Baixe 220558812 - John - Grainger - Author - Jr - William - Stevenson - Author - Power - System - Analysis - Solution - Manual e outras Manuais, Projetos, Pesquisas em PDF para Engenharia Elétrica, somente na Docsity! Chap 1 Problem Solutions 1.1 [fev — 141 4sin(it + 30º) V and é = 11.81 cos(wt — 30º) A. find for each (a) the maximum value, (b) the rms value and (c) the phasor expression ir polar . . to s Ds . and rectangular form if voltage is the 'Teferençe* Is the circuit inductive or º capacitive? no . Ss ; T- IH 4 Sm (E soda e co , 2 “ Solution: = (4,4 Cos (ustt3E KO) - | - ” (a) Maximum values ico -Çe Vmaz = 1414 V Imar = HI A (b) rms values: 4 141.4 11.31 Vi= DD = 100V l=-"= =8A — vi= = | (c) Phasor expressions in polar and rectangular form: v = 100,25 100+;0V Quito 1 = 8Mf-a-joa The circuit is inlogúive as I legs V. +30 lecas 1.2 If the circuit of Prob. 1.1 consists of a purely resistive and a purely reactive element, find R and X, (a) if the elements are in series and (b) if the elements are in parallel, e Ã Y a MAC — Solution: Z2= FA º R (a) Elements in series: €, ça gem a e Ô E 7 - oo, o pos Sade aro . . “8 R = 680 Xt = 5868 O SG 3 6,89 A (b) Elements in parallel: N = G de 1 - ez Ot 808780" "="0-04+—s0-0603 Y = 5=eez=8= TE 27 Rox 42698 . plo xe = R = “ade "q = GQ) nie: Rtg> “ass 26 (a yemaral) 1.3 In a single-phase circuit V, = 120745º V and vu = 100,-15º V with respect to a reference node o. Find Vie in polar form. H Solution: Via = Vio-Vo Voa = 100,2-15º — 120,745º = 96,59 — 925.88 — (84.85 +;84.85) 11.74 = 110.73 = 111.357-83.95º V 2 1.4 A single-phase ac voltage of 240 V is applied to a series circuit whose impedance is 10/60º O. Find R, X, P, Q and the power factor of the circuit. Solution: R = Wecos60 = 50 9 X = 1sin60º = 866 O 240,0º = É = mp-ç0 , 10,60 — 24-80 A P = (24)x5 = 2880 W Q = (24) x8.66 = 4988 var pf = cos (tan) = 0.50 or cos (tan! F) = 0.50 1.5 If a capacitor is connected in parallel with the circuit of Prob. 1.4 and if this capacitor supplies 1250 var, find the P and Q supplied by the 240-V source, and find the resultant power factor. Solution: P = 2880 W Q = 4988-1250 = 3738 var 3738 = 10008) O pf. = cos (tan a) 0.61 1.6 A single-phase inductive load draws 10 MW at 0.6 power factor lagging. Draw the power triangle and determine the reactive power of a capacitor to be con- nected in parallel with the load to raise the power factor to 0.85. 13.33 Solution: 10. q = dg Sin (cos 06) = 13.33 a cos-i0.85 = 31.79º 10tan31.79º = 6.2 var Ha Q = -(1333-6.2) = —7.13 Mvar o | i | | | | | | | 5 (d) ja+ 02 Solution: (aje-1 = —0.5+70.866—-1 = 1.732/7150€ (b)l-02+a = 1-(-0.5 - 50.866) — 0.5+ 90.866 = 1+51.732 = 2,00,60º (c)a+a+j = —0.5-50.866-0.5+90866+31 = —1+91= 14147135 (d)jata? = 1/2102 41/240º = —0.866- 05-05-0866 = —1.366— 51.366 = 1.932,/225º 1.13 Three identical impedances of 10/—15º 9) are Y-connected to balanced three- phase line voltages of 208 V. Specify all the line and phase voltages and the currents as phasors in polar form with Vo as reference for a phase sequence of abe. Solution: Van = 1202210º V Vap = 208/240º V Voa = 120/90 V Voc = 208/120º V Vea = 120/-30º V Va = 208/00 V O Van 120/2100; =o h= Sea 2285 A O Vim 120/90 o 1, = Zoo 12,105º A O Va 120/5300 + o L= = 5 - lzmi5 A 1.14 alatrced-three-phase system the Y-connected impedances are 10730º O. e KH Ve = 4167 SO V; specify Tm in polar form. A nm vem Solution: 416 Gy Va = 240/-60º V la = e = 2M,-90º A A . G —s Ven = 246 (768º —— me as —6&” “ To Na ZitCS = 24 (MA E E 6 1.15 The terminals of a three-phase supply are labeled a, b and c. Between any ã pair a voltmeter measures 115 V. A resistor of 100 12 and a capacitor of 100 Q H at the frequency of the supply are connected in series from a to b with the 3 resistor connected to a. The point of connection of the elements to each other is labeled n. Determine graphically the voltmeter reading between c and n if phase sequence is abc and if phase sequence is acô. Solution: Sequence a-b-c 115 o ke = 116sin60º = 996 V meter reading = 57.5+996 = 157.1] V nk = = 575 V 9 Sequence a-c-b E Fe = 99. nk 7 meter reading = 99, H e i 1.16 Determine the current drawn from a three-phase 440-V line by a three-phase H 15-hp motor operating at full ioad, 90% efficiency and 80% power factor lagging. Find the values of P and Q drawn from the line. Solution: 15 x 746 = OLD = 29.39 A ui V3 x 440 x 0.9x 0.8 V3x 440 x 20.39 x 0.8 = 12,431 W drawn from line V3 x 440 x 20.39 x 0.6 V h O ] 9,324 var drawn from line 1.17 1.18 1.19 7 H the impendance of each of the thres lines connecting the motor of Prob. 1.16 toa bus is 0.3 + j1.0 9), find the line-to-line voltage at the bus which supplies 440 V at the motor. Solution: I = 20.39(0.8- 40.6) = 16.31 — j12.23 A When the reference is voltage to neutral of the motor at the terminal where J is calculated, or 440/V3 = 254/05 V, the supply bus voltage to neutral is 254 + 30 + (0.3 + 91.0)(16.31 — 712.23) Line-to-line voltage |V| = v3 [271.1 +912.64] 271.1+512.64 470 V A balanced-A load consisting of pure resistances of 15 $) per phase is in par- alle! with a balanced-Y load having phase impedances of 8 + 56 92. Identical impedances of 2 + j5 $2 are in each of the three lines connecting the combined loads to a 110-V three-phase supply. Find the current drawn from the supply and line voltage at the combined loads. Solution: Convert A to equivalent Y having 15/3 = 5 9/phase 5(8+76) — 40+530 13-56 7004350 5+8+76 — 18+76 13-96 * 205 341+70.732 = 349/121º Q Current drawn at supply: Z = 2+75+341+30.73 = 54149573 = 7.88/46.65º Q 110/v3 Hi = ca — 806 A from supply Letting V; equal voltage at the load, line-to-line voltage: V = 806x349 = 28.13 Vto neutral Linetoline > = v3x2813 = 4872 V A three-phase load draws 250 kW at a power factor of 0.707 lagging from a 440-V line. In parallel with this load is a three-phase capacitor bank which draws 60 kVA, Find the total current and resultant power factor. Solution: Letting S; and 5, represent the load and capacitor bank, respectively, Si = 25047250 10 Solution: Per-unit base calculations: (0.44)? x 1000 Base Z = 3 = 9.68 per unit R = Sã = 0.031 per unit X = E = 0.1033 per unit Basel = oa = 26.24 A [= e = 0.777 per unit Voltage calculations: V = 10+0.777(08 — 50.6) (0.031 + 50.1033) = 10+0.777 x 0.10797 36.43º = 10+0.0674+50.0498 = 1068672.97º per unit lVzL] = 1.0686x440 = 470 V 1.25 Write the two nodal admittance equations, similar to Egs. (1.57) and (1.58), for the voltages at nodes (2) and (2) of the circuit of Fig. 1.23. Then arrange the nodal admittance equations for ali four independent nodes of Fig. 1.23 into the Vous form of Eq. (1.61). Solution: dsO V-VKh+HM-VDN+M-V)Y = 0 ds O o UGHM-MDE+M-Wr = 1 Rearranging equations for bus (2) and bus (3) yields bus OD NR +HV(K+HNAY)- MANY =0 bs o VN -KLAV(G+HYA+4Y) = The Yous form is D 2 g a D [MW +r+Y) =—Ya —. —Y Y 0 2 —Ya G+R+Y) -—h -Y. hl|.|o 9 -Y. —% C+H+Y) o nl ix 9 —Yr -— 0 Cry +49] | y k 1.26 The values for the parameters of Fig. 1.23 are given in per unit as follows: Ya=-708 Y=-740 V=-780 “jo Y=-450 Y=-;25 Y=-;08 h6=10 (90º 1 =0.68/-135º OO 11 Substituting these values in the eguations determined in Prob. 1.25, compute the voltages at the nodes of Fig. 1.23. Numerically determine the corresponding Ztus matrix. Solution: Using the Ypus solution of Problem 1.25, substitute the given admittance values: 45 580 j40 925][W o jo -HTO 440 350 [| Vo | 0 jo 40 -j88 50 vw |T| 10,-98º 525 550 30 983] [Vs 0687-135 Compute voltages: YesV = I Vous Y us V Vous 1 jOTI8T j0.6688 30.0307 70.619 ao = 1506688 707045 50.6242 50.6258 where You = Zu = | 06307 07045 306840 30.5660 506194 06258 70.660 50.6840 Vo Yos Wu jO.TIBT 406688 30.6307 j0.6194 o vw | — | 506688 07045 506242 50.6258 0 vw | * | 906307 07045 40.6840 0.5660 1.0,-90º Va 506194 406258 30.5660 50.6840 | | 0.68,-135º Vi 0.9285 — 30.2978 0.97502-17.78º v | | 09251-503009 | — | 0.9728/-1802º w| = |osse2-gozo | * | og ,iõeo Va 0.8949 — 50.3289 0.9534, —20.18º Chapter 2 Problem Solutions 2.1 A single-phase transformer rated 7.2 KVA, 1.2kV/120 V has a primary wind- ing of 800 turns. Determine (a) the turns ratio and the number of turns in the secondary winding, (b) the currents carried by the two windings when the transformer delivers its rated kVA at rated voltages. Hence, verify Eq. (2.7). Solution: (a) Mo Mn 12x168º Mo Wo Tao NM 80 Therefore, No = EE 5200 Page lofl To: ee5200-|8 mtu.edu Subject: sign convention, conjugate, cosine reference Glad to see this kind of exchange on the e-mail list, that is what | have been hoping for. Some comments that could help with the Ch.1 review problems: 1) As I mentioned in the first lecture when we discussed Euler's identity, it is standard practice to define phasors according to the cosine (real) component and this is termed "cosine reference." Therefore, when converting from time domain to phasor domain, we must first convert all sinusoidal functions to equivalent cos functions. By sketching out a sine and a cosine function, it becomes clear that a sine is just a cosine that has been delayed by 90º. Therefore, sin(wt) = cos(wt -90º) or cos(wt) = sin(wt + 90º). 