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Exam 1 Solutions

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Name: __________________________________________________________________________
ECE 5377
Exam 1 – Solutions
Wednesday February 27th, 1-2:30 PM
Closed book, Closed notes
One double-sided sheet allowed
1. __24___/24
2. __20___/20
3. __24___/24
4. __32___/32
Score: __100__/100
1
Name: __________________________________________________________________________
Problem 1 – 24 Points
A new three-phase, 60 Hz, 200-km transmission line is to be built using Bluebird conductor. Bluebird
conductors have a GMR of 0.0588 feet at 60 Hz, a 60-Hz resistance of 0.505 Ω/mi at 50°C, and an outer
diameter of 1.762 in. Each phase conductor is comprised of a bundle of three conductors spaced
symmetrically 1 feet apart. A flat horizontal tower configuration is used, with a distance of 20 feet
between adjacent phases. Assuming that the line is uniformly transposed, draw the nominal-π equivalent
circuit for the line, showing the total (not per-length) admittances and impedance of the line.
 = 4 x 10-7 H/m
0 = 8.854 x 10-12 F/m
1.609 km = 1 mi
The resistance is in per-unit length but doesn’t account for bundling and is not in per-km:
0.505 Ω 1
1 mi
𝑅=(
)( )(
) = 0.105 Ω/km
mi
3 1.609 km
The GMD is calculated using the horizontal phase spacing:
3
𝐷𝑒𝑞 = √(20 ft)(20 ft)(40 ft) = 25.2 ft
The outer diameter needs to be a radius in feet (to match the Deq units):
1.762 in
1 ft
𝑟=(
)(
) = 0.0734 ft
2
12 in
The conductors are bundled in an equilateral triangle with d = 1 ft:
9
3
𝐷𝑆𝐵 = √(𝐷𝑆 ∗ 𝑑 ∗ 𝑑)(𝑑 ∗ 𝐷𝑆 ∗ 𝑑)(𝑑 ∗ 𝑑 ∗ 𝐷𝑆 ) = √𝐷𝑆 𝑑 2
3
𝐷𝑆𝐿 = √(0.0588 ft)(1 ft)2 = 0.389 ft
3
𝐷𝑆𝐶 = √(0.0734 ft)(1 ft)2 = 0.419 ft
The x and y are therefore:
25.2 1000 m
𝑥 = 𝑗2𝜋(60)(2 × 10−7 ) ln (
)(
) = 0.314 Ω/km
0.389
1 km
𝑦 = 𝑗2𝜋(60) (
1000 m
2𝜋𝜀0
)
= 5.12 × 10−6 S/km
25.2
1 km ln (
0.419)
The nominal-π circuit is modeled using:
𝑍̅ = (200 km)(𝑅 + 𝑗𝑥) = (200)(0.105 + j0.314) = 21 + j62.8 Ω
𝑌̅
5.12 × 10−6
= (200) (
) = 5.12 × 10−4 S
2
2
2
Name: __________________________________________________________________________
Nominal-π sketch:
3
Name: __________________________________________________________________________
Problem 2 – 20 Points
A balanced 3-phase, Y-connected, 480-VLL source supplies two balanced three-phase loads connected in
parallel. One is Y-connected and draws 15 kVA at 0.8 power factor lagging. The other is ∆-connected
with impedance (30-j45) Ω per-phase. Determine:
a) The total line current
b) The three-phase power delivered by the source
Remember that single-phase modeling requires wye connections, so the line and phase currents are equal.
The line current through load 1 is found using the complex power:
̅ =[
𝐼𝐿1
∗
̅
𝑆3𝜙
√3(𝑉𝐿𝐿 )
] =
15 ∠ −cos −1(0.8)
√3(480 V)
= 18.04∠ − 36.87° A
The line current through load 2 is found using ohm’s law:
̅ = 𝐼𝜙2
̅
𝐼𝐿2
480
∠0°
𝑉̅𝜙
=
= √3
= 15.37∠56.31° A
𝑍̅∆𝑌2 (10 − 𝑗15)
The total line current is therefore:
̅ + 𝐼𝐿2
̅ = 23.04∠4.89° A
𝐼𝐿̅ = 𝐼𝐿1
The total three-phase power delivered by the source can be found using V and I:
̅ = √3𝑉̅𝐿𝐿 𝐼𝐿∗̅ = √3(480∠0° V)(23.04∠4.89° A)∗ = 19.16∠ − 4.89° kVA
𝑆3𝜙
4
Name: __________________________________________________________________________
Problem 3 - 24 Points
A single-phase, 50-kVA, 2400/240-V, 60-Hz distribution transformer has winding resistances R1 = 0.75
Ω, R2 = 0.0075 Ω and leakage reactances X1 = 1.2 Ω, X2 = 0.02 Ω.
