Killowatt/hours Conversion
- Thread starter DEC01
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1) There is none. You could have a 100 W lightbulb on the whole month and have a peak load of 0.1 kW with 72 kWh energy. You could have the 100 W lightbulb on for only one hour in the month and have a peak load of 0.1 kW with 0.1 kWh energy.DEC01 said:
2) Not unless you have a demand charge on your utility bill.
1) The Utility bill will show the total kWh for the month.
2) The Utility will often base part of their fee on the peak kW in a month. Some times it's called Billing Demand or Capacity Charge. They also often show a 12-month history on the bill of the total kWh and peak kW.
If all that the Utility bill shows is the total kWh then there's no good rule of thumb way to estimate the peak kW. Occupancy type (school, office, industrial), owner habits and month will vary any estimates widely.
Here's an example from one Utility bill I have on file.
kWh kW Ratio
Jan: 27,505___54___509
Feb: 23,062___64___360
Mar: 26,433___58___455
Apr: 23,022___56___411
May: 10,114__32___316
Jun: 8,431____31___271
Jul: 8,479_____28___302
Aug: 8,226____29___283
Sep: 9,177____29___316
Oct: 10,178___31___328
Nov: 11,176___33___338
Dec: 23,493___56___419
With the ratio changing monthly from 271 to 509, any estimate on peak demand based on the total kWh for a month can fluctuate by 90% for this customer. I suggest going thru the Utility bill looking for anything that might be the peak kW. Also, don't be scared to call the Utility. You might be transfered 3-4 times before you get to a person that can help you, but it's better then guessing.
2) The Utility will often base part of their fee on the peak kW in a month. Some times it's called Billing Demand or Capacity Charge. They also often show a 12-month history on the bill of the total kWh and peak kW.
If all that the Utility bill shows is the total kWh then there's no good rule of thumb way to estimate the peak kW. Occupancy type (school, office, industrial), owner habits and month will vary any estimates widely.
Here's an example from one Utility bill I have on file.
kWh kW Ratio
Jan: 27,505___54___509
Feb: 23,062___64___360
Mar: 26,433___58___455
Apr: 23,022___56___411
May: 10,114__32___316
Jun: 8,431____31___271
Jul: 8,479_____28___302
Aug: 8,226____29___283
Sep: 9,177____29___316
Oct: 10,178___31___328
Nov: 11,176___33___338
Dec: 23,493___56___419
With the ratio changing monthly from 271 to 509, any estimate on peak demand based on the total kWh for a month can fluctuate by 90% for this customer. I suggest going thru the Utility bill looking for anything that might be the peak kW. Also, don't be scared to call the Utility. You might be transfered 3-4 times before you get to a person that can help you, but it's better then guessing.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
080716-1613 EST
DEC01:
You need to tell us why you want to known peak power.
Power, watts, horsepower, is the rate of doing work (expending energy). This is equivalent to how fast you drive.
Energy is the integral of power over time (integration is the summing of small increments over some interval). The distance you travel in a certain amount of time is the integral of speed over that time period. Some of the units of energy are BTU, KWH, watt-hours, joules.
My electric bill for last month showed 1147 KWH. My average residual load at night I estimate at 100 to 200 watts. Assume 7 hours for night and 200 watts, or 1.4 KWH and 30 days. Thus my daytime average KWH load is (1147 - 42)/30 = 1105/30 = 36.8 KWH/day excluding nighttime. Or 2.16 KW average power usage during the day. For short times there will be peaks of maybe 8 to 10 KW.
If I wanted to use wind power I might design for 100 KWH of energy storage, and generator capability of maybe 50 KWH/day. So the generator might be 6 to 10 KW capability. A single 12 V 80 AH car battery can store about 0.96 KWH. If these were used I would need 100 batteries. Just on that basis it is not cost effective. We have a local builder demonstrating wind power at his home, but no battery storage for cost and maintenance reasons. His energy storage is the grid and he only generates part of his energy needs. I believe his is a 2 KW generator at 10 MPH wind speed and a 60 to 70 ft pole.
