Download Chapter 4 Additional Problems

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
a
V1 240

2
V2 120
Based on [4.27],
M
Xm
400

 0.5305 H
 a 2  60  2 
By use of [4.25] and [4.26],
L1 
X1
 aM 
L2 
X2



0.18
 2  0.5305  1.0615 H
2  60 
M
0.045 0.5305


 0.2654 H
a 2  60 
2
Using the conclusion of Problem 3.15,
k
M

L1L2
0.5305
1.0615 0.2654
 0.999
X4.2 For the ideal transformer circuit of Fig. X4.1, Rp  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 
 2400 W
RL
6
2
PRL
PR p  PS  PRL  5600  2400  3200 W
V1  PRp Rp 
a
 320018  240 V
V1 240

2
V2 120
(b)
I2 
V2 1200

 200 A
RL
6
I1 
1
1
I 2   200  100 A
a
2
I S  I1 
V1
2400
 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)
PFS 
PS
5600

 0.999 lagging
VS I S  240.28  23.33
X4.3 For the circuit
of Fig. X4.1, a  10 , RL  24  , Rp  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
I1 
1
1
I 2  100  10 A
a
10
I S  I1 
V1
24000
 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)
 2400   2400  4000 W
V2
PS  1  PL 
Rp
3600
2
X4.4 For the circuit of Fig. X4.1, a  2 , R p  20  , and RL  10  . Determine the percentage of input power PS that is dissipated by R p regardless of the voltage values.
% PS 
% PS 
V12 / R p 100 
V12 / R p  V22 / RL
a 2 RL 100 
a 2 RL  R p


a 2V22 / R p 100 
a 2V22 / R p  V22 / RL
 2 2 10 100   66.7%
 2 2 10   20
X4.5 For the circuit of Fig. X4.1, let RL  0 , Rp  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, respectively, is
m 
Vad
240
2


4.44  N  N  f 4.44  2 N  60 4.44 N
m 
Vcd
120
2


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
m 
Vab
120
2


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 disconnected, 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 determine the minimum kVA rating of a 2400:240/120 V transformer required to sustain this
load without risk of winding over-temperature.
4
(a)
I2 
2400
 120 A
20
I3 
1200
 I 2  120  120  240 A
10
I1 
120
120
120
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 secondary winding; thus,
SR  2V3 I3  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
I X1 
VX 1 120

 20 A
R2
6
IH 3 
VH 3 120

 12 A
Z1
10
(b) Letting R denote rated,
I H 1R  I H 2 R 
I X 1R 
SR
VH 1H 4 R

3000
 12.5 A
240
SR 3000

 25 A
VXR 120
From the results of part (a), it is seen that the X 1 -X 2 coil and the H 3 -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
X eq
a
   XC
1
2
1
4
2 1
 

 Ceq  1   C  C
or
Ceq 
1
4
C
X4.11 The three-winding ideal transformer of Fig. X4.6 has N1  N 2  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  N2 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,
I2 
V2 V1

R R
(2)
I3 
V3 V1

R 2R
(3)
Use (2) and (3) in (1) to find
I1 
V1 V1 5V1


R 4R 4R
Hence,
Z1 
V1 4
 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 transformer, 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,
I H  I2 
15,000
 125 A
120
VH  V1  V2  2400  120  2520
S X  SH  VH I H   2520 125  315 kVA
(b) The core and copper losses are unchanged from the two-winding transformer.
Po  SH PF   315,000  0.8  252 kW

Po 100 
Po  losses

 252,000 100 
252,000  280  300
8
 99.77 %
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 supplying 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,
cos   
V1 2  V2' 
2

 I 2' Z eq

2
2V1V2'
From Example 4.8, I 2'  20.83 A , and
Zeq  Req  j X eq  0.12  j 0.36  0.379571.56 
9
Then,
  240 2   240 2   20.83  0.3795 2 
  1.887
  cos 
2  240  240 


1
By application of KVL,
I 2' 
V1  V2' 2401.887  2400

 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,
I L1 
I L2 
ST
S VL1
ST
3 VL 2

150,000
 6.94 A
3 12, 470 

150,000
 12.03 A
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
m 
V1
4.44 N1 f
Since the ratio of voltage to frequency
V1 2400

f
60

60 Hz
2000
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,
ST 
50
 25 kVA   20.83 kVA
60
11