Teaser transformer , with 0.866 tapped windings (Vertical member of T) A1
A2
B1
C1
B2
d1
d2
Main transformer , with center‐
tapped windings (Horizontal member of T) C2
Supplying Three phase load
Fed from Three phase source
Scott Connection (T‐T)
Advantages: ‐ Compared to open delta and open Y ‐open delta, a neutral can be connected to both of the primary and secondary sides
Primary coils
A1
A2
3V2
2
3V1
2
V1
Secondary coils
3 N1 3 N 2
:
2
2
d1
V2
d2
N1 : N 2
V1
C1
V1
B1
V1
2
V2
2
V1
2
Vab = V∠60
V2
2
V2 =
N2
V1
N1
V2
KVL in loop d 2 a2 c2 d 2 :
Va1d1 = Vb1d1 + Va1b1
Vc2 a2 = −Va2 d 2 + Vc2 d 2
⎝2⎠
Vad
C2
KVL in loop d1b1a1d1 :
Vca = V = V ∠60 + ⎛ V1 ⎞∠180
⎜ ⎟
1
Vab = V
Vbc = V∠180
V2
B2
=−
3
⎛V ⎞
V2 ∠90 + ⎜ 2 ⎟∠0
2
⎝ 2⎠
3
) + (−0.5 + j 0)]V1 = [− j 3 + (+0.5 + j 0)]V
2
2
2
3
3
3
= j
V1 =
V1∠90
= [0.5 − j
]V2 = V2 ∠ − 60
2
2
= [(0.5 + j
Vbd
Vbc = V
Vca = V∠ − 60
Vcd
2
VAB
Vad
VA
Vab = V
Vca = V
φ
Vad
Vbd
Vbc = V
Vcd
VBC
Ia
Vab
I aupf
Ib
Vcd
VB
VC
Vbc
Vbd 30
I db =
Ic
I bupf
30
I cupf
Vcd
= I dc
VCA
UPF load
For a balanced load P.F. of cosφ, the teaser current lags or leads the voltage by φ, while one half of the main transformer has a P.F. of cos(30‐φ), and the other half has a P.F. of cos(30+φ )
Vca
Location of neutral point “n”:
A1
A1
n
3V1
2
V1
d1
V1
V1
B1
V1
2
QVa1d1
V1 / 2
V1
2
Va1n = Vb1n = Vc1n
3V1
=
2
N
Vnd1
d1
Va1d1 =
3V
2
C1
V1 / 2
V1
=
3
⇒
Vnd1 = Va1d1 − Va1n = 0.2886V1
⇒
Vnd1
Va1d1
=
0.2886 1
=
0.866 3
Three phase to two phase conversion:
Via modified Scott connection:
A1
Turns ratio: Main :
B1
Teaser :
C1
d1
i.e :
N1 : N 2
3
N1 : N 2
2
N1 : 1.15 N 2
Taps are applied in primary windings only, while full secondary windings are utilized. The teaser primary tap is adjusted to 86.6% to provide equal in magnitude with phase quadrature two phase secondary voltages . 3φ→2φ conversion (cont.):
Secondary coils
Primary coils
A1
d1
V1
V2
C1
V1
B1
2V2
V2
3V1
2
V1
A2
V1
2
V1
2
( )
1
I
= I M1 ∠0 − IT1 ∠90
2
N
N2
1
= ( 2 ) I 2∠0 − (
) I 2∠90
N1
2
3
( ) N1
2
1
= I1∠0 −
I1∠90
3
→
d1c1 total
1
2
= I12 + ( I1 )2 ∠30 =
I1 ∠30
3
3
B2, d2
C2
V2
B2, d2
IT1
Vteaser
Vmain
IM1
(I )
0.577IM1
d1c1 total
Balanced primary currents
Secondary V, Ι (for UPF load)
Two transformers fed from a 440 V 3phase supply in a Scott fashion to operate two single phase 200 V loads with a total demand of 150 kVA. Find the necessary turns ratios and the secondary and primary currents.
(N 2 / N1 )main = 200 / 440 = 0.4545
(I main )Secondary = (I teaser )Secondary = 150000 / 2 = 375 A
200
(I main )PrimaryBalancingComponent = (N 2 / N1 )main × (I main )Secondary
= 0.4545 × 375 = 170.45 A
(N 2 / N1 )Teaser =
N 2 Main
IM1
≈ 1.15 × ( N 2 / N1 )main = 0.5227
3
(Id c )total
N1Main
2
= ( N 2 / N1 )Teaser × (I teaser )Secondary = 0.5227 × 375 = 196 A
0.577IT1
1 1
(I teaser )Primary
1
Ineffective teaser current comonent in primary main = (I teaser )Primary = 98 A
2
Total current in each half of primary main = (98) 2 + (170.45) 2 = 196A
A two phase, 7.5 kW , 240 V, 60 Hz , 0.8 power factor motor has a 83% efficiency. The motor is fed from 600V three phase distributor through a Scott connected transformer bank. Find the following: ‐ The current in each 2‐ phase line ‐ The current in each 3‐ phase line