Chapter 4 Additional Problems by a48nB0

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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  1200 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        320018  240 V
                   V1 240
              a         2
                   V2 120

     (b)
                     V2 1200
              I2              200 A
                     RL    6

                     1      1
              I1      I 2   200  100 A
                     a      2

                           V1           2400
              I S  I1        100          23.330 A
                           Rp             18

             VS  Z I S  V1   0.590  23.330   2400  240.282.78 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  2400  24000 V

                     1       1
              I1      I 2  100  10 A
                     a      10
                           V1          24000
              I S  I1        10           1.6670 A
                           Rp            3600




                                                  2
               VS  Z I S  V1  10090 1.6670   24000  2405.83.97 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)
                      2400
               I2            120 A
                        20

                      1200
               I3            I 2  120  120  240 A
                        10
                      120       120       120
               I1         I2       I3        360  1.80 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  0.379571.56 


                                                                    9
      Then,
                        240  2   240  2   20.83  0.37952 
                         1
                cos                                               1.887
                                      2  240  240             
                                                                  
      By application of KVL,
                       V1  V2' 2401.887  2400
              I 2'                                 20.8219.38
                         Zeq       0.379571.56


                                             
              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




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