Calculating load for lighting

[websparky]

George,

Here is the proper way to calculate 3 phase loads.

1.) Office building 24000 square feet
2.) from table 220.12, 24000 x 3.5 = 84000VA
3.) 84000 x 1.25 = 105000 total lighting load after all demand factors
4.) 105000VA / system voltage 480Y/277
5.) 480Y x 1.732 = 831V for three phase systems
6.) 105000VA / 831V = 126A total amp load
Prove math below:
7.) 24, 20A circuits, 277V, 16A each
8.) 277V * 16A = 4432VA
9.) 4432VA * 24 circuits = 106368VA (105000VA)

 

[George Stolz]

That darn square root of three biting me in the rear again.

Thanks, Dave. That matches the technique shown in Example D3 in Annex D.

I think this could be made clearer in 220, it almost seems to dance around that. There doesn't seem to be a section that clearly states that.

 

[Smart $]

...

I think this could be made clearer in 220, it almost seems to dance around that. There doesn't seem to be a section that clearly states that.

Sure would be nice if there was... but I wouldn't restrict that wish to only this article

Regardless, there's one mistake in Dave's calculations. He's already applied 125% to the 220.12 load, so to determine circuits required to be available he should be calculating with 20A instead of 16A. Therefore it should be:

Prove math below:
7.) 19, 20A circuits available @ 277V
8.) 277V * 20A = 5540VA
9.) 5540VA * 19 = 105260VA​

PS: BTW, this correlates to the 20 circuits I determined for 4 panels in my earlier post.

 

[George Stolz]

Regardless, there's one mistake in Dave's calculations. He's already applied 125% to the 220.12 load, so to determine circuits required to be available he should be calculating with 20A instead of 16A.

No, he's correct. Each circuit breaker is continuously loaded, therefore they cannot carry more than 16A apiece.

PS: BTW, this correlates to the 20 circuits I determined for 4 panels in my earlier post.

BTW, it's still wrong.

It's okay to be wrong, we're all learning here. You have nothing to prove.

 

[Smart $]

No, he's correct. Each circuit breaker is continuously loaded, therefore they cannot carry more than 16A apiece.


BTW, it's still wrong.

It's okay to be wrong, we're all learning here. You have nothing to prove.

No, he's wrong. The 16A is prior to applying a 125% BC "correction factor" for not-a-100%-rated assembly. After applying 125% the number of circuits is calculated at 20A.

Check the math:

1) 84000 VA ? 277V = 304A
2) 304A ? 16A/circuit = 19 circuits
3) 84000 VA x 125% = 105000 VA
4) 19 circuits x 277V x 20A/circuit = 105260 VA

 

[websparky]

Dumb $ and Dumber $

Dumb $ and Dumber $

Dumb $

My calculations are 100% acurate.

If you do not understand them or how the code is applied in this case, then open a book and read or post a question.

Your calculations are 100% wrong. Just as they were in the post about phases. Simply restating them will not make them right.

Constructive comments:
Please do not keep replying to this thread and repeating your inacurate claims.

 

[hardworkingstiff]

my 2 cents

my 2 cents

If you add 25% to the beginning calculation then the "total load calculated" has the 25% already calculated. In this case putting 20 amps of the "total load calculated" on a circuit you are really only putting 16 amps of the beginning calculation on the circuit.

Smart$ is correct on this.

 

[George Stolz]

I still think Dave is correct.

The first part of the math is determining the lighting load on the service.

The "proving" second part is determining the number of circuits needed.

The service is required to be considered at the 105kVA number.

The branch circuits can't be loaded individually over 16A.

So those impaired circuits must eventually add up to the service amount.

At first glance, it does look like it's been compensated for twice, but in reality each component has to be respected individually, at it's level, IMO.

 

[Smart $]

I think you're compounding the load with the branch circuit ampacity adjustment.

The service or feeder is deetermined at the 84kVA number. Its ampacity is then adjusted to 125% of that number—105kVA in this case—for continuous loads. When demand factors apply, the adjustment for such should be done before adjusting for ampacity as demand factors may not apply equally in all load subcategories.

The minimum number of branch circuits is determined at the 84kVA number using 80% of the branch circuit's ampacity—84000VA ? 277V ? 16A = 19—or the 84kVA is adjusted to 125% then using 100% of the circuit's ampacity—84000kVA x 125% ? 277V ? 20A = 19.

For an added level of certainty, review Example D3 in the 2005 NEC. Everywhere in the example, replace 8500 VA actual lighting load with 9000 VA (and recalculate as appropriate). This is just to make the actual and calculated loads the same and help avoid confusion (well at least that's my intent). Now if you follow the adjusted example you will see the 9000 VA is only multiplied once for the minimum number of branch circuits and only once for feeder size, and quite independent of one another. For the feeder it is only done in the overcurrent protection section and uses 16,200 VA x 1.25. In combining the continuous loads earlier, note the 16,200 VA subtotal is arrived at using 9000 VA, not 11,250 VA.

Question: What is the minimum number of branch circuits for general lighting in Example D3 if the lighting is all on 1-pole, 20A amp OCPD's?

 

[dahualin]

For electrical load calculation, the total lighting load will be 2400x3.5x1.25=10500VA that is calcuated load for sizing service. The connected load is 84000VA (We assume that the actual connected lighting load is 84000VA). Every time I do circuiting for lighting is based on connected load, each circuit load will not be bigger than 16A that has considered the continuous load).

So, I agree with hardworkingstiff. Smart $ is correct on this.