A three - phase, 33 kV oil circuit breaker is rated 1200 A, 2000 MVA, 3s. The symmetrical breaking current is -
35 kA
The symmetrical breaking current is a crucial parameter for any circuit breaker, especially for an oil circuit breaker, as it defines the maximum symmetrical fault current that the breaker is designed to successfully interrupt. This value is directly derived from the breaker's MVA (Mega Volt-Ampere) rating and the system voltage. The MVA rating signifies the total apparent power the circuit breaker can safely handle during fault conditions.
We are provided with the following key specifications for the three-phase, 33 kV oil circuit breaker:
Our goal is to accurately determine the symmetrical breaking current using the given MVA rating and the system voltage.
For a three-phase electrical system, the relationship between apparent power (MVA), line-to-line voltage (kV), and line current (kA) is given by the formula:
$$S = \sqrt{3} \times V_{L-L} \times I$$
Where:
To calculate the symmetrical breaking current (\(I\)), we can rearrange the formula as follows:
$$I = \frac{S}{\sqrt{3} \times V_{L-L}}$$
Now, let's substitute the provided values into the rearranged formula:
Performing the calculation:
$$I = \frac{2000 \text{ MVA}}{\sqrt{3} \times 33 \text{ kV}}$$
$$I = \frac{2000}{1.732 \times 33}$$
$$I = \frac{2000}{57.156}$$
$$I \approx 34.99 \text{ kA}$$
When rounded to the nearest whole number, the symmetrical breaking current is approximately 35 kA.
Let's check our calculated value against the provided options:
| Option Number | Value | Matches Calculation |
|---|---|---|
| 1 | 1200 A | No |
| 2 | 35 kA | Yes |
| 3 | 104.8 kA | No |
| 4 | 3600 A | No |
Based on our calculation, the symmetrical breaking current for the 33 kV oil circuit breaker with a 2000 MVA rating is approximately 35 kA, which corresponds to option 2.
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