Actual electrical meter reading.

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michalspike

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brooklyn, new york
How I can convert Amperage reading on my clamp meter to watts what we are actually paying to our electricity provider.
I know how to convert amperage to watts.
For example:
I have three phase services: each leg drawn 20Amp( this is what its show on clamp meter) how I'm converting it? correct me if I'm wrong.
20x120v=2400W
2400X3= 7200w=7,2kw

How I know how much is it in an hour?
 
You really can't do it with any accuracy unless you are measuring power factor, voltage and current all at the same time.

You could make an estimate.
 
There are quite a few cheap kwh meters on Amazon you could pick up.

Read after 30 days and multiply by your kwh rate posted on your bill.

Sent from my SAMSUNG-SM-T337A using Tapatalk
 
michalspike said:
20x120v=2400W
2400X3= 7200w=7,2kw

How I know how much is it in an hour?
Click to expand...

Multiply by one hour to get kWh. The 'h' at the end of kWh is key, it means your counting energy not power.
If it runs for two hours then multiply by 2. 7.2kW * 2hrs = 14.4kWh.
This assumes the power stays exactly constant at 7.2kW for that time period, which in the real world is never the case, but if you're just trying to understand the theory it's fine.

Most basic:

amps*voltage = power
power * time = energy
Example: 10A x 240V = 2400W (or 2.4kW)
2400W for 1 hour = 2400W watt hours (or 2.4kWh)

When I say 'power' above' I mean real power. Now add power factor.

Power factor refers to when peak amps and peak volts do not match in the AC cycle. If they don't match in the cycle then measuring amps and voltage and multiplying them gives you 'apparent power' rather than real power. The power factor is expressed as the ratio of real power to apparent power. For pure resistive loads (e.g. old fashioned light bulbs), the load responds to the voltage wave instantly so apparent power is real power and power factor is 1. For inductive and some electronic loads, peak amps may 'lag' or 'lead' peak voltage and the power factor becomes a fraction of 1. The math is complicated.

apparent power * power factor = real power

A proper energy meter constantly monitors the relationship of amps and voltage continuously (or close enough) and so thus accounts for power factor.
 
160318-0941 EDT

michalspike:

What is the real question you are trying to ask?

Do you want try to check whether your power company meter is reading correctly, or is it some other question?

.
 
michalspike said:
I have three phase services: each leg drawn 20Amp( this is what its show on clamp meter) how I'm converting it? correct me if I'm wrong.
20x120v=2400W
2400X3= 7200w=7,2kw
Click to expand...

Since no one actually said it: The example that you give is a correct approximation. As others have said, you need to measure power factor to be more accurate, and you need to measure voltage and current simultaneously on all phases to be still more accurate.

michalspike said:
How I know how much is it in an hour?
Click to expand...

The conversion from power to energy requires multiplying by time. So say you run 7.2kW for 1/10 hour, you simply multiply to get 0.72 kWh. But with a single amp clamp you will only be able to get a very rough idea of what your energy consumption is; there are simply too many variables that you are not seeing with the tool.

-Jon
 
winnie said:
Since no one actually said it: The example that you give is a correct approximation. As others have said, you need to measure power factor to be more accurate, and you need to measure voltage and current simultaneously on all phases to be still more accurate.



The conversion from power to energy requires multiplying by time. So say you run 7.2kW for 1/10 hour, you simply multiply to get 0.72 kWh. But with a single amp clamp you will only be able to get a very rough idea of what your energy consumption is; there are simply too many variables that you are not seeing with the tool.

-Jon
Click to expand...

Why not just look at the bill?

Mine shows that, in addition to the $7 per month system access fee, I paid 13.73 cents per kWh last month. That's the total paid, including taxes.

I also learned how to read the new meters as they 'ticked off' the kWhs. That way I could turn on some of the loads and see the impact on the meter reading.
 
You're not going to get an accurate KWh reading with just amps. You must assume voltage remains constant, current remains constant and power factor remains 1.0

Not gonna happen. Even a ballpark figure won't be much good because current is seldom steady state. Loads turn on and off. You'd have to use a recording device with channels for 3 phase voltage, current and PF over the period of time (hours) you are comparing to. If it's the utility meter, you'd have to take a "snapshot" at the same time the meter was read. Pretty hard to do. If your meter is solid state, it's very accurate, which your setup wouldn't be. The only thing you might find close is max demand (KW), but even then, demand is averaged over a 15 min period. Can your ammeter do that?

Your calculation is correct, though. KWh is Volts X Amps X PF X 3 X time (3600 seconds) / 1000 if volts is measured line to neutral. If volts is line to line, you multiply by 1.732, since 120 X 1.732 is 208. KWh meters have a Kh or Ke factor, which is WATTHOURS per revolution or per pulse. A standard residential meter has a Kh of 7.2, so 1 KWh is 1000/7.2 revolutions. That means the meter has to rotate 138.8 times to measure 1 KWh. If your billed for several hundred KWh, you'll be there for a while......

