2.1 A transformer is made up of a 1200-turn primary coil and an open-circuited 75-turn
secondary coil wound around a closed core of cross-sectional area 42 cm2. The
core material can be considered to saturate when the rms applied flux density
reaches 1.45 T. What maximum 60-Hz rms primary voltage is possible without
reaching this saturation level? What is the corresponding secondary voltage?
How are these values modified if the applied frequency is lowered to 50 Hz?
Ans：
（1）根據 1.50 式
E max = ωNφ max = 2πfNAc Bmax
得
E max, primary = 2πfNAc Bmax
= 2π (60)(1200)(42 × 10 −4 )(1.45)
= 2755.12 V
⇒ E rms , primary =
E max, primary
2
= 1948.16 V
E max,sec ondary = 2πfNAc Bmax
= 2π (60)(75)(42 × 10 −4 )(1.45)
= 172.19 V
⇒ E rms ,sec ondary =
E max,sec ondary
2
= 121.76 V
（2） f = 60Hz → 50Hz
根據 1.50 式﹐ E max = ωNφ max = 2πfNAc Bmax
得
E max, primary = 2πfNAc Bmax
= 2π (50)(1200)(42 × 10 − 4 )(1.45)
= 2294.71 V
⇒ E rms , primary =
E max, primary
2
= 1622.61 V
E max,sec ondary = 2πfNAc Bmax
= 2π (50)(75)(42 × 10 −4 )(1.45)
= 143.42 V
⇒ E rms ,sec ondary =
E max,sec ondary
2
= 101.41 V
2.2 A magnetic circuit with a cross-sectional area of 15 cm2 is to be operated at 60 Hz
from a 120-V rms supply. Calculate the number of turns required to achieve a
peak magnetic flux density of 1.8 T in the core.
Ans：
根據 1.50 式
E max = ωNφ max = 2πfNAc Bmax
得
120 2 = 2π (60 )N (15 × 10 −4 )(1.8)
⇒N=
120 2
= 167 turns
2π (60 )(15 × 10 −4 )(1.8)
2.3 A transformer is to be used to transform the impedance of a 8- Ω resistor to an
impedance of 75 Ω . Calculate the required turns ratio, assuming the transformer to
be ideal.
Ans:
假設此變壓器為理想的變壓器
N
Z1 =  1
 N2
2

 Z 2

75
= 3.061
8
= 3 turns
匝數比: N =
2.4 A 100- Ω resister is connected to the secondary of an idea transformer with a turns
ratio of 1:4 (primary to secondary). A 10-V rms , 1-kHz voltage source is connected
to the primary . Calculate the primary current and the voltage across the 100- Ω
resister.
Ans:
此電壓器為理想變壓器
一個 10-V rms ,1-k Hz 電壓源連接在一次測，二次側兩端接一負載 100 歐姆的阻
抗 ， 匝數比 1 : 4
ㄧ次側阻抗
N
R1 =  1
 N2
2

 R2

2
1
=   × 100 = 6.25Ω
4
V1
10
=
= 1.6 A
R1 6.25
一次側電流
I1 =
二次測電壓
N 
4
V2 =  2 V1 =   × 10 = 40V
1
 N1 
2.7 A single-phase 60Hz transformer has a nameplate voltage rating of 7.97kv:266v ,
which is based on it winding turns ratio. The manufacturer calculates that the
primary (7.97kv) leakage inductance is 165mH and the primary magnetizing
inductance is 135H. For an applied primary voltage of 7970v at 60Hz,calculate the
resultant open-circuit secondary voltage.
Ans:
Ll1 =165mH
7.97kv : 266v
+
Vp=7970 V
Voc
Lm=135H
-
Voc = (
135 × 7970
266
)×(
) = 265.67V
135 + 0.165
7970
2.8 The manufacturer calculates that the transformer of Problem 2.7 has a secondary
leakage inductance of 0.225mH.
a. Calculate the magnetizing inductance as referred to the secondary side.
b. A voltage of 266v,60Hz is applied to the secondary. Calculate
(i) the resultant open-circuit primary voltage and
(ii) the secondary current which would result if the primary were short-circuited.
