PROBLEM SOLUTIONS: Chapter 1

1 PROBLEM SOLUTIONS: Chapter 1 Problem 1.1 Part (a): R c = l c l c = =0 A/Wb µa c µ r µ 0 A c R g = g µ 0 A c =1.017 10 6 A/Wb Φ= NI R c + R g =1.4 10 4 Wb part (c): λ = NΦ =1.016 10 Wb part (d): L = λ I =6.775 mh Problem 1. R c = l c µa c = l c µ r µ 0 A c =1.591 10 5 A/Wb R g = g µ 0 A c =1.017 10 6 A/Wb Φ= NI R c + R g =1.059 10 4 Wb part (c): λ = NΦ =8.787 10 3 Wb part (d): L = λ I =5.858 mh

Problem 1.3 Lg N = = 110 turns µ 0 A c Problem 1.4 N = Problem 1.5 I = B core µ 0 N/g =16.6 A L(g + l c µ 0 /µ) L(g + l c µ 0 /(µ r µ 0 )) = = 11 turns µ 0 A c µ 0 A c I = B core µ 0 N/(g + l c µ 0 /µ) =18. A µ r =1+ 3499 1+0.047(.) 7.8 =730 ( ) g + µ0 l c /µ I = B =65.8 A µ 0 N

3 part (c): Problem 1.6 Equations H g = NI g ; B c = ( Ag A c ) B g = B g (1 xx0 ) gh g + H c l c = NI; B g A g = B c A c and B g = µ 0 H g ; B c = µh c can be combined to give B g = NI ( )( ) = g + µ0 Ag µ A c (l c + l p ) g + ( µ0 µ NI ) )(1 xx0 (l c + l p ) Problem 1.7 ( ) I = B g + µ0 µ (l c + l p ) =.15 µ 0 N A µ = µ 0 (1+ ) 1199 = 101 µ 0 1+0.05B 8 ( ) I = B g + µ0 µ (l c + l p ) =3.0 µ 0 N A

4 part (c): Problem 1.8 Problem 1.9 ( µ0 N A c g = L ) ( ) µ0 l c =0.353 µ mm l c =π(r o R i ) g =3.57 cm; A c =(R o R i )h =1. cm part (c): R g = g µ 0 A c =1.33 10 7 A/Wb; R c =0 A/Wb; L = N R g + R g =0.319 mh part (d): part (e): I = B g(r c + R g )A c N =33.1 A λ = NB g A c =10.5 mwb Problem 1.10 Same as Problem 1.9 R g = g µ 0 A c =1.33 10 7 A/Wb; R c = l c µa c =3.16 10 5 A/Wb

5 part (c): L = N R g + R g =0.311 mh part (d): I = B g(r c + R g )A c N =33.8 A part (e): Same as Problem 1.9. Problem 1.11 Minimum µ r = 340. Problem 1.1 Problem 1.13 L = µ 0N A c g + l c /µ r L = µ 0N A c g + l c /µ r =30.5 mh Problem 1.14 V rms = ωna cb peak =19. V rms I rms = V rms ωl =1.67 A rms; W peak =0.5L( I rms ) =8.50 mj

6 Problem 1.15 R 3 = R 1 + R =4.7 L = µ 0A g N ( ) = 51 g + µ0 µ l c cm mh part (c): For ω =π60 rad/sec and λ peak = NA g B peak =0.45 Wb: (i) V rms = ωλ peak = 171 V rms (ii) I rms = V rms ωl =1.81 A rms (iii) W peak =0.5L( I rms ) =0.817 J part (d): For ω =π50 rad/sec and λ peak = NA g B peak =0.45 Wb: Problem 1.16 (i) V rms = ωλ peak = 14 V rms (ii) I rms = V rms ωl =1.81 A rms (iii) W peak =0.5L( I rms ) =0.817 J

7 E max =4fNA c B peak =345 V Problem 1.17 N = LI = 99 turns; g = µ 0NI µ 0l c =0.3 6 mm A c B sat B sat µ From Eq.3.1 W gap = A cgbsat =0.07 J; W core = A cl c Bsat µ 0 µ =0.045 J Thus W tot = W gap + W core =0.5 J. From Eq. 1.47, (1/)LI =0.5 J. Q.E.D. Problem 1.18 Minimum inductance = 4 mh, for which g = 0.067 mm, N = 0 turns and V rms =6.78V Maximum inductance = 144 mh, for which g = 4.99 mm, N = 1078 turns and V rms = 4 V Problem 1.19 L = µ 0πa N πr =56.0 mh Core volume V core (πr)πa =40.0 m 3.Thus ( ) B W = V core =4.87 J µ 0 part (c): For T = 30 sec, di dt = (πrb)/(µ 0N) =.9 10 3 A/sec T Problem 1.0 v = L di dt = 163V A cu = f w ab; Vol cu =ab(w + h +a)

8 ( ) Jcu A cu B = µ 0 g part (c): J cu = NI A cu part (d): part (e): P diss =Vol cu ( ρj cu ) ( ) B W mag =Vol gap µ 0 ( ) B = gwh µ 0 part (f): ( L 1 ) R = LI ( 1 ) = W mag ( RI 1 ) = W mag = µ 0whA cu Pdiss P diss ρgvol cu Problem 1.1 Using the equations of Problem 1.0 P diss = 115 W I =3.4A N = 687 turns R = 10.8 Ω τ =6.18msec Wire size = 3AWG Problem 1. (i) B 1 = µ 0N 1 I 1 g 1 ; B = µ 0N 1 I 1 g (ii) λ 1 = N 1 (A 1 B 1 + A B )=µ 0 N 1 (iii) λ = N A B = µ 0 N 1 N ( A g ) I 1 ( A1 + A ) I 1 g 1 g

9 (i) B 1 =0; B = µ 0N I g (ii) λ 1 = N 1 A B = µ 0 N 1 N ( A (iii) λ = N A B = µ 0 N ( A g ) I g ) I part (c): (i) B 1 = µ 0N 1 I 1 g 1 ; B = µ 0N 1 I 1 g + µ 0N I (ii) λ 1 = N 1 (A 1 B 1 + A B )=µ 0 N 1 (iii) part (d): L 11 = N 1 Problem 1.3 λ = N A B = µ 0 N 1 N ( A g ( A1 + A ) ; L = µ 0 N g 1 g g ( A1 + A g 1 ) I 1 + µ 0 N ( A g ) I 1 + µ 0 N 1 N ( A g ( A g ) I g ) ( ) A ; L 1 = µ 0 N 1 N g ) I R A = L 11 = l A µa c ; R 1 = l 1 µa c ; R = l µa c ; R g = g µ 0 A c N 1 R 1 + R + R g + R A / = N 1 µa c l 1 + l + l A /+g (µ/µ 0 )

10 L AA = L BB = L AB = L BA = N R A + R A (R 1 + R + R g ) = N [ ] µa c la + l 1 + l + g (µ/µ 0 ) l A l A +(l 1 + l + g (µ/µ 0 )) N (R 1 + R + R g ) R A (R A +(R 1 + R + R g )) = N [ ] µa c l 1 + l + g (µ/µ 0 ) l A l A +(l 1 + l + g (µ/µ 0 )) L A1 = L 1A = L B1 = L 1B = part (c): NN 1 R A +(R 1 + R + R g ) = NN 1 µa c l A +(l 1 + l + g (µ/µ 0 )) Q.E.D. Problem 1.4 v 1 = d dt [L A1i A + L B1 i B ]=L A1 d dt [i A i B ] L 1 = µ 0N 1 N g [D(w x)] v = dλ ( ) dl 1 N1 N µ 0 D dx = I 0 = dt dt g dt ( ) N1 N µ 0 D ( ɛωw ) = cos ωt g Problem 1.5 H = N 1 i 1 π(r o + R i )/ = N 1 i 1 π(r o + R i ) part (c): v = d dt [N (tn )B] =N tn db dt v o = G v dt = GN tn B

