New Restaurant calculations help

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Cory@MAHC

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Dubuque, IA, USA
I'm working on Unit 11 in the 2017 version of Mike's electrical exam prep and I'm having some trouble with questions 25 and 26 in the 'Challenge Questions'. The answer key doesn't contain the work to get to the answer for these 2 questions and I'm wondering if I'm doing the math wrong or if the answer is wrong. Anyone able to help me through how you would do these 2 questions?
 
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Cory@MAHC

Member
Location
Dubuque, IA, USA
jumper said:
Can you post the questions, verbatim please, and the answer?

Errata and corrections for your book.

https://www.mikeholt.com/documents/catalog/17_Exam_Prep/17_EP_Correx_Directory.pdf
Click to expand...


Using the optional method, what's the service load for a 4,000 sq. ft. not all-electric restaurant containing the following loads?

Service 120/240V, single-phase
*Lighting actual, 6kVA
*Receptacles, sixty at 120V
*Gas furnaces, one 5A at 120V
*A/C, 50A at 240V
*Heat, 10 kW at 240V
*Sign, 1.2kVA at 120V

Kitchen Loads:
*Mixer, 10A at 240V
*Gas Ovens, two 250W at 120V
*Gas Steamer, 350W at 120V
*Electric Warmers, four 5kW at 240V
*Electric Steam Table, 20kW@240V
*Appliances, 5,000VA at 120V
*Electric Deep Fryer, 15,000VA at 240V

(a) 312A
(b) 327A
(c) 337A
(d) 408A
 
jumper

jumper

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3 Hr 2 Min from Winged Horses
Cory@MAHC said:
Using the optional method, what's the service load for a 4,000 sq. ft. not all-electric restaurant containing the following loads?

Service 120/240V, single-phase
*Lighting actual, 6kVA
*Receptacles, sixty at 120V
*Gas furnaces, one 5A at 120V
*A/C, 50A at 240V
*Heat, 10 kW at 240V
*Sign, 1.2kVA at 120V

Kitchen Loads:
*Mixer, 10A at 240V
*Gas Ovens, two 250W at 120V
*Gas Steamer, 350W at 120V
*Electric Warmers, four 5kW at 240V
*Electric Steam Table, 20kW@240V
*Appliances, 5,000VA at 120V
*Electric Deep Fryer, 15,000VA at 240V

(a) 312A
(b) 327A
(c) 337A
(d) 408A
Click to expand...

Okay.

First none of the kitchen loads/appliances are continuous.

Recs are figured at 180VA per device. Non continuous.

Lighting, sign, and largest HVAC load are considered continuous.

HVAC load is the larger of either the heating or cooling. Certain systems use both at the same time, but that is an exception and not used on tests unless specified.

What did the key give as an answer? I ask because I may have to do a little reverse engineering on the problem, continuous loads are not technically part of 220, but have to be considered for conductor and OCPD sizing for 230. 210 and 215 also.
 
C

Cory@MAHC

Member
Location
Dubuque, IA, USA
jumper said:
Okay.

First none of the kitchen loads/appliances are continuous.

Recs are figured at 180VA per device. Non continuous.

Lighting, sign, and largest HVAC load are considered continuous.

HVAC load is the larger of either the heating or cooling. Certain systems use both at the same time, but that is an exception and not used on tests unless specified.

What did the key give as an answer? I ask because I may have to do a little reverse engineering on the problem, continuous loads are not technically part of 220, but have to be considered for conductor and OCPD sizing for 230. 210 and 215 also.
Click to expand...

So here is how I figured it:

Lighting: 6,000VAx1.25=7,500VA
Recept.:60x180VA=10,800VA(no demand since this is in Part IV)
Gas Furnace: 5Ax120V=600VA
A/C: 50Ax240V=12,000VAx1.25=15,000VA
Heat: 10,000Wx1.25=12,500W
Sign: 1,200VAx1.25=1,500VA
Mixer: 10Ax240V=2400VA
Gas Ovens: 2x250W=500W
Gas Steamer: 350W
Electric Warmers: 4x5,000W=20,000W
Electric Steam Table: 20,000W
Appliances: 5,000VA
Electric Deep Fryer: 15,000VA

Total Connected Load of: 111150VA or 463A
Demand Load would be 100% of the Connected Load since this is a "Not All Electric Restaurant" and is less than 200kVA

220.88 New Restaurants. Calculation of a service or feeder load, where the feeder serves the total load, for a new restaurant shall be permitted in accordance with Table 220.88 in lieu of Part III of this article.
The overload protection of the service conductors shall be in accordance with 230.90 and 240.4
Feeder conductors shall not be required to be of greater ampacity than the service conductors.
Service or feeder conductors whose calculated load is determined by this optional calculation shall be permitted to have the neutral load determined by 220.61

Table 220.88 Note reads:
Add all electrical loads, including BOTH heating and cooling loads, to calculate the total connected load. Select the one demand factor that applies from the table, then multiply the total connected load by this single demand factor.

