ELG3311: Tutorial 2
Problem 2-1:
The secondary winding of a transformer has a terminal voltage of vs (t) = 282.8 sin 377t V.
The turns ratio of the transformer is 50:200 (a = 0.25). If the secondary current of the
transformer is iS (t) = 7.07 sin (377t – 36.87o) A, what is the primary current of this
transformer? What is the voltage regulation and efficiency? The impedance of this
transformer referred to the primary side are:
RT = 0.005 Ω
XT = 0.225 Ω
RC = 75 Ω
XM = 20 Ω
Solution: The following equivalent circuit is referred to the primary.
IP
RT
XT
Ie
VP
RC
jX
aVS
The secondary voltage and current are
VS =
IS =
282.8
2
7.07
2
∠0o V = 200∠0o V
∠ - 36.87 o A = 5∠ - 36.87 o A
The secondary voltage referred to the primary side is
VSP = aVS = 0.25 × 200∠0o = 50∠0o
The secondary current referred to the primary side is
ISP =
IS 5∠ − 36.87 o
=
= 20∠ − 36.87 o A
a
0.25
The primary circuit voltage is given by
VP = VSP + ISP (RT + jX T )
(
)
VP = 50∠0o + 20∠ - 36.87 o (0.05 + j 0.225) = 53.6∠3.2o V
The excitation current of this transformer is
1
Ie = IC + I m =
53.6∠3.2o 53.6∠3.2o
+
= 0.7145∠3.2o + 2.675∠ - 86.8o = 2.77∠ - 71.9o
75
j 20
The total primary current is
I P = ISP + I e = 20∠ - 36.87 o + 2.77∠ - 71.9o = 22.3∠ - 41o A
The voltage regulation of the transformer
VR =
VP - aVS
53.6 − 50
× 100% =
× 100% = 7.2%
50
aVS
The input power is
(
)
Pin = VP I P cosθ = 53.6 × 22.3 × cos 3.2o + 41o = 857 W
The output power is
Pout = VS IS cos θ = 200 × 5 × cos 36.87 o = 800 W
The efficiency
η=
Pout
800
× 100% =
× 100% = 93.4%
Pin
857
2
Problem 2-2:
A 20-kVA 8000/277-V distribution transformer has the following resistances and
reactances:
RP = 32 Ω
XP = 45 Ω
RC = 250 kΩ
RS = 0.05 Ω
XS = 0.06 Ω
Xm = 30 kΩ
The excitation branch impedances are given referred to the high-voltage side of the
transformer
a.
b.
c.
d.
Find the equivalent circuit of this transformer referred to the high-voltage side.
Find the per-unit equivalent circuit of this transformer.
Assume that this transformer is supplying rated load at 277 V and 0.8 PF lagging. What
is the transformer’s input voltage? What is its voltage regulation?
What is the transformer’s efficiency under the conditions of part (c)?
Solution: The turns ratio a = 8000/277 = 28.89. The secondary impedances referred to the
primary side are
RSP = a 2 RS = (28.89)2 (0.05) = 41.7 Ω
X SP = a 2 X S = (28.89 )2 (0.06) = 50.1 Ω
IP
41.7 Ω
j45 Ω
32Ω
j50.1Ω
IS/a
Ie
j30 Ω
250 kΩ
VP
aVS
(b) The rated KVA of the transformer is 20 kVA, and the rated voltage on the primary side
is 8000 V, therefore, the rated current in the primary side is 20 kVA/8000 V = 2.5 A.
Accordingly, the base impedance on the primary side is
Z base =
Vbase 8000
=
= 3200 Ω
I base
2.5
Since Zpu = Zactual/Zbase, the resulting per unit equivalent circuit is
IP
0.01
0.013
j0.0141
J0.0157
IS/a
Ie
VP
78.125
J9.375
aVS
3
( c ) The equivalent circuit referred to the primary side is shown in the following diagram:
IP
41.7 Ω
j45 Ω
32Ω
j50.1Ω
IS/a
Ie
VP
j30 Ω
250 kΩ
IS =
20kVA
∠ − 36.87 o = 72.2∠ - 36.87 o
277V
ISP =
72.2∠ - 36.87 o
= 2.5∠ − 36.87 o A
28.89
aVS
The primary voltage is
VP = VSP + (R T + jX T ) ISP
(
)
= 8000∠0o + (73.7 + j95.1) 2.5∠ - 36.87 o = 8290∠0.55o
The voltage regulation is
VR =
8290 - 8000
× 100% = 3.63%
8000
(d) The losses are
( )
2
Pcopper = ISP RT = (2.5)2 (73.7 ) = 461 W
Pcore =
(VSP ) 2
RC
=
82902
= 275 W
250,000
The efficiency is
η=
Pout
20kVA × 0.8
× 100% = 95.6%
× 100% =
20kVA × 0.8 + 461 + 275
Pout + Pcopper + Pcore
4
Problem 2-3:
A 2000-VA, 230/115-V transformer has been tested to determine it equivalent circuit. The
results of the test are shown below:
OC
VOC = 230 V
IOC = 0.45 A
POC = 30 W
SC
VSC = 13.2 V
ISC = 6 A
PSC = 20.1 W
All data given were taken from the primary side of the transformer.
a.
b.
Find the equivalent circuit of the transformer referred to low-voltage side.
Find VR at rated conditions and (1) 0.8 PF lagging, (2) 1.0 PF, (3) 0.8 PF lagging.