11 Problem 1.6 R g = g =4.4 10 5 A/Wb; R c = l c = 333 µ 0 A g µa g µ A/Wb Want R g 0.05R c µ 1. 10 4 µ 0. By inspection of Fig. 1.10, this will be true for B 1.66 T (approximate since the curve isn t that detailed). Problem 1.7 N 1 = V peak ωt(r o R i )B peak = 57 turns V o,peak (i) B peak = GN t(r o R i ) =0.833 T (ii) V 1 = N 1 t(r o R i )ωb peak =6.5 V, peak Problem 1.8 From the M-5 magnetization curve, for B =1. T,H m =14A/m. Similarly, H g = B/µ 0 =9.54 10 5 A/m. Thus, with I 1 = I = I I = H m(l A + l C g)+h g g =38. A N 1 W gap = ga gapb =3.1 Joules µ 0 part (c): λ =N 1 A A B =0.168 Wb; L = λ I =4.3 9 mh Problem 1.9

1 Area = 191 Joules part (c): Core loss = 1.50 W/kg. Problem 1.30 B rms =1.1 Tandf =60Hz, V rms = ωna c B rms =46.7 V Core volume = A c l c =1.05 10 3 m 3. Mass density = 7.65 10 3 kg/m 3. Thus, the core mass = (1.05 10 3 )(7.65 10 3 ) = 8.03kg. At B = 1.1 T rms = 1.56 T peak, core loss density = 1.3W/kg and rms VA density is.0 VA/kg. Thus, the core loss = 1.3 8.03= 10.4 W. The total exciting VA for the core is.0 8.03= 16.0 VA. Thus, its reactive component is given by 16.0 10.4 =1. VAR. The rms energy storage in the air gap is W gap = ga cb rms µ 0 =3.61 Joules corresponding to an rms reactive power of VAR gap = ωw gap = 1361 Joules Thus, the total rms exciting VA for the magnetic circuit is VA rms = sqrt10.4 + (1361 + 1.) = 1373 VA and the rms current is I rms =VA rms /V rms = 9.4 A. Problem 1.31 part(a): Area increases by a factor of 4. Thus the voltage increases by a factor of 4 to e = 1096cos377t. l c doubles therefore so does the current. Thus I =0.6 A. part (c): Volume increases by a factor of 8 and voltage increases by a factor of 4. There I φ,rms doubles to 0.0 A. part (d): Volume increases by a factor of 8 as does the core loss. Thus P c = 18 W. Problem 1.3 From Fig. 1.19, the maximum energy product for samarium-cobalt occurs at (approximately) B =0.47TandH = -360 ka/m. Thus the maximum energy product is 1.69 10 5 J/m 3. Thus, ( ) 0.8 A m = cm =3.40 cm 0.47 and

13 ( ) 0.8 l m = 0.cm µ 0 ( 3.60 10 5 =0.35cm ) Thus the volume is 3.40 0.3 5=1.0 cm 3, which is a reduction by a factor of 5.09/1.1 = 4.9. Problem 1.33 From Fig. 1.19, the maximum energy product for neodymium-iron-boron occurs at (approximately) B =0.63TandH = -470 ka/m. Thus the maximum energy product is.90 10 5 J/m 3. Thus, ( ) 0.8 A m = cm =.54 cm 0.63 and ( ) 0.8 l m = 0.cm µ 0 ( 4.70 10 5 =0.7 cm ) Thus the volume is.54 0.5 = 0.688 cm 3, which is a reduction by a factor of 5.09/0.688 = 7.4. Problem 1.34 From Fig. 1.19, the maximum energy product for samarium-cobalt occurs at (approximately) B =0.47TandH = -360 ka/m. Thus the maximum energy product is 1.69 10 5 J/m 3. Thus, we want B g =1. T,B m =0.47 T and H m = 360 ka/m. ( ) ( ) Hg Bg h m = g = g =.65 mm H m µ 0 H m ( ) ( ) Bg Bg A m = A g =πrh =6.0 cm B m B m Am R m = =.87 cm π Problem 1.35 From Fig. 1.19, the maximum energy product for neodymium-iron-boron occurs at (approximately) B m =0.63TandH m = -470 ka/m. The magnetization curve for neodymium-iron-boron can be represented as B m = µ R H m + B r where B r =1.6 T and µ R =1.067µ 0. The magnetic circuit must satisfy

14 H m d + H g g = Ni; B m A m = B g A g For i =0andB g =0.5 T, the minimum magnet volume will occur when the magnet is operating at the maximum energy point. ( ) Bg A m = A g =4.76 cm B m ( ) Hg d = g =1.69 H m mm i = For B g =0.75, i = 17.9 A. For B g =0.5, i = 6.0 A. [ ( ) ] dag B g µ RA m + g µ 0 Brd µ R Because the neodymium-iron-boron magnet is essentially linear over the operating range of this problem, the system is linear and hence a sinusoidal flux variation will correspond to a sinusoidal current variation. N

15 PROBLEM SOLUTIONS: Chapter Problem.1 At 60 Hz, ω = 10π. primary: (V rms ) max = N 1 ωa c (B rms ) max = 755 V, rms secondary: (V rms ) max = N ωa c (B rms ) max = 17 V, rms At 50 Hz, ω = 100π. Primary voltage is 95 V, rms and secondary voltage is 143 V, rms. Problem. N = Vrms ωa c B peak = 167 turns Problem.3 75 N = =3 turns 8 Problem.4 Resistance seen at primary is R 1 =(N 1 /N ) R =6.5Ω. Thus I 1 = V 1 R 1 =1.6 A and V = ( N N 1 ) V 1 =40 V Problem.5 The maximum power will be supplied to the load resistor when its impedance, as reflected to the primary of the idealtransformer, equals that of the source ( kω). Thus the transformer turns ratio N to give maximum power must be Rs N = =6.3 R load Under these conditions, the source voltage will see a total resistance of R tot = 4 kω and the current will thus equal I = V s /R tot =ma.thus,thepower delivered to the load will equal P load = I (N R load )=8 mw

16 Here is the desired MATLAB plot: Problem.6 The maximum power will be supplied to the load resistor when its impedance, as reflected to the primary of the idealtransformer, equals that of the source ( kω). Thus the transformer turns ratio N to give maximum power must be Rs N = =6.3 R load Under these conditions, the source voltage will see a total impedance of Z tot = +j kω whose magnitude is kω. The current will thus equal I = V s / Z tot = ma. Thus, the power delivered to the load will equal P load = I (N R load )=16 mw Here is the desired MATLAB plot:

17 Problem.7 ( ) X m V = V 1 = 66 V X l1 + X m Problem.8 Referred to the secondary L m, = L m,1 = 150 mh N part(b): Referred to the secondary, X m = ωl m, =56.7Ω, X l and X l1 =69.3 mω. Thus, ( ) X m (i) V 1 = N V = 7960 V X m + X l =84.8 mω and Problem.9 I 1 = (ii) I sc = V X sc = V 1 X l1 + X m =3.47 A; V = NV 1 V X l + X m X l1 = 1730 A ( ) X m = 398 V X l1 + X m Let X l = X l /N and X sc = X l1 + X m (X m + X l ). For I rated = 50 kva/10 V = 417 A V 1 = I rated X sc =3.1 V I = 1 N ( ) X m I rated =15.7 X m + X l A Problem.10 I L = P load V L =55.5 A and thus I H = I L N =10.6 A; V H = NV L + jx H I H = 381 9.6 V The power factor is cos (9.6 )=0.986 lagging.

18 Problem.11 30 kw Î load = 30 V ejφ =93.8e jφ A where φ is the power-factor angle. Referred to the high voltage side, Î H = 9.38 e jφ A. ˆV H = Z H Î H Thus, (i) for a power factor of 0.85 lagging, V H = 413 V and (ii) for a power factor of 0.85 leading, V H = 199 V. part (c):

19 Problem.1 Following methodology of Problem.11, (i) for a power factor of 0.85 lagging, V H = 4956 V and (ii) for a power factor of 0.85 leading, V H = 4000 V. part (c): Problem.13 I load = 160 kw/340 V = 68.4 Aat =cos 1 (0.89) = 7.1 ˆV t,h = N( ˆV L + Z t I L ) which gives V H =33.7 kv. ˆV send = N( ˆV L +(Z t + Z f )I L )