2) The negative sign associated with | is most likely due to how | is defined on the circuit, i.e. the assumed reference direction of current flow that is marked on the circuit. 3) To correctly calculate complex power consumed by (or flowing in to) a circuit element, Sin = VI” = P + jQ, where V and | have reference polarity/direction according to passive sign convention. 4) Careful with conjugates: remember that the conjugate of a complex number has the same magnitude, but the sign of its angle is changed. For example, ifl = 10/30º A, then |* = 10/-30º A. Thus, negating a complex number is not the same as taking its conjugate. Thanks for the comments on the Ch.1 problems, | will try to go thru these and then issue any corrections that may be required. Looks like we are off to a good start, good to be thinking about these details and doing some review/refreshing. Dr. Mork Printed for Bruce Mork <bamork Gmtu.edu> 9/9/2003 EE 5200 Page2of2 and figure it out. Any more points of uncertainty or possible errors? Please go ahead and start the discussion here, hopefully this is helpful. See you all in class tomorrow morning, we will go through some more per unit things. Dr Mork Printed for Bruce Mork <bamork Emtu.edu> 9/9/2003 EE 5200 To: ee5200-10mtu.edu Subject: Chapter one problem 1.7 At 12:02 PM 9/3/2008 -0500, you wrote: 1.7: | believe that they drew the correct conclusion about the efficiency but for the wrong reason. Anybody care to comment? The authors rationalization seems to be sound: Adding shunt capacitors (shunt compensation) reduces the inductive component of the current being drawn from the mains, i.e. flowing down the line, thus reducing the net current flowing in the line. This reduces the |"2 R line losses. The current flowing into the motor, however, is unchanged (assuming the capacitor placement has not changed the terminal voltage). This is essentially a power factor correction situation, no internal changes have been made to the motor, it is still operating in the same way and with the same efficiency. Since efficiency is related only to real power P, the Q that is produced by the caps has no effect on motor efficiency. There are some devices, in cyclic loading applications, that increase overall motor efficiency by reducing the source voltage to the motor when the mechanical load on the motor is removed/reduced, and then restores full voltage when the motor is loaded down again. Not restoring full voltage, or operating a loaded induction motor at reduced voltage will draw excessive current, resulting in a very low efficiency and extreme I/2 R heating of the armature windings, thus buming itout. One basic type of motor protection is thus to trip the motor off line if the voltage is too low and/or the current is too high, and/or if the winding temperature gets too high. Dr. Mork Printed for Bruce Mork <bamork Gmtu.edu> Page lofl 9/9/2003 11 Substituting these values in the equations determined in Prob. 1.25, compute the voltages at the nodes of Fig. 1.23. Numerically determine the corresponding Zipus Matrix. Solution: Using the Ypus Solution of Problem 1.25, substitute the given admittance values: -jj4s j80 440 $25 Vu 0 j80 —fT0 40 350 VW lo 0 j40 j40 -;88 30 Y 1.07-90º 25 45D 30 383 Va 0.68,-135º Compute voltages: YesV = 1 Vas bus Y Yu TI jOTIB7 $0.6688 70.6307 506194 ao = 506688 J0.7045 0.6042 30.6258 where Yous = Zous = | 06307 50.7045 306840 30.5660 06194 06258 0.5660 50.6840 V= Yo Y j0.7187 506688 40.6307 30.6194 o Y . 506688 40.7045 30.6242 j0.6258 o Va - 50.6307 30.7045 30.6840 30.5660 1.07-90º Va 40.6194 0.6258 50.5660 30.6840 0.682—135º Yi 0.9285 — 50.2978 0.9750,—17,78º Y a 0.9251 —;0.3009 | — | 0972871802 Va E 0.9562 — 40.272] * | 0.9941,-15.89º Va 0.8949 — 50.3289 0.9534,--20.18º Chapter 2 Problem Solutions 2.1 A single-phase transformer rated 7.2 kVA, 1.2kV/120 V has a primary wind- ing of 800 turns. Determine (a) the turns ratio and the number of turns in the secondary winding, (b) the currents carried by the two windings when the transformer delivers its rated KVA at rated voltages. Hence, verify Eq. (2.7). Solution: (a) Mo MN L2xiêo Moo wo CID o Mi 80 Therefore, No = = — erefore, No 10 10 14 Solution: (a) From Eq. (2.22) and (2.23), [4] = alt! Bol[h] = omo[48 65 1[%] so? [ TES 3.003 ] [ h ] 3.393 1.885 (b) From Eq. (2.25). [8] = [8 EE) - suo Sa 435 )14] to (1) [ioioe ] = 100008 [ 738 ]n hence A = 0.295,2-90º A A = mio v (ii) [ade] = tooxacoe| 2 )r hence Vi = 117.30/0º V h 1.11,-90º A 2.5 Forthe pair of mutually coupled coils shown in F ig. 2.4, develop an equivalent-T network in the form of Fig. 2.5. Use the parameter values given in Prob. 2.4 and assume that the turns ratio a equals 2. What are the values of the leakage Feactances of the windings and the magnetizing susceptance of the coupled coils? Solution: li = Ly-ely =19-2x09H=01H Lo Loa-Lioja = 05-2x09/2 H = 005 E Ly = 4x06H=02H Im = alo) =2x09H=18H w = 1207 rad/sec iss iss Eh Leakage reactances: = 3770 15 2 = Tá O 5.4 = mw = SÊ o = 88 q . . E E Magnetizing susceptance: Bm = wlm | 1207 XLE 5 = 1474 x10? 5 2.6 A single-phase transformer rated 1.2 kV /120 V, 7.2 kVA has the following wind- ing parameters: 7, = 0,8 9,17, = 1.2 92,1, = 0.01 Nand x, = 0.01 9. Deter- mine (a) the combined winding resistance and leakage reactance referred to the pri- mary side, as shown in Fig. 2.8, (b) the values of the combined parameters referred to the secondary side (c) the voltage regulation of the transformer when it is delivering 7.5 kVA to a load at 120 V and 0.8 power factor lagging. Solution: (a) With turns ratio a = 1.2 x 103/120 = 10, Rj = m+otr = 0:8+100x001 9 = 18 Q X = mteiz = 12+100x001 0 = 220 (b) R, 2 Ri/0? = 18/10 2 = 0018 Q X & Xe = 22/1000 = 0022 Q Z, à R 5X E e IA DAN po te) mes m = (18+922) Q + 7200 o - Le = |S9/Valz-8 = “150 (038.8) A = 60,-36.9º A - I . ha = EE =60,-369º A ” aVopr = 1200 V Vim = ab +hoeZo 16 4 1200 +60/-36.9 (184722) V = 1216 57/0198 V 120 V W.ri/a = 121.66 V (121.66 — 120) /120 = 1.38 % IFo.rc] H Box] % Regulation “ 2.7 A single-phase transformer is rated 440/220 V. 5.0 EVA. When the low-voltage side is short circuited and 35 V js applied to the high-voltage side, rated current flows in the windings and the power input is 100 W. Find the resistance and Teactance of the high- and low-voltage windings if the power loss and ratio of Teactance to resistance is the same in both windings. Solution: Rated / 22.73 A (low voltage) Za9 = 11.36 A (high voltage) 2 = 3 & 308 N(RZ,X high-voltage) X = 3082-0.7742 = 298 Q ã - 28 = 385 For equal loss in high- and low-voltage windings, 77 High voltage:r = e = 0.887 Q 5 = 385x0387 = 149 Q 2 Low voltage: r = 0.387x (3) = 0097 Q 220)? = 149[) «0 z 14 (8) 0.373 9 2.8 A single-phase transformer rated 1.2 kV (120 V, 7.2 KVA yields the follówing test results: Open-Cireuit Test (Primary Open) Voltage V> = 120 V; Current L = 1.2 A; Power W, = 40 W Short-Circuit Test (Secondary Shorted) Voltage Vi = 20 V; Current = 604: Power W = 36 W 19 Output = 120x63245x0.8W = 607157 W 6071.57 aorrs7 sao, ao — 8700 % 120 x 63.245 VA = 7.589 kVA mar Corresponding KVA level 2.10 A single-phase system similar to that shown in Fig. 2.10 has two transformers A-B and B-C connected by a line B feeding a load at the receiving end C. The ratings and parameter values of the components are Transformer 4-B: 500 V/1.5 kV, 9.6 kVA, leakage reactance = 5% Transformer B-C: 1.2 kV/120 V, 7.2 EVA, leakage reactance = 4% Line B: series impedance = (0.5 + 93.0) Q Load €: 120 V, 6 kVA at 0.8 power factor lagging (a) Determine the value of the load impedance in ohms and the actual ohmic impedances of the two transformers referred to both their primary and secondary sides. (b) Choosing 1.2 kV as the voltage base for circuit B and 10 KVA as the systemwide KVA base, express all system impedances in per unit. (c) What value of sending-end voltage corresponds to the given loading condi- tions? Solution: (e) Ohmic impedances 2 Transformer A-B Primary: FE x 40.05 = 1.302 Q 152 x 108 . Secondary: FEx 10 x 70.05 = ;11.719 Q 2 6 Transformer B-C Primary: ea xj004 = j80 Q 2 Secondary: =p x 40.04 = ;0.08 Q vp 2 Load: fa 26 = Es cestos = 24,369º 9 : cs y (b) Impedance bases E ! o 1.22 x 108 Na Circuit B: 0x Io = 144 Q 2 Circuit €: 120 = 144 0 10 x 103 20 Per unit impedances on new bases: “11.719 “as 8 “ias (15+730) 144 24 . Load: Tag (88 Transformer A-B: = 30.08138 per unit Transformer B-C; 40.0556 per unit Line B: H 9.0104 + 50.0208 per unit H 1.667736.9º per unit (e) Sending-end voltage calculations sE 0.0104 + j 0.0208 j 008138 100556 — (0.0104 +; 0.15778) pu 1.567 .369º pa Ve = 120V = 10 per unit 1.667736.9º + (0.0104 + 20.15778) . = 1. = 1.06 Vs 10x 166773695 1.0642 per unit The sending-end voltage base is Vobae = x12xHº = 400 V 500 15x 103 Therefore, the required sending-end voltage is Vo = 400x 10642 = 425.69 V 2.11 À balanced A-connecied resistive load of 8000 KW is connected to the low- voltage, A-connected side of a Y-A