Ans:
part(a): Referred to the secondary
Lm , 2 = Lm ,12
(
N1
)
N2
135
=
(
7970 2
)
266
= 150.38mH
part(b): Referred to the secondary
Xm = ωLm , 2 = 2πf ×150.38mH = 56.7Ω
Xl 2 = ωLl 2 = 2πf × 0.225mH = 84.82mΩ
Xl1 = ωLl1 = 2πf ×165mH = 62.2mΩ
N1
Xm
(
)V 2
N 2 Xm + Xl 2
7970
56.7
=
×
× 265.67 = 7948 V
266 56.7 + 0.0848
(i ) V 1 =
7.97kv : 266v
Ll2 =0.225mH
V1
AC 265.67V
Lm=150.38mH
(ii) Isc =
V2
V2
=(
)
Xsc
Xl 2 + Xl1 // Xm
265.5
=
= 1807.35 A
0.0848 + (56.7 // 0.0622)
Xl2 =84.82mΩ
Xl1 // Xm
AC 265.67V
Xm=56.7Ω
2.9 A 120-V:2400-V, 60-Hz, 50-KVA transformer has magnetizing reactance (as
measured from the 120-V terminals) of 34.6Ω. The 120-V winding has a leakage
reactance of 27.4 mΩ and the 2400-V winding has a leakage reactance of 11.2Ω.
a. With the secondary open- circuited and 120V applied to the primary (120-V) winding,
calculate the primary current and secondary voltage.
b. With the secondary short - circuited, calculate the primary voltage which will result
in rate current in the primary winding. Calculate the corresponding current in the
secondary winding.
Ans:
(a)
I1 =
V1
120
=
= 4.655 A
X l1 + X m 34.6 + 27.4 * 10 −3
 Xm
V2 = NV1 × 
 X l1 + X m
 2400
34.6


 =
× 120 × 
−3 
 34.6 + 27.4 × 10 
 120
= 2398.1 V
(b)
Let X l' 2 = X l 2 / N 2 =
X m' =
Xm
N2
=
11.2
= 0.028Ω
20 2
34.6
= 0.0865Ω
20 2
X sc = X 1 + X m' //( X m' + X l' 2 ) = 0.0767Ω
50kva
= 416.6667 A
120
V1 = I rated × X sc = 416.6667 × 0.0767 = 31.9584V
I rated =
I 2=
1  34.6 
×
 × 417 = 15.7513 A
20  34.6 + 11.2 
2.10 A 460-V:2400-V transformer has a series leakage reactance of 37.2Ω as referred to
the high-voltage side. A load connect to the low-voltage side is observed to be
absorbing 25 kW, unity power factor, and the voltage is measures to be 450 V.
Calculate the corresponding voltage and power factor as measured at the high-voltage
terminals.
Ans:
IL =
PLoad 25kw
=
= 55.5556 A
450
VL
N=
2400
= 5.217
460
高壓側的電流為:
I L 55.5556
=
= 10.6481 A
N
5.217
V H = NVL + jX H I H = 2380.7∠9.5339 V
IH =
The power factor is cos (9.5339) = 0.9941 lagging
2.13 A single-phase load is supplied through a 35-kV feeder whose impedance is
95+j360Ω and a 35-kV:2400-V transformer whose equivalent impedance is
0.23+j1.27Ω referred to its low-voltage side. The load is 160kW at 0.89 leading
power factor and 2340V.
A. Compute the voltage at the high-voltage terminals of the transformer.
B. Compute the voltage at the sending end of the feeder.
C. Compute the power and reactive power input at the sending end of the feeder.
Ans：
(a) 此時要求計算機高壓側電壓
Iload = 160kw / 2340V=68.4
∵此時功率因數為 0.89，可得角度 27.1
阻抗 Zt=0.23+j1.27，角度 79Vt,H =N(VL +
VH=33.7kV
Zt IL )
(b)
^
Vsend = N (V L + ( Z t + Z f ) I L )
饋線送電端電壓
代入可得 33.4kV
(c)
饋線視在功率=實際功率+無效功率
^
^*
S send = Psend + jQsend = V send I send = 164 kw − j 64.5kVAR
因此實際功率為 164kw，無效功率為-64.5kVAR
2.17 The following data were obtained for a 20-kVA , 60-Hz,2400:240-V distribution
transformer tested at 60Hz :
Voltage,
Current,
V
A
Power,
With high-voltage winding open-circuited
122
With low-voltage terminals short-circuited
257
240
61.3
W
1.038
8.33
a. Compute the efficiency at full-load current and the rated terminal voltage at
0.8 power factor.
b. Assume that the load power factor is varied while the load current and
secondary terminal voltage are held constant. Use a phasor diagram to
determine the load power factor for which the regulation is greatest. What is
this regulation?
