how many kilowatts makes a watt?
152831
1000 watts make a kiolwatt
1030224
The answer in SI units is 1200 joules or 1.2 kilojoules. the definition of a joule is one watt second. in Calories, or English units, which shouldn't really be used with watts as it mixes systems of measurements, is 286.806883 calories or 0.286806883 kilo calories.
280553
You cannot compare those two together. One watt-hour is equivalent to one watt of power used for one hour. Watt alone refers to the power produced which equals the voltage multiplied by current therefore:\n\n1 megawatt hour = 1000 Kwh\n1 megawatt  = 1000 Kw
783813
Ampacity is the amount of current a wire or device can handle. A watt hour meter socket records how much current has passed through that socket.\n\nWatt (power)= voltage x Current\nWatt-hour (energy) = voltage x current x time\n\nVoltage is constant. So if you have something in a socket that records 1 watt hour, it used 1 watt for one hour. or 1 watt/120V = .008 amp hours.\n\nbut to answer your question a device can only handle so much current, a typical socket is rated for 15-20 amps.
776640
Well wattage is Volts times Amps. 120X0.12=14.4Watts
906536
These are a published value.  The actual voltages vary as you can see from this quote from a power plant site \n\n"POWER PLANT CAPABILITY: The two hydroelectric power generators boast a generating capacity of 92,000 kilowatts. The estimated annual energy output is 205 million kilowatt-hours."\n\nAnd none of the values quoted are a multiple of 11.  It appears that the source your using just happened to have values that were multiples of 11.\n\nA quick look on Google listed quite a number of plants that produced values that had 13800kv to 205 million kv.\n\nChance that the ones you looked at were multiples.\n\nBest of luck in your project.
162775
One kilowatt used over a one hour period is one kilowatt-hour.  To find out how many kilowatt hours an electrical appliance uses, look on the label (wattage - say 300 watts).  Divide that by 1000 to get kilowatts (0.3 kilowatts).  Now multiply it by the time you are going to use it for (say 20 minutes [6 kilowatt-minutes] - divided by 60 minutes in an hour [0.1 kilowatt-hours]).  The result should be the number of kilowatt-hours used.
346550
kingbodonti is correct...\n\nPower in watts = Work in joules / time in seconds\n\nPower = 1,000 joules/50 seconds\n\nPower = 20 watts ANSWER
908174
When I ride a bike at the gym it says I generate about 100-150 watts.  Now that is enough to power a couple light-bulbs, or a computer, but only as long as you keep peddling.  It probably isn't worth researching or looking into since not enough power is being generated to be worth it.  I'm sure the ISS gets plenty of power from its solar panels.
1041665
When 1 Joule of energy is spent in 1 second, the power is said to be 1 Watt.\nIn everyday household use, Watt is used to measure the power of electrical appliances. That is to say, if an appliance is labeled as 220W, it means that the appliance spends 220 Joules of energy(electrical) every second.
897585
It depends on how low low is and how high high is. Kilowatt hours are the standard measurements for power expendature pricing. The bill is calculated by multiplying power (kilowatts) by time (hours). If there is twice as much power for half as long, your bill will not change.
362669
Kilowatt-hour is the commercial unit of electrical energy. Joule is the S.I. unit. Kilowatt-hour is the energy consumed when an appliance with one kilowatt rating is used for an hour. It is 3.6 x e6 J.
1028188
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.
1015451
Power = work done per unit of time.\n\nP=W/T
928079
There are no watts in one volt.\n\nWatts is a measure of the rate of energy transfer. If you only have voltage, you have no energy transfer. Your wall outlet has voltage, but no electricity is actually flowing through it (until you plug something in), so there is no wattage.\n\nVoltage is the 'pressure' of the electricity - how hard it pushes to get through the circuit. Amperage is the actual quantity of electricity moving through the wire. Multiply the two of them together, and you get your wattage.
175366
961 kcal=1118 watts\nI=P/E\nWich come out to about 22.6 amps.\nSo you would have to go with the 32.8 amp fuse. Also the math I did was based on 100% efficiency. So the 32.8 amp fuse would also make up for that.
657313
W/m*K = Watt/(meter*Kelvin)\n\nPower per unit length per unit temperature\n\nWhere power is defined as energy per unit time.