Need Advice for Residential Load Calc

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msext

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Ok I have taken the test way to many times now. I am taking the admin test section 3 for residential in Washington State. The exam has 4 questions of load calcs. I have followed the book and still can not seem to pass. I think they are trick questions. Here is an example of a question can someone please help me work through to find out what I am doing wrong please.

Question:
2300 Sq Foot Duplex, each unit has the following

2 - 8KW Ranges
1 - 5 Amp Garbage Disposal
1 - 11 Amp Dishwasher
1 - 3KW Hot Water tank
1 - 15kw Heat
5 - 21 Amp 240V A/C

Using the Standard method (or maybe Optional I can not remember now which was used) what is the Load in Amps?

Answers:
360
375
325
349


Question 2
Optional method, 2000 Sq Ft, 10 Unit multi family building, Each unit has the following

7 KVA Range
1.2 KVA Dishwasher
.6 KVA Garbage Disposal
18 Amp 120 Volt A/C
2.5 KVA 240 Volt Hot Water Tank

Do not incluse any house or laundry loads

Answers:

381
390
402
413


IM curious as to the step by step so I can figure out where and what I am doing wrong. Thank You to all who can help.

Mike
 
Re: Need Advice for Residential Load Calc

question 1 i cameup with 348 amp using standard calc i also added 2 5kva dryers to the problem

question 2
2000sf times 3va =6000 plus 3000va for 2 small applicance circuits =9000va x 10units x .43 (220-32)= 38700
7kva x10 x .43= 30100
1.2kva x10 x .43= 5160
600va x10 x .43 = 2580
18amps x 120 v = 2160va x 10 x .43= 9288
2.5kva x 10 x .43 = 10750
_____
96578
96578 divided by 240 = 402amps
hope that helps
 
Re: Need Advice for Residential Load Calc

M,

You asked for step by step so here we go:

I'm going to start with question #2. Rick worked it through correctly and I'm going to give you the explanation and the order of steps & code references.

Optional calculation for multifamily dwellings Section 220.32

used when:
-no dwelling is supplied by more than one feeder
-each dwelling is equipped with electric cooking equipment
-each dwelling has electric space heating, air conditioning or both.

If these conditions aren't met use part II of art. 220

Remember to convert everything to volt-amps.

1. 1500VA for each 2 wire small appliance branch circuit - a total of 3000VA; remember,these are your kitchen counter appliance circuits. Also 1500VA for the laundry circuit - your question specifically dismisses the laundry load. (sec 220.16)

2. 3VA per square foot for general lighting;
3x2000=6000VA

3. Nameplate rating of all fastened in place appliances (ranges, wall mounted ovens and counter mounted cooking units - no demand factors applied at this time).

4. Nameplate rating of any motor of low power factor loads.

5. Use the larger of the A/C or space heating load, nameplate rating only

range: 7000va
dishwasher: 1200va
disposal: 600va
A/C: 2160va
hot water: 2500va

TOTAL: 22460VA

multiply total by number of units
22460x10=224600

multiply that total by the demand factor in table 220.32 for the number of dwelling units. In this case it is 43% (.43)

224600x.43=96578VA

divided by 240V=402.41Amps ---> 402Amps


Now to question 1.

Using the optional calculation for multfamily dwellings again, since it's a duplex (two units)

small appliance load:
1500x2=3000va
laundry: 1500va
general lighting: 2300 sq. ft.x3=6900va

Total general load: 11,400va

Fixed Appliances:
range: 16000
disposal: 600 (5Ax120v)
dishwasher: 1320 (11Ax120v)
hot water: 3000va
heat: 15000va (largest of heat & A/C - nameplat rating a/c is only 21Ax240v=5040va)
Total: 35920va

general load 11,400va
fixed apps. +35920va
---------
47320va
x2
---------
94640va/240v=394.33 --> 394A

Is there any chance that when typing in your post you transposed the digits of one of your answers? --- Perhaps 349 was meant to be 394? If so this may be your answer. Charlie B, if you're out there, please check my arithmetic.

Good Luck,
Rod
 
Re: Need Advice for Residential Load Calc

Thank You!!

So when the question states that each unit has 5 A/C units I am only supposed to account for 1 of them and not the 5 in each of the sides of the duplex or the house? I think that is where I have been making my mistake I have been using all 5 of them and then if it is a duplex I have used 10. It didnt sound right to me and how many duplexes have you seen with 5 A/C units in each unit.

Thank again
Mike
 
Re: Need Advice for Residential Load Calc

Mike,
I've seen questions where they throw in loads just to see if you diseregard them or not. It seems these five A/C would most likely be window units which are cord & plug connected, not fixed appliances therefore can be disregarded for purposes of your calculations. The 3VA per sq. ft. is intended to take care of these types of loads. However, refer to sections 210.23 and 440.62 (B) & (C) for real world purposes. Good luck!

