View Full Version : "calrod" resistance heaters
kengineering
05-06-2006, 10:35 PM
Electric condensate evaporator pans use a resistance "calrod" heater to boil away the ice that melts from the evap coil during defrost. My question is this; If I have to replace a 120 volt pan that draws 8.4 amps and I only have a 240 volt pan that draws 4.2 amps on the truck can it be sustituted? It would seem that Ohms law would just cause a higher amp draw to compensate for the lower voltage but the end result would be the same total of 1000 Watts. Thanks , Ken
NoNickName
05-06-2006, 10:44 PM
The power is the same, but the joule effect accounts for i squared r, that is the square of the amps times the resistance.
You do the math...
Your 208 volt heater will provide only 252 watts of power connected to the 120 volt supply, and draw only 2.1 amps. First you establish the resistance of the heater either by measuring, or calculation. You have amps and voltage, so
120 x 8.4 = 1008 Watts. Now you want to substitute a 240 volt heater.
240 X 4.2 = 1008 Watts. But we need to find the resistance of this heater.
(240 squared)/1008 Watts = 57.14 ohms.
(120 squared)/57.14 ohms = 252 watts.
Disappointing, isn't it? :)
http://ourworld.compuserve.com/homepages/Bill_Bowden/ohmslaw.htm
kengineering
06-06-2006, 12:35 AM
Thanks for clearing that up. Ken
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