ECE 476 – Power System Analysis Fall 2013
Exam #1, Thursday, October 3, 2013. 8:00-9:20AM
Name: Solutions
Problem 1 (10 points - 1 point each)
Circle TRUE / FALSE:
(1) TRUE / FALSE — In order to reduce the consumption of reactive power of an industrial load, banks of capacitors
are placed in parallel with the load.
(2) TRUE / FALSE — In a balanced 3-φ abc positive sequence system, the average power is given by
2π
2π
P = 3Va Ia cos θ + Va Ia cos (θ + 2ωt) + Va Ia cos θ + 2ωt +
+ Va Ia cos θ + 2ωt −
3
3
(3) TRUE / FALSE — With load convention, where the current enters the positive terminal of the circuit element,
if Q is negative then positive reactive power is absorbed.
(4) TRUE / FALSE — In a purely resistive single-phase load fed by a sinusoidal voltage of period 0.02 s, the period
of the instantaneous power drawn by the load is 0.04 s.
(5) TRUE / FALSE — For a lossy transmission line, the propagation constant is a purely imaginary number.
(6) TRUE / FALSE — For a 1-φ ideal transformer, the relations between primary and secondary voltages and
currents are
V1
V2
=
N1
N2
and
I1
I2
=
N1
N2 .
(7) TRUE / FALSE — In a lossless transmission line loaded with its surge impedance, the voltage phase angle
remains constant along the transmission line.
(8) TRUE / FALSE — When computing the flux linkages for a single-wire line of infinite length and finite radius,
the linkages due to the flux inside the wire can be neglected an the only contribution to the flux lineages is due
to the flux outside the wire.
(9) TRUE / FALSE — For a Y -connected load, the magnitudes of the phase and line voltages are the same.
(10) TRUE / FALSE — The line-to-neutral capacitance of a single-phase transmission line is one-half of the line-toline capacitance.
1
Problem 2 (30 points)
I¯1
+
−
−
+
−
+
c
Z̄ L
+
V̄1
−
n
Z¯L
a
Z̄L
b
The 3-φ, 60-Hz system above consists of a Y -connected voltage source with a ∆-connected load. Based on oscilloscope
readings, it is determined that ∠V̄1 − ∠I¯1 = 10◦ . Also, the magnitudes of V̄1 and I¯1 are measured to be 208 V and
10 A, respectively. Assume positive sequence.
(a) Determine the per-phase impedance Z̄L . (15 points)
√
I¯1
I¯1 = I¯ab 3∠ − 30◦ =⇒ I¯ab = √ ∠30◦
3
√
√
V̄1
3V̄1
3(208∠10◦)
Z̄L = ¯ = ¯
= 36.03∠ − 20◦ Ω
=
10∠30◦
Iab
I1 ∠30◦
(b) Determine the power factor of the load. (5 points)
p.f. = cos(−20◦ ) = 0.9397 leading
(c) We would like to add a device in parallel with the load so that the system will have a net power factor of unity.
Should this device be capacitive or inductive? (5 points)
This device should be inductive.
(d) Determine the per-phase rating (in kVA) of the device from part (c). (5 points)
∗
= 3V̄1
S̄φ = V̄1 I¯ab
¯
∗
I1
10
◦
◦
◦
√ ∠30
= (208∠10 ) √ ∠ − 30 = 1128 − j411 VA
3
3
Therefore, the rating of the device should be 0.411 kVA.
2
Problem 3 (30 points)
A 60-Hz 138-kV 3-φ transmission line is 362 km long. The distributed line parameters are
r = 0.1 Ω/km
l = 1.3 mH/km
c = 9 nF/km
The transmission line delivers (receiving end) 40 MW at 132 kV with 0.95 power factor lagging.
IN ALL SUBSEQUENT CALCULATIONS, USE THE LONG-LINE MODEL.
(a) Find the sending-end voltage and current. (20 points)
Step 1: Find z and y for ω = 2π60 rad/s.
z = 0.1 + j2π60(0.0013) = 0.1 + j0.49 = 0.5∠78.46 Ω/km
y = j2π60(9 × 10−9 ) = j3.39 × 10−6 = 3.39 × 10−6 ∠90◦ S/km
Step 2: Find Zc and γl.
z
=
y
r
p
0.5∠78.46
1.475 × 105 ∠ − 11.54◦ = 384.0∠ − 5.77◦ Ω
=
3.39 × 10−6 ∠90◦
p
p
√
γl = zyl = 362 (0.5∠78.46)(3.39 × 10−6 ∠90◦ ) = 362 1.695 × 10−6 ∠168.46◦
Zc =
r
= 0.4713∠84.23◦ = 0.04738 + j0.4689 per unit
Step 3: Compute eγl and e−γl .
eγl = e0.04738 ej0.4689 = 1.049∠0.4689 rad = 0.9358 + j0.4740
e−γl = e−0.04738 ej−0.4689 = 0.9537∠ − 0.4689 rad = 0.8508 − j0.4310
Step 4: Compute cosh γl and sinh γl.
eγl + e−γl
1.049∠0.4689 + 0.9537∠ − 0.4689
=
= 0.8935∠1.381◦
2
2
eγl − e−γl
1.049∠0.4689 − 0.9537∠ − 0.4689
sinh γl =
=
= 0.4545∠84.63◦
2
2
cosh γl =
Step 5: Compute ABCD parameters.
