Question

. The neutral of 10 MVA, 11 KV alternator is earthed through a resistance of 5 ohms. The earth =fault relay is set to loperate at 0.75 A. The CT's have a ratio of 1000/5. What percentage of the alternator winding is protected?

Answer 1

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Answer 2

The percentage of the alternator winding is protected 88.2%.

Explanation:

Given,

Resistance = $5 ohm$

Value of fault relay = $0.75 A$

CT's ratio = $\frac{1000}{5}$

We have to find the percentage of the alternator winding is protected.

Pick up value of relay

=  $\frac{1000 \times 0.75}{5}A$

= $150A$

Let the percentage of winding unprotected is $x\%$.

Voltage induced in this section, = $(\frac{11 \times 10^3}{\sqrt{3} } \times \frac{x}{100} )$

⇒ $150 = \frac{\frac{11 \times 10^3}{\sqrt{3} } \times \frac{x}{100} }{5}$

⇒ $150 \times 5 = \frac{11 \times 10}{\sqrt{3} } \times x }$

⇒ $x = 150 \times 5 \times \frac{ \sqrt{3}}{11 \times 10}$

⇒ $x = 75 \times \frac{ \sqrt{3}}{11}$

⇒ $x = 11.8 \%$

The percentage of winding protected = $(100 - 11.8)\%$

                                                               = $88.2 \%$

Hence, the percentage of the alternator winding is protected 88.2%.

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