transformer rated 10,000 kVA, 138/13.8 kV. Find the load resistance in ohms in each phase as measured from line to neutral on the high-voltage side of the transformer. Neglect transformer impedance and assume rated voltage is applied to the transformer primary. Solution: 8,000 Bino) = 2 = 3347 À Vinel = Ja xi R 138,000/v3 = 2380 9 33.47 2.12 Solve Prob. 2.11 if the same resistances are reconnected in Y. Solution: Tfthe A-connected resistors are Teconnected in Y, then the resistance to neutral will be three times as great and R = 3x2380 = 7140 Q 21 2.13 Three transformers. each rated 5 kVA, 220 V on the secondary side, are conected A-A and have been supplying a balanced 15 kW purely resistive load at 220 V. A change is made which reduces the load to 10 KW, still purely resistive and balanced. Someone suggests that, with two-thirds of the load, one transformer can be removed and the'system can be operated open-A. Balanced three-phase voltages will still be supplied to the load since two of the line voltages (and thus also the third) will be unchanged. To investigate further the suggestion (a) Find each of the line currents (magnitude and angle) with the 10 KW load and the transformer between a and c removed. (Assume V5, = 22070º V, sequence q bc.) (b) Find the kilovoltamperes supplied by each of the remaining transformers. (c) What restriction must be placed on the load for open-A operation with these transformers? (d) Think about why the individual transformer kilovoltampere values include a Q component when the load is purely resistive. Solution: a + 1, R R n R b — Aro nn —— (a) Vas and Vs. remain the same after removing the third transformer, so Vi, is also the same and we have a three-phase supply, and these voltages are: Vap = 220/0º V, Voo = 2207 240º V and Voa = 2207 120º V. Then, Van = 1272-30º V, Vin = 127/210º V and Ven = 1277890º V. The line currents are 10,000 = SD 430º = 26.24/-30º é v3 x 220 A h = 2624/210º A L = 2624/90 A (b) KVAsupplica = 220 x 26.24 x 103 = 5.772 KVA (c) The load must be reduced to (5.0/5.772) x 100 = 86.6% or 4.33 KW for each transformer. (d) The current and voltage in each of the remaining two transformers are not in phase. Output of each transformer before the reduction in load is, S4 = Valj = 220/0º x 2624/30" = 5000 +72886 VA Sa = Volj = 220/60º x 26.24/270º = 5000 - ;2886 VA | 24 Solution: i ! (a) Y-Y connection: i i 1MTRvI nov | = 12x100xv3 V 1X 13x 3EVA Wal = 120/83 V 2 12003 - r r Rj = 5x = 500 9 , . 1º ( 1205 ) a R R 2 (1.23) x 108 Z = taxas 08 o X, = 005 perunit = 200x0050 = 10 9 Z; = (500+310) Q (b) Y-A connection: Lata rizov 1% 12x 3 EVA Vil = 1200xv3 V coma Val = 120 V r rm 2 na 120003) a R R Rj = = (mê) = 1500 Q X, = 10 8 from part (a) = Z, = (1500+710) 9 (c) A-Y connection: 12ev ops v 1X T2XIKVA Vizl = 1200 V — vn [Val = 1203 V r 1200 42 500 Rj = 5x(D>—=) = >> = 18667 0 q R' R r (558) 3 12002 2 = qaxsrie — 6880 = X, = 005 perunit = 66.67x0.059 = 3.33 Q Zi = (16667+7333) Q (8) A-A connection: 1x a viva |Vzil = 1200 V gd ——S Mil = 120 r r . 120032. a R R Rj = 5x E) = 500 Q X = 333 9 from part (c) = Zi = (50043333) Q 2.17 Figure 2.174 shows a three-phase generator supplying a load through a three- phase transformer rated 12 KVA/600 V Y, 600 kVA. The transformer has per- phase leakage reactance of 10%. The line-to-line voltage and the line current at 25 the generator terminais are 11.9 kV and 20 À, respectively. The power factor seen by the generator is 0.8 lagging and the phase sequence of supply is ABC. (a) Determine the line current and the line-to-line voltage at the load, and the per-phase (equivalent-Y) impedance of the load. (b) Using the line-to-neutral voltage Vá at the transformer primary as refer- ence, draw complete per-phase phasor diagrams of all voltages and currents. Show the correct phase relations between primary and secondary quanti- ties. (c) Compute the real and reactive power supplied by the generator and con- sumed by the load. Solution: (a) Voltage ratio = Current ratio = Xo= LetVs = Then, !s = Il = vi = w = Line voltage at the load Line current at the load Load im-pedance - Zr x w h mA » » 4 dE da a:l 2x1 e a= o CS = 20/80 = = 0.057.308 12x 1087 fai x01 = 240 9 o E Rv = 687 EV Hj = 20,-36.9º A lja” = 20x20/-369º - 30º A = 400,-66.9º A Ve jXils = 68770º— (Fes) kV 6.503,-3.34º KV 6.593, 3.340 Vija = ese = - o Tia 50, 506 kV 329.65/-33.34º V vBlMl=5n V Hrl = 400 4 Vil = 329.65,—-33.34 400,568.9º = 0.824/836º Q H k e ed rmontenane eqre 26 6) ref, ref 3a Vs =6.87kV 36.9º .593 KV 30º E VL=329.65V Is=h'=204 ; h =400 A ! (c) Pg + 305 from the generator is 3VsT5, where IV) = 3x6.87/0ºx20/36.0º KVA = 412.2/36.9º KVA | = 329.8 kW+ 247.3 kvar Í Pr +5Q1 by the load is 3VL I7, where =. 3x 329:65/-33.34º x 4007 66.9º 3VL Is Dr E 1000 329.7 kW + 5218.7 kvar kVA = 395.67 33.56º kVA 2.18 Solve Prob. 2.17 with phase sequence ACB. Solution: (a) Final answers remain the same except for the following intermediate results: a = 20/-30º 1/a* = 0.057-30º Il = Ia = 400,/-360º 1300 A = 400,-6.9º A VL = Vija=32965,/-334+30ºV = 32065/067º V 6) Vi = 329.65 V ao ref zef, +. ar Vç= 6.87 EV Vi = 6.593 kV v=6.585 L =400A (c) Same results as in Problem 2.17. 2.19 A single-phase transformer rated 30 KVA, 1200/120 V is connected as an auto- transformer to supply 1320 V from a 1200 V bus. (a) Draw a diagram of the transformer connections showing the polarity marks on the windings and directions chosen as positive for current in each wind- ing so that the currents will be in phase. 29 and into bus (3 thru Xs, P+5Q = 4h = 0.289+ 50.167 per unit 03 0. h = 0.577-50.333-(-301) = 0.577 50.233 h = 0.289-50.167+(-50.1) = 0.289 — 50.267 Into bus (& thru X, o (b) AV =003; La= = —j01 ta to: P+5Q = 0.577 470.233 per unit and into bus (&) thru Xa, P+5Q = 0.2894 50.267 per unit (e) AV = 102210 = 0.99944+500349-10 = —0.0006 + 50.0349 —0.0006 + ;0.0349 . = DTL = 0116+70002 Loire 703 +3 h = 0577-0333 - (0.116+4 50.002) = 0.461 — 50.235 Jo = 0.289-50.167+0.116+4 40.002 = 0.405 — 50.165 Into bus (3) thru X;, P+5Q = WIº = 0.461 +70335 per unit and into bus (3) thru Xs, P+5Q = WI" = 0.405430.165 per unit Note: Compare P and Q found in parts (b) and (c) with part (a). 2.22 Two reactances X, = 0.08 and X> = 0.12 per unit are in parallel between two buses (2) and (3%) in a power system. If V, = 1.05/10º and V, = 1.020º per unit, what should be the turns ratio of the regulating transformer to be inserted in series with X> at bus (&) so that no vars flow into bus (2) from the branch whose reactance is X,? Use the circulating-current method, and neglect the reactance of the regulating transformer. P and Q of the load and V, remain constant. Solution: In reactance X,, 1.05210º — 1.0 1.034 + 50.1823 - 1.0 Io = 008 DO J008 = 2.279 — 50.425 To eliminate vars to bus (3) thru X;, we need in the Xo branch Jabeiro = —j0.425 AV . gergi - 4 a-1 = AV = -50.425(70.08 + 70.12) = 0.0850 a 1.085 turns ratio “YAM eve = (201/66) 03 peop sonpay ro -S To a SM SOL = a 0g0'T TO VAR 65 = sexgrxrero = Iºs] VAW FO = sexotxesgo = Ilsl vero = legrof—serol = |" - “Il Qom eeso = Lrrrof—20r0) = [mp4 A 3€ el aro - (SLOAEEOLOL: sus 1 13€ : 9g0'0 Av “9s00q apnyuBem %9'g UNA sovol-esro = (gal-go)FÉ = q Til se sogrot-z0r0 = (gol-go = 1 (go: sds 1 I “Proj VAN-SE UMA sjueimo “YAN S0g=SEx(Z4T/SI) 03 peo] sonpol soar) pue 'pepeojrao si 7 WUN ES9T0+SLST'O VAW CL = cex = E, AR g SESTO Fesl EE9TO + LST O = ex Sal = | VAN gui £ SESTO Psi mun sed cego = (sr/eg)x1go = “x Mun rd c;sTO = (02/sg)x600 = Ix “aseq USsoYo au) 03 SurgIsatos 4 :uoLynjoS “UIqOId SIW9 IO] ÁIOJIBISIYES SI PORjOUI qUSimo-SuryemoIr au, “opis aBeIjoA-MO] Sy3 UO AX Z'ET VA SE JO sseq € asn) “SIQULIOJSUBI) SU PBO[ISAO OU [LM WollMa prol [8]09 sy] JO saradureyjoreaur UIMUIIXEU SU) pue peol [240] VAIX S£ [eulfio ag) J0J IullojsueI? qoea jo qnd -Mmo sxodureyjoseSou at) pur “dey AX SIT SW) UO Suprema ota “7 IouLIoysUeI O] PaledUIOS IULIOJSURI) JE) JO 9pIs SSLIOA-MO| 9UY PIEMO3 SBEYOA UI 9500q i Y9E RB O AY LIT IP PES dE T JoULIOJsWeI) UO são) SUJ 3] "PopeopsaMo st ISULIOJSUBI) IOUMSU TEU OS poartury aq 1SNUI Pro [ej01 SU) YIIMA 09 saladureyos -eSoUI SW) PUB 'ISULIOJSUBI) ovo JO Indiano aladureajoseSsur auy 'IouLojsueI Uva WBNOLYI MUM Iod UI JUSLITO au) Jo opngufem oq3 purg gum sed 400 =X Uma VAIX CI poYel st 7 IouojsueI) pre gum Jd 600 = X qu VAN OE peyei st | JouLIojsucIT, SuiBBe] IOjoey Lomod 90 Ye AX ZEI VA CE Jo pro e 4tddns 09 famered us axezado AY Va SI/ASII PaYel WpPra SIDULIOJSUBI) OM] EZ'T 0s PE 31 Chapter 3 Problem Solutions 3.1 Determine the highest speed at which two generators mounted on the same shaft can be driven so that the freguency of one generator is 60 Hz and the frequency of the other is 25 Hz. How many poles does each machine have? Solution: Let P = number of poles: 2x60x60 2x60x25 speed = Po = Px Po (80 Ps 3 Peo and Pos must be even integral numbers lowest value where Psy = 2.4Pos. Thus, Pos =10 Pro = 24 3.2 The three-phase synchronous generator described in Example 3.1 is operated at 3600 rpm and supplies a unity power factor load. If the