I want to know the rate people use the energy (fuel, wood, Gas Or LNG, Electrical stove).
125373
Not quite the answer you are looking for but, there is a great article in this month's Popular Mechanics about the truth about BioFuels.  It has a great comparison of the above...
674514
its some type of energy thing for some
625746
Probably you would not notice much of a difference. In fact if you used electric heat you wouln't notice any difference at all (at least in the winter). The reason is the amount of electricity an appliance uses is pretty much proportional to the amount of heat it generates. If you really want to save electricity get rid of your oven, refrigerator, toaster, hair dryer, basically any appliance which generates a lot of heat.
138820
kWh is NOT Kilowatts per hour. It is Kilowatt Hours and is the measure of the amount of energy drawn in one hour hour if it is drawn at a rate of one Kilowatt.\n\nKilowatts are a measure of Energy per unit time ie. Joules/s so kWh are J/s x 3600 s ie. measured in Joules.\n\nThe peak is just the highest energy usage rate in a given time period
557533
This website might help: http://www.worldenergy.org/wec-geis/edc/open.plx?file=default/default.htm \nIt has LOTS of info on which countries use which resources to get energy. Hope it helps!
651676
i gues.....um.......Electricity
211413
approx. 1600000000Kw hrs of electricity per hour you kept it on\n\nthat would cost a lot of money by the way and you might blow a fuse
572706
Power = work done per unit of time.\n\nP=W/T
340879
If you mean overall efficiency, that is defined as the ratio of energy out to energy in.  Just put a fixed load across the fuel cell (a resistor), log the voltage over time and the mass consumption of hydrogen.  \n\nThe gross heat of combustion of hydrogen is 142 kJ/gm.  This is also called the upper heat.  It accounts for the energy resulting from condensation of the water produced.  Since the water is in liquid form this is correct.  \n\nYou do need to select a load resistor that is well matched to the cell to get maximum power out.  Call this resistor R.  \n\nTo get the efficiency, square all the voltage readings, then add them up, divide by the resistance and multiply by the time between samples.  This gives the total energy out of the cell in joules (if resistance is in ohms and time is in seconds).  \n\nTake the grams of hydrogen used and multiply by the 142000 joule/gm.  This gives the energy in.\n\nDivide energy out by energy in to get the overall efficiency of the cell.  You should expect a number around 50%.
179570
I assume you mean electrolysis?  I don't know the exact ratio of energy out to energy in, but I do know that it's not very efficient.  There will be a lot of lost energy.  What kind of H2 on demand system?  Are you trying to run an automobile?  If so, you'd need more than about 10 kilowatts of electricity.  Might as well just build an electric car if you have that much electricity to waste.
276488
kilowatt it 1000 watts of power, electrical power. not gas. ur converting apples to oranges cant b done
448177
You measure the amount of fuel put in.  Using the flow of that fuel (cubic feet per hour), you then determine the amount of energy in each cubic foot of that fuel (or look it up - Natural gas is 1020 btu/cubic foot).  Then measure the amount of heat given off and the amount of electric produced.  \n\nEnergy out (Btu's) / Energy in (Btu's) = efficiency.  \n\nYou need to convert the electric out to btu's by multiplying the kW to Btu's (multiply by 3412).  With the heat given off, you need to determine the temperature of the exhaust and the amount of air flow of that temperature and that will give you the thermal side of the output.
65866
This really depends on your age, weight, height and gender.\n\nYou may find BMR (Basal Metabolic Rate) & AMR (Active Metabolic Rate) calculators helpful to find your Total Energy Requirement.\n\nhttp://www.preventdisease.com/healthtools/articles/bmr.html
360713
I kept reading anyway.\n\nThe heater is already telling you the power consumption.\n900W = 0.9 kW\nTherefore running the heater for one hour uses 0.9 kWH.\n\nYour formula is wrong. Watt/1000  * cost/kWH = cost per hour\n\nFor heaters, it is true that Volts * Amps = Watts.\nTherefore if you know the supply is 120V and the power is 900W, you can calculate that the current load is 7.5 A.
580515
This is a very good question.  Basically, the reason that we say kW and do not say kW/hr, is that calories and British Thermal Units (BTUs) are units of ENERGY, while the Watt is a unit of POWER.  The unit of time is already incorporated in the Watt.\n\nPOWER is defined as the rate at which WORK is done.  WORK is, in general, a transfer of ENERGY.  So POWER is the rate at which ENERGY is transfered from one object (or system) to another.\n\nThe capacity of a burner is stated in terms of POWER, that is, the rate at which it can transfer energy from the fuel (chemical potential energy) to the objects being heated.\n\nA Watt is defined as 1 Joule / second, (the Joule (J) is the basic SI unit of ENERGY).  1 Watt (again, a unit of POWER) is therefore equivalent to 3600 J/hr, since there are 60*60 = 3600 seconds in one hour.\n\n-------------------------\n\nFor your example, the power rating of the burner is 30kW.\n\nPower = 30kW = (30,000 W)*[(3600 J/hr)/(1 W)] = 108,000,000 J/hr = 108MJ/hr\n\nWe now use two conversion factors:\n\n1 kilocalorie = 4184 joules\n1 BTU = 1055.056 joules\n\nUsing these equivalencies, we can calculate the power in kcal/hr or BTU/hr:\n\n(108,000,000 J/hr)*[(1 kcal)/(4184 J)] = 25,812.6 kcal/hr\n       (this is pretty close to 25,800 kcal/hr stated on the burner)\n\n(108,000,000 J/hr)*[(1 BTU)/(1055.05585 J)] = 102,364.2 BTU/hr\n\n-------------------------\n\nYou stated that you are using Natural Gas with calorific value 40MJ/m^3.  (I found 38MJ/m^3 on Wikipedia, but this rounds to 40, so I'll go with your number for the example.)\n\nWe have:\n40MJ/m^3 = 9560.2 kcal/m^3  (using the above conversion factor)\n\nSo the consumption of Natural Gas is\n\n(25,812.6 kcal/hr)/(9560.2 kcal/m^3) = 2.7 m^3/hr\n\n(This rounds up to 3 m^3/hr, so your answer above is correct.)\n\nPlease note that 40MJ/m^3 is NOT equal to 11 kW/m^3, because the former is an ENERGY density, and the latter is a POWER density.\n\nHowever, 40MJ/m^3  IS approximately equal to 11 kWH/m^3, which I assume is what you meant to write.\n\nUsing this method gives the same answer:\n\n(30 kW) / (11 kWH/m^3) = 2.7 m^3/hr  (approx. 3 m^3/hr)\n\nAlso, as a matter of notation, the term "LPG" (Liquid Petroleum Gas) is typically used for either Propane or Butane in liquid form, NOT for Natural Gas.\n\nI hope this explanation is helpful for you.
33777
Alternative forms of energy are very much in the news. I think that would go over well. Start with http://en.wikipedia.org/wiki/Green_electricity  and go from there.
155906
There are 24 x 31 = 744 hours in a 31 day month.  Check your fan name tag for the rated watts (single speed?).  Multiply the watts times 744 and divide by 1,000 to get KW-Hr (kilowatt-hours) of power per month.  Check your power bill for the charge per KW-Hr.  It should be about 10 to 15 cents per KW-Hr (or KWH), depending upon where you live.  For example if a small fan draws say 20 watts, the calculation would be:\n\n(20 x 744)/1,000 = 14.88 KW-Hr @ $0.10 per KW-Hr = $1.49 per month or about $17.85 per year.\n\nGood luck.