You are correct Ken. That is where I went wrong. I knew something didn't seem correct.
On Thu, Oct 4, 2012 at 10:41 AM, ken deboer <barlaz...@gmail.com> wrote: > Jack, > I think you went the wrong way with the total BTU inputted. should be 36 > wh X ~3 btu/w = ~100 btu. > ken > On Thu, Oct 4, 2012 at 5:13 AM, Jack Cole <jcol...@gmail.com> wrote: > >> I think I have pretty high heat loss as it is open to the air. Here are >> some pretty conservative calculations assuming no heat loss and complete >> conversion of electrical input to heat. Please check my math / conversions >> to see if I am doing this correctly. >> >> 1 BTU is the amount of heat needed to raise 1 lb of water 1 degree F. >> >> A power supply at 12 V and 1 amp gives 12 wh. >> >> 1 BTU = .293 wh (see wikipedia BTU) >> >> A temperature change of 60F for 3.718 oz requires the following BTUs. >> >> 3.178 oz / 16 oz = .199 >> >> 60 * .199 = 11.92 BTUs required to change the temp 60F assuming no heat >> loss. >> >> Running 3 hours gives a total input of 36 wh. So convert 36 wh to BTU. >> >> 36 * .293 = 10.5 BTUs total input >> >> I calculate COP by BTUs required to raise the temp 60F / input BTU. >> >> COP = 11.92 / 10.5 = 1.135 >> >> Have I done the correct process with these calculations? >> >>
