ECE473
EXAM #2 – SOLUTIONS
FALL 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:
/20
Problem 2:
/20
Problem 3:
/20
Problem 4:
/40
Total:
Infinitely-Long Wire Carrying a Current i (A) or Charge q (C/m):
B = µo H =
q
E =
2πϵo x
µo i
2πx
Transmission Line Parameters:
l
Ω
A
T2 + T
T1 + T
R = ρ
R2
R1
=
L = 2 × 10−7 ln
GMD
H/m
GMR’
2πϵo
F/m
ln GMD/GMR
= re−1/4 ∼
= 0.7788r
∼
= 8.854 × 10−12 F/m
C =
r′
ϵo
µo = 4π × 10−7 H/m
1 mi = 1609 m
1 ft = 12 in
1 in = 2.54 cm
1
/100
Problem 1 (20 Points)
Conductor
Size
(cmil)
700 000
Resistance (Ω/mi)
25 o C
dc
0.0836
25 Hz
0.0842
50 Hz
0.0861
Resistance (Ω/mi)
50 o C
60 Hz
0.0871
dc
0.0914
25 Hz
0.0920
50 Hz
0.0937
60 Hz
0.0947
A 700,000-cmil copper conductor with 37 strands is being considered for a single-phase two-wire
transmission line. Refer to the above table for the characteristics of this hard-drawn copper conductor with 97.3% conductivity?.
(a) What is the per-wire 60-Hz AC resistance of this transmission line in Ω/mi at 25 o C?
R1 = 0.0861 Ω/mi
(b) What is the per-wire 60-Hz AC resistance per of this transmission line in Ω/mi at 50 o C?
R2 = 0.0947 Ω/mi
(c) By extrapolation, calculate the per-wire 60-Hz AC resistance of this transmission line in Ω/mi
at 75 o C?
R
R3
R2
R1
T1
R2
R1
R3
R2
R3
T3 + T
=
=
=
R3 =
=
=
T2 + T
T1 + T
T3 + T
T2 + T
R2
T2 + T
T2
T3
T
R2
R1
=
T2 + T
T1 + T
R3
R2
=⇒
=
T3 + T
T2 + T
R1
R3 − R2
R2 − R1
=
=⇒
=
T1 + T
T − T2
T2 − T1
) 3
(
T3 − T2
R2 + (R2 − R1 ) ×
T2 − T1
(
)
75 − 50
0.0947 + (0.0947 − 0.0871) ×
50 − 25
0.0947 + (0.0947 − 0.0871) = 0.0947 + 0.0076 = 0.1023 Ω/mi
=⇒
2
Problem 2 (20 Points)
a1
d
a2
b1
b2
d
d
b3
d
a3
D
For the above single-phase transmission line configuration, calculate the GMR of each bundle if
d = 0.5 m and the outer diameter of each stranded conductor is 2r = 1.5 cm.
√
GMR =
=
=
9
√
9
√
9
(Da1a1 Da1a2 Da1a3 )(Da2a1 Da2a2 Da2a3 )(Da3a1 Da3a2 Da3a3 )
(r · d · 2d)(d · r · d)(2d · d · r)
√
4r3 d6
=
9
4×
(
0.015
2
)3
× 0.56 ∼
= 0.1438 m
3
Problem 3 (20 Points)
a1
d
a2
b1
b2
d
d
b3
d
a3
D
For the above single-phase transmission line configuration, calculate the exact GMD between phases
if D = 10 m and d = 0.5 m and compare it to the distance D = 10 m between the centers of
the two bundles.
√
GMD =
9
(Da1b1 Da1b2 Da1b3 )(Da2b1 Da2b2 Da2b3 )(Da3b1 Da3b2 Da3b3 )
√
√
√
√
√
=
( 90.5 100.25 110.5)(9.5 · 10 · 10.5)( 90.5 100.25 110.5)
√
9
90.5 · 100.25 · 110.5 · 9.5 · 10 · 10.5 ∼
=
= 10.00 m ∼
= D
√ √
9
4
Problem 4 (40 Points)
D
D
d
d
d
The conductors of a bundled and completely-transposed three-phase 500-kV transmission line are
arranged in a plane as shown above. The separation between subconductors is d = 18 in and
the center-to-center distance between bundles is D = 50 ft. Each bundle has three 636,000-cmil,
ACSR Rook conductors with parameters given below:
Code
Word
Outside
Diameter
Rook
(in)
0.977
Geometric
Mean
Radius
(ft)
0.0329
Resistance (Ω/mi)
50 o C
dc
0.1618
25 Hz
0.1638
50 Hz
0.1678
60 Hz
0.1688
(a) Determine the GMR of each bundle in ft.
GMR =
√
3
√
r·d·d =
3
0.977/2
× 1.52 ∼
= 0.4508 ft
12
(b) Determine the GMR’ of each bundle in ft.
√
√
3 ′
3
GMR’ =
r ·d·d =
0.0329 × 1.52 ∼
= 0.4199 ft
(c) Determine the equivalent GMD of this three-phase transmission line in ft.
√
√
√
3
3
3
GMD =
D · D · 2D = D 2 = 50 2 ∼
= 63.00 ft
(d) Calculate the line-to-neutral capacitance per mile C in nF/mi.
C =
2πϵo
ln(GMD/GMR)
F
∼
= 11.26 × 10−12 × 1609 × 109 ∼
= 18.12 nF/mi
m
(e) Calculate the 60-Hz line-to-neutral capacitive susceptance per mile B in µS/mi.
B = ωC
= (2π × 60)(18.12 × 10−9 ) ∼
= 6.831 µS/mi
(f) Calculate the per-phase inductance per mile L in mH/mi.
GMD ∼
H
= 10.02 × 10−7 × 1609 × 103 ∼
= 1.612 mH/mi
GMR’
m
(g) Calculate the 60-Hz per-phase inductive reactance per mile X in Ω/mi.
L = 2 × 10−7 ln
X
= ωL = (2π × 60)(1.612 × 10−3 ) ∼
= 0.6079 Ω/mi
(h) Calculate the 60-Hz per-phase AC resistance per mile R in Ω/mi at 50 o C.
R =
Rc
3
=
0.1688
3
∼
= 0.05627 Ω/mi
5
(i) Calculate an approximate value of the electric field in kV(rms)/cm at the surface of a conductor
in each bundle and comment on this transmission line design.
q = Can Van = 11.26 × 10−12 ×
E ∼
=
500 × 103
√
3
∼
= 3.250 × 10−6 C/m
q/3
2πϵo r
3.250 × 10−6 /3
2π × 8.854 × 10−12 × 0.977/2 × 2.54
= 15.70 kV(rms)/cm < 21.1 kV(rm)/cm
=
More exactly,
E ∼
=
q/3
2πϵo r
(
√ )
r 3
1+
d
(
√ )
3.250 × 10−6 /3
(0.977/2) 3
=
1+
2π × 8.854 × 10−12 × 0.977/2 × 2.54
18
∼
= 15.70 × 1.047 = 16.44 kV(rms)/cm < 21.1 kV(rm)/cm
6