Open delta

A1

B1

C1 C2

B2

A2

Fed

from

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Open delta

A1

B1 B2

A2

b2

a1 a2

b1

c1 c2

Fed

from

Thr

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sou

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Supp

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Open delta

A1

B1 B2

A2

b2

a1 a2

b1

c1 c2

b2

b1

a1

c1

c2

a2

Fed

from

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Supp

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Open delta

A1

B1 B2

A2

b2

a1 a2

b1

c1 c2

b2

b1

a1

c1

c2

a2

0∠=VVab

120−∠=VVbc

Fed

from

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Open delta

A1

B1 B2

A2

b2

a1 a2

b1

c1 c2

b2

b1

a1

c1

c2

a2

0∠=VVab

120−∠=VVbc

[ ]

120 )866.05.0(

866.05.0 1200

2222

∠=+=

++−=−∠−∠−=

+−=

VVj

VjVVVVVVV

:acb aVL in loopApplying K

bcabca

Fed

from

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Open delta

A1

B1 B2

A2

b2

a1 a2

b1

c1 c2

b2

b1

a1

c1

c2

a2

0∠=VVab

120−∠=VVbc

[ ]

120 )866.05.0(

866.05.0 1200

2222

∠=+=

++−=−∠−∠−=

+−=

VVj

VjVVVVVVV

:acb aVL in loopApplying K

bcabca

Fed

from

Thr

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022

∠=VV ba

120−∠=VVbc

120∠=VVca

b2

b1

a1

c1

c2

a2

022

∠= II ba

12022

−∠= II cb

12022

∠= II ac

Phase load currents and line currents in open delta:

022

∠=VV ba

120−∠=VVbc

120∠=VVca

b2

b1

a1

c1

c2

a2

022

∠= II ba

12022

−∠= II cb

30120022222

−∠=∠−∠=−= IIIIII acbaa

12022

∠= II ac

Phase load currents and line currents in open delta:

022

∠=VV ba

12022

−∠=VV cb

120∠=VVca

b2

b1

a1

c1

c2

a2

a2

b2 c2

022

∠= II ba

12022

−∠= II cb

30120022222

−∠=∠−∠=−= IIIIII acbaa

12022

∠= II ac

9012012022222

∠=−∠−∠=−= IIIIII cbacc

Phase load currents and line currents in open delta:

N.B.: Line currents have the same values and angles even if the load was star connected

302

−∠= IIa

0:1

22∠=VV

rTransforme

ba

st

902

∠= IIc

120:2

−∠=VVrTransforme

bc

nd

902

−∠=− IIc

( )( )30

300

*1 222

∠=∠∠=

=

VIIV

IVS abaT

( )( )30

90120

)( *2 222

−∠=∠−∠=

−=

VIIV

IVS ccbT

LineLine I VSSS TTVV ×=+=→→→

321)(

21

21 LineLine IV QQTT

×=−=→→

VVS→

1TS→

2TS→

Active & reactive power of open delta:

2aI

2cI

°−30

2cI−

°+ 90 022

∠=VV ba

12022

−∠=VV cb

De‐rating of open delta

A1

B1 B2

A2

PhaseLine II =

%7.573

1

3

)( 3

3

3

21

≈≈∴

×=

×=

×≈+=

×=

∆∆

→→→

SS

IV

IV

I VSSS

IVS

VV

PhaseLine

PhaseLine

LineLine

Phasephase∆∆

TTVV

Fed

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A1

B1 B2

A2

PhaseLine II =

%7.573

1

3

)( 3

3

3

21

≈≈∴

×=

×=

×=+=

×=

∆∆

→→→

SS

IV

IV

I VSSS

IVS

VV

PhaseLine

PhaseLine

LineLine

Phasephase∆∆

TTVV

De‐rating of open delta Fe

d fr

om T

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pha

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Supp

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A1

B1 B2

A2

%7.573

1

3

)( 3

3

3

21

≈≈∴

×=

×=

×=+=

×=

∆∆

→→→

SS

IV

IV

I VSSS

IVS

VV

PhaseLine

PhaseLine

LineLine

Phasephase∆∆

TTVV

PhaseLine II =

De‐rating of open delta Fe

d fr

om T

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pha

se s

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Open delta

A1

B1 B2

A2

PhaseLine II =

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

IBC

IAB

IB

Fed

from

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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:

( )

%./% DeratingkVArstransformetwoofratingAvailable

kVAVISLoad

6863002603001502

26025060033 maximum

===×=

=××==

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%

Open Y – Open Delta

A1

B1 n

B2

A2

C1 C2 Fed

from

Thr

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Open Y –Open Delta

A1

B1 n

B2

A2 Disadvantages:

- Very large return current flow in neutral of primary

Fed

from

Thr

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sou

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Supp

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