Appendix A
Phasor Analysis
A.l Vector Representation of Sinusoids: The Concept of Phasors
Consider the harmonic, or sinusoidal, time functions
u(t) = urnax sin wt; (A.l) v(t) = vrnax sin(wt - a). These functions may represent steady-state voltages, currents, velocities, or any other physical variables. Graphically, the functions may be depicted (Figure A.I) as the vertical projections of the rotating vectors U and V. These vectors have the lengths urnax and vrnax ' respectively, and rotate with the angular velocity, w, [rad/s]. The algebraic sum
wet) = u(t) + vet) (A.2) is also sinusoidal and can be represented by the projection of a third rotating vec• tor, W. This vector is obtained as the vectorial sum of the vectors U and V as shown in Figure A. I. It is clear that analysis of sinusoidally varying quantities can be carried out by a geometric study of vectors. The studies can be further simplified by "freezing" the rotating vectors in a given position. Such "time-frozen" vectors are often referred to as phasors. One may think of a phasor as a snapshot of a rotating vector.
Example A-I Find the sum w of the two sinusoids
u(t) = 6 sinwt, (A.Ll) v(t) = 5 sin(wt - 37°). Solution: The two vectors U and V are shown in an x-y coordinate system (Fig• ure A.2). For simplicity they have been "frozen" at the instant when the U vector
405 406 Appendix A Phasor Analysis
FigureA.l
y
u x ------v ---- w
FigureA.2 A.1 Vector Representation of Sinusoids: The Concept of Phasors 407
coincides with the x axis. According to our definition, the rotating vectors have been turned into phasors. The U phasor, which coincides with the x axis, is referred to as the reference phasor. If the sum of the two phasors U and V is the phasor, W we first determine its com• ponents, Wx and Wy along the x and y axes, respectively, and obtain
Wx = 6 + 5 cos 37° = 9.993, (A.1.2) Wy = 0 - 5 sin37° = -3.009. Iwi is given by Iwi = \1'9.993 2 + 3.009 2 = 10.436. (A. 1.3) The angle I W of phasor, W relative to the reference phasor is
~= tan-I ( -Wy) = tan-I (-3.009) = -16.76°. (A. 1.4) Wx 9.993 For the time function, wet), we can write wet) = 10.436 sin(wt - 16.76°). (A. 1.5)
ExampleA.2 A sinusoidal current,
i = imax sin wt [A], (A.2.1) is injected into a circuit consisting of a resistor, R, in series with an inductor, L, and a capacitor, C. Describe the steady-state voltages v R , v L , and Vc defined in Figure A.3. Solution: The voltage across the resistor, according to Ohm's law, equation (3.23) is
V R = Ri = Rimax sin wt [V]. (A.2.2) The voltage across the inductor, from equation (3.70), is
i L C -=----A.M.J\r--.rmnI~---1~ - y vc
, y v FigureA.3 408 Appendix A Phasor Analysis
di vL = L- [V], dt
= wLimax cos wt [V], (A.2.3)
= wLimax sin (wt + 90°) [V]. The voltage across the capacitor, from equation (3.11) is
1 1 Vc = CQ = C f idt
= -1 f i sin wt dt = - -imax cos wt (A.2.4) C max wC
= imax sin (wt - 90°) [V]. wC
Summary: The resistor voltage vR is a sinusoid and in phase with the current i. The inductor voltage vL and capacitor voltage Vc are likewise sinusoidal but, respectively, leading and lagging the current by 90°. If we represent the above time variables by the phasors VR , VL , Vc ' and I, respectively, we obtain the pha• sor diagram shown in Figure A.4.
y
I x
FigureA.4 A.l Vector Representation of Sinusoids: The Concept of Phasors 409
ExampleA.3 For the circuit in Example A2, we have the following numerical values:
imax = 10 [A], R = 20 [0], w = 377 [rad/s] (60 Hz), (A3.1) L = 0.040 [H], e = 300 [JLF]' Find the amplitudes of the voltages across R, L, and e, and also the total voltage across the circuit. Solution: From equation (A2.2) we get,
vR max = Rimax = 20 . 10 = 200 [V]. (A3.2) From equation (A2.3),
vLmax = wLimax = 377 . 0.040 . 10 = 150.8 [V]. (A3.3) From equation (A2.4),
v = imax = 10 = 88.4 [V]. (A3.4) C max we 377 . 300 . 10-6 The total voltage v across the circuit in Figure A3 is the sum of the individual voltages. By representing v by its phasor V, we obtain the latter simply by vecto• rial addition (Figure AS) of the phasors VR , VL , and Vc. We get
y
v Vc
FigureA.5 410 Appendix A Phasor Analysis
Ivl = Y2002 + (150.8 - 88.4)2 = 209.5 [V]. (A.3.5)
I 150.8 - 88.4) fl: = tan- ( = l7.33°. (A.3.6) 200 For the total voltage we can write,
v = 209.5 sin(wt + 17.33°) [V]. (A.3.7)
A.2 Phasor Representation Using Complex Numbers
Complex algebra is a most valuable analytical tool for the study of phasors. We first give a very brief expose of complex algebra followed by examples.
A.2.1 Complex Numbers: Definition Figure A.6 shows a complex number plane. The complex number z is defined by its coordinates, x andjy, along the real and imaginary axes, respectively. We write the complex number z in the cartesian form:
z = x + jy. (A.3) The factor j is defined by (A.4)
We say that x is the "real part of z," and y the "imaginary part." The distance Iz I from the origin to the coordinate point z is referred to as the modulus or mag• nitude of z.
jy-axis
~~------~z
Izl
x x-axis
Figure A.6 A.2 Phasor Representation Using Complex Numbers 411
We have (A.5) The angular orientation I..!:. of z in reference to the real axis is obtained from
I..!:. = tan- 1 C) (A. 6)
From Figure A.6 we have x = Izl cos I..!:., (A.7) y = Iz I sin I..!:..
Euler's theorem 1 states that
ej ~ = cos I..!:. + j sin I..!:. (A.8) The complex number z can be written in the alternate polar form:
z = x + jy, Iz I cos I..!:. + j Iz I sin I..!:., (A.9)
Izl(cosl..!:. + jsinl..!:.) = Izle j / z•
A.2.2 Complex Algebra Addition, subtraction, multiplication, and division of complex numbers are defined in terms of the same operations valid for real numbers. We give one example of each operation.
