Newby here,

I want to confirm my understanding for this?

480V load (solar) 208 Utility

I have a (480V * 316A * 1.732) = 262.7k kVa / (208 v x 1.732) = 729A 208V

For transformer sizing?

208V * 729A * 1.732 / 1000 = 262.70 kVA transformer

Thanks

]]>I want to confirm my understanding for this?

480V load (solar) 208 Utility

I have a (480V * 316A * 1.732) = 262.7k kVa / (208 v x 1.732) = 729A 208V

For transformer sizing?

208V * 729A * 1.732 / 1000 = 262.70 kVA transformer

Thanks

PF 70e.

I'm supplying the drive with 0-20 ma.

8 should = 30 hz if vfd set for 4-20

10ma=30 hz if set for vfd set for 0-20

81 & 82 equal 0&60

91 & 92 eqaul 60&0

My drive does not give me either at the given Input.

About 80%. ?

Which parameter am I missing?

Why is available network always furthest from project?

]]>I'm supplying the drive with 0-20 ma.

8 should = 30 hz if vfd set for 4-20

10ma=30 hz if set for vfd set for 0-20

81 & 82 equal 0&60

91 & 92 eqaul 60&0

My drive does not give me either at the given Input.

About 80%. ?

Which parameter am I missing?

Why is available network always furthest from project?

This note is referencing a MWBC, correct?

I have a location that wants a 800 Amp 120/208v of current 480v gear.

They have drawn up and requested a 480V 350Amp breaker to feed 300KVA transformer feeding a main lug 800-Amp panel.

They didnt account for 125%

Calculated at 350A 480V primary i get a KVA of 290.976, that gives me a 807.69 Amp on secondary, can this still feed a 800-amp main lug panel or do i need 800-amp main breaker to protect the buss,

feeders are listed as parallel 500 MCM 90*C for 430 amps

Thanks in advance

]]>They have drawn up and requested a 480V 350Amp breaker to feed 300KVA transformer feeding a main lug 800-Amp panel.

They didnt account for 125%

Calculated at 350A 480V primary i get a KVA of 290.976, that gives me a 807.69 Amp on secondary, can this still feed a 800-amp main lug panel or do i need 800-amp main breaker to protect the buss,

feeders are listed as parallel 500 MCM 90*C for 430 amps

Thanks in advance

Hi

My question is related to the derating rules applied to 90 degC cables.

When I am using a 90 degC cable in 40 degC ambient temp, I will apply correction factor 0,91 and check if this complies with 75degC conductor of same size.

Example:

Amb = 40 degC

Non-continous current = 100 A

90degC conductor - 3AWG : 115A x 0,91 = 104,65A

Check with 75 degC conductor 3AWG: 100A ampacity = OK to use 90degC 3AWG conductor.

For parallel scenario is:

6 current carrying conductors in conduit = 0,8 adjustment factor

30 degC amb = 1 correction factor

Non-continous current: 650A

2 x 500kcmil 90 degC conductors are chosen.

Total ampacity = 430*2*0,8 = 688A

Check with 75degC conductor: 380 x 2 = 760A --> OK

My question is, should I do the 75degC check as above or should I also add adjustment factors to the check as below?

380 x 2 x 0,8 = 608A --> NO OK.

Hope that my question makes sense to you.

Thanks in advance

]]>My question is related to the derating rules applied to 90 degC cables.

When I am using a 90 degC cable in 40 degC ambient temp, I will apply correction factor 0,91 and check if this complies with 75degC conductor of same size.

Example:

Amb = 40 degC

Non-continous current = 100 A

90degC conductor - 3AWG : 115A x 0,91 = 104,65A

Check with 75 degC conductor 3AWG: 100A ampacity = OK to use 90degC 3AWG conductor.

For parallel scenario is:

6 current carrying conductors in conduit = 0,8 adjustment factor

30 degC amb = 1 correction factor

Non-continous current: 650A

2 x 500kcmil 90 degC conductors are chosen.

Total ampacity = 430*2*0,8 = 688A

Check with 75degC conductor: 380 x 2 = 760A --> OK

My question is, should I do the 75degC check as above or should I also add adjustment factors to the check as below?

380 x 2 x 0,8 = 608A --> NO OK.

Hope that my question makes sense to you.

Thanks in advance

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.