ECE574
EXAM #4
SPRING 2011
Instructions:
1. Closed-book, closed-notes, open-mind exam.
2. Work each problem on the exam booklet in the space provided.
3. Write neatly and clearly for partial credit. Cross out any material you do not want graded.
Name:
Problem 1:
/15
Problem 2:
/25
Problem 3:
/15
Problem 4:
/15
Problem 5:
/30
Total:
A
/100
A
−
ZA
B
−
ZAB
−
ZB
−
ZBC
−
ZC
C
C
Z̄AB
Z̄BC
Z̄CA
−
ZCA
B
Y-∆ Transformation
Z̄A Z̄B + Z̄B Z̄C + Z̄C Z̄A
=
Z̄C
Z̄A Z̄B + Z̄B Z̄C + Z̄C Z̄A
=
Z̄A
Z̄A Z̄B + Z̄B Z̄C + Z̄C Z̄A
=
Z̄B
Z̄A
Z̄B
Z̄C
1
∆-Y Transformation
Z̄AB Z̄CA
=
ZAB + Z̄BC + Z̄CA
Z̄BC Z̄AB
=
ZAB + Z̄BC + Z̄CA
Z̄CA Z̄BC
=
ZAB + Z̄BC + Z̄CA
Problem 1 (15 points)
An unsymmetrical fault draws the following phase-a sequence currents:
I˜a0 (kA)
5
̸
0o
I˜a1 (kA)
5
̸
120o
I˜a2 (kA)
5
̸
− 120o
(a) Find the following phase-b sequence currents:
I˜b0 (kA)
I˜b1 (kA)
I˜b2 (kA)
(b) Find the following phase-b sequence currents:
I˜c0 (kA)
I˜c1 (kA)
I˜c2 (kA)
(c) Find the abc phase currents I˜a , I˜b , and I˜c .
(d) Indicate the type of fault (SLG, LL, DLG) and identify the faulted phase or phases.
2
Problem 2 (25 points)
1
2
3
4
The above two generators are each rated 100 MVA, 20-kV, with reactances XG1 = XG2 = 20% and
XG0 = 4%. Their neutrals are each grounded through a reactance Xn = 4%. Each three-phase
transformer is rated 100 MVA, 345Y/20∆ kV, with a leakage reactance of XT 0 = XT 1 = XT 2 = 8%.
On a base of 100 MVA, 345 kV, the reactances of the transmission are XL1 = XL2 = 15% and
XL0 = 50%. The positive-sequence voltages are both equal to 1 ̸ 0o pu so that no current flows
initially in the network.
(a) Finish drawing the per-unit sequence networks below and write down all voltage and impedances
values.
3
~
I1
+
~
V1
−
~
I0
~
I2
+
~
I1
+
+
~ ~
V2 V0
−
~
V1
−
−
~
I0
~
I2
+
+
~ ~
V2 V0
−
−
(b) Double Line−to−Ground Fault at Bus 3
(a) Double Line−to−Ground Fault at Bus 2
The above figure shows the interconnection of the Thevenin equivalent sequence networks for a
double line-to-ground fault at bus 2 and at bus 3.
(b) Show that the fault current flowing to ground in a double line-to-ground fault is equal to
I˜F = I˜b + I˜c = 3I˜0 .
(c) Finish the circuit of Figure (a) by writing down all missing voltage and impedance values.
(d) Solve the circuit in Figure (a) for I˜1 and then find I˜F = 3I˜0 by current division in per-unit
(pu) and in amperes (A).
(e) Finish the circuit of Figure (b) by writing down all missing voltage and impedance values.
(f) Solve the circuit in Figure (b) for I˜1 and then find I˜F = 3I˜0 by current division in per-unit
(pu) and in amperes (A).
4
Problem 3 (15 points)
jX’=j0.2
1.05
d
θ jX T=j0.1
CB1
jX L=j0.4
CB3
E
δ
+
−
PG =1.75
QG
CB2
CB4
jX L=j0.4
+
−
1
0o
The above system shows a synchronous generator connected to an infinite bus through a step-up
transformer and a double-circuit transmission line. The generator supplies PG = 1.75 pu real
power with a terminal voltage Vt = 1.05 pu.
Find the steady-state values of θ (deg), QG (pu), E (pu), and δ (deg).
5
Problem 4 (15 points)
A
CB1
jX’
d
jX T
j0.3
j0.1
B
j0.05
j0.05
C
j0.05
CB2
j0.05
CB4
CB3
E
δ
+
−
j0.05 D
j0.05
E
+
j0.05
−
V
0o
The above system shows a synchronous generator connected to an infinite bus through a step-up
transformer and a double-circuit transmission line. Points A through E designate different possible
fault locations. A three-phase bolted short circuit occurs at one of these designated fault locations
and the arc fault extinguishes itself after some time without any breaker tripping.
(a) Compute the transfer reactance X1 between the back emf Ẽ = E
voltage Ṽ = V ̸ 0o for each of the five fault locations.
̸
δ and the infinite bus
(b) Which fault location will yield the smallest critical clearing time? (Explain with words.)
(c) Which fault location will yield the largest critical clearing time? (Explain with words.)
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Problem 5 (30 points)
CB1
jX’=j0.2
d
jX L1=j0.4
CB3 F
1
δ
CB2
jX T=j0.1
+
CB4
+
jX L2=j0.4
−
−
1
0o
Te ,Tm (pu)
2
Te0 =Te2
Tm =1
Te1 =0
0
δs
π
π−δ s
π
δ
2
The classical swing equations of a synchronous generator connected to an infinite bus are given by
dδ
2H dω
= ω − ωs
and
= Tm − Te
dt
ωs dt
where Tm = 1 pu and the electromagnetic torque


 Tmax sin δ = 2 sin δ
Te =
for t < 0
0
for 0 ≤ t < tcl

 T
sin
δ
=
2
sin
δ
for t ≥ tcl
max
where the prefault and postfault electrical characteristics are identical for a self-clearing fault at
t = tcl when the angle has reached a value δ(tcl ) = δcl = 80o .
(a) Find the prefault and postfault steady-state angle δs in degrees.
(b) Draw the accelerating area A1 and the maximum decelerating area A2 on the above graph.
7
(c) Compute numerically area A1 .
(d) Compute numerically area A2 .
(e) Is the system stable or unstable? (Explain).
(f) Find the clearing time tcl in milliseconds if H = 2 s and fs = 60 Hz.
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