## Let’s choose 138 kV line #3 circuit breaker

To better understand a process of high voltage equipment selection, let’s discuss an example of choosing **138 kV Line #3 Circuit Breaker**, shown in **Figure 1**, assuming that the following information about the system is available:

**Continuous current for all the lines:**

- Line #1 – 1,000 A,
- Line #2 – 1,500 A,
- Line #3 – 800A

**3-phase fault current on the bus is 46.5 kA, contributions from the lines:**

- Line #1 – 24.5 kA,
- Line#2 – 12kA,
- Line#3 – 10kA

**Projected Substation load growth is 25%.**

**Available breaker ratings:**

- Continuous current: 1,200 A, 2,000 A, 3,000 A
- Short circuit interrupting capability: 40 kA, 50 kA

### Let’s see the solution…

**Future load considering 25% growth:** (1000 A + 1500 A – 800 A) x 1.25 = **2,125 A.**

This load may be fed from Line #3 through its circuit breaker if lines #1 and #2 are switched off. So, the breaker should be sized to carry at least **2,125 A**. Closest available continuous rating meeting this requirement is **3,000 A**.

**Maximum 3-phase fault current** that Line #3 breaker needs to interrupt may be calculated by deducting from a total bus fault current a contribution from line #3, i.e. 46.5 – 10 = **36.5 kA**

**Closest short circuit current interrupting rating and meeting requirement to be at least 36.5 kA is 40 kA.**

**This is a summary of the selected circuit breaker ratings:**

- Rated maximum voltage –
**138 kV** - Rated continuous current –
**3,000 A** - BIL –
**650 kV** - Rated short circuit current –
**40 kA**

**power flows, load projection, short circuit current calculations**, etc., which are usually performed by utility company planners.

**Reference:** Fundamentals of Modern Electrical Substations (Part 3: Electrical Substation Engineering Aspects) – Boris Shvartsberg, Ph.D., P.E., P.M.P.

yuvaraj

Very nice explaination

methaq

Edvard in calculating Future load considering 25% growth: (1000 A + 1500 A – 800 A) x 1.25 why you use -800 its not clear. if this calculation for line #3, we have Continuous current for Line #3 = 800A!

muthukumar vr

hello methaq…i think apply KCL in that grid…you will get (1000 A + 1500 A – 800 A) these value.