0 which gives V send =33.4 kv. part (c): S send = P send + jq send = ˆV send Î send = 164 kw j64.5 kvar Thus P send = 164 kw and Q send = 64.5 kvar. Problem.14 Following the methodology of Example.6, efficiency = 98.4 percent and regulation = 1.5 percent. Problem.15 Z eq,l = V sc,l I sc,l = 107.8 mω and thus X eq,l = R eq,l = P sc,l I sc,l =4.78 mω Z eq,l Req,L = 107.7 mω Z eq,l =4.8+j108 mω R eq,h = N R eq,l =0.455 Ω X eq,h = N X eq,l =10.4 Ω Z eq,h =10.3+j0.46 mω part (c): From the open-circuit test, the core-loss resistance and the magnetizing reactance as referred to the low-voltage side can be found: and thus R c,l = V oc,l P oc,l = 311 Ω S oc,l = V oc,l I oc,l = 497 kva; Q oc,l = Soc,L P oc,l =45. kvar

1 X m,l = V oc,l Q oc,l = 141 Ω The equivalent-t circuit for the transformer from the low-voltage side is thus: part (d): We will solve this problem with the load connected to the highvoltage side but referred to the low-voltage side. The rated low-voltage current is I L =50MVA/8 kv=6.5 ka. Assume the load is at rated voltage. Thus the low-voltage terminal voltage is V L = V load + Z eq,l I L =8.058 kv and thus the regulation is given by (8.053-8)/8 = 0.007 = 0.7 percent. The total loss is approximately equal to the sum of the open-circuit loss and the short-circuit loss (393 kw). Thus the efficiency is given by η = P load = 50.0 =0.99 = 99. percent P in 50.39 part (e): We will again solve this problem with the load connected to the high-voltage side but referred to the low-voltage side. Now, ÎL =6.5 5.8 ka. Assume the load is at rated voltage. Thus the low-voltage terminal voltage is V L = V load + Z eq,l Î L =7.758 kv and thus the regulation is given by (7.758-8)/8 = -0.030 = -3.0 percent. The efficiency is the same as that found in part (d), η =99. percent. Problem.16 The core length of the second transformer is is times that of the first, its core area of the second transformer is twice that of the first, and its volume is times that of the first. Since the voltage applied to the second transformer is twice that of the first, the flux densitities will be the same. Hence, the core loss will be proportional to the volume and Coreloss = 340 = 9.67 kw

The magnetizing inductance is proportionalto the area and inversely proportionalto the core length and hence is times larger. Thus the no-load magnetizing current will be times larger in the second transformer or I no load = 4.93 = 6.97 A Problem.17 Rated current at the high-voltage side is 0 kva/.4 kv = 8.33 A. Thus the total loss will be P loss = 1 + 57 = 379 W. The load power is equal to 0.8 0 = 16 kw. Thus the efficiency is η = 16 =0.977 = 97.7 percent 16.379 First calculate the series impedance (Z eq,h = R eq,h + jx eq,h )of the transformer from the short-circuit test data. R eq,h = P sc,h I sc,h =3.69 Ω S sc,h = V sc,h I sc,h =61.3 8.33 = 511 kv A Thus Q sc,h = Ssc,H P sc,h = 44 VAR and hence X eq,h = Q sc,h I sc,h =6.35 Ω The regulation will be greatest when the primary and secondary voltages of the transformer are in phase as shown in the following phasor diagram Thus the voltage drop across the transformer will be equal to V = I load Z eq,h = 61. V and the regulation will equal 61. V/.4 kv = 0.06 =.6 percent. Problem.18 For a power factor of 0.87 leading, the efficiency is 98.4 percent and the regulation will equal -3.48 percent. Problem.19 The voltage rating is 400 V:640 V. The rated current of the high voltage terminal is equal to that of the 40-V winding, I rated =30 10 3 /40 = 15 A. Hence the kva rating of the transformer is 640 15 = 330 kva.

3 Problem.0 The rated current of the high voltage terminal is equal to that of the 10-V winding, I rated =10 4 /10 = 83.3 A. Hence the kva rating of the transformer is 600 83.3 =50kVA. part (c): The full load loss is equal to that of the transformer in the conventionalconnection, P loss =(1 0.979) 10 kw = 10 W. Hence as an autotransformer operating with a load at 0.85 power factor (P load =0.85 50 kw = 4.5 kw), the efficiency will be 4.5 kw η = =0.995 = 99.5 percent 4.71 kw Problem.1 The voltage rating is 78 kv:86 kv. The rated current of the high voltage terminal is equal to that of the 8-kV winding, I rated =50 10 6 /8000 = 6.5 ka. Hence the kva rating of the transformer is 86 kv 6.5 ka = 537.5 MVA. The loss at rated voltage and current is equal to 393 kw and hence the efficiency will be 537.5 MW η = =0.9993 = 99.93 percent 538.1 MW Problem. No numerical result required for this problem. Problem.3 7.97 kv:.3 kv; 191 A:651 A; 1500 kva 13.8 kv:1.33 kv; 109 A:1130 A; 1500 kva part (c): 7.97 kv:1.33 kv; 191 A:1130 A; 1500 kva part (d): 13.8 kv:.3 kv; 109 A:651 A; 1500 kva Problem.4 (i) 3.9 kv:115 kv, 300 MVA (ii) Z eq =0.0045 + j0.19 Ω (iii) Z eq =0.104 + j4.30 Ω

4 (i) 3.9 kv:66.4 kv, 300 MVA (ii) Z eq =0.0045 + j0.19 Ω (iii) Z eq =0.0347 + j1.47 Ω Problem.5 Following the methodology of Example.8, V load = 36 V, line-to-line. Problem.6 The totalseries impedance is Z tot = Z f + Z t = j11.7 +0.11 + j. Ω = 0.11 + j13.9 Ω. The transformer turns ratio is N =9.375. The load current, as referred to the transformer high-voltage side will be ( ) 35 MVA I load = N e jφ =7.81e jφ ka 34kV where φ = cos 1 0.93 = 1.6. The line-to-neutral load voltage is V load = 4 3kV. At the transformer high-voltage terminal V = 3 NV load + I load Z t = 31.7 kv, line-to-line At the sending end V = 3 NV load + I load Z tot = 33.3 kv, line-to-line Problem.7 Problem.8 First calculate the series impedance (Z eq,h = R eq,h + jx eq,h )ofthetransformer from the short-circuit test data. Z eq,h =0.48 = j1.18 Ω

5 The totalimedance between the load and the sending end of the feeder is Z tot = Z f + Z eq,h =0.544+j.058Ω. The transformer turns ration is N = 400:10 3 = 11.6. The referred load voltage V load and current I load will be in phase and can be assumed to be the phase reference. Thus we can write the phasor equation for the sending-end voltage as: ˆV s = V load + I load Z tot We know that V s = 400/sqrt3 = 1386 V and that I load = 100 kva/( 3.4kV). Taking the magnitude of both sides of the above equation gives a quadradic equation in V load whichcanbesolvedforv load Vload +R tot I load V load + Z tot Iload Vs V load = R tot I load + V s (X toti load ) =1.338 kv Referred to the low-voltage side, this corresponds to a load voltage of 1.338 kv/n = 116 V, line-to-neutral or 01 V, line-to-line. Feeder current = 400 = 651 A 3Ztot HV winding current = 651 3 = 376 A LV winding current = 651N = 7.5 ka Problem.9 The transformer turns ratio is N = 7970/10 = 66.4. The secondary voltage will thus be ˆV = V 1 N ( ) jx m = 119.74 0.101 R 1 + jx 1 + jx m Defining R L = N R L = N 1kΩ=4.41 MΩ and Z eq = jx m (R + R L + jx ) = 134.3+j758.1 kω the primary current will equal Î 1 = 7970 R 1 + jx 1 + Z eq =10.3 79.87 ma

6 The secondary current will be equal to ( ) jx m Î = NÎ1 R + R L + j(x = 119.7 m + X ) 0.054 ma and thus ˆV = R L Î = 119.7 0.054 V part (c): Following the methodology of part (b) ˆV = 119.6 0.139 V Problem.30 This problem can be solved iteratively using MATLAB. The minimum reactance is 91 Ω. Problem.31

7 Problem.3 The transformer turns ratio N = 00/5 = 40. For I 1 = 00 A I = I ( ) 1 jx m N R + j(x m + X ) =4.987 0.04 Defining R L = N 50µΩ =0.4Ω I = I ( ) 1 jx m N R + R L + j(x m + X ) =4.987 0.10 Problem.33 Problem.34 Z base,l = V base,l =1.80 Ω P base Z base,h = V base,h = 45 Ω P base

