minimum size service

Thread Starter

albn

Joined Jul 6, 2012
12
An apartment building has 12 dwelling units with 1640 square feet of space plus the FF: equipment, 12kw water heater,5 kw dyer, 10.5 kw baseboard heater. what is the minimum service permitted for this building @ 120/240 volts ?

A. 859 amperes
B. 1074 amperes
C. 1110 amperes
D. 2096 amperes

Please help I've tried already so many times still could not arrived to any numbers mentioned above, your kind assistance is highly appreciated.:confused:<SNIP> Thanks
 
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THE_RB

Joined Feb 11, 2008
5,438
The correct answer is E;

A. 859 amperes
B. 1074 amperes
C. 1110 amperes
D. 2096 amperes
E. Do your own homework
 

takao21203

Joined Apr 28, 2012
3,702
27*12 I don't know the answer I need to press the button on my keyboard for Windows calculator. Sometimes using Microsoft hardware can be of advantage.
 

Thread Starter

albn

Joined Jul 6, 2012
12
thanks for you reply but like what you mentioned, I did tried and still confused about the calculation that's why I posted in here! to find somebody who can show how to solve it right. Thank you
 

GetDeviceInfo

Joined Jun 7, 2009
2,196
you still haven't told us what code your applying. Generally, you find your base area wattage, add your other loads as per demand allowance, compensate for range/no range, strip out your heat, calculate the unit contribution to the overall, add back in your heat as per demand allowance, add common area, parking, lighting, etc.
 

mlog

Joined Feb 11, 2012
276
An apartment building has 12 dwelling units with 1640 square feet of space plus the FF: equipment, 12kw water heater,5 kw dyer, 10.5 kw baseboard heater. what is the minimum service permitted for this building @ 120/240 volts ?

A. 859 amperes
B. 1074 amperes
C. 1110 amperes
D. 2096 amperes

Please help I've tried already so many times still could not arrived to any numbers mentioned above, your kind assistance is highly appreciated.:confused:<SNIP> Thanks
You really need to do your own work, but we'll help you get started.

Pardon my ignorance of the subject, but I'll start by asking some questions.

1. What is FF?

2. Does the 1640 sq ft of living space affect the lighting and other electric usage? I think so. What is the equation for this? Is it linear? Is it of the form y=mx+b?

3. Is there a water heater, dryer, and baseboard heater in each of the 12 units?

4. Is there any additional power requirement for the building that is not contained within the 12 units? What about hall lights, outside lights, heating for the hallways (are there hallways?), etc.?

Only baseboard heating...I guess there's no air conditioning....

Making my own assumptions, I get something like this.

Total power per unit = Water heater + Dryer + Baseboard heater + General lighting and other usage

Total building power = 12 units x Total power per unit

Total building current = Total building power / Voltage

After you get the current, is there a margin applied to it, say 50% above the calculated load?

Just a quick check on the back of a napkin suggests to me that only one of the answers can possible be correct.
 

Thread Starter

albn

Joined Jul 6, 2012
12
Sorry again each unit is 1640 sq.feet and equipt with range,water heater, baseboard heater,and dryer. There's no formula like the one you mentioned like: y=mx+b, it's just what the NEC Code book calculation which is subject to different adjustment per location, and application. What I typed in my latest post are the same what the code given samples. Thanks again
 

GetDeviceInfo

Joined Jun 7, 2009
2,196
obviously I can't help you much, as I work by the CEC. They are not greatly different, and becoming closer with each revision.

I noticed that you did add a range.

why dont' you show us some of your work;
what's your basic (area) calculated load per/unit,
what's your calculated range per/unit (I assume there all the same rating?)
What's your other load calculations per/unit
Are you stripping out your heat?
How does each unit contribute to the overall?
Are you adding back in your heat?
Are there undisclosed loads in common areas?

Help us help you , or not.
 

BSomer

Joined Dec 28, 2011
434
Do you have an up to date code book available? In the back (Annex D) are examples of how to calculate your loads, feeder conductor sizes, etc... The main article of concern for these calculations is 220.

Specific tables that would be used:
Table 220.42 - Lighting Load Demand Factors
Table 220.54 - Electric Dryer Demand Factors
Table 220.55 - Electric Range/Cooking Appliance Demand Factors

There are others but if you read through Article 220 and pay attention to the "multifamily dwelling" sections I think that you can work your way through it.
 

GetDeviceInfo

Joined Jun 7, 2009
2,196
(27.5KW x 12) /240V = 1375Amps
This is greater than A, B, or C
So, the answer must be ?
In multi family, units are derated by a unit number/factor.

I doubt it matters though, as the OP has not given us any indication as to his ability, and his assignment date has likely gone by.
 

Thread Starter

albn

Joined Jul 6, 2012
12
we'll your calculation is the same as mine but still not matched the answer, it must be the the right answer otherwise it's wrong. thanks anyway.
 

Thread Starter

albn

Joined Jul 6, 2012
12
we'll I've checked that out already still I could not arrive to the right answer, thanks anyway for your input I really appreciated.
 

Thread Starter

albn

Joined Jul 6, 2012
12
I responded with complete computation according to the sample on annex D of NEC cod book but I don't know to whom I responded cuz' it's not showing here. thanks anyway for your input I really appreciated.
 
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