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kvv77 [185]
3 years ago
12

A 1000-MVA, 20-kV, 60-Hz, three-phase generator is connected through a 1000-MVA, 20-kV, Dy345-kV, Y transformer to a 345-kV circ

uit breaker and a 345-kV transmission line. The generator reactances are X0 d 5 0.17, X9 d 5 0.30, and Xd 5 1.5 per unit, and its time constants are T0 d 5 0.05, T9 d 5 1.0, and TA 5 0.10 s. The transformer series reactance is 0.10 per unit; transformer losses and exciting current are neglected. A three-phase short-circuit occurs on the line side of the circuit breaker when the generator is operated at rated terminal voltage and at no-load. The breaker interrupts the fault three cycles after fault inception. Determine (a) the subtransient current through the breaker in per-unit and in kA rms and (b) the rms asymmetrical fault current the breaker interrupts, assuming maximum dc offset. Neglect the effect of the transformer on the time constants

Engineering
1 answer:
aniked [119]3 years ago
7 0

Answer:

(a) the subtransient current through the breaker in per-unit and in kA rms =   71316.39kA

(b) the rms asymmetrical fault current the breaker interrupts, assuming maximum dc offset. = 152KA

Explanation:

check the attached files for explanation

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Answer:

The shear strain is 0.05797 rad.

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Step1

Given:

Height of the pad is 1.38 in.

Deformation at the top of the pad is 0.08 in.

Calculation:

Step2

Shear strain is calculated as follows:

tan\phi=\frac{\bigtriangleup l}{h}

tan\phi=\frac{0.08}{1.38}

tan\phi= 0.05797

For small angle of \phi, tan\phi can take as\phi.

\phi = 0.05797 rad.

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Answer:

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Explanation:

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state & prove parallelogram law of vector addition &Also determine magnitude &direction of resultant vector.​
ludmilkaskok [199]

Answer:

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Answer:

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