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# How to select a breaker??

How to select a breaker??

#1

09-17-2013 07:06 AM

Top #2

For a feeder or a branch circuit? Single phase or 3-phase? At what voltage? What type/size of load? You'll need all this information to make the right selection.

09-17-2013 09:19 AM

Top #3

Muhammad,

You have to know the available short circuit (kA) in your system where you will place your breaker. It is included in the selection of breakers to choose from.

You have to know the available short circuit (kA) in your system where you will place your breaker. It is included in the selection of breakers to choose from.

09-17-2013 11:59 AM

Top #4

Can you provide any key to select a breaker if i know the ampere of a pump or total load of an area?

09-17-2013 02:51 PM

Top #5

Breaker must interrupt the short-circuit current and allow continuous supply of load. For motor you have to consider start-up current.

If you are not sure, hire a professional since there are severe safety issues.

If you are not sure, hire a professional since there are severe safety issues.

09-17-2013 05:05 PM

Top #6

For a branch circuit feeding a single pump, you would generally size the circuit at 125% of the pump's full-load amperage. If you're not using a variable-frequency drive or soft-starter (which have built-in overload protection), you would use a Motor-circuit protector (MCP) breaker that has both thermal and magnetic trip capability. Sizing would be according the breaker manufacturer's recommendations for a motor of a given horsepower, but not larger than would be required to protect the circuit conductors.

"The total load of an area" is much too ambiguous to answer. If you have lighting and receptacles, you're going to need a different type of breaker than if you have motors or mixed types of load. There is no general approach. Circuit breaker types are very specific to the application.

Mr. Desrosiers' point about safety should not be taken lightly. Installing the wrong type of breaker could result in equipment damage and/or physical harm.

"The total load of an area" is much too ambiguous to answer. If you have lighting and receptacles, you're going to need a different type of breaker than if you have motors or mixed types of load. There is no general approach. Circuit breaker types are very specific to the application.

Mr. Desrosiers' point about safety should not be taken lightly. Installing the wrong type of breaker could result in equipment damage and/or physical harm.

09-17-2013 08:01 PM

Top #7

Before breaker's selecting for your electrical system, you need to calculate value of expected short circuit current at the place of breaker's installation. Then you need to calculate value of heat pulse and 1s current (expected value of current during one second). After that you need to calculate power of breaker and finally, after all, you can select appropriate breaker. Values of characteristics of selected breaker need to be higher from calculated values of characteristics of your power system...

09-17-2013 10:12 PM

Top #8

There are instantaneous breakers as well as time delay breakers. For time delay breaker, for example, you go 250% maximum of the rated current based upon the HP of a motor (look in the NEC), not on the nameplate label. The nameplate current value is for overload protection. Also try to size the breaker so that the conductors are protected.

09-18-2013 12:28 AM

Top #9

Available types are MCB,MCCB,MPCB,ACB,VCB,SF6 CB and their sub divisions. You should specify type of load (lights,motors etc),category(constant,fluctuating,periodical etc) SC level,starting/switching inrush current ,Voltage,rated current etc to 3 or more manufacturers and ask them to recommend.

09-18-2013 03:08 AM

Top #10

thanks every one for your suggestions

09-18-2013 05:29 AM

Top #11

You can calculate operational current of breaker using this expression:

Inp=SnT/((sqrt(3))*Un)

After that, you need to calculate expected value of surge current:

kud=1+e(-0,01/Tae)

Iud=(sqrt(2))*kud*I'

After that, you need to calculate expected value of heat impulse:

A=(sqr(I0"))*Tae*(1-e(-2*ti/Tae))+(sqr(I'))*(ti+Td")

And finally, you need to calculate 1s current (expected value of current during 1s):

I1s=sqrt(A/1s)

So, current of interruption of your breaker and power of interruption of your breaker are:

Ii=I'

Si=(sqrt(3))*Un*Ii

Additional expressions that you can use during your calculation:

I0"=Un/((sqrt(3))*Ze");

I"=1,1*Un/((sqrt(3))*Ze");

I'=1,15*Un/((sqrt(3))*Ze');

where are:

ti-time of interruption

Inp-operational current of breaker

SnT-rated power of transformer

Un-rated voltage

kud-surge coefficient

Tae-time constant of aperiodic component of short circuit current

Iud-surge current

A-heat impulse

I0"-short circuit current in subtransient period (generators are in no-load conditions)

I'-short circuit current in transient period

Td"-time constant of subtransient component of short circuit current

I1s-current during one second

Ii=expected value of current of interruption of your breaker

Si=expected value of power of interruption of your breaker

Ze"-equivalent impedance of power system in the place of fault (subtransient period)

I"-short circuit current in subtransient period (generators are in full-load conditions)

I'-short circuit current in transient period

Ze'-equivalent impedanse of power system in the place of fault (transient period)

