We know that cable selection is mainly based on circuit breakers. The circuit breaker rating is a little higher than what it is or the cable rating is required at the above rating. However, cable selection is not made based on the circuit breaker on the high voltage line.

High voltage line cable selection is mainly based on the short circuit current of the line. Our high voltage line current is low but voltage is high, so cable selection is done based on short circuit current.

33 From KV 11 The line up to kV is called the high voltage line.

In regards to cable selection of high voltage lines

there are 5 things to keep in mind:

1. You need to know the short-circuit current of the line.
2. Safety factor with short-circuit current 10 Have to multiply.
3. We have to make cable selection using formula.

Above 3 High voltage line, cable selection is done on the basis of T. But the question may be what is the short circuit current and why the safety factor with a short circuit current  10 Have to multiply.

What is Short Circuit Current?

If for some reason there is a fault in the line / short circuit then the current flow for a very short time is called short circuit current.

Why the safety factor with short-circuit current 10 Have to multiply?

If the current is low in our high voltage line, if the current is increased when the short circuit current is organized, the safety factor is calculated to protect the cable from this excess current. In addition to the safety factor 1o Who is caught according to the standard.

Cable Selection Sources on High Voltage Line:

S = (Ish * √t) / K

Where

Ish =Short circuit current (kilo-umpire)

S =Cable Size (Square mm)

K=Cable factor (copper factor =0.143 , Aluminum Factory = 0.098)

t = Fault Clearing Time (When the short circuit current is organized in line, the time selected to remove it is called fault clearing time).

Short Circuit Current Ish Extraction Source:

First we need to find the short circuit current of the high voltage line. The formula is-

Ish = P / ( √3 * V * % of Z )

Here,

% of zIs the impedance of the transformer which acts as a barrier. The lower the impedance of the transformer, the better. However, in the case of cable selection, its value 7.5% Get caught. How is this standard 7.5%This will be discussed in the testing phase of the transformer. Notice the following single line diagram:

This diagram is basically a high-voltage line diagram. In this diagram 3 Transformers (5MVA, 3000KVA, 2000KVA) There is Since our high-voltage line 33/11 Up to KV so we will make cable selection based on any transformer rating. We are 5MVA We will select the cable with the rating of the transformer.

Power, voltage and so on to determine the short circuit current %of Z Have to know The transformer rating (power) from the diagram above 5MVA, Since the transformer T is a step down transformer so when 33 The voltage will be when I select the cable from the KV side 33 When KV again 11 The voltage on the cable side of the KV side will be voltage 11 Impedance of KV and transformer (%of Z) 7.5 %

33 Cable selection for KV line 

33 Short circuit current of the KV site

P = 5 MVA

V = 33 KV

% of Z = 7.5%

Ish = P / ( √3 * V * % of Z )

Ish = ( 5 * 10 ^ 6 ) / ( √3 * 33 * 10 ^3 * 7.5% )

Ish = 1166 A

Safety factor with this 10 Multiply

Ish = 1166 * 10 = 11660 A

The value of the short circuit will be taken in kilo amperes

Ish=11660/1000

Ish=11.66 KA

We know that the cable selection of high voltage lines is done with the formula. The formula,

S = (Ish * √t) / K

From the above formula, t is the fault clearing time. What is this fault clearing time?

The short circuit current caused by the short circuit current in the line is called fault clearing time. This is fault clearing time 1 From second to highest 3 Calculates in seconds. However, during cable selection 1 If you want to calculate the seconds and select the copper cable, then the value of the copper 0.143 Have to take

According to the sources for cable selection –

Short circuit current Ish = 11.66 KA

Fault Clearing Time  t = 1 sec

Copper factor K = 0.143

S = (Ish * √t) / K

S = (11.66 * √1 ) / 0.143

S = 81 rm

S = 95 rm (95A)

Cable size 3 * 1 * (1 * 95) rm 

[3 If the three-phase line 1 Is a cable (1 * 95) rm Cable rating]

If there is no cable size as per the value obtained, then you have to select it with the cable size at or near the top rating. This is how the cable selection of the high voltage side is done. Likewise now 11 I will make cable selection of KV side.

11 Cable selection of KV HT side

11 Short circuit current on the KV side

P = 5 MVA

V = 11 KV

%of Z = 7.5%

Ish = ( 5 * 10^6 ) / ( √3 * 11 * 10 ^3 * 7.5% )

Ish = 3499 A (With safety factor 1x)

Ish = 3499 * 10 = 34990 A

The value of the short circuit will be taken in kilo amperes

Ish=34990/1000

Ish=34.99 KA

Now the cable selection of high voltage line will be done using the formula.

According to the sources for cable selection –

Short circuit current Ish = 34.99 KA

Fault Clearing Time t= 1 sec

Copper factor K= 0.143

The formula

S = (Ish * √t) / K

S = 244.68 rm

S = 240 rm (555A)

Cable size 3 * 1 * (1 * 240) rm

The values here and the size of the cable are very close 11 Cable Size of KV Side 240 rm

2000 KVM Cable selection

P = 2000 KVA

V = 11 KV

%of Z = 7.5%

Ish = ( 2000 * 10^3 ) / ( √3 * 11 * 10 ^3 * 7.5% )

Ish = 1397 A (Along with the safety factor 10 Fold)

Ish = 1397 * 10 = 13970 A

The value of the short circuit will be taken in kilo amperes

Ish = 13970/1000

Ish = 13.97 KA

According to sources, cable selection –

Short circuit current Ish = 13.97 KA

Fault Clearing Time t= 1 sec

Copper factor K= 0.143

The formula

S = (Ish * √t) / K

S = 97 rm

S = 95rm (300A)

Cable size 3 * 1 * (1 * 95) rm

3000 KVM Cable selection

P = 3000KVA

V = 11 KV

%of Z = 7.5%

Ish = ( 3000 * 10^3 ) / ( √3 * 11 * 10 ^3 * 7.5% )

Ish = 2099 A (Along with the safety factor 10 Fold)

Ish = 2099 * 10

Ish = 20990 A

The value of the short circuit will be taken in kilo amperes

Ish = 20990 / 1000

Ish = 20.99 KA

According to sources, cable selection –

S = (Ish * √t) / K

Short circuit current Ish = 20.99 KA

Fault Clearing Time  t = 1 sec

Copper factor  K = 0.143

S = 146 rm

S = 150 rm (405A)

Cable size 3 * 1 * (1 * 150) rm

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