ExampleA.4 Given the two complex numbers
z\ = 3 + jl and Z2 = 4 - j2, (A.4.1)
I This theorem can be proved as follows: From the series
x 2 x 3 X4 eX = 1 + x + 2! + 3! + 4! + ... , we obtain
. . ~2 .~3 ~4 e1
= cos~ + jsin~ 412 Appendix A Phasor Analysis find:
(i) ZI + Z2' (ii) ZI - Z2' (iii) ZI Z2' (iv) ZI/Z2' Solution: (i) The sum of the complex numbers is obtained as follows:
ZI + Z2 = (3 + jl) + (4 - j2) (A.4.2) = (3 + 4) + j(l - 2) = 7 - jl. Note that the real and imaginary parts are added separately. Note also (Figure A.7) that if each complex number is associated with a phasor, then the complex sum of the numbers corresponds to a vectorial addition of the phasors. (ii) The difference:
ZI - Z2 = (3 + j1) - (4 - j2) (AA.3) =(3-4)+j(1 +2)= -1 +j3. (iii) The product:
Z,Z2 = (3 + j1)(4 - j2) = 3 ·4 + (j 1)( - j2) + j(1 ·4 - 3 . 2) (A.4A)
= 14 - j2.
jy t
---x
Figure A.7 A.2 Phasor Representation Using Complex Numbers 413
(iv) The quotient:
Zl _ 3 + j1 _ (3 + jl)(4 + j2) Z2 - 4 - j2 - (4 - j2)(4 + j2) 3·4 + (j I) (j2) + j(1. 4 + 2·3) (A.4.5) 16 + 4 10 10 = 20 + j 20 = 0.5 + jO.5.
By using the polar fonn of complex numbers, we obtain (v) The product:
ZlZ2 = (3 + j1)(4 - j2) = 3.162ejI8 .435° ·4.472e-j26.5W = 14.140e-j8.13Qo = 14.140(0.990 - jO.141) (A.4.6)
= 14.00 - j2.oo (as before). (vi) The quotient: 3 162ej 18.4350 Zl = . _ j45° Z2 4.472e-j26.5W - 0.707e , (A.4.7) = 0.5 + jO.5 (as before).
ExampleA.5 Use complex algebra to solve Example A.3.
Solution: If we express the voltage phasors, VR , VL , and Vc as complex num• bers we get
VR = 200 [V].
VL = j 150.8 [V]. (A.5.1) Vc = - j88.4 [V]. The applied voltage represented by the phasor V is then obtained by the applica• tionofKVL:
V = VR + VL + Vc = 200 + j150.8 - j88.4, (A.5.2) = 200 + j62.4 = 209.5ejI7.33° [V]. When the total voltage is expressed as a function of time, we get vet) = 209.5 sin(wt + 17.33°) [V]. (A.5.3) Compare this to equation (A.3.7). 414 Appendix A Phasor Analysis
A.3 Impedances
Consider the phasor P and the complex number z. The product zp can be written as zp = Iz kilo Ip ki.e = Iz lip k(il+ i.e). (A 10) We make the following observation: Multiplication by z results in a magnitude amplification of Iz Iand a phase advancement by L.!:. degrees. In particular, when a phasor is multiplied by the factor j, it is rotated counter• clockwise through the angle 90° with an unchanged magnitude. Multiplication by the factor -j results in a clockwise rotation through 90°. Consider the four pha• sors shown in Figure A4. In view of equations (A.2.2), (A2.3), and (A2A), and in consideration of the multiplication rules above, we can write:
VR = RI [V]; [V]; (All)
I I Vc = -j-I = - [V]. wC jwC
The voltage phasor V, representing the applied voltage across the circuit, is the sum (Figure A.5) of the individual voltage phasors, that is,
1 V = RI + jwLI + - I [V] jwC (AI2)
= (R + jwL + -.1_) I [V]. }wC
The multiplication factors R,jwL, and l/jwC, which when multiplied by the cur• rent phasor, I give us the voltage phasors, are referred to as the impedances of the individual elements (Ohm's law applied to impedances). We use the symbol, Z for impedances and for the resistor, inductor, and capacitor, respectively, we get
ZR == R [0] ZL ==jwL [0] (A 13) 1 Z =- c - jwC [0].
The total impedance, Z, for the series circuit is, according to equation (AI2),
I Z = R + jwL +• [0]. (A14) jWC A.3 Impedances 415
v FigureA.8
In symbolic form, the relationships between impedances and current and voltage phasors are shown in Figure A8.
ExampleA.6 Find the impedance of the R-L-C circuit assuming the numerical values given in Example A3. Solution: We obtain
~=R=20 [0], (A6.1)
~ = jwL = j377· 0.040 = j15.08 [0], (A 6.2) I I Zc = jwC = j377. 300 . 10-6 [0]; (A6.3) 8.84 [ ] =-. =-j8.84 O. ] The total impedance is equal to Z = 20 + j15.8 - j8.84 = 20 + j6.24 [fl]. (A6.4)
ExampleA.7 If a voltage of 100 V rms is applied across the circuit of Example A6, find the rms values of the current and voltages across each of the three circuit elements and write all voltages and current as functions of time. Solution: If we chose the applied voltage Vas our reference phasor, that is,
V = looeW = 100 [V]. (A7.1) (Note that the "length" of the phasor has been assumed to be proportional to its rms value, that is, vrnaxtV2. This is practical because we are usually interested in the rms values of ac variables rather than their peak values.) From equation (AI2) we then get 416 Appendix A Phasor Analysis
V 100eW 1= - = [A]; Z 20 + j6.24 (A.7.2) .00 100e} _ -j17.330 20.95ej17.330 - 4.773e [A].
