The size of a single-family dwelling is 25,117 square feet. The service is 120/240 volts single-phase. Without application of demand factors, what is the minimum number of general use, noncontinuous 120-volt, 15-ampere branch circuits?

1. 45

2. 32

3. 42

4. 14

See steps below:

Step_1: Table 220.12 lists 3VA/sq-ft. Calculate; 25,117 sq-ft x 3 VA/sq-ft = 75,352 VA. (Note: Lighting is considered a continuous load, however, do not apply this demand adjustment as stated for this specific question.)

Step_3: Calculate the power per circuit; P = I x E = 15 amperes x 120 volts = 1,800 VA / circuit.

Step_4: Calculate the number of 15 A circuits; Total Power / Power per Circuit = 75,352 VA / 1,800 VA = 41.86. Therefore, size up to 42 Circuits.

Do y'all agree that 42 circuits is the correct answer to this question?:?

]]>1. 45

2. 32

3. 42

4. 14

See steps below:

Step_1: Table 220.12 lists 3VA/sq-ft. Calculate; 25,117 sq-ft x 3 VA/sq-ft = 75,352 VA. (Note: Lighting is considered a continuous load, however, do not apply this demand adjustment as stated for this specific question.)

Step_3: Calculate the power per circuit; P = I x E = 15 amperes x 120 volts = 1,800 VA / circuit.

Step_4: Calculate the number of 15 A circuits; Total Power / Power per Circuit = 75,352 VA / 1,800 VA = 41.86. Therefore, size up to 42 Circuits.

Do y'all agree that 42 circuits is the correct answer to this question?:?

An office building to be constructed will have two hundred and fifty, 20 Amp, 120 Volt duplex receptacles installed. Determine the MINIMUM number of 120 Volt, 20 amp circuits required to supply the receptacles.

This question is on my study material and my answer was:

250 X 180 VA = 45000 VA - 10000 VA = 35000 VA X .5 = 17500 VA + 10000 VA = 27500 VA / 2400 VA = 11.45 (12) branch circuits.

The study material I'm using is not using the Table 220.44 Demand Factor and is dividing 45000 VA by 2400 VA to get 18.75 branch circuits.

Can someone help me understand why the demand factor from Table 220.44 wouldn't be used? Or if it is an incorrect solution on the author of the study material?

]]>This question is on my study material and my answer was:

250 X 180 VA = 45000 VA - 10000 VA = 35000 VA X .5 = 17500 VA + 10000 VA = 27500 VA / 2400 VA = 11.45 (12) branch circuits.

The study material I'm using is not using the Table 220.44 Demand Factor and is dividing 45000 VA by 2400 VA to get 18.75 branch circuits.

Can someone help me understand why the demand factor from Table 220.44 wouldn't be used? Or if it is an incorrect solution on the author of the study material?

When calculating the arc flash hazard at a DC switchboard, being fed from an AC switchboard via rectifier circuit, the SKM software does not recognize the upstream AC device, so the arcing time at the DC swbd runs to the max 2 seconds. SKM support staff seem to agree with this, as the AC and DC systems are calculated separately. However, would the clearing time of the upstream AC feeder breaker relay have any bearing on the DC system?

]]>I believe the answer in my workbook is incorrect so I just wanted verification.

Question is:

what size service and service conductors for a multi family dwelling (paralleled in two raceways)?

load calculation comes out to 575 amps

answer sheet says 600 amp service which is correct but says the service conductors should be 350 kcmil at 310 amps. (2 conductors X 310=620 amps)

Shouldnt the service conductors be 300 kcmil rated at 285 amps because paralleled you get an equivalent of 570 amps which the next breaker size up is 600?

so couldn’t you use 300 kcmil conductors as opposed to 350 kcmil?

