Lighting and receptacle load calculation

[Angel Electric]

A five story office has a footprint of 200 x 150.
Each floor will have 117 receptacle outlets.
What is the VA Load for the service of the building.

The Answer provided on the Test answers is 806,250 VA... I keep getting different at 525,000 VA for the Lighting.
Then for the receptacle load I get 66,000 VA.


Lighting Calculation

200 X 150 =30000 SQ. FT. x 5 Floors=150,000 SQ. Ft. Total x 3.5 from Table 220.12 For the Lighting Load. I got 525,0000 VA

Receptacle Calculation

Then Per 220.14K I tried 180VA X 117 Outlets x 5 Floors worth =105,300 Then The larger being 150,000 X 1 VA.Per Sq. FT. =150,000VA.
Now I derate The receptacle VA 150,000-10000=140000 X.40= 56,000Va

Now I add 525,000 VA From Lighting Load to 66,000 VA For the Receptacle Load. I get 591,000 VA Not 806,250 VA.

I really need to find what I'm doing Wrong...
Please any help is appreciated!

 

[Dennis Alwon]

I get exactly what you got.

 

[Smart $]

...

Receptacle Calculation

Then Per 220.14K I tried 180VA X 117 Outlets x 5 Floors worth =105,300 Then The larger being 150,000 X 1 VA.Per Sq. FT. =150,000VA.
Now I derate The receptacle VA 150,000-10000=140000 X.40= 56,000Va

...

Not that it gets you a whole lot closer to the "answer", but demand over 10kVA on receptacles is 50%... not 40%

 

[Dennis Alwon]

Not that it gets you a whole lot closer to the "answer", but demand over 10kVA on receptacles is 50%... not 40%

Good point-- I just used his numbers but never realized it was 50%... I don't see how they got that high- even with no demand it would only be 625,000

 

[jumper]

I think I can see how author got answer, not saying I agree with author.

Lights 525,000 x 125%= 656,250. Author may have used continuous demand factor.

Receptacles 150,000

Total=806,250

Author may have not T 220.44, but 220.42 for demand factor of receptacles-which is 100%.

220.44 Receptacle Loads ? Other Than Dwelling
Units. Receptacle loads calculated in accordance with
220.14(H) and (I) shall be permitted to be made subject to
the demand factors given in Table 220.42 or Table 220.44.

 

[Dennis Alwon]

I think I can see how author got answer, not saying I agree with author.

Lights 525,000 x 125%= 656,250. Author may have used continuous demand factor.

Receptacles 150,000

Total=806,250

Author may have not T 220.44, but 220.42 for demand factor of receptacles-which is 100%.

Very good - I forgot about the 125% for continuous duty on the lights

 

[Smart $]

I think I can see how author got answer, not saying I agree with author.

Lights 525,000 x 125%= 656,250. Author may have used continuous demand factor.

Receptacles 150,000

Total=806,250

Author may have not T 220.44, but 220.42 for demand factor of receptacles-which is 100%.

Appears to be correct, as far as the author's premise...

However, there is technically no continuous load demand factor for Article 220 calculations.

And to top that off, there's a good chance not all the lighting in a 5-story office building will be continuous loads. Some of it will likely even be at reduced demand, such as closets, halls, and stairways.

I realize its a general question, but the author is violating the first rule of the electrical profession ? Never assume anything!

 

[Angel Electric]

Excellent!!

Excellent!!

Atleast I'll be able to accurately give them their correct answer... I appreciate the efforts and expertise.. This stuff gets overwhelming after question # 56 Ect...

 

[Dennis Alwon]

However, there is technically no continuous load demand factor for Article 220 calculations.

This is true. In fact in the example in the book Annex D3 for store buildings it shows the sq.ft load vs the actual connected load. They took the actual load and then multiplied that by 125% but it was lower than the 3 watts/sq. ft so they used that.

It seems like the watts/sq.ft incorporates that 125% . Now if the actual load times 125% was larger than the watts/sq.ft then you would have to use the watts/sq.ft.

Given this example I would say the author is incorrect. NO?

 

[jumper]

Dennis, are you saying the example is wrong?