If you want to estimate peak power, then put a recording wattmeter on the input lines, and monitor this for a month. Or you can probably estimate by an experiment of turning everything on in the house and read both input currents. You might assume a 0.95 power factor. This worst case of everything on probably would be higher than the strip chart recorder value.
.
DEC01:
You need to tell us why you want to known peak power.
Power, watts, horsepower, is the rate of doing work (expending energy). This is equivalent to how fast you drive.
Energy is the integral of power over time (integration is the summing of small increments over some interval). The distance you travel in a certain amount of time is the integral of speed over that time period. Some of the units of energy are BTU, KWH, watt-hours, joules.
My electric bill for last month showed 1147 KWH. My average residual load at night I estimate at 100 to 200 watts. Assume 7 hours for night and 200 watts, or 1.4 KWH and 30 days. Thus my daytime average KWH load is (1147 - 42)/30 = 1105/30 = 36.8 KWH/day excluding nighttime. Or 2.16 KW average power usage during the day. For short times there will be peaks of maybe 8 to 10 KW.
If I wanted to use wind power I might design for 100 KWH of energy storage, and generator capability of maybe 50 KWH/day. So the generator might be 6 to 10 KW capability. A single 12 V 80 AH car battery can store about 0.96 KWH. If these were used I would need 100 batteries. Just on that basis it is not cost effective. We have a local builder demonstrating wind power at his home, but no battery storage for cost and maintenance reasons. His energy storage is the grid and he only generates part of his energy needs. I believe his is a 2 KW generator at 10 MPH wind speed and a 60 to 70 ft pole.
If you want to estimate peak power, then put a recording wattmeter on the input lines, and monitor this for a month. Or you can probably estimate by an experiment of turning everything on in the house and read both input currents. You might assume a 0.95 power factor. This worst case of everything on probably would be higher than the strip chart recorder value.
.
tallgirl
Senior Member
- Location
- Glendale, WI
- Occupation
- Controls Systems firmware engineer
The best answer depends on why you want to know the answer to the question in the first place.
As gar wrote above, if you're trying to size for some sort of standby power, well, things get complicated.
What gar wrote about knowing overnight loads, then computing daytime averages is a good start. Beyond that, you can start with known loads and estimated duty cycles -- a fridge is so many watts, and runs such-and-such percent. For example, my fridge draws about 250 watts and runs about half the time. That gives 125 watts times 24 hours times 30 days, or 90KWH per month. My old A/C unit ran 8 hours per month per average degree above 72F, or a total of 14 degrees (86F - 72F) * 8 hours * 3.5KW for last month when the average temperature was 86F, giving a total of 392KWH. I don't know what the new A/C unit draws, but if I knew that, I could give you that estimated usage as well. And so on, going through each load.
There's another approach and that's a gadget called a "Kill-A-Watt". It's a device you plug into the wall, then plug the appliance or whatever into. It reads out in kilowatt hours, kilo-volt-amp hours, and so on.
As gar wrote above, if you're trying to size for some sort of standby power, well, things get complicated.
What gar wrote about knowing overnight loads, then computing daytime averages is a good start. Beyond that, you can start with known loads and estimated duty cycles -- a fridge is so many watts, and runs such-and-such percent. For example, my fridge draws about 250 watts and runs about half the time. That gives 125 watts times 24 hours times 30 days, or 90KWH per month. My old A/C unit ran 8 hours per month per average degree above 72F, or a total of 14 degrees (86F - 72F) * 8 hours * 3.5KW for last month when the average temperature was 86F, giving a total of 392KWH. I don't know what the new A/C unit draws, but if I knew that, I could give you that estimated usage as well. And so on, going through each load.
