Consultant

[icu4prod]

I happened to go over some electrical questions from the Craftsman Electrician?s Exam Preparation Guide for the 2008 NEC.

Run into this question below ?

A department store is illuminated with 215 fluorescent lighting fixtures and connected to a 120-volt supply. Each lighting unit draws 2.2 amperes. How many 20-ampere branch circuits are necessary to feed these fixtures if each branch circuit must not exceed 80% of the branch circuit?

The solution given: 215 fixtures x 2.2 = 473 amp, total load. A 20- amp branch circuit load to 80% of its capacity = 20 x.80 =16 amperes. 477 amperes (total load) / 16 = 29.56 or 30 circuits.

The answer (30 circuits) does not seem correct,

Without the 80% de-rating): I total = 215 units x 2.2 amps = 473A/20 = 23.65 (24 circuits)
This was my answer - with the 80% de-rating: this seems more seems logical: 473 x 80% = 378.4A/20 = 18.92 (19 circuits).

This question is a bit confusing, because, de-rating means to decrease this method seems to add.

This question does not sound logical ? 30 different circuits for a 20 amp branch circuits for a light that only consumes (2.2amps x 120V) 264 watts?
Not as per reference rule 220.18(B), and 210.11, Also, that means 30-2 pole circuit breakers will be needed and that?s equivalent to 60 breaker, which means a least two panels ? due to the 42 max per panel as per rule- 408.54 Maximum Number of Overcurrent Devices. A panelboard shall be provided with physical means to prevent the installation of more overcurrent devices than that number for which the panelboard was designed, rated, and listed.

For the purposes of this section, a 2-pole circuit breaker or fusible switch shall be considered two overcurrent devices; a 3-pole circuit breaker or fusible switch shall be considered three overcurrent devices.

To de-rate means to decrease ? not increase. As per rule- 210.23 (A) (1) Cord-and-Plug-Connected Equipment Not Fastened in Place. The rating of any one cord-and-plug-connected utilization equipment not fastened in place shall not exceed 80 percent of the branch-circuit ampere rating.

What a big difference 30 circuits (11 more), which seems to spell out an overload.
What is the ampacity needed to size the conductor for each branch circuit? Is it 2.2 amps for each branch circuit ? is it a minimum of 18 AWG as per 600.31 (B)?


Can someone please explain this better?

 

[Dennis Alwon]

Try it this way. Each light draws 264 watts. Max wattage on a 20 amp circuit is 80% of 2400= 1920. Now how many lights can you put on the circuit?

Now divide that into the number of fixtures

 

[david luchini]

I happened to go over some electrical questions from the Craftsman Electrician?s Exam Preparation Guide for the 2008 NEC.

Run into this question below ?

A department store is illuminated with 215 fluorescent lighting fixtures and connected to a 120-volt supply. Each lighting unit draws 2.2 amperes. How many 20-ampere branch circuits are necessary to feed these fixtures if each branch circuit must not exceed 80% of the branch circuit?

The solution given: 215 fixtures x 2.2 = 473 amp, total load. A 20- amp branch circuit load to 80% of its capacity = 20 x.80 =16 amperes. 477 amperes (total load) / 16 = 29.56 or 30 circuits.

The answer (30 circuits) does not seem correct,

Without the 80% de-rating): I total = 215 units x 2.2 amps = 473A/20 = 23.65 (24 circuits)
This was my answer - with the 80% de-rating: this seems more seems logical: 473 x 80% = 378.4A/20 = 18.92 (19 circuits).

For what it's worth, 30 is not the correct answer. The correct answer is 31.

You multiplied by 80%, when you should have divided by 80%. 473/80% = 591A/20 = 29.56 (30 circuits - more on why 30 is wrong later.)

This question is a bit confusing, because, de-rating means to decrease this method seems to add.

This question does not sound logical ? 30 different circuits for a 20 amp branch circuits for a light that only consumes (2.2amps x 120V) 264 watts?

Each fixture only consumes 264 watts. There are 215 fixtures total...that is 56.76kW.

Not as per reference rule 220.18(B), and 210.11, Also, that means 30-2 pole circuit breakers will be needed and that?s equivalent to 60 breaker, which means a least two panels ? due to the 42 max per panel as per rule- 408.54 Maximum Number of Overcurrent Devices. A panelboard shall be provided with physical means to prevent the installation of more overcurrent devices than that number for which the panelboard was designed, rated, and listed.

The supply voltage is 120V. You would need 30 single pole circuit breakers, not 2 pole circuit breakers. Also, note that the 42 max per panel rule went away in the 2008 Code.

To de-rate means to decrease ? not increase. As per rule- 210.23 (A) (1) Cord-and-Plug-Connected Equipment Not Fastened in Place. The rating of any one cord-and-plug-connected utilization equipment not fastened in place shall not exceed 80 percent of the branch-circuit ampere rating.