8 Thus R 1 =0.0095Z base,l =17.1 mω; X 1 =0.063Z base,l = 113 mω X m = 148Z base,l = 66 Ω R =0.0095Z base,h =.33 Ω; X =0.063Z base,h =15.4 Ω Problem.35 (i) Z base,l = (7.97 103 ) 75 10 3 =0.940 Ω; X L =0.1Z base,l =0.113 Ω (ii) Z base,h = (7970) 75 10 3 = 847 Ω; X H =0.1Z base,h = 10 Ω (i) 797 V:13.8 kv, 5 kva (ii) X pu =0.1 (iii) X H = 10 Ω (iv) X L =0.339 Ω part (c): (i) 460 V:13.8 kv, 5 kva (ii) X pu =0.1 (iii) X H = 10 Ω (iv) X L =0.113 Ω Problem.36 In each case, I pu =1/0.1 = 8.33 pu. (i) I base,l = P base /( 3 V base,l ) = 5 kva/( 3 797 V) = 163 A I L = I pu I base,l = 1359 A (ii) I base,h = P base /( 3 V base,h ) = 5 kva/( 313.8 kv)=9.4 A I H = I pu I base,h =78.4 A In each case, I pu =1/0.1 = 8.33 pu. (i) I base,l = P base /( 3 V base,l ) = 5 kva/( 3 460 V) = 8 A I L = I pu I base,l = 353 A (ii) I base,h = P base /( 3 V base,h ) = 5 kva/( 313.8 kv)=9.4 A I H = I pu I base,h =78.4 A

9 Problem.37 On the transformer base ( ) ( ) Pbase,t 800 MVA X gen = 1.57 = 1.57 = 1.7 pu 850 MVA P base,g On the transformer base, the power supplied to the system is P out = 700/850 = 0.84 pu and the totalpower is S out = P out /pf =0.85/0.95 = 0.868 pu. Thus, the per unit current is Î =0.868 φ, whereφ = cos 1 0.95 = 18.. (i) The generator terminalvoltage is thus ˆV t =1.0+ÎZ t =1.03 3.94 pu = 6.8 3.94 kv and the generator internalvoltage is ˆV gen =1.0+Î(Z t + Z gen )=.07 44.3 pu = 53.7 44.3 kv (ii) The totaloutput of the generator is given by S gen = ˆV t Î =0.86 + 0.3361. Thus, the generator output power is P gen =0.86 850 = 70. MW. The correspoinding power factor is P gen / S gen = 0.96 lagging.

30 PROBLEM SOLUTIONS: Chapter 3 Problem 3.1 By analogy to Example 3.1, T =B 0 Rl [I 1 sin α + I cos α] =6.63 10 [I 1 sin α + I cos α] N m Thus T =0.530 cos α N m T =0.530 sin α N m part (c): T =0.530 [I 1 sin α + I cos α] N m Problem 3. T =0.5304 N m Problem 3.3 Can calculate the inductance as Thus L = Nφ I = 1000 0.13 10 =13 H W fld = 1 LI = 650 Joules Problem 3.4 For x =0.9 mm, L =9.5 mh and thus, for I =6A,W fld = 0.531 Joules. For x =0.9 mm, L =19.6 mh and thus, for I =6A,W fld = 0.35 Joules. Hence, W fld = 0.179 Joules. Problem 3.5 For a coil voltage of 0.4 V, the coil current will equal I =0.4/0.11 = 3.7 A. Under the assumption that all electrical transients have died out, the solution will be the same as that for Problem 3.4, with a current of 3.7 A instead of 6.0 A. W fld =0.0 Joules W fld = 0.068 Joules. Problem 3.6 For x = x 0, L = L 0 = 30 mh. The rms current is equal to I rms = I 0 / and thus <W fld >= 1 LI rms =0.7 Joules

31 <P diss >= I rmsr =1.63 W Problem 3.7 B g = µ 0Ni g ( ) ( ) B g B g W fld = Air-gap volume = ga g µ 0 µ 0 ( = µ 0N ( ) ) A 0 4θ 1 i 4g π L = W fld i = µ 0N A 0 g ( 1 ( ) ) 4θ π Here is the MATLAB plot: Problem 3.8 v C (t) =V 0 e t/τ ; τ = RC W fld = q /(C) =CvC /. Thus W fld (0) = CV 0 ; W fld( ) =0

3 part (c) i R (t) = v C(t) R ; P diss(t) =i R (t)r = V 0 e t/τ R Problem 3.9 W diss = 0 P diss (t) dt = CV 0 i L (t) = V 0 R e t/τ ; τ = L R W fld (0) = V 0 L R ; W fld( ) =0 part (c) P diss (t) =i L (t)r = V 0 e t/τ R Problem 3.10 Given: W diss = 0 P diss (t) dt = V 0 L R τ = L R =4.8 sec; I R =1.3 MW Thus 1 Li = 1 ( i ) L R ( τ ) = i R =6.4 MJoules R Problem 3.11 Four poles T fld = W fld θm = d [ ] I 0 dθm (L 0 + L cos θ m ) = I0 L sin θ m

33 Problem 3.1 B g = µ 0 Ni g + g 1 R/(h) where g 1 is the length of the fixed gap, l is its length and R is the radius of the solenoid. Here is the MATLAB plot: W fld = πr g ( ) B g µ 0 Here is the MATLAB plot:

34 part (c): L =W fld /i. Here is the MATLAB plot: Problem 3.13 If the plunger is moved very slowly (i.e. idl/dt << Ldi/dt, the current will be essentially constant and all of the change in stored energy will come from the mechanical work applied to the plunger. Thus, Work = W fld (g =0. cm) W fld (g =.5 cm) = 46.7 µjoules The battery will supply only the energy dissipated in the coil. Problem 3.14 The coil inductance is equal to L = µ 0 N A c /(g) and hence the lifting force is equal to f fld = i dl dg = ( µ0 N A c 4g ) i where the minus sign simply indicates that the force acts in the direction to reduce the gap (and hence lift the mass). The required force is equal to 931 N (the mass of the slab times the acceleration due to gravity, 9.8 m/sec ). Hence, setting g = g mim and solving for i gives i min = and v min = i min R =1.08 V. Problem 3.15 ( gmin N ) f fld µ 0 A c = 385 ma a 1 = 9.13071 10 5 a =0.1409 a 3 =8.1089 a 4 = 10558. b 1 =9.68319 10 11 b = 1.37037 10 7 b 3 =6.3831 10 5 b 4 =1.71793 10 3

35 (i) Here is the MATLAB plot: (ii) W fld =13.0 Joules and W fld =13.7 Joules Assuming no core relctance, W fld =11.8Joules and W fld =13.0Joules part (c): (i) Here is the MATLAB plot: (ii) W fld = 14 Joules and W fld = 148 Joules Assuming no core relctance, W fld = 139 Joules and W fld = 147 Joules Problem 3.16 L = µ 0N A c ; f fld = g ( i ) dl dg = L i g The time-averaged force can be found by setting i = I rms where I rms = V rms /(ωl). Thus <f fld >= I rms gω L = Vrms ω µ 0 N = 115 N A c

36 Because the inductor is being driven by a voltage source, the gap flux density remains constant independent of the air-gap length and hence the force also remains constant. Problem 3.17 B s = µ 0i s φ s = B s xl = µ 0xl s part (c): Note that as the coil moves upward in the slot, the energy associated with the leakage flux associated withing the coil itself remains constant while the energy in the leakage flux above the coil changes. Hence to use the energy method to calculate the force on the coil it is necessary only to consider the energy in the leakage flux above the slot. W fld = B s dv = µ 0xli µ 0 s Because this expression is explicity in terms of the coil current i and becasue the magnetic energy is stored in air which is magnetically linear, we know that W fld = W fld. We can therefore find the force from f fld = dw fld dx = µ 0li s This force is positive, acting to increase x and hence force the coil further into the slot. part (d): f fld =18.1 N/m. Problem 3.18 W fld = ( µ0 H ) coil volume = ( µ0 πr 0 N h ) i Thus f = dw rmfld dr 0 = ( µ0 πr 0 N h ) I 0 and hence the pressure is P = f πr 0 h = ( µ0 N h ) I 0