Inp=SnT/((sqrt(3))*Un)

After that, you need to calculate expected value of surge current:

kud=1+e(-0,01/Tae)

Iud=(sqrt(2))*kud*I'

After that, you need to calculate expected value of heat impulse:

A=(sqr(I0"))*Tae*(1-e(-2*ti/Tae))+(sqr(I'))*(ti+Td")

And finally, you need to calculate 1s current (expected value of current during 1s):

I1s=sqrt(A/1s)

So, current of interruption of your breaker and power of interruption of your breaker are:

Ii=I'

Si=(sqrt(3))*Un*Ii

Additional expressions that you can use during your calculation:

I0"=Un/((sqrt(3))*Ze");

I"=1,1*Un/((sqrt(3))*Ze");

I'=1,15*Un/((sqrt(3))*Ze');

where are:

ti-time of interruption

Inp-operational current of breaker

SnT-rated power of transformer

Un-rated voltage

kud-surge coefficient

Tae-time constant of aperiodic component of short circuit current

Iud-surge current

A-heat impulse

I0"-short circuit current in subtransient period (generators are in no-load conditions)

I'-short circuit current in transient period

Td"-time constant of subtransient component of short circuit current

I1s-current during one second

Ii=expected value of current of interruption of your breaker

Si=expected value of power of interruption of your breaker

Ze"-equivalent impedance of power system in the place of fault (subtransient period)

I"-short circuit current in subtransient period (generators are in full-load conditions)

I'-short circuit current in transient period

Ze'-equivalent impedanse of power system in the place of fault (transient period)

09-18-2013 07:59 AM

Top #12

Dear Mr. Muhammed

I found difficuilty to contribute in this issue since to my opinion your query is vague, sensless and ambiguous. It may be prepared in hurry or you are far away. I would like to say you need to read moreabout LV System.Please refer to IEE or BS 7671 for further information.

Nevertheless, the following formulae to be met.

A)-

Ib ≤ In ≤Iz …………………………… (1)

Where

Ib = Load Design current

In = Protective Device rated current.

Iz = Current carrying Capacity of the Down stream cable. (all .correction factors to be applied as applicable).

B)- Operating Current of CB

I2 ≤ 1.45In …………………………… (2)

Where

I2 = The Protective device operating current.

If the protection is mainly against over load, In = Ib

C)- Thermal Constraint, that the protected cable shall capable to bear the let through energy within the prescribed time without insulation failure.

I²t ≤ K²S² …………………………… (3)

Where

I² = (Minimum Short curren)²

t = Time to trip.

S = Cable Cross section area.

K = Cable Materia Factor.

D)- maximum fault current shall be less than Makinig / Braking current of CB.

E)- Descrimination with up/down stream breakers if "any" to be verified by time or current grade.

F)- manufacutrurer Technical Data Sheet to be present.

Pls review ,meanwhile I remain with best regards.

I found difficuilty to contribute in this issue since to my opinion your query is vague, sensless and ambiguous. It may be prepared in hurry or you are far away. I would like to say you need to read moreabout LV System.Please refer to IEE or BS 7671 for further information.

Nevertheless, the following formulae to be met.

A)-

Ib ≤ In ≤Iz …………………………… (1)

Where

Ib = Load Design current

In = Protective Device rated current.

Iz = Current carrying Capacity of the Down stream cable. (all .correction factors to be applied as applicable).

B)- Operating Current of CB

I2 ≤ 1.45In …………………………… (2)

Where

I2 = The Protective device operating current.

If the protection is mainly against over load, In = Ib

C)- Thermal Constraint, that the protected cable shall capable to bear the let through energy within the prescribed time without insulation failure.

I²t ≤ K²S² …………………………… (3)

Where

I² = (Minimum Short curren)²

t = Time to trip.

S = Cable Cross section area.

K = Cable Materia Factor.

D)- maximum fault current shall be less than Makinig / Braking current of CB.

E)- Descrimination with up/down stream breakers if "any" to be verified by time or current grade.

F)- manufacutrurer Technical Data Sheet to be present.

Pls review ,meanwhile I remain with best regards.

09-18-2013 10:21 AM

Top #13

Muhammed, the senario is slighting different between selecting CB for existing Installation, new installations, feeding one circuit or mutiple branch circuits where the maximum demand (applying diversity factor)also to be calculated,,,,etc.

09-18-2013 01:10 PM

Top #14

I have some thumb rules.

0 to 63 A - MCB

63A to 800 A - MCCB

above 800 A - ACB

0 to 63 A - MCB

63A to 800 A - MCCB

above 800 A - ACB

09-18-2013 03:20 PM

Top #15

Factors to consider are:

1. System Voltage

2. Rated load current

3. Fault level

1. System Voltage

2. Rated load current

3. Fault level

09-18-2013 05:58 PM

Top #16

Thank you all for these knowledgeable infos.