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Chapter 4 Additional Problems X4.1 Determine the value of the coefficient of coupling (k) for the transformer of Example 4.7 of the text. The turns ratio is V1 240 a 2 V2 120 Based on [4.27], Xm 400 M 0.5305 H a 2 60 2 By use of [4.25] and [4.26], X1 0.18 L1 aM 2 0.5305 1.0615 H 2 60 X2 M 0.045 0.5305 L2 0.2654 H a 2 60 2 Using the conclusion of Problem 3.15, M 0.5305 k 0.999 L1 L2 1.0615 0.2654 X4.2 For the ideal transformer circuit of Fig. X4.1, R p 18 , RL 6 , and X 0.5 . If V2 1200 V and PS 5600 W , (a) determine the turns ratio a, (b) the source voltage VS , and (c) the input power factor PFS . 1 (a) V 2 120 2 PRL 2 2400 W RL 6 PR p PS PRL 5600 2400 3200 W V1 PRp R p 320018 240 V V1 240 a 2 V2 120 (b) V2 1200 I2 200 A RL 6 1 1 I1 I 2 200 100 A a 2 V1 2400 I S I1 100 23.330 A Rp 18 VS Z I S V1 0.590 23.330 2400 240.282.78 V (c) PS 5600 PFS 0.999 lagging VS I S 240.28 23.33 X4.3 For the circuit of Fig. X4.1, a 10 , RL 24 , R p 3.6 k , X 100 , and PL 2400 W . Calculate (a) VS and (b) PS . (a) V2 PL RL 2400 24 240 V I 2 PL / RL 2400 / 24 10 A Assume V2 on the reference. V1 aV2 10 2400 24000 V 1 1 I1 I 2 100 10 A a 10 V1 24000 I S I1 10 1.6670 A Rp 3600 2 VS Z I S V1 10090 1.6670 24000 2405.83.97 V (b) V2 PS 1 PL 2400 2400 4000 W 2 Rp 3600 X4.4 For the circuit of Fig. X4.1, a 2 , R p 20 , and RL 10 . Determine the percent- age of input power PS that is dissipated by R p regardless of the voltage values. V12 / R p 100 a 2V22 / R p 100 % PS V12 / R p V22 / RL a 2V22 / R p V22 / RL a 2 RL 100 2 2 10 100 % PS 66.7% a 2 RL R p 2 2 10 20 X4.5 For the circuit of Fig. X4.1, let RL 0 , R p 10 , X 1 , and a 2 . Determine the power factor PS . With RL 0 , V2 0 and V1 aV2 0 ; thus R p is shorted and the input impedance is j X . Since I S must lag VS by 90º, PFS cos 90 0 lagging X4.6 The non-ideal transformer of Fig. X4.2 has three coils each with identical number of turns N. Terminal pair e-f is open circuit. If terminals b and c are connected together and Vad 240 V (60 Hz), the input current is 1 A. If terminal pairs a-b and e-f are open circuit and Vcd 120 V (60 Hz), predict the value of input current. Leakage flux and coil resistance are to be neglected. 3 Based on [4.7], the maximum value of mutual flux for the two cases, respec- tively, is Vad 240 2 m 4.44 N N f 4.44 2 N 60 4.44 N Vcd 120 2 m 4.44 Nf 4.44 N 60 4.44 N Or, the mutual flux has the same value for both cases. Thus, the magnetizing current and core losses have identical values for both cases. Consequently, the input current for the second case must also be 1 A. X4.7 The non-ideal transformer of Fig. X4.2 has three coils each with identical number of turns. Terminal pair e-f is open circuit. Leakage flux, coil resistance, and core losses can be neglected. When the two left-hand coils are additively connected in parallel and Vab 120 V (60 Hz), the input current is 2 A. If the lower coil (terminal pair c-d) is disconnected (open circuit) and Vab remains 120 V (60 Hz), predict the value of input current to the single coil. Based on [4.7], the maximum value of mutual flux for both cases is given by Vab 120 2 m 4.44 Nf 4.44 N 60 4.44 N Hence, the mmf established for both cases must be identical. For the first case of parallel-connected coils, the 2 A input current divides equally between the two coils to produce an exciting mmf of 1 N 1 N 2 N . When one of the two coils is discon- nected, the current through the remaining coil must increase to 2 A to produce the required mmf of 2N. Thus, input current remains 2 A. X4.8 For the ideal residential distribution transformer of Fig. X4.3, (a) determine current I1 . (b) Assume that the two series-connected secondary windings are identical and deter- mine the minimum kVA rating of a 2400:240/120 V transformer required to sustain this load without risk of winding over-temperature. 4 (a) 2400 I2 120 A 20 1200 I3 I 2 120 120 240 A 10 120 120 120 I1 I2 I3 360 1.80 A 2400 2400 2400 (b) Since I 3 is the larger secondary current, the rating is dictated by the lower secon- dary winding; thus, S R 2V3 I 3 2 120 24 5.76 kVA X4.9 The transformer of Fig. X4.4 is rated as 3 kVA, 240:120 V, 60 Hz if H 2 is connected to H 3 with 240 V applied between H1 and H 4 . However, it is connected as shown in Fig. X4.4 where Z1 10 . (a) Find the values of I X 1 and I H 3 . (b) Are all windings operating within rated current values? 5 (a) Conclude from the voltage rating that three coils have identical number of turns. VX 1 VH 3 VH 1 120 V VX 1 120 I X1 20 A R2 6 VH 3 120 IH 3 12 A Z1 10 (b) Letting R denote rated, SR 3000 I H 1R I H 2 R 12.5 A VH 1H 4 R 240 S R 3000 I X 1R 25 A VXR 120 From the results of part (a), it is seen that the X1 -X 2 coil and the H3 -H 4 coil are operating within rated current. 