Ch2_soln - +1; Substituting these values in the equations...

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Unformatted text preview: +1; 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 Zbus matrix. Solution: Using the Ybus solution of Problem 1.25, substitute the given admittance values: —314.5 38.0 34.0 32.5 v1 0 38.0 —jl7.0 340 35.0 V2 _ 0 34.0 34.0 -38.8 30 V3 1.0 3—90° 32.5 35.0 30 —-3‘8.3 V4 0.68/~135" Compute voltages: YbusV = I Ybus-lybusv = Y'bus”:lI 30.7187 30.6688 30.6307 30.6194 -1 __ 30.6688 30.7045 30.6242 30.6258 Where .Ybus ‘ zbus ‘ 30.6307 30.7045 30.6840 30.5660 30.6194 30.6258 30.5660 30.6840 v = Ybus'lI V1 30.7187 30.6688 30.6307 30.6194 0 V2 _ 30.6688 30.7045 30.6242 30.6258 0 V3 — 30.6307 30.7045 30.6840 30.5660 101—90" V4 30.6194 30.6258 30.5660 30.6840 0.68 —135° V1 0.9285 — 30.2978 0.9750 (—17.780 V2 _ 0.9251 —jO.3009 _ 0.9728 g —18.02° V3 — 0.9562 — 30.2721 - 0.9941 5—15.89" V4 0.8949 — 30.3289 0.9534 (—20.18c’ 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 (2.7). transformer delivers its rated kVA at rated voltages. Hence, verify Eq. Solution: (0) N1 __ V1 __ N3 _ V2 — N1 II T herefore, N2 11 A V MAMHW.—~mmwmw- 12 Srnwd = lvllratedlIllrated = ll/eratedllerated 7.2x103 = 1.2x103|111,a,ed = 120|121md 7.2 x 103 Ull'md = 1.2x103 = 6 A 7.2 x 103 IIQIrated = 120 = A 11 6 - ' . .7: — = -— = 0.] Left hand Slde of Eq (2 ) 12 60 N2 1 Right-hand side of Eq. (2.7): Left-hand side of Eq. (2.7) -N—1 —— 10 = 0.1 Right-hand side of Eq. (2.7) 2.2 The transformer of Prob. 2.1 is delivering 6 kVA at its rated Voltages and 0.8 power factor lagging. (a) Determine the impedance 22 connected across its secondary terminals. (b) What is the value of this impedance referred to the primary side (i.e. Z5)? (c) Using the value of Z; obtained in part ((2), determine the magnitude of the primary current and the kVA supplied by the source. Solution: (0) $2 = 152mg = 6x103436.9° VA 52 ‘ 12 r (72) 22 = n = V2 = MK 12 Sgt/V; 35 _ (120)2 Q ’ 6x1034—36.9° = 2.44369" :2 = (1.92+j1.44) Q , N1 2 V1 2 22 = — Z1 = —. Zl = 100x2.4g36.9° 0 N2 V2 = 2404369" Q = .192+j144 n (e) W 1.2 x 103 I = — = ———— = ' I ll | él 240 A o A 151) = mm = 1.2x 103x5 VA = 6 kVA 13 2.3 With reference to Fig. 2.2, consider that the flux density inside the center—leg of the transformer core, as a function of time t, is B (t) = Bm sin (27rft) where Bm is the peak value of the sinusoidal flux density and f is the operating frequency in Hz. If the flux density is uniformly distributed over the cross-sectional area A m2 of the center-leg, determine (a) the instantaneous flux <25 (t) in terms of Bm, f , A and t, ( b) the instantaneous induced—voltage 61 (t), according to Eq. (2.1). (c) Hence show that the rms magnitude of the induced voltage of the primary _ is given by lEll = fifileBmA. (d) If A = 100 ch, f = 60 Hz, Bm = 1.5 T and N1 = 1000 turns, compute [E1[. Solution: (0) ¢o(t) = B(t)A = BmAsin(27rft) (5) e1 (t) = NI d2?) = NleAg {sin(27rft)} = 27rleBmAcos(27rft) (C) E1 = figklfiflmaz = Eff—JE—T—M = fiflvale-A (c) With given values, E1 firrx60x1000x15x100x10-4 v = 4.0 kV 2.4 For the pair of mutually coupled coils shown in Fig. 2.4, consider that L11 = 1.9 1'1,ng 