(Note that the value of the current is in rms because the voltage was expressed in rms.) For the voltages across the components, we have VR = ZR 1 = 20. 4.773e-j17 .33° = 95.47e-jI7.33° [V];
VL = ZL 1 = j15.0S. 4.773e-jI7.33° = 71.9Sej72.67° [V]; (A.7.3) Vc = Zcl = -jS.S4·4.773e-jI7 .33° = 42.1gejI07.33° [V]. Note that if we show the above phasors in a phasor diagram, we obtain the same diagram as in Figure A.5 but reduced in scale and rotated clockwise by 17.33°. If we express the voltages and current as functions of time we obtain v(t) = 100\12 sin wt = 141.4 sinwt [V], i(t) = 6.750 sin(wt - 17.33°) [A], vR(t) = 135.0 sin(wt - 17.33°) [V], (A.7.4) vL(t) = I01.S sin (wt + 72.67°) [V], vC ExampleA.8 The spring-mass-dashpot system shown in Figure A.9 is subjected to a sinu• soidally varying force, f(t). Find the displacement, x, and the velocity, s, of the mass when it is in the steady state. Solution: The restraining forces of the spring and dashpot are fsp = kx [N] (A.S.I) and [N], (A.S.2) respectively, where k and kd are constants. According to Newton's laws of motion, the acceleration, d 2x/dt 2, of the mass will be proportional to the force acting on it, hence dx d 2x f(t) - kx - kd dt = m dt 2 [N], (A.S.3) A.3 Impedances 417 Equilibrium __-.• level f(t) = f max sin wt FigureA.9 or, in terms ofthe velocity, ds f(t)=fmaxsinwt=k sdt+kds+m• [N]. (A.8.4) f dt In steady state, the velocity will be sinusoidal and of the form: s = Smax sin(wt + i..!..). [mls]. (A.8.S) By substitution of equations (A.8.S) into (A.8.4) we obtain k 0 f max sin wt = - smax sin (wt + i..!.. - 90 ) [m/s]; w + kdsmax sin(wt + i..!..) [m/s]; (A.8.6) 0 + mwsmax sin (wt + i..!.. + 90 ) [m/s]. We now represent the force,f(t), by the phasor, F, and the velocity, s(t), by the phasor, S, in accordance with F = f maxejOO [N]; (A.8.7) [m/s] These phasors are shown in Figure A.I 0, where we have also shown the displace• ment phasor, X. (X lags S by 900 • Why?) In terms of these phasors we can write equation (A.8.6) as follows: 418 Appendix A Phasor Analysis s F (Reference phasor) FigureA.I0 k F = ---:- S + kdS + jwmS [N]. (A8.8) JW This equation is similar to equation (AI2) and we can therefore define the mechanical impedance, Zm as k Zm == kd + jwm + ---:• [N· s/m]. (A8.9) JW We then can write equation (A8.8) in the shorter form: F S=- [mls]. (A8.1O) Zm ExampleA.9 For Example A8 we have the following numerical values: f max = 10 [N], m = 1 [kg], (A9.1) k = 100 [N/m] , kd = I [N· s/m]. Determine the amplitude of the velocity, smax' as a function of w. Solution: We have 100 Zm = I + jw + -.• [N· s/m]. (A9.2) JW A.3 Impedances 419 Combining equations (A8.?) and (A8.1O) we obtain lOew [mls] (A9.3) 1 + jew - 100/w) that is 10 s = --;======"':: [m/s]. (A9.4) max VI + (w - 100/w)2 I.!. = -tan-1 (w _ 1~0) (A9.5) We have plotted Smax and I.!. against w in Figure All. Note that the peak: of the velocity occurs at w = 10 radls. This is the resonance frequency. It is interesting to note that I.!. = 0 at resonance; that is, at resonance the velocity and the force are in phase: f res = f max sin wt [N]; (A9.6) [m/s]. Smax mist tI!.. degrees 5 smax 20 -w rad/s FignreA.ll 420 Appendix A Phasor Analysis A.4 Admittances Consider the parallel circuit shown in Figure A12. For the currents, II and 12 flowing in the impedances ZI and Z2 we have [A]. (A15) The total current, I is I = I + I = ~ + ~ = V(~ + ~) [A]. (AI6) I 2 ZI Z2 ZI Z2 We introduce the concept of admittance as follows: [S]. (AI7) and the total admittance, [S]. (AI8) Equation (A16) can then be written in the form: 1= V(Y1 + Y2 ) = VY [A]. (AI9) Example A.tO Connect the three impedances in Example A6 in parallel and feed this circuit from an ac generator delivering 100 V rms, at 60 Hz. Find the current drawn from the generator. Solution: The three parallel elements represent a total admittance of 1 1 1 Y=- + --+ [S] 20 j15.08 -j8.84 = 0.0500 - jO.06631 + jO.l131 [S] (A1O.1) = 0.0500 + jO.0468 [S] -I v Figure A.12 A.4 Admittances 421 Equation (A. 19) then gives 1= lDO(0.0500 + jO.0468) = 5.00 + j4.68 = 6.85ej43.1l" [A]. (A.lD.2) Note: The current I leads the voltage V by 43°, that is, the parallel circuit draws a leading current-it is capacitive. When the three circuit elements were connected in series across a lDO-V source (Example A.7), the current was 4.7 A, and it lagged the voltage by 17°-the series circuit was inductive. The explanation of this phenomenon is left as an exercise for the reader. Appendix B Spectral Analysis B.1 Periodic Waveforms A function or "wave," f(t), is said to be periodic if, for all values of t, it satisfies the equation f(t) = f(T + t), (B.l) where Tis the period (see Figure B.l). According to Fourier's theorem, it is possible to express a periodic function as an infinite sum of harmonic components in accordance with 00 f(t) = Ao + ~ Av sin (vwt + cf», (B.2) v=) where 27T w=- [radls]. T is the base orfundamental radian frequency. If we make use of the trigonometric identity sin (a + f3) = sin a cos f3 + cos a sin f3 , (B.4) we can write the series (B.2) in the alternate form: 00 f(t) = Ao + ~ (Bvsinvwt + Cvcosvwt). (B.5) v=) The amplitudes Ap and phase angles CPv in the series (B.2) are related to the coef• ficientsBvand C)n the series (B.5) by A v = VB v2 + C v2 (B.6) 422 B.2 Finding the Amplitudes of the Harmonics 423 f(t) t , , ,, --•t FigureB.l The constant Ao represents the "de" component or average value of the periodic wave. The coefficients AI' A2 , ••• are the amplitudes of the first, second, ... har• monies. The "first" harmonic (of frequency w) is also referred to as the "funda• mental" or "base" component. B.2 Finding the Amplitudes of the Harmonics It can be shown 1 that the coefficients in equation (B.5) may be derived from the following: Bv = -2 IT f(t) sin vwt dt; T 0 (B.7) Cv = ~ IT f(t) cos vwt dt. T 0 The de component is computed from Ao = -1 IT f(t) dt. (B.8) T 0 1 The proof is carried out as follows: 1. Multiply both sides of equation (B.5) by the factor sin (11M). 