Electricity price

I want to share some of my shop electricity cost with my brother. since he lives on the properity, and pays all the electric bill. So If a 3hp 22v 1750 rpm compressor runs for about one hour solid. What would it cost? I know every state has different prices for electricity. I just need a average of some sort. We probable live in one of the most expensive states, Calif Stan

Well, duh!! Don't ask us. Read your electric bill or telephone the utility company, they'll tell you. Also don't forget to figure in so called "peak" or "demand" times.

Electric rates range from about 8 cents to 16 cents per kilowatt hour. Maybe more or less depending on area of country. Amps x volts divided by 1000 equals kilowatts. Find the running amps from the tag on the electric motor, multiply x 240 volts x power factor 1.0 for single phase. Unless you're not in the US, the utility thinks 240 not 220 volts. Sometimes, it isn't just the kilowatts, its all the taxes and add ons that run the bill up. So just picking out one compressor might not be the most neighborly.

Well you really need the real amps it draws. (not the spec, the fact. This will be several numbers because the amps change as the load goes up. So once that is determined amps times volts equals watts. figure the average watts for the run time, then turn that into a 1000 watt Hour number. Then divide that into your electric rate per KWhr. And charge him $1.45 per hour (just joking on the amount it is probably less) JimN

For ballpark figuring, that 3HP motor would draw about 3KW per hour on three phase. And 4KW per hour on single phase.
So in addition to the monthly base rate for being connected to the grid. Plus any delivery etc charges.The motor will cost 80 cents per hour at 20 cents per KW plus.........
Seems expensive but try running a 3HP engine on gasoline or diesel for an hour and figure the cost(s).

In round figures, the motor LIKELY draws about 3000 Watts.

Run steady for one hour, 3 KHW.

3 KWH @ $.08/KWH = $0.24.

3 KWH @ $.16/KWH = $0.48.

(quoted from post at 19:52:29 12/13/09) Well you really need the real amps it draws. (not the spec, the fact. This will be several numbers because the amps change as the load goes up. So once that is determined amps times volts equals watts. figure the average watts for the run time, then turn that into a 1000 watt Hour number. Then divide that into your electric rate per KWhr. And charge him $1.45 per hour (just joking on the amount it is probably less) JimN

Click to expand...

None of this correct. AC motors must be measured in true watts, Google "power factor" or "Ac motor theory." the only time AC POWER (watts) is equall to I X E as it is in DC circuits is when you have a purely resistive load, such as an electric heater.

Some estimate of motor wattage use CAN indeed be had by figuring DC power (I X E) and then just applying a "blanket" power factor of 80%

I'll dare to place a little bet that my round numbers are very close.

What will people ask next ???????????????
when they start asking about nnalert with your women, that's when it gets interesting,

Ya can find lots of different people on this forum --- like any other, they need to write Dear Abby,,,

I have a rental that we share the electricty at. I would have just install a meter for my share but there are other issues. Not to mention the well pump wiring is located in the shop. I guessing if you like me you generally do not work in the dark, so there are lights involved. I keep a radio on most of the time I am in there. Grinders, drills, and such all use their share. Then there's the deleivery charge (ours is seperate from the cost of the electricity).

I set it up so that I would front the minimum charge (40 bucks). That way if I used nothing and had the service myself, I would be paying that minimumn amount anyway. If I happen to be doing a lot of welding over a weekend, the 40 bucks at 10 cent a KW equates to 400KWHs of electricity. I have exceed that maybe one time, building a cow feeder, but figure all the others more than made up for it. Renter agreed and turned down my offer to kick in a little more.

Just isn't worth the expense (meter and wiring) trying to be any closer. But it helps that I know what the usage should be in that house and can guess within 50KWH what the bill will read any given month (lived in it for 7 years).

WOW the guy asked a simple question and Id think if a person provided him a well thought out "ballpark estimate" people should be pleased.....

The ONLY way to give a "perfect" answer would be to measure the actual load and actual current draw and then see how the utility charges including the KWH rate and any poor power factor penalties and time of use (peak times may charge more). I view the utility as billing for Killowatt Hours NOTTTTTTTT Killo Volt Amp Hours HAVE TO ASK THEM THAT IM NOTTTTTTTTT SPECULATING ANY ANSWER HERE. (arent those things outside our house called Watt Hour meters instead of Volt Amp Hour meters???)

I see the way Janicholson and the buick man (he works in this field you know) approached the problem AS ACCURATE (considering the question given allllllll the unknowns such as power factor and any penalties and KWH rate etc)

Where I worked as an engineer we were charged for KWH use (amps x volts x time, it was not Killo Volt Amp Hours) but had an additional penalty if we were running a lousy power factor.

That whole Power factor thing is too complex to set out here in a few paragraphs what takes a book to fill and beyond what the poster asked in my opinion. Theres power and apparent power and Power Factor has to do with the degree of lag angle between the voltage and current. If the load is pure resisitve, current and voltage are in phase and if P = E x I x Cosine of the phase angle and the cosine of 0 = 1 THEN ITS A UNITY 1 POWER FACTOR (Volt x Amps = Watt)

HOWEVER a motor is an inductive load and some of the energy the utility provided goes to creating a magnetic field in the motor windings but you get no actual "work" out of it. The utility has to provide the amps x volts (watts) to make the motor turn (thats work). In the motor inductive load the current lags the voltage so theres a phase angle between the voltage and current legs and if P = E x I x Cosine of that phase angle THE POWER FACTOR IS NOTTTTTTTT GOING TO EQUAL ONE and the utility may or may not charge differently HAVE TO ASK THEM... Theres still the Voltage and Current BUTTTTTTTT the phase angle between them changes and thats what power factor is all about.....


Sooooooooo my answer would be similar to Buicks and Janicholsons Id ESTIMATE the volts x amp draw (thats ONLY an estimate, we dont know the load mind you) and estimate the running time and convert that to KWH and see what the KWH charge is and that would be my answer. Then if that wasnt good enough for the poster he can see if and how the utility deals with power factor AND IF theres any penalty and/or if they bill for KWH or KVAH etc BUT I BET BUICK AND JANICHOLSONS ANSWERS WOULD BE REASONABLY CLOSE

God Bless yall, this may help or may confuse lol power factor is beyond the posters question but KWH and the utilitys charge for a KWH he can likely comprehend and REGARDLESS I bet Buciks answer will be darn close.......

An early Merry Christmas

John T (Tooooo long retired and rusty EE to explain all this stuff as well as Buick could)

I was avoiding a high science answer that served no one well. The physics answer if actually computed with certified measuring equipment would cost 5 years worth of the consumed/disputed electricity. Watts the matter with that? Jim

I agree, see my response above, to paraphrase you WATTS all the fuss

John T

How do you work your numbers without allowing for efficiency and power factor?No mention of three or single phase either.
My rough numbers are very close for use by non electrical personal.

As a retired physics teacher I would use the KISS theory. Keep It Simple Stupid. Most electric meters today are digital. Turn everything off except for your compressor. Run your compressor for an hour and simply read the energy you use. Take your monthly electric bill, divide the total cost by the kw-hrs and that is your cost per kw-hr. Multiply your cost times the energy your compressor uses in an hour. Is this Simple enough?