terminal voltage of the machine is 22 kV and the field current is 2500 A, determine the line current and the total power consumption of the load. Solution: Using the values in the solution of Example 3.1, 45855 Cota = 3855 * 2500 V = 20869.1 V Given: Vir = 22kV Van = (V2/V3)x 22000 V = 179629 V H ua — 17062.9 coswt, then ia = ia mus COS WÊ and e = 17962.9cosut -- 4.1484 x 1072 x 1207 X ia nus SÍNwt = 17962.9coswt — 15639 is... Sin wt Cata = 17962924 (15639500,,)) = 20869.1 V Hence, lou = 152594 A Li = ioualV2 = I079KA Po = v3x22x10.79x1MW = 411.2 MW 3.3 A three-phase round-rotor synchronous generator has negligible armature re- sistance and a synchronous reactance Xg of 1.65 per unit. The machine is connected directly to an infinite bus of voltage 1.070º per unit. Find the in- ternal voltage E; of the machine when it delivers a current of (a) 1.07230º per 34 3.5 A three-phase round-rotor synchronous generator, rated 16 KV and 200 MVA, has neghgible losses and synchronous reactance of 1.65 per unit. It is operated on an infinite bus having a voltage of 15 kV. The internal emf E; and the power angle ô of the machine are found to be 24 kV (line-to-line) and 27.4º, respectively. (a) (d) (c) Determine the line current and the three-phase real and reactive power being delivered to the system. If the mechanical power input and the field current of the generator are now changed so that the line current of the machine is reduced by 25% at the power factor of (a), find the new internal emf E; and the power angle õ. While delivering the reduced line current of (b), the mechanical power input and the excitation are further adjusted so that the machine operates at unity power factor at its terminals. Calculate the new values of E; and ô. Solution: (a) (66) Using 16 kV, 200 MVA base; V. = 15/16 per unit = 0.9375 per unit E/6 = E 274º per unit = 1.5/27.4º per unit E6-VizO = LX/90º —8 152274º -0.0375,0º = 1,x165/900-0 1/9008 = 0.4818/60.27º per unit Tas-8 = 0.4818,-29.73º per unit 200 x 102 Basel = 2" KA = T27KA aee v3 x 16 Thereiore, 1 = 048]8x7T.217KA = 3477 KA S = 0.9375x0.4818 per unit = 0.4517 per unit = 90.34 MVA Thus, P = 90.34c0529.73º MW = 78.45 MW Q = 90.34sin29.73º Mvar = 44.80 Mvar New L, = 0.75 x 0.4818 per unit = 0.3614 per unit 90-68 = 60.27º EiL6 = Viz0C+LX90 -8 H 0.937570º +0.3614 x 1.65260927º = 1.337/92.8º per unit = 214/2928 kV LL New ly = 0.3614 per unit 8 = 0 E/6 = 0.9375/0º+40.3614 x 1.65/90º 1111/32.5º per unit = 17.87/32,5º kV 3.6 The three-phase synchronous generator of Prob. 3.5 is operated on an infinite bus of voltage 15 kV and delivers 100 MVA at 0.8 power factor lagging. (a) Determine the internal voltage E;, power angle é and the line current of the machine. (b) Ifthe field current of the machine is reduced by 10%, while the mechanical power input to the machine is maintained constant, determine the new value of ô and the reactive power delivered to the system. (c) The prime mover power is next adjusted without changing the excitation so that the machine delivers zero reactive power to the system. Determine the new power angle é and the real power being delivered to the system. (d) What is the maximum reactive power that the machine can deliver if the level of excitation is maintained as in (b) and (c)? Draw a phasor diagram for the operation of the machine in cases (2), (b) and ; (o). Solution: (a) From Prob. 3.5, Valor Vi) = 0.9375 per unit 5 = 05perunit Xa = 1.65 per unit 6 = -369º k = S/V; = 0.5/0.9375 per unit EL6 = Vi +1Xa 90 +08 0.5 = 0,9375720 + ————= x 1.65 º — 36.9º i 520º + Toa * 1.65, 90º — 36.9º per unit = 16258/25.7º perunit = 26.0/25.7º kV 36 New E; = 09x 1.6258 per unit = 1.46322 per unit P = 05x0.8perunit = 0.4 per unit . PXa Co 0,4 x 1.65 = [PÃO dn 1f o MáXibo 10 o é =sn (E) sn (rs tasm) 28.76 V New Q = S*(Ecos6-V) . Xa = am (1.46322 cos 28.76º — 0.9375) = 0.196 per unit = 39.2 Mvar (e) When Q=0, Y 0.9375 = Aa = = 50.15º 5 = sig cs e) 50.15 EV. 0.0375 x 1.46322 : = —— >DDw—— 0.15º t P a sinó 65 sin 5 per uni = 0.638 per unit = 127.65 MW (d) For V., E; and Xg fixed, Qmez occurs when é = 0º. Hence, V. 0.9375 - . Qmos = X (E-Vi) = 65 (1.46322 — 0.9375) per unit = 0.2787 per unit = 59.74 Mvar E=1.626 pu. V4=0.9375 pu. 3.7 Starting with Eq. (3.31), modify Ea. (3.38) to show that IK . P = esq lBI(Reos6 + Xesmô) — MIR) [Ml Q E+5 (Xa (IE;| cosó — |Vil) — RIEilsin 6) when the synchronous generator has non-zero armature resistance R. Lobo, r —1.0057 39 257705 —1.57135 —1.57135 —1,57135 3.1427 He 257705 —1.0057 —1,57135 mH Ley H cos (—180º) HD iabe Withiy=4kA and Ly, = 433.6569 mH, Los cos dy Los = Ms | cos(8s — 120º) cos (fg — 240º) cos6oº 15.8475 31.695 | cos(—60º) mH = 15.8475 | mH 81.695 ia sin (30º) 10 tb = 20000 | sin (—90º) A= -20 | kA te sin (—210º) 10 As Loo Lab Loc Loy ia A | | Lo Lo Lie Ly io Xe Loo Lob Le Ley te As Lia Ly Le Ly ir 257705 —1.0057 —157135 15.8475 10 o | —10057 257705 157135 15.8475 - | er O | 157185 -1.57135 31427 —31.6050 10 15.8475 158475 816950 433.6569 4 93.5610 —13.9215 = | 96305 | WbT 1258.2026 (b) When 6; = 60º, 1 1 1 1 2 5 3-1 Wo q Á -vê P = 2 v3 vê o = 1 4 0 3,7 2 vw e 1 1 1 1 1 2 2 A vã 3 va Aa wo -y5 93.5610 97.5381 A, = F— o —13.9215 | = | 760016 | Wb-T do » 3 —79.6395 0 mma re” Ê dade ia a a 10 -12.2474 al=1%-k olj-0|= 21.2182 | KA : 1 1 io ao a E] 20 o Il TDl—— P dade te) La = Lo+M,+4Lm = 4.71405mH 40 Lo = Li+M,- iLm = 3.58275mH VEM, = 388188mH Xe = Lyig+VEMiy = 471405 x (—12.2474)+ 38.8183x4 WbT = 97.538] WbT A = Lig = 3.58275x 212132 WbT = 76.0016 WbT ão = Loio = O (sinceig=0) My = VEMpis+Leyiy = 38.8183 x (—12.2474) + 433.6569 x 4 Wb-T = 1259.20 WbT 3.10 The armature of a three-phase salient-pole generator carries the currents ia = V2x 1000sin(8a— 84) À 2 x 1000sin (84 — 120º — 84) A V2 x 1000sin (8; — 240º — 6.) A %& te (a) Using the P-Transformation matrix of Eq. (3.42), find the direct-axis cur- Tent iy and the quadrature-axis current iq. What is the zero-sequence cur- rent io? (b) Suppose that the armature currents are 2x 1000sin (a — 84) A v=%w=0 H ia Determine is, à, and io. Solution: (a) ig ia iq = Pli ig te ' i ja sin (8a — 6) - 3/8 -43 o | x 10002 | sin(, — 0, — 120º) | A *% & > sin (84 — 6, — 240º) sin (8a — 8, — 60º) 10003 | sin (84 — 04 +30º) | À h 0 41 (5) ia 5 à + + sin(0; — 05) ia = V3 g -38 o | x 10002 o A i + LA o * ã sa sin (8a — 8.) 20 - 4 sin(64—-0,) | À 1 va sin (Ba — 8) 3.11 Calculate the direct-axis synchronous reactance Xa, the direct-axis transient reactance X4 and the direct-axis subtransient reactance X4 of the 60 Hz salient- pole synchronous machine with the following parameters: L, = 2.7656 mH Lys = 433.6569 mH Lp = 4.2898 mH M, = 1.3828 mH M,= 31.6950 mH Mp = 3.1523 mH Lm = 0.377] mH M,= 37.0281 mH Solution: La = EL+M-— Sm = 2.7656+ 1.3828 5 x 0.377imH = 4.71405 mH Xa = 1207 x471405x 1030 = 17779 a 3Mjo . 3 369500 o Lã = la-5 Tg — AT1405 5 x asa MH = 12303 mH X4 = 1207 x1.2393x 109 = 04879 W-L- M$Lp + MBLe, — 2MyMbM, 4a — “ca Lolp-M = andas o (Ses x 4.2898 + 3.15232 x 433.6569 — 2 x 31.6950 x 3.1523 x 37.0281 nE 2 4336569 x 4.2808 — 3702817 ' = 09748 mH X& 1207 x 0.9748 x 10290 = 0.367 Q 3.12 The single-line diagram of an unloaded power system is shown in Fig. 3.22. Reactances of the two sections of transmission line are shown on the diagram. The generators and transformers are rated as follows: 44 (a) Draw the impedance diagram for the power system. Mark impedances in per unit. Neglect resistance and use à base of 50 MVA, 138 kV in the 40-0 line. (b) Suppose that the system is unloaded and that the voltage throughout the system is 1.0 per unit on bases chosen in part (a). Ifa three-phase short cir- cuit occurs from bus C'to ground, find the phasor value of the short-circuit current (in amperes) if each generator is represented by its subtransient reactance. (c) Find the megavoltamperes supplied by each synchronous machine under the conditions of part (b). Solution: (a) Base voltages are: 409 lines 138kV 2089 lines 138kKV Gen. 1&2 20XkV Motor 3 13.8 kV . . 138? Base impedance in lines = = 381 9 40 40 9 line: = = =0. H ine: Z 381 0.105 per unit R 20 . W9Mline:Z = CET 0.053 per unit Transformers: 5 YY = 01x 5. 