2. Integrate both sides over one full time period. In so doing, one finds that all integrals on the right side are equal to zero except the integral: T Lsin 211wtdt which is equal to BJ/2. The first line of equation (B.7) is obtained. The second line of equation (B.7) can likewise be confIrmed by following the above steps, but using cos (11M) as a multiplication factor. Finally, equation (B.8) is obtained by integrating both sides of equation (B.5) over one full cycle. 424 Appendix B Spectral Analysis If the average value of the periodic wave is zero, it clearly has no de component. Certain features of the wave (symmetry) may simplify the task offinding the har• monics, as demonstrated in the following example. Example B.1 Consider the periodic "triangular" wave shown in Figure B.2. Find the funda• mental component and all higher harmonics of this wave. Solution: We first explore the symmetry features of this wave. We note in par• ticular that the wave is characterized by f(t) = - f(T- t), (B. 1.1) and (B.1.2) On the basis of equations (B.l.l) and (B.1.2) we can make the following observations: Observation 1. Write the integral (B.8) in two parts: Ao = ! fT f(t) dt = ! [fT!2 f(t) dt + ( f(t) dt] (B.1.3) ToT 0 JT/2 Because of equation (B .1.1), the two parts of the integrals are equal but of oppo• site sign. Therefore, Ao = 0; the wave contains no de component; its average value is zero. Observation 2. For a cosine wave we have, cos vwt = cos vw(T - t). (B.IA) In view of equation (B.I.I), it is clear that the second set of integrals (B.7) van• ishes for all v. Therefore, C v =0 for v=I,2,3, ... ,oo, (B.1.5) that is, the wave contains no cosine terms. Observation 3. For a sinewave we have sin vwt = sin vw(f - t) for odd v's, (B.1.6) sinvwt = -sinvw(f - t) for even v's. (B.1.7) B.2 Finding the Amplitudes of the Harmonics 425 Because of equation (B.l.2) it is clear that the first set of integrals of (B.7) van• ishes for all even values of P; Bp = 0 for p = 2,4, 6, ... , 00. (B. 1.8) It follows that the wave contains only odd harmonics. Observation 4. In view of equation (B.l.2), (B. 1.6), and (B.l. 7) the product f(t) sinpwt (B.1.9) attains identical values for t, (f - t), (f + t) and (T - t) (assuming p is odd). Thus we need to perform the integration of (B. 7) over only one-quarter cycle; that is, 8 LT/4 Bp = - f(t) sin pM dt for P = 1,3, 5, ... ,00 (B.LlO) T 0 In the interval 0 < t < T/4 the functionf(t) (see Figure B.2) is a straight line of the form: A f(t) = 4-t. (B.Lll) T Substituting into equation (B.LlO) we get " T/4 Bp = 32 A2 L t sin PlLIt dt for p == 1,3,5, .... 00. (B.Ll2) T 0 f(t) t / ,/ , -+• ,, t FigureB.2 426 Appendix B Spectral Analysis The integral gives the following for Bp : A B =~2A2(-1)(P+3)/2 for v=1,3,5, ... ,oo. (B.1.13) p 7T v The triangular waveform is described by the infinite series j(t) = -28 AA[ sinwt --1 sin3wt + - 1 sin5wt - ...J (B.1.14) 7T 9 25 In practice, one obtains very good accuracy by including only the first few terms in the series. Figure B.3 shows the original triangle compared to the waveform obtained by including only the first three terms in the series. B.3 Spectral Analysis by Numerical Integration In many practical cases one cannot perform the analysis as neatly as the previous example would have us believe. Often the periodic wave is obtained experimen• tally and the functionj(t) is available not in analytic form but as a graph. In such situations, the spectral analysis must be carried out numerically using approximations for the integrals (B.7) and (B.8). f(t) _41------"""7'...... ---Triangular wave .. Figure B.3 B.3 Spectral Analysis by Numerical Integration 427 i(/) amps 9 t 8 7 6 5 ~------~------~-----T------~ 4 3 6 12 14 16 20_ 1 millisecs FigureB.4 ExampJeB.2 Figure B.4 shows the magnetization current in a power transformer as recorded in an experiment (see Figure 5.9). The base frequency is 50 Hz, which means that the period, T is equal to 20 ms. Carry out a spectral analysis of the waveform. Specifically, find the amplitude of the 50-Hz component. Solution: Because the functionJ(t) cannot be expressed in an analytical form, we must calculate BI and CI by using the J(t) values obtained from the graph. We can write the integrals (B.7) as the approximations: 211t n BI "'" --~ ir sinwtr; T r~1 (B.2.1) We have divided the time interval 0 < t < Tinto n time segments of width I1t. In Figure B.4, we have chosen n = 20, which corresponds to I1t = 1 ms. The value 428 Appendix B Spectral Analysis Table B.t r ir wtr number amperes degrees sinwtr ir sinwtr coswt ir coswt 1 2.1 9 0.156 0.329 0.988 2.074 2 2.9 27 0.454 1.317 0.891 2.584 3 3.3 45 0.707 2.333 0.707 2.333 4 4.5 63 0.891 4.010 0.454 2.043 5 6.2 81 0.988 6.124 0.156 0.970 6 8.0 99 0.988 7.902 -0.156 -1.251 7 9.0 117 0.891 8.019 -0.454 -4.086 8 6.0 135 0.707 4.243 -0.707 -4.243 9 0.6 153 0.454 0.272 -0.891 -0.535 10 -1.8 171 0.156 -0.282 -0.988 1.778 Sum 34.267 1.667 of the current, ir, in the center of each time segment is recorded. We then compute and tabulate wtr, sin wtr, and cos wtr for each value of tr in the interval. Finally the products ir sin wtr and ir cos wtr are computed. The results are shown in Table B.I. Because of the symmetrical nature of the waveform, we can use f(t) = -f(T/2 + t) (B.2.2) and perform the summation over half the period. This means that we compute B1 and C1 from the following expressions: 4 dt ./2 BI = --L ir sin wtr, T r=1 (B.2.3) 4 dt ./2 C1 = --L ir cos wtr· T r=1 By using the values obtained from Table B.l we get 4· 0.001 BI = 0.020 ·34.267 = 6.853, (B.2.4) 4·0.001 C1 = 0.020 . 