 

[Dennis Alwon]

Dennis, are you saying the example is wrong?

Yes, I am saying that 125% should not be used since the 3va/sq,ft was used and no actual load was given. The example in the NEC book is correct not the OP's answer.

 

[jumper]

Yes, I am saying that 125% should not be used since the 3va/sq,ft was used and no actual load was given. The example in the NEC book is correct not the OP's answer.

But the example in the book uses 125% for 3va/sq,ft, look at the bottom. It just incorporates it later when sizing the feeder.

 

[Smart $]

Yes, I am saying that 125% should not be used since the 3va/sq,ft was used and no actual load was given. The example in the NEC book is correct not the OP's answer.

General Lighting is listed under Continuous Loads in Example D3. The value in the Calculated Load section is simply 3000 ft? ? 3 VA/ft?.... 9,000 VA.

The footnote* explains that this value at 125% is greater than the actual load at 125%.

*In the example, 125% of the actual connected lighting load (8500 VA ? 1.25
= 10,625 VA) is less than 125% of the load from Table 220.12, so the minimum
lighting load from Table 220.12 is used in the calculation. Had the actual
lighting load been greater than the value calculated from Table 220.12, 125%
of the actual connected lighting load would have been used.

The red text is in error. I have no idea why they even used 125% factoring at this stage of the calculation... it's moot at this stage. There's not even need to factor by 125%. The same one is greater than the other without any factoring.

 

[Dennis Alwon]

General Lighting is listed under Continuous Loads in Example D3. The value in the Calculated Load section is simply 3000 ft? ? 3 VA/ft?.... 9,000 VA.

The footnote* explains that this value at 125% is greater than the actual load at 125%.

​

The red text is in error. I have no idea why they even used 125% factoring at this stage of the calculation... it's moot at this stage. There's not even need to factor by 125%. The same one is greater than the other without any factoring.

Yes, but I read that to mean if the actual continuous load is known then you must use 125%. If thjat number is greater than the va/sq.ft then you would use that... No?

I don't understand why they use 125% atr the bottom. If the actual load *125% was larger than the va/sq.ft they expect you to factor in another 125% at the end????

Something is not right...

 

[Dennis Alwon]

I agree with Smart the red in his post is incorrect but you are correct jumper--125% needs to be used-- I assumed it didn't with that text written there

 

[Smart $]

I agree with Smart the red in his post is incorrect but you are correct jumper--125% needs to be used-- I assumed it didn't with that text written there

I submitted a proposal to remove all the 125% factoring in the footnote.

 

[Dennis Alwon]

I submitted a proposal to remove all the 125% factoring in the footnote.

Good-- I was going to do that--you should get the credit for that

 

[MasterTheNEC]

Less us not forget that they are just footnotes to an Annex D anyway so just examples. I don't see anything in the footnote that would have confused me. Just Sayin !

 

[SPARKLEY]

full General Lighting load in Dwellings will never be energized >3 hours?

full General Lighting load in Dwellings will never be energized >3 hours?

So the Examples D1 through D3 treat General Lighting as continuous in commercial and noncontinuous in residential. But is this based on assumption that the full General Lighting load in Dwellings will never be energized >3 hours? It would seem beneficial to the trade if NFPA clarified this, either in the Examples or perhaps via footnote to Table 220.12. I?m not a NFPA member yet so I can?t post a proposal, but may I ask you gents? thoughts?

 

[Smart $]

So the Examples D1 through D3 treat General Lighting as continuous in commercial and noncontinuous in residential. But is this based on assumption that the full General Lighting load in Dwellings will never be energized >3 hours? It would seem beneficial to the trade if NFPA clarified this, either in the Examples or perhaps via footnote to Table 220.12. I?m not a NFPA member yet so I can?t post a proposal, but may I ask you gents? thoughts?

They are just examples... and I'd say the intent is to just give typical scenarios. There's nothing in Code that says you can't have continuous lighting in residential or noncontinuous lighting in commercial/industrial. If you're doing the load calc', call 'em like you see 'em.

And you don't have to be an NFPA member to submit a proposal. Public Input is exactly that. But if you use the internet method, you must establish a user account on the NFPA website.