Rod
 
Re: Need Advice for Residential Load Calc

rod i have a few questions on your calculations on the number one question. i hope i don't come across as a know it all (i'm not)but where in the code does it say you can omit 25.2 kva of ac and use 15 kva of heat instead, all i read is use the largest of the 2.and in your calculations you use no demand factor. the optional calc way to do this problem is 220.33(i think) this would give you 266 amps using heat and 323.5 amps using ac.
 
Re: Need Advice for Residential Load Calc

Rick in question #1 there is only 5040 watts of A/C load for each unit.

Roger

[ August 28, 2004, 08:57 AM: Message edited by: roger ]
 
Re: Need Advice for Residential Load Calc

Hey Roger & Rick,

I think Rick is right and I'm wrong. ---Mike, I hope this doesn't confuse the h*&^ out you.---- I don't mind it being pointed out that I'm wrong Rick, especially when we all learn something and I think I just did and I certainly hope Mike is. And Roger thanks for coming to my rescue but I think he's got us, this is a pretty advanced question, at least in my experience and a really tough one for a license exam. Here's why:

First, in my calculations I considered the A/C units as window units i.e. cord&plug connected. The 21 amps should have been a tip off that I overlooked. I think the author of the question probably had in mind those split units with the small rectangular outdoor units. With a 21 amp draw these clearly are not your basic window a/c units and do not meet the criteria of article 440.62 (now that I go back and re-read it)and obviously exceeds the load capacity of any standard residential branch circuit. With that in mind I went on to calculate this as a two family dwelling unit for which there is no demand factor in Table 220.32 for a 2 family unit. Then Rick pointed out article 220.33 which I had read before but since I wasn't including the a/c units I disregarded. When you take the a/c units times 5 and then calculate the whole thing based on THREE units (as per art. 220.33) and apply the demand factor for three units, I then came up with 323.55 amps taking the smaller of the two (2 or 3 dwelling units permitted by the article).

Now I'm making an assumption about the type of a/c units the author is talking about but given the values provided I think it's a pretty safe one. As a little aside, the extent of my familiarity with these types of units is that these are the type that were installed in Afghanistan to provide heat and a/c to the rooms in the old Soviet hangars that we occupied since all the infrastructure there was completely bombed out-literally.

I had fun with this one and I'm not sure it's over yet, keep 'em coming!

Rod
 
Re: Need Advice for Residential Load Calc

No wonder they say in order to pass this administator test you need the billion dollar class. So can someone reshow question 1 for me? I just want to make sure i have it down before I head off to take my next test for the 5th time next weekend. Thanks Again

Mike
 
Re: Need Advice for Residential Load Calc

hey msext
the optional calc on this problem is just like rod did it except i use the ac instead of heat then times total va(57520) by 3 (code 220.33)
=172560va then times 172560va by .45 (code 220.32&.33)= 77652va then divide 77652va by 240 = 323.5
my advice to you is to work out the problem in the forum using the standard and optional calc then maybe somebody can point out where you went wrong
good luck

roger the question says 5 ac per unit
rod this is a tough exam question constering you only have 15 minutes per question
 
Re: Need Advice for Residential Load Calc

Rick and Rod, I flat missed the 5 per unit. :eek: I guess that is why they say misreading a question is one of the biggest causes for failing an exam.

Roger
 
Re: Need Advice for Residential Load Calc

Mike,Rick & Roger:

Rick is right, this is an extremely tough exam question. Ya Roger, when I was listing out my loads the other day to do this one I just plain left out the 5 x a/c units.I'm licensed in two states, NC and Mass, and I don't think I would have picked up on all the little twists & turns in this problem. Not unless I had been exposed to it prior. Mike, now you have been. Just look at Rick's last post and substitute the a/c value for the heat in the calculation. Read article 220.33. I know it's a little unclear but it basically says that if the load for two units exceeds the load of three units before the demand factor is applied for three units, then you may use the lesser of the two loads. You see, there is no demand factor for two units in table 220.32; it starts at 3-5 and works up from there. When you come across a two unit situation with an unusually high computed load (in this case due to those a/c units) 220.33 tells you that if you multiply that computed (pre-demand factor) load by 3 instead of two, then apply the 45% demand factor - go ahead and use the smaller of the two possible demand factors. I truly hope that clears it up. There's a whole bunch of code articles applied to this problem. The author of this question clearly set out to challenge you. As far as taking an exam prep course, I'd recommend it. If you haven't already, look for one in your area that is focused on the exam that you're taking instead of a general NEC course. Good luck
 
Re: Need Advice for Residential Load Calc

Slight correction to last post, explanation of 220.33: when the load for two units exceeds the load for three units AFTER the demand factor is applied to the three unit load total, then you may use the lesser of the two loads ---- WHEW!
 
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