Heater: Sheath Type,Current:1.Amp,Output200W,Insulation Glass"E type,Encosure:1P 20,Weight:10 Kgs(including all accessors)Frame Keeper:AC7A-F,Screen Holder:Cirumeremce 80,Screen Frame:AC7A-F CIR 417 (FULL SET PRICE)
131404
what are you saying? CLEAR VIEW SCREEN Type Weater Proofmodelno:KYJ-350-H350-H supply source:AC/220 size"DIA350 speed 1800 200RPM \nHeater: Sheath Type,Current:1.Amp,Output200W,... Glass"E type,Encosure:1P 20,Weight:10 Kgs(including all accessors)Frame Keeper:AC7A-F,Screen Holder:Cirumeremce 80,Screen Frame:AC7A-F CIR 417 (FULL SET PRICE)
291827
u can try www.hotstuffworks.com which can guide u to the basics of any latest electrical instrument/appliances.its a site of great help..best of luck..
555168
i aever heard of a heated vest biu u might have luck searching on the internet
271135
Try http://www.shareware-catalog.net/home-education/science/fahrenheit-converter-download-23656.htm
188369
Chokes don't like heat, but I suspect something more. There's a charging circuit, the choke acts like a filter for the storage battery. The charging circuit is pulsing the battery. This pulsing is killing the choke, but without it it will cause damage to the battery and/or the light.\nYou will want to increase the ampacity of the choke, but keep the same mH value if you can.
274740
The first ans.\n1.ice at 0c to water at 0\nlatent heat of fussion is 80 cal/g\nheat that must be given to the system is=80*225=18000 cal\n \n2.water at 0c to water at 100c \nspecifi heat of water is 1 cal/g/c\nheat that must be given for the conversion is =100*225*1=22500 cal\n\n3.water at 100c to steam at 100c\nlatent heat of vap. is 540 cal\nheat for this con. is =540*225\n\nadd all the amounts to get req. ans.
341365
The heater part will work fine, the problem will be with the pump motor. If the motor is actually 220 volt it will run ok but 12.5% slower (less water pressure) A lot of things in the US that are called 220 volt will only be using 220 for the high power part. I.E. a 220 window air conditioner often has a 110 volt fan, it only uses one hot leg and neutral. If your hot tub has anything (pump motor, lights, air compressor) that uses 110 volts, that part won't work without being modified. There is no neutral in European 220. In some cases you will get something to work anyway but it is using earth leakage on other wiring and will not last for long, the voltage isn't steady. It will also get you killed in a worst case. You will not have any trouble getting/finding a good electrician there to make the changes needed and it won't cost nearly what it would here.
525074
The first method would be to compare you utility bills to see if there is a drastic change (or any) between last year at this time and the most recent bill with the insulation on.  \n\nFor an exact calculation, you need the following:\n\nThe R value of the insulation\nLength of pipe insulated\nTemperature of the water inside the pipe\n\nThen plug them into this equation:\n\nQ=UA(Tinside - Toutside)\nwhere Q is the amout of Btus \nU is the thermal insulation (or lack there of) expressed as =1/R\nTinside = water temperature\nToutside = air temperature (average) where the pipe is installed\nA = the surface area of the pipe\n\nOnce you get that for the insulation, do it once with out the insulation.  For the second R value without the insulation, use a R-value of 0.5.  The two different Q's you get will be the amount of Btu's you saved.  \n\nTo convert that to $, you need to convert the Btu's (Q) to kW.  Divide the Btu's by 3.412 to get watts.  That will be the amount watts you saved at your hot water heater/boiler or whatever is heating the water (assuming it is electric).  You then look at your electric bill to determine the cost $/W or $/kWh.  Take the number you got for watts and convert to kilo Watts by dividing by 1000.  Then multiply it by the cost off of you utility bill for $/kWh.  \n\nThis is all for a one hour period.  You need to calculate that for a month, day, or year (720, 24, 8765 hours respectively).  You now have what your savings for this project was.
549835
The math is correct for the above answers but since you asked the question with regards to overheating there are other considerations.\nThe major enemy of all electrical and electronic components is heat. \nResistors are rated at ambient temperature of 25 degrees C. This rating is in free air with component leads connected to a substantial terminal that can help carry heat away.\nIf you operate this resistor at maximum rating and put it in an enclosure it will fail long before expected. \nManufacturers tend to exaggerate their ratings and test them in ideal but not realistic conditions. A resistor that may hold up running at maximum ratings will soon fail when subjected to the thermal cycling in and out of this extreme temperature condition.\n\nA rule of thumb that is practical is to size a resistor at 1/2 of its wattage rating unless there are only short duration high wattage peaks. Damage doesn't occur until it has had time to come up to temp. If you are interested in long component life operate components at well below their maximum.
254937
Well, get out your heat transfer text book and start calculating. You should have all the info you should need.
427719
I tried to import a spreadheet for you but I coouldn't make it come through, if you email me I'll send it to you, like1m@yahoo.com  \n\nYou can fill in the two variables, just write it in excel and you will be able to calculate your values for voltage and current dependent on the resistance for any given values.  All answers here are correct as it is the basic premise of \nOhms law, but you asked for a chart.  You do need to know the resistance though.  The interrelation is such that for at a given voltage with infinate resistance there will be no current (flow), for that same voltage with minimum resistance there will be maximum current (flow) this would be referred to as a dead short and is comparable basically to welding. Either extreme is not normally practical unless you are turning the power "off" in one case or welding in the other (dead short).  Dependent on the requirement usually the interrelation is desired to be somewhere in the middle.  There is also the element of reactance which is slightly more dynamic that static resistance. but for now this is what your auto calc chart should look like....  \n\nInput ohms & volts to autocalc Amps\n\nInput OhmsInput VoltsI=V/R\n      200                     24                 0.12\n\nInput ohms & Amps to autocalc Volts\n\nInput OhmsInput AmpsV=I*R\n    200                     0.12                   24
554931
Unit  Armorer 45B series
408166
OF2 (~104), NF3(~107), CF4 (~109.5),  BF3 (120), BeF2(~180)
464521
First, what is the input voltage and the desired output voltage?  Is the output voltage fixed, or do you need a range of adjustment (taps)?  Do you need isolation between the input and the output?  If not, an autotransformer (single winding) might be acceptable.  What is the frequency, and is it single phase, three phase Y, or three phase delta?\n\nNext, what is the power rating?  Small transformers are usually rated in output current, whereas larger transformers are rated in KVA.  This is the maximum load;  a smaller load would not be a problem.\n\nFinally, what are the mechanical factors?  Does it need to be hermetic or can it be open frame?  Do you want terminals or wire leads?  What shape and size?  What is the mounting?\n\nIt's not really that hard.  Usually you can browse through a catalog and find the transformer you need in minutes.
297276
(m)\n\ngoto the following website for your question with neat diagrams\n\nhttp://hyperphysics.phy-astr.gsu.edu/HBASE/electric/gaulaw.html
380816
the answer is \n basic equaion man you must know this off head ...V=IR\nVOLTAGE =CURRENT(I) *RESISTANCE (R) \nSO THE ANSWER IS \nI=V/R\nTHEN ITS 24/200\n=0.12
605169
cop = (heat absorbed)/(work) = heat absorbed)/(heat delivered-heat absorbed)\n2.2 = (3.545*10^4)/(work)\n\ntherefore mech energy =work = 16113.63 J\n nd heat rejected =  (3.545*10^4)/(2.2)  + 3.545*10^4\n                            =  515636.63 J