2 L21 =1 H, L22 2 H and T1 = T2 = 0 The system is operated at 60 Hz. ' ((2) Write the impedance form [Eq. (2.24)] of the system equations (1)) Write the admittance form [Eq. (2.26)] of the system equations (c) Determine the primary voltage V1 and the primary current 11 when the secondary is (2') open circuited and has the induced voltage V2 2 100 fl V (ii) short circuited and carries the current 12 = 21 90° A 14 Solution: (a) From Eq. (2.22) and (2.23), V __ ‘ L11 L12 [1 _ - 1.9 0. ll’él * le L21 L22ll12l _ 31207Tl0-9 0. 3102 [ $83 i233? l l l; l ONO |-—_J rfi 3‘5 I—g (6) From Eq. (2.25), [5;] = ale-we Harm] = ale-Mas; Ham] (CW) 3 [100%] = mossy] I hence 11 = 0.295i0o A 1 V1 = 211.114): V l (ii) [2&] = 10‘2x1fllifiéélvl hence V1 = 117.30L0° V 11 = 1.111—90" A 2.5 For the pair of mutually coupled coils shown in Fig. 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 reactances of the windings and the magnetizing susceptance of the coupled coils? Solution: L1, = LH—aL21 =1.9—2x0.9 H = 0.1 H L2; = L22—L12/a = 0.5—2x0.9/2 H = 0.05 H a2L21 = 4x005 H = 0.2 H Ln = aL21= 2x0.9 H = 1.8 H w = 1207 rad/sec j 31m j 75.4 n Leakage reactances: x1 = 37.? Q 15 $2 = 75.4 9 TA $2 = .3—9=18.850 4 1 Magnetizing susceptance: Brrl = 'wL'm = 12O7rlx 1.8 s = 1.474x10"3 s 2.6 A single—phase transformer rated 1.2 kV / 120 V, 7.2 kVA has the following wind- ing parameters: r1 = 0.8 0, x1 = 1.2 9, 7‘2 : 0.01 Qand x2 = 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, R1 = n+8” = 0.8+100x001 n = 1.8 :2 X1 = x1+a2z2 = 1.2+100x0.01 n = 2.2 n (b) 32 2 Rl/a2 = 1.8/100 n = 0.018 :2 X2 2 Xl/a2 = 22/100 0 = 0.022 n 21 11 R. 5x1 ‘3; h—p—WYWW Iv! 13V: (c) o———_______¢ 21 = (1.8+j2.2) n 7200 o 4 IQ,FL = Isg/VQ; 5—9 = fig—369 A = 605-36.? A z I \ Irn = 2'0” = 6.05—369" A J OVQ’FL = V V1,FL = 0V2,FL+Ii,i-*LZ1 16 ll 1200+6.0/—36.9‘(1.8+j2.2) V = 1216.57/O.19° v 120v l/le/a = 121.66 v (l21.66-—120)/120 = 1.38% lV2.FLl IVQ, NLl % Regulation 2.7 A single—phase transformer is rated 440/220 V. 5.0 kVA. When the low—voltage side is short circuited and 35 V is applied to the high-voltage side, rated current flows in the windings and the power input is 100 W. Find the resistance and reactance of the high- and low-voltage windings if the power loss and ratio of reactance to resistance is the same in both windings. :1 Solution: i '000 Rated] = —-—0220 = 22.73 A (low voltage) ~ = % = 11.36 A (high voltage) 100 = = 0.774 9 R 11.362 2 = Egg = 3.08 n (R, z, X high-voltage)- X 2.98 = . 2- . 2 = . — = _ = , X v308 0774 298 Q R 0774 385 For equal loss in high- and low-voltage windings, ."7 High voltage: 7- = 213—4 = 0.387 52 :c = 3.85 x0.387 = 1.49 9 Low volta e' r — 0387x 220 2 — 0097 Q g ‘ _ ' 440 7 ' 220 2 a: 14 (440) 0373 $2 2.8 A single-phase transformer rated 1.2 kV/ 120 V, 7.2 kVA yields the follOwing test results: Open-Circuit Test (Primary Open) Voltage V2 = 120 V; Current 12 z 1.2 A; Power W2 2 40 W Short-Circuit Test (Secondary Shorted) Voltage V1 = 20 V; Current I] = 6.0 A; Power W1 2 36 W 17 Determine (a) the parameters R1 = r; + azrgT X1 = 3:; + a222, GC and B7" referred to the primary side, Fig. 2.7 (b) the values of the above parameters referred to the secondary side (c) the efficiency of the transformer when it delivers 6 kVA at 120 V and 0.9 power factor. Solution: (a) From open—circuit test, G; = Wg/Vg = 40/12023 = 2.78x10’3 5 11;] = IQ/VQ = 12/1205 = 0.01 s B; = Inf—cg? = 9.606x10—3 s a = 1.2x103/120 = 10 Therefore, cc = (ye/cl2 = 2.78x10-5 3 Bm Ema? = 9.606x10'5 s From the short-circuit test, R = Wl/If = 36/002 :2 = 1.0 9 |Z| = V1/11 = 20/609 = 3.33 o X = \/|Z|2—R2 = 3.18 n (b) R’ = R/a2 = 0.01 9 X’ = X/a2 = 0.0318 (2 G; = 2.78x10'3 8 3;" = 9.606x10-3 s (c) When SQ = 6.0 kVA and V2 = 120 V, 6 x 103 , I2 — 120 ‘A — 00 A Core loss ath = 120V = 40W Winding loass at 12 50 A = |1212R' = 502 x0.01W = 25 W 1+" Power output at 52 = 6.0 kVA at 0.9 pf. = 6 x 103 x 0.9 W = 5400 W 5 5400 ’7 ' 5400+40+25 ” 98'81 % 2.9 A single-phase transformer rated 1.2 kV/ 120 V, 7.2 kVA has primary—referred parameters R1 = T1+Q2T2 = 1.0 Q and X1 = x1+a2$2 = 4.0 $2. At rated voltage its core loss may be assumed to be 40 W for all values of the load current. 18 Solution: (a) (2') c059 # 0.8, lagging 6 = —36.9° V2 = 120i° V I — 7200 36 9° - 60 36 9° A 2 ’ mg~ - g’ Total loses = 40 + 602 x W = 76 W Output power = 7.2 x 103 x 0.8 W = 5760 W 5760 77 _— m -- 98.698% I . Vi,FL = aV2,FL+ 2:1 (R1+JX1) VLFL = 120 x 10L0°+m4—36.9° (1.0+j4.0) V = 1219.3¢O.73° lV2,FLl = 120 V lV2,FLl = lV1,FLl/a = 121.93 V %Reg‘ulation = 121.323120 = 1.61 % (ii) 0050 = 0.8, leading 0 = 36.9° 1; = 9869896 because it does not depend on whether 9 is leading or lagging. I . V1,FL — 0V2,FL+ 25L (31+JX1) Vin = 120x10fl+f—3436.9° (1.0+j4.0) V = 1190.611.1° % Regulation = = —0.78% (b) Load current at which 77 is maximum is given by , R lI2l2 a7 = Pcore Therefore, . 40 [12] = 10/100 A = 63.245 A 40W Winding loss at [12‘] = 1;}? 19 Output = 120 x 63.245 x 0.8 W = 6071.57 W 6071.57 "m r — 9870093 Corresponding kVA level = 120 x 63.245 VA = 7.589 kVA 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 A-B: 500 V/ 1.5 kV, 9.6 kVA, leakage reactance = 5% Transformer 3—0: 1.2 kV/ 120 V, 7.2 kVA, leakage reactance = 4°70 Line B: series impedance = (0.5 + 33.0) 9 Load C: 120 V, 6 kVA at 0.8 power factor lagging ((1) 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: (a) Ohmic impedances Transformer A-B Primary: $11-65 x j0.05 = 3' 1.302 9 1.52 x 106 , . Secondary: W x_70.05 = 311.719 9 2 6 Transformer B-C Primary: x j0.04 = 3'80 9 2 _ Secondary: $21—iomg x 30.04 = 30.08 0 V 2 02 Load: -li?ll/_9 = 61:1034cos”10.8 = 2.4436.9° Q a ~46 Q) (b) Impedance bases . _ 1.22 x 106 “J” Circu1t B: W (2 = 144 Q 2 Circuit C: fl— 9 = 1.44 0 10 x 103 20 2.11 V 2.12 Per unit impedances on new bases: . 11.719 Transformer A-B. j 144 . 8 J 144 (1.5+j3.0) 144 2.4 : — 369° Load 1.44 [—— = 30.08138 per unit Transformer B—C: 30.0556 per unit Line B: 0.0104 + 30.0208 per unit 1.66M 369° per unit (c) Sending-end voltage calculations 5.5- 0.0104 + j 0.0208 R.E. j 0.08138 1.667 4369" p.u. (0.0104 +j 0.15778) p.u. VR = 120 V = 1.0 per unit . 