Solution: From OC Test, we get
RC = 1763 Ω
X m = 534 Ω
From SC, we get
RT = 0.588 Ω
X T = j 2.128 Ω
The equivalent circuit referred to the secondary side is shown in the following diagram
0.140 Ω
a IP
j0.532 Ω
Ie
VP/a 441 Ω
134 Ω
VS
(b) The reated secondary current is
1000
= 8.7 A
IS =
115
(1) When 0.8 PF lagging
(
VPS = VS + ZT IS = 115∠0o + (0.140 + j0.532) 8.7∠ − 36.87 o
)
= 118.8∠1.4o
VR =
118.8 - 115
× 100% = 3.3%
115
(2) When 1.0 PF
(
VPS = VS + ZT IS = 115∠0o + (0.140 + j0.532 ) 8.7∠0o
)
= 116.3∠2.28o V
5
VR =
116.3 - 115
× 100% = 1.1%
115
(3) When 0.8 PF leading
(
VPS = VS + ZT IS = 115∠0o + (0.140 + j0.532 ) 8.7∠36.87 o
)
= 113.3∠2.24o
VR =
113.3 - 115
× 100% = −1.5%
115
The copper and core losses of the transformer are: 10.6 W and 32 W.
The efficiency is
115 × 8.7 × 0.8
η=
= 94.9%
115 × 8.7 × 0.8 + 10.6 + 32
6
Problem 2-4:
A single-phase power system is shown in Figure P2-1. The power source feeds a 100-kVA
14/2.4-kV transformer through a feeder impedance of 38.2 + j140 Ω. The transformer’s
equivalent series impedance referred to its low-voltage side is 0.12 + j0.5 Ω. The load on the
transformer is 90 kW at 0.85 PF lagging and 2300 V.
Zload
VG
Source
a.
b.
c.
Feeder
Transformer
Load
What is the voltage at the power source of the system?
What is the voltage regulation of the transformer?
How efficient is the overall power system?
Solution: We will refer this circuit to the secondary. The feeder’s impedance referred to the
secondary side is
 2.4 
ZSline = 
(38.2 + j140) = 1.12 + j 4.11Ω
 14 
The secondary current is given by
IS =
90
= 43.48 A
2300 × 0.9
IS = 43.48∠ - 25.8o
(a) The power at the power source of this system (referred to the secondary lab) is
S
Vsource
= VS + IS ZSline + IS ZT
(
)
(
)
= 2300∠0o + 43.48∠ − 25.8o + (1.12 + j 4.11) + 43.48∠ − 25.8o (0.12 + j 0.5)
o
= 2441∠3.7 V
Now find the voltage at the power source
(
)
 14 
o
Vsource = 2441∠3.7 o 
 = 14.24∠3.7 kV
 2.4 
(b) To find the VR, we must find the voltage at the primary side referred to the secondary
under full load condition
VPS = VS + IS ZT
(
)
= 2300∠0o + 43.48∠ − 25.8o (0.12 + j 0.5) = 2314∠0.43o
7
Therefore
VR =
2314 - 2300
× 100% = 0.6%
2300
(c) The power supplied by the load is
(d)
S
Pin = Vsource
IS cosθ = 2441 × 43.48 × cos 29.5o = 92.37 kW
The efficiency is
η=
Pou t
Pin
× 100% =
90
× 100% = 97.4%
92.37
8
Problem 2-14:
A 12.4-kV single-phase generator supplies power to the load through a transmission line.
The load’s impedance is Zload = 500∠36.87o Ω, and the transmission line’s impedance is Zline
= 60∠60o Ω.
Zline
Zload
VG
(a)
1:10
Zline
10:1
Zload
VG
(b)
a.
b.
If the generator is directly connected to the load (Figure 2-3a), what is the ratio of the
load voltage to the generator voltage? What are the transmission losses of the system?
If a 1:10 step-up transformer is placed at the output of the generator and a 10:1
transformer is placed at the load end of the transmission line, what is the new ratio of
the load voltage to the generated voltage? What are the transmission losses of the
system now?
Solution: In case of the directly-connected load, the line current is
I line = I load =
12∠0o
o
60∠60 + 500∠36.87
o
= 22.32∠ − 39.3o A
The load voltage is
(
)(
)
Vload = I load Zload = 22.32∠ − 39.3o 500∠36.87 o = 11.16∠ − 2.43o kV
The ratio of the load voltage to the generated voltage is 11.16/12.4 = 0.9. The transmission
losses are
2
Ploss = I line
Zload = (22.32)2 (30) = 14.9 kW
(b) Two transformers are used. The impedance of the transmission line becomes
9
2
2
(
)
1
1
o
o
Z load
 Zline =   60∠60 = 0.6∠60 Ω
line = 
 10 
 10 
I load
line = I load =
12.4∠0o
0.6∠60o + 500∠36.87 o
= 24.773∠ − 3690 o A
The load voltage is
(
)(
)
Vload = I load Zload = 24.773∠ - 36.90 o 500∠36.87 o = 12.386∠ − 0.03o kV
The ratio of load voltage to generated voltage is 12.386/12.4=0.999. The current in the
transmission line is
1
1
I line =   I load =   (24.77 ) = 2.477 A
10
 
 10 
The losses in the transmission line are
2
Ploss = I line
Rline = (2.477 )2 (30 ) = 184 W
Transmission losses have decreased by a factor of more than 80.
10