Calculations have me confused

[lew3611]

hey guys, I have three practice questions that have me confused. I have the correct answer, but dont know how I managed to do it at least not totally. So I will throw them @ you guys and please share with me the correct formula 1. A store has 120 feet of show window, what is the demand on the service for the show windows. Service is 1ph, 120/240 volts. Hint. NEC 220.12 & 220.14(G). (my main concern on this one is where does 125% have a play in this. book has 30,000 va for answer 2. This next question is based off of question #1, the total ampacity required by the show window is:____________ book has 125 amps for answer 3. The VD for two #10 solid uncoated THWN conductors, 320 feet long one way, connecting a 12.4 amp load to a 120 volt source is:____________. every form of VD calculations I have learn I was still not able to come up with what the book came up with. I used VD=(2K*L*I)/Cm. what am I doing wrong here. by the way on this one the book has 9.6 for the correct answer

 

[tallgirl]

hey guys, I have three practice questions that have me confused. I have the correct answer, but dont know how I managed to do it at least not totally. So I will throw them @ you guys and please share with me the correct formula

3. The VD for two #10 solid uncoated THWN conductors, 320 feet long one way, connecting a 12.4 amp load to a 120 volt source is:____________. every form of VD calculations I have learn I was still not able to come up with what the book came up with. I used VD=(2K*L*I)/Cm. what am I doing wrong here. by the way on this one the book has 9.6 for the correct answer

I couldn't tell if you understood your first two questions, or what. So I'm only going to tackle #3.

Table 8 in Chapter 9 lists the resistance of uncoated #10 at 1.21 ohms per kilofoot. You have .64 kilofeet (2 * 320 feet equals 0.640 kilofeet ...), so that 0.640 * 1.21 = .7744 ohms. Your load is 12.4 amps. Dragging out Ohm's Law, E = I * R = 12.4 * .774 = 9.6 volts.

You might want to check out 210.19 (A) (1) FPN 4, while you're in the neighborhood ...

 

[Rockyd]

About the load being contiuous -

See Contiuous load in article 100.

210.19(A)(1) covers protecting the concuctors.

210.20(A)(1) covers the OCPD.

220.14(G) calls out the requirements. Barebone minimum allowed is found in 210.62.

Sometimes K is not derived from the correct approach. That is the correct wire size from the chart. Sometimes just the K is based on 1000' of 1000kcmil(per chapter9 table8 ), which is 12.9 - try the math both ways. I see the formula as correct -

VD = 2KID
CM​

After factoring down .320 of the 1000' if it's still not right, contact the agency that wrote the test questions for a methodology as to how they worked the problem through.

 

[Smart $]

1. A store has 120 feet of show window, what is the demand on the service for the show windows. Service is 1ph, 120/240 volts. Hint. NEC 220.12 & 220.14(G). (my main concern on this one is where does 125% have a play in this. book has 30,000 va for answer

The book's 30,000VA answer is wrong based on the description provided. First, 220.12 has no part in this as that involves general lighting load, of which show windows are not. 220.14(G), and more specifically 220.43(A) regarding service load, specify show window loads to be calculated at 200 VA/ft. That value is of course the minmum permitted calculated load. It can be more... in which case one would use the actual value, perhaps even pad it to the higher side for future expansion. Nevertheless...

120ft x 200VA/ft = 24,000VA

There is no demand factor that can be applied to show windows!

2. This next question is based off of question #1, the total ampacity required by the show window is:____________ book has 125 amps for answer

24,000VA / 240V = 100A

Show windows are considered a continuous load. Applying 210.19(A)(1) or 215.2(A)(1)...

100A x 125% = 125A

 

[lew3611]

Originally posted by Smart$:
The book's 30,000VA answer is wrong based on the description provided. First, 220.12 has no part in this as that involves general lighting load, of which show windows are not. 220.14(G), and more specifically 220.43(A) regarding service load, specify show window loads to be calculated at 200 VA/ft. That value is of course the minmum permitted calculated load. It can be more... in which case one would use the actual value, perhaps even pad it to the higher side for future expansion. Nevertheless...

120ft x 200VA/ft = 24,000VA

There is no demand factor that can be applied to show windows!


Once I saw 220.14(g) this is what I came up with however I thought I was wrong because of what the book actually had, Thanks everyone

 

[tallgirl]

Originally posted by Smart$:
The book's 30,000VA answer is wrong based on the description provided. First, 220.12 has no part in this as that involves general lighting load, of which show windows are not. 220.14(G), and more specifically 220.43(A) regarding service load, specify show window loads to be calculated at 200 VA/ft. That value is of course the minmum permitted calculated load. It can be more... in which case one would use the actual value, perhaps even pad it to the higher side for future expansion. Nevertheless...

120ft x 200VA/ft = 24,000VA

There is no demand factor that can be applied to show windows!

Of course not -- they tend to be either all on or all off. Demand factor is for when some percentage of "stuff" tends to be on and some other percentage tends to be off.

If you look at the percentage values in the various demand factor tables, the percentage goes down as the number of items goes up. That's not because operating 10 clothes dryers (50%) is more efficient, say, than operating 5 (85%), it's because people are less likely to have all 10 going at once than they are to have all 5 going at once.

Once I saw 220.14(g) this is what I came up with however I thought I was wrong because of what the book actually had, Thanks everyone

Right, but it's continuous (when it's on ... ), so you have to multiply by 1.25 (125%). 1.25 * 24kVA = 30kVA (there's the answer to #2).

30kVA / 120V = 250A (it's a 1-phase 120/240), so I'm guessing they are looking for a MWBC, divide the 250A by 2, and that's 125A per phase (there's the answer to #3).

 

[RB1]

Rockyd,

The K factor for uncoated solid No. 10 from Table 8 = 10380/1000*1.21=12.56

 

[Rockyd]

My implication of the value of K when it is an absolute unknown on a test has to have a value. That value when an x factor is required, without any other clues, is commonly taken from chapter9 table8 based on 1000' of 1000kcmil.


Edit - Where do I get a SWAG answer like that? Can't remember where I exactly picked it up... but even George Hart (the guy who writes Ugly's Electrical Reference book and is one of my tool box items) agrees to it's 12.9 for Cu, and 21.2 for Al (I know, Ugly's isn't the code book, but the info in it is good). See the section on Voltage Drop Calculations Inductance Negligible