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# Calculate Electric Cost

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This is incorrect. Residential power meters do not measure reactive power -- ever. They only measure real power. The electric company only applies a surcharge to large industrial installations that consume large amounts of reactive power but in most cases these factories will install power factor correction to avoid a surcharge. So power factor is irrelevant to the question at hand.

(By the way, a refrigerator induction motor is not "entirely inductive". If it were, it could do no work. An induction motor is about 90% inductive only when disconnected from any load. Under load, the power is primarily resistive ranging from 50% to 90%, depending on the motor.)

About all you can say about power factor is that the actual maximum power used by the refrigerator will be some fraction (say 50% to 90%) of the volts times amps on the nameplate. If you plug in one of the watt meters suggested, it will give you the true, real power consumed that appears on your bill.

posted by JackFlash at 5:06 PM on November 13, 2011

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# Calculate Electric Cost

November 13, 2011 10:59 AM Subscribe

How much does my refrigerator cost to run per month? The info tag on the inside reads; 110-127 VAC, 60 HZ, 11.2 AMPS. It runs 24/7 and my NYC electric is charged at .1999 per KWH. Including the calculation in your answer would be appreciated. Thanks.

What other info would be required to make an educated guess?

posted by mbx at 11:17 AM on November 13, 2011

posted by mbx at 11:17 AM on November 13, 2011

Your refrigerator really should not be running 24/7. If it is, you should get it fixed.

The number of KWH something uses at a steady rate is easy to calculate: Watts = Volts * Amps. So at 110 VAC, 11.2 amps is 1232 Watts, or 1.232 KW. In one hour, it would use 1.232 KWH.

But as themel said, the true cost is impossible to calculate with the information you've given us. First, the fridge shouldn't be running full blast all the time. Second, when it is running full blast, it still may not use exactly the rated amount of power. Finally, I'm pretty sure everywhere I've lived, there have been different power billing rates at different times and for different usage tiers, so a flat price per KWH over the course of the year might not be right.

posted by primethyme at 11:19 AM on November 13, 2011 [1 favorite]

The number of KWH something uses at a steady rate is easy to calculate: Watts = Volts * Amps. So at 110 VAC, 11.2 amps is 1232 Watts, or 1.232 KW. In one hour, it would use 1.232 KWH.

But as themel said, the true cost is impossible to calculate with the information you've given us. First, the fridge shouldn't be running full blast all the time. Second, when it is running full blast, it still may not use exactly the rated amount of power. Finally, I'm pretty sure everywhere I've lived, there have been different power billing rates at different times and for different usage tiers, so a flat price per KWH over the course of the year might not be right.

posted by primethyme at 11:19 AM on November 13, 2011 [1 favorite]

Well, here in euro-country, we have mandatory energy efficiency labels on refrigerators that have to contain the annual energy consumption. There might be a US equivalent somewhere?

posted by themel at 11:23 AM on November 13, 2011

posted by themel at 11:23 AM on November 13, 2011

The easiest way to get this is to get a energy usage meter. As themel and primethyme correctly indicate, there are many variables here that prevent a useful educated guess. Another variable here that could conceivably change your billing is power factor since a fridge is (more or less) an entirely inductive load, with older fridges being even worse for power factor. If, for some reason, your only load at your residence was your fridge, your electric company would start to bill you more for consuming too much reactive power.

posted by saeculorum at 11:27 AM on November 13, 2011

posted by saeculorum at 11:27 AM on November 13, 2011

To elaborate: most heating and cooling devices don't actually RUN 24/7. The fridge has an on and off state which is triggered by an internal temperature sensor. When it goes above the set temperature, the thing turns on and starts pumping heat from the inside of the fridge to the outside (your kitchen.)

So that 11.2 amp number is likely the power draw when it is actively trying to cool the inside. IF it were actually running all the time, you would find the cost by:

Find power draw (in watts): Watts = Volts x Amps, so W=~120V x 11.2A = 1344 W = 1.3 kW

Then, kWH= W x Hours, so we find the hours/month: H=30 days x 24 hr/day = 720 H/Month

Then KWH/M = 1.3kW x 720H/M = 936 kWH/M

Then multiply by the billing rate (which IS variable on the season) to get a maximum cost/month of ~$187

Now hopefully this number seems high - it should be, because your fridge shouldn't be actively cooling all the time. To find the true cost, estimate the percentage of time its actively cooling.

* In general, heating and cooling appliances don't just blow air of exactly the desired temperature 100% of the time - they blow extra hot (ovens, central heat) or extra cold (fridge, a/c) air SOME of the time, and rely on insulation to maintain the temperature. (example: if you want to warm your house from 70 degrees to 75, turning up the heat to 85 won't make it heat up any faster!)

posted by Wulfhere at 11:28 AM on November 13, 2011 [1 favorite]

So that 11.2 amp number is likely the power draw when it is actively trying to cool the inside. IF it were actually running all the time, you would find the cost by:

Find power draw (in watts): Watts = Volts x Amps, so W=~120V x 11.2A = 1344 W = 1.3 kW

Then, kWH= W x Hours, so we find the hours/month: H=30 days x 24 hr/day = 720 H/Month

Then KWH/M = 1.3kW x 720H/M = 936 kWH/M

Then multiply by the billing rate (which IS variable on the season) to get a maximum cost/month of ~$187

Now hopefully this number seems high - it should be, because your fridge shouldn't be actively cooling all the time. To find the true cost, estimate the percentage of time its actively cooling.

