On 04/08/2011 07:39 PM, [email protected] wrote: > In reply to Stephen A. Lawrence's message of Fri, 08 Apr 2011 12:43:52 -0400: > Hi, > [snip] > >> At the point where the effluent is at 60 degrees, it's taking 300 watts >> just to heat the flowing water. However, at that point "ignition" takes >> place, and another heat source begins to warm the device. Since the >> joule heater is taking care of warming the water (at 60C), all the >> *additional* heat will go into warming the device. >> > I think some of the additional heat also has to warm the water to more than > 60 ºC (since the temperature of the water also goes up with the temperature of > the device). >
Yes, but that's taken into account automatically -- the thermal mass of the water in the device won't be changing (until it starts vaporizing), just as the thermal mass of the device itself can be assumed (nearly) constant. Consequently, the heat it takes to warm the (device + contained water) from 20C to 21C should be about the same as the heat it takes to warm the (device + contained water) from 60C to 61C. In the latter case, there's a steep thermal gradient in the flowing water between the inlet and the outlet, but that gradient may be assumed to be maintained by the 300 watt heater. The heat from the reaction, consequently, should be going entirely into warming the (device + contained water) system, so how fast that system warms up should tell us the power being generated. Or, anyhow, I think it should... hmmm. > > >> So, the slope of the >> curve at that point corresponds to however much heat is being generated >> by the reaction, applied solely to warming the device (as, again, the >> heater's output is sufficient to warm the water to that temperature). >> I've drawn a tangent line to the curve at 60 degrees in blue. >> > Regards, > > Robin van Spaandonk > > http://rvanspaa.freehostia.com/Project.html > > >