7 6. ° 0.0104 0.15778 VS = x W = per unit 1.667 g 36.9° The sending-end voltage base is 500 3 vsybne = mxl.2x10 = V Therefore, the required sending-end voltage is V5 = 400x 1.0642 = 425.69 V A balanced A-connected 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 I“, = ’ = 33.47 A I 1 9' «3 x 138 138,000N5 R _ 33-47 _ 2380 n Solve Prob. 2.11 if the same resistances are reconnected in Y. Solution: If the A-connected resistors are reconnected in times as great and Y, then the resistance to neutral will be three 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. fit 5 Z 2 3. ‘1“ a’. 'i To investigate further the suggestion ((1) Find each of the line currents (magnitude and angle) with the 10 kW load and the transformer between a and c removed. (Assume Val, = 220 fl V, sequence a b c.) (27) 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? ((1) Think about why the individual transformer kilovoltampere values include a Q component when the load is purely resistive. Solution: —> 1b (a) Vab and V4,c remain the same after removing the third transformer, so Vm is also the same and we have a three-phase supply, and these voltages are: Vab = 220 L_° V, Vbc = 22012400 V and V“ = 2201 120° V. Then, V,m = 1275—30° V, V5,, = 127/ 210° V and Va, = 127430: V. The line currents are 10 000 I = ———’—— —30° = 26.24 —30° A “ \/§ x 2201—— L—~ 1,, = 26241210" A Ic = 26.24490° A (b) kVAsupplied = 220 x 26.24 x 10-3 = 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, 51 = VabI; = 220flix2624430° = 5000 + 32886 VA 52 = VcbI; = 220mm x26.244270° = 5000—32886 VA 22 Note that Q is equal in magnitude but opposite in sign. There is no Q output from the open delta. After the load reduction, 51 = 4333+j2500 VA 52 = 4333—32500 VA 2.14 A transformer rated 200 MVA, 345Y/20.5A kV connects a balanced load rated 180 MVA, 22.5 kV, 0.8 power factor lag to a transmission line. Determine (a) the rating of each of three single-phase transformers which when properly connected will be equivalent to the above three-phase transformer and (b) the complex impedance of the load in per unit in the impedance diagram if the base in the transmission line is 100 MVA, 345 kV. Solution: (a) Each single—phase transformer is rated 200/3 = 66.7 MVA. Voltage rating is (345/ \/§) / 20.5 or 199.2/20.5 kV. (b) 2 LoadZ = (2128:) (cos—10.8 = 2.81436.87° Q (low-voltage side) At the load, BaseV = 20.5 kV _ (20.5)? _ BaseZ — 100 — 4.20 O .8 LoadZ = 373436.87o = 0.669436.87° per unit 2.15 A three-phase transformer rated 5 MVA, 115/132 kV has per-phase series impedance of (0.007 + 3' 0.075) per unit. The transformer is connected to a short distribution line which can be represented by a series impedance per phase of (0.02 + 30.10) per unit on a base of 10 MVA, 13.2 kV. The line supplies a bal- anced three—phase load rated 4 MVA, 13.2 kV, with lagging power factor 0.85. ((1) Draw an equivalent circuit of the system indicating all impedances in per unit. Choose 10 MVA, 13.2 kVA as the base at the load. (b) With the voltage at the primary side of the transformer held constant at 115 kV, the load at the receiving end of the line is disconnected. Find the voltage regulation at the load. 23 Solution: (a) Base voltages are shown on the single—line diagram. 115 kV 13.2 RV Load Transformer Z = 2 (0.007+j0.075) = 0.014 +j0.150 per unit 0 V5 = 1.0 per unit Line Z = 0.02+j0.10 per unit (13.2)2 x 1000 _ = —---—-—— = 43. 