1.667 = 0.333. By using equation (B.6) we compute the amplitude and phase angle of the base harmonic or the fundamental as 8.4 Periodic Waveforms in the Space Domain 429 Al = Y6.853 2 + 0.333 2 = 6.853; (B.2.5) 4> = tan- 1 (0.333) = 278° 1 6.853·· The fundamental component of the current in the wavefonn is i(t) = 6.861 sin (wt + 2.78°) [A], (B.2.6) where 27T W = -- = 314 [rad/s]. (B.2.7) 0.020 B.4 Periodic Waveforms in the Space Domain So far, we have assumed that the independent variable is time, t. This is nonnally the case, as harmonic analysis is most often used in communication theory where time is the variable. However, there are areas of science and technology where the periodic phe• nomena to be analyzed involves space variables. For example, the periodic cur• rent, emf, and magnetic flux found in the air gaps of electric machines belong to this category. ExampleB.3 Figure B.5 shows the "sheet of current" in one phase of the distributed stator winding of a three-phase induction machine. We remember from Section 8.4.2 [(x) t1-+1 -----;~~~~~-__l One (distance between pole pain) = 2frD A" p , 1-+-__ frD ____I P = Distance between adjacent poles FigureB.S 430 Appendix B Spectral Analysis that this sheet current was created by representing a macroscopic (or "smeared") view of the current distribution in the stator slots. We also remember that, because the current is ac, the sheet current will pulsate in time. Figure B.5 is therefore a snapshot taken at a given moment in time (e.g, when the current has reached its peak). Find the amplitude of the fundamental waveform of the sheet current. Solution: By placing the origin as shown in Figure B.5, the waveform will have the same symmetrical features as the triangular wave in Example B.l. The obser• vations made concerning the harmonics of the triangular wave apply here. The fundamental waveform must therefore be a sinusoid, as shown by the dashed line in Figure B.5. We want to find its amplitude, B,. The equations derived earlier were in terms of the independent variable, t, and the period, T. Now the independent variable is x and the period is 27TD/p.2 We can therefore use the previous formula after making the following changes to the variable: t~x 27TD T~~- (B.3.l) p 27T P w=~~~. T D Substituting equation (B.3.1) into (B.1.1O), we get Bv = 4p ('TD/2P f(x) sin (v p x) dx (B.3.2) 7TD Jo D In this case the functionf(x) has the following values (see Figure B.5): f(x) = 0 for 0 < x < 7TD/3p; (B.3.3) f(x) = A for 7TD/3p < x < 7TD/2p. (B.3.4) The integral (B.3.2) gives P 1TD 2P B, = 4p [r1TD/3 o. sin (px) dx + f / A . sin (px) dx], 7TD Jo D 1TD/3p D (B.3.5) = 4p [0 + ~D] = 1 A [A/m]. 7TD 2p 7T 2 The expression 211D/p is the peripheral width of one pole pair. B.4 Periodic Waveforms in the Space Domain 431 Example B.4 Use the result obtained in Example B.3 to show analytically that the flux in all three stator phases of an induction motor as well as a synchronous machine, jointly create a rotating "flux wave." (In Sections 4.7.1 and and 8.3, this was shown graphically.) Solution: The result in Example B.3 can be stated as follows: [Wb]. (B.4.l) This represents the fundamental of the stator flux due to the current in phase a as viewed at a particular instant in time. If we multiply equation (B.4.I) by sin wt, we express the pulsating nature of the wave as 2 A The currents in phases band c give rise to similar fluxes, but they are shifted both in space and time by 27r/3 and 47r/3 radians, respectively, that is, 2 A • 27r). ( 27r) 2 A • 47r). ( 47r) The total flux is obtained by the sum 3 A • This is the equation for a wave revolving with constant speed and constant ampli• tude in the positive x direction. Appendix C The SI Unit System C.1 General The British System of Units, at present the most popular in the United States, has roots that go back to Roman times. Internationally, it is losing ground very fast to the Metric System introduced by the French Academy of Sciences, in the eigh• teenth century. C.2 Basic Units We have summarized in Table C.I the most important features of the SI system as it is related to electric energy engineering. C.3 Derived Units Other units derived from the above are called "derived units" of the SI. For exam• ple, the unit of force in the SI system is the newton [N]. It is derived from New• ton's Second Law (force = mass X acceleration) and is defined as the force Table C.I Basic SI Units Physical Quantity Name of Unit Symbol Length meter m Mass kilogram kg Time second Electric Current ampere A Temperature degree Kelvin K (Other basic units for radioactivity, luminous intensity, and "amount of substance" exist.) 432 C.S Conversion Between Unit Systems 433 Table C.2 Derived SI Units Physical Quantity Name of Unit Symbol Force newton N = kg· mls2 Work (or energy) joule [J] = [N' m] Power watt [W] = [J/s] Pressure pascal [Pal = [N/m2] Electric charge coulomb [C] = [A' sl Electric potential volt [V] = [J/C] = [W/A] Electric capacitance farad [F] = [eN] Electric resistance ohm [0,] = [VIA] Magnetic flux weber [Wb] = [V, s] Magnetic flux density tesla [T] = [Wb/m 2] Magnetic inductance henry [H] = [Wb/A] needed to impart an acceleration of 1 [m/s 2 ] to a mass of 1 [kg]. Force therefore has the dimension "kilogram times acceleration," which is written as [kg· m/s 2 ]. Consider another example. The SI unit for energy is the joule [J]. It is derived (see Section 2.3) from the basic equation for energy (energy = force X distance) and is defined as the work done or energy generated by a force of 1 [N] acting over a distance of 1 [m]. Energy has the dimension "force times distance," which is written as [N . m]. Table C.2 shows some of the more important derived SI units. C.4 Multiplication Factors and Prefixes The SI system is based on decimal units, formed by multiplying or dividing a sin• gle base unit by powers of 10. In this manner, widely different unit ranges can be accommodated. For example, a communications engineer is interested in microwatts [J.L W] of power, and his power colleague works with megawatts [MW]. These two units differ by a factor of 10 12. Table C.3 shows the prefixes and letter symbols for the unit multipliers. C.S Conversion Between Unit Systems Conversion between other unit systems and SI units is possible from a knowledge of the basic conversion constants. The most useful ones are given in Table CA. The conversion factors for the most commonly used energy and power units are given in Appendix D. 434 Appendix C The SI Unit System Table C.3 Multiplication Factors