A = D = cosh γl = 0.8935∠1.381◦
B = Zc sinh γl = (384.0∠ − 5.77◦)(0.4545∠84.63◦) = 174.53∠78.86◦
1
0.4545∠84.63◦
C=
sinh γl =
= 1.18 × 10−3 ∠90.4◦
Zc
384.0∠ − 5.77◦
Step 6: Compute receiving end voltage V̄2 .
132
V̄2 = √ ∠0◦ = 76.21∠0◦ kV
3
Step 7: Compute receiving end current I¯2 .
S̄load =
40
+j
3
40
3
tan cos−1 0.95 = 13.33 + j4.382 MVA
S̄load = 13.33 + j4.382 = V̄2 I¯2∗ = (76.21∠0◦)I¯2∗
∗
13.33 + j4.382
¯
=⇒ I2 =
= 0.1842∠ − 18.19◦ kA
76.21∠0◦
3
Step 8: Compute sending end voltage V̄1 and current I¯1 .
V̄1 = AV̄2 + B I¯2 = (0.8935∠1.381◦)(76.21 × 103 ∠0◦ ) + (174.53∠78.86◦)(0.1842 × 103 ∠ − 18.19◦)
= 88.92∠19.49◦ kV
I¯1 = C V̄2 + DI¯2 = (1.18 × 10−3 ∠90.4◦ )(76.21 × 103 ∠0◦ ) + (0.8935∠1.381◦)(0.1842 × 103 ∠ − 18.19◦)
= 162.53∠15.10◦ A
√
The magnitude of the sending end line-to-line voltage is 88.92 3 = 154 kV.
(b) Find the transmission-line efficiency, i.e., find η = −P21 /P12 , where P21 is the active power flowing from the
receiving end to the sending end of the line and P12 is the active power flowing from the sending end to the
receiving end of the line. (10 points)
P12 = R{V̄1 I¯1∗ } = R{(88.92∠19.49◦)(0.16253∠ − 15.10◦ )} = 14.41 MW
P21 = −13.33 MW
13.33
−P21
=
= 0.925
η=
P12
14.41
4
Problem 4 (30 points)
A 3-φ transmission line is mounted on the tower as shown below. For (a)-(c), assume that the line is not transposed
anywhere along its length.
b
• The radius of each conductor is r meters.
• The three line currents, denoted by ia , ib , and ic , respectively, are defined
as positive into the paper and they sum up to zero.
• The relative permeability of the material comprising the wires is denoted
by µr .
3D
a
c
4D
4D
(a) Find the distributed flux (per meter) linking conductor a in terms of ia , ic , µr , r, and D. (5 points)
λa =
=
=
λa =
µ0
4π
µ0
4π
µ0
4π
µ0
4π
1
1
1
ia ln ′ + ib ln
+ ic ln
r
dab
dac
1
1
1
ia ln
+ ic ln
− (ia + ic ) ln
5D
8D
re−µr /4
1
1
1
1
ia ln
+ ic ln
− ln
− ln
5D
8D
5D
re−µr /4
5D
5
ia ln
+ ic ln
8
re−µr /4
(b) Find the distributed flux (per meter) linking conductor b in terms of ib , µr , r, and D. (5 points)
λb =
=
=
=
λb =
µ0
4π
µ0
4π
µ0
4π
µ0
4π
µ0
4π
1
1
1
+
i
ln
+
i
ln
a
c
r′
dab
dbc
1
1
1
ib ln
+
i
ln
+
i
ln
c
a
5D
5D
re−µr /4
1
1
ib ln
+ (ia + ic ) ln
5D
re−µr /4
1
1
ib ln
−
i
ln
b
5D
re−µr /4
5D
ib ln
re−µr /4
ib ln
(c) Find the distributed flux (per meter) linking conductor c in terms of ia , ic , µr , r, and D. (5 points)
λc =
=
=
λc =
µ0
4π
µ0
4π
µ0
4π
µ0
4π
1
1
1
ic ln ′ + ia ln
+ ib ln
r
dac
dbc
1
1
1
ic ln
− (ia + ic ) ln
+ ia ln
8D
5D
re−µr /4
1
1
1
1
ic ln
+ ia ln
− ln
− ln
5D
8D
5D
re−µr /4
5D
5
ic ln
+ ia ln
8
re−µr /4
5
(d) Assume that the length of the line is d meters, and that the line is transposed twice along its length. Draw a
diagram showing the relative position of the phases along the length indicating the distances from the end of the
line at which the transpositions are located. (5 points)
See GS&O, page 190, Fig. 4.13.
(e) Find the average inductance (per meter) of the transposed line. (10 points)
p
p
√
√
3
3
3
dab dbc dac = 3 (5D)(5D)(8D) = 200D = 2D 25
!
√
2D 3 25
µ0
GM D
−7
=
2
×
10
ln
ln
Lφ =
H/m
2π
r′
re−µr /4
GM D =
6