6 Q L°ad '2' 3400/0235 ° .2 2 BaseZat load = (13 ) = 17.42 9 10 43.56 _1 , o LoadZ = 1742[cos 0.85 = 250(318 ll 2.125 +j1.317 per unit 0.014 + p.150 0.02 + 10.10 2.125 11.317 (values are in per unit) (b) Voltage regulation calculations 1.0 _ 1.0 0.014 + 0.02 + 2.125 +j(0.150 + 0.10 + 1.317) 2.6684 35.97° = 0.375 {—35.97° per unit VRYFL = 0.3755—35.97° x 2.5131.8° = 09371—417" per unit VR'NL = V5 = 1.0 1 -— 0.937 I = 2.16 Three identical single-phase transformers, each rated 1.2 kV/ 120 V. 7.2 kVA and having a leakage reactance of 0.05 per unit, are connected together to form a three-phase bank. A balanced Y-connected load of 5 9 per phase is connected across the secondary of the bank. Determine the Y-equivalent per— phase impedance (in ohms and in per unit) seen from the primary side when the transformer bank is connected (0) Y-Y, (b) Y-A, (c) A-Y and (d) A—A. Use Table 2.1. < W12“? (LE 1 24 Solution: (a) Y—Y connection: 1.213'kw12013'v IVLLI = 1.2x 103x\/§ v jx‘ 1mm Wu] = 120\/§V '——"“"——§ 1, 2 3 mom/5 r r R, = — x __ = Q , , g L O (um/3) ZL Rt R‘ ' 2 . 06 i Zb = M = 200 g _ 7.2 x 103 x 3 X, = 0.05 per unit = 200x0.05§2 = 10 Q E Z'L = (500+j10) Q 5 (b) Y—A connection: 1.213'kV/120v ix, 7.2x3kVA i |VLL| = 1200xx/5 v ___nrvwx__§ I qul = 120 V ' I“ , I" E ' 2 1200¢§ _ 2‘. a, E ’ = = 00 9 RL 5 X ( ) 10 X1 = 10 erom part (a) _ Z’L = (1500+j10) o (c) A—Y connection: 12 kV/izorz'v jx1 7.2x3kVA [VLLI = 1200 V ___nmm___§ 114,] = 120%)? v |—~ r 1200 2 500 R’ = 5x = —- = 166.67 n 25 ac R L (120%) 3 12002 = ————-— = , 7. _ Z" 7.2x3x103 666 Q X; = 0.05 per unit 2 66.67x0.05§2 = 3.33 0 Z}, = (166.67+j3.33) Q (d) A~A connection: jx, 1va:41:2va [VLLI = 12oov ‘—‘ “H—g qul = 120 V r r ’ 1200 2 ' 21” RL. 12L RL — 5 X -—— 000 X; = 3.33 9 from part (c) _ Zi = (500 +3333) {2 2.17 Figure 2.17a 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 terminals are 11.9 kV and 20 A, respectively. The power factor seen by the generator is 0.8 lagging and the phase sequence of supply is ABC. ((1) Determine the line current and the line—to—line voltage at the load, and the per—phase (equivalent-Y) impedance of the load. (1)) Using the line-to—neutral voltage VA 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. (6) Compute the real and reactive power supplied by the generator and con- sumed by the load. Solution: (0) 12 103 Voltage ratio = a = —6):)T—130° = 20430" Current ratio = (—11: = 0.054 30° 2 (12 x 103) X‘ 600x103 x01 2 LetV - 11'9/0° kV - 687 kV s «5* — ~ Then, 15 = I’L = 201—36.9° A IL _ I’La‘ = 20x20/—36.9°—-30° A = 4005—66.? A VL' = VS—leIs = ssng— (31%) kV l = 6.5935—3.34° kV VL = me; = M kV = 329655—3334“ v 20 4 30° Line voltage at the load @1le = 571 v Line current at the load = [Id = 400 A - 32. _ , o L°ad1mpedance- ZL = VL/IL = .365— L—z’fl- 9 4005—6690 r V 54:41. = 0.8241336" Q 26 (b) ref .34= v5 = 6.87 kV 30n VL’ = 6.593 kV VL: 329.65 V (c) Pg + ng from the generator is 3VsIg, where 31/51; = 3x 6.87./_Q‘:x 20436.9c kVA = 412.24 369° kVA = 329.8 kW+ 3247.3 kvar PL +jQL by the load is 3VLIZ, where 3 x 329.65 (—33.340 x 4004 66.9" 1000 329.7 kW + j218.7 kvar ll 3VLIZ kVA = 395643356" kVA 2.18 Solve Prob. 2.17 with phase sequence ACB. Solution: (a) Final answers remain the same except