Fed from Three phase source
A1
A2
B1
B2
C1
C2
Supplying Three phase load
Open delta
Fed from Three phase source
a1
c1
a2
A1
B1
A2
b1
b2
B2
c2
Supplying Three phase load
Open delta
Fed from Three phase source
a1
c1
a2
A1
B1
A2
b1
b2
B2
c2
Supplying Three phase load
Open delta
a1
b1
a2
b2
c1
c2
Open delta
a2
a1
A1
A2
b2
b1
B1
B2
c1
c2
Vab = V∠0
Vbc = V∠ − 120
Supplying Three phase load
Fed from Three phase source
a1
b1
a2
b2
c1
c2
Open delta
a2
a1
A1
A2
b2
b1
B1
B2
c1
c2
Vab = V∠0
Vbc = V∠ − 120
Supplying Three phase load
Fed from Three phase source
a1
b1
a2
b2
c1
c2
Applying KVL in loop a2b2 c2 a2:
Vca = −[Vab + Vbc ]
= −V∠0 − V∠ − 120
= −V + 0.5V + j 0.866V
= (0.5 + j 0.866)V
= V∠120
Open delta
a2
a1
A1
A2
b2
b1
B1
B2
c1
c2
Vab = V∠0
Vbc = V∠ − 120
Supplying Three phase load
Fed from Three phase source
a1
b1
a2
b2
c1
c2
Applying KVL in loop a2b2 c2 a2:
Vca = −[Vab + Vbc ]
= −V∠0 − V∠ − 120
= −V + 0.5V + j 0.866V
= (0.5 + j 0.866)V
= V∠120
Phase load currents and line currents in open delta:
Vca = V∠120
a1
b1
I c2 a2 = I∠120
a2
Va2b2 = V∠0
I a2b2 = I∠0
I b2c2 = I∠ − 120
Vbc = V∠ − 120
b2
c1
c2
Phase load currents and line currents in open
delta:
Vca = V∠120
a1
b1
I c2 a2 = I∠120
a2
Va2b2 = V∠0
I a2b2 = I∠0
I b2c2 = I∠ − 120
b2
c1
Vbc = V∠ − 120
I a2 = I a2b2 − I c2 a2 = I∠0 − I∠120 = I∠ − 30
c2
Phase load currents and line currents in open delta:
Vca = V∠120
a1
b1
a2
Va2b2 = V∠0
I c2 a2 = I∠120
I a2b2 = I∠0
I b2c2 = I∠ − 120
b2
c1
Vb2c2 = V∠ − 120
c2
I a2 = I a2b2 − I c2 a2 = I∠0 − I∠120 = I∠ − 30
I c2 = I c2 a2 − I b2c2 = I∠120 − I∠ − 120 = I∠90
a2
b2
N.B.: Line currents have the same values and angles even if the load was
star connected
c2
Active & reactive power of open delta:
I c2
+ 90°
− 30°
st
1 Transformer :
2 nd Transformer :
Vbc = V∠ − 120
I c2 = I∠90
I a2
Va2b2 = V∠0
I a2 = I∠ − 30
Va2b2 = V∠0
− I c2
Vb2c2 = V∠ − 120
− I c2 = I∠ − 90
→
S T1
→
ST 1 = Va2b2 I
= (V∠0 )(I∠30 )
= VI∠30
→
→
→
S VV
*
a2
S VV = S T 1 + S T 2 = 3 VLine × I Line
ST 2 = Vb2c2 (− I c2 )*
= (V∠ − 120 )(I∠90 )
→
S T2
= VI∠ − 30
→
→
Q T1 = − Q T 2
1
= (VLine × I Line )
2
I Line = I Phase
A1
B1
A2
B2
Supplying Three phase load
Fed from Three phase source
De‐rating of open delta
S ∆∆ = 3 V phase × I Phase
→
→
→
S VV = S T 1 + S T 2 ≈ 3 VLine × I Line
= 3 VLine × ( I Phase )
= 3 VLine × I Phase
∴
SVV
1
≈
≈ 57.7%
S ∆∆
3
I Line = I Phase
A1
B1
A2
B2
Supplying Three phase load
Fed from Three phase source
De‐rating of open delta
S ∆∆ = 3 V phase × I Phase
→
→
→
S VV = S T 1 + S T 2 = 3 VLine × I Line
= 3 VLine × ( I Phase )
= 3 VLine × I Phase
∴
SVV
1
≈
≈ 57.7%
S ∆∆
3
I Line = I Phase
A1
B1
A2
B2
Supplying Three phase load
Fed from Three phase source
De‐rating of open delta
S ∆∆ = 3 V phase × I Phase
→
→
→
S VV = S T 1 + S T 2 = 3 VLine × I Line
= 3 VLine × ( I Phase )
= 3 VLine × I Phase
∴
SVV
1
≈
≈ 57.7%
S ∆∆
3
I Line = I Phase
A1
A2
B1
B2
IAB
IB
IBC
Supplying Three phase load
Fed from Three phase source
Open delta
Disadvantages:
- De-rated operation, 57.7% of ∆∆
bank; × (3/2) = 86.6% of the capacity
of the remaining two transformers
-Average P.F. of V bank < Load P.F.
(∼ 86.6% of balanced load P.F.);
The two transformers operate at
different power factors; for a
balanced load P.F. of cosφ, one
transformer has a P.F. of cos(30-φ)
while the other has a P.F. of
cos(30+φ),
accordingly
voltage
regulation differs.
- Accordingly, increasing load (even
if perfectly balanced), causes
terminal secondary voltages to be
unbalanced
Two single phase 150 kVA, 7200/600 V transformers are connected in open
delta configuration. Find the maximum three phase load supplied.
Rated Secondary current of either transformer = 150 000 / 600 = 250 A
This current should not be exceeded whatever was the type of connection,
Accordingly:
(S Load )maximum =
3VI = 3 × 600 × 250 = 260kVA
Available rating of two transformers = 2 × 150 = 300kVA
% Derating = 260 / 300 = 86.6%
One transformer is producing reactive power, while the other is
consuming reactive power. This energy exchange between the
two transformers is the reason for limited power output of 57.7%
instead of an expected 66.7%
Fed from Three phase source
n
A1
A2
B1
B2
C1
C2
Supplying Three phase load
Open Y – Open Delta
A1
B1
n
A2
B2
Supplying Three phase load
Fed from Three phase source
Open Y –Open Delta
Disadvantages:
- Very large return current flow
in neutral of primary