Download CHAPTER 2 TRANSFORMERS THREE

CHAPTER 2
TRANSFORMERS
THREE-PHASE TRANSFORMERS: PHASE SHIFT AND EQUIVALENT CIRCUITS
Positive sequence components
Negative sequence components
N1 and N2 represent the number of turns in the high-voltage and low-voltage windings,
N1 (1) N1
Vab 
3Va(1)30
N2
N2
N1 (2) N1

Vab 
3Va(2)  30
N2
N2
VA(1) 
VA(2)
for positive-sequence voltages
for negative-sequence voltages
Line currents in the Y- transformer are displaced by 30° in the direction of the voltages.
Voltage bases:
N1
3Va(1) 30
VA(1)
N2

(H )
N ( L)
Vbase
3 1 Vbase
N2
Complex power flow:
(H )
Vbase
N
 3 1
( L)
Vbase
N2

(H )
( L)
Vbase
, Vbase
(line-to-line)

VA(1)(pu)  Va(1)
(pu) 30 ,
I A(1)(pu)  I a(1)(pu) 30
VA(1) I A(1)  Va(1) 30  I a(1)  30  Va(1) I a(1)
Example 2.9 : Figure: 2.17 shows a three– phase generator rated 300 MVA, 23 kV supplying a system
load of 240 MVA , 0.9 power-factor lagging at 230 kV through a 330-MVA 23 / 230Y – kV step-up
transformer of leakage reactance 11 %. Neglecting magnetizing current and choosing base values at
the load of 100 MVA and 230 kV, find IA, IB, and IC supplied to the load in per unit with VA as
reference. Specifying the proper base for the generator circuit, determine Ia, Ib, and Ic from the
generator and its terminal voltage.
FIGURE 2.17
(a) Single-line diagram; (b) per-phase equivalent circuit for Example 2.9, all parameters in per unit.
Base current on the Y side =
100 000
 251.02 A
3  230
The power-factor angle of the load current is
  cos1 (0.9)  25.84
lagging
VA  1.00 reference; Line currents
602.45
  25.84  2.4  25.84 p.u.
251.05
I B  2.4  25.84  120  2.4  145.84 p.u
IA 
I C  2.4  25.84  120  2.494.16 p.u
Low-vol tage side currents further lag by 30
I a  2.4  55.84 p.u.
The transformer reactance  0.11
I b  2.4  175.84 p.u
I c  2.464.16 p.u
100 1

p.u.
330 30
the terminal voltage of the generator
1
 2.40  55.84
30
 0.9322  j 0.4551  1.0374  26.02 p.u.
Vt  VA  30  jXI a  1.0  30  j
 terminal voltage of the generator = 1.0374  23  23.86 kV
The real power supplied by the generator
P  Re Vt I a   1.0374  2.4  cos( 26.02  55.84 )  2.160 p.u.