0.250 per unit 20 ' 50 YA = 01x T- 0.333 per unit Gens. 14 2: 183250 X" = 020x (5) x % = 0.405 per unit 45 Motor 3: 50 X" = 020x FT = 0.833 per unit (b) Ifa fault occurs at C, by symmetry equal currents are input from generators 1 and 2. Moreover, no current should exist between busses Á and B through the j0.105 per unit branch. If this branch is omitted from the circuit, the system simplifies to SH E, = E, = E, = 1070 per unit X = X = 0.405+0.25040.053 40.333 per unit = 1.041 per unit Xas = 0.333 per unit o “E] e : 5] = lh] = TT Toa per unit = 0.9606 per unit lb] = E = sm per unit = 3.0 per unit Hj = 49212 per unit 50 x 108 J LC = DDD A =W9BA case É V5x 138 x 10º Hj = 4922x20018A = 1029KA (e) IS = [89] = 10x 50x 0.9606 MVA = 48.03 MVA Sa) = 10x50x30MVA = 150 MVA Chapter 4 Problem Solutions 4.1 The all-atumintm conductor identified by the code word Bluebell is composed of 37 strands ekch having a diameter of 0.1672 in. Tables of characteristics of all-aluminum conductors list an area of 1,033,500 emil for this conductor (1 cmil = (r/4)x 107º in?). Are these values consistent with each other? Find the overall areal of the strands in square millimeters. Solution: diameter = 0.1672x 1000 = 167.2 mils/strand 45 Motor 3: 50 X" = 0.20x 3 = 0.333 per unit (b) Ifa fault occurs at C, by symmetry equal currents are input ffom generators 1 and 2 Moreover, no current should exist between busses 4 and B through Lhe 70.105 per unit branch. If this branch is omitted from the circuit, the system simplifies tO 3x L E) c k O SH E, = E, = E = 1.070 per unit X = Xbo= 0.405 + 0.250 + 0.053 + 0.333 per unit = 1.041 per unit Xs = 0.333 per unit o El 10 eo . &l = lhl= E = Tó per unit = 0.9606 per unit IEis] 1.0 . . = 2 = t = 30 135] Dol Ta per uni 3.0 per unit Hj] = 48222 per unit 50 x 108 h tC = SDDD—— À = 291.8 A case d Vê x 13.8 x 10º gl = 49212x 2091.8 A = 10.29 KA Sil [So = 10x 50x 0.0606 MVA = 48.03 MVA Ss = 10x50x30 MVA = 150 MVA Chapter 4 Problem Solutions 4.1 The all-aluminum conductor identified by the code word Bluebell is composed of 37 strands each having a diameter of 0.1672 in. Tables of characteristics of allaluminum conductors list an area of 1,033,500 emil for this conductor (1 emil = (7/4) x 10º in?). Are these values consistent with each other? Find the overall area of the strands in square millimeters. Solution: diameter = 0.1672x 1000 = 167.2 mils/strand 48 where, for example, D,a denotes the distance in meters between conductors a and d. (b) Hence compute the mutual inductance per kilometer between the power line and the telephone line. (c) Find the 60 Hz voltage per kilometer induced in the telephone line when the power line carries 150 A. oe Solution: (a) Let circuit a-b carry the current 1, so that l=-h=IA(andi=i=0) since 3) 1 = O for the group, Eq. (4.36) remains valid. = 2x1077 —— Xe x (rem; Do » Z th 5) Therefore, x = 2x 107 xIn Dee Wi-t/m oc Similarly, - Dia A = 2x 1077 é Das Aca (linkage of the loop) is given by DicDas Aok = 2x1077 e Da. e — Ag x x ln DacDs Mutuai Inductance = xin DicDaa DacDoa = H/m Die = /(125-052 4182? = 195m Dad (1.25+0.5)2 41.82 = 25im E 49 Flux linkages with e-d: — 5 Note that flux through due to do Goa = 2x 10 lniis e-d due to 1, is dueto lb de = —2x 10-"I in fl Ê opposite that due to Jp Note also that J, and 1, are 180º out of phase. So, due to 1, and 4, 2.51 Pa = 4x107"1 nto 2.51 = "Ino = dl 0 M 4x 107“In TOS 101x1 H/m (e) Via =wMI=877x101x10-7 x 10º x 150=5.71] V/km 4.7 If the power line and the telephone line described in Prob. 4.6 are in the same horizontal plane and the distance between the nearest conductors of the two lines is 18 m, use the result of Prob. 4.6(a) to find the mutual inductance between the power and telephone circuits. Also find the 60 Hz voltage per kilometer induced in the telephone line when 150 A flows in the power line. Solution: o o o o | 25 | 18 | 1 | | sm | m (tm | 215 dueto Ja da = 2x10K Insos dueto Db da = —2x101 nã since bh = La , - 215x 18 7 dueto I and h da = 2x10 ns = —0,01288 x 107" 1, M = Se - oorsBx10* H/m ha Va = wMI = 377x150M x 10º = 0.0728 V/km 4.8 Find the GMR cf a three-strand conductor in terms of 7 of an individual strand. Solution: Given this bundie: & GMR = (0.779 x2rx2r)) = rYAx 0770 = 146r 4.9 Find the GMR cf each of the unconventional conductors shown in Fig. 4.15 in terms of the radius r of an individual strand. Solution: (a) Bundie: BB 50 cMR = “or [(2 x2x 22) sl" = 1.72%r (b) Bundle: & 2 cMR = 0779)! (2 x 2x 23) (2x2x2)r2 = "VERONA x 3/18 = 169% (c) Bundle: Goo GMR = (0.779) x Br? x Br? x ar? = 1.704r (d) Bundie: (O, 3 3 GMR = Virirx W(4x4x2x2x2V3) (24x 2,3) 130 rx2x342 x 07798 = 2.20 4.10 The distance between conductors of a single-phase line is 10 ft. Each of its conductors is composed of six strands symmetrically placed around one center strand so that there are seven equal strands. The diameter of each strand is 0.1 in. Show that D, of each conductor is 2.177 times the radius of each strand. Find the inductance of the line in mH/mile. Solution: ; Outside conductors are counter-clockwise numbered 1 through 6. The center conductor is number 7. Each radius is r and the distances between conductors are: Do = %W Du = 4r Ds = Dik -Dã = 2rv3 DD = Wr)(DEDLDaD (2) = x N/(2r2 x 3x 2r2 x Dr? x Dr x DE ETTA, UVB) or vV6 o ct 10x12 o , L = 4x10h ZIF x 00 * 1000 x 1609 = 4.51 mH/mi 4.11 Solve Example 4.2 for the case where side Y of the single-phase line is identical to side X and the two sides are 9 m apart as shown in Fig. 4.9. 53 4.18 For short transmission lines if resistance is neglected the maximum power wkich can be transmitted per phase is equal to [Vel x [Val 1] where Vs and Va are the line-to-neutral voltages at the sending and receiving ends of the line and X is the inductive reactance of the line. This relationship will become apparent in the study of Chap. 6. If the magnitudes of Vs and Va are held constant and if the cost of a conductor is proportional to its cross-sectional area, find the conductor in Table A.3 which has the maximum power-handling capacity per cost of conductor at a given geometric mean spacing. Note to Instructor: The purpose of this problem is to stimulate the student's examination of Table A.3 and is worthwhile in introducing class discussion of conductor selection. Solution: Power transmission capability per conductor cost if resistance is neglected is |Vs]|Vai/(X 4) based on our cost assumption where A is the cross-sectiona! area of the conductor. There fore, the product X - 4 must be minimized. Assuming De is fixed, examining the Table shows that in comparing any two conductors the percent difference in À is much greater than that of X. So, A is the controlling factor, and Partridge or Wazuwing would be selected. However, resistance cannot be neglected. A “conduetor must be large enough in cross sec- tion that melt-down caused by |I[2R loss will not occur under the most extreme operating conditions. The reference (Aluminum Electrical Conductor Handbook) gives information on thermal effects. If reaciance causes too high a voltage drop on a line, double-circuit lines or bundied conductors must be provided. The reference (Analytical Development of Loadability Characteristics for EHV and UHV Transmission Lines) contains information on maximum transmission capability of lines. 4.19 A three-phase underground distribution line is operated at 23 KV. The three conductors are insulated with 0.5 cm solid black polyethylene insulation and lie fiat, side by side, directly next to each other in a dirt trench. The conductor is circular in cross section and has 33 strands of aluminum. The diameter of the condutor is 1.46 cm. The manufacturer gives the GMR as 0.561 em and the cross section of the conductor as 1.267 cm?. The thermal rating of the line buried in normal soil whose maximum temperature is 30º C is 350 A. Find the de and ac resistance at 50º C and the inductive reactance in ohms per kilometer. To decide whether to consider skin effect in calculating resistance determine the percent skin effect at 50º C in the ACSR conductor of size nearest that of the underground conductor. Note that the series impedance of the distribution line is dominated by R rather than X; because of the very low inductance due to the close spacing of the conductors. A 54 4.20 Note to Instructor: When assigning this problem, it may be advisable to outline part of the procedure. Solution: Rev de 228 +50 o. lis Roo, de 228 +20 o plo 283x 1008 Roca = T= Tori — 0228 0/km Rsoc de = 1,121x0.223 = 0.250 0/km Skin effect can be estimated from the values in Table A.3. The area 1.267 cm? is 2 1.267 x (53) x í x 10º = 250,000 cmils Wezwing has an area of 266,800 cmils and for this conductor Rsos ac 0.388 Foo 00046 x 5.28 = 1123 Since temperature rise would account for a factor of 1.121, skin effect is only about 0.2%. With insulation thichness of 0.5 em center-to-center conductor spacing is 2 x 0.05 +1.46 = 2.46 cm. So, Dea = VZ40x246x2x 246 = 3.099 Xr = 87 x100x2x107"m cs = 0.129 0/km The single-phase power line of Prob. 4.6 is replaced by a three-phase line on a horizontal crossarm in the same position as that cf the original single-phase line. Spacing of the conductors of the power line is Dis = 2Di2 = 2D»a, and equivalent equilateral spacing is 3 m. The telephone line remains in the position described in Prob. 4.6. If the current in the power line is 150 A, find the voltage per kilometer induced in the telephone line. Discuss the phase relation of the induced voltage with respect to the power-line current. Solution: YDxDx2D = “2 =3 = = = 238m [— s28m — o o Os— 1.8m 55 The center conductor of the 3-phase line causes no flux linkages with d-e since the conductor is at an equal distance from d and e. De = De = YRA4(238-0.5) = 2.60m De = Da = (/182+(238+05/ = 340m 1, 340 dueto Ja, da = 2x10 E] 1,340 dueto Jo, Que = 2x10 into 3.40 Total flux linkages = 2x10"(L-h)n 500 Since Jp lags Ja by 120º, OS V3 La 