Energy usage drawn from the grid is measured in kWh. What does the word "peak" mean ?
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
480440
ohms law...\n\nfind total wattage \ntimes cost per kwh\n so\n21 x 220 =\n4620 watts per hr\nx 5.8 cents per 1k watts per hr
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%.
137963
yup...energy (Joules) multiplied by time (seconds)
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.
607758
W=Qh-Ql=(1.23-0.986)*1e5\n=2.44*10^4  (...using 1st Law of Thermodynamics)\nP=W/t=2.44*10^4/60 =406.7 watt\n\nConsidering COP(coefficient of performance) as the energy efficiency rating,\nCOP=Ql/W\n=9.86/2.44 = 4.04\n\nHere, Ql stands for heat taken from source at low temperature(room), Qh for heat rejected to hot sink(outside air) and W is the work done by the unit.
409313
Volts.  The the force behind electricity. Ampere is the mount of current. Ohm is the unit of measure for resistance.
677011
Kilowatt\nCost depends on method of generation and location
314775
1. volts\n2. (b) negatively charged.\n3.  short.  Could also be "conductor," but the lack of control seems to indicate an unwanted short.\n4.  watt-hours.\n5.  \n6.  (a) The formula is P=V*V/R.  Where V is the voltage and R is the resistance.  So P=45*45/15=135 watts.  Actually the object is consumming the power, not producing it.\n    (b)  135*10 = 1350 watt-hours, or 1.35 kw-hr.\n    (c) 0.08*1.35 = $1.08\n7.  (a)  V=IR, so if the voltage is increased while the resistance is kept constant, then the current will also increase. (I=V/R)\n    (b)  If the resistance is increased and the voltage is kept constant, then the current has to decrease. (I=V/R)
279561
Tidal power is a means of electricity generation achieved by capturing the energy contained in moving water mass due to tides. Two types of tidal energy can be extracted: kinetic energy of currents between ebbing and surging tides and potential energy from the difference in height (or head) between high and low tides. The former method - generating energy from tidal currents - is considered much more feasible today than building ocean-based dams or barrages, and many coastal sites worldwide are being examined for their suitability to produce tidal (current) energy.
351771
If what you say is true, then a Kw is costing about 10p which is on a par with electricity?
346097
Joules (even though "s" is the last letter, this seemed to be the most reasonable guess)
402656
A joule is "One Watt Second"  It is how we convert Power to Energy.  Energy is Power used over a Period of time. 2 hours = 7200 Seconds.\nA Kilowatt is 1000 watts.  3 Kilowatts will be 3000 watts.\nSo we have 3000 Watts used over a period of 7200 seconds.  Joules = 3000 X 7200\nJoules = 216,000,000 Final answer
296531
Lower level that is energy of producers. We get 100% energy from lower level but as level goes up energy level decreases due consumption of energy.
116519
It appears to be in relation to energy production.  I have referenced a hyperlink to a PDF file that refers to this term with in its context...the file does have some English translation.
472266
kWh is Kilowats/hour - like MPH in a car.  kW is Kilowats.  Figure out the speed, time and distance formula for the problem and you will have your answer.
339136
Power is the rate at which work is done. It is the work/time ratio. \nWork = force * distance\n\nWhat force was used to stop the car?\nWhat distance did it travel?