## Incorrectly specified CT?

When a CT manufacturer says that he is unable to manufacture the requested CT, nine times out of ten this is because the CT has been incorrectly specified.

To eliminate all the cumulated safety margins taken by all the people involved, the CT must be redefined on the basis of real needs:

- Real currents in the installation,
- Types of protection, required power,
- Discrimination study and protection plan (settings).

This approach must be adopted whenever the specification leads to a non standard CT.

Costs, lead times and safety are the factors at stake.

### Let us take an example:

We have calculated the **class X** of a **1000/5 CT** for a generator differential protection, assuming that **X’’ = 15 %**. Not knowing the exact characteristics of the generator, we have assumed that the **generator I _{n} equals the CT I_{n}** .

This results in **I _{f} = 100/15 × I_{n}** of the CT i.e. V

_{k}≥ 2 × 6.7 × 5 (R

_{ct}+ 2 R

_{L})

We have rounded off 6.7 as **7**. Then we assumed:

**2 R _{L} = 300 m of 2.5 mm^{2}**, i.e. 2.4 Ω hence: V

_{k}

**≥**70 R

_{ct}+ 168.

^{2}connections and by requesting the generator characteristics.

**Then:**

- 2 R
_{L}= 1.5 Ω - Generator I
_{n}= 830 A. - X’’ = 25 %, hence:

The difference is marked and shows the importance of obtaining the right information and of knowing the safety margins.

If the CT is declared impossible to manufacture, a solution, i.e. a compromise, must be found between all those involved. There is always a way out, which can be found with the help of specialists.

**As an example, here are a few leads:**

- Play on the equivalences between CTs,
- Reduce the safety coefficient (for instance 2 to 1.5 for an overcurrent protection),
- Change the secondary from 5 to 1 A (see table 1), c increase wiring cross-section,
- Overrate the CTs (primary I
_{n}), - Move the relay with respect to the CT,
- Use matching CTs with low consumption,
- and so on…

Table 1 – Losses in wiring for a 2.5 mm^{2} cross-section (8 Ω/km at 20° C). With 1 A, losses are 25 times less.

Length (m) |
5 | 10 | 20 | 50 | 100 | 200 | 400 |

Wiring losses (VA) for: |
|||||||

I_{n }= 1 A |
0.04 | 0.08 | 0.16 | 0.4 | 0.8 | 1.6 | 3.2 |

I_{n }= 5 A |
1 | 2 | 4 | 10 | 20 | 40 | 80 |

### The overrating of a CT can solve a manufacturing problem

**Let us take two examples:**

Example #1 – A 100/1 CT with a load of 2.5 VA requires an ALF of 25 for an overcurrent protection.

**The standard CTs proposed are 2.5 VA-5P20.** If a CT with a ratio of 150/1 – 2.5 VA-5P20 is proposed, the accuracy limit factor (ALF) need will be reduced in the CT primary ratio, i.e. necessary

**An ALF of 20 is thus sufficient!**

If the class X requested for a CT is proportional to a through current or a primary Isc, these values are multiplied by the CT ratio. Thus, the required knee point voltage will be less for an overrated CT, unless its increasing resistance R_{ct} starts to neutralise the ratio benefit.

In all cases, it will be possible to create a higher knee point voltage than with a CT of lower ratio, as it is proportional to the number of secondary turns.

**Globally, the chance of obtaining workable characteristics will be greater.**The same reasoning can be made for a 1 A secondary CT compared with a 5 A CT. However, the factor gain of 5 obtained on the formula by the CT ratio is often completely erased, if not reversed, by a far greater increase of secondary winding resistance.

Indeed, the space required for the number of turns x 5 results in reduction in wiring cross- section, thus naturally increasing its linear resistance. **The new resistance can thus be amply multiplied by 10 with respect to the 5 A CT.**

Example #2 – If you are tempted to impose a CT overrating, you must check the repercussions of the change in ratio.

**For example:**

- If the CT supplies a pilot wire differential protection, you must ensure that the corresponding CT at the other end of the line has also the same ratio change.
- In the case of a restricted earth protection, you must ensure that:
- the CT installed on the neutral point is also modified,
- the earth fault detection is not compromised by the overrating.

- For all protection types, you must check that setting of the protection is still possible.

### Optimisation of the differential protection CTs

Let us take the example of a transformer differential protection (see figure 1).

#### Calculating the through current

The transformer impedance limits the through current to:

Short-circuit power becomes:

**The through current at the secondary is:**

at 11 kV side:

at 3.3 kV side:

#### Formulas to be applied for V_{k} (standard protection):

Calculating the matching CTs with a ratio of **5/(5/√3)**:

**Calculating main CTs:**

at 11 kV side: **300/5**

at 3.3 kV side: **1000/5**

#### Optimisation approach

Let us examine the case of the **300/5 CT placed in the 11 kV switchboard**.

##### First hypothesis

The matching **5/(5/√3)** is the one proposed as standard by the relay manufacturer. It is located **with the relay on the 3.3 kV side**. Wiring is **2.5 mm ^{2}** throughout.

- R
_{L1}= 4 Ω - R
_{L2}= 0.08 Ω - R
_{L3}= 0.024 Ω - R = 0.25 Ω, secondary winding resistance of the matching CT,
- R
_{sp}= 0.15 Ω, primary winding resistance of the matching CT, - R
_{p}= 0.02 Ω, relay resistance.

**We find:**

- V
_{ka mini}= 43.7 V (standard V_{ka }= 58 V), - V
_{kp1 mini}= 198 R_{ct}+**847**

##### Second hypothesis

**The same as the first, except that R _{L1} wiring is in 10 mm^{2}, hence R_{L1}=1 Ω**

The result is:

V_{kp1 mini} = 198 R_{ct} + **243**

##### Third hypothesis

The matching CT is on the 11 kV side as well as the relay: R_{L1} = 0.08 Ω

V_{kp1 mini} = 198 R_{ct} + **61**

##### Fourth hypothesis

**Same as the third hypothesis, except for the matching CT which is not standard, but which is imposed on the CT manufacturer where:**

R_{s} ≤ 0.1 Ω,

R_{p} ≤ 0.1 Ω,

which results in:

V_{ka mini} = 26.5 V

V_{kp1 mini} = 198 R_{ct} + **41**

_{k}of the 300/5 CT is around 800 V.

The same approach adopted for the 1000/5 CT, placed on the 3.3 kV side, yields results that are fairly similar concerning V_{k}.

However, in view of the fact that a 1000/5 CT is easier to manufacture than a 300/5 CT, it is more advantageous to place the relay and the matching CT on the 11 kV side. If a 1 A CT is used, the same hypotheses as above enable a move from:

**V _{kp1} = 39.6 R_{ct} + 249 to V_{kp1} = 39.6 R_{ct} + 17**

The 1 A CTs may be easier to manufacture than the 5 A CTs, but all depends on the relative weight of the R_{ct} and the wiring in the V_{k} expression.

**Reference //** Current transformers: specification errors and solutions by Paola FONTI (Schneider Electric)

Why in PS class CT Imag current is specified at Vk/2 ?

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