a) Using the transformer ratings as the base ratings, determine Sbase, Vbase1, Vbase2, Zbase1 and Zbase2.
b) The transformer is supplying rated load at rated secondary voltage and 0.8 power factor lagging.
Neglecting the excitation current, determine the input terminal voltage of the transformer on the
high-voltage side.
You are free to solve (b) in per-unit or using actual values but the final answer needs to be the actual
voltage.
Using actual values:
Using per-unit values:
The values may differ slightly due to intermediate rounding.
5
Name: __________________________________________________________________________
Problem 4 – 32 Points
Short Answer, eight points each
a) Assume a 3φ 230/138 kV transformer has a per unit reactance of 0.1 using a 300 MVA base. What is
the transformer’s per unit reactance on a 100 MVA base?
In the case where the voltages are unchanged, the Zbase conversion reduces to a ratio of the MVAs:
𝑋𝑛𝑒𝑤
𝑍𝑏𝑎𝑠𝑒,𝑜𝑙𝑑
𝑉𝑏𝑎𝑠𝑒,𝑜𝑙𝑑 2
𝑆𝑏𝑎𝑠𝑒,𝑛𝑒𝑤
𝑆𝑏𝑎𝑠𝑒,𝑛𝑒𝑤
= 𝑋𝑜𝑙𝑑 (
) = 𝑋𝑜𝑙𝑑 (
)(
) = 𝑋𝑜𝑙𝑑 (
)
2
𝑍𝑏𝑎𝑠𝑒,𝑛𝑒𝑤
𝑆𝑏𝑎𝑠𝑒,𝑜𝑙𝑑
𝑆𝑏𝑎𝑠𝑒,𝑜𝑙𝑑
𝑉𝑏𝑎𝑠𝑒,𝑛𝑒𝑤
Therefore, the transformer per-unit reactance is:
100 MVA
𝑋𝑛𝑒𝑤 = (0.1 pu) (
) = 0.033 pu
300 MVA
b) A 300-km three-phase overhead transmission line, assumed to be lossless, has a series inductance of
1.0 mH/km per phase and a shunt capacitance of 0.01 mF/km per phase. Determine the phase
constant β and the wavelength λ of the line.
𝛽 = 𝜔√𝐿𝐶 = 2𝜋(60)√(1 × 10−3
𝜆=
2𝜋
𝜔√𝐿𝐶
=
H
F
) (1 × 10−5
) = 0.0377 km−1
km
km
2𝜋
= 166.67 km
𝛽
6
Name: __________________________________________________________________________
Ω
c) A 280-km, three-phase overhead transmission line has a series impedance 𝑧̅ = 0.531∠79.04° km
and series admittance 𝑦̅ = 4.105 × 10−6 ∠90°
S
.
km
Calculate ZC and 𝛾ℓ.
0.531∠79.04°
𝑍𝐶 = √
= 359.5∠ − 5.43° Ω
4.105 × 10−6 ∠90°
𝛾ℓ = ℓ√𝑧̅𝑦̅ = (280 km)√(0.531∠79.04°)(4.105 × 10−6 ∠90°) = 0.295∠84.56° pu
d) What is the surge impedance loading (SIL) of a 500 kV line with 𝑧̅ = 0.03 + 𝑗0.35 Ω/km and 𝑦 =
𝑗4.4 × 10−6 S/km?
Surge impedance is the ZC for a lossless line (𝑅 = 0, 𝑧̅ = 𝑗0.35 Ω):
SIL =
|𝑉𝑟𝑎𝑡𝑒𝑑 |2 |𝑉𝑟𝑎𝑡𝑒𝑑 |2
(500 kV)2
=
= 886.41 MW
𝑍𝐶
𝑗0.35
𝑗𝑋
√ ⁄𝑗𝐵 √
⁄
𝑗4.4 × 10−6
7
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