Circuit Breaker Selection of LT and HT Side | Circuit Breaker Calculations

There are some circuit breaker calculations for the circuit breaker selection on the LT and HT side. We have already published a lot of articles about circuit breakers on this blog.

There is no need for calculation for the circuit breaker selection of the HT section and it is very easy to select, but it is necessary to calculate the circuit breaker selection on the LT side.

• HT Side Is used.
• LT Side  A  ACB, MCB, MCCB, Etc. are used.

The Breakers Available in the Market

VCB

ACB

MCCB: Adjustable Type

Circuit breaker calculations in high tension (circuit breaker calculations)

33 KV to 11 KV  In the case of transformers 1250A Ratings VCB  Is selected and 11 KV to 0.4 KV  In the case 630A  Ratings VCB Is selected. Below is the latest one  0.4 KV Low tension is associated with side, low tension is connected VCB That’s it 630 A The rating is VCB Have to choose

• HT section A 1250 A VCB
• HT Section Associated with its LT section On top of that 630 A VCB Is selected.

Notice the image below:

In the figure above  5 MVA, A step-down transformer of rating has been used. In its primary  33 KV The line enters the secondary 11 KV  Has come out This is a high tension section where 1250 A  The rating is VCB Has been used.

In the figure above 2500 KVA, A transformer of rating has been used. In its primary 11 KV  Entered and into the secondary 0.4 KV Got out That is, its primary section HT, The secondary section of it LT. HT  From the section LT The section is connected via a transformer, in this case, the upper HT Section VCB Will be 630A The rating is.

LT Side Breaker Selection (Circuit Breaker Calculations)

Circuit breaker selection requires a little more current than the line current, or the circuit breaker selection of the current quality.

Breaker 1 Selection

We need to calculate the value of the transformer above the breaker 1.

P = 2500 KVA

V2 (Secondary) = 0.4 KV

I = ?

Notice one thing from the image above, 2500 KVA There is no resistive, inductive, capacitive load on the transformer. Do you remember the power factor in this case ???

Apparent Power, P = √3 V I (KVA)

Active Power, P = √3 V I Cosθ (KW)

Re-Active Power, P = √3 V I Sinθ (KVAR)

Guess which formula to use ????

Sure  P = √3 V I Because 2500 KVA There is no active power and reactive power in the line from above. In that case, what will be transmitted from the grid is what we get.

Then I = P / √3 V

I = (2500 * 10^3) / √3 * 0.4 * 10^3

I = 3608.43 A

Then select the circuit breaker 3608  A Or maybe a little more around. Directly 3608 A Circuit breakers are not available in the market. Now look at the market above the table

Then from the chart above 3608 A  For that we 4000 The rating is ACB I can use a circuit breaker.

BREAKER 2 & 3 SELECTION

Notice the image above, obtained from the output of the transformer 3608 Empire but breaker for bus 2, 3, 4, 5, 5 It has been divided. In this case, the current must be calculated separately on each breaker. We need to know the power to find this current. Besides LT Voltage 0.4 KV We’ve already got it.

Breaker 2 and 3 Capacitor banks Are connected. This part provides reactive power. To get the power out Re-Active Power, P = √3 V I Sinθ (KVAR) Have to apply.

We know that in terms of power factor,

Cosθ = 0.8, Sinθ = 0.6

We already have, V = 0.4 KV, I = 3608 A

P = √3 V I Sinθ

P = √3 * 0.4 * 10^3 * 3608 * 0.6

P = 1499817 ≅ 15,00000

P = 1500 KVA

Breaker 2 and 3 There are two capacitor banks in the part where the power 750+750=1500 KVAR Can be used

Breaker 2 Power, in this case, P = 750 KVAR

I = P / √3 V

I = 750 * 10^3 / √3 * 0.4 * 10^3

I = 1082 A

This quality circuit breaker will not be available in the market. So ours from the chart ACB / MCCB 1250 A I can use anyone. ACB 1250 A It’s the tariff  MCCB 1250 A Much more than that So  MCCB 1250A Rating  We can select the unemployed.

Then the breaker 2 and 3 for this MCCB 1250A breaker selected.

Breaker 4,5,6 Determine the power of the total load

Here you have to calculate the active power, whereV = 0.4 KV, I = 3608A, Cos θ = 0.8

Active Power, P = √3 V I Cosθ (KW)

P = 1999756 ≅ 2000000 W

P = 2000 KW

Then the breaker 4,5,6 Power of load in parts = 1000 + 500 + 500 Take it If you want 1000 +750+250 Can do with it.

Breaker 4 Diagnosis

P = 1000 KW, V = 0.4 KV, I = ?

I = P / √3 V Cosθ

I = 1804 ≅ 1800 A

Breaker 4,5,6 As it is associated with the load, in accordance with international standards 20%Extra Breaker  Have to choose

Then: 1800 * 20% = 360

I = 1800 + 360

I = 2160 A

The breaker of this Ampere rating is near 2500 A ACB. Then the breaker 4 for this 2500 A  Selected breaker of rating. 

Breaker just like that 3 and 5 Have to choose And your comment on it. We will inform you by answering the answer. Be sure to comment if you have any questions about circuit breaker calculations.

Owner Of ICEEET

Engr. Mizan

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