1 1 I H 1 I H 3 I X 1 12 20 32 A 1 1 Thus, the current of the H1 -H 2 coil significantly exceeds the rated value of 12 A. 6 X4.10 Determine the value of Ceq for the ideal transformer of Fig. X4.5. 2 a X eq X C 1 2 1 2 1 4 Ceq 1 C C or 1 Ceq C 4 X4.11 The three-winding ideal transformer of Fig. X4.6 has N1 N2 2 N3 and identical load resistors (R ) connected across coils 2 and 3. Determine the input impedance Z1 as indicated on Fig. X4.6. 7 MMF balance requires that N1I1 N 2 I 2 N3 I3 1 N1I1 N1I 2 N1I3 2 or 1 I1 I 2 I3 (1) 2 Since the value of flux through all three coils is identical, V1 V2 2V3 . By Ohm's law, V2 V1 I2 (2) R R V3 V1 I3 (3) R 2R Use (2) and (3) in (1) to find V1 V1 5V1 I1 R 4R 4R Hence, V1 4 Z1 R I1 5 X4.12 A 15 kVA, 2400:240/120 V, 60 Hz, two-winding transformer is to be reconnected as a 2400:2520 V step-up autotransformer. From test work on the two-winding trans- former, it is known that its rated voltage core losses and coil losses are 280 W and 300 W, respectively. For this autotransformer, (a) determine the apparent power rating and (b) the full-load efficiency if supplying 2520 V to a 0.8 PF lagging load. (a) The connection is similar to Fig. 4.31b except that the upper coil consists of the parallel additive connection of the two 120 V secondary windings. Following the procedure of Example 4.14, 15,000 I H I2 125 A 120 VH V1 V2 2400 120 2520 S X S H VH I H 2520 125 315 kVA (b) The core and copper losses are unchanged from the two-winding transformer. Po S H PF 315,000 0.8 252 kW Po 100 252,000 100 99.77 % Po losses 252,000 280 300 8 X4.13 An autotransformer is frequently used as a variable voltage supply in the laboratory. The construction is a single coil wound on a toroidal core. A common lead exists between the input and output as shown by Fig. 4.32. The other output lead makes sliding contact with the coil. If the input voltage is impressed across the total span of the coil, then the two coil sections are always additive. If such an autotransformer is rated for 20 A output, what must be the current rating of the winding? Let be the per unit portion of the N turn coil between output leads. Then mmf balance is given by I1 1 N I 2 N (1) Referring to Fig. 4.32, I1 I 2 I X 20 (2) Simultaneous solution of (1) and (2) yields I1 20 I 2 1 20 Whence it is seen that as 1, I1 20 A and I 2 0 A . Conversely, as , I1 0 A and I 2 20 A . Thus, the coil must be rated for 20 A to handle the extremes in output voltage. X4.14 The transformer of Example 4.8 (using the approximate equivalent circuit) is sup- plying rated current and voltage to a load. For the load point, V1 2V2 V2' . Determine the load PF. This described condition can only occur for a leading PF as illustrated by the phasor diagram of Fig. X4.7. By the Law of Cosines, V1 2 V2' 2 2 I 2' Z eq cos 2V1V2' From Example 4.8, I 2' 20.83 A , and Z eq Req j X eq 0.12 j 0.36 0.379571.56 9 Then, 240 2 240 2 20.83 0.37952 1 cos 1.887 2 240 240 By application of KVL, V1 V2' 2401.887 2400 I 2' 20.8219.38 Zeq 0.379571.56 PFL cos V2' I 2' cos 19.38 0.943 leading X4.15 Three 100 kVA, 12,470:7200 V, 60 Hz, two-winding transformers are to be connected to provide 300 kVA, 7200/4160 V service to an industrial customer from the three-phase 7200 V distribution mains. (a) Determine the connection arrangement for this transformer bank. (b) For balanced operation, with an apparent power load of 150 kVA, determine the values of line current on both sides of the transformer bank. (a) To meet the service agreement of 300 kVA with rated voltages, the transformers must be connected on the primary side. A wye connection on the secondary side provides for 7200 V line-to-line and 4157 V (nominal 4160 V) line-to-neutral. (b) With the transformer bank operating at half rated apparent power capability and referring to Fig. 4.33d, ST 150,000 I L1 6.94 A S VL1 3 12, 470 ST 150,000 I L2 12.03 A 3 VL 2 3 7200 X4.16 A 25 kVA, 2400:240 V, 60 Hz, two-winding transformer is to be applied in a distribution system with service at 2000:200 V, 50 Hz. (a) Is there any fundamental problem with this application? (b) Determine the apparent power rating in the 50 Hz application. (a) Based on [4.7], the maximum value of mutual flux is V1 m 4.44 N1 f Since the ratio of voltage to frequency V1 2400 2000 f 60 60 Hz 50 50 Hz 10 is unchanged, the magnetic core will operate at the design level of flux density with the magnetizing current unchanged. The core losses, being frequency dependent, will reduce in value, resulting in a cooler operating temperature. Operationally, there is no problem. (b) The transformer can still only thermally handle the rated current for the 60 Hz design case. However, voltage has reduced. Thus, 50 ST 25 kVA 20.83 kVA 60 11