4,30% = 3.40 E W due = 2x10 V3loln 55230 Wim M = 929x 19º H/m V = uMx15 = 37x 10x 9.29x 150x1000 = 5.25V/km e t & 1 The induced voltage leads 1, by 90º + 30º = 120º; that is, V is in phase with Z.. 4.21 A 60-Hz three-phase line composed of one ACSR. Bluejay conductor per phase has fiat horizontal spacing of 1! m between adjacent conductors. Compare the inductive reactance in ohms per kilometer per phase of this line with that of a line using a two-conductor bundle of ACSR 26/7 conductors having the same total cross-sectional area of aluminum as the single-conductor line and 11 m spacing measured from the center of the bundies. The spacing between conductors in the bundile is 40 em. Solution: De = Vilx1]1x22 = 13.86m Bluejay: D, = 00415(254x12x102) = 00126m 13.86 X = -" 3 qe = 0 2x107 x 10 x37in gia = 0.528 0/km DR» 00... Md 57 5.2 The 60-Hz capacitive reactance to neutra! of a solid conductor, which is one conductor of a single-phase line with 5 ft spacing , is 196.1 K9-mi. What value of reactance would be specified in a table listing the capacitive reactance in ohrn-miles to neutral of the conductor at 1-ft spacing for 25 Hz? What is the cross-sectional area of the conductor in circular mils? Solution: At 5ft spacing, Xo = 2965x10t mó = 196,10 2 mi m> = 6614 7 = 000870 ft, or 0.0805 in = (2x 0.0805 x 1000)? = 25, 992 circ mils From Eq. (5.12), at 1-ft spácing and 25 Hz, 1779 1 25 0.00870 c= x 108in = 356,200 2 mi 5.3 Solve Example 5.1 for 50 Hz operation and 10 ft spacing. Solution: 1.779 x 108 10 Xe = O om O mi = 02115 MO mi Bo = Lam usmi Xe O . Xi = 59 X0.1074 MO mi 8 Xi = 59X 00683 MO mi X = SS (01074 + 0.0688) MO mi = 0.2109 MO -mi Bo = 4742 uS/mi i : 5.4 Using Eq. (5.23), determine the capacitance to neutral (in uF/km) of a three- phase line with three Cardinal ACSR conductors equilaterally spaced 20 ft apart. What is the charging current of the line (in A/km) at 60 Hz and 100 kV line to line? Solution: For Cardinal conductors, p= 1196 (1 "2 "2 27 x 8.85 x 10-12 ' G = ESSO pm=ogx102 Fjm = 9.276x 10 uF/km 2 t In qrróesar 58 100 x 10? lag = 27x60x9.276x 107º x A A/km = 0.202 A/km 5.5 À three-phase 60-Hz transmission line has its conductors arranged in a triangu- lar formation so that two of the distances between conductors are 25 ft and the third is 42 ft. The conductors are ACSR Osprey. Determine the capacitance to neutral in microfarads per mile and the capacitive reactance to neutral in ohm- miles. If the line is 150 mi long, find the capacitance to neutral and capacitive Teactance of the line. Solution: Osprey diam. = 0.879in Da = VBxbx4 = 2972f 27 x 8.85 x 10712 € = mma Fim n(667%9)/2 = 8301x 102 F/m = 8.301 x 108 x 1.609 uF/m = 0.01336 uF/mi 108 = = —— = 0. 8 Q.mi Xe = 37x ODISH 0.1985 x 10º Q -mi From Table A.3, X/ = 0.0981. Interpolation from Table A .4 yields X4 = 0.0999+0.72(0.1013— 0.0999) = 0.1006. From Table 4.4, Xc = 0.1987 x 108 9. mi, For 150 miles, Cr = 150x0.01336 = 2.004 yF 0.1987 —-— 0º = 5 Q XKc EO xl 132: 5.6 À three-phase 60-Hz line has flat horizontal spacing. The conductors have an outside diameter of 3.28 cm with 12 m between conductors. Determine the capacitive reactance to neutral in ohm-meters and the capacitive reactance of the line in ohms if its length is 125 mi. Solution: De ViZx13x24 = 1512m Tr = 00328/2 = 00164 X = 2.862 x 10% 15.12 = 3.95 Bo. & dora — 3.256 x 10º OQ -m For 125 miles, 3.256 x 10º Xe = ax165 — 16198 5.7 (a) Derive an equation for the capacitance to neutral in farads per meter of a single-phase line, taking into account the effect of ground. Use the same nomenclature as in the equation derived for the capacitance of a three-phase E 59 line where the effect of ground is represented by image charges. (b) Using the derived equation, calculate the capacitance to neutral in farads per meter of a single-phase line composed of two solid circular conductors each having a diameter of 0.229 in. The conductors are 10 ft apart and 25 ft above ground. Compare the result with the value obtained by applying Eg. (3.10). Solution: Do —+ O -m Fio due to eonnnllo nn image charges —Ja & To (a) Due to charges on a, b: Due to image charges: - Fiz H, Va = 4 ln ape + ga ln q Due to image and actual charges: 2 G% hp Do Et) q Do Ho o» = Bk nã Da Pora 27k Cn = 2Coh = 7 EinEg Fm z (6) By Eq. (5.10), 5 12 C = 2x x8.85x 10 = 7.996 x 10-12 F/m 10x 12 In (dez) And from part (a) above, 2x x 8.85 x 10-12 tn (BA) — Im (E) 5.8 Solve Prob. 5.6 while taking account of the effect of ground. Assume that the conductors are horizontally placed 20 m above ground. Cn = = 8018x 102 F/m 62 In original positions in the transposition cycle, Vis TI = 1a43t distance a-b' vVIg5285 = 31.75 distance aa” = v252+282? = 37.54 ft Dê, = Dp = 41443 x3175) = 21046 De = V2x2B = 2646 ft Dea = VU042x2646 = 27 k ; D, = [(vomrixarsa)' voosa=m]" = riste 22.71 distance a-b H = — = 5. 7 L=ox0inds = 5688x107H/m = 5.963 x 1077 x 10º x 1609 = 0.959 mH/mi XL = 377x0.959x 10? = 0.362 0/mi/phase (6) r = EB = 0.0462 ft asin part (a) above, except that 7 is substituted for D,: Do = [(VOMBIxIT3A) VODAD= =]! = 1282 R From part (a) above, Deq = 22.71 ft and Xc = 2965x10-tn o = 85,225 Q-mi/phase to neutral Lg = iss o0o 8 = 0.935 A/mi/phase = 0.467 A/mi/conductor ; Chapter 6 Problem Solutions 6.1 An 18-km 60-Hz single circuit three-phase line is composed of Partridge conduc- tors equilaterally spaced with 1.6 m between centers. The line delivers 2500 kW at 11 kV to a balanced load. Assume a wire temperature of 50ºC. (a) Determine the per-phase series impedance of the line. (b) What must be the sending-end voltage when the power factor is (:) 80% lagging (ii) unity (iii) 90% leading? (c) Determine the percent regulation of the line at the above power factors. (d) Draw phasor diagrams depicting the operation of the line in each case. Solution: (a) 18 0.37 3792 x Tao 0.465 9/mi R From Table A.3, X, = 42420 " ss 62 in original positions in the transposition cycle, distance ab = 1424352 = 1443ft distance ab” = v142+2852 = 3175 f distancea-a” = 252428? = 37.54 ft Dk = Di = Viá4xaris = 2n04k De = 422x28 = 2646f Dea = V21042x2646 = 2271 R D. = [vans X 3754)" VOONAIx | = 1152f L = ox nda = 5.602 x 107 H/m = 5.963x 107? x 10º x 1609 = 0.959 mH/mi X = 37x0959x 10º = 0,362 0/mi/phase Es O)r = 5 = 0.0462 ft asin part (a) above, except that r is substituted for Ds: Do = [(VEDaa Xa754)" VOA BJ* = 12804 From part (a) above, Deq = 22.71 ft and 22n Xe = 2.965x10“h Tm — 85, 225 2. mi/phase to neutral Long Tb orol vê = 0.935 A/mi/phase = 0.467 A/mi/conductor Chapter 6 Problem Solutions 6.1 An 18-km 60-Hz single circuit three-phase line is composed of Partridge condue- tors equilaterally spaced with 1.6 m between centers. The line delivers 2500 KW at 11 kV to a balanced load. Assume a wire temperature of 50ºC. (a) Determine the per-phase series impedance of the line, tb) What must be the sending-end voltage when the power factor is (i) 80% lagging (ii) unity (iii) 90% leading? (c) Determine the percent regulation of the line at the above power factors. (d) Draw phasor diagrams depicting the operation of the line in each case. Solution: (a) 18 0.3792 3792 x T60s 0.465 9/mi R From Table 4.3, X, = 4.242 0 " 63 and since 1.6 m = (1.6 x 100)/(2.54 x 12) = 5.25 £t, Xa = 0.2012 (Table A 4,53”) X = 0465402012 = 0.666 9/mi Forl8km, X = UBx7 0.666 Es = 7410 Z = 4MP+jTAS = 857/6035 0 (b) For power factor = 1.0, 1, = 20 11,000 R = x = 1812 A 1 v3 = 6350 V Vs = 6350+131.2(4.04 + 97.451) 6906 + 3977.6 = 69757 8.06º sending-end line voltage = V3Vs = v3x6975 = 12,081 V For power factor = 0.8 lagging, Hal = Vs = sending-end line voltage = For power factor = 0.9 leading, Hal = Vs = sending-end line voltage = % Regulation at pf.=0.8 lagging, % Reg. at unity pf., % Reg. at pf =0.9 leading, % Reg. (d) For pf.=0.8 lagging, 2500 V3x11x0.8 6350 + 164/-36.87º x 8.577 60,88º 7639 475.60 = 7660/4.19º V3Vs = V3x7660 = 13,268 V = 164 A 2500 V3x11x09 6350 + 145.8/25,84º x 8.57760.85º 6433 +j1247 = 6553710.97º V3Vs = V3x 6553 = 11,350 V = M58A [sl = Vel = = x 100% [Val . 7660 — 6350 => ——D——— O! mo X 100% 6975 — 6350 - 00% 6553 — 6350 6350 20.63% 9.84% h x 100% 3.20% k ciano indo 66 Solution: 80 9 600 2 1000 9 1000 = 1.08VrR+ 80 Va 1.08VR + 807r Ig = Teto 0 = 0.0028Vp + 1133Fp V; Vo = VR+ (1a+ vo) x80 = Va +80/p+0.08VR = 0.001VR + 0.0018VR + Ir + 0.133/r The ABCD constants are 4 B 108 CC = 800 D= 1133 6.4 The ABCD constants of a three-phase transmission line are A=Da=gÕ936+70016 = õ0.936/0.98º B = 335+j188 = 142/7640 0 C=(-518+7914)x10? gs The load at the receiving end is 50 MW at 220 kV with a power factor of 0.9 lagging. Find the magnitude of the sending-end voltage and the voltage regulation. Assume the magnitude of the sending-end voltage remains constant. Solution: 50, 000 la = SD 25848 = 1458/-25.84º À * = VEx 2x0 És e 220, 000 VR = = 127,0 R V3 127,00070º V Vs = 0.936/0.98º x 127,00070º + 1427 764º x 145.8,--25.84º = 118,855+ 52033 + 13,153 + 915,990 = 133.23/7.77º kV With line-to-line sending-end voltage [Vs] = V3 x 133.23 ='230.8 kV, 230.8 . Veni = ag = 2465kV 246.5 — 220 % Reg. o x 100 = 120% 6.5 A 70 mi, single-circuit, three-phase line composed of Ostrich conductors is ar- ranged in flat horizontal spacing with 15 ft between adjacent conductors. The line delivers a load of 60 MW at 230 kV with 0.8 power factor lagging. 