and Prefixes for Forming Decimal Multiples and Submultiples of the SI Units Multiplier Exponent Prefix Symbol 1 000 000 000 000 000 000 10 18 exa E I 000 000 000 000 000 10 15 peta P 1 000 000 000 000 10 12 tera T 1000 000 000 10 9 giga G 1000000 10 6 mega M 1000 10 3 kilo k 100 10 2 hecto h 10 10 1 deca da 0.1 10-1 deci d 0.01 10-2 centi c 0.001 10-3 milli m 6 0.000 001 10- micro J.L 0.000 000 001 10-9 nano n 0.000 000 000 001 10- 12 pico p 0.000 000 000 000 00 1 10- 15 femto f 0.000 000 000 000 000 001 10- 18 atto a Table C.4 Some Useful Conversion Constants I m = 3.28083990 ft = 39.37007874 inches I ft = 0.3048000 m I kg = 2.204622621b = 35.27396195 oz I Ib = 0.45359237 kg 1 m 3 = 264.1720 gallons (U.S.) 1 gallon (U.S.) = 0.003785412 m 3 degrees Kelvin = 5/9 (degrees Fahrenheit) + 255.37 degrees Fahrenheit = 9/5 (degrees Kelvin) - 459.67 References ASTM/IEEE Standard Metric Practice, IEEE Standard STD 268-1986. This Booklet may be obtained from IEEE Service Center, 445 Hoes Lane, Piscataway, NJ 08854. "'0» "'0 CD ::J 0- >< o Table D.l Conversion of Units of Energy Unit Symbol eV J Btu Wh kcal toe electron-volt eV 1 1.602 X 10- 19 1.519 X 10- 22 4.448 X 10- 23 3.826 X 10- 23 3.81 X 10- 30 joule J 6.24 X 10 18 1 9.479 X 10- 4 0.278 X 10- 3 0.239 X 10- 3 2.38 X 10- 11 British thermal unit Btu 6.586 X 10 21 1.055 X 10 3 1 0.2931 0.252 2.51 X 10- 8 watt-hour Wh 2.247 X 10 22 3.60 X 10 3 3.412 1 0.860 8.57 X 10- 8 kilocalorie kcal 2.614 X 10 22 4.185 X 10 3 3.968 1.163 1 9.97 X 10- 8 tonnes oil equiv. toe 0.262 X 10 30 4.2 X 1010 3.985 X 10 7 1.166 X 10 7 1.003 X 10 7 1 .j>. '"(J1 ~ ~ Table D.2 Conversion of Units of Power Unit Symbol W hp Btu/sec kcallsec "0» "0 watt W I 1.34 X 10- 3 0.948 X 10-3 0.239 X 10-3 CD a.::l horsepower hp 746 I 0.707 0.178 x' British thermal units per second Btu/sec 1.055 X 103 1.414 1 0.252 0 kilocalories per second kcal/sec 4.184 X 103 5.607 3.966 1 Appendix D 437 To convert British thermal units to joules, one must go to the left-hand side of Table D.I and find the row for Btu (row 3). One then follows the row to the col• umn corresponding to joules (column 4). The number given there (1.055 X 10 3) is the multiplicant for the conversion. To convert kilocalories per second to horsepower, one must go to the left-hand side of Table D.2 and find the row for kilocalories per second (row 4). One then follows the row to the column corresponding to horsepower (column 4). The num• ber given there (5.607) is the multiplicant for the conversion. Answers To Selected Exercises Chapter 1 1.1 m = 445.1 kg 1.2 About 18.31 billion dollars Chapter 2 2.1 a) 16.5% (or 1.62 b) 1.62 [J/leg] 2.2 a) 0 b) 17,900 [MW] (or 24.0 million hp) 2.3 678.6 [kW] 2.4 The kinetic energy of the elevator must also be included. This energy component amounts to 2 W kin = t .5000.5 = 62,500 [1]. 2.5 9.26 miles per gallon .. (M - m)go 2.7 x= rmlsl m + M + I/R 2 2.10 !1T = 1.55°C 2.12 a) 30.5 [MJ/kg] b) 4.8% and 95.2%, respectively Chapter 3 3.1 a) 3vo [V]. b) we = (3/2) Cv2 [1]. c) A force is required to pull the plates apart (the opposite charges attract). The work done by this force adds to the energy of system. 438 Answers To Selected Exercises 439 V 3.3 a) i(t) = - e-2t/ RC [A] R b) The two capacitor voltages can be written as f (1 + e-2t/ RC ) and f (1 - e-2t/ RC ), respectively. c) Wo = ~CV2 [J] d) Note that Wo is independent of the R value. Thus, charge redistribu• tion via a very small R (short circuit) results in the same heat loss as a large R. (How is this possible?) 3.7 a) 15.6' 10 -3 [N]. (The force on each charge is directed outward in a direction perpendicular to the line between the other two charges.) b) 11.98 [k Vim] (directed perpendicular to the base) c) Each charge contributes a field component vector directed away from the charge. As the three-component vectors have equal magni• tudes they cancel. 3.9 a) f= 50 [Hz] b) d = 0.00398 [mm] 3.10 a) i = 0.686 [A] b) p = 8.23 [W] c) p = 0.00823 [W/m] 3.12 a) 1.333 [kA] b) 44.8 [MW] c) 633.6 [kV] d) 844.8 [MW] e) 94.7% 3.15 T = 1000 [N 'm] 3.18 L = 62.5 [mH] (without air gap) L = 2.42 [mH] (with 2-mm air gap) Chapter 4 4.1 Q = ±13.85 [kVAr] 4.3 Induced voltage = O. (Note that flux linked with stator coil is zero for all rotor positions.) 4.4 a) 459.3 [kW] b) 1377.8 [kW] c) 11022 [kWh] 4.5 a) 15.29 [kV] b) 7.647 [kV per phase] (or 13.24 [kV line to lineD 440 Answers To Selected Exercises 4.8 a) 84.8 [A] b) 146.9 [A] c) S = 215.9 + j 134.7 (3-phase, kilo values) 4.10 a) 0 = 38.68° b) III = 918 [A] (leading the voltage by 19.34°) c) Q = -6.319 MVAr. (Generator absorbs reactive power, acting like a shunt reactor.) . PXs 4.12 a) smo = IvllEI If P, Ivi, and Xs are constants, it follows that sin 8 (and thus 8) will decrease if IE I is increased. Iv II E I cos <5 - IV 12 bQ=-'------.!--'-----'~~----'------'-) Xs The generator absorbs reactive power if Q < 0; that is, if IE I cos 0 < Ivi· If IE I is increased, <5 will decrease (see part a) and cos 8 will increase. Thus the product IE I cos {) will increase, meaning that II E I cos 8 - Ivii will decrease, thus decreasing IQ I c) By 17.9% 4.13 a) 0 = 16.48° b) cf> = 26.57° c) lEI = 14.69 [kV] 4.14 lEI = 10.76 [kV] Chapter 5 5.1 a) 60 [A] and 150 [A], respectively b) Izi = 1.33 [0] 5.2 a) 120% of normal flux value b) 100% (200 [YD. The high flux may possibly result in core losses (and temperatures) that may damage the transformer. 5.3 a) Zs = 0.0806 + j0.418 [0] (on HV side) Zs = 0.0129 + jO.0669 [0] (on LV side) b) 5.11 % of normal value c) Because the core flux is only 5% of normal value, the core losses likewise are minute (actually less than 5% of normal values because core losses increase almost by the square of the flux). 5.5 a) 63.66 [A] and 159.2 [A], respectively; Secondary voltage = 200 [V] b) 61.23 [A] and 153.1 [A], respectively; Secondary voltage = 192.4 [V] c) Yes, slightly. 5.7 105 [kV A] Answers To Selected Exercises 441 5.8 a) 1167 [V] b) Due to the high flux densities (233% of normal. Why?) the core losses will be very high with overheating as a result. 