* In general, heating and cooling appliances don't just blow air of exactly the desired temperature 100% of the time - they blow extra hot (ovens, central heat) or extra cold (fridge, a/c) air SOME of the time, and rely on insulation to maintain the temperature. (example: if you want to warm your house from 70 degrees to 75, turning up the heat to 85 won't make it heat up any faster!)

posted by Wulfhere at 11:28 AM on November 13, 2011 [1 favorite]

The rating on the plate is a maximum— it's there so you can know whether you'll overload your wiring (etc) with the fridge. So calculating based on it will give you a very generous upper bound on the cost. It's like calculating the cost of gas for your car based on its top speed and MPG and assuming it's being driven 100% of the time. At any rate, to continue primethyme's calculation: if it uses 1.232 kWh per hour, and there are about 730 hours in a month, at .1999 that's $180/month absolute max if you just left the fridge's door open all the time.

If you want to know how much it's

posted by hattifattener at 11:30 AM on November 13, 2011

If you want to know how much it's

*actually*charging you, pretty much have to meaure it. (Make sure the wattmeter you use can handle a 11.2-amp load!)posted by hattifattener at 11:30 AM on November 13, 2011

If you want to know for sure you should get a watt meter.

I used one of these to monitor how much ConEd is charging me to use random things* in my house. They aren't expensive and I know that J&R has them in stock.

*I now know that I pay $37.24 per year to run my alarm clock.

posted by ooklala at 11:31 AM on November 13, 2011

I used one of these to monitor how much ConEd is charging me to use random things* in my house. They aren't expensive and I know that J&R has them in stock.

*I now know that I pay $37.24 per year to run my alarm clock.

posted by ooklala at 11:31 AM on November 13, 2011

If your fridge was made before 2001 you should almost certainly trade it in - online calculator based on age/size

posted by Lanark at 11:34 AM on November 13, 2011

posted by Lanark at 11:34 AM on November 13, 2011

(Also, FWIW, my local library allows you to check out wattmeters— part of a city conservation program, I assume— you could check for some similar program in your area if you don't want to buy the meter outright.)

posted by hattifattener at 11:37 AM on November 13, 2011

posted by hattifattener at 11:37 AM on November 13, 2011

One thing to keep in mind is that your fridge will also require more or less cooling depending on how much is inside. If the fridge is empty, every time you open the door you are letting out a large amount of cool air, and replacing it with the warm air from your room. If it is filled to the brim, there is little air that can escape, and little space for warm air to fill it up, so you will lose less of your cold air every time it is opened when it is full.

posted by markblasco at 11:45 AM on November 13, 2011

posted by markblasco at 11:45 AM on November 13, 2011

Like mentioned above, measuring is always a good route.

But if you just want to get some estimate about roughly how much power your fridge uses you could make a few assumptions:

- That the temperature difference between the inside of your fridge and the outside is relatively constant

- That you can ignore the contents of the fridge (or assume what you have in it now is roughly the level of loading it will have on average).

Then you could just count the number of times your fridge turns on in some period of time (say an hour) and how long it is running for. This would allow you to calculate how much of the day your fridge is actually running then use the above maths to get an estimate of how much that is costing you.

Though it would be much easier to use a meter if you can get your hands on one (as opposed to hanging out in the kitchen with a stop watch).

posted by selenized at 12:18 PM on November 13, 2011

But if you just want to get some estimate about roughly how much power your fridge uses you could make a few assumptions:

- That the temperature difference between the inside of your fridge and the outside is relatively constant

- That you can ignore the contents of the fridge (or assume what you have in it now is roughly the level of loading it will have on average).

Then you could just count the number of times your fridge turns on in some period of time (say an hour) and how long it is running for. This would allow you to calculate how much of the day your fridge is actually running then use the above maths to get an estimate of how much that is costing you.

Though it would be much easier to use a meter if you can get your hands on one (as opposed to hanging out in the kitchen with a stop watch).

posted by selenized at 12:18 PM on November 13, 2011

I don't know if this helps, but I measured my fridge's consumption with one of those meters (left attached for 24 hours) recently, and I calculated it used $0.16 per day, so around $4.80 a month.

Mine is a Westinghouse 221, and super old - so old that the sticker with energy information has bleached out so you can't read it. Yours is likely to be more efficient (unless it's a lot bigger, or you live in a much hotter climate, or you leave it mostly empty). I am charged for electricity at a rate of $0.1515 per kWh.

posted by lollusc at 2:07 PM on November 13, 2011

Mine is a Westinghouse 221, and super old - so old that the sticker with energy information has bleached out so you can't read it. Yours is likely to be more efficient (unless it's a lot bigger, or you live in a much hotter climate, or you leave it mostly empty). I am charged for electricity at a rate of $0.1515 per kWh.

posted by lollusc at 2:07 PM on November 13, 2011

*Another variable here that could conceivably change your billing is power factor since a fridge is (more or less) an entirely inductive load, with older fridges being even worse for power factor. If, for some reason, your only load at your residence was your fridge, your electric company would start to bill you more for consuming too much reactive power.*

This is incorrect. Residential power meters do not measure reactive power -- ever. They only measure real power. The electric company only applies a surcharge to large industrial installations that consume large amounts of reactive power but in most cases these factories will install power factor correction to avoid a surcharge. So power factor is irrelevant to the question at hand.

(By the way, a refrigerator induction motor is not "entirely inductive". If it were, it could do no work. An induction motor is about 90% inductive only when disconnected from any load. Under load, the power is primarily resistive ranging from 50% to 90%, depending on the motor.)

About all you can say about power factor is that the actual maximum power used by the refrigerator will be some fraction (say 50% to 90%) of the volts times amps on the nameplate. If you plug in one of the watt meters suggested, it will give you the true, real power consumed that appears on your bill.

posted by JackFlash at 5:06 PM on November 13, 2011

This thread is closed to new comments.

posted by themel at 11:10 AM on November 13, 2011