for the following intermediate results: D a = 201—30 l/a‘ = 0.051—30° I], = Iia' = 4005—36.9°+30° A = 4004-69" A VL = Vé/a = 329655—334o + 30° V = 32965126.?" V (b) vL= 329.65 V 30° ref “' ref .340 V = 6.87 kV W = :593 kV 30" XL A IL' =15 = 20A (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. "av-WM u-AW-r 1" ,1 K. . “Wash-e ‘»’:"-r 27 (6) Mark on the diagram the values of rated current in the windings and at the input and output. ( 0) Determine the rated kilovoltamperes of the unit as an autotransformer. (d) If the efficiency of the transformer connected for 1200/ 120 V operation at rated load unity power factor is 97%, determine its efficiency as an autotransformer with rated current in the windings and operating at rated voltage to supply a load at unity power factor. Solution: 30: 000 rated IHV = 1200 = 25 A 30, 000 , rated ILV — 120 — 250 A Connected for 1200/ 120-V operation (regular transformer), Pam = 30,000 W En =' 30, 928 W Loss = 928 W ' Loss remains the same in the autotransformer because current in the windings and voltage across the windings are unchanged. For the autotransformer, Pm,t = 250x 1320 = 330,000 W H = 330,928 W 330 000 = a = . V = 7; 330,928 x100 99 7% Ratedlr A 330,000» Note that, once we consider loss, we no longer have an ideal transformer; and both winding resistance and reactance as well as magnetizing current and core loss must be considered. The applied voltage and input current will be greater than the values shown to achieve rated output, in which case the equivalent circuit corresponding to Fig. 2.7 would be used. 28 2.20 2.21 Solve Prob. 2.19 if the transformer is to supply 1080 V from a 1200 V bus. Solution: 250 A As in Prob. 2.19, Loss = 928 W. As an autotransformer, Pm.t = 250x 1080 = 270,000 W 1% =: 270,928 \v 270.000 x 100 = 99.7% '7 270,928 Rated kVA = 270, 000, but see the note which accompanies the solution of Problem 2.19. Two buses @ and @ are connected to each other through impedances X1 = 0.1 and X2 = 0.2 per unit in parallel. Bus 0 is a load bus supplying a current I 2 105—30" per unit. The per-unit bus voltage Vb isl.0fl. Find P and Q into bus 0 through each of the parallel branches (0) in the circuit described, (b) if a regulating transformer is connected at bus b in the line of higher reactance to give a boost of 3% in voltage magnitude toward the load (a = 1.03), and (c) if the regulating transformer advances the phase 2° (a = ejflgo). Use the circulating-current method for parts (6) and (c), and assume that V; is adjusted for each part of the problem so that Vb remains constant. Figure 2.26 is the single-line diagram showing buses a and b of the system with the regulating transformer in place. Neglect the impedance of the transformer. Solution: ® X1 =j0.l @ 11.04—30° (a) Thru X1 the current is 11 = g— x 1.0 5—30" = 0.577 —j0.333 and thru X2 the current is 12 = % x 1.05—30" = 0.289 -j0.167. Into bus @ thru X1, P+jQ = Vb]; = 0.577+j0.333 per unit 29 and into bus @ thru X2, p+jQ = Vb]; = 0.289+j0.167 per unit 0.03 =. - . =———=—'0.1 (b) AV 003,1m JM 2 I1 = 0.577—j0.333—(—j0.1) = 0577—20233 12 = 0.289—j0.167+(—j0.1) = 0289—30267 Into bus @ thru X1, P +jQ = 0.577 +j0.233 per unit and into bus @ thru X2, P +jQ = 0.289 +j0.267 per unit (6) AV = 1.042 -— 1.0 = 0.9994 +j0.0349 — 1.0 = —0.0006 +j0.0349 Im = 33w = 0.116 + 30.002 30.3 I1 — 0.577 -j0.333 — (0.116 +j0.002) = 0.461 -—j0.335 I2 = 0.289 —j0.167 +0.116+j0.002 = 0.405 —j0.165 Into bus @ thru X1, P + .70 and into bus ([2) thru X2, P+jQ = VbI‘ = 0.405+j0.165 per unit Note: Compare P and Q found in parts (b) and (c) with part (a). 