]]>Question is:

what size service and service conductors for a multi family dwelling (paralleled in two raceways)?

load calculation comes out to 575 amps

answer sheet says 600 amp service which is correct but says the service conductors should be 350 kcmil at 310 amps. (2 conductors X 310=620 amps)

Shouldnt the service conductors be 300 kcmil rated at 285 amps because paralleled you get an equivalent of 570 amps which the next breaker size up is 600?

so couldn’t you use 300 kcmil conductors as opposed to 350 kcmil?

Hello,

I have an application where I am designing a feeder rated for 4.0MW (197 amps) at 13.8kV from an existing switchgear.

The cubicle that I am utilizing is a 1200 amp fully equipped space for which I have specified a 15kV, 1200 amp breaker and SEL-351 relay.

NEC 215.3.B seems unclear to me about how to size the cables. For 13.8kV (medium voltage) the cables can be sized for load only, correct?

I was originally going to use a 3/C - 2/0 with ground cable, but I am looking on the existing one-line and all the feeders are 3 - 500kCMIL or 6 - 500kCMIL.

]]>I have an application where I am designing a feeder rated for 4.0MW (197 amps) at 13.8kV from an existing switchgear.

The cubicle that I am utilizing is a 1200 amp fully equipped space for which I have specified a 15kV, 1200 amp breaker and SEL-351 relay.

NEC 215.3.B seems unclear to me about how to size the cables. For 13.8kV (medium voltage) the cables can be sized for load only, correct?

I was originally going to use a 3/C - 2/0 with ground cable, but I am looking on the existing one-line and all the feeders are 3 - 500kCMIL or 6 - 500kCMIL.

I'm designing the electrical for an ISO 7 clean room. the conduit and wire unlike many such rooms will not be exposed on the walls but run in a wall cavity. I believe I can just use typical NEC chapters 1-4 wiring. i.e. MC cable throughout.

Is this correct?

And regardless, does the NEC define special requirements for Clean Rooms?

Is there any other standard defining electrical requirements for the various levels of clean rooms?

]]>Is this correct?

And regardless, does the NEC define special requirements for Clean Rooms?

Is there any other standard defining electrical requirements for the various levels of clean rooms?

Say I have two boxes some distance apart, and one of them is next to the service. I upsized the CCC's between the boxes for voltage drop, so I must upsize the EGC commensurately. The CCC's between the box next to the service and the service, however, are sized only for minimum ampacity. Do I need to run the upsized EGC between that box and the service or can I size it to T250.122?

This is not an exam question.

]]>This is not an exam question.

A three phase 4 wire Y system has 270 amps on each leg. If all loads are linear loads, what is the size of the neutral in amps? |

A. 270

B.249

C.810

D.567

I thought it was 270. book says its 249 and i have no idea how to come up with that. any help

]]>A. 270

B.249

C.810

D.567

I thought it was 270. book says its 249 and i have no idea how to come up with that. any help

when calculating the max voltage drop of 3% for feeder (or branch circuit) for a 120/208V, do I want the Vd to be 3% of 120V or 208V? And does that change for the 5% of the total circuit?

]]>I am trying to solve Problem "B"

I got I

The power formula I am also using is, P =

What is the correct value of "θ" should I be using?

In other problems, I always used the value of θ from the current angle.

I have a customer contact me about installing a Nema 14-50 outlet on the side of his house. He said he will be purchasing an electric vehicle soon. I plan to install a 50 amp two pole breaker and run NM 6-3 to this outlet.

https://www.homedepot.com/p/Talon-50...GP1S/206184036

I don't want to over think this--but after I complete the work everything is on him. Right?

]]>https://www.homedepot.com/p/Talon-50...GP1S/206184036

I don't want to over think this--but after I complete the work everything is on him. Right?

I'm being tasked with designing and installing the branchcircuit wiring for the following scenario:

I am installing 4power distribution units (PDU) (which is IT lingo for fancy power strips) ratedat 30 amps each in a data server room. The total load being served currently is 19600watts but I want to size everything for the full 30 amps and I'm going to divide this up between the 4 PDU units on 4 separate circuits.I'm going to supply this load with 208 volts from a 120/208 volt 3-phase panel.Which calculates to 23.6 amps per PDU.