Ans:
(a)
視在功率 S=VI
20kVA/2.4kV=8.33A
Ploss = 122+257=379W
P = S *
cosθ
P=0.8*20=16kW
有效功率 = 視在功率 * 功率因素
η=
16
= 0.977 = 97.7 percent
16.379
我們知道阻抗公式 Z eqH = ReqH + jX eqH
ReqH = PscH / I 2 SCH , PscH = 257 w, I scH = 8.33 A, Req = 3.69..歐姆
視在功率：Sec , H = Vsc, H = 61.3 * 8.33 = 511kVA
因此無效功率
Qsc , H = S 2 sc , H − Psc2, H = 442VAR and
X eq , H =
Qsc , H
I sc2 , H
= 6.35Ω
hence
2.20 A 120:480-V, 10-KVA transformer is to be used as an autotransformer to supply a
480-V circuit from a 600-V source. When it is tested as a two-winding transformer
at rated load, unity power factor, its efficiency is 0.979.
n A. Make a diagram of connections as an autotransformer.
n B. Determine its KVA rating as an autotransformer.
n C. Find its efficiency as an autotransformer at full load, with 0.85 power factor
lagging.
Solve:
(a)
(b)
I rated = 10 4 / 120 = 83.3 A
The transformer is 600 × 83.3 = 50 KVA
(c)
Ploss = (1 − 0.979)10 KW = 210W
又因 power factor=0.85
Ploss = 0.85 × 50 KW = 42.5 KW
η=
42.5 KW
= 0.995
2.71KW
2.24 Three 100-MVA single-phase transformers, rated at 13.8KV:66.4KV, are to be
connected in at three-phase bank. each transformer has a series impedance of
0.0045+J0.19Ω referred to its 13.8-kV winding.
A. If the transformers are connected Y-Y, calculate (i) the voltage and power
rating of the three-phase connected, (ii) the equivalent impedance as referred to
its high-voltage terminals, and (iii) the equivalent impedance as referred to its
high-voltage terminals.
B. Repeat part (a) if the transformer is connected Y on its low-voltage side and
△ on its high-voltage side.
Solve:
n part (a):
(i) 23.9 kV:115 kV, 300 MVA
(ii) Zeq = 0.0045 + j0.19 Ω
(iii) Zeq = 0.104 + j4.30 Ω
n part (b):
(i) 23.9 kV:66.4 kV, 300 MVA
(ii) Zeq = 0.0045 + j0.19 Ω
(iii) Zeq = 0.0347 + j1.47 Ω
2.26 A three-phase Y-△ transformer is rated 225-kV:22-kV,400MVA and has a series
reactance of 11.7Ω as referred to its high-voltage terminals. The transformer is
supplying a load of 325 MVA, with 0.93 power factor lagging at a voltage of 24 kV
(line-to-line) on its low-voltage side. It is supplied from a feeder whose impedance
is 0.11+j2.2Ω connected to its high-voltage terminals. For these conditions,
calculate (a) the line-to-line voltage at the high-voltage terminals of the
transformer and (b) the line-to-line voltage at the sending end of the feeder.
Ans:
總阻抗 Ztot= Zf+Zt =j11.7+0.11+j2.2 = 0.11+j13.9
這個變壓器的轉換比率
N = 225k/24k = 9.375
負載電流
I load = N 2 (
325MVA jφ
)e = 7.81e jφ kA
3 * 24kV
φ = cos −1 0.93 = −21.6°
負載電壓
Vload = 24 3kV
(a)
高壓側線對線的電壓
V = 3 NVload + I load Z t = 231.7 kV
(b)
二次側的電壓
V = 3 NVload + I load Z tot = 233.3kV
2.32 A 200-A:5-A,60-Hz current transformer has the following parameters as seen from
the 200-A (primary) winding:
X 1 = 745µΩ X 21 = 813µΩ X m = 307 mΩ
R1 = 136 µΩ R21 = 128µΩ
a. Assuming a current of 200 A in the primary and that the secondary is
short-circuited, find the magnitude and phase angle of the secondary current.
b. Repeat the calculation of part (a) if the CT is shorted through a 250 μΩ
burden.
