Flaviano D. Dula
Δ
-Y Connection
Current ratio of each phase of the transformer
a
=
I
ØS
/I
ØP
Line - to – current ratio of the 3Ø transformer
a
T
=
I
S
/I
P
=
I
ØS
/(√3I
ØP
)
=
I
S
/(I
P
/√3
)
=
(√3I
S
)/I
P

Three-Phase Transformer

Flaviano D. Dula
Problem 70
Three
10:1
transformers are
connected
Δ-Y
for stepping up the
2300-
volts
3-Ø source. Calculate the secondary
line voltage.
Ans.
V
S
=
39.84kV

Flaviano D. Dula
Solution
V
P
=
2300V
V
ØP
V
P
/a
√3V
P
a
Primaries
Secondaries
∆ - Y
step-up transformer
=
V
S

Flaviano D. Dula
Solution
V
S
√3
V
P
a
√3(2300)
1/10
=
=
39.84 kV
Answer
Or
a
V
ØP
V
ØS
=
V
P
V
S
/
√3
Ratio of transformation of each transformer:
a
V
S
√3
V
P
a
then
=
=
=

Flaviano D. Dula
Problem 71
A
2000-kW,
2400-V,
75%
power
factor load is to be supplied from a
34,500
-volt, 3Ø
line through a single bank
transformers. Give the
secondary/primary line currents in
amperes for the
Y-Y
connections.
REE April 2006
A.50/200
B. 48/650
C. 60/800
D. 642/45

Flaviano D. Dula
Solution
I
P
P
load
√3
V
P
cosθ
P
=
2000kW
√3(34.5kV)(0.75)
=
44.626A
Answer
Primary line current:
I
P
=

Flaviano D. Dula
Solution
I
S
P
load
√3
V
S
cosθ
=
2000kW
√3(2.4kV)(0.75)
=
641.5A
Answer
Secondary line current:
I
S
I
S
=
aI
P
=
(34500/2400)
x
44.626A
=
641.5A
or
=

Flaviano D. Dula
Problem 72
A
13.8-kV/480-V,
10-MVA
3-Ø
transformers has
5%
impedance. What
is the impedance in ohms referred to
primary?
REE March 1998
A.
0.952Ω
B. 0.03Ω
C. 5.125Ω
D. 9.01Ω

Flaviano D. Dula
Solution
Per Unit Quantity
=
Actual Quantity
Base Quantity
Z
p.u.
=
Z
actual
Z
base
Z
e
(V
base
)
2
/S
base
=
=
Z
e
S
base
(V
base
)
2
or
Per unit impedance: Z
p.u.

Flaviano D. Dula
Solution
Z
p.u.
Z
e
I
rated
V
rated
=
Z
e
S
rated
(V
rated
)
2
%Z
Z(MVA)
(kV)
2
Percent impedance:
%Z
Per unit impedance:
Z
p.u
=
Multiply
each by V
rated
=
X
100%

Flaviano D. Dula
Solution
herefore, the impedance in ohms referred to primar
Z
P
%Z(kV
L
)
2
(MVA
3Ø
)
x
100%
(0.05)
(
13.8
)
2
10
=
0.952 Ω
Note:
Z
actual
=
Z
p.u
.
Z
base
,
Z
base
=
(kV)
2
/MVA
Actual
impedance
=
=

Flaviano D. Dula
Problem 73
A transformer rated
2000-kVA, 34500/240-
volts has
5.75%
impedance. What is the per-
unit impedance?
REE April 2004
A. 0.0635
B. 0.0656
C. 0.0575
D. 34.2

Flaviano D. Dula
Solution
Z
p.u.
=
%Z/100
=
5.75% /
100
=
0.0575
Answer
Per unit impedance:
Z
p.u

Flaviano D. Dula
Problem 74
A
Y-
Δ
transformation is used to take
power from a 3-phase supply at a line
voltage of 13,200-volts and deliver a 3-Ø
load at a line voltage of 480-volts. The total
load is 2500-kVA. At rated load: determine:
(i)
ratio of transformation
a
,
(ii)
high and low sides line and phase currents,
(iii)
the transformer coil
(or winding)
voltages,
(iv)
kVA rating of each transformer, and
(v)
bank
or
overall ratio.

Flaviano D. Dula
Solution
V
P
=
13200V
V
P
3
√
I
P
I
ØS
=
a
I
ØP
a
(
√3I
P
)
=
I
S
V
P
a√3
Primaries
Secondaries
I
ØP
=
I
P
2500
kVA
Load,
480V
(3Ø)
=
V
S
V
ØP
=

Flaviano D. Dula
Solution
(i)
a
V
ØP
V
ØS
=
V
P
/√3
V
S
=
13200/√
3
480
=
15.877
Answer
=
Ratio of transformation of each transformer or
phase transformation ratio:
a
Y
P
-Δ
S

Flaviano D. Dula
Solution
(ii)
High or primary rated line current:
I
P
I
P
S
Load
√3V
P
2500
x
1000VA
√3
x
13200V
=
=
109.35A
Answer
=
Y
P
-Δ

Flaviano D. Dula
Solution
Low or secondary rated line current:
I
S
I
S
S
Load
√3V
S
2,500
,
000VA
√3
x
480V
=
=
3007.03A
Answer
a(√
3
I
P
)
=
13200/√3
480
(√3
x
2500
x
1000
√3
x
13200
)
=
3007.03A
or
=
=
Y-Δ
s

Flaviano D. Dula
Solution
=
I
ØS
a
I
ØP
=
I
P
=
109.35A
and
and
high or primary rated phase current:
I
ØP
I
ØP
S
Load
3V
ØP
2,500
,
000
3(13200/√3)
=
=
109.35A
Answer
Note:
Y-connection: I
L
=
I
Ø
or
=
Y
P
-Δ

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- Flaviano D. Dula