68 6.6 6.7 Qs = 1.086x0.742 xsin(2.125º —(-36.54º)) = 0.503 per unit 100 x 0.503 = 50.3 Mvar A = 1+ A = 1+ 5 (0.1166767.5º x 0.192/90º) = 0.990, 0,248º [Vel /LAl> [Va re] [Va gil (1.086/0.980) — 1.0 = DL xz10% = 97%3% 10 x 100% A single-circuit, three-phase transmission line is composed of Parakeet con- ductors with flat horizontal spacing of 19.85 ft between adjacent conductors. Determine the characteristic impedance and the propagation constant of the line at 60 Hz and 50ºC temperature. Solution: At 50ºC and 60 Hz, from Table 4.3, for Parakeet conductors, r 0.1832 Q/mi Xa = 0423 Q/mi Da = Vi9o85x2f = 25f At 25 ft, Xo(inductive) 0.3906 N/mi Therefore, z = 01832+45(0.423+0.3906) 2/mi = 0834/7731º Q/mi Xi = 00969x 108 OQ mi Xalcapacitive) = 0.0955x 10º Q.mi o j 08,80% = Zi+X T quo ,0g0 S/Mi = 51975 x 10"8,90º S/mi Characteristic impedance: 2. E. BAT . Ze = v V5igsx10- SN = 400.6/-6.346º 0 Propagation constant: v= Vzy = 4/0.834x5.1975x 10-62 77.312 +90 mi"! = 208x 1072 mi! Using Eos. (6.23) and (6.24) show that, if the Teceiving-end of a line is ter- minated by its characteristic impedance Z., then the impedance seen at the sending end of the line is also Ze regardless of line length. 69 Solution: H Za = Z.. then Ip = Vp/Zo: and Va — TaZ.=0. Y From Ea. (6.23) Vs Ya + InZe Vet IaZo ns From Ea. (6.24) Is = 2% e where L is the length of the line. Finally, Z, = Vs/ls = Ze (which is independent of L) 6.8 A 200-mi transmission line has the following parameters at 60 Hz resistance r = 0.21 Q/mi per phase í series reactance x = 0.78 Q/mi per phase i shunt susceptance b = 5.42 x 10º S/mi per phase : (a) Determine the attenuation constant a, wavelength À and the velocity of propagation of the line at 60 Hz. (b) If the line is open circuited at the receiving end and the Teceiving-end voltage is maintained at 100 kV line-to-line use Egs. (6.26) and (6.27) to determine the incident and reflected components of the sending-end voltage and current. (c) Hence determine the sending-end voltage and current of the line. Solution: (a) r = 6219/mi = = 078Q/mi 2 = (021+70.78) Q/mi = 0.808277.31º Q/mi y = 542x 10"f277.31º S/mi 4 = VZ = 2092x 102/8247 mi! = a+j8 =: (2.744 x 107º 4 92,074 x 107%) mir" Atrenuation-constant a = 2.744 x 107º nepers/mi 27 2mx108 Wavel hAÃ==>=—"-=—mi = i avelengt| 2 TOA mi 3030 mi m 3 Velocity of propagation Af = = = REA mijs = 181770 mi/s (6) Characteristic impedance: Z, = 386.05,-7.53º O 72 Is = 300/-cos 09 A = 300,-25.84º À 5 cosa Va = 0.8904/134º x 150.11 ,0º — 188.825.70.450 x 800,-25.84º |, 1000 = 108.85,-22.76º kV Val = v3x10885kV = 188.5 kV line-to-line Ir = -1.131x 102,90.41º x 150.11 x 1073209 + 0.8904/1.34º x 300,-25.84º A = 3720/-48.95º A Hal = 324 The receiving-end power factor is then PÉ. = cos(-22.76º 448.95º) = 0.897 lagging 6.12 A 60 Hz three-phase transmission is 175 mi long. Tt has a total series impedance of 35 + 140 9) and a shunt admittance of 930 x 1078790º S. H delivers 40 MW at 220 kV, with 90% power factor lagging. Find the voltage at the sending end by (a) the short-line approximation, (b) the nomial-z approximation and (c) the long-line equation. Solution: 1 = Ymmi Z = 35+340 = 1443/7596º 9 Y = 930x10!s 40, 000 la = >—2>—— = 116.6,-2584º A R V5x 220 x 0.9 deito (a) Using the short-line approximation, Vs = 127,017+116.6,-25.84º x 144.3/75.96º = 127,017 + 10, 788 +512,912 = 138,408/535º V IV = 3x 138,408 = 239.73 kV (b) Using the nominal-m approximation and Eq. (6.5), o.1342 o Vs = tarot ( ; 18590 +1) + 1443/7590 x 116.6,-25.,84º = 127,017 (0.935 + 90.0163) + 10,788 + 912,912 = 129,5494 14,982 130,41276.6º v3x 130,412 = 225.88 kV IVsl (c) Using the long-line equation, 144.37 75.96º * o o Ze (sao ig esagEE Z obs ) = 884 ,-7,02 q = v144.3x930x 10-8,165.06º = 0.3663/83.0º = 0.0448 + 30.364 ObMB DIA — 1 045872086" — 097734 j0.3724 E T8 ec OAB Ç-50.364 — 09562,-20.86º = 0.8935 — 50.3405 cosha! = (0.9773 + 70.3724 + 0.8035 — 50.3405)/2 = 0.9354 4 500160 sinhyl = (0.9773 4 70.8724 — 0.8935 + 50.3405)/2 = 0.0419+ ;0.3565 Vs = 127,017(0.9354 + 50.0160) + 116.6,-25.84º x 394, —7.02º (0.0419 + 40.3565) ” = 118,812+92,0324 10,563+912,715 = 129,315+714, 747 : = 130,15326,5º V sl = v3x130,153 = 2254kV 6.13 Determine the voltage regulation for the line described in Prob. 6.12. Assume that the sending-end voltage remains constant. Solution: By Problem 6.12, volt-to-neutral results, Vs = 130.15 kV Va = 127.02kV , For Jr=0, Vs = Vpcosh4l, 130.15 | Wa nel = sas eso Uia] 139.12kV 139.12 — 127.02 - % Reg. = rm *100 = 953% 6.14 A three-phase 60-Hz transmission line is 250 mi long. The voltage at the sending É end is 220 kV. The parameters of the line are R = 0.2 9/mi, X = 0.8 N/mi í and Y = 5.3 uS/mi. Find the sending-end current when there is no load on the line. Solution: i Z = (02+708)x250 = 206.1/75.96º Y = 250x5.3x 10º = 1,305x 10-2,90º ' ql = VZY = v206.1x1.325x 10-3,165.96º = 0.5226,82.08º | 0.0639 + 70.5187 206.17 75.96º Ze = VE = ED = 394/1020 1.325 x 10-32, 900 By Eq (6.39) for Jp = 0, inh Il = (Vs) cod Bl = 05187 rad = 29.72º extçil — 0.9258+ 30.5285 ecale-i8l = 08147 50.465] 1 coshyl = 5 (0.9258 + 0.8147 4 50.5285 — 50.4651) = 0.8709/2.086º 74 sinhyl l 0.9258 — 0.8147 + 5 (0.5285 + 0.4651)] = 0.4999,83.61º 3 Lob] — 20,000//3 O4mmorasero o = 3947-702 *garoo,20ge — 1850/88,54º A 6.15 If the load on the line described in Prob. 6.14 is 80 MW at 220 kV, with unity power factor, calculate the current, voltage and power at the sending end. Assume that the sending-end voltage is held constant and calculate the voltage regulation of the line for the load specified above. Solution: 220 80,000 Va= = =127kV Ir = = = 209.95 A =] "Ex With values of cosh! and sinhYl from Problem 64, Vs = 127,017(0.8703 + 90.0817) + 209.95 x 394, -7.02º x 0.4999,7 83.61º 110,528 + 74,026 + 9,592 + 940,9232 = 128,014,720.23º V to neutral 4 [Vel = 3x 128,014 = 21.7kV Is = mosstostas+so0317) + DO o o, 83.610 s — O HEsid: me 3947-7020 * 040082 88.61 = 182.72+56.66-1.77+4161.13 = 2468/4284 A Ps = v3x2217x2468c0s (20.3º — 42.84º) = 87,486 kW (or 87.5 MW) AtIr=0, 127,000 Dr = t 0.8709 145,826 V to neutral 145.8 — 127 127 [Vel % Reg. = 148% 6.16 A three-phase tranmission line is 300 mi long and serves a load of 400 MVA, 0.8 lagging power factor at 345 kV. The ABCD constants of the line are A = D=oQBI80/13 B = W22/842º O C = 0.001933,90.4º S (a) Determine the sending-end line-to-neutral voltage, the sending-end current and the percent voltage drop at full load. (6) Determine the receiving-end line-to-neutral voltage at no load, the sending- end current at no load and the voltage regulation. Solution: 345.000 400,000 Va = = 199,18620º V Ir = Si = 6694,-3687º À R vã FO VEx3s 77 or, inversely proportional to X if we assume constant |Vs| and |VR|. Additionaliy, De, = Y5x5x10 = 6.30 m, or 6.30/0.3048 = 20.67 f (a) 211 2087 For Partridge: X = D0754in Traz = 0.5172 9/km 0 2087 For Osprey: X = 0074n a = 0.4969 0/km Ratio of Pr (new/old): 0.5172 0.4969 1.041 (41% increase) 6) D. = 0.0217 x (0.4/0.3048) = 0.1688 ft 20.67 : X = 0.0754tn pas = 0.3625 N/km 0.5172 > = 1427 Ti 5675 1.427 (427% increase) (c) Pa increases by factor of soy? = 2.78 dueto increased V. Pp decreases due to increase of X. Dea = VBXExD = 8807R 33.07 = 0.0754 = 0.5526 X 0.0754ln Vo; 0.5526 km Decrease factor = fair? — 05526 Resultant factor of increase = 2.78x O.ir2 = 2.602 0.5526 Increase = 1603% However, in addition to the increase in conductor spacing and insulation, larger conduc- tors wili probably be required since current wil! increase by a factor of about 230/138 and |F|2R loss in the line by a factor of about 2.78 for the increase in load at the same power factor. 6.21 Construct a receiving-end power-circle diagram similar to Fig. 6.1) for the line of Prob. 6.12. Locate the point corresponding to the load of Prob. 6.12 and locate the center of circles for various values of |Vs| if |Va| = 220 kV. Draw the circle passing through the load point. From the measured radius of the latter circle determine |Vs] and compare this value with the values calculated for Prob. 6.12. 78 Solution: Use scale of 1” = 50 MVA. By comparing the work in Problem 6.12(c) with the equation Vs = AVR + Bla we find A = 0.9354+700160 = 0.936,0,08º B = 394,-7.02º (0.0419+ 903565) = 141.4/76.28º Q B-a = 76.28º -0.98º = 75.3º (AJVRÍÊ 0.9354 x 220? E] 141.4 Use above data to construct load line through origin at cos! 0.9 = 25.8º in the first quadrant. Draw a vertical line at 40 MW. The load point is at the intersection of this line and the load line. The radius of the circle through the load point is 7.05”, 705x50 = 3525 [Vel Val . e = 8525 tB] 320.2 MVA 352.5 x 141.4 [Vs] 20 = 226.5 kV 6.22 A synchronous condenser is connected in parallel with the load described in Prob. 6.12 to improve the overall power factor at the receiving end. The sending- end voltage ís always adjusted so as to maintain the receiving-end voltage fixed at 220 kV. Using the power-cirele diagram constructed for Prob. 6.21, determine the sending-end voltage and the reactive power supplied by the synchronous condenser when the overall power factor at the receiving end is (a) unity (b) 0.9 leading. Solution: On the diagram for Problem 6.21 draw a new load line in the fourth quadrant at cos-10.9 with the horizontal axis. Draw power circles at radii [Vs||VR|/|B| = 311, 327, 342, 358, 373 and 389 MVA for |Vs| = 200, 210, 220, 230, 240 and 250 kV, respectively. This provides the power circle diagram that we can use for paris (a) and (b). For pf.