Is there a way to find out the requirement power (watts) for a room? (including lights, air condition etc)?
970880
Power requirements for a room are determined by the total heat load. This heat load consists of solar load (if the room has an exterior window), people load (how many people use the room at a time, and how active are they), appliance loads (lights, equipment, etc.) and the ambient conditions surrounding the room and insulation (r-value for walls separating room from hot and cold areas).\n\nIf you need more specific help for a commercial project - e-mail me.
975896
Energy is measured in watts, so I assume you mean a voltage source of 200v.\n\n(1)\nThe voltage source of 200v applied to the light bulb (laminar) will be 200v.  Since the light bulb is designed for 130v, the supply voltage will likely blow the bulb.\n\n(2)\nHere you have two possibilities, depending on your circuit.\n(a) If you have a parallel circuit, then each light bulb will have 200v across it, and again the bulbs will likely blow because the voltage is to high.\n(b) If you have a series circuit, the voltage across the light bulbs will be subject to their resistance.  Assuming the resistance is the same, then the voltage will be the same, that is, 100v each.\nThe voltage across each light bulb will be equally proportional to it's resistance ratio.
914617
Hi there:\n\nYou can estimate your power consumption using calculations for posted " average" ratings for all the parts of your computer, however, nothing is real, and the power supply rating has nothing to do with the power you are actually using.\n Only the electrical meter on the main line is of any value...\n\nI have answered this question in different forms a few times, and a typical answer is listed below:\n\nHow can you find out how many Watts your PCs power supply uses?\n\nLet's say in an hour, how many watts have been used...\n \n\n\nBest Answer - Chosen By Voters\n\nHi there:\nI looked at the answers above, and only one, CJMT had any inkling of the correct answer \n\n- The only way to know is to MEASURE IT!\nYou can calculate thousands of calculations all day, but until you actually measure YOUR own computer, you are just guessing!\n\nNew power supplies now come WITH and actual Wattage power meter, which is on the back or mounted on a front panel with a plug in cable. Some Motherboards, have a power meter built in as well, and, considering these " new and improved" features, it clearly shows that you are among thousands of computer users who are interested in this factor...\n\nIn the " old " days, you would just get a power " Clamp on " meter of some sort, and clamp over ONE wire of the AC power cord ( this is a bit difficult with normal computer power cords, so you plug in an extension which has the the wires separated, and clamp the meter over ONE of the wires.)\n\nNow you can purchase easy-to-use plug in-the-wall meters that have digital readouts. \n\nSimple.\n\nYOU MUST actually measure your wattage - for example, if you have an 800 WATT power supply, this means nothing, except that the power supply " CAN " supply 800 Watts if it " has to " -- you could actually only be using 289 Watts of the available 800.\n\nIn order to really know what your " computer " is costing in terms of power, you should plug in the monitor, the CPU box, the speakers, the printer, the modems, power adapters etc. etc. TO A POWER BAR, and then measure the actual total to the wall plug.\n\nListed below are links on the web to typical measuring tools that are commonly available. You can check out stores and electronics and computer outlets in your area, once you know what you are looking for. If you know an electronics guy or a hydro guy, he could just let you use a clamp meter in a few seconds to get a reading, and save yourself a few bucks - but the $30 home units would come in handy to measure anything in \nthe whole house -- well worth it !\n\n\nPower Supplies with WATT METERS\nhttp://www.highpowersupply.com/product-ps-hpc480102df.html\nhttp://www.newegg.com/product/product.asp?item=N82E16817171001\n\nKill a Watt Meter\n\nhttp://www.gooddeals.com/shopexd.asp_Q_id_E_119\n\nBriggs and Stratton Meter\n\nhttp://www.northerntool.com/webapp/wcs/stores/servlet/NTEprodprint?productId=200326720&p\n\nrnbr=1704700&storeId=6970&catalogId=4006970&langId=-1\n\nSAFE HOME 3 products\nhttp://www.safehomeproducts.com/shp2/shpsearch.asp?source=adwords&kw=wattage+meter\n\nProfessional Watt meters\nhttp://www.testequipmentdepot.com/amprobe/clampmeters/digitalac.htm\nhttp://www.testequipmentdepot.com/fluke/currentclamps/330series.htm\nhttp://www.testequipmentdepot.com/hioki/clampmeters/3280.htm\n\nThere are hundreds of other models and manufacturers available, and since millions of people are interested in actually knowing where the high cost of power is being used in their homes, I expect to see hundreds of more products available\nat lower and lower prices as manufacturing gears up.\n\nHope this helps\n\nrobin
602903
You don't, any more than you convert between volts and watts.  You can read about dBm in the article below, but simply put, it is a measure of power.
