# Calculate daily KW usage?

July 17, 2009 10:35 PM Subscribe

If I know my current power usage, how do I calculate my daily kw?

I'm monitoring my power and have a CurrentCost device providing current KW usage.

I'll sample the usage every minute, and store it in a db, but how do I calculate my totaly daily usage?

Use the average kw per hour and add them up for the day?

(this feels like a question with an obvious answer and I'm just not getting it today)

I'm monitoring my power and have a CurrentCost device providing current KW usage.

I'll sample the usage every minute, and store it in a db, but how do I calculate my totaly daily usage?

Use the average kw per hour and add them up for the day?

(this feels like a question with an obvious answer and I'm just not getting it today)

Maybe I'm unclear on what a CurrentCost device is. Won't it directly output in kWhr? In that case you just leave it on for however long you want, take the figure it gives you in kWhr and divide by the number of days you had it running. The more days you leave it running the more accurate your result will be. I'd go at least a week since power consumption will vary based on what day of the week it is.

It will also vary based on what month it is and how it varies will depend on where you live. If you're in hot climate with mild winters and scorching summers your power consumption will be much higher in summer (assuming you use AC). If you live in a northerly clime with mild summers and frigid winter your consumption will be much higher in winter.

So the best way to do it depends on why you want the figure. But at a minimum I'd calculate average consumption per day for a week and then decide if you want to do it again at various times of the year.

posted by Justinian at 10:45 PM on July 17, 2009

It will also vary based on what month it is and how it varies will depend on where you live. If you're in hot climate with mild winters and scorching summers your power consumption will be much higher in summer (assuming you use AC). If you live in a northerly clime with mild summers and frigid winter your consumption will be much higher in winter.

So the best way to do it depends on why you want the figure. But at a minimum I'd calculate average consumption per day for a week and then decide if you want to do it again at various times of the year.

posted by Justinian at 10:45 PM on July 17, 2009

I googled CurrentCost device and a quick glance says that it does already output in kwWhr.

CurrentCost Classic.

Unless I'm making some sort of obvious error, you just leave it going for like a week and then divide the figure it gives you in kWhr by 7 days. Or leave it on for 30 days and divide by 30 days. You get the idea.

posted by Justinian at 10:47 PM on July 17, 2009

CurrentCost Classic.

Unless I'm making some sort of obvious error, you just leave it going for like a week and then divide the figure it gives you in kWhr by 7 days. Or leave it on for 30 days and divide by 30 days. You get the idea.

posted by Justinian at 10:47 PM on July 17, 2009

Our electricity bill shows usage patterns per day and per month in a little graph at the bottom of the monthly statement. Maybe yours does too?

posted by amyms at 10:59 PM on July 17, 2009

posted by amyms at 10:59 PM on July 17, 2009

*Use the average kw per hour and add them up for the day?*

"Kilowatt per hour" is a nonsensical unit. A watt (power) is a joule (energy) per second (time). A kilowatt is one thousand watts. A kilowatt-hour (power x time) is a unit of energy, equal to 3,600,000 joules.

To calculate your average power usage, find the amount of energy you use over a given time period and divide it by that time.

*Unless I'm making some sort of obvious error, you just leave it going for like a week and then divide the figure it gives you in kWhr by 7 days.*

Well, you would divide it by the number of hours in seven days (24 x 7 = 168), yielding your average power usage in units of kilowatts.

posted by mr_roboto at 11:10 PM on July 17, 2009

Simple explanation: If a kW is analogous to a mile/hour (speed), then a kWh is analogous to a mile (distance travelled).

If you're sampling kW, and you want to find out the total amout of energy used, find the average power used (in kW) and multiply by 24 hours. For example, if you use 0.8 kW on average, you would use 0.8 kW * 24 hr = 19.2 kWh in one day.

posted by alexei at 11:59 PM on July 17, 2009 [1 favorite]

If you're sampling kW, and you want to find out the total amout of energy used, find the average power used (in kW) and multiply by 24 hours. For example, if you use 0.8 kW on average, you would use 0.8 kW * 24 hr = 19.2 kWh in one day.

posted by alexei at 11:59 PM on July 17, 2009 [1 favorite]

Sorry I should have explained.

Yes CurrentCost devices do output the data in kw/h for days, weeks and months, but I don;t really care for the device's display. It's tethered for one thing, and displays cost in UKP, I don't live in the UK.

It does however output xml data via a serial port.

I'm caputing that data for analysis.

Also, when I said "Use the average kw per hour and add them up for the day?", I should have been clearer.

I should perhaps of said "Use the average of the samples (kw) in an hour, and sum each hour's average, for each day?"

Thanks everyone. Some really useful answers.

posted by matholio at 12:14 AM on July 18, 2009

It't not kW/hr, it's kWhr — it's the amount of energy you use if you run a 1-kilowatt device for one hour. It's not kilowatts per hour.

posted by hattifattener at 12:27 AM on July 18, 2009

posted by hattifattener at 12:27 AM on July 18, 2009

kW is the amount of power your using. Your daily kW is just the average of all the measurements.

1 Watt = 1 Joule of energy per second.

Imagine a hose, which outputs one gallon per second. If you want to know how much water flows through the hose in a day, you just multiple by the number of seconds in a day, which is 86400.

It's the same with electricity. You just multiply the wattage by the number of seconds in the day to find out how many Joules of energy your using. But you probably want kWh, which is how your electricity bill is usually figured.