Yes, you are derating the rating of the branch circuit. In other words, each 20A branch circuit can be loaded to 16A. 473A /16 = 29.65 (30 circuits- more on why 30 is wrong later.)

What a big difference 30 circuits (11 more), which seems to spell out an overload.
What is the ampacity needed to size the conductor for each branch circuit? Is it 2.2 amps for each branch circuit ? is it a minimum of 18 AWG as per 600.31 (B)?

Why would having more circuits spell out an overload? The load is the same.

The ampacity of the conductors would have to be at least 15.4A, which would be the load on each circuit. 7*2.2=15.4 (8*2.2=17.6, which would exceed the derated branch circuit.) But the conductor can't be smaller than #12 for a 20A circuit (240.4(D).)

So you have 215 fixtures with 7 max fixtures per branch circuit, which would give you 215/7=30.7 = 31 circuits required.

 

[steve66]

I basically agree that if you follow Dennis's method, you get 31 circuits.

The solution in the book lumps the entire load together, and assumes it can be almost evenly distributed among 30 circuits. The problem is, that you can't put 7.4 lights on each circuit.

Not unless each light has 3 separate ballasts. And without being told each light has 3 ballasts, I would assume each light must go on one single circuit, and you can't have a fraction of a light on one circuit.

 

[Dennis Alwon]

This type of question trips up many people. I remember one many years ago with 500 watt metal halide lights on a 20 amp circuit. Same scenario-- how many circuits would you need for 40 lights. Your tendency is to lump it all together but as stated you can't have more than 3 of these lights on one 20 amp circuit so the results become very different from the lump sum which would allow more than 3 of the lamps on a cir.

 

[infinity]

Am I the only one who finds using the term derating in this example to be misleading?

 

[James L]

The solution given: 215 fixtures x 2.2 = 473 amp, total load. A 20- amp branch circuit load to 80% of its capacity = 20 x.80 =16 amperes. 477 amperes (total load) / 16 = 29.56 or 30 circuits.

The answer (30 circuits) does not seem correct,

Without the 80% de-rating): I total = 215 units x 2.2 amps = 473A/20 = 23.65 (24 circuits)
This was my answer - with the 80% de-rating: this seems more seems logical: 473 x 80% = 378.4A/20 = 18.92 (19 circuits).

With 19 circuits, you would have 25 amps on each breaker (473/19=24.89)

If you want to start your calculation from the total load, you do not multiply by it by .8
Either divide by .8 (473/.8) or you multiply by 1.25 (473x1.25) =591.25, then divide by the 20 amp rating of the breaker.

It's probably easier to leave the total load figure alone, and calculate from the breaker rating instead (20x.8=16), then (473/16)

But, as everyone else has mentioned, that doesn't calculate evenly with 2.2 amps per fixture

You could only have 7 per circuit
31 circuits needed

 

[icu4prod]

Terms: Derating in electrical application.

Terms: Derating in electrical application.

Am I the only one who finds using the term derating in this example to be misleading?

Thanks for your response,

Am I missing something ... I always thought - Derating - is the reduction of the maximum capacity (load [a unit: conductors-utilization equipment-panels-devices -etc..]) can reliably handle when fins/side sections are removed.

please explain ... your response.


Thank you,



.

 

[don_resqcapt19]

Am I the only one who finds using the term derating in this example to be misleading?

Not really...you just have to apply it to the ampacity of the circuit and not to the load.

 

[Smart $]

Am I the only one who finds using the term derating in this example to be misleading?

Not really...you just have to apply it to the ampacity of the circuit and not to the load.

Thanks for your response,

Am I missing something ... I always thought - Derating - is the reduction of the maximum capacity (load [a unit: conductors-utilization equipment-panels-devices -etc..]) can reliably handle when fins/side sections are removed.

please explain ... your response.


Thank you,



.

Derating is slang trade terminology meaning ampacity adjustment of insulated wire conductors for the conditions of use, specifically for ambient temperature and the number of current-carrying conductors in close proximity.

In the case presented, there is no derating issue. The consideration being made is simply the branch-circuit rating. Refer to 210.19(A)(1). By stating the circuits can only be loaded to 80% is implying the loads are continuous by NEC definition. It is derived from the requirement that a branch circuit's ampacity be not less than the noncontinuous load (none in this case) plus 125% of the continuous load (the entire load in this case)... before adjustment and correction factors are applied.

 

[infinity]

Derating is slang trade terminology meaning ampacity adjustment of insulated wire conductors for the conditions of use, specifically for ambient temperature and the number of current-carrying conductors in close proximity.

In the case presented, there is no derating issue. The consideration being made is simply the branch-circuit rating. Refer to 210.19(A)(1). By stating the circuits can only be loaded to 80% is implying the loads are continuous by NEC definition. It is derived from the requirement that a branch circuit's ampacity be not less than the noncontinuous load (none in this case) plus 125% of the continuous load (the entire load in this case)... before adjustment and correction factors are applied.

I agree and like your definition.