414310
You cannot determine KWH usage with AC in that way.\n\nIf you looked at the voltage, it would be a sinewave.\nIf you could look at the current, it might also be a sinewave, but it is very unlikely that it would be in perfect phase with the voltage.\n\nOnly if the current and the voltage are in perfect phase with each other can you determine watts by multiplying current times voltage.\n\nThe arithmetic you are suggesting will work only when the power factor is exactly 1.00, which could be close to correct if the load is all incandescent lighting, but will not be correct if the load is inductive, such as electric motors, or capacitive.\n\nTo measure power usage, you will need a watt-hour meter, similar to what the utility companies provide.\n\nThe following source (found via Google) explains it:\n\n"A common misconception is that power is simply the product of voltage times current, implying that any ammeter can be used to determine power usage. This is only true in  a few special cases."
790311
To get the most out of your room air conditioner, you'll need to consider the size of the unit that best suits your needs.\nThe cooling capacity of room air conditioners is measured in British thermal units per hour, or Btu/h. To figure out the maximum cooling capacity for a room, you will first need to calculate the size of the room where the air conditioner will be placed.\nTo do this, simply multiply the width of the room by the length of the room.  Use the following information and formulars to help you with calculating your air conditioning requirements:\n\n\nHeat Load\n\nThe amount of heat generated is known as the heat gain or heat load. Heat is measured in either British Thermal Units (BTU) or Kilowatts (KW). 1KW is equivalent to 3412BTUs.  The heat load depends on a number of factors, by taking into account those that apply in your circumstances and adding them together a reasonably accurate measure of the total heat can be calculated. Factors include: \n\nThe floor area of the room\n The size and position of windows, and whether they have blinds or shades \nThe number of room occupants (if any) \nThe heat generated by equipment \nThe heat generated by lighting \nArea of Room\n\nThe amount of cooling required depends on the area of the room. To calculate the area in square metres:\n\n\nRoom Area BTU = Length (m) x Width (m) x 337\n\n\nWindow Size and Position\n\nIf, your room has no windows, you can ignore this part of the calculation. If, however there are windows you need to take the size and orientation into account.\n\n\nSouth Window BTU = South Facing window Length (m) x Width (m) x 870\nNorth Window BTU = North Facing windows Length (m) x Width (m) x 165\nIf there are no blinds on the windows multiply the result(s) by 1.5.\nAdd together all the BTUs for the windows.\n\nWindows BTU = South Window(s) BTU + North Window(s) BTU\n\n\nOccupants\n\nYou will have to take that into account people who normally working in the space. The heat output is around 400 BTU per person.  \n\nTotal Occupant BTU = Number of occupants x 400\n\n\nEquipment\n\nClearly most heat in a room is generated by the equipment. This is trickier to calculate that you might think. The wattage on equipment is the maximum power consumption rating, the actual power consumed may be less. However it is probably safer to overestimate the wattage than underestimate it.\n\n\nEquipment BTU = Total wattage for all equipment x 3.5\n\n\nLighting\n\nTake the total wattage of the lighting and multiply by 4.25.\n\n\nLighting BTU = Total wattage for all lighting x 4.25\n\n\nTotal Cooling Required\n\nAdd all the BTUs together.\n\nTotal Heat Load = Room Area BTU + Windows BTU + Total Occupant BTU + Equipment BTU + Lighting BTU\n\nThis is the amount of cooling required so you need one or more air conditioning units to handle that amount of heat.\n\n--------------------\n\nGeneral thumb rule \n\n100 sqr ft area with roof height of 9 - 10 ft will need 1 Ton AC unit.