Ans:
這個變壓器的轉換比率
N = 200/5 = 40
(a) 由(2.45)式可求得 2 次側的電流
I2 =
I1
jX m
( 1
)
N R2 + j ( X m + X 21 )
200
j 307 m
(
)
40 128µ + j (307m + 813µ )
= 4.987∠0.02°
=
(b) RL1 = N 2 250 µΩ = 0.4Ω
解答所算之答案有誤
I2 =
I1
jX m
( 1
)
1
N R2 + RL + j ( X m + X 21 )
j 307 m
200
(
)
40 128µ + 0.4 + j (307 m + 813µ )
= 3.04∠52.43°
=
2.34 A 15-kV:175-kV, 125MVA, 60Hz single-phase transformer has primary and
secondary impedances of 0.0095 + j0.063 per unit each. The magnetizing
impedance is j148 per unit. All quantities are in per unit on the transformer
base. Calculate the primary and secondary resistances and reactances and the
magnetizing inductance (referred to the low-voltage side) in ohms and henrys.
Ans:
Z base ,L =
2
Vbase
,L
Pbase
=
15 k 2
125 M
= 1.8 Ω
Z base ,H =
2
Vbase
,H
Pbase
=
175 k 2
125 M
= 245Ω
Q p .u =
實際值
基準值
∴ R1 = 0.0095 Z base ,L
= 17.1mΩ
X 1 = 0.063 Z base ,L
= 113mΩ
R2 = 0.0095 Z base ,H
= 2.33Ω
X 2 = 0.063 Z base ,H
= 15.4 Ω
X m = 148 Z base ,L
= 266 Ω
X m = ωL
Xm Xm
266
=
=
ω
2πf 2π • 60
= 0.71Η
L=
2.35 The namplate on a 7.97-kV:460-V, 75kVA, single-phase transformer indicates
that it has a series reactance of 12 percent (0.12 per unit) .
a. Calculate the series reactance in ohms as referred to (i) the low-voltage
terminal and (ii) the high-voltage terminal.
b. If three of these transformers are connected in a three-phase Y-Y connection,
calculate (i) the three-phase voltage and power rating, (ii)the per unit
impedance of the transformer band, (iii) the series reactance in ohms as
referred to the high-voltage terminal, and (iv) the series reactance in ohms
as referred to the low-volatge terminal.
c. Repeat part (b) if the three transformers are conneted in Y on their HV side
and Δ on their low-voltage side.
Ans:
(a)
460 2
=
Z
(i) base ,L
75 k
= 2.82Ω
X L = 0.12 Z base ,L = 0.12 • 2.82
= 0.339 Ω
7970 2
=
Z
base
,
H
(ii)
75k
= 847 Ω
X H = 0.12 Z base ,H = 0.12 • 847
= 102Ω
(b)
(i)
QY − Y
∴ 460 • 3 = 797V
7.97 k • 3 = 13.8 kV
797V : 13.8kV
75 k • 3 = 225 kV
(ii)
X p .u = 0.12
(iii)
X H = 102Ω
(iv)
X L = 0.339Ω
(c)
(i)
Δ-Y
460:13.8kV
225kVA
(ii)
X p .u = 0.12
(iii)
X H = 102Ω
(iv)
X L = 0.113Ω
2.36 a. Consider the Y-Y transformer connection of Problem 2.35, part (b). If the rated
voltage is applied to the high-voltage terminals and the three low-voltage
terminals are short-circuited, calculate the magnitude of the phase current in per
unit and in amperes on (i) the high-voltage side and (ii) the low-voltage side.
b. Repeat this calculation for the Y-△ connection of Problem 2.35, part (c).
[ 註 ] 2.35 The nameplate on a 7.97-kV:460-V, 75-kVA, single-phase transformer
indicates that it has a series reactance of 12 percent (0.12 per unit).
b. If three of these transformers are connected in a three-phase Y-Y
connection, calculate (i) the three-phase voltage and power rating, (ii)
the series reactance in ohms as referred to the high-voltage terminal,
and (iv) the series reactance in ohms as referred to the low-voltage
terminal.
c. Repeat part (b) if the three transformers are connected in Y on their HV
side and △ on their low-voltage side.
