new guy here,,,,, ive been a journeyman for about 10 years now but ive never had to do load calcs ,, im gonna have to take a test for electrical inspector and i know for sure theres gonna be some load calc questions on the test,, I NEED HELP ...lol... i seen one of the questions that read a multi family dwelling has 6 dryers rated at so many Kvas (cant remember how many) how many amps would you need to run the dryers,,,my code book 08 is pretty confussing when it comes to examples ,,, so all load calc examples would be helpfull,, also is there any known web sites that give examples ????

The first trick is to figure out how all the tables in 220 work, then coordinate that with the examples.
I know ranges was my stumbling block. Once I figured that out, the rest of the tables started making sense.
That also seems to be the objective most of the questions on the test address.
Do you have the handbook? That explains a lot of these things.

no hand book but i do have the uglys ,, ive taken this test once and the calcs killed me ,, the state is giving me a second chance if i pass the test i have the job but again its the calcs that are hurting me

Stallcup has a book out, calc's made easy (or somethimg like that) 21, helped me out a lot

check the store here....

~S~

romex2121,

Welcome to ECN!

Here is a quick and dirty response to the Load Calculation for Demand Factor Scenario of Six Electric-Only Clothes Dryers in Dwelling Unit(s):

*** NOTE: For this example, I am basing the Dryers at 6.0 KVA, 240V 1 Phase ***

6000 VA (6.0 KVA) = 25.0 Amps at 240V 1 Phase 2 Wire

6,000 VA (6.0 KVA) x 6 Dryers = 36,000 VA (36.0 KVA)

Per Table 220.54, Demand Factor for 6 Dryers = 75% (0.75)

36,000 VA x 0.75 = 27,000 VA (27.0 KVA)

27,000 VA ÷ 240V = 112.5 Amps

Demand for this part of a Load Calculation Database:

* 27.0 KVA
* 112.5 Amps

If you need practice, make up a few project examples and compile Load Calcs for each one - via what is outlined in NEC Article 220, along with the example Project Calculations found in Annex D (Annex D is at the back of the NEC Codebook); then post your scenario Projects' information + Load Calcs in this Thread, for review.

Good luck.

Scott

thanks guys,,, i think im getting the hang of it a little better now, it was figuring out the formulas to find the answers that i didnt understand,, theres still alot i need to work on but i only have till weds. thats the day i test for the job,,,,

thanks scott,, ive been going over the examples in the NEC
(anex D) and its starting to make since to me now,, like i said this is something ive never had to do so its all new to me,, as a journyman i should know this stuff tho...lol...
im gonna throw another one out there for ya,,
one range at 12,000Va ,, one wall oven at8,000Va,, and one wall oven at 9,000Va ,, this is a 120/240 residential home
what size service would be needed to run this home ??? and also what size neutral would be needed for the service per calculation ????

on a side note what the hell is up with artical 220 ?? in annex D it says to refer to artical 220.6 for the demand factor for ranges,in my 2008 NEC code book there is no artical 220.6 in fact if you look thru the whole artical there seems to be alot of it missing

rx:
IF you are refering to the neutral calcs in Annex D...it says...

"Calculation for Neutral for Feeder and Service

Lighting and Small-Appliance Load 5,100 VA
Range: 8000 VA at 70% (see 220.61) 5,600 VA
Dryer: 5500 VA at 70% (see 220.61) 3,850 VA
Total 14,550 VA"

220.61

romex2121;

Glad to see the Load Calcs are making sense now!

As to your example:



There are Two "Options" for the total Demand Calculation of these Household Electric Cooking Units.

*** OPTION #1 ***

1st Option is to use Table 220.55 (2005 NEC Article 220, Part III).

(1) 8.0 KVA Oven x 0.8 = 6,400 VA Demand
(per Column "B");
(1) 9.0 KVA Oven + (1) 12.0 KVA Range = 11,550 VA Demand
(per Column "C").

Total Electric Cooking Equipment Demand: 17,950 VA (74.79 Amps @ 240V 1PH)

------------------------------------
------------------------------------

*** OPTION #2 ***

2nd Option is to use Section 220.82(B) (2005 NEC Article 220, Part IV).
"First 10 KVA @ 100%, remaining at 40%"

8.0 KVA + 9.0 KVA = 12.0 KVA = 29.0 KVA

First 10 KVA at 100%: 10.0 KVA x 1.0 = 10,000 VA
Remaining 19 KVA at 40%: 19.0 KVA x 0.4 = 7,600 VA

Total Electric Cooking Equipment Demand: 17,600 VA (73.34 Amps @ 240V 1PH)

------------------------------------
------------------------------------


*** NOTE: The "Electric Cooking Equipment Demand" is only One Part of the complete Load Calculation figure.
There are still other parts to include, such as:
* General Lighting (Va per square foot),
* Small Appliance Branch Circuits,
* Laundry Circuit,
* Other Fixed Appliances,
* HVAC,
* Other Motor Loads - if used,
* Electric Water Heating - if used.






Please refer to the note above.




I am working off 2005 NEC - actually 2007 California Electrical Code, which uses 2005 NEC as the "Model Code"; so the Sections may be different between the Two NEC Versions.

Per the 2005 NEC, Household Electric Cooking Equipment Demand Factors are found in Table 220.55, which is in Section 220.55 (Part III).
This is Option #1

Option #2 is Section 220.82

Good luck!

Scott

thanks for all the help Scott,,, but as luck would have it i faild the test by 3 questions,, the bad thing is i know the general codes like the back of my hand but it was the calcs. that i did'nt know that well,, some of the questions on the test on calcs. were pretty intense (for me anyway)im a little bummed but now i can proceed with obtaining my electrical contractors card, which was the plan intill the inspector job came up,, again thanks for taking the time to help me out,,