E3 and E4 solutions

```Exercises 3 and 4 Solutions Electric Power System 2015
1(4)
Exercises 3 and 4
3.1 V0=0, V1=2200 V, V2=0 Only positive sequence!
3.2 V0=0, V1=0, V2=2200 V Only negative sequence!
3.3 V0=2200 V, V1=0, V2=0 Only zero sequence!
3.4
I1
ZTH1
Pos
seq
X eq
VF
+
V1
–
I2
ZTH2
+
X eq
Neg
seq
3R
V2
–
I0
ZTH0
Zero
seq
X eq
3Xn
V0
–
a)
b)
+
Determine sequence diagrams impedances that will carry current when the
Z1=ZTH1+jXeq=j0.12 p.u. Z2=ZTH2+jXeq=j0.12 p.u.Z0=3jXn+jXeq=j0.25 p.u.
Zbase=0.42/10 =0.016 R=0.016/0.016 p.u.=1 p.u.
VTH 1
1
I1  I2  I0 

 0.329  9.3p.u.
Z1  Z2  Z0  3R j 0.12  j0.12  j0.25  3 1
c)
There are no zero-sequence currents on the 10kV-side since these are blocked
by the -winding.
I1
X”d
X eq
Znet1
Bus 1
VF
Z1

VF
VF
+
V1
–
I2
X2
X eq
Znet2
Z2

Bus 1
+
V2
–
I0
X0
X eq
Znet0
Z0

Bus 1
V0
–
3.5
OS
+
IEA
Exercises 3 and 4 Solutions Electric Power System 2015
2(4)
Max. initial fault current = Subtransient fault current at three-phase short-circuit
Least severe fault type = Single-line to ground
Convert all values to common base Sbase=100MVA:
Network: Vbase=50kV  Zbase=(50kV)2/100MVA=25, Rfault=10/25=0.4p.u.,
Znet1=Znet2=Znet0=ZBUS(1,1)=j0.2p.u., Ibase=100MVA/( 3 50kV)=1.15kA
Transformer: Change from 50MVA to 100MVA: Xeq=100/50&middot;0.1=0.2p.u.
Generator: From 25MVA to 100MVA: X&quot;d=0.4p.u., X2=0.4p.u., X0=0.2p.u.
a)
Positive sequence Th&eacute;venin equivalent: VF=55kV/50kV=1.1p.u.,
Z1=(jX&quot;d+jXeq)//Znet1=j(0.4+0.2)//j0.2=j0.15p.u.
Three-phase fault: I1=VF/Z1=1.1/j0.15p.u.=-j7.33p.u., I2=I0=0
Ia 
I 0  7.33  90
Phase currents are: I b   A I1    7.33150 p. u.
Ic 
I 2   7.3330 
The breaker must be able to interrupt a current of 7.33p.u. or 8468A.
b)
Negative sequence Th&eacute;venin equivalent: Z2=(jX2+jXeq)//Znet2=j0.15p.u.
Zero sequence Th&eacute;venin equivalent: Z0=Znet0=j0.2p.u. (X0 and Xeq not
involved since -winding gives open-circuit)
I0=I1=I2=VF/(Z0+Z1+Z2+3Rfault)=1.1/(j0.15+j0.15+j0.2+1.2)=0.84622.6&deg;p.u.
Ia 
I 0  2.54  22.6

A
0
I
Phase currents are:  b 
I1  
p.u.
0

Ic 
I 2  
A phase current of 2.54p.u. or 2930A should be classified as a fault.
3.6
a) SLG fault phase A at bus 2 with ZF=0:
I0  I1  I2 
VF
Z 0  Z1  Z 2
where VF=1.05 and Z0 is element (2,2) of Zbus0, Z1 is element (2,2) of Zbus1, Z2
is element (2,2) of Zbus2:
I0  I1  I2 
1.05
  j 4.71
j 0.1089  j 0.057  j 0.057
Ia=I0+I1+I2=3I1=14.13 p.u., which agrees with Table 9.4.
b) LL fault phases B and C at bus 2 with ZF=0:
I0=0, I1  I2 
I1  I2 
VF
where VF=1.05, Z1 and Z2 as in a):
Z1  Z 2
1.05
  j 9.21
j 0.057  j 0.057
c) LLG fault phases B and C at bus 2 with ZF=0:
OS
IEA
Exercises 3 and 4 Solutions Electric Power System 2015
I1 
3(4)
Z0
Z2
VF
; I2  I1
; I0  I1
;
Z2Z0
Z

Z
Z

Z
0
2
0
2
Z1 
Z2  Z0
where VF=1.05, Z0, Z1 and Z2 as in a):
1.05
  j11.12
j0.057 j0.1089
j0.057 
j0.057  j0.1089
j0.1089
I2  j11.12
 j7.30
j0.1089  j0.057
j0.057
I0  j11.12
 j3.82
j0.1089  j0.057
I1 
3.7
Sequence current characteristics and A matrix from formula sheet.
Double line-to-ground (I):
I1= - (I0+I2) (b),
Ib2
Two non-zero phase currents that
need not balance each other (C).
Ib1
Ic1
Ib
Ic
I0
Ia=0
Ia1 Ia2
Line-to-line fault (III):
Ib2
Ic2
Ib1
Ic1
Ic2
I1= - I2 and I0=0 (a),
Ib
Two non-zero phase currents must
balance each other (B).
Ic
Ia1 Ia2
Single line-to-ground fault (II):
Ib2
I0=I1=I2 (c)
Ic2
I0
Ib1
One non-zero phase current (A).
Ic1
Ia1
Ia2
Ib=Ic=0
OS
IEA
I0=Ia=0
Exercises 3 and 4 Solutions Electric Power System 2015
3.8 a
ZTH1
Pos.
seq.
Xeq
3.8 b
VTH
B1
ZTH2
VTH
B1
Xeq
Neg.
seq.
B2
ZTH0
Zero
seq.
3.8 c
4(4)
B2
Xeq
3Xn
3Xn
B0
B0
B0=2504.410-940=5610-6 -1.
20
3
√3
∙ 10 ∙ 56 ∙ 10
0,65
Ic=|Ia|=3|I1|=1,9 A
3.8 d Z0= 3
∙ ∙
//
∙
∙
∙ ∙
&gt;106  which makes
the fault current (very close to) zero.
3.8 e (20/3)/6 kV/k=1,9 A This is equal to Ic, which makes sense: To cancel the
fault current, the coil should produce a current equal to Ic but with opposite
phase.
3.9
B0=2500.3310-640=4,110-3 -1 3
20
3
3
∙ 10 ∙ 4.1 ∙ 10
144
3
√3
Changing from overhead line to cable changes Ic by a factor of 48/1.9=75
=(0,33 F/km)/(4.4 nF/km)
3.10 a. IR = (20/3)/3 kV/k=3.8 A b.
OS
IEA
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