एक 2-ध्रुव, 50 Hz, 11 kV, 100 MVA वाले आवर्तित्र में 10,000 kg.m2 का जड़त्वाघूर्ण है। तो जड़त्व स्थिरांक का मान, H क्या है?

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ESE Electrical 2016 Paper 2: Official Paper
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  1. 3.9 s
  2. 4.3 s
  3. 4.6 s
  4. 4.9 s

Answer (Detailed Solution Below)

Option 4 : 4.9 s
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ST 1: UPSC ESE (IES) Civil - Building Materials
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Detailed Solution

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संकल्पना:

जड़त्व स्थिरांक, H = (MJ में संग्रहित गतिज ऊर्जा)/(मशीन की रेटिंग)

संग्रहित गतिक ऊर्जा \(= \frac{1}{2}I{\omega ^2}\)

I जड़त्वाघूर्ण है। 

ω = 2πf

गणना:

P = 2, f = 50 Hz, V = 11 kV, P = 100 MVA

जड़त्वाघूर्ण, I = 10,000 kg-m2

ω = 2πf = 2π (50) = 100 π

= 314.15 रेडियन/सेकेंड

संग्रहित गतिक ऊर्जा\(= \frac{1}{2} \times 10,000 \times {\left( {314.15} \right)^2}\)

= 4.93 × 108J = 493 MJ

\(H = \frac{{493\;MJ}}{{100\;MVA}}\)

⇒ H = 4.93 सेकेंड

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