which i much greater than the metered 233 kw. The kw would then be ued to determine the bill a follow:

27 P q COMPUTER ANALYSIS kw 240 kw kw (109.2 kw)($12.05/kw) $ which i ignificant. The total bill can then be determined a follow: Cot $14, $1, $15, Thu, the power factor of the load dropped to 233 kw/388 kva 0.6 which would put an unneceary additional load on the power plant. It i certainly time to conider the power-factor-correction option a decribed in thi text. It i not uncommon to ee large capacitor itting at the point where power enter a large indutrial plant to perform a needed level of power-factor correction. All in all, therefore, it i important to fully undertand the impact of a poor power factor on a power plant whether you omeday work for the upplier or for the conumer COMPUTER ANALYSIS PSpice Power Curve: Reitor The computer analyi will begin with a verification of the curve of Fig which how the in-phae relationhip between the voltage and current of a reitor. The figure how that the power curve i totally above the horizontal axi and that the curve ha a frequency twice the applied frequency and a peak value equal to twice the average value. Firt the imple chematic of Fig mut FIG Uing PSpice to review the power curve for a reitive element in an ac circuit.

28 876 POWER (ac) P q be et up. Then, uing the Time Domain(Tranient) option to get a plot veru time and etting the Run to time to 1 m and the Maximum tep ize to 1 m/ m, we elect OK and then the Run PSpice icon to perform the imulation. Then Trace-Add Trace-V1(R) will reult in the curve appearing in Fig Next, Trace-Add Trace- I(R) will reult in the curve for the current a appearing in Fig Finally the power curve will be plotted uing Trace-Add Trace- V1(R)*I(R) from the baic power equation, and the larger curve of Fig will reult. The original plot had a y-axi that extended from 50 to 50. Since all of the data point are from 20 to 50, the y-axi wa changed to thi new range through Plot-Axi Setting-Y Axi- Uer Defined-( 20 to 50)-OK to obtain the plot of Fig FIG The reulting plot for the power, voltage, and current for the reitor of Fig You can ditinguih between the curve by imply looking at the ymbol next to each quantity at the bottom left of the plot. In thi cae, however, to make it even clearer, a different color wa elected for each trace by clicking on each trace with a right click, electing Propertie, and chooing the color and width of each curve. However, you can alo add text to the creen by electing the ABC icon to obtain the Text Label dialog box, entering the label uch a P(R), and clicking OK. The label can then be placed anywhere on the creen. By electing the Toggle curor key and then clicking on I(R) at the bottom of the creen, we can ue the curor to find the maximum value of the current. At A1 250 m or 1 4 of the total period of the input voltage, the current i a peak at 3.54 A. The peak value of the power curve can then be found by right-clicking on V1(R)*I(R), clicking on the graph, and then finding the peak value (alo available by imply clicking on the Curor

29 P q COMPUTER ANALYSIS 877 Peak icon to the right of the Toggle curor key). It occur at the ame point a the maximum current at a level of 50 W. In particular, note that the power curve how two cycle, while both v R and i R how only one cycle. Clearly, the power curve ha twice the frequency of the applied ignal. Alo note that the power curve i totally above the zero line, indicating that power i being aborbed by the reitor through the entire diplayed cycle. Further, the peak value of the power curve i twice the average value of the curve; that i, the peak value of 50 W i twice the average value of 25 W. The reult of the above imulation can be verified by performing the longhand calculation uing the rm value of the applied voltage. That i, P V R 2 (1 0 V) 2 25 W R 4 Power Curve: Serie R-L-C Circuit The network of Fig , with it combination of element, will now be ued to demontrate that, no matter what the phyical makeup of the network, the average value of the power curve etablihed by the product of the applied voltage and reulting ource current i equal to that diipated by the network. At a frequency of 1 khz, the reactance of the mH inductor will be 8, and the reactance of the capacitor will be 4, reulting in a lagging network. An analyi of the network will reult in Z T 4 j with and E 10 V 0 I Z T A 45 P I 2 R (1.768 A) W FIG Uing PSpice to examine the power ditribution in a erie R-L-C circuit.

30 878 POWER (ac) P q FIG Plot of the applied voltage e, current i R i, and power delivered p e i for the circuit of Fig The three curve of Fig were obtained uing the Simulation Output Variable V(E: ), I(R), and V(E: )*I(R). The Run to time under the Simulation Profile liting wa 20 m, although 1 m wa choen a the Maximum tep ize to enure a good plot. In particular, note that the horizontal axi doe not tart until t 18 m to enure that we are in a teady-tate mode and not in a tranient tage (where the peak value of the waveform could change with time). The horizontal axi wa et to extend from 18 m to 20 m by imply electing Plot-Axi Setting-X Axi-Uer Defined-18m to 20m-OK. Firt note that the current lag the applied voltage a expected for the lagging network. The phae angle between the two i 45 a determined above. Second, be aware that the element were choen o that the ame cale could be ued for the current and voltage. The vertical axi doe not have a unit of meaurement, o the proper unit mut be mentally added for each plot. Uing Plot-Label-Line, a line can be drawn acro the creen at the average power level of 12.5 W. A pencil will appear that can be clicked in place at the left edge at the 12.5-W level. The pencil can then be dragged acro the page to draw the deired line. Once you are at the right edge, remove the preure on the moue, and the line i drawn. The different color for the trace were obtained imply by rightclicking on a trace and reponding to the choice under Propertie. Note that the 12.5-W level i indeed the average value of the power curve. It i intereting to note that the power curve dip below the axi for only a hort period of time. In other word, during the two viible cycle, power i being aborbed by the circuit mot of the time. The