= 1.0 read [Vs] = 214 kV at 40 MW on the horizontal axis. The vertical distance between the horizontal axis and the load line in the first quadrant respresents the kvar of the capacitors needed. The value is 19.3 kvar. For p.f.= 0.9 leading, read [Vs] = 202 kV where the vertical line through 40 MW intersects the load line in the fourth quadrant. The vertical distance between the two load lines at 40 MW represents the kvar of capacitors needed. The value is 38.6 kvar. 6.23 A series capacitor bank having a reactance of 146.6 9 is to be installed at the midpoint of the 300-mi line of Prob. 6.16. The ABCD constants for each 150 mi portion of line are D = 0.9534,70,8º 90.337284,1º 9 0.001014/90.1º S A B Cc 79 (a) Determine the equivalent ABCD constents of the cascade combination of the line-capacitor-line. (See Table A.6 in the Appendix.) (b) Solve Prob. 6.16 using these equivalent ABCD constants. Note to Instructor: This problem is somewhat long, but the solution is interesting to show that the ABCD constants of networks in series as given in Table 4.6 can be calculated by matrix multiplication. The problem also shows the large reduction in voltage accomplished by series capacitors in the middle of the line. Compare results of Problems 6.16 and 6.23. Solution: (a) LetÃ = 0.953420,3º 90.33284.1º - 0.001014790.1º 0.9584,0.3º ã 1020º 146/-900 ) + Ax [ 0 10,0” xA 0.953420.3º 50.91,-78.65º 0.9534/0.3º 90.33/841º 0.001014/90.1º 1.102270.27% |* 0.001014/90.1º 0.9534/0.3º . 0.959721,18º 42.307/645º — | 0.002084/90.4º 0.9597,1.18º l c»D lu (b) For Va and Ta from Problem 6.16, Vs = 0.9597/L18º x 199, 18670 +42.30/645º x 6604/-36,87º 216,870245º Is = 0.002084/90.4º x 199,186/0º + 0.95977 118º x 669.4 [36.87º = 5204/4.44º Voltage drop = pele x 100 = 8.15% (Compare this voltage drop with that of Problem 6.16) VanL = A6 soa | 225,977/332º V 0.95977 1.180 Isnz = 0.002084/90.4º x 225,977/332º = 470.9/987º A 225,977 — 199,186 % Reg. osso x 100 = 13.45% (without capacitors 57.6%) 6.24 The shunt admittance of a 300-mi transmission line is Ve = 0+356.87x 108 S/mi ET 82 Solution: Imagine a vertical line on the diagram of Fig. 6.15(b) at one-fourth the line length from the sending end toward the receiving end. Intersections of this line and the siant lines occur at T = 0.257, 1.757, 2.257, 3.757, etc. Changes in voltage occur at these times. The sum of the incident and reflected voltages are shown between slanted lines and determine the values Plotted below. 120V 10 v | | LJ sv cal 60 V : tlm o T 2 3 sT 6.28 Solve Example 6.8 is a resistance of 54 Q is in series with the source. Solution: For voltage, - 54-30 2 * É 54300 7 = 90-30 1 ºR É ara O 2 Initial voltage impressed on line: 30 e — = 4286V 34H x 120 42. inal value: Final value: . 0 sv 90 + 54 poê Po 75 V 7 v Tha2.86 73.47 V 7 v— 2” 68 v— 54.29 v arf-zo.4 60 v ar f 74.34 0 T 27 37 4r 5” 6.29 Voltage from a de source is applied to an overhead transmission line by closing a switch. The end cf the overhead line is connected to an underground cable. Assume both the line and the cable are lossless and that the initial voltage along the line is v*. If the characteristic impedances of the line and cable are 400 9 and 50 12, respectively, and the end of the cable is open-circuited, find in terms of vt A 83 (a) the voltage at the junction of the line and cable immediately after the arrival of the incident wave and (b) the voltage at the open end of the cable immediately after arrival of the first voltage wave. Solution: (a) The initial wave of voltage vj arriving at the juction with the cable “sees” the Z, of the cable. So, at the end of the overhead line: 50 — 400 = 0.777 50 + 400 PR = and the voltage at the juction is (1-0.777)v* = 022%7 which is the refracted voltage wave travelling along the cable. (b) At the end of the cable pp = 1.0 and vp = (0.223+0.223)0* = 044607 6.30 A de source of voltage Vs and internal resistance Rs is connected through a switch to a lossless line having characteristic impedance R.. The line is terminated in a resistance Ra. The travelling time of a voltage wave across the line is 7. The switch closes at t = 0. (a) Draw a lattice diagram showing the voltage of the line during the period t=O0tot=7T. Indicate the voltage components in terms of Vs and the reflection coeficients pa and ps. (b) Determine the receiving-end voltage at t = 0, 27, 4T and 6T, and hence att =2nT where n is any non-negatíve integer. (c) Hence determine the steady state voltage at the receiving end of the line in terms of Vs, Rs, Ra and R.. (d) Verify the result in Part (c) by analyzing the system as a simple de circuit in the steady state. (Note that the line is lossless and remember how inductances and capacitances behave as short circuits and open circuits to de.) Solution: (a) vovl — Rr-R. 2 R$-R 1= "SRA Rs PRE RAIA. “E RAR. 84 PsoBvs o t=0> Va(0) t=2T >» Va(2T) t=4T7 > Va(4T) H Val +Ve+orVo = (1+on)Vo VR(2T) + pspRV; + ps ohVs 1 +pn)V; + psortl + pr)Ve (+ or)L+ ospr)Vr Val4T) + RVE + PORV; ++ ospr)Ve + 2oRU + pR)VE = (1428) [L+ mer + (papa) Vi ' t=6T > Va(6T) Hence at any given t = 2n7T, n-l Te 1- ValênT) = (1+08)4 5 (soa) | Vy = (1+ 09) Tleenl y, 5=0 PsPR (c) At the steady state, n — 00. If Rs or Ra £0, [psprl < 1 and (pspr)” — 0 as n — 00. Hence, V V; = pd 'R(oo) G+pr) ID pspR j Ra-R Rs- R Since, pr = = hos, PR RRTÃ, Ml = Dk 2Rp item = Fes 10 (Rs+RM(RR+RO) 1- pspR 2R.(Rs + Ra) R Vo = Envy. 1 R+R f 2Rp (Rs + RJ(Rr+ Ro) Re Ra V; = . = . Po) = mrR O RAs cho) “Ro+A OT HRS (d) 86 7.2 Using the Ypus modification procedure described in Sec. 7.4 and assuming no 7.3 mutual coupling between branches, modify the Ypus obtained in Prob. 7.1 to refeci removal of the two branches DG) and (OG) from the circuit of Fig. 7.18. Solution: To remove branches (D-) and GQ), we add the following blocks to Ypus : oo oo Sliciwo Sli This results in the following modified Y pus : Do o ao -;35 925 jo 50 30 js js já 0 30 jo ja -jN2 GB 30 JO 50 je -H0 52 jd JO JO jG2 28 eseso The circuit of Fig. 7.18 has the linear graph shown in Fig. 7.19 with arrows indicating directions assumed for the branches a to h. Disregarding all mutual coupling between branches (a) determine the branch-to-node incidence matrix A for the circuit with node O as reference. (b) find the circuit Ypus using Eq. (7.37). Solution: (a) The branch-to-node incidence matrix is found to be DOS Do Oofi-lr o oo Ol-1 0 1 0/0 Oojoºo-1 1 00 Aa-O|/0-1 0 01 oOoto o 11/06 Dlo o o 1 Oil-1 0 0 0 0 OLo 0 o 0 87 (6) Yor is given by o 0 ooo oo -;25 . . . . . . : 2 . . . . Ê =; 99090000 d % . : - - —j0.8 D Q 9 O o Dl-jps 925 92 50 30 Olss 15 do 30 35 Yo = ATypA= 8] 52 ja js js 350 9| 2% 30 8 —0 32 ol so 55 j 2 -j78 7.4 Consider that only the two branches D-Q) and Q-Q) in the circuit of Fig. 7.18 are mutually coupled as indicated by the dots beside them and that their mutual impedance is 0.15 per unit (that is, ignore the dot on branch 2-6). Determine the circuit Ybus by the procedure described in Sec. 7.2. Solution: The primitive impedance matrix for the mutually coupled branches (D-(3) and OQ is in- verted as a single entity to yield the primitive admittance matrix = 2a DB DS DB [E gas - [rasa Sites) OO [5015 50.25 5146341 —j4.87805 Building blocks of the two mutually coupled branches D-) and O) are Do 29 2 [4 “t-szanoos) 8 [4 “a nasean DO 2 q 2 [4 “Jonas 5 E a t-samen) Building blocks of the remaining branches are determined as o DO 9 o oia Slácjess Blu o o 3 o 8 [4 “ift-sa) 2 [4 “ileso & [1)-;80) ss Combining all the above building blocks gives o 2 G o o O [-5593902 ;306M1 G097561 50 50 O | j3963841 —j1237805 341464 30 35 O | 097561 ;341464 51239025 98 30 a 30 jo 58 -nNo 32 o jo 55 50 j2 8 7.5 Solve Prob. 7.4 using Eq. (7.37). Determine the branch-to-node incidence ma- trix A from the linear graph of Fig. 7.19 with node O as reference. Solution: The branch-to-node incidence matrix found in Prob. 7.3 can be used here. Ypr is obtained by inverting Zpr as follows. o ooo vo o” O [504 . . . . . . . 6 505 j015 O JO 15 j025 -721.0 : a Ypr = Zpr “o . 0.125 . o 05 g E O - j1.25 O O e Do vo oO O [-525 . . . . . O -j243902 j1.46341 O 5146341 —j4.87805 -O : 8 o . —38 . o 2 O “jd . O “ —308 Using Eq. (7.27), we have D o B D o OD [593902 ;396341 gogmse O º O | 5396841 —j1237805 ;341464 O 55 O | jo97561 341464 —9123005 ;8 O q 0 o 8 -pjo 2 o 0 js o 52 -j78