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3 years ago

6

Find the inductive reactance per mile of a single-phase overhead transmission line operating at 60 Hz, given the conductors to b

e Patridge and the spacing between centers to be 20 ft

1 answer:

spin [16.1K]3 years ago

6 0

Answer:

The inductive reactance is 0.8281 Ω/mile

Explanation:

Given;

frequency, f = 60 Hz

space between the center, GMD = 20 ft

The inductive reactance per mile is calculated as;

$X_L =X_a + X_d\\\\X_L = 2.022*10^{-3} *f *ln\frac{1}{GMR} + 2.022*10^{-3} *f *ln\ GMD$

Where;

GMR is geometric mean radius (obtained from manufacturer's table)

GMD is geometric mean distance

For Partridge conductor, GMR = 0.0217 ft;

Xa = 2.022 x 10⁻³ x 60 x ln (1 / 0.0217)

Xa = 0.4647 Ω/mile

$X_d$ = 2.022 x 10⁻³ x f x ln GMD

$X_d$ = 2.022 x 10⁻³ x 60 x ln(20)

$X_d$ = 0.3634 Ω/mile

Inductive reactance;

$X_L = X_a + X_d\\\\X_L = 0.4647 \ ohms/mile \ + 0.3634 \ ohms/mile\\\\X_L = 0.8281 \ ohms/mile$

Therefore, the inductive reactance is 0.8281 Ω/mile

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One kilogram of air, initially at 5 bar, 350 K, and 3 kg of carbon dioxide (CO2), initially at 2 bar, 450 K, are confined to opp

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Answer:

Check the explanation

Explanation:

Energy alance of 2 closed systems: Heat from CO2 equals the heat that is added to air in

$m_{a} c_{v,a}(T_{eq} -T_{a,i)} =m_{co2} c_{v,co2} (T_{co2,i} -T_{eq)}$

1x0.723x $(T_{eq} -350)$ =3x0.780x $(450-T_{eq} )$ ⇒ $T_{eq}$ = 426.4 °K

The initail volumes of the gases can be determined by the ideal gas equation of state,

$V_{a,i} = \frac{mRT_{a,i} }{P_{a,i} }$ = $\frac{1x (8.314 28.97 kJ kg • °K)x 350°K}{5 bar x 100KPa bar}$ = 0.201 $m^{3}$

The equilibrium pressure of the gases can also be obtained by the ideal gas equation

$P_{eq=\frac{(m_{a}R_{a}T_{eq})+(m_{a}R_{a}T_{eq} ) }{(V_{a,eq}+V_{CO2,eq)} } =\frac{(m_{a}R_{a}T_{eq})+(m_{a}R_{a}T_{eq} ) }{(V_{a,i}+V_{CO2,i)} }$

$P_{eq}$ = 1x(8.314 28.97)x426.4+3x(8.314 44)x426.4

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3 years ago

To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a dislocation density of 105

GenaCL600 [577]

Answer:

$62.14\ \text{miles}$

$6213727.37\ \text{miles}$

Explanation:

The distance of the chain would be the product of the dislocation density and the volume of the metal.

Dislocation density = $10^5\ \text{mm}^{-2}$

Volume of the metal = $1000\ \text{mm}^3$

$10^5\times 1000=10^8\ \text{mm}\\ =10^5\ \text{m}$

$1\ \text{mile}=1609.34\ \text{m}$

$\dfrac{10^5}{1609.34}=62.14\ \text{miles}$

The chain would extend $62.14\ \text{miles}$

Dislocation density = $10^{10}\ \text{mm}^{-2}$

Volume of the metal = $1000\ \text{mm}^3$

$10^{10}\times 1000=10^{13}\ \text{mm}\\ =10^{10}\ \text{m}$

$\dfrac{10^{10}}{1609.34}=6213727.37\ \text{miles}$

The chain would extend $6213727.37\ \text{miles}$

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Burn rate can be affected by: A. Variations in chamber pressure B. Variations in initial grain temperature C. Gas flow velocity

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In this assignment, you will demonstrate your ability to write simple shell scripts. This is a cumulative assignment that will c

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Answer:

Explanation:

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-d = convert file(s) to MS-DOS/Windows newline format (linefeed + newline)

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What is the maximal coefficient of performance of a refrigerator which cools down 10 kg of water (and then ice) to -6C. Upper he

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Given:

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Coefficient of Performance (COP) for refrigerator is given by:

Max $COP_{refrigerator}$ = $\frac{T_{2}}{T_{2} - T_{1}}$

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4 years ago