Question

In a 33kv OH line,there are three units in the string of insulators.If the capacitance between each insulator pin and earth is 11% of self capacitance of each insulator,Find (1) The distribution of voltage over 3 insulators and (2) String efficiency​

Answer 1

Answer:

33kv +64 jab dacsnoa

Explanation:

plz mark me as brainlist

Answer 2

The distribution of voltage over 3 insulators and String efficiency​ is 85.74%.

Step-by-step explanation:

Given:

The string of insulators in an $33kv$OH line is made up of three pieces.

The capacitance between each insulator pin and earth is 11%.

To find:

The distribution of voltage over 3 insulators and String efficiency​.

Solution:

Step 1:

The voltage across each insulator is

$V_{2} > V_{1}$

$V=V_{1}+V_{2}$⇒$(1)$

Let us assume that$C_{1} =KC$

Apply $KCL$ at node $A$

$I_{2} =I_{1} +i_{1}$

$V_{2}wc=V_{1}wc +V_{1}wc_{1}$

$V_{2}wc=V_{1}wc +V_{1}wKC$

Therefore, $V_{2} =V_{1}(1+k)$

And, $V_{1} =V_{2}(1+k)$

Step 2:

From the equation (1)

$V=V_{1} +V_{2}$

If we considered the distribution of voltage over 3 insulators

then, $V_{2}=V_{1} (1+K)$

$V_{3}=V_{2}+(V_{1} +V_{2})K$

Step 3:

String efficiency $n_{st} =\frac{V_{1}+V_{2}+V_{3} }{nV_{3}} \times 100$

$K=\frac{C_{1} }{C}$

$K=0.11$

$V_{2} =V_{1} (1+0.11)=1.11 V_{1}$

$V_{3} =V_{2}(V_{1} +V_{2} )K$

$=1.11V_{1} +2.11V_{1} (0.11)$

$=1.342V_{1}$

$n_{st} =\frac{V_{1}+V_{2}+V_{3} }{nV_{3}} \times 100$

$=\frac{V_{1}+1.11V_{1}+1.342V_{1} }{3\times 1.342V_{1}}}$

$=\frac{3.452}{4.026}$

$=85.74$%

As a result, the voltage distribution over three insulators and the string efficiency is 85.74 percent.

#SPJ3