14,? 0.461 +jO.335 per unit 2.22 Two reactances X1 = 0.08 and X2 = 0.12 per unit are in parallel between two buses @ and @ in a power system. If V, = 1.05 410: and Vb = 1.0Llfi per unit, what should be the turns ratio of the regulating transformer to be inserted in series with X2 at bus @ so that no vars flow into bus @ from the branch whose reactance is X1? Use the circulating-current method, and neglect the reactance of the regulating transformer. P and Q of the load and Vb remain constant. Solution: In reactance X], 1.05( 10° — 1.0 1.034 +j0.1823 -1.0 I = ————————— = ———————- = , 7 — ' , ab 1.008 1.008 2 2 9 30 425 To eliminate vars to bus (9 thru X1, we need in the X2 branch Iab,circ = AV . j0.8+j0.12 ' "30425 a—l = AV = —j0.425(j0.08+ 20.12) = 0.0850 a 1.085 turns ratio 'VAW are = (ZO'I/QS) 0: p901 aanpau 91 oz I 'I = —— 'I = —— 9 0 6I‘SI 030 1703 VAW 6I'QI = QEXO‘IXVEVO = I‘sl VAN v'oz = 98x0'tx989'0 = I‘SI 17970 = Isst'oF—EGS‘OI = |°“~‘°I'-‘I| 8890 = luv'of-Lov'ol = 1°"°I+‘Il (gm-o + 99910)! ' ‘ [— = ——---—-——-——— = 3'45 mo. 920.0 I fisooq apnnufivw %9's Imm QBZ'OF-SBS'O = (9'of—s‘o)-5§- = ‘I an 98 QOS‘OF—L V = -;- ' __ = I o o (90 so)“I I fpeol VAW'QS mm memo 'VAW 9'08 = 98 X (Z‘LI/QI) 03 p201 aonpaJ aJogaxaqa ptm ‘pap'eopaAo s; z# nun x 98910 + QLQI'O v Z'LI = 9 = 3 AW 9 9mm IS' 98910 + 92.91'0 v 811 = 9 x——-——- = I M 8 mm '5‘ nmnzad gggm = (91/92)><2.0‘0 = zX nunmd 9510 = (oz/99)><60'o = IX ‘aseq uasoqo sq: on gums/\qu :uounlog 'UISIqOId sun 10; Moqoegsm'es s; poqzam qualms—Buuemolp a»qu 'epgs QS’BQIOA-MOI mp, 110 A31 5'91 ‘VAW 99 go eseq 9 egg E ‘SJSmlogsuen sq: p’QOIIQAO qou mm qomm p201 ream sq: 30 SQJSdUIBQIOAEBSLU [unume sq; pm; p901 [2309, VAN gg reufiuo aqq Jo; Immogsuen 1.1329 30 1nd I —1no SlSdUl’BQIOA’BESUI sq; pug ‘d'eq Ax 911 91.1: no supemex qogqm ‘3 Jaunogsmen 01 peredmoo 19m10;sumn 12m 50 epgs SBBQIOA-AAOI sq: pmmoz SSBCLIOA u; 1sooq %9'9 12 9M? 01 A31 {11 12 gas 311; I Jauu0}su12n uo sd‘eq aqa, 31 'pep'eopaAo s; Jaunogsuen Iaqqgeu 1121p 03 pawn“ sq 13mm p201 12101 sq: qogqm on SQIQdUI'BQIOA 439m 31p pm; ‘Jeunogsman goes go andzmo QIQdLII'BQIOA’BBBUI sq: ‘Jaunogsumq goes qfinmqq mm 19d u; manna sq: go apnqgufi’eux sq: pug; “mm 19d 1.0'0 = X (MM VAN 91 PalEl 5! Z IBWJOJSWH Pm? ‘Wm 19d 600 = X THEM VAW OZ para; 8; I marmogsueq; 'BULBBBI 101312} ISAAOCI 8'0 1'8 A31 3‘31 ‘VAW 98 30 P301 2 AIddns 01 [anemd u; sawedo A31 V391 / Agn pare; goes Slamogsman om; gz'z 08 ...
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This note was uploaded on 02/22/2009 for the course EE ELEC 349 taught by Professor Ahmad during the Spring '09 term at Qatar University.

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Ch2_soln - +1; Substituting these values in the equations...

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