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3 years ago

Which career best fits the group of the words below? High voltage, lines

Engineering

2 answers:

Mrac [35]3 years ago

7 0

Answer:

Electrical Engineering, and, more specifically, Power Systems Engineering (or simply, Power Engineering).

Explanation:

The words high voltage and lines refer to power transmission lines (transmission of electricity). Power transmissions lines are the main concern of the electric power transmission, which is, at the same time, part of the subfield of the Electrical Engineering called Power Systems Engineering or simply Power Engineering, which "deals with the generation, transmission, distribution and utilization of electric power, and the electrical apparatus connected to such systems" Power Engineering (2020), in Wikipedia.

Then, the career that best fits this group of words is, therefore, Electrical Engineering, and, more specifically, as a subfield of it, the Power Systems Engineering.

Llana [10]3 years ago

5 0

Answer:

I forget the word for it, but probably the guys who set up the power lines in the city.

Explanation:

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A program is seeded with 30 faults. During testing, 21 faults are detected, 15 of which are seeded faults and 6 of which are ind

Vesna [10]

Answer:

Estimated number of indigenous faults remaining undetected is 6

Explanation:

The maximum likelihood estimate of indigenous faults is given by,

$N_F=n_F\times \frac{N_S}{n_S}$ here,

$n_F$ = the number of unseeded faults = 6

$N_S$ = number of seeded faults = 30

$n_s$ = number of seeded faults found = 15

So NF will be calculated as,

$N_F=6\times \frac{30}{15}=12$

And the estimate of faults remaining is $N_F-n_F$ = 12 - 6 = 6

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Un mol de gas ideal realiza un trabajo de 3000 J sobre su entorno, cuando se expande de manera isotermica a una temperatura de 5

Shkiper50 [21]

Answer:

74,4 litros

Explanation:

Dado que

W = nRT ln (Vf / Vi)

W = 3000J

R = 8,314 JK-1mol-1

T = 58 + 273 = 331 K

Vf = desconocido

Vi = 25 L

W / nRT = ln (Vf / Vi)

W / nRT = 2.303 log (Vf / Vi)

W / nRT * 1 / 2.303 = log (Vf / Vi)

Vf / Vi = Antilog (W / nRT * 1 / 2.303)

Vf = Antilog (W / nRT * 1 / 2.303) * Vi

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3 years ago

A three-phase wye-connected synchronous generator supplies a network through a transmission line. The network can absorb or deli

Amanda [17]

Answer:

the graph and the answer can be found in the explanation section

Explanation:

Given:

Network rated voltage = 24 kV

Impedance of network = 0.07 + j0.5 Ω/mi, 8 mi

Rn = 0.07 * 8 = 0.56 Ω

Xn = 0.5 * 8 = 4 Ω

If the alternator terminal voltage is equal to network rated voltage will have

Vt = 24 kV/√3 = 13.85 kV/phase

The alternative current is

$I_{a} =\frac{40x10^{6} }{\sqrt{3}*24x10^{3} } =926.2A$

$X_{s} =0.85\frac{13.85}{926.2} =12.7ohm$

The impedance Zn is

$\sqrt{0.56^{2}+4^{2} } =4.03ohm$

The voltage drop is

$I_{a} *Z_{n} =926.2*4.03=3732.58V$

$r_{dc} =\frac{voltage}{2*current} =\frac{13.85}{2*926.2} =7.476ohm$

rac = 1.2rdc = 1.2 * 7.476 = 8.97 Ω

The effective armature resistance is

$Z_{s} =\sqrt{R_{a}^{2}+X_{s}^{2} } =\sqrt{8.97^{2}+12.7^{2} } =15.55ohm$

The induced voltage for leading power factor is

$E_{F} ^{2} =OB^{2} +(BC-CD)^{2}$

if cosθ = 0.5

$E_{F} =\sqrt{(13850*0.5)^{2}+(\frac{3741}{2}-926.2*12.7)^{2} } =11937.51V$

if cosθ= 0.6

EF = 12790.8 V

if cosθ = 0.7

EF = 13731.05 V

if cosθ = 0.8

EF = 14741.6 V

if cosθ = 0.9

EF = 15809.02 V

if cosθ = 1

EF = 13975.6 V

The voltage regulation is

$\frac{E_{F}-V_{t} }{V_{t} } *100$

For each value:

if cosθ = 0.5

voltage regulation = -13.8%

if cosθ = 0.6

voltage regulation = -7.6%

if cosθ = 0.7

voltage regulation = -0.85%

if cosθ = 0.8

voltage regulation = 6.4%

if cosθ = 0.9

voltage regulation = 14%

if cosθ = 1

voltage regulation = 0.9%

the graph is shown in the attached image

for 10% of regulation the power factor is 0.81

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