5.11 150 [kVA] tertiary resistive load in combination with 150 [kVA] sec• ondary resistive load will result in 300 [kV A] primary [kV A]. This is the limit of the primary. 5.12 a) R = 538.2 [0] b) 74.7 [A] c) 545.4 [A] (primary); 129.3 [A] (secondary) 5.14 By 0.953% Chapter 6 6.1 The primary winding carries 0.358 [A]. The secondary winding carries 22.6 [A] in one section and 13.5 [A] in the other. 6.3 a) 12.0 [MW] (3-phase) b) 10.83 [MW] (3-phase) 6.4 Increase by 0.604% 6.7 a) 149.48 [kV] b) Sending end powers: 102.48 [MW], 17.85 [MVAr]. Receiving end powers: 98.01 [MW],O [MV Ar]. 6.8 a) 27.1 [A per phase] b) Line consumes 5.04 [kW] and generates 6.58 [MV Ar] c) Iv 2 1 = 141.19 [kV] 6.10 a) C = 106.1 [,uF per phase] b) 31.18 [kV] 6.13 The load flow will appear as follows: ~ 3 + j3.202 3 + j1.202~ .. 2 - jO.202 2 + jO.202 5 + j3 1 + j1 Figure Ans.1 442 Answers To Selected Exercises 6.15 The load flow will appear as follows: ~ 0 + j3.230 6 + j1.020 ~ l I 5 - jO.230 5 + jO.020 5 + j3 1 + jl Figure Ans.2 Chapter 7 7.1 a) Maximum force occurs for S = 0 BLV b) fmmax = R [N] 7.2 a) Maximum power occurs for s = ~ So 1 V2 b) P mmax - 4R 1 V2 c) Po = 4R d) Tf = 50% _ 1 V2 a 7.5 Pmmax - 4R a No. It would be overheated. For example, the motor in Example 7.8 would deliver 13.7 [kW] and dissipate an equal amount in ohmic heat. (Its normal heat losses are only 256 [W] according to Example 7.8.) 7.6 x = 2.193 7.7 a) e = 498.0 [V]; ia = 9.30 [A] b) e = 523.0 [V]; ia = 9.86 fA] 7.9 The motor will deliver 86.79 [kW] (116 [hp]) to the load. Operating effi• ciency = 84.67%. Shaft torque = 361 [N . m]. 7.12 Rmin = 2.29 [0]. Generated power = 9l.56 [kW]. Diesel output = 105.55 [kW]. ChapterS 8.2 a) 3.17% b) 196.83% c) 156.8 [A per phase] Answers To Selected Exercises 443 8.3 a) PI = 12.01 [kW] Pm = 10.64 [kW] Po = 1.37 [kW] Tm = 88.39 [N . m] b) PI = -13.98 [kW] (delivers power to netvork) Pm = -15.86 [kW] (draws this power from prime mover) Po = 1.88 [kW] Tm = -121.2 [N· m] (instead of delivering torque to load, will now require torque from a prime mover to run at this speed) c) PI = 27.33 [kW] Pm = -5.54 [kW] (needs power from prime mover) Po = 32.87 [kW] (With these losses the motor would burn up in a hurry. Note that 83.2% of this lost power is drawn from the net• work and 16.8% from prime mover.) Tm = -44.1 [N . m] d) PI = 23.57 [kW] Pm = -11.99 [kW] Po. = 35.56 [kW] (see comment under part c) Tm =0 8.S Pm = 75.15 [kW] = 100.7 [hpJ 1111 = 107 [A per phase] cos f/J = 0.988 'YJ = 93.3% Tm=823 [N·m] 8.6 635 [A] (if 1M model is used); 646.5 [A] (if magnetization current is included) 8.7 Stator current = 111 [A per phase] Motor output power = 73.30 [kW] = 98 [hpJ Efficiency = 89.5% Power factor = 0.970 8.9 93.07 [kW] 8.11 Let the transformer series impedance be [0 per phase]. (We neglect its magnetizing impedance.) Then the formula for maxi• mum torque will be T = 45 Ivl2 max 7Tns V(RI + RT)2 + (XI + X~ + XT)2 + RI + RT where Ivi is the primary transformer voltage (which we consider constant). INDEX Alternating current Compensation windings, 299 comparison to direct current, 126 Complex algebra, 131 generation of, 90 Complex power, 134 in dc machine, 290 Computer disc drive, 392 Amber, 53 Computer dispatched power, 251 Ammeter, 121,209 Conductance, 73 Arnortisseur, 186 Conjugate current, 134 Amplifier, 249 Control rods, 45 Apparent power, 134 Controlled-rectifier, 317 Armature, 137 Cooling capacity, 280 current, 296 Core losses, 185,205,297 reaction, 299 Corona discharge, 59 resistance, 295 Coulomb, 54 winding, 287 Coulomb's law, 54 Asynchronous machine, 335 Crawling, 328 Atomic lattice, 73 Critical .mass, 45 Automatic excitation control (AEC), 256 Automatic Load-Frequency Control Damper, 186 (ALFC),248 A-phase currents, 160 Autotransformers, 218 Dielectric constant of vacuum, 55 Average power, 127-130 Differential equation, 81 "Digital" motor, 391 Back emf, 279, 300 Diode, 315 Balanced load, 245 Direct current Baseload, 13 comparison to alternating current, 126 Benjamin Franklin, 53 Direct current machine Boiling-water reactor (BWR), 45 back emf, 292 Bound currents, 114 core losses, 297 Boyle's law, 34 equivalent circuit, 295 British thermal unit (Btu), 40 generator, 300 Brushless dc motor, 396 linear motor, 276 Bus, 241 no-load speed, 278, 303, 306 Bushings, 214 prototype, 286-287 speed control of dc motor, 306 Capacitance, 63 speed-torque characteristics, 302-308, Capacitor-Start Motor, 379 313-315 Capacitor, 66 windage, brush and bearing friction, 297 Carbon brushes, 281, 286, 287, 337 Distribution network, 239 Centrifugal switch, 378 parameters, 252 Chemical energy, 39 transformers, 193 Chemical potential energy, 78 Drift velocity, 73 Chernobyl, 4 Dynamic braking, 302, 337 Cheval vapeur, 26 Circle Diagram: Induction machine, 345 Eddy current, 185, 205 Coercive intensity or force, 112 Effective or rms values, 130 Cogging, 328, 397 Einstein's equation, 45 Coil, 83 Electric rotation, 90 charge, 53 Commutator, 287, 288, 396 current, 70 445 446 Index Electric (cont.) direct current, 300 energy storage, 13 induction machine, 336 field, 56 Geometric series, 147 field intensity, E, 109 Gravitation, 16 potential, 58 force, 17 power network, 239 Grid or power grid, 52, 169,222,239 solar panels, 285 loop structure, 241 sources, 76 radial structure, 241 traction, 390 Electrical degrees, 139 Half-wave Rectifier, 318 Electromagnet, 286 Hall effect, 398 Electromagnetic force law, 93 Harmonic analysis, 207 Electromagnetic induction, 86 High-grade heat, 13, 39 Electromechanical torque, 163,288 Hoists, 274 Electromotive force (emf), 76 Homo-Polar Machine, 281 Electron, 53 Horse-power, 2, 5, 26 Electrostatic energy, 56 HVDC transmission, 129 Electrostatic force, 55 Hybrid solar-electric energy, 47 Elektron, 53 Hydro-storage, 31 Elevators, 274 Hydrocarbon fossils, 3 End rings, 326 Hydroelectric Power, 10 Energy, 20 Hydroturbine, 274 Entropy, 38 Hysteresis, 205 Equipotential, 59 losses, 185 Error detector, 398 Exciter, 137 Induced Electromotive Force, 76, 292 Exponential growth, 6, 7, 15 Induction machine