Heavy duty lampholder

[Curtiszeph]

I am working through a exam test prep book, and I came across this question that asked me to calculate the demand for a restaurant. Part of the calculation included 10 heavy duty lampholders. I calculated these as 10 X 600 X 125% = 7500va. When looking at the test solution I was slightly off on my amperage because the book didnt calculate the heavy duty lampholders as a continous load. Am I missing something? TIA

 

[JFletcher]

I am working through a exam test prep book, and I came across this question that asked me to calculate the demand for a restaurant. Part of the calculation included 10 heavy duty lampholders. I calculated these as 10 X 600 X 125% = 7500va. When looking at the test solution I was slightly off on my amperage because the book didnt calculate the heavy duty lampholders as a continous load. Am I missing something? TIA

I'll take a stab at this...

While commercial lighting is usually a continuous load, it is not required to be treated as such unless it actually meets the definition of a continuous load and is on for more than 3 hours.

That said, I have not done a commercial load calculation in quite a while. I would have made the same assumption that you did, that the heavy duty lamp holders are Mogul base, and probably for exterior parking lot lights, which would definitely be continuous loads. I am curious as to where in the book it would tell you to use 600 watts per lamp holder. unless it was supposed to be part of the 3 volt amp per square foot, 600 watts per lamp holder is archaic with the advent of LED lights.

Could you print the exact parameters of the test question, and the book answer?

If anyone can point the two of us in the right direction, I'm all eyes.

 

[Curtiszeph]

The requirement to use 600VA is found at 220.14(E). I think it may have been a mistake as continous loading was used for the general lighting but not the show window in the answer key.

 

[Curtiszeph]

Edit

 

[Curtiszeph]

Question asks:
Calculate the demand on the OCD for a 15000 sq ft restaurant, 208/120 3 ph. Use general method, the restaurant has the following:
2-35kw ranges
1-15kw booster heater
1-20kw water heater
1-6kw dishwasher
10-heavy duty lampholders
65 duplex receptacles
20kva parking lot lighting
4-5hp a/c units 308v 3 ph
20kw electric heat


The answer is as follows:
15000 sq ft x 2va x 125% = 37500
2-35kw ranges 70kw x 70% =49000
15kw heater 15kw x 70% =10500
20kw heater 20kw x 70% =14000
6kw dishwasher 6kw x 70% = 4200
10 heavy duty lamps 10 x 600 =6000
65 receptacles 65 x 180va After demand = 10850
20kva parking lot lights 20kva x 125%=25000
4-5hp A/C units = 16.7a x 208v x 1.732 = 6016 x 4= 24064
20kw electric heat (omit)
Largest motor 6016 x 25% = 1504
Total = 182618 / 208 /1.732 = 507A demand

It is tom henrys 2014 NEC exam calculations prep book. I wouldn't be surprised if it was a mistake, I have found others. I just wanted to be sure I wasnt missing something simple.