Voltage drop

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Jpflex

Jpflex

Electrician big leagues
Location
Victorville
Occupation
Electrician commercial and residential
Here is a practice NEC test question but I feel the people who made this question are wrong, the question is as follows:

12.You need to install a feeder to a subpanel in a newly constructed remote workshop at a single family dwelling. The source power is 120/240V single phase. The load in the workshop is engineered to be 85 amperes continuous at 240V, and a non-continuous 240V load of 35 amperes. The wire length will be 245' from the main service termination point to the workshop termination point. Using the Informational notes in Article 215 and the following formula, size the THWN copper feeder wires to provide reasonable efficiency of operation by preventing excessive voltage drop (keeping the wires as small as possible). The formula is: VD = 2 x L x R x I ÷ 1000. VD = voltage drop, L = length, R = resistance in /1000'.
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a. #1/0 THWN copper
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b. #4/0 THWN copper
check
c. #350 kcmil copper
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d. #500 kcmil copper
215.2(A)(1)(a,b) directs the load to be calculated as follows: 35A @ 100%, 85A @ 125%. 35 x 106.25 = 141.25A. Inf. note 2 shows the need to maintain the voltage drop to less than 3%. The wire resistance is found in Ch. 9 Table 8, and the wire ampacity is from Table 310.15(B)(16). Using the resistance from #350 kcmil: 2 x 245' x 0.0382 x 141.25A = 2643.9175 ÷ 1000 = 2.643V, just under the 3V maximum.
 
My calculations do not coincide with the answer expected by the question

Total calculated continuous and non continuous load is correct at 141.25 amperes

However, their wire size selected at 350 kcmil is wrong because

VD = 2 x 0.0382 ohms per 1,000 ft x (245 distance feet per wire run 1 way/ 1,000 ft) x 141.25 amperes

Is 2.64 volts

However 2.64 volts /240 volt source is is only a 1.1% voltage drop meaning that these people went over kill on too large of a wire
 
Using the formula, and "cheating" using an app I bet 2/0 (on both)
 
Tulsa Electrician said:
Is 350 your ansawer?
Click to expand...
My answer is a 2/0 wire

(0.101 ohms per 1,000 ft / 1,000 ft = 0.000101 ohms per foot)

2 x 0.000101 ohms x 245’ feet distance run (one way) x 141.25 amperes = 6.99 volts

6.99 volts / 240 volt source = 0.029 percent drop or 2.9 percent drop within 3 percent target
 
Here is where I took the test

Free Electrician Math Practice Test

Take a free Electrician Math Practice Test to see what kind of math questions are on actual electrician license exams. Most states require an electrician to pass an exam to receive a journeyman or master electrician license. State exams cover such areas
www.tests.com
 
Voltage drop calculations should not use a 125% factor for continuous loads, as the extra 25% current never occurs. In fact even 100% of the calculated load is properly too high; what matters would be the real maximum current.

Cheers, Wayne
 
augie47 said:
Using the formula, and "cheating" using an app I bet 2/0 (on both)
Click to expand...
I got the same answer as you so those idiots who made the test mistakes 3 volts for 3 percent and did this wrong? Which has thrown me off
 
wwhitney said:
Voltage drop calculations should not use a 125% factor for continuous loads, as the extra 25% current never occurs. In fact even 100% of the calculated load is properly too high; what matters would be the real maximum current.

Cheers, Wayne
Click to expand...
Can’t you either use figures with factoring in wire derating at 100 percent of continuous load plus non continuous load or not factoring conductor derating x 120 percent of continuous load plus 100 percent of non continuous load?
 
wwhitney said:
Voltage drop calculations should not use a 125% factor for continuous loads, as the extra 25% current never occurs. In fact even 100% of the calculated load is properly too high; what matters would be the real maximum current.

Cheers, Wayne
Click to expand...
Excellent point !!!! and source of my error... with that in mind 1/0
(I'm surprised they didn't give a possible 2/0 for folks like me)
 
wwhitney said:
Voltage drop calculations should not use a 125% factor for continuous loads, as the extra 25% current never occurs. In fact even 100% of the calculated load is properly too high; what matters would be the real maximum current.

Cheers, Wayne
Click to expand...
I see your point and kinda suspected that before but when applying a formula what will NEC /CM require?

VD = 2 KID/CM
Or
VD = I x conductor resistance x distance x 2
 
augie47 said:
Excellent point !!!! and source of my error... with that in mind 1/0
(I'm surprised they didn't give a possible 2/0 for folks like me)
Click to expand...
It can’t be 1/0 wire because that would be a 3.6% voltage drop
 
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