I guess I don't really see what it is you are trying to do. KWh meters are pretty sophisticated nowadays, so I agree with others...go by your bill.
 
Sorry..I didn't see the post by jaggedben....

Sorry..I didn't see the post by jaggedben....

He said it better than I can. Don't confuse VA (Volts X Amps) with Watts. We all learned basic electricity using DC theory...then Westinghouse had to come along and convince the country to use AC and transformers (thank God). Edison wanted DC with wire as big as your leg for longer distance distribution. But then again, maybe we wouldn't have had to figure out what to do with all that "apparent power"! I'm SOOOOO... glad I'm out of college!
 
160318-2022 EDT

michalspike:

In post #7 I ask you to try to tell us what is your real question. You have not responded.

For some actual data see my website page http://beta-a2.com/energy.html .

The un-averaged graph, PE1, at my website has a resolution of 1 second. The averaged graphs are based on 1 hour periods. One, PE2, is for a single day. The others, PE3 and PE4, combine days.

A TED-1000 System was used for the measurements.

.
 
michalspike said:
I know.
But how to calculate if you have central air, plus electrical water heaters.
You cant plug this in
Click to expand...

Electric water heater (assuming resistance type heating element) is 1.0 power factor, you can get pretty close to actual energy used just from current reading and assuming voltage remains pretty constant, all you need is to know run time.

The central air however will have a power factor less then 1.0, the actual load level can also vary depending on conditions as well, some of those conditions can be temperature (inside and outside), air flow restritions (plugged air filters vs clean) or mechanical conditions.
 
I'm sorry I couldn't replay earlier because I was sick.
We have a job where client ask us to check real electric usage. He just want to make sure that he is not getting rip off or his meter is good.
I tried explain him that might be a true. Its a 3500SQ/foot apartment with 2 electric 80 gallons water heaters, central air etc.
How I can deal with this kindly way?
 
michalspike said:
I'm sorry I couldn't replay earlier because I was sick.
We have a job where client ask us to check real electric usage. He just want to make sure that he is not getting rip off or his meter is good.
I tried explain him that might be a true. Its a 3500SQ/foot apartment with 2 electric 80 gallons water heaters, central air etc.
How I can deal with this kindly way?
Click to expand...
POCO is not likely ripping him off. He sounds like he has a large enough space that the consumption may add up faster then one might think just depends on what there is and how it is used. If mechanical meter, if they are in error is likely in customer's favor. Since is an apartment - maybe make certain someone else isn't tied into your customer's line somehow if theft of service is a potential concern.
 
kwired said:
POCO is not likely ripping him off. He sounds like he has a large enough space that the consumption may add up faster then one might think just depends on what there is and how it is used. If mechanical meter, if they are in error is likely in customer's favor. Since is an apartment - maybe make certain someone else isn't tied into your customer's line somehow if theft of service is a potential concern.
Click to expand...

Theft of the services is not possible this is a new construction. Dedicated meter and disconnect switch everything locked and band .
 
michalspike said:
Theft of the services is not possible this is a new construction. Dedicated meter and disconnect switch everything locked and band .
Click to expand...
OK starting to make more sense now. Had a guy about 25 years ago that built a new house on a farm. Fairly large home, lots of windows and high cathedral ceilings, can't recall square footage, but in following years he complained how much it cost to heat it - we were like what do you expect with a large home and on top of that no matter how much insulation you put in walls or ceilings a lot of windows loses a lot of heat.

Same guy also complained about having mice in the garage. Every time you drove by that place at least one of the overhead doors would typically be open but he couldn't figure out how they get in there :happysad:.
 
I know its really hard to deal with some clients.
But I just don't want to be rude.
I thought there is a way to calculate it over a actually amperage usage.
But I see this is not easy as I thought.
 
michalspike said:
I know its really hard to deal with some clients.
But I just don't want to be rude.
I thought there is a way to calculate it over a actually amperage usage.
But I see this is not easy as I thought.
Click to expand...

As far as heating/cooling - talk to an HVAC guy that knows how to do heating and cooling load calculations. You are not going to get true results but you can get a comparison of your 3500 SF to a smaller say 500 -1000 SF space in similar weather conditions and hopefully that will help them understand some that it is going to take some energy to heat or cool a large space.

Same with the dual 80 gallon water heaters, they likley are there because there is some higher then usual demand for hot water. On top of that the water heaters just being in the space do lose some heat to the space - which increases load on the AC, but does lessen heating load during heating season.
 
michalspike said:
I know its really hard to deal with some clients.
But I just don't want to be rude.
I thought there is a way to calculate it over a actually amperage usage.
But I see this is not easy as I thought.
Click to expand...
No, it's not. In addition to the difficulties others have mentioned, I believe that perhaps the greatest difficulty is trying to show a movie with a snapshot. Even if you could get an accurate reading of kW at some moment in time, the story is told by what happens over time, not what is happening at any particular instant.

One quick and dirty test you can perform, though, is to turn off all his loads and make sure the meter is not still registering power consumption.
 
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