NEC EXAM PREP QUESTIONS 2008 ADDITION

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In Unit 9 question 51 is an optional method calculation. The correct answer is 1/0 awg AL according to the instructor, although I can only come up with 103A which would be #1 awg Al?

The question and my calculations;

A dwelling unit has 2,330 qu ft of living space. In addition, there is 900 sw ft of porch and 400 sw ft of carport (not a garage) with the following;

Waste Disposal ( 1 KVA)
Dishwasher (1.5 KW)
Trash Compactor (1.5 KVA)
Water Heater (6 KW)
Range (14 KW)
Dryer (4.5 KW)
A/C Unit (5 hp 230V)
Baseboard Heat (Four at 2.5 KW each)

Use the optional method to determina what size aluminum conductors are required for the 120/240V, single-phase service?

General lighting and rec (2,330 x 3) = 6,990VA (porch and carport are not included)
Small appliance and laundry = 4,500
Waste Disposal = 1000VA
Dishwasher = 1500VA
Trash Compactor = 1500VA
Water Heater = 6000VA
Range = 14,000VA
Dryer = 4500VA

Total VA = 39,990

A/C Unit (5 hp at 230V) Table 430.248 5hp = 28A = 6,440VA 220.82(C)(1)
Baseboard Heat (Four at 2.5 KW each) 10,000VA x 40% = 4,000VA 220.82(C)(5)
Use the largest of the 2 loads 6,440VA

NEw Total VA = 46,430VA

First 10,000VA at 100%
Balance at 40% (36,430 x 40% = 14,572VA)
Calculated net load 24,572VA / 240V = 102.4A

Table 310.15(B)(6) 110A = #1awg AL

Is this correct or am I missing something?
 

Smart $

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...

Is this correct or am I missing something?
Looks good to me. Perhaps instructor forgot about porch and carport exclusion under 220.12.


And sometimes test authors trip on their own creation!!!
 
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david luchini

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I come up with 119.65 Amps, #1/0 AL would be correct.

I see two issues with your calculation, though one is minor and would not affect the outcome.

The minor one is the A/C calculation. You figure 28A @ 230V = 6440VA. Because of 220.5(A), I would figure 28A @ 240V = 6720VA.

Where you went wrong is adding the A/C load to the rest of the load and then applying the demand factors.

What you should have done is add up the loads without the heating or A/C (those listed in 220.82(B)), applied the demand factor to that, and then add the heating or A/C (from 220.82(C)).

So you have 39,990VA total general loads.

The 1st 10kVA at 100% = 10,000VA
the remainder at 40%... = 11,996VA
The A/C @ 100% .........= . 6,720VA
Total demand .............. = 28,716VA

28,716VA @ 240V = 119.65 Amps. Using a 125A service, you would need #1/0 AL per T310.15(B)(6).
 

infinity

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I see two issues with your calculation, though one is minor and would not affect the outcome.

The minor one is the A/C calculation. You figure 28A @ 230V = 6440VA. Because of 220.5(A), I would figure 28A @ 240V = 6720VA.

I agree, unless 230 volts were given in the question 220.5(A) would tell you to use 240 volts.
 
230V was given to him in his original question.

A dwelling unit has 2,330 qu ft of living space. In addition, there is 900 sw ft of porch and 400 sw ft of carport (not a garage) with the following;

Waste Disposal ( 1 KVA)
Dishwasher (1.5 KW)
Trash Compactor (1.5 KVA)
Water Heater (6 KW)
Range (14 KW)
Dryer (4.5 KW)
A/C Unit (5 hp 230V)
Baseboard Heat (Four at 2.5 KW each)

However, I think they wanted to say 240V for the A/C Unit....However, it is possible the test provider wanted it at 230V so you have to ask the test provider of their intent.

Regardless, since it is the exam providers question we can answer it .

2,330 sq. ft x 3VA = 6,990VA
SABC = 3,000VA
LC = 1,500VA
All Appliances = 28,500 VA
Totals Above = 39,990VA - 10,000VA @ 100% and remainder at 40% = 21,996VA
Then add :
Larger of the two AC v. Heat = (lets go with 230V since it is shown) 28A x 230V = 6,440 VA

Total VA is : 28,436 / 240V = 118.48A requires 1/0 AL ( so either way using 230V or 240V would not change the outcome of the question. so always use what the instructor provides you unless expressed otherwise. )


 

david luchini

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The examples in Annex D (D2a) call for 28A x 230V.

That's interesting...In Example D2(b) they include the 230V loads at 240V, and in D3(a) they include the 460V loads at 480V.

In D2(b) they have six (6) 7A, 230V room air conditioners that they figure at: 6*7A=42A*240V=10.08kVA.

In D3(a) they figure a 7.5HP, 460V, 3 ph Air Compressor at 11A*480V*1.732 and a 1.5HP, 460V, 3 ph grinder at 3A*480V*1.732.

This seems the correct way to me given 220.5(A).
 
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