37 The pressure is positive and hence acts in such a direction as to increase the coil radius r 0. Problem 3.19 W fld (q, x) = q 0 v(q,x)dq f fld = W fld x q part (c): W fld = vq dw fld dw fld = qdv + f fld dx Thus Problem 3.0 W fld = v 0 q(v,x)dv ; f fld = W fld x v W fld = q 0 v(q,x)dq = q C = xq ɛ 0 A W fld = v 0 f fld = W fld x q(v,x)dv = Cv = ɛ 0Av x = Cv v = ɛ 0Av x and thus Problem 3.1 ( V T fld = f fld (V 0,δ)= ɛ 0AV0 δ dc ) dc dθ = ( ) Rd Vdc g In equilibrium, T fld + T spring = 0 and thus

38 ( ) Rd θ = θ 0 + Vdc gk Here is the plot: Problem 3. L 11 = µ 0N 1 A g 0 ; L = µ 0N A g 0 L 1 = µ 0N 1 N A g 0 ; part (c): W fld = 1 L 11i 1 + 1 L i + L 1 i 1 i = µ 0A 4g 0 (N 1 i 1 + N i ) part (d): f fld = W fld g 0 i1,i = µ 0A 4g 0 (N 1 i 1 + N i ) Problem 3.3 W fld = 1 L 11i 1 + 1 L i + L 1i 1 i = I (L 11 + L +L 1 ) sin ωt T fld = W fld θ = 4. 10 3 I sin θ sin ωt N m i1,i

39 T fld =.1 10 3 I sin θ N m part (c): T fld = 0.1 N m. part (d): part (e): The curve of spring force versus angle is plotted as a straight line on the plot of part (d). The intersection with each curve of magnetic force versus angle gives the equilibrium angle for that value of current. For greater accuracy, MATLAB can be used to search for the equilibrium points. The results of a MATLAB analysis give: part (f): I θ 5 5.5 7.07 35.3 10 1.3

40 Problem 3.4 T fld = i 1 i dl 1 dθ =.8i 1i sin θ N m λ =0 i = Therefore, for i 1 = 10 sin ωt, ( L1 L ) i 1 = 1.1i 1 cos θ T fld = 3.14i 1 sin θ cos θ = 314 sin (ωt) sin θ cos θ = 78.5(1 cos (ωt)) sin (θ) N m <T fld >= 78.5 N m part (c): It will not rotate. It will come to rest at angular positions where <T fld >= 0 and d<t fld > =0 dθ i.e. at θ =90 or θ = 70. Problem 3.5 Winding 1 produces a radial magnetic which, under the assumption that g<<r 0, B r,1 = µ 0N 1 g The z-directed Lorentz force acting on coil will be equal to the current in coil multiplied by the radial field B r,1 and the length of coil. f z =πr 0 N B r,1 i = πr 0µ 0 N 1 N g The self inductance of winding 1 can be easily written based upon the winding-1 flux density found in part (a) i 1 L 11 = πr 0lµ 0 N 1 g The radial magnetic flux produced by winding can be found using Ampere s law and is a function of z. 0 0 z x B z = µ0ni(z x) gh x z x + h µ0ni g i 1 i x + h z l

41 Based upon this flux distribution, one can show that the self inductance of coil is L = πr 0µ 0 N ( l x h ) g 3 part (c): Based upon the flux distribution found in part (b), the mutual inductance can be shown to be L 1 = πr 0µ 0 N 1 N (x + h ) g l part (d): f fld = d dx [ 1 L 11i 1 + 1 ] L i + L 1i 1 i = πr 0µ 0 N g i + πr 0µ 0 N 1 N i 1 i g Note that this force expression includes the Lorentz force of part (a) as well as a reluctance force due to the fact that the self inductance of coil varies with position x. Substituting the given expressions for the coil currents gives: f fld = πr 0µ 0 N g I cos ωt + πr 0µ 0 N 1 N I 1 I cos ωt g Problem 3.6 The solution follows that of Example 3.8 with the exception of the magnet properties of samarium-cobalt replaced by those of neodymium-boron-iron for which µ R =1.06µ 0, H c = 940 ka/m and B r =1.5 T. The result is { -03 N at x = 0 cm f fld = -151 N at x = 0.5 cm Problem 3.7 Because there is a winding, we don t need to employ a fictitious winding. Solving H m d + H g g 0 = Ni; B m wd = B g (h x)d in combination with the constitutive laws B m = µ R (H m H c ); B g = µ 0 H g gives B m = µ 0(Ni+ H c d) dµ 0 µ R + wg0 (h x) Note that the flux in the magnetic circuit will be zero when the winding current is equal to I 0 = H c d/n. Hence the coenergy can be found from

4 integrating the flux linkage of the winding from an initial state where it is zero (i.e. with i = I 0 ) to a final state where the current is equal to i. The flux linkages are given by λ = NwDB m and hence W fld(i, x) = The force is then (i) for i =0, i I 0 λ(i,x)di = µ 0wDN dµ 0 µ R + wg0 (h x) f fld = dw fld dx = µ 0w DNg 0 ( µ0d(h x) µ R + wg 0 ) [ Ni [ Ni + H c + H c f fld = dw fld dx = µ 0w Dg 0 (H c d) ( µ0d(h x) µ R + wg 0 ) ( i + H )] cd N ( i + H )] cd N where the minus sign indicates that the force is acting upwards to support the mass against gravity. (ii) The maximum force occurs when x = h f max = µ 0wD(H c d) = M max a where a is the acceleration due to gravity. Thus Want M max = µ 0wD(H c d) a f(i min,x=h = a M max = µ 0wD(H c d) 4 Substitution into the force expression of part (a) gives Problem 3.8 Combining I min =( )( H c d)= 0.59H c d H m d + H g g =0; πr 0 B m =πr 0 lb g B g = µ 0 H g ; B m = µ R (H m H rmc ) gives

43 H c dµ 0 B g = ( )( ) g + µ0 ld µ R r 0 The flux linkages of the voice coil can be calculate in two steps. First calculate the differential flux linkages of a differential section of the voice coil of dn turns at height z above the bottom of the voice coil (which is at z = x). l dλ = dn B g (πr 0 )dz = ( H cdµ 0 )(πr 0 )(l z ) ( )( ) dn z g + µ0 ld µ R r 0 Recognizing that dn =(N /h)dz we can now integrate over the coil to find the total flux linkages λ = x+h x dλ = N ( H c dµ 0 )(πr 0 )(l x h/) ( )( ) g + µ0 ld µ R r 0 part (c): Note from part (a) that the magnet in this case can be replaced by a winding of N 1 i 1 = H c d ampere-turns along with a region of length d and permeability µ R. Making this replacement from part (a), the self inductance of the winding can be found and thus λ 11 = N 1 Φ 11 =πr 0 hn 1 B g = πr 0hN1 dµ ( )( 0 )i 1 g + µ0 ld µ R r 0 L 11 = πr 0hN1 dµ 0 ( )( g + µ0 ld µ R Similar, the mutual inductance with the voice coil can be found from part (b) as r 0 ) L 1 = λ = N 1λ i 1 H c d = N N µ 0 (πr 0 )(l x h/) ( )( ) g + µ0 ld µ R r 0 We can now find the coenergy (ignoring the term L i /) W fld = 1 L 11i 1 + L 1i 1 i = µ 0(H c d) πr 0 h µ 0 N ( H c d)(πr 0 d)(l x h ( )( ) + ( )( ) ) i g + µ0 ld µ R r 0 g + µ0 ld µ R r 0

44 part (d): f fld = dw fld dx = µ 0N ( H c d)(πr 0 d) ( )( ) g + µ0 ld µ R r 0 Problem 3.9 H m t m + H x x + H g g =0; π(r 3 R )B m = πr 1 B x =πr 1 hb g B g = µ 0 H g ; B x = µ 0 H x ; B m = µ R (H m H c ) where µ R =1.05µ 0 and H c = 71 ka/m. Solving gives B g = µ 0 R 1 ( H c t m ) =0.56 T hx + gr 1 + µ0r 1 htm µ R(R 3 R ) and ( ) h B x = B g =0.535 T R 1 We can replace the magnet by an equivalent winding of Ni = H c t m. The flux linkages of this equivalent winding can then be found to be λ = N(πR 1 h)b g = πµ 0 hr1 N i = Li hx + gr 1 + µ0r 1 htm µ R(R 3 R ) The force can then be found as part (c): f fld = i = dl dx = πµ 0 (hr 1 ) (Ni) ( hx + gr 1 + µ0r 1 htm µ R(R 3 R ) ) πµ 0 (hr 1 ) ( H c t m ) ( ) = 0.0158 N hx + gr 1 + µ0r 1 htm µ R(R 3 R ) X 0 = x f K =4.0 mm