There's another approach and that's a gadget called a "Kill-A-Watt". It's a device you plug into the wall, then plug the appliance or whatever into. It reads out in kilowatt hours, kilo-volt-amp hours, and so on.
charlie
Senior Member
- Location
- Indianapolis
With everyone's comments on target, I hate to pick on you.rattus said:
W = P x T is just about right. Wh or what is normally used, kWh (thousands of watt hours) = P (in kilowatts X T (time in hours)
We use the energy usage of customers in a formula that we have derived from testing to get a rough idea of the demand on our transformers. We use this over a four month period in the summer and a different four month period in the winter. Since we use test data and generally have multiple customers on each transformer, we can guess at the loading on our transformers and be fairly accurate. There is no way to predict with any accuracy what the load is for a single customer.
tallgirl
Senior Member
- Location
- Glendale, WI
- Occupation
- Controls Systems firmware engineer
charlie said:We use the energy usage of customers in a formula that we have derived from testing to get a rough idea of the demand on our transformers. We use this over a four month period in the summer and a different four month period in the winter. Since we use test data and generally have multiple customers on each transformer, we can guess at the loading on our transformers and be fairly accurate. There is no way to predict with any accuracy what the load is for a single customer.
I must be driving the guys at TXU Energy absolutely batty. July '06 I used 1800KWH, July '07 I used 970KWH, and it looks like about 540KWH for July '08.
Charlie, where you been?
Charlie, where you been?
Charlie, "W" is commonly used as the symbol for "WORK" (ENERGY) and is consistent with "P" for "POWER" and "T" for "TIME". This is a generic formula which does not specify units.
Strictly speaking, the use of "Wh" is incorrect because the generic symbols "P" and "T" represent physical effects not units of measure.
What I am trying to say is that formulas should be written in terms of generic symbols, not units of measure.
Charlie, where you been?
charlie said:With everyone's comments on target, I hate to pick on you.
W = P x T is just about right. Wh or what is normally used, kWh (thousands of watt hours) = P (in kilowatts X T (time in hours)
Charlie, "W" is commonly used as the symbol for "WORK" (ENERGY) and is consistent with "P" for "POWER" and "T" for "TIME". This is a generic formula which does not specify units.
Strictly speaking, the use of "Wh" is incorrect because the generic symbols "P" and "T" represent physical effects not units of measure.
What I am trying to say is that formulas should be written in terms of generic symbols, not units of measure.
charlie
Senior Member
- Location
- Indianapolis
Sorry, I was learned that W = watts. With your explanation, I concur but I will still hold on to kWh which you will find on the face of almost every electric utility meter.rattus said:
True:
True:
Charlie,
Right, "KWh" is the unit of "WORK" done or "ENERGY" transferred.
It is a bit confusing that sometimes physical quantities and units of measure use the same symbols.
True:
charlie said:Sorry, I was learned that W = watts. With your explanation, I concur but I will still hold on to kWh which you will find on the face of almost every electric utility meter.
Charlie,
Right, "KWh" is the unit of "WORK" done or "ENERGY" transferred.
It is a bit confusing that sometimes physical quantities and units of measure use the same symbols.
quogueelectric
Senior Member
- Location
- new york
Dont forget to double or more your bill for the oil surcharge.DEC01 said:
Thank you all so much for all your thoughts and input.
I was asked this question by a freind and collegue and didn't ask him at the the time his purpose for knowing. Therefore I'm not in a position at this time to answer the why's or what fors. I do appreciate the quick and thoughtful insight though. I'll pass it along. Thank you!
I was asked this question by a freind and collegue and didn't ask him at the the time his purpose for knowing. Therefore I'm not in a position at this time to answer the why's or what fors. I do appreciate the quick and thoughtful insight though. I'll pass it along. Thank you!
weressl
Esteemed Member
DEC01 said:
1./ Kilowatt is a magnitude and kilowatthour is quantity.
2./ Not from the data supplied. There are demand registering kilowatthour meters and the demand is usually measured within a predetermined demand period, say 15 or 30 minutes. It actually does not measure the peak momentary kilowatt but rather the average in the deman period. It will measure the kWhrs used in the erpiod and divide it by 4, 15 min. being 1/4th of an hour and so on.
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