A kilowatt hour is the equivalent to using 1 kilowatt for 1 hour, (or using 1 watt for a thousand hours).

Think about it this way: You have a measure of something over time, and then you multiply time again. So you have time*energy/time = energy.

so just like a Joule, a kWh is a measure of energy. The ratio is the same as the ratio of 1 second to 1000 hours, or 3,600,000

--

So to summarize. Just average your energy use over the day, and then multiply by the amount of time you've been calculating the average.

posted by delmoi at 2:59 AM on July 18, 2009

1 Watt = 1 Joule of energy per second.

Imagine a hose, which outputs one gallon per second. If you want to know how much water flows through the hose in a day, you just multiple by the number of seconds in a day, which is 86400.

It's the same with electricity. You just multiply the wattage by the number of seconds in the day to find out how many Joules of energy your using. But you probably want kWh, which is how your electricity bill is usually figured.

A kilowatt hour is the equivalent to using 1 kilowatt for 1 hour, (or using 1 watt for a thousand hours).

Think about it this way: You have a measure of something over time, and then you multiply time again. So you have time*energy/time = energy.

so just like a Joule, a kWh is a measure of energy. The ratio is the same as the ratio of 1 second to 1000 hours, or 3,600,000

--

So to summarize. Just average your energy use over the day, and then multiply by the amount of time you've been calculating the average.

posted by delmoi at 2:59 AM on July 18, 2009

If you're sampling power usage, in kilowatts, every minute, what you can do is assume the power usage was about the same for that whole minute. So the numbers you get can be interpreted as energy consumption in kilowatt-minute (e.g. 4 kilowatts for 1 minute = 4 kilowatt-minutes).

Just divide these by 60 (to get kilowatt-hours) and then add them all up over the period of interest (1 day) and you'll have your answer in kilowatt-hours. That's the unit the electric company probably uses to bill you, so I'm pretty sure that's what you're trying to calculate.

This won't be quite as accurate as an electric meter, and may not be as good as a device that reads in kilowatt-hours directly (although some of them use exactly this algorithm, too). But it'll be pretty close.

Beware that if any of the once per minute samples are missing, the final number will be too low. If that can potentially happen, there's a way to adjust for it. In one day, you should have 1440 samples. Let's say 10 are missing, so you only have 1430. Just multiply the final result by 1440 then divide it by 1430.

For any other time period, it's always just "multiply by expected number of samples" then "divide by actual number of samples". This should work even if you somehow ended up with too many samples in a day, as well.

posted by FishBike at 5:56 AM on July 18, 2009

Just divide these by 60 (to get kilowatt-hours) and then add them all up over the period of interest (1 day) and you'll have your answer in kilowatt-hours. That's the unit the electric company probably uses to bill you, so I'm pretty sure that's what you're trying to calculate.

This won't be quite as accurate as an electric meter, and may not be as good as a device that reads in kilowatt-hours directly (although some of them use exactly this algorithm, too). But it'll be pretty close.

Beware that if any of the once per minute samples are missing, the final number will be too low. If that can potentially happen, there's a way to adjust for it. In one day, you should have 1440 samples. Let's say 10 are missing, so you only have 1430. Just multiply the final result by 1440 then divide it by 1430.

For any other time period, it's always just "multiply by expected number of samples" then "divide by actual number of samples". This should work even if you somehow ended up with too many samples in a day, as well.

posted by FishBike at 5:56 AM on July 18, 2009

Maybe I'm over simplifying:

A kilowatt is 1000 watts. So, a 100 watt lightbulb is .1 kw. 10 100 watt lightbulbs is 1kw. If you run those 10 lightbulbs for an hour, you have consumed 1 kwh of energy.

The question is, is the device outputting an instantaneous reading, or is it outputting a total of usage for the sample period? If it's summing up total usage for each minute, you should have a fairly accurate answer.

But if it's just an instantaneous sample, you end up with gaps in the data. Assuming it is accurate to the second, for each minute, you will know your correct usage for at one second, but you won't know what's been going on for those other 59 seconds. I don't know how meaningful this inaccuracy is in the real world.

posted by gjc at 7:11 AM on July 18, 2009

A kilowatt is 1000 watts. So, a 100 watt lightbulb is .1 kw. 10 100 watt lightbulbs is 1kw. If you run those 10 lightbulbs for an hour, you have consumed 1 kwh of energy.

The question is, is the device outputting an instantaneous reading, or is it outputting a total of usage for the sample period? If it's summing up total usage for each minute, you should have a fairly accurate answer.

But if it's just an instantaneous sample, you end up with gaps in the data. Assuming it is accurate to the second, for each minute, you will know your correct usage for at one second, but you won't know what's been going on for those other 59 seconds. I don't know how meaningful this inaccuracy is in the real world.

posted by gjc at 7:11 AM on July 18, 2009

*Yes CurrentCost devices do output the data in kw/h for days, weeks and months, but I don;t really care for the device's display.*

I'd be willing to bet cash money that the device

**does not**display the data in kilowatts per hour. Kilowatts per hour is a nonsensical unit. It probably displays energy used in units of kilowatt-hours (kilowatts x hours, i.e. power x time = energy).

posted by mr_roboto at 10:21 AM on July 18, 2009

This thread is closed to new comments.

posted by zippy at 10:40 PM on July 17, 2009