Extra-high-voltage (EHV), 62 circle diagram, 346 equivalent circuit, 339 Farad [Fl, 63 external rotor resistance, 361 Faraday's Law, 86,194 generator, 336 Fault conditions, 244 modified circle diagram, 355 Ferromagnetism, 104 rotational losses, 329 Ferrum, 104 rotor current referred to the stator, 344 Field coils, 108, 137 speed-torque characteristics, 335 Filter, 318 wound-rotor, 337 Fission, 45 Y-6. starting method, 362 Forced cooling, 168,215 Infinite network bus, 186 Fossil fuel, 2, II Insulator strings, 252 Fractional-horsepower motor, 365 Internal combustion engine (ICE), 274 Free electrons, 72 Inverter, 390 Frequency sensor-comparator, 249 Iron or core losses, 185 Frequency error, 249 Isotopes, 54 Full-wave Rectifier, 319 Fusion, 46 louie, 4, 21 louIe's constant, 40 Gate, 317 Generating stations, 241 Kilo-calorie, 40 Generator Kinetic energy, 2, 14,36,248 alternating current, 90, 135, 148 Kirchoff's Voltage Law (KVL), 100,295 Index 447 Lagging phase angle, 129 Ohm's law, 73, 277 Lamination, 138, 185,287 Ohm's Law for a magnetic circuit, 105 Lap winding, 289 Ohmic power dissipation, 75 Leading phase angle, 129 Open-loop control systems, 392 Leakage reactance, 208 Optimum generation, 250 Left-Hand-Rule, 94 Optimum power dispatch, 250 Lenz's law, 86, 87, 329 Line current, 160 Pantograph, 390 voltage, 154 Peak load, 14 voltage profile, 256 Peaking generator, 31 Linear induction motor, 386 power, 14 Linear de motor, 276 Pelton turbine, 28, 74 Linked flux, 87 Periodic wave, 91 Load flow analysis, 250, 266 Periodic Table of Elements, 104 Load-frequency control (LFC), 244 Permanent magnet, 121,286,329,392,396 Loss1ess elements, 129 Permeability, 194,341,354 Low grade heat, 12,39 of vacuum, 83 Phase current, 157 Phase sequence, 148, 152, 171 Magnetic field, 81 Phase voltage, 154 field intensity, H, 110 Phase-to--ground voltage, 154 Magnetic flux, 84 Phase-to-phase or line voltage, 154 distribution, 291 Phasor diagram, 130 Magnetic moment, 102 Phasor representation, 129 orbital moment, 113 Positive ions, 53 saturation, 168, 188,310 Potential energy, 10, 22 spin moment, 113 Power angle, 3, 176,258 Magnetization Power grid or grid 52, 169,222,239 current, 202 Power demand, 246 curves, 110 stations, 239 impedance, 355 transmission, 75 reactance, 203 Power or rate of energy, 25 Magnetohydrodynamic generators, 285 Pressurized-water reactor (PWR), 45 Magnetomotive force (rnrnf), 106 Primary winding, 193, 216 Mechanical degrees, 139 Protons, 53 Moment of inertia, 37, 286 Multi-phase alternating current, 127 Radioactive leaks, 13 Mutual inductance, 97 Reactive loads, 246 Reactive power, 130 Negative ion, 54 Real power, 130 Neutral conductor 148 Rectifier circuits, 316-322 Neutrons, 53 Regenerative braking, 30, 280, 302, 336-337 Newton's law, 36 Relative permeability, 105 Nuclear energy, 44 Reluctance, 106, 381 fission 45 torque, 117 fusion 46 Residual flux, 311 reaction, 13 flux density, 112 Nuclear-powered generator, 190 magnetism, 114 Nucleus, 53 Resistance, 74 Numerical control machinery, 395 Resistance split-phase motor, 378 448 Index Resistivity, 74, 191 Supercooling, 168 Right-hand rule, 82 Surface current density, 166 Right-hand screw, 103 Surface or "sheet" current, 164 Road vehicles, 274 Switching stations, 241 Robotics, 392 Synchronization, 156, 171 Rotational Loss, 297, 329 coefficient, 260 Synchronous machine, 135 Saliency effect, 185 balanced loading, 156 Salient rotor, 139,381 condenser, 169 Satellite dish antenna, 392 distribution effects, 145 Seebeck effect, 39 field---<:urrent control, 176 Self-inductance, 98 impedance, 177 Separately excited dc machine, 295, 302 inductor, 169 Series--excited dc machine, 313 machine control, 174 Servo systems, 395 overexcitation, 176 Servomechanisms, 398 pull-out power, 181 Shaded-pole motor, 380 reactance, 178 Shaft power, 297 reaction torque, 163 Sheet current, 331 speed,335 Shunt dc motor, 305-306 stator current wave, 164 Shunt dc generator, 309-312 three-phase winding design, 151 SI unit system, 16 underexcitation, 176 Single-phase Synchro, 398 ac generator, 135 ac power, 129 Tesla [T], 83 current, 127 Thermal energy, 38 transformer, 193 Thermodynamics, 4, 29, 38 Slip, 329, 333 nonreversible transformation, 39 Slip frequency, 343 reversible transformation, 39 Slip rings, 337, 382 Third harmonic current Slots, 137 in transformers, 224 Solar energy, 46 Three-Mile Island, 4 Solar--electric cells, 9 Three-phase transformer Sparkover, 299 power, 219 Squirrel---<:age motor, 326, 340 three-phase core, 223 Stall,336 Three-phase Standard surface gravity, 20 generator or synchronous machine, 148 Standstill, 300, 330, 334 rectifier, 321 torque, 352 power system, 127 Starter Thunderstorm, 68 winding, 377 Thyristors, 315 compensator, 361 Time---<:onstant, 81 resistor, 277, 300 Timing diagram, 395 Stator, 135, 287, 326 Toroid,82 flux wave, 328 coils, 96 Stepper motor, 391 Torque, 94 Stray Loss, 297 reluctance, 117,381 Sub-synchronousspeed,365 Transformer Super-flywheels, 37 /l-Y Connection, 223 Superconducting magnets, 14 ideal,193 Index 449 impedance, 199 loss, 76 laminations, 213 load characteristics, 245 open-circuit (OC) test, 209 network, 239 power, 198 reactive line power, 261 primary winding, 193,216 real line power, 257 ratings, 215 voltage, 76 turns ratio, 194, 399 radial structure, 241 secondary winding, 193,216 secondary current referred to the Unbalanced grid loading, 246 primary, 200 Underground cables, 252 secondary voltage referred to the Universal gravitational constant, 17 primary, 200 Universal motor, 315, 365 short-circuit (SC) test, 209 Uranium isotope U-235, 45 step--down, 218 step-up, 218 Vector cross product, 94 tertiary winding, 193,216 Viscous damper, 42 voltage-per-turn (VTP), 195 Volt [V], 58 Transistors, 315 Volt-amperes reactive (V Ar), 131 Transmission line Voltage profile, 261, 256 collapse, 259 electrically short line, 252 Waterwheel,2, 17 power limit, 259 Windage friction, 297 line parameters, 252 links, 239 Y -connected 3-phase generator, 148, 156 loop structure, 241 Y-Y connected transfonner, 221