Peak current draw on a residential service

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codeunderstanding

Senior Member
I was trying to find out what a average peak current draw would be on a 200 anp service based upon KW hours used in a pay period. Heres a example.

1000KW hours used in 28 day period
1000KW*1000=1,000,000 Watts
1,000,000/240v=4,167amps
4,167amps/28days=148.82amps per day.

Would this be somewhat right or am I totally off.
 
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Bob NH

Senior Member
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The "Live Free or Die" State
You are totally off.

You don't have correct units for the formulas. There is no such thing as Amps per day.

You haven't applied correct rationale for determining peak current.

It is not clear that you have enough information for making more than a wild estimate of peak current.

You should start out by writing down what you think "Average peak current draw" is, and then try to figure out a logical way of finding it from the data.
 
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rattus

Senior Member
Way off!

Way off!

codeunderstanding said:
I was trying to find out what a average peak current draw would be on a 200 anp service based upon KW hours used in a pay period. Heres a example.

1000KW hours used in 28 day period
1000KW*1000=1,000,000 Watts
1,000,000/240v=4,167amps
4,167amps/28days=148.82amps per day.

Would this be somewhat right or am I totally off.
Click to expand...

Average power is:

Pavg = 1000KWH/672Hrs = 1.49KW

Average current is:

Iavg = 1.49KW/240V = 5.79A

Peak current would be much greater and would depend on the actual loads such as AC, electric heat, range, oven, clothes dryer, etc. Some of the NEC experts can undoubtedly point you to a code reference which provides a method for predicting the peak current.
 
steve66

steve66

Senior Member
Location
Illinois
Occupation
Engineer
There isn't actually any code method to determine the peak. I assume that the variables would change so much from one building to the next (and from one season to the next) that there is just no single formula that would be accurate.

The code does allow using the peak if it is measured over a 15 min. interval using a demand meter, and if you have data from an entire year.

Steve
 
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rattus

Senior Member
Easier still:

Easier still:

NoVA Comms Power said:
Wouldn't it be easier (and far more accurate) just to turn on the loads you'd consider to be included in the "average peak load" and figure out the KW consumed based the meter's "spin rate" ?

Page 3 of this document explains the math involved: http://www.csupomona.edu/~pbsiegel/www/sci210/electricmeter.pdf
Click to expand...

If you can get to the service drop easily, it is a simple thing to slip an Amprobe around the entrace cables.
 
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