75746
You'll need to provide a little more information in order to have your question answered properly.  AC or DC? Input voltage?  Fixed or variable output voltage? If variable, over what range? If fixed, what is the desired output voltage?  Output current?  How critical is the load regulation?\n\nAs you can see, the answer to your simple question can be a bit complicated :)
1166636
Actually, there is an easy way to approximate your answer. A closet should reach steady-state equilibrium before the end of the day - that is, the heat added by the bulb would equal the heat lost by conduction through the ceiling, walls, etc. Once steady-state equilibrium is reached, it doesn't matter what the heat capacity of the air or the closet is - only the thermal conductivity of the walls, door, etc. and the surface area of each will be important.\n\nHere is where effective "R-value" of the walls, ceiling, door, floor is useful. R-value is just the rate of heat transfer in units of K-m^2/Watt (or in British units ft^2 F/(BTU/hr)). The metric "R-value" is about 1/6 of the corresponding R-value in British units due to the conversion factors.\n\nThe formula for difference in heat at steady state (i.e., heat rise) would be approximately:\n\nDelta-T = 100 Watts / (R1A1 + R2A2 + R3A3 + R4A4 + R5A5 + R6A6) where each of the 6 surfaces of the room are treated separately. \n\nYou need to know approximately what your insulation is for each surface - but a little research on the internet and comparison based on your closet's insulating properties should get you a halfway decent approximation.
851868
IT depends.\n\nIf you need to run a hard drive, CD drive, and the rest of a traditional PC, then it will be more than enough -- most power supplies are 200 watts.\n\nI'd upgrade if you hook a water heater, water bed, space heater, etc tho.
360016
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.
844099
It depends on the applied voltage.  Divide the voltage by the resistance which gives you the amperage.  Now multiply the amperage  with the voltage, and that is your wattage.\nW = V²/R\nThis is straight forward for DC; with AC things become a little more difficult, as because of the fluctuation of the voltage, the average is only .707 of the nominal voltage.
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.
1026240
less than 0.3  for power sations\n0.5ohm for substation\n1 to 2 for homes
1186419
That would depend on the voltage. Current = Watts / Volts\nA couple of examples:\n110V - 90.9 Amps\n220V - 45.45 Amps\n240V - 41.67 Amps\n440V - 22.7 Amps\n10,000V - 1 Amp
865595
A typical computer uses between 100-150 watts per hour.  But your computer usage will vary depending on how many peripherals and components are installed.  In the United States, rates vary wildly.  I'm in the more expensive area of the country where we pay about 20 cents per Kilowatt Hour.  This amounts to about $15-20 per month per computer.\n\nAnd to the guy who said the TV uses more... you're WRONG.  TVs use between 50 and 100 watts. (except for the larger TVs).\n\nI'm not sure if they make a UK equivalent, but you might want to invest in a device like the Kill-A-Watt meter (google it) - it will tell you how much electricity is being used. \n\nAlso, idle time vs. use time can see the wattage change significantly.  Using my the network can increase wattage use by about 15 watts...\n\nAsking the electric company sounds good but is silly - I've done that, they tend to tell you the computer uses 300 watts (because that's the average power supply in a computer).  Problem is, just because the power supply can provide 300 watts does NOT mean that that's what it uses.  It's kinda like saying your light socket uses 150 watts because it can use a 150 watt bulb... even when you only use a 60 watt bulb.
627569
It's a simple matter of adding up the power required by everything, EVERYTHING, you want connect to your supply.\n\nMachines, tools, heating, cooling/air conditioning, ventilation fans, lighting, computers, printers, copiers, telephones, fax machines, answering machines, refirgerators, freezers, water coolers, security devices, etc, etc, etc.\n\nChoose a generator with a capacity at least a bit greater than your total load, but preferably quite a bit more to allow for future load increase.
172746
Check with several home heating specialists.  Recently talked to someone who just had NA furnace installed.  Requires no electricity.