Exam Prep

[swhansss]

am studying for the Oklahoma Unlimited contractor license exam. I am going through the Mike Holt Exam Prep book. I'm in Dwelling unit Calculations.

Why in the example calculations do you sometimes use 115 or 230 volts and sometimes you use 120 or 240 volts. I just completed some optional dwelling calculations and used 230 volts for an A/C load which is the voltage used on table 430.248. The sample problem calculated it using 240 volts. The difference made the total VA wrong on the problem I did. Will they be that close in calculating loads on the test?

 

[Dennis Alwon]

Look at art. 220.5. IMO you should be using those voltages. It used to be 115/230 so I am guessing that some of the examples were from long ago and never got changed

220.5 Calculations.
(A) Voltages. Unless other voltages are specified, for purposes
of calculating branch-circuit and feeder loads, nominal
system voltages of 120, 120/240, 208Y/120, 240, 347,
480Y/277, 480, 600Y/347, and 600 volts shall be used.
(B) Fractions of an Ampere. Calculations shall be permitted
to be rounded to the nearest whole ampere, with decimal
fractions smaller than 0.5 dropped.

 

[texie]

am studying for the Oklahoma Unlimited contractor license exam. I am going through the Mike Holt Exam Prep book. I'm in Dwelling unit Calculations.

Why in the example calculations do you sometimes use 115 or 230 volts and sometimes you use 120 or 240 volts. I just completed some optional dwelling calculations and used 230 volts for an A/C load which is the voltage used on table 430.248. The sample problem calculated it using 240 volts. The difference made the total VA wrong on the problem I did. Will they be that close in calculating loads on the test?

As Dennis said, for Art 220 load calcs you use the values shown in 220.5.
The confusion comes with the Art. 430 motor amperage tables. The table headings tell you that the amperage shown for a given motor can be assumed to be the same for the range of voltage indicated. The key here is to use the tables only for amperage. When you convert that to VA for an Art. 220 calc you refer to the voltages shown in 220.5.
And yes, you are correct that misunderstanding this could give you an incorrect answer on an exam load calc question if the test writers are so inclined to offer a choice based on, say 230, volt.

 

[swhansss]

Equipment bonding jumper

Equipment bonding jumper

Another Question:

I'm confused concerning the sizing of the supply side equipment bonding jumper. Article 250 102C. Example: If I have three parallel 400 conductors and the the total kcmil is 1,200 I can use 12.5 percent of that for the bonding jumper size if the kcmil is over 1,100. My grounding electrode conductor can't be reduced from table 250.66 but the jumper can. Is that correct? Also when I take 12.5 percent of 1,200 I get 150. I go to table 8 in chapter 9 and it shows the kcmil for 1/0 at 105,600. The sample test question said that 1/0 is the correct answer. What am I missing here?

 

[swhansss]

Marinas Demand Factor

Marinas Demand Factor

Mike Holt Question:

What is the load: 20 - 30A 250 volt, 20 - 20A 125 volt. Use table 555-12 Demand Factors. Why in the answer does the book show 30 as the number of receptacles. There are 30 on one line but the table doesn't say anything about how many are on one line. There are 40 total why wouldn't the demand be on 40? Load is 800A X demand of .7 or 560A according to the book. I would think the demand should be .6

Thanks,

 

[GoldDigger]

Mike Holt Question:

What is the load: 20 - 30A 250 volt, 20 - 20A 125 volt. Use table 555-12 Demand Factors. Why in the answer does the book show 30 as the number of receptacles. There are 30 on one line but the table doesn't say anything about how many are on one line. There are 40 total why wouldn't the demand be on 40? Load is 800A X demand of .7 or 560A according to the book. I would think the demand should be .6

Thanks,

As I recall, there is a confusing graphic to go with that one. Since you will be calculating amperes of load, and assuming a balanced load on both sides of the 120/240 3 wire, looking at each phase conductor separately, each line has 20 240 volt receptacles pulling current from it and half of the 120 volt receptacles. So 20 + 10 = 30. The two types will have different full load amps, but the count will still be 30.