Ans:
(a) 三相 Y 接： VL = 3VP IL = IP
Ipu =
1
= 8.33
0.12
pu
(i)因為三相 P = 3VI
I=
所以
P
3V
所以低壓側電流
Ibase, L =
Pbase
3 × 75kVA
=
= 163
( 3 × Vbase, L )
3 × 460V × 3
(
IL = Ipu × Ibase, L = 8.33 × 163 = 1357.8
)
A
A
(ii) 同理；高壓側電流
Ibase, H =
(
Pbase
3 × 75kVA
=
= 9.41
3 × Vbase, H
3 × 7.97kV × 3
)
(
IH = Ipu × Ibase, H = 8.33 × 9.41 = 78.39
)
A
A
(b) 三相 Y 接： VL = VP
Ipu =
IL = 3 IP
1
= 8.33
0.12
pu
(i) 因為三相 P = 3VI
I=
所以
P
3V
所以低壓側電流
Pbase
3 × 75kVA
=
= 282.4
3 × 460
( 3 × Vbase, L)
A
IL = Ipu × Ibase, L = 8.33 × 282.4 = 2352.4
A
Ibase, L =
(ii) 同理；高壓側電流
Ibase, H =
(
Pbase
3 × 75kVA
=
= 9.41
3 × Vbase, H
3 × 7.97kV × 3
)
(
IH = Ipu × Ibase, H = 8.33 × 9.41 = 78.39
)
A
A
2.37 A three-phase generator step-up transformer is rated 26-kV:345-kV, 850 MVA and
has a series impedance of 0.0035+j0.087 per unit on this base. It is connected to a
26-kV, 800 MVA generator, which can be represented as a voltage source in series
with a reactance of j1.57 per unit on the generator base.
a. Convert the per unit generator reactance to the step-up transformer base.
b. The unit is supplying 700 MW at 345 kV and 0.95 power factor lagging to the
system at the transformer high-voltage terminals. (i) Calculate the transformer
low-side voltage and the generator internal voltage behind its reactance in kV.
(ii) Find the generator output power in MW and the power factor.
Ans:
 Pbase, t 
 800 MVA 
pu
(a) Xg = 
 × 1.57 = 
 × 1.57 = 1.48
 850 MVA 
 Pbase, g 
(b) Pout =
700 MVA
= 0.824
850 MVA
Sout =
Pout 0.824
=
= 0.867
cos θ
0.95
因為 S = VI
所以
pu
V =1
pu
pu
I = 0.867∠φ
φ = cos −1 0.95 = −18.2°
所以
I = 0.867∠ − 18.2°
(lagging )
pu
(i) Vˆt = 1.0 + Iˆ × Zt
又 Zt = 0.0035 + j 0.087 = 0.087∠87.7 °
pu
所以 Vˆt = 1.0 + (0.867 ∠ − 18.2 ° )(0.087 ∠87.7 ° )
= 1.0 + 0.0754∠69.5°
= 1.0 + 0.0264 + j 0.0706
= 1.0264 + j 0.0706
= 1.029∠3.93°
pu
= 1.029∠3.93° × 26kV
= 26.754
kV
Vˆg = 1.0 + Iˆ(Zt + Zg )
又 Zt + Zg = 030035 + j 0.087 + j1.57
= 0.0035 + j1.657
= 1.657∠89.9°
(
pu
)(
所以 Vˆg = 1.0 + 0.867∠ − 18.2° 1.657∠89.9°
)
= 1.0 + 1.437∠71.7 °
= 1.0 + 0.4512 + j1.364
= 1.4512 + j1.364
= 2.0∠43.23°
pu
= 2.0∠43.23° × 26kV
= 52∠43.23°
kV
(ii) Sˆg = Vˆt × Iˆ∗
(
)(
(
)(
= 1.029∠3.93° 0.867∠ − 18.2°
= 1.029∠3.93° 0.867∠18.2 °
)
∗
)
= 0.892∠22.13°
= 0.8263 + j 0.336
= P + jQ
所以 Pg = 0.8362 × 850 MVA = 702.36
MW
Sg = 0.892 × 850 MVA = 758.2
pf = cos θ =
Pg 702.36
=
= 0.926
Sg
758.2
MVA
lagging