My confusion isthis: These PDU units come with a NEMA 30amp plug to go into a standard 30 ampNEMA receptacle. The code is clear that I have to upsize both my conductor andbreaker 125% because it’s a continuous load and because its IT equipment645.5(A)….which I interpret as meaning conductor capable of 37.5amps (8 gauge)and a breaker rated at 40 amps….but the code also states at 210.21 (B) (1) thata single receptacle on an individual branch circuit cannot have an amp ratingless than the circuit. In addition Table 210.21 (B)(3) seems to back this upthat a 30 amp receptacle must be installed on a 30 amp circuit.

How can I upsize125%? And still meet the requirements of 210.21 (B)(1)

I'm sure this is a probably a dumb question andan easy answer and I'm just not grasping the whole concept, but if you couldset me straight I would sure appreciate it!!

]]>I am installing 4power distribution units (PDU) (which is IT lingo for fancy power strips) ratedat 30 amps each in a data server room. The total load being served currently is 19600watts but I want to size everything for the full 30 amps and I'm going to divide this up between the 4 PDU units on 4 separate circuits.I'm going to supply this load with 208 volts from a 120/208 volt 3-phase panel.Which calculates to 23.6 amps per PDU.

My confusion isthis: These PDU units come with a NEMA 30amp plug to go into a standard 30 ampNEMA receptacle. The code is clear that I have to upsize both my conductor andbreaker 125% because it’s a continuous load and because its IT equipment645.5(A)….which I interpret as meaning conductor capable of 37.5amps (8 gauge)and a breaker rated at 40 amps….but the code also states at 210.21 (B) (1) thata single receptacle on an individual branch circuit cannot have an amp ratingless than the circuit. In addition Table 210.21 (B)(3) seems to back this upthat a 30 amp receptacle must be installed on a 30 amp circuit.

How can I upsize125%? And still meet the requirements of 210.21 (B)(1)

I'm sure this is a probably a dumb question andan easy answer and I'm just not grasping the whole concept, but if you couldset me straight I would sure appreciate it!!

Admittedly I am not an expert in either generators or SKM software. I am modeling a power system in SKM and I have been told that our generator is 1750kVA, 0.7PF rated. This results in 1225KW but I do know that the engine powering the gen is only 1000kW. Generator model in SKM only accepts kVA and PF; should I adjust the kVA for the engine and enter 1428kVA, 0.7PF?

Thanks,

]]>Thanks,

I design electrical controls for large conveyor systems. In our power distribution, 480VAC 3PH, we feed up to 4 units (1.7 FLA) with a 45A breaker (this 45A CB scheme has been passed down for years and I don't know the origins of it). Each unit has their own local protection. We size our conductors for voltage drop. In my mind there is no issue because the 45A CB protects the conductor and since each unit has its own local CB, there isn't a risk of it shorting out pulling 45A.

I have searched all over to find out if there is anything in NEC that says that a CB can be too large in this case. Can someone please help me put this question to bed?

]]>I have searched all over to find out if there is anything in NEC that says that a CB can be too large in this case. Can someone please help me put this question to bed?

If I have a moter FLC 10hp 208 volt is 30.8, sizing inverse time breaker 2.50% table 430.52.

FLC 38.8x2.50%= 77 amps next size up 80 amp. My wire size would be FLC amps 30.8x1.25=

38.5 number 8 awg wire, I’m I doing this right it seems wrong 8 awg wire on a 80amp breaker?

]]>FLC 38.8x2.50%= 77 amps next size up 80 amp. My wire size would be FLC amps 30.8x1.25=

38.5 number 8 awg wire, I’m I doing this right it seems wrong 8 awg wire on a 80amp breaker?