440.32 Single Motor-Compressor. Branch-circuit conductors supplying a single motor-compressor shall have an ampacity not less than 125 percent of either the motor-compressor rated-load current or the branch-circuit selection current, whichever is greater.

The answer book lists the answer as:

(d) 402A

Go figure.
 
jumper

jumper

Senior Member
Location
3 Hr 2 Min from Winged Horses
Was the answer 402A or 408A? Above you said (d)408A.

If 408A, then removing the heat loads, despite the foot note, gives this:

Lighting: 6,000VAx1.25=7,500VA
Recept.:60x180VA=10,800VA(no demand since this is in Part IV)
A/C: 50Ax240V=12,000VAx1.25=15,000VA
Sign: 1,200VAx1.25=1,500VA
Mixer: 10Ax240V=2400VA
Gas Ovens: 2x250W=500W
Gas Steamer: 350W
Electric Warmers: 4x5,000W=20,000W
Electric Steam Table: 20,000W
Appliances: 5,000VA
Electric Deep Fryer: 15,000VA

Total kVA=98,050
Amps=408.5, fractions .5 or less can be rounded down for service calculations.

Best I could come up with.
 
C

Cory@MAHC

Member
Location
Dubuque, IA, USA
jumper said:
Was the answer 402A or 408A? Above you said (d)408A.

If 408A, then removing the heat loads, despite the foot note, gives this:

Lighting: 6,000VAx1.25=7,500VA
Recept.:60x180VA=10,800VA(no demand since this is in Part IV)
A/C: 50Ax240V=12,000VAx1.25=15,000VA
Sign: 1,200VAx1.25=1,500VA
Mixer: 10Ax240V=2400VA
Gas Ovens: 2x250W=500W
Gas Steamer: 350W
Electric Warmers: 4x5,000W=20,000W
Electric Steam Table: 20,000W
Appliances: 5,000VA
Electric Deep Fryer: 15,000VA

Total kVA=98,050
Amps=408.5, fractions .5 or less can be rounded down for service calculations.

Best I could come up with.
Click to expand...

The answer book says (d) 402A. I agree that removing the heat load gives the 408A.
 
C

Cory@MAHC

Member
Location
Dubuque, IA, USA
jumper said:
Okay, did a search and found this link.

https://www.ecmag.com/section/codes...cuit-feeder-and-service-calculations-part-lxi

It says not to factor in extra 25% for continuous, I mentioned earlier that 220 doesn’t.
It does say use both heat and AC. Dehumidification may be involved.

The numbers still not come out though.
Click to expand...

And I'm sure I'm over-thinking it here as I would be surprised to see a question like this on my PSI Master's Exam, but it's fun to try and suss it out.

Yes, the link you posted is the book I have. there are no corrections for this problem listed.
 
Dennis Alwon

Dennis Alwon

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Staff member
Location
Chapel Hill, NC
Occupation
Retired Electrical Contractor
One has to assume the heat and the a/c will not be on together. If that is the case you need to remove the heat load because the a/c load is larger
 
jumper

jumper

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Location
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Dennis Alwon said:
One has to assume the heat and the a/c will not be on together. If that is the case you need to remove the heat load because the a/c load is larger
Click to expand...

The footnote and the article says to use both.

I just want to make sure the load data values and answer are correct.

Then if so, I intend to rerun my calcs until I get it correct. It is a simple calc, I should not keep getting it wrong. Frustrating.
 
jumper

jumper

Senior Member
Location
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Dennis Alwon said:
BTW, I come up with 400.2 amps. I get 402.7 if I leave the gas heater in and just take out the electric heat
Click to expand...

Dennis Alwon said:
BTW, I also used 4000 *2 watts/sq.ft then multiplied by 1.25= 10,000 va not 7500va
Click to expand...

I can fudge numbers and get close also, but by following the rules as I interpreted them I cannot exactly match the data given and the answer.

Need to know if question has these load values and what the answer is.

Something seems wrong - could be typo, wrong answer in key, wrong load values or we are all screwing up this calculation.
 
Dennis Alwon

Dennis Alwon

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Staff member
Location
Chapel Hill, NC
Occupation
Retired Electrical Contractor
I am not fudging numbers. You need to use 2 va/sq.ft rather than the actual lighting load as it is higher. You also use the larger of the a/c vs heater. Now here is an assumption you need the blower unit on the gas furnace for the a/c to work so you include that????? This comes to 402.7.

I will bring it to the attention of Brian House.
 
jumper

jumper

Senior Member
Location
3 Hr 2 Min from Winged Horses
Dennis Alwon said:
I am not fudging numbers. You need to use 2 va/sq.ft rather than the actual lighting load as it is higher. You also use the larger of the a/c vs heater. Now here is an assumption you need the blower unit on the gas furnace for the a/c to work so you include that????? This comes to 402.7.

I will bring it to the attention of Brian House.
Click to expand...

Where does it say in 220.88 that I disregard the large heat load? Could easily be dehumidification heat. 10 kW is a common strip heat number.
 
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