Given:

• The system line voltage is $V_{\text{line}} = 132 \, \text{kV}$.
• The voltage distribution ratio across the insulator string is $1:0.8:0.6$ (from top to bottom).

Step 1: Convert Line Voltage to Phase Voltage

For a 3-phase system, the relationship between the line voltage and the phase voltage is given by: $V_{\text{phase}} = \frac{V_{\text{line}}}{\sqrt{3}}$ Substitute $V_{\text{line}} = 132 \, \text{kV}$: $V_{\text{phase}} = \frac{132}{\sqrt{3}} = 76.22 \, \text{kV}$

Step 2: Determine Voltage Across Each Insulator

The total voltage across the string is equal to the phase voltage. Let the voltages across each insulator from top to bottom be $V_1$, $V_2$, and $V_3$.

The voltage distribution ratio is given as $1:0.8:0.6$. Let the common factor of proportionality be $k$, such that: $V_1 = k \times 1$ $V_2 = k \times 0.8$ $V_3 = k \times 0.6$ Thus, the total voltage across the string is: $V_{\text{total}} = V_1 + V_2 + V_3 = k(1 + 0.8 + 0.6) = k \times 2.4$ But $V_{\text{total}} = V_{\text{phase}} = 76.22 \, \text{kV}$, so: $k \times 2.4 = 76.22 \, \text{kV}$ $k = \frac{76.22}{2.4} = 31.76 \, \text{kV}$

Now, calculate the voltage across each insulator: $V_1 = k \times 1 = 31.76 \, \text{kV}$ $V_2 = k \times 0.8 = 31.76 \times 0.8 = 25.41 \, \text{kV}$ $V_3 = k \times 0.6 = 31.76 \times 0.6 = 19.06 \, \text{kV}$

Thus, the voltage across each insulator is:

• $V_1 = 31.76 \, \text{kV}$
• $V_2 = 25.41 \, \text{kV}$
• $V_3 = 19.06 \, \text{kV}$

Step 3: Calculate String Efficiency

String efficiency is the ratio of the voltage across the string to the sum of the voltages across the insulators, assuming uniform voltage distribution.

The total voltage across the string is $V_{\text{total}} = 76.22 \, \text{kV}$.

If the voltage was uniformly distributed across the three insulators, each insulator would have a voltage of: $V_{\text{uniform}} = \frac{V_{\text{total}}}{3} = \frac{76.22}{3} = 25.41 \, \text{kV}$

String efficiency ($\eta_{\text{string}}$) is given by: $\eta_{\text{string}} = \frac{\text{Sum of individual voltages}}{\text{Total voltage across the string}} \times 100$ In our case, the sum of the individual actual voltages is $31.76 + 25.41 + 19.06 = 76.22 \, \text{kV}$. Therefore: $\eta_{\text{string}} = \frac{25.41 \times 3}{76.22} \times 100 = 33.33 \%$

Thus, the string efficiency is approximately $33.33\%$.

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Summary:

• Voltage across the insulators: $V_1 = 31.76 \, \text{kV}$, $V_2 = 25.41 \, \text{kV}$, $V_3 = 19.06 \, \text{kV}$.
• String efficiency: $33.33\%$.

Would you like further clarification or more details on this? Here are some related questions you might find interesting:

1. How does the voltage distribution affect the performance of suspension insulators?
2. What factors influence the string efficiency in real-world applications?
3. How can the string efficiency of a suspension insulator be improved?
4. What are the consequences of low string efficiency in high-voltage systems?
5. How do weather conditions affect the performance of insulator strings?

Tip: To improve string efficiency, grading rings or capacitive grading can be added to distribute the voltage more uniformly across the insulators.