Reading current usage from utility meter

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Newer Meters...

Newer Meters...

Many of the newer digital meters have a function built in that will allow you to read the current, voltage and watts. The cheaper residential meters in our municipal don't have this function - but if the meter can record demand, they usually do have this function. If so, there will be a reed switch installed behind the display. If you pass a magnet above the display, the normal display will change to "ALT". Then, it will scroll through voltage on each leg, current through each leg, instantaneous demand, number of power outages, etc. Just don't leave the magnet sitting there - this will change the meter to another mode, and I don't know if the meter will return to normal if you do this. No guarantee that your meter will have this function - but might be worth a try :)
 
That's not going to work.

You have 3 options.

1. Call the power company and ask them what the peak demand has been in the last 12 months. Have them email this record over to you so you can show your inspector.

2. Put a recording meter on the service with a 15 minute average sampling rate for at least 30 days. Be sure to adjust your amperage totals for any HVAC loads that may not be at full load yet.

3. Do a load calc.
 
My last and the best option is renting an amp meter and implementing 220.87. I do not own arc flash suit so I am going to hire a testing company to install this at the main 600A service. I dont think it is safe for me to do it with no protection on.

At the same time customer is not too educated about electric and I dont want him to think I am doing something unnecessary.
 
Cow said:
That's not going to work.

You have 3 options.

1. Call the power company and ask them what the peak demand has been in the last 12 months. Have them email this record over to you so you can show your inspector.

2. Put a recording meter on the service with a 15 minute average sampling rate for at least 30 days. Be sure to adjust your amperage totals for any HVAC loads that may not be at full load yet.

3. Do a load calc.
Click to expand...

if they give me a peak demand, that is going to be kwh right? so how do we go Amp from kwh?
 
de2 said:
if they give me a peak demand, that is going to be kwh right? so how do we go Amp from kwh?
Click to expand...
Peak demand will be kW not kWH, and likely largest peak for any 15 minute period, actual peak could be different, but didn't last very long. Divide by volts to get amps, and divide again by square root of three if three phase (won't be completely accurate if unbalanced phases, but hopefully close enough).
 
181006-1106 EDT

kwired and de2:

If total kW load is resistive, and voltage is line to neutral of a wye system, then line current in amps is (kW/3)*1000/V where V is line-to-neutral voltage in volts. If kWH is known over a known time range, then an average current can be obtained from kWH/Hours to get an average current in the above equation.

.
 
gar said:
181006-1106 EDT

kwired and de2:

If total kW load is resistive, and voltage is line to neutral of a wye system, then line current in amps is (kW/3)*1000/V where V is line-to-neutral voltage in volts. If kWH is known over a known time range, then an average current can be obtained from kWH/Hours to get an average current in the above equation.

.
Click to expand...
kwh is;
30 days 33,680
29 days 36,080
31 days 27,200
This is a known time range.
I am going to write a formula, but I dont believe this is correct, here it goes;

36,080 is the max over 3 months
36,080kwh=kw * 24*29
kw=51.84kw

I= (51.84*1000) / ( 1.73*208) = 144A, do you think it is a educated guess?
 
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181006-1917 EDT

de2:

I get the same answer.

To this there probably needs to be a multiplier of an unknown value because this is an average value.

Part of that multiplier is 24/actual hours used in a day.

The other part is based on how power consumption varies with time. Actual power measurements are really needed others have pointed this out.

.
 
de2 said:
kwh is;
30 days 33,680
29 days 36,080
31 days 27,200
This is a known time range.
I am going to write a formula, but I dont believe this is correct, here it goes;

36,080 is the max over 3 months
36,080kwh=kw * 24*29
kw=51.84kw

I= (51.84*1000) / ( 1.73*208) = 144A, do you think it is a educated guess?
Click to expand...
An educated guess but of little value. The time range is too long. There are too many long peaks and valleys in that average. A shorter time range like the 15 minutes discussed above, while not perfect, is closer to what you need.
 
Here is a link for the meter you questioned:

https://deckmonitoring.zendesk.com/...2788-Complete-GE-Manual-for-kV2c-Socket-Meter

Or, you could try searching the GE website - this is just the first thing I came up with that shows you how to get the readings you're looking for. This meter is different than ours - the magnet is utilized differently, and on the side of the meter (rather than the top), at about the 4-5 O'clock position (I think). Activating the test switch (with a magnet) for more than 6 seconds will put the meter in the "Site Genie" mode - it will scroll through current, voltage, etc.

I didn't read the whole thing, but I'm sure you'll have to take that multiplier into account.

Hope this helps.
 
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