45 Problem 3.30 If the plunger is stationary at x =0.9a, the inductance will be constant at L =0.1L 0. Thus where τ = L/R. The force will thus be f fld = i i(t) = V 0 R e t/τ dl dx = L 0 a ( ) V0 e t/τ R X 0 =0.9a + f K 0 =0.9a L 0 ak 0 ( ) V0 R Problem 3.31 Since the current is fixed at i = I 0 = 4 A, the force will be constant at f = I 0 L 0 /()a = 1.45 N. Thus X 0 =0.9 a + f K 0 =1.56 cm M d x dt = f + K 0(0.9a x) 0. d x dt =5.48 350x N dl v = I 0 R + I 0 dt = I 0R L 0 dx v =6 0.18dx a dt dt part (c): The equations can be linearized by letting x = X 0 + x (t) and v = V 0 + v (t). The result is and d x dt = 1750x v = 0.18 dx dt part (d) For ɛ in meters, x (t) =ɛ cos ωt m where ɛ = 1750 = 41.8 rad/sec and

46 v (t) =7.61ɛ sin ωt V Problem 3.3 For a dc voltage of V 0 = 6 V, the corresponding dc current will be I 0 = V 0 /R = 4 A, the same as Problem 3.31. Hence the equilibrium position will be the same; X 0 =1.56 cm. For a fixed voltage, the dynamic equations become: or and or V 0 = ir + d dt (Li) =ir + L 0 ( 1 x ) di a dt ( L0 6=1.5i +4 10 3 (1 40x) di (0.18) idx dt dt M d x dt = f K 0(0.9a x) = ( i a ) i dx dt )( ) L0 + K 0 (0.9a x) a 0. d x dt = 0.0909i +6.93 350x part (c): The equations can be linearized by letting x = X 0 + x (t) and i = I 0 + i (t). The result is or and or ( 0=i R + L 0 1 X ) 0 di a dt ( ) L0 dx I 0 a dt 0=1.5i 3 di +1.5 10 dt 0.78dx dt M d x dt = ( ) I0 L 0 i K 0 x a 0. d x dt = 0.77i 350x Problem 3.33 Following the derivation of Example 3.1, for a rotor current of 8 A, the torque will be give by T = T 0 sin α where T 0 = 0.0048 N m. The stable equilibrium position will be at α =0.

47 J d α dt = T 0 sin α part (c): The incremental equation of motion is and the natural frequency is J d α dt = T 0α ω = T0 J =0.6 rad/sec corresponding to a frequency of 0.099 Hz. Problem 3.34 As long as the plunger remains within the core, the inductance is equal to L = µ 0dπN ag ( (a ) x ) where x is the distance from the center of the solenoid to the center of the core. Hence the force is equal to f fld = i dl dx = µ 0dπN i x ag Analogous to Example 3.10, the equations of motor are f t = M d x dt B dx dt K(x l 0) µ 0dπN i x ag The voltage equation for the electric system is v t = ir + µ 0dπN ( (a ) x ag ) di dt µ 0dπN x ag These equations are valid only as long as the motion of the plunger is limited so that the plunger does not extend out of the core, i.e. ring, say, between the limits a/ <x<a/. dx dt

48 PROBLEM SOLUTIONS: Chapter 4 Problem 4.1 ω m = 100 π/30 = 40π rad/sec 60 Hz; 10π rad/sec part (c): 100 5/6 = 1000 r/min Problem 4. The voltages in the remaining two phases can be expressed as V 0 cos (ωt π/3) and V 0 cos (ωt +π/3). Problem 4.3 It is an induction motor. parts (b) and (c): It sounds like an 8-pole motor supplied by 60 Hz. Problem 4.4

49 part (c): part (d): Problem 4.5 Under this condition, the mmf wave is equivalent to that of a single-phase motor and hence the positive- and negative-traveling mmf waves will be of equal magnitude. Problem 4.6 The mmf and flux waves will reverse direction. Reversing two phases is the procedure for reversing the direction of a three-phase induction motor. Problem 4.7 F 1 = F max cos θ ae cos ω e t = F max (cos (θ ae ω t ) + cos (θ ae + ω t )) F = F max sin θ ae sin ω e t = F max and thus (cos (θ ae ω t ) cos (θ ae + ω t )) F total = F 1 + F = F max cos (θ ae ω t )

50 Problem 4.8 For n odd β/ β/ π/ π/ cos (nθ)dθ cos (nθ)dθ = sin (nθ ) For β =5π/6, sin ( nθ )= 0.97 n = 1 0 n = 3 0.6 n = 5 Problem 4.9 Rated speed = 100 r/min I r = πgb ag1,peak(poles) 4µ 0 k r N r = 113 A part (c): Φ P = ( ) lrb ag1,peak =0.937 3 Wb Problem 4.10 From the solution to Problem 4.9, Φ P =0.937 Wb. V rms = ωnφ =8.4 kv Problem 4.11 From the solution to Problem 4.9, Φ P =0.937 Wb. V rms = ωk wn a Φ =10.4 kv Problem 4.1 The required rms line-to-line voltage is V rms =13.0/ 3=7.51 kv. Thus Vrms N a = ωk w Φ Problem 4.13 The flux per pole is = 39 turns Φ=lRB ag1,peak =0.0159 Wb The electrical frequency of the generated voltage will be 50 Hz. The peak voltage will be

51 V peak = ωnφ = 388 V Because the space-fundamental winding flux linkage is at is peak at time t = 0 and because the voltage is equal to the time derivative of the flux linkage, we can write v(t) =±V peak sin ωt where the sign of the voltage depends upon the polarities defined for the flux and the stator coil and ω = 10π rad/sec. In this case, Φ will be of the form Φ(t) =Φ 0 cos ωt where Φ 0 =0.0159 Wb as found in part (a). The stator coil flux linkages will thus be λ(t) =±NΦ(t) =NΦ 0 cos ωt = ± 1 NΦ 0(1 + cos ωt) and the generated voltage will be v(t) = ωφ 0 sin ωt This scheme will not work since the dc-component of the coil flux will produce no voltage. Problem 4.14 Similarly, we can write F a = i a [A 1 cos θ a + A 3 cos 3θ a + A 5 cos 5θ a ] = I a cos ωt[a 1 cos θ a + A 3 cos 3θ a + A 5 cos 5θ a ] F b = i b [A 1 cos (θ a 10 )+A 3 cos 3(θ a 10 )+A 5 cos 5(θ a 10 )] = I a cos (ωt 10 )[A 1 cos (θ a 10 )+A 3 cos 3θ a + A 5 cos (5θ a + 10 )] and F c = i c [A 1 cos (θ a + 10 )+A 3 cos 3(θ a + 10 )+A 5 cos 5(θ a + 10 )] = I a cos (ωt + 10 )[A 1 cos (θ a + 10 )+A 3 cos 3θ a + A 5 cos (5θ a 10 )] The total mmf will be

5 F tot = F a + F b + F c = 3 I a[a 1 cos (θ a ωt)a 5 cos (5θ a + ωt)] = 3 ( I a[a 1 cos (θ a ωt)a 5 cos 5 θ a +( ωt ) 5 ) ] We see that the combined mmf contains only a fundamental space-harmonic component that rotates in the forward direction at angular velocity ω and a 5 th space-harmonic that rotates in the negative direction at angular velocity ω/5. Problem 4.15 The turns must be modified by a factor of ( )( ) 18 100 = 9 4 1400 14 =0.64 Problem 4.16 Φ p = 30E a N(poles)n =6.5 mwb Problem 4.17 ( Φ p = poles ) B peak lr = ( ) 1.5 0.1 (.095/) = 1.5 mwb 4 N ph = V rms poles (30/ 3) 4 = = 43 turns πfme k w Φ p π 60 0.95 0.015 From Eq. B.7 Problem 4.18 L = 16µ 0lr πg ( ) kw N ph =1. mh poles Φ p = V rms πnph =10.8 mwb B peak = Φ p lr =0.53 T

53 I f = πb peakg µ 0 k r N r =0.65 A part (c): L af = λ a,peak Vrms /ω = =0.69 H I f I f Problem 4.19 No numerical solution required. Problem 4.0 Φ peak = ( ) Dl B peak poles F r,peak = 4k rn r I r,max π poles ( ) poles Φ peakf r,peak =4.39 10 6 N m T peak = π P peak = T peak ω m = 88 MW Problem 4.1 Φ peak = ( ) Dl B peak poles F r,peak = 4k rn r I r,max π poles ( ) poles Φ peakf r,peak =16.1 N m T peak = π P peak = T peak ω m =6.06 kw Problem 4. dm af dm bf T = i a i f + i b i f dθ 0 dθ 0 = Mi f (i b cos θ 0 i a sin θ 0 )

54 This expression applies under all operating conditions. T =MI 0 (cos θ 0 sin θ 0 )= MI 0 sin (θ 0 π/4) Provided there are any losses at all, the rotor will come to rest at θ 0 = π/4 for which T = 0 and dt/dθ 0 < 0. part (c): part (d): T = MI a I f (sin ωtcos θ 0 cos ωt sin θ 0 ) = MI a I f sin (ωt θ 0 )= MI a I f sin δ v a = R a i a + d dt (L aai a + M af i f ) = I a (R a cos ωt ωl aa sin ωt) ωmi f sin (ωt δ) Problem 4.3 v b = R a i b + d dt (L aai b + M bf i f ) = I a (R a sin ωt + ωl aa cos ωt)+ωmi f cos (ωt δ) T = MI f (i b cos θ 0 i a sin θ 0 ) = MI f [(I a + I /) sin δ +(I /) sin (ωt + δ)] The time-averaged torque is thus Problem 4.4 <T >= MI f (I a + I /) sin δ T = i a dl aa dθ 0 + i b dl bb dθ 0 + i a i b dl ab dθ 0 = I a I f M sin δ +I al sin δ + i a i f dm af dθ 0 + i b i f dm bf dθ 0 Motor if T>0, δ>0. Generator if T<0, δ<0. part (c): For I f = 0, there will still be a reluctance torque T =I a L sin δ and the machine can still operate.

55 Problem 4.5 v = f λ =5 m/sec The synchronous rotor velocity is 5 m/sec. part (c): For a slip of 0.045, the rotor velocity will be (1 0.045) 5 = 3.9 m/sec. Problem 4.6 ) (π ) ( ) p 4 k w N ph = 1.45 ( )( ) 9.3 10 3 (π 3 4 I rms = B peak ( g µ 0 )( 3 µ 0 Problem 4.7 Defining β = π/wavelength ) ( 7 0.91 80 ) = 18 A Φ p = w π/β 0 B peak cos βxdx = wb peak β = 1.48 mwb Since the rotor is 5 wavelengths long, the armature winding will link 10 poles of flux with 10 turns per pole. Thus, λ peak = 100Φ p =0.148 Wb. part (c): ω = βv and thus V rms = ωλ peak =34.6 V, rms

56 PROBLEM SOLUTIONS: Chapter 5 Problem 5.1 Basic equations are T Φ R F f sin δ RF. Since the field current is constant, F f is constant, Note also that the resultant flux is proptoortional to the terminal voltage and inversely to the frequency Φ R V t /f. Thuswecanwrite Reduced to 31.1 Unchanged part (c): Unchanged part (d): Increased to 39.6 T V t sin δ RF f P = ω f T V t sin δ RF Problem 5. The windings are orthogonal and hence the mutual inductance is zero. Since the two windings are orthogonal, the phases are uncoupled and hence the flux linkage under balanced two-phase operation is unchanged by currents in the other phase. Thus, the equivalent inductance is simply equal to the phase self-inductance. Problem 5.3 L ab = 1 (L aa L al )=.5 mh L s = 3 (L aa L al )+L al =7.08 mh Problem 5.4 L af = Vl l,rms 3ωIf =79.4 mh Voltage = (50/60) 15.4 kv = 1.8 kv. Problem 5.5 The magnitude of the phase current is equal to 40 10 3 I a = 0.85 3 460 =59.1 A and its phase angle is cos 1 0.85 = 31.8.Thus

57 Then Î a =59.1e j31.8 Ê af = V a jx s Î a = 460 3 j4.15 59.1e j31.8 = 136 56.8 V The field current can be calculated from the magnitude of the generator voltage I f = Eaf ωl af =11.3 A Ê af = 66 38.1 V; I f =15.3 A part (c): Ê af = 395 7.8 V; I f =0. A Problem 5.6 The solution is similar to that of Problem 5.5 with the exception that the sychronous impedance jx s is replaced by the impedance Z f + jx s. Ê af = 106 66.6 V; I f =1. A Ê af = 61 43.7 V; I f =16.3 A part (c): Ê af = 416 31. V; I f =.0 A Problem 5.7 L af = Vl l,rms 3ωIf =49.8 mh Î a = 600 103 3 300 = 151 A Ê af = V a jx s Î a =1.77 41.3 V I f = Eaf ωl af = 160 A

58 part (c): See plot below. Minimum current will when the motor is operating at unity power factor. From the plot, this occurs at a field current of 160 A. Problem 5.8 Z base = V base P base = (6 103 ) 750 10 6 =0.901 Ω part (c): L s = X s,puz base ω L al = X al,puz base ω =4.88 mh =0.43 mh L aa = 3 (L s L al )+L al =3.40 mh Problem 5.9 SCR = AFNL AFSC =0.50 Z base =(6 10 3 ) /(800 10 6 )=0.845 Ω X s = 1 =.19 pu = 1.85 Ω SCR part (c): X s,u = AFSC =1.9 pu = 1.6 Ω AFNL, ag

59 Problem 5.10 SCR = AFNL AFSC =1.14 Z base = 4160 /(5000 10 3 )=3.46 Ω part (c): X s = 1 =1.11 pu = 3.86 Ω SCR X s,u = Problem 5.11 No numerical solution required. AFSC =0.88 pu = 3.05 Ω AFNL, ag Problem 5.1 The total power is equal to S = P /pf = 400 kw/0.87 = 488 kva. The armature current is thus Î a = 488 103 3 4160 (cos 1 0.87) = 670 9.5 A Defining Z s = R a + jx s =0.038 + j4.81 Ω E af = V a Z s I a = 4160 3 Z s I a = 4349 V, line to neutral Thus ( ) 4349 I f =AFNL 4160/ = 306 A 3 If the machine speed remains constant and the field current is not reduced, the terminal voltage will increase to the value corresponding to 306 A of field current on the open-circuit saturation characteristic. Interpolating the given data shows that this corresponds to a value of around 4850 V line-to-line.

60 Problem 5.13 Problem 5.14 At rated power, unity power factor, the armature current will be I a = 5000 kw/( 3 4160 V) = 694 A. The power dissipated in the armature winding will then equal P arm =3 694 0.011 = 15.9 kw. The field current can be found from E af = V a Z s I a = 4160 Z s I a = 3194 V, line-to-neutral 3 and thus ( ) 3194 I f =AFNL 4160/ = 319 A 3 At 15 C, the field-winding resistance will be ( ) 34.5 + 15 R f =0.79 =0.34 Ω 34.5+75 and hence the field-winding power dissipation will be P field = I f R f = 1.1 kw. The total loss will then be P tot = P core + P arm + P friction/windage + P field = 10 kw Hence the output power will equal 4880 kw and the efficiency will equal 4880/5000 = 0.976 = 97.6%.

61 Problem 5.15 AFNL = 736 A. AFSC = 710 A. part (c): (i) SCR = 10.4, (ii) X s = 0.964 per unit and (iii) X s,u =1.17 per unit. Problem 5.16 For V a = 1.0 per unit, E af,max =.4 per unit and X s = 1.6 per unit Q max = E af,max V a X s =0.875 per unit Problem 5.17 Z base = V base P base =5.9 Ω X s = 1 =0.595 per-unit = 3.15 Ω SCR

6 Using generator convention for current part (c): For V a = 1.0 per-unit, E af = 150 =0.357 per-unit 40 Î a = E af V a =1.08 jx s 90 per-unit = 1.36 90 ka using I base = 155 A. part (d): It looks like an inductor. part (e): For V a = 1.0 per-unit, E af = 700 =1.67 per-unit 40 Î a = E af V a jx s =1.1 90 per-unit = 1.41 90 ka In this case, it looks like a capacitor.

63 Problem 5.18 Problem 5.19 It was underexcited, absorbing reactive power. It increased. part (c): The answers are the same. Problem 5.0 X s = 6 =0.68 per-unit 84 P =0.875 and S = P/0.9 =0.97, both in per unit. The powerfactor angle is cos 1 0.9 = 5.8 and thus Îa =0.875 5.8. Ê af = V a + jx s Î a =1.15 11.6 per-unit The field current is I f =AFNL Êaf = 958 A. The rotor angle is 11.6 and the reactive power is Q = S P =4.4 MVA part (c): Now E af = 1.0 per unit. ( ) δ =sin 1 PXs E af V a and thus Êaf =1.0 13.6. =13.6 Î a = Êaf V a jx s =0.881 6.79 Q = Imag[V a Îa ]= 0.104 per-unit = 1.04 MVAR

64 Problem 5.1 Î a = Êaf V a jx s = j V a X s + E af X s (sin δ j cos δ) The first term is a constant and is the center of the circle. The second term is a circle of radius Eaf/X s. Problem 5. (i) (ii) V t = V = 1.0 per unit. P = 375/650 = 0.577 per unit. Thus ( ) δ t =sin 1 PX =1.6 V t V

65 and Î a = V te jδt V jx =0.578 3.93 per-unit I base = P base /( 3 V base ) = 15.64 ka and thus I a =9.04kA. (iii) The generator terminal current lags the terminal voltage by δ t / and thus the power factor is (iv) pf = cos 1 δ t /=0.998 lagging Êaf = V + j(x + X s )Îa =1.50 per-unit = 36.0 kv,line-to-line (i) Same phasor diagram (ii) Îa =0.98 6.3 per-unit. I a = 14.5 ka. (iii) pf = 0.994 lagging (iv) E af =.06 per unit = 49.4 kv, line-to-line. Problem 5.3

66 part (c): Problem 5.4 From the solution to Problem 5.15, X s =0.964 per unit. Thus, with V = E af = 1.0 per unit

67 P max = V E af =1.04 per-unit X s noindent Hence, full load can be achieved. This will occur at ( ) δ =sin 1 Xs =74.6 E af V infty The generator base impedance is 1.31 Ω. Thus, X =0.14/1.31 = 0.107 per unit. Now P max = V E af =1.04 per-unit = 0.934 per-unit = 135 MW (X + X s Problem 5.5 Follwing the calculation steps of Example 5.9, E af =1.35 per unit. Problem 5.6 Now X d =.964 per unit and X q = 0.73 per unit. Thus P = V E af sin δ + V ( 1 1 ) sin δ =1.037 sin δ +0.173 sin δ X d X q X d An iterative solution with MATLAB shows that maximum power can be achieved at δ =53.6. Letting X D = X d + X and X Q = X q + X P = V E af X sin δ + V ( 1 1 ) sin δ =0.934 sin δ +0.136 sin δ X Q X D An iterative solution with MATLAB shows that maximum power that can be achieved is 141 Mw, which occurs at a power angle of 75. Problem 5.7

68 Problem 5.8 Problem 5.9 Problem 5.30 For E af =0, P max = V t ( 1 X q 1 ) =0.1 = 1% X q This maximum power occurs for δ =45. I d = V t cos δ =0.786 per-unit X d I q = V t sin δ X q =1.09 per-unit

69 and thus I a = Hence Id + I q = 1.34 per unit. S = V t I a =1.34 per-unit Q = S P =1.3 per-unit Problem 5.30 P = V E af sin δ + V ( 1 1 ) sin δ X d X q X d The generator will remain synchronized as long as P max >P. An iterative search with MATLAB can easily be used to find the minimum excitation that satisfies this condition for any particular loading. For P =0.5, must have E af 0.37 per unit. For P =1.0, must have E af 0.87 per unit. Problem 5.3 We know that P =0.95 per unit and that and that P = V V t X bus sin δ t Î a = ˆV t V jx t It is necessary to solve these two equations simultaneously for ˆV t = V t δ t so that both the required power is achieved as well as the specified power factor

70 angle with respect to the generator terminal voltage. This is most easily done iteratively with MATLAB. Once this is done, it is straightforward to calculate V t =1.0 per-unit; E af =.05 per-unit; δ =46.6 Problem 5.33 Define X D = X d + X bus and X Q = X q + X bus. (i) E af,min = V bus X D =0.04 per-unit (ii) E af,max = V bus + X D =1.96 per-unit

71 part (c): Problem 5.34 f = n poles 10 = 3000 6 10 = 150 Hz Problem 5.35 Because the load is resistive, we know that I a = P = 4500 =13.5 A 3V a 319 We know that E af = 08/ 3 = 10 V. Solving for X s gives E af = V a +(X si a ) E X s = af Va =3.41 Ω I a part (c): The easiest way to solve this is to use MATLAB to iterate to find the required load resistance. If this is done, the solution is V a = 108 V (line-to-neutral) = 187 V (line-to-line). Problem 5.36 Thus Î a = E a R a + R b + jωl a = ωk a R a + R b + jωl a Îa = ωk a (Ra + R b ) +(ωl a) = K a 1+( ) L Ra+R b a ωl a Clearly, I a will remain constant with speed as long as the speed is sufficient to insure that ω>>(r a + R b )/L a

7 PROBLEM SOLUTIONS: Chapter 6 Problem 6.1 Synchronous speed is 1800 r/min. Therefore, 1800 1755 s = =0.05 =.5% 1800 Rotor currents are at slip frequency, f r = s60 = 1.5 Hz. part (c): The stator flux wave rotates at synchronous speed with respect to the stator (1800 r/min). It rotates at slip speed ahead of the rotor (s1800 = 45 r/min). part (d): The rotor flux wave is synchronous with that of the stator. Thus it rotatesat synchronous speed with respect to the stator (1800 r/min). It rotates at slip speed ahead of the rotor (s1800 = 45 r/min). Problem 6. The slip is equal to s =0.89/50 = 0.0178. The synchronous speed for a 6-pole, 50-Hz motor is 1000 r/min. Thus the rotor speed is n =(1 s)1000 = 98 r/min The slip of a 4-pole, 60-Hz motor operating at 1740 r/min is 1800 1740 s = =0.0333 = 3.33% 1800 The rotor currents will therefore be at slip frequency f r =60 0.0333 = Hz. Problem 6.3 The synchronous speed is clearly 100 r/min. Therefore the motor has 6 poles. The full-load slip is 100 111 s = =0.0733 = 7.33% 100 part (c): The rotor currents will be at slip frequency f r =60 0.0733 = 4.4 Hz. part (d): The rotor field rotates at synchronous speed. Thus it rotates at 100 r/min with respect to the stator and (100-111) = 88 r/min with respect to the rotor. Problem 6.4 The wavelenth of the fundamental flux wave is equal to the span of two poles or λ =4.5/1 = 0.375 m. The period of the applied excitation is T =1/75 = 13.33 msec. Thus the synchronous speed is

73 v s = λ T =8.1 m/sec = 101.3 km/hr Because this is an induction machine, the car in this case) will never reach synchronous speed. part (c): 101.3 95 s = =0.06 = 6.% 101.3 The induced track currents will be a slip frequency, f = s75 = 4.66 Hz. part (d): For a slip of 6.%and a car velocity of 75 km/hr, the synchronous velocity must be v s = 75 =80.0 km/hr 1 s Thus the electrical frequency must be ( ) 75 f =80 =59. Hz 101.3 and the track currents will be at a frequency of sf =3.68Hz. Problem 6.5 For operation at constant slip frequency f r, the applied electrical frequency f e is related to the motor speed in r/min n as f e = n ( poles 10 ) + f r and thus, since the slip frequency f r remains constant, we see that the applied electrical frequency will vary linearly with the desired speed. Neglecting the voltage drop across the armature leakage inductance and winding resistance, the magnitude of the armature voltage is proportional to the air-gap flux density and the frequency. Hence the magnitude of the armature voltage must vary linearly with electrical frequency and hence the desired speed. The electrical frequency of the rotor currents is equal to the slip frequency and hence will remain constant. Since the rotor will be operating in a constant flux which varies at a constant frequency, the magnitude of the rotor currents will be unchanged. part (c): Because the rotor air-gap flux density and the rotor currents are unchanged, the torque will remain constant. Problem 6.6 Since the torque is proportional to the square of the voltage, the torque-speed characteristic will simply be reduced by a factor of 4. Neglecting the effects of stator